INDEX
S. No
Topic
Week 1
Page No.
1
Rolle’s Theorem
1
2
Mean Value Theorems
17
3
Indeterminate Forms (Part ‐1)
36
4
Indeterminate Forms (Part ‐2)
52
5
Taylor Polynomial and Taylor Series
70
Week 2
6
Limit of Functions of Two Variables
86
7
Evaluation of Limit of Functions of Two Variables
101
8
Continuity of Functions of Two Variables
118
9
Partial Derivatives of Functions of Two Variables
138
10
Partial Derivatives of Higher Order
158
Week 3
11
Derivative & Differentiability
186
12
Differentiability of Functions of Two Variables
206
13
Differentiability of Functions of Two Variables (Cont.)
226
14
Differentiability of Functions of Two Variables (Cont.)
247
15
Composite and Homogeneous Functions
266
Week 4
16
Taylor’s Theorem for Functions of Two Variables
286
17
Maxima & Minima of Functions of Two Variables
303
18
Maxima & Minima of Functions of Two Variables (Cont.)
320
19
Maxima & Minima of Functions of Two Variables (Cont.)
336
20
Constrained Maxima & Minima
358
Week 5
21
Improper Integrals
376
22
Improper Integrals (Cont.)
393
23
Improper Integrals (Cont.)
408
24
Improper Integrals (Cont.)
427
25
Beta & Gamma Function
448
Week 6
26
Beta & Gamma Function (Cont.)
466
27
Differentiation Under Integral Sign
483
28
Double Integrals
501
29
Double Integrals (Cont.)
522
30
Double Integrals (Cont.)
542
Week 7
31
Integral Calculus –Double Integrals in Polar Form
559
32
Integral Calculus –Double Integrals: Change of Variables
573
33
Integral Calculus –Double Integrals:Surface Area
588
34
Integral Calculus –Triple Integrals
604
35
Integral Calculus – Triple Integrals (Cont.)
617
Week 8
36
System of Linear Equations
633
37
System of Linear Equations –Gauss Elimination
648
38
System of Linear Equations –Gauss Elimination (Cont.)
663
39
Linear Algebra - Vector Spaces
680
40
Linear Independence of Vectors
695
Week 9
41
Vector Spaces –Spanning Set
706
42
Vector Spaces –Basis and Dimension
723
43
Rank of a Matrix
741
44
Linear Transformations
758
45
Linear Transformations (contd.)
776
Week 10
46
Eigenvalues & Eigenvectors
795
47
Eigenvalues & Eigenvectors (Cont.)
811
48
Eigenvalues & Eigenvectors (Cont.)
829
49
Eigenvalues & Eigenvectors (Cont.)
841
50
Eigenvalues & Eigenvectors:Diagonalization
863
Week 11
51
Differential Equations - Introduction
881
52
First Order Differential Equations
899
53
Exact Differential Equations
920
54
Exact Differential Equations (Cont.)
938
55
First Order Linear Differential Equations
956
Week 12
56
Higher Order Linear Differential Equations
974
57
Solution of Higher Order Homogeneous Linear Equations
992
58
1007
59
Solution of Higher Order Non-Homogeneous Linear Equations
Solution of Higher Order Non-Homogeneous Linear Equations
(cont.)
60
Cauchy-Euler Equations
1045
1026
Engineering Mathematics - I
Prof. Jitendra Kumar
Department of Mathematics
Indian Institute of Technology, Kharagpur
Lecture – 01
Rolle’s Theorem
Welcome to the first lecture on Engineering Mathematics, I am Jitendra Kumar from the
Department of Mathematics. And, today we will be discussing the Rolle’s Theorem from
differential calculus of variable one.
(Refer Slide Time: 00:35)
So, these are the topics we will be covering today. So, starting with the Rolle’s Theorem. And
since it is a very fundamental theorem so we will also go through the detail proof because this
will be used for various other results in other lectures and then some worked examples.
1
(Refer Slide Time: 00:56)
So, what is Rolle’s Theorem? So, Rolle’s Theorem if a function f of single variable is
continuous in closed interval [a , b] and differentiable in open interval (a ,b). And, there is a
one more condition which says that the function value at a is equal to the function value at
the point b. So, the two end points the function is taking same value.
In that case, the theorem says that there exist a number c in open interval (a ,b ¿ such that
( c ) =0 , meaning that there will be a point c where the slope of the tangent will be 0 . So, if we
go through the geometrical interpretation, so, let us consider this is the function which is
plotted in this x−axis and the y−axis here. So, this is the point at a so, the function value at
this point a is here and the same value the function is taking at b. So, the function is
continuous and differentiable everywhere then this theorem says that there will be at least one
point where the derivative will vanish or the tangent will be parallel to x-axis. So, the slope
of the tangent is 0 meaning it is parallel to thex−axis.
So, clearly we can observe that there is a point here somewhere next to this a, where the
tangent is parallel to the x−axis. Indeed in this situation there are more points one I can see
here where tangent is again parallel to the x−axis. And, there is another point where the
tangent is parallel to the x−axis, but the theorem says that there will be at least one point
where the tangent will be parallel to the x−axis. So, in this particular situation we are getting
more than one point where the tangent is parallel to the x−axis.
2
(Refer Slide Time: 03:14)
So, if we go through; now the proof which is pretty simple so, let us go step by step. So, here
we assume that the function takes the maximum and the minimum value and they are denoted
by big M as maximum and small m as minimum in this interval [a ,b], which is guaranteed
due to the extreme value theorem because the function is continuous in the closed interval.
And therefore, a maximum and minimum will be reached in this interval at some point.
So, now we consider the following situation a particular situation the case-I when M =m. So,
the maximum value is equal to the minimum value. Now, think about the situation, then the
where the function is having maximum and the minimum value as same. So, in this situation
clearly there is no change in the function value and therefore, the minimum value is equal to
the maximum value.
So, basically if we plot f ( x )=m= M , it is a constant function and that would be the situation
when the minimum value will be equal to the maximum value. So, in this particular case f ( x)
is the constant function, and since f ( x) is the constant function naturally whatever point you
take the derivative is going to be 0. So, the theorem is proved in this case when M =m. So,
the maximum is equal to the minimum the function value.
3
(Refer Slide Time: 04:52)
So, now the second case when the maximum value of the function is not equal to the
minimum value of the function. So, in this case we will consider three situations or three
cases again. The first one let us assume that the maximum value in this situation here which
is clearly can be observed that it is different than the function value at a and b. The function
value at a and b are equal as per the assumptions of the theorem.
So, here we assume that the function the maximum value of the function is different than the
function value at a and b. The second case when we take the minimum value is different than
the function value at a and b and the third situation that for some functions both maybe
different. So, in this case the minimum or rather I would say the local minimum in each case
or local maximum. So, which is here and this is different than the equal values at a and b.
And, here as well the local these two local maximum are also different than the equal value at
a and b.
4
(Refer Slide Time: 06:10)
So, both are different in this case. So, in either situation let us consider the case we suppose
that the maximum value of the function is different from the equal values of f at a and b
which are the same here. So, the M is different the big M the maximum is different.
Similarly we will consider later on if M is not different then the small m should be different
at least one of them will be different because M is not equal to small m.
So, we take that the function is taking this value M at a point c. So, f ( c ) =M, the function is
having this local maximum at the point c. So, if this f (c) is the local maximum then we have
f ( c+ Δ x )−f (c); just considered the situation f (c+ Δ x). So, this point here c+ Δ x is in the
close vicinity of c just assume this Δ x is close to 0.
So, in that case since f (c) is the maximum value of the function then f ( c+ Δ x ) and this
difference; so, f ( c+ Δ x )will be smaller than the this f ( c ) because f ( c ) is the local maximum.
So, in this case there will be a such a Δ x definitely because f ( c ) is the maximum value that,
this expression here f ( c+ Δ x )−f (c) will be less than equal to 0 whether this Δ x is positive or
Δ x is negative. That means, any point you take in the vicinity of this point c then this
difference here f ( c+ Δ x )−f ( c ) ≤ 0 . And, now if I divide this expression here by Δ x and if I
in the first case I take Δ x as positive, then the sign of this expression will not change. And, it
will remain as less than equal to 0 if Δ x is positive.
5
On the other hand if I take Δ x as a negative number then this expression will change the sign
and this f ( c+ Δ x )−f (c) divided byΔ x will become greater than equal to 0. And now, I will
take in this first case when I have taken here the Δ x positive the limit that Δ x goes to 0 and
this expression and the less than 0.
So, if you take a close look at this one this is the right hand derivative of the function f and
since f is differentiable this will be equal to the derivative of the function. So, we have here
this inequality that the derivative will be less than equal to 0 in the situation. On the other
hand when you divide this by Δ x which is negative and take the limit again the same similar
case here. Since the function is differentiable that left derivative will be also positive because
this inequality is greater than equal to 0.
So, in this case we got f ' ( c )=0 whereas, they we have f ' ( c ) ≤0 . So, out of these two we
conclude that the f ' ( c )has to be 0 because it cannot be less than equal to 0 or greater than
equal to 0 at the same time. So, there only possibility is that f ' (c) has to be 0. So, in this way
we have proved this that there is a point in this interval c, in the open interval c where the
derivative vanishes.
(Refer Slide Time: 10:05)
There are few remarks which are of great importance. So, here the hypothesis of the Rolle’s
Theorem are sufficient, but not necessary for the conclusion. What do we mean by this? So,
what we have seen that this continuity of the function in the close interval [a, b] and the
differentiability in the open interval (a, b) and there was a third condition that f (a) is equal to
6
f (b). So, if these three conditions are satisfied then there will exist a point c where the
derivative will vanish. So, these conditions are sufficient meaning that these conditions here
all these three conditions implies that f ' ( c )=0. But not the other way around that f ' ( c )=0
does not imply that the function will be continuous, differentiable and we will take these
equal values at some points a and b.
So, in other words if all these hypothesis these three hypotheses are met then the conclusion
is guaranteed; conclusion means the f ' ( c )=0 that is guaranteed. However, if the hypothesis
are not met then you may or may not reach the conclusion which we will see with the help of
some examples now. Let us consider this example
2
f ( x )= x ;−2≤ x ≤1
3 x−2; 1< x ≤ 2
{
So, this function here the clearly if we see that the function is continuous, the function value
at 1 is here f at 1 which is we can substitute directly the function is defined until 1.
checked
So, f (1) is 1 and then if you take the right limit so, f (1+0) the right limits. So, the limit
Δ x → 0 , and this f (1+ Δ x) and minus this f (1) or just the limits we are not going to get the
derivative now. So, this just this expression here f (1+ Δ x) and Δ x → 0 and we take here the
Δ x positive. So, the right limit of this function as Δ x → 0 . So, this will be simply the limit
Δ x → 0 , the Δ x we are taking as positive here. And, then since Δ x is positive, 1+ Δ x we will
be calculated from this here 3 x−2. So, you have the 3 and x means 1+ Δ x the argument and
minus 2 and this is nothing but 3 and minus 2 1. So, 1+3 Δ x and Δ x → 0 .
So, this is 1 and which is equal to the f (1). So, the function is naturally continues in this case
and if we check the differentiability; that means, the right derivative first. So, the f (1+0) the
right derivative means the limit Δ x → 0 and the Δ x is positive. So, f (1+ Δ x) minusf (1) and
divided by Δ x this goes not here. So, in this case the limit Δ x → 0 the Δ x is positive; so,
here 1+ Δ x again will be calculated from 3 x−2 which we have just done before. So, it was
1+3Δ x was coming and divided by Δ x and then here minus this f (1) is 1.
7
So, this gets cancelled and then this value here is nothing, but 3. So, the right side derivative
of this function is 3 where as the left hand derivative. So, f (1−0) which is the notation and
here the limit if you compute Δ x → 0, Δ x negative, what will happen to this one. So, here
you have again
f ( 1+ Δ x ) −f ( 1 )
.
Δx
But now this f ¿ + Δ x) and Δ x is negative will be computed from x2 . So, meaning we have
here Δ x → 0 and this is
2
( 1+ Δx ) −1
Δx
So, this one when we expand this there will be 1+ Δ x2 +2 Δ x terms so, 1 1 will get cancel and
this 2 Δ x and divided by Δ x will give you a 2 and the rest because of the limit will go to 0.
So, here the derivative is 2 whereas, there the left side derivative is 3 and the right side
derivative is 2. So, the function is not differentiable in this case.
(Refer Slide Time: 15:12)
And we can plot this one and then again you can see that at this point 1 here the function
breaks its differentiability. So, the right side derivative which we have just seen was minus
the left side derivative so was 2 and the right side was 3. So, there is a point here where the
8
function is not differentiable. But, what is interesting in this case the all the hypothesis are not
made because the function is not differentiable at this point.
But there is a point here 0 which you can easily compute again from this x2 is a derivative is
2 x and x is equal to 0 the derivative will become 0. So, here the f ' ( 0 ) =0. So, the derivative
vanishes or the tangent is parallel to the x−axis in this case though the function was not
differentiable here. So, exactly what we have said if the hypotheses are not met the function
may or may not reach the conclusion. So, in this case it is reaching the conclusion, but this is
not because of the Rolle’s Theorem.
(Refer Slide Time: 16:20)
Another example if we take that we have
f ( x )=
{2−xx ; 0≤;1<x x≤1≤ 2
So, again the similar situation one can easily float this function and one clearly sees that at
this point 1 the function is not differentiable. And, in this case we are not getting any point
between this 0 and 1 where the function is taking over the derivative is vanishing. So, in this
situation f ' (x )≠ 0 at any point in the given interval. So, we have seen these two examples the
other one was this one, the previous example where the function was not differentiable, this is
also not differentiable.
9
But in 1, f ' ( 0 ) =0. So, there is a point where the derivative vanishes whereas, in this case the
derivative does not vanish at any point in the interval. So, therefore, these conditions these
three hypotheses of the Rolle’s Theorem are sufficient conditions and they are not the
necessary conditions. So, under those conditions it is guaranteed that the function will
derivative of the function will vanish at least at one point in the open interval (a, b).
(Refer Slide Time: 17:44)
Another remark that the continuity condition which we have seen the continuity in the closed
interval for this function is essential, if it is not met then we may not that the theorem may not
guarantee the existence of such a c where f ' ( c )=0. So, for example, if you look at this
function
f ( x )= x ;0 ≤ x<1
0; x=1
{
So, what do we see here the function is continuous and differentiable on (0, 1) and also
f ( 0 ) =f (1). So, this condition is met differentiability condition is met, but the function is not
continuous at 1. We should note that because the function is x from 0 to 1 and then it is x is
equal to 1.
So, there is jump here which we can see. So, at x is equal to 1 the function is taking value as
0 and otherwise its taking here as x. So, the function is not differentiable at oh sorry
continuous at 1, otherwise all other conditions are met in this case of the Rolle’s Theorem.
10
And, then we clearly see the derivative is 1 everywhere here between these two 0 and 1 open
interval 0 and 1. And therefore, the f ' (x )≠ 0 at any point in this interval x 0 to 1.
(Refer Slide Time: 19:12)
Another example we will discuss now the applicability of the Rolle’s theorem for this
function
2
f ( x )= x +1; x ∈[0 , 1]
3−x ; x ∈ ¿
{
So, again if the continuity is concerned then the function is continuous because it is taking
like f (1) is f (1) is 2 and f if we take the right limit here f (1+0). So, the limit Δ x → 0, and
this f (1+ Δ x) will be this is limit Δ x → 0 and Δ x positive because the right limit we are
taking here. And, in this case this will be 3−(1+ Δ x); that means, it is a 2−Δ x.
So,Δ x → 0, this is 2 and the value is equal to 1. So, the function is continuous in this interval
0 to 2 and what is about the differentiability. If you look at the differentiability is pretty
similar to the earlier case. So, if you compute the right derivative so,1+ 0; that means, the
Δ x → 0 and Δ x is positive because the right limit I am talking about. And, in this case again
you have take the
f ( 1+ Δ x ) −f ( 1 )
.
Δx
11
So, limit Δ x → 0 andf ( 1+ Δ x ) .
So, f ( 1+ Δ x ) we have computed here this is 2−Δ x and f (1) is 2 again and divided by Δ x.
So, this limit will be coming as −1 because this will got cancelled and then you will get −1
there. So, the right derivative is −1 and the left derivative f (1−0) which is limit Δ x → 0
again with Δ x negative.
So, in this case f (1+ Δ x) will be computed from here. So, ( 1+ Δ x ) 2 +1 minus f (1) which is 2
divided by Δ x. So, Δ x → 0 and here you will get 1+ Δ x2 +2 Δ x; so, 1+ Δ x2 +2 Δ x+1−2. So,
this will cancel out and then here also so, you will get and this power. So, Δ x → 0 this will be
coming as 2. So, in this case the left derivative is 2 and the right derivative is −1. So, the
function is not differentiable at the point 1. So, the Rolle’s Theorem is not applicable in this
case.
(Refer Slide Time: 22:10)
And if we take a look here at this floor, then you again see that at 1 here the function is not
differentiable which we have just seen.
12
(Refer Slide Time: 22:23)
So, moving further this is another example which says the using Rolle’s Theorem show that
the equation this x power x13 +7 x 3−5=0 has exactly one real root in [0, 1], in the closed
interval [0, 1]. So, this is another kind of application which where we can use the Rolle’s
Theorem to show that this equation has exactly one real root. So, if we move further suppose
that this f ( x) this function here x power x13 +7 x 3−5 has more than one real root in [0, 1]. So,
we assume that this function f ( x) has more than one real root. So, if it has more than root
then we can take any two roots let us say alpha and beta.
So, you have taken two roots and since this alpha and beta are the roots so, f (α ) will be 0 and
that will be also equal to f ( β). So, α and β both are roots so, the function will be 0 at α and
as well as at β. So, here we just for the convenience we have assume that α is smaller than β
and naturally these two will fall between 0 and 1; because 0 and 1 are is not the root of the
equation which clearly we can see there. So, this α β these two roots because, we have
assume that this function has more than two roots so, these α and β will be between less
between 0 and 1.
So, both have the positive number here α and β and less than 1. So, what Rolle’s Theorem
says, if we apply the Rolle’s Theorem to this interval α and β. If we apply we apply this
Rolle’s Theorem to the interval α and β in that case the Rolle’s Theorem says that there will
be a point f ' ( c )will be 0; there will be a point c where f ' (c) will be 0. Because, of the reason
because the function is taking now equal value at α and β, function is differentiable, it is a
13
polynomial function, there is no problem, the it is continuous naturally and it is taking the
same value at α and β.
So, if we apply in this interval Rolle’s Theorem that will give us that f ' ( c )=0 for some c in
the interval (α , β). So, this implies so, what is this f ' ( c )? So, f ' ( c )is13 c 12 +21 c2 =0; for some
c in the interval (α , β). Again note that the (α , β) both are positive number and now which
you see because the c is positive here, then this expression here 13 c 12 +21 c2 ≠ 0, because this
is a power 12, here the even number also c 2 and this c is positive.
So, this is a positive quantity, this is a positive quantity. So, it cannot be equal to 0, but the
Rolle’s Theorem says that it will be equal to 0; that means, we have a assumption which was
that the function has more than two real roots is wrong. So, it contradicts our assumption of
more than one real root. But, now the question is whether there is a root in this case, because
we have just proved that there cannot be more than two roots.
So, if you take a close look at this function here at 0 the value is a −5 somewhere here and if
you put this 1 there the other end then we will get 3. So, the value will be 3 at 1 so, if this is 1
here. So, at 1 the value is 3 and the 0 the value is −5 and function is continuous. So,
definitely to reach to this point it will cross somewhere the real axis and so, that proves the
existence of one root in this case which confirms the existence of one root because this is
changing its sign.
(Refer Slide Time: 27:00)
14
So, f (0) is −5 and f (1) is 3.
(Refer Slide Time: 27:14)
Now, there are the references which we are used to prepare this lecture, the book by the
Piskunov, Differential and Integral calculus, Volume 1 and also the Kreyszig Advanced
Engineering Mathematics.
(Refer Slide Time: 27:28)
So, again the conclusion here we have a studied the Rolle’s Theorem which says that if the
function is continuous and differentiable having the same value at this a and b, then there will
15
be a point c somewhere in the open interval ¿,b), where the tangent to this function will be
parallel to the x−¿ axis.
So, this is the Rolle’s Theorem which is a particular case of the mean value theorem which
we will discuss in the next lecture. And, basically this assumption of having the equal values
will be removed and then we will get more general results. And, those are the mean value
theorem the topic of the next lecture.
Thank you.
16
Engineering Mathematics – I
Prof. Jitendra Kumar
Department of Mathematics
Indian Institution of Technology Kharagpur
Lecture – 02
Mean Value Theorem
Hi. So, welcome to the second lecture on Engineering Mathematics – I and today, we will
discuss Mean Value Theorems.
(Refer Slide Time: 00:23)
So, let us go through the concepts covered. So, we will discuss the Lagrange mean value
theorem; a very important concept which is the extension of the previous lecture where we
have a studied Rolle’s theorem and there is another generalized a mean value theorem or the
Cauchy mean value theorem which will be also discussed in today’s lecture.
17
(Refer Slide Time: 00:45)
So, let me just recall from the previous lecture. So, if a function f is continuous in a closed
interval [a , b] and differentiable in open interval (a , b) and the function value at the point a
and the point b. So, the end points of the interval is equal, then there exist a number c in the
open interval (a ,b) such that the derivative vanishes at this point.
The geometrical interpretation is as clear from this figure. So, we have a function f which is
continuous and differentiable and the function value at a and b both are equal. Then, there
exist a point c here where the tangent is parallel to the x−¿ axis.
(Refer Slide Time: 01:37)
18
So, now coming to the Lagrange mean value theorem we have the function f which is
continuous similar to the previous conditions of the Rolle’s theorem and differentiable in the
open interval (a ,b).
The third condition where the function was equal at the two end points is not required here.
So, it is more general and less restrictive and in that case again there exist at least one number
c in the open interval (a ,b) such that this quotient here
f ( b ) −f ( a )
b−a
is equal to the derivative at a point c. So, let us first discuss the geometrical interpretation of
this Lagrange mean value theorem.
So, if we have a function which is continuous and differentiable in some interval a ,b and
then let us take a look what is this quotient here. So, if you join these two points f at a and f
at b then we get this line segment. So, what is the slope of this line segment? Let us compute.
So, in this case if I draw this triangle here the height here will be f (b) because the distance
from here to this point is f (b) and the distance from this point to this point here is f (a). So,
f ( b )−f (a) is this height of this triangle and the base here is b−a , because up to this point is
b and up to this point here the co ordinate of this point is a f (a).
So, here this distance is b−a and this one is f ( b )−f (a) and. So, this quotient here f (b)
minus f (a) were b−a this perpendicular divided by the space will be the tangent of this
angle. So, basically this expression here f ( b )−f (a) were b−a is the slope of this line
segment which we have drawn by meeting these two end points of the a curve and now, what
this theorem says that this will be equal to f prime c. So, the slope at some point in the
domain a to b.
So, the geometrical meaning is that they will be at least one tangent; in this particular case we
can see these three tangents which are parallel to this line segment. So, this Lagrange mean
value theorem says that there will be at least one point where the tangent will be parallel to
this line segment joining these two points, the end points of the curve. So, as I have written
here in other words there is at least one tangent in this interval that is parallel to the line
segment that goes through the end points of the curve.
19
(Refer Slide Time: 04:41)
The proof is very simple if we consider this function
ϕ ( x ) =f ( x )−
[
f ( b ) −f ( a )
x.
b−a
]
So, if you take a close look at this function it is a basically difference of two functions f ( x)
and minus some constant times x. So, if f is continuous in the closed interval [a , b] and
differentiable in the open interval (a ,b) and x is also a function which is continuous and
differentiable in those intervals, then this difference will be also continuous in closed interval
[a , b] and differentiable in the open interval (a ,b).
So, for the setting of this function is done because if you compute here for example, the ϕ at
the point a and ϕ at the point b then we will realize that these two values are also equal. So, ϕ
at a is nothing, but
f ( a )−
[
f ( b )−f ( a )
a
b−a
]
So, this if I simplify then and this will become b f ( a )−a f (a) and then minus a f ( b )−a f (a)
and divided by this b−a. So, this a f (a) will get cancelled and then we will get b f ( a )−a
f (b) over b−a and now, if I compute here f (b). So, here you have then f (b) minus this
course in f ( b )−f (a) and divided by this b−a and then here b.
20
So, if I simplify now this so,
b f ( b ) −a f ( b ) −b f ( b ) +b f ( a )
b−a
So, in this case this b f (b) gets cancel and we get
b f ( a ) −a f ( b )
.
b−a
In the earliest case also we got
b f ( a ) −a f ( b )
.
b−a
So, the function is taking same value at a and b and if we recall again the condition was for
Rolle’s theorem other than the continuity and differentiability that the function should be
having the same value at the two end points. So, in this case this function f satisfies all the
conditions of the Rolle’s theorem.
(Refer Slide Time: 07:35)
And therefore, we can apply the Rolle’s theorem to this function ϕ ( x). So, what will now
give us if we take the derivative here the ϕ '( x) is equal to the derivative of x minus this is a
constant. So, here f ( b )−¿ f (a) and divided by this b−¿ a and the derivative of x will be 1.
21
And the Rolle’s theorem says that the there will be a point where the function will be the
derivative of the function will be 0.
(Refer Slide Time: 08:15)
So, in this case now if we apply the Rolle’s theorem then ϕ '(c) will be 0 and for some c in
the interval in the open interval (a , b) and which implies precisely that this is 0 and that is the
Lagrange mean value theorem that f ' (c) will be equal to
f ( b ) −f (a)
=0.
b−a
So, the construction of this function here was important to prove the Lagrange mean value
theorem and this ϕ here satisfy all the properties of the Rolle’s theorem and we can apply the
Rolle’s theorem to this function and we got the desired result of the Lagrange mean value
theorem.
22
(Refer Slide Time: 08:51)
So, there is another one the generalized mean value theorem which is also called the Cauchy
mean value theorem. So, here we will consider two functions instead of one. So, if f ( x) and
g( x) are two functions continuous in closed interval [a ,b] and differentiable in open interval
(a , b) and there is another condition on g that g ' the derivative of g does not vanish
anywhere inside the interval then there exist a point c in the open interval ( a, b )such that this
Cauchy theorem
f ( b )−f (a)
g ( b )−g(a)
is equal to the ratio of the derivative of this f and g at the point c.
So, the proof is again pretty similar to the earlier proof of the Lagrange mean value theorem
and in this case we set this function or define a function in such a way that this ϕ ( x) is equal
to f ( x )−¿ f (a) minus this quotient here which will be coming in the result of this Cauchy
mean value theorem and multiplied by g ( x ) −¿ g(a). So, again the similar argument since f
and g they are continuous in closed interval and differentiable in the open interval (a , b).
Student: (Refer Time: 10:15).
So, the ϕ is also differentiable and continuous in the given intervals. Moreover if we see here
that what is the ϕ at a, that is f ( a )−¿ f (a) here this is 0 and g ( a ) −¿ g(a) is also 0. So,
23
everything is 0. So, the ϕ (a) is 0 and the ϕ (b) which is f ( b )−¿ f (a) and ( b )−¿ g(a). So, this
g ( b ) −¿ g(a) will get cancel with this g ( b ) −¿ g(a) and then we will get f ( b )−¿ f (a) minus
this f ( b )−¿ f (a) which is again 0.
So, in this case the ϕ (a) is 0 and ϕ (b) is 0 and ϕ satisfies all the conditions of the Rolle’s
theorem and therefore, we can apply Rolle’s theorem to this function ϕ ( x).
(Refer Slide Time: 11:15)
So, applying the Rolle’s theorem, but before that there is a point here that we have to tell that
this ϕ is well define because this g ( b ) −¿ g(a) should not go to 0; that means, g(b) should
not be equal to g(a). The question is why g(b) cannot be equal to g(a)? We have not made
such a restriction directly in the assumptions of this Cauchy mean value theorem , but again
there was an additional condition that g '( x) does not vanish anywhere inside the interval.
So, if this g(b) is equal to g(a) in this case we can again apply the Rolle’s theorem to the
function g which will say that there will be a point c in the open interval (a , b) where the
derivative will vanish. But, as per the assumption of the theorem g '
does not vanish
anywhere inside the interval. So, this cannot be equal. So, there will be never such a situation
that this g(b) will become equal to g(a) and this will become infinity.
So, the function is well defined the function is differentiable, it is continuous and g(b) ϕ (a)
is equal to ϕ (b). So, all the conditions of the Rolle’s theorem are satisfied for this function
ϕ ( x).
24
(Refer Slide Time: 12:49)
So, if we apply the Rolle’s theorem now, to the function then we will get exactly the result
which is given here because the ϕ '( x) will be the derivative of f and then this is a constant
here minus again this expression and the derivative of g.
(Refer Slide Time: 13:11)
So, what will be this again? So, let me just come to this point. So,
ϕ ' ( x )=f ' ( x )−
[
f ( b )−f ( a )
g ' ( x)
g ( b )−g ( a )
]
25
and the Rolle’s theorem says that at the point x is equal to c this is equal to 0.
So, what do we get then? The ϕ '(c) and divided by this g '(c) is equal to the
f ( b )−f (a)
.
g(b)−g (a)
So, that is the Cauchy mean value theorem or the generalized mean value theorem.
So, if you now we discuss the geometrical meaning or the of this Cauchy mean value
theorem. So, here now we consider this parametric curve which is given by x=g¿); g is the
function the given function there, but I have introduced this parameter t which is commonly
used for the parametric curves and the y is equal to the other function f (t ) and t varies from
a to b in this close interval.
So, this parametric curve you can trace by varying the values of t. So, if for example, t=a,
then we have here the x co-ordinate g(a) and the y co-ordinate f (a)of this point. So, this
point is g(a), f (a) and then if we vary t we will basically move on this curve we will trace
this curve and till we reach the end point here, t is equal to b which is given by g(b) , f (b).
So, now the geometrical meaning is similar to the earlier result on Lagrange mean value
theorem. So, they will if we join these two points by this line segment, then this theorem
says; so, first of all this the slope of this line segment will be given by this
f ( b )−f (a)
g ( b )−g(a)
because of the same argument as we have discussed earlier. The height will be f ( b )−f (a)
and this the base of this triangle will be g ( b ) −g (a). So, this is the slope of this line segment
and then the right hand side here says that there will be at least one point on this curve where
the tangent will be parallel to this quotient line.
So, if you take a close look this f ' (c)/ g' ( c ) is nothing, but the slope of the tangent line at
some point c, because the slope will be calculated as at some point here the dy /dx is equal to;
for the parametric curve, so, this will be dy /dt and divided by dx /dt or the y ' (t ) divided by
the x ' (t). And, this y is basically the f , so, here you have the f ' (t) over g '(t ) and this
26
theorem says that there will be a point somewhere in the interval. So, t is equal to c. So, we
will get this slope here of this tangent line as f 'at c divided by g' ( c ) .
(Refer Slide Time: 16:55)
And, now let me just quickly summarize at this point what we have learnt today. So, we
discuss the generalized mean value theorem which was this
'
f ( b )−f (a) f ( c )
=
g ( b )−g(a) g' ( c )
27
(Refer Slide Time: 17:15)
And, now what will happen if ( x )=¿ x. So, g ( x ) =¿ x meaning that here you have this g ( b ) =¿
b and this g ( a ) =¿ a and g '( x) here will be just 1. So, what do we get in this case the
Lagrange mean value theorem because that conclusion will be
f ( b ) −f (a) '
=f ( c ) .
b−a
So, in this particular case when we take g ( x ) =x we will get the Lagrange mean value
theorem.
And, what will happen to this Lagrange mean value theorem if we put ( b )=¿ f (a), the
additional condition what we have for the Rolle’s theorem. So, f ( b )−f (a) this quantity here
will become 0 and then we will get f ' ( c )=0. So, this is the generalized mean value theorem
and as a particular case if we take the function g ( x ) =x we will get the Lagrange mean value
theorem and again if we add another condition that f ( b )=¿ f (a) we will get the Rolle’s mean
value theorem which is f ' ( c )=0.
28
(Refer Slide Time: 18:31)
Now, we go to the we will go to this some examples the first one that using mean value
theorem we will show that this inequality cos e x−cos e y is less than equal to x− y for x, y ≤ 0.
So, first we note that when both are equal x, y are equal then naturally this cos e x−cos e was
0 and is equal to 0. So, then in equality is naturally satisfied when x and y both are same. So,
we will consider the case when they are not same.
And, now we consider the f ¿) another function cos et because clearly we can see that we
want to prove this cos e x−cos e y using mean value theorem. So, if you consider this function
t
f ( t )=cos e in this interval x , y and naturally we have assume that x ≠ y and now, we apply
the mean value theorem to this result what we will get? cos e x−cos e y x minus y is equal to
there will be some point in this interval open interval ( x , y) and the value of this quotient will
be is equal to f ' (c).
So, this is the Lagrange mean value theorem and now we will estimate this derivative because
the derivative we can compute the f ¿) is cos et . So, taking the absolute value both the sides
we get this cos e x−cos e y and this absolute value will take to the right hand side. So, the x−¿
y absolute value and the absolute value of these f ' . So, f ' is nothing, but the n −sin et × e t .
So, because of the absolute value we have not taken this −sin into consideration. So, we
have e c sin ec because this the derivative has to be evaluated at point c, ok.
29
Now, this implies, so, if I we take the maximum value of this expression here. So, the c
varies from x to y. So, we have taken the c from this x to y and we will take the maximum
value of this one. And, note that the c belongs to this ( x , y) open interval, so, it is basically a
negative number the c <0 because x and y both are less than equal to 0 and therefore, the c
will be strictly less than 0 in the open interval.
So, the sin is always founded by one. So, we have less than equal to one the sin function and
the e c the exponential function for this negative argument c will be always less than 1
because e 0 is 1 and o for all a negative values it takes value less than 1 for positive values it
will take more than 1.
So, this is strictly less than 1, this is less than equal to 1. So, this expression here or the
maximum value of this derivative is bounded by a strictly bounded by 1. So, we got this
inequality cos e x−cos e y is less than the absolute value of x− y which we want to prove in
this result.
(Refer Slide Time: 22:01)
The second example we will consider that this f is the differentiable function on the closed
interval −2 to 2 and such that the value is given as −2 is equal to 1, f is given as 2 as 5 and
there is another information here that f ' (x ); f ' (x ) is bounded by 1 for all values of x in this
interval −2 to 2. And, using mean value theorem we want to find the value of the function at
0.
30
So, if we take a look at this problem and we want to find the value of f (0), so, we need to
apply the mean value theorem or Lagrange mean value theorem in the interval −2 to 0 and 2
to 0 and then we will get some estimate on this f (0). So, if we apply the Lagrange mean
value theorem on −2 to 0 interval what we get the
f ( 0 ) −f (−2)
0−(−2)
is equal to there will exist some c 1 in the open interval −2 to 0 so that this value will be equal
to the derivative at that point c 1.
Now, this derivative here f prime c 1 is bounded by 1. So, we know the estimate of this f ' (c1 ).
This is always between −1 and 1. So, what is this expression here? The f (−2) is 1. So,
f ( 0 ) −1
2
f (0) minus 1 divided by 2 lies between −1 and 1 because this is equal to the derivative and
the derivative is bounded by less than equal to 1 the absolute value. So, this expression here
lies between −1 and 1.
Now, if we multiply this 2 to both the sides or that we can multiply 2 to this inequality here
we will get −2≤ f ( 0 ) −1≤2 and then we can add this 1 to the inequality. So, we will get here
the 3 less than equal to f (0) and less than equal to so, here minus 2 was there plus 1, so,
minus 1 and then here 2 and then plus 1 we will get 3. So, out of this inequality we will get
that −1≤ f ( 0 ) ≤3. Again if you use the Lagrange mean value theorem in the interval 0to 2; in
the interval 0 to 2 we will get
f ( 2 )−f (0)
2−0
is equal to the first derivative at some pointc 2.
So, again here the f (2) is known the f (2) is 5; so,
5−f (0)
.
2
31
So, what do we have here? We have f (2); f (2) is given as 5 and minus this f (0) divided by 2
and this value again is bounded by minus 1 and 1. So, we got this one here
−2≤ 5−f ( 0 ) ≤ 2.
So, this implies that this
−7 ≤−f ( 0 ) ≤−3
So, if you multiply by minus 1 here, so, the inequality will change; so,
7 ≥ f ( 0 )≥ 3
(Refer Slide Time: 25:57)
So, this inequality we will get now that f ( 0 ) ≥ 3 and ≤7 this one which says that the f ( 0 ) ≥ 3,
but ≤7. The earlier inequality says that f ( 0 ) ≤3.
So, by these two inequalities here f ( 0 ) ≤ 3 and f ( 0 ) ≥ 3 what we will conclude that f (0) has to
be 3. So, f (0) has to be 3. So, we got the value using the mean value theorem of the function
at 0 given that those derivatives and the end points value was given.
32
(Refer Slide Time: 26:35)
The last example here, the function f which satisfies now that the derivative is
1
5−x2
and ( 0 )=¿ 2. Now, we want to use the Lagrange mean value theorem to estimate the bounds
on f (1). So, in this case the exact value of f (1) is not possible so, we will estimate the lower
and the upper bound for f (1).
So, again if we use the Lagrange mean value theorem in the interval 0 to 1 because you want
to estimate 1. So, 0 to 1 then we will get f ( 1 )−¿ f (0) divided by this difference 1 and there
will be some point whose value will be equal to the derivative at that point. So, out of this
inequality1)
f ( 1 )−f (0) '
=f ( c ) .
2
Now, what is the derivative? Derivative is
1
.
5−x2
'
So, f ( c )=
1
.
5−c2
33
Now, just note that c here is between 0 and 1. So, the lower bound of this
1
5−c 2
will be obtained when we then the c this c approaches to 0; that means, this value is always
greater than 1/5 and when the c approaches to this maximum value in the interval as 1 in that
case this will become 4, and the maximum value of this 1 over 5 minus c square will be 1 by
4. So, now we know that the derivative lies between 1/5 and 1/4.
So, what is the derivative f ' (c) is here? f ( 1 )−2. So, with this we got the inequality that
f ( 1 )−2 lies between 1/5 and 1/4 and this implies. So, to we can take to the other side and
also it has to be added to the right side here. So, this we will get 11/5 less then f (1) and then
if we add 2 here this will be 9/ 4. So, we got the estimate on f (1) that it lies between 11/5
and 9/4.
(Refer Slide Time: 28:55)
So, these are the references we used here. The Piskunov, Differential and Integral Calculus
and Kreyszig, Advanced Engineering Mathematics.
34
(Refer Slide Time: 29:07)
So, the conclusion for today’s lecture that we have learnt the generalize mean value theorem
and as a special case when we substitute this g( x) the other function the second function as x,
we will get the Lagrange mean value theorem and if we take another assumption that f (a) is
equal to f ( v) then this will be the Rolle’s theorem which says that f ' ( c )=0, ok.
Thank you, for today’s lecture.
35
Engineering Mathematics - I
Prof. Jitendra Kumar
Department of Mathematics
Indian Institute of Technology, Kharagpur
Lecture - 03
Indeterminate Forms Part - 1
Welcome to the lectures on Engineering Mathematics-I. and today’s topic is Indeterminate
Forms.
(Refer Slide Time: 00:26)
So, these are the concepts we will be covering today. So indeterminate forms, L'Hospital's
rules which is very fundamental principle to determine a such forms and some worked out
examples.
36
(Refer Slide Time: 00:40)
So, before I start to indeterminate forms let me just introduce your recall from the previous
lecture, the generalized mean value theorem or the Cauchy mean value theorem which was
discussed in previous lecture.
So, there we have seen that if there are two functions f and g continuous in closed interval
and differentiable in open interval (a, b) and g ' the derivative of g does not vanish anywhere
inside the interval then there exist a point c in open interval (a, b) such that this quotient here
f ( b )−f (a)
g ( b )−g(a)
is equal to the ratio of the derivatives at that point c. So, this generalized mean value theorem
will be used today to prove sum of the results.
37
(Refer Slide Time: 01:36)
And what are the indeterminate forms. So, for example, if we consider this
sin ( x−1 )
x−1
this ratio of sin and x minus 1 or we can consider like
2
x −1
.
x−1
So, we want to evaluate these when x=1. So, if we substitute x=1 simply we are getting here
0 and divided by0; similarly here as well we are getting 0/ 0. So, in these cases we cannot just
simply substitute x=1 and get the value of these expressions given here or for example, we
have
1−cos x
x
and we want to see that what will happen to this expression when x=0.
So, if we put x=0 here. So, 1−cos 0 is again 1 so, 0 and divided by 0. So, we have another
0/ 0 form which cannot be evaluated directly by substituting x=0. So, the question is that
when f and g both tend to 0; what happened to the ratio f ( x ) / g( x). These are the situations
which we have considered in these examples in each of them f ( x) and g( x) both tend to 0.
38
And now we want to see that what will happen to these expressions. And, in today’s lecture
we will see that for example, this form here
sin
(x ¿−1)
¿
x−1
when we take the limit as x goes to 1. Because we cannot simply substitute as x=1 in this
expression, but we can talk about the limit.
So, the
lim sin
x →1
( x¿−1)
⁡¿
x−1
will be coming as 1 whereas, in the second case when
2
x −1
x−1
this 1 can simply get by cancelling this x−1 from the numerator, because this numerator one
can write like x−1 and x+1. So, this x−1 will get cancel with this x−1 and then this limit
will be simply 2. This one
1−cos x
x
one can again evaluate we will see later in the lecture that this limit is 0. So, in these all three
cases we have seen that these forms were 0/ 0 forms, but their limits are different; in the first
case it is 1, here it is 2 and the third one is 0.
39
(Refer Slide Time: 04:12)
So, what are these indeterminate expressions; we will see now. So, they may appear like in
0/ 0 we have just seen the other possibility is that the numerator and denominator both are ∞.
So, we have the form ∞ / ∞ or the 0 ×∞. There are other indeterminate forms like ∞−∞or in
these exponent form 00, ∞ 0 power infinity. So, all these are the indeterminate form and we
do not know what is the value of for example, ∞ 0 or 1∞ ∞−∞.
So, there was a remark here that these expressions which are different then these which we
are calling indeterminate forms. For example, 0∞ , ∞ ×∞, ∞ +∞, ∞ ∞ or ∞−∞ and note that these
forms are not indeterminate forms and we can directly find the value of these expressions.
For example, this 0∞ : what is the value of 0∞ it is just 0 and ∞ ×∞ will the two very big
numbers when you multiply naturally you will get ∞.
Again here the ∞ +∞again will become ∞ and ∞ ∞ with the same reason this will be also ∞.
And, ∞−∞ we can write rewrite it as 1 over ∞ ∞ and then ∞ ∞ is ∞. So, this will become 1∞
which is 0. So, these forms are not indeterminate forms. So, if we find such expressions
during the calculations we can directly substitute these values. But, we have to be more
careful for such 0/ 0 ∞ ×∞ is 0 ×∞ all these cases. Because, we have to evaluate by some
rules those values and it is not clear that; what is the value of ∞ ×∞ for example.
40
(Refer Slide Time: 06:29)
So, there is a concept here the L’Hospital’s rule a very fundamental rule for determining such
a indeterminate forms. So, what is this here let us go through the, suppose this f ( x) and g( x)
are two functions and continuous in closed interval [a ,b] and differentiable in open interval
(a , b) and this g prime the derivative of g does not vanish anywhere inside the interval. So,
all these conditions are the conditions of the Cauchy mean value theorem or the generalized
mean value theorem we have just seen today. And, in addition to those conditions if we have
like f ( a )=0 the function is taking value as 0 at a and the second function g is also taking the
value as 0 at this point a. Then we will see in the proof of this L’Hospital’s rule that if you
want to evaluate the limit as x → a, naturally in the setting a from the right hand side the limit
the right limit as x → a f ( x)/ g(x) this ratio which directly we see here which says f (a) is 0.
So, it is like 0/ 0 form.
But if you take this limit and this rule says that this limit, this limiting value is equal to this
limiting value which is the ratio of the derivatives. So, it is a another application of the
derivatives and naturally when this limit the limit of the derivatives exist, otherwise this does
not make sense if the this does not exist. We will come to this point little later and we will see
one example where this limit does not exist, but it does not mean that the limit of f ( x)/ g(x)
does not exist. So, this is the rule here that if such limits exists the limit of the derivatives,
then we this will be equal to the limit of this ratio of the functions.
41
(Refer Slide Time: 08:31)
So, the proof is pretty simple if we use the L if you use the Cauchy mean value theorem so,
which was summarize today. So, this is the Cauchy mean value theorem when we have two
functions f and g then f ( x )−f (a) and g ( x ) −g(a) will be f ' (ξ)/ g ' (ξ) and the ξ lies
somewhere between a and x. So, we have taken a point here x in the interval, x is naturally
not equal to a and then we have applied this generalized mean value theorem in the interval a
to x ok. So, we know that the value of the function at a; and the value of the function g. So,
for both the functions at a is 0.
So, this here with this expression left hand side will become f ( x)/ g(x). So, what we have
here the f ( x)/ g(x) is equal to f ' (ξ) over g '(ξ) and this ξ belongs to a to x interval. So,
now, you note that if this limit here we take as a x → a and since this is ξ belongs to the open
interval a to x. So, if we x goes to a naturally the ξ will also go to a. So, this is what the next
here. So, if we take the limit here x → a f ( x)/ g(x) and then this will be equal to the limit ξ
goes to a plus because the ξ belongs to the interval a → x.
So, the ξ will go toa+¿. So, from the right side and this derivative here f ' / g ' . And now we
can replace just this ξ by some other name or the most suitable is x in the setting here. So,
lim ¿ this ratio f ( x)/ g(x) is nothing, but the
what we have seen that this x →a+¿
¿
lim
this is the proof of this L’Hospital’s rule using the generalized mean value theorem.
42
'
x →a+¿ f / g' ¿
¿. So,
(Refer Slide Time: 10:47)
There is a slightly more general form of this L’Hospital’s rule, because if we note here that
we have taken x → a from the right side because we had taken this interval for the functions a
to b. So, this is a more general form this if f and g are two functions differentiable on open
interval I and naturally they are also continuous on the ξ containing this a and this f (a) is 0.
So now, a is somewhere inside the interval not at the boundary. So, if we have f ( a )=0=g(a)
and g' ( x ) ≠ 0 in this interval, if x ≠ a; x=a anything can happen then we do not need such
restrictions on g '. But, other than this x=a the g does not a vanish.
So, in this case also one can easily prove that limit x → a now there is no left or right concept
'
here. So, the limit simply x → a f ( x)/ g is equal to the lim f (x)/ g' ( x) provided the limit on
x →a
the right hand side exist. And, the proof is similar to what we have already done before
because, now we can consider two intervals here in this I. So, a → x when x and taking x> a
and we can also consider another interval x → a when x< a.
So, in these two intervals we will apply the previous result which will establish there that
x → a+¿ in this case is equal to
lim ¿. And, then when we apply that result to this interval
x →a+¿ ❑¿
and we will get that x → a from the negative sides of both the limits from the right side. And
f (x)/ g( x) is
from the left side we will get the same result which will conclude that the lim
x →a
f ' ( x)/g ' ( x).
equal to the lim
x →a
43
(Refer Slide Time: 13:02)
So, another important remark so, this L’Hospital’s rule also hold for the case when the
functions f and g are not defined at x → a. So, what we have taken in the previous two results
that f (a) is 0 and g(a) is 0. So, those at that point the function values was 0, but the same
rule one can apply if for sample function these two functions are not defined exactly at a, but
g( x)=0. So, in this case also we can apply
their limits are 0. So, the limit x → a is 0 and lim
x →a
the same result what we have established earlier another remark.
So, if we realize that the first derivatives are also 0 at a and the derivatives f ' and g ' they
satisfy the conditions that were imposed earlier on functions f and g mainly the continuity
differentiability then applying the L’Hospital’s rule to this ratio. So, we can do we can apply
the L’Hospital’s rule again to this f ' over g ' because the similar situations happening now for
f ' and g '.
Because they both are 0 and then the rule says that this limit here f ' g ' will be equal to the
double derivatives, the ratio of the double derivatives of f and g as x → a. So, the this is
again more generalized form that this limitf / g can be evaluated by the limit of the ratio of f '
g ', but if these two f ' and g ' become 0 as x → a or x=a then we can again apply the
L’Hospital’s rule.
So, the same limit will be equal to the limit of f the second derivative divided by g, the
second derivative as x → aor we can continue this further if for example, the f ' sorry f ' '
44
here. So, the double derivative of f also vanish at x is equal to a and this double derivative of
g also vanishes at x is equal to a. So, we can further apply this L’Hospital’s rule together
limit of this f double derivative divided by g double derivative.
(Refer Slide Time: 15:32)
So, L’Hospital’s rule is also applicable. So, another generalization here that not necessarily
that x → awe have just discuss that x → a was some finite number, but one can also apply this
result when limitx →+∞ or x →−∞. So, this is a very general rule which we are not proving
here for example, this infinity case, but one can apply the L’Hospital’s rule their too.
So, now the extension of this L’Hospital’s rule to the infinity by infinity form. So, suppose
this f ( x ) → ∞ and g ( x ) → ∞ as x → a or x → ± ∞ similar to the earlier case. But, the now the
differences that we have instead of f ( x ) → 0 g ( x ) → 0 they both are tending to ∞. And, in this
case also we have the same rule that this limit of the ratio of these two functions will be the
limit of the ratio of their derivatives; when this f ( x) and g( x) goes to 0 provided this right
that the limit at the right hand side here this exists.
So, limit f ' / g ' exist. So, what is the general rule now if we include those all results what we
have discussed so far, that they are two they are could be two forms. So, either 0/ 0 form or
∞ / ∞ form. In either case whether x → a or x → ± ∞ the limit of the ratio of the two functions
will be equal to the limit of the ratios of their derivatives.
45
(Refer Slide Time: 17:20)
So, this is another important remark which I have mentioned before. So, if this limit here
does not exist, if the limit of the derivatives does not exist; it does not mean that the limit of
f / g does not exist which we can see by the simple example. So, if we consider this
lim
x →∞
x+1
⁡
x
So, in this case what is happening if we just see what is the form here; so, x goes to ∞ plus
here something finite. So, the numerator is ∞ and divided by again x goes to ∞. So, we have
the ∞ / ∞form and in this case if we simply apply the L’Hospital’s rule what will happen.
So, here if you take the derivative of the numerator it is a 1+sin x will become cos x and the
limit of the derivative of thisx is 1 and the limit x → ∞. So, here the limit x → ∞ 1+ cos x
since this cos x when x → ∞ is not defined. So, basically this limit here 1+ cos x and as x → ∞
is not defined. So, this limit does not exist, but if we evaluate this in some other ways like
x → ∞ x+sin x and we rewrite this as1+
and separate it. So, we have
now this limit; so, 1+
sin x
. So, we divide this x here to x and then sin x
x
x sin x
sin x
+
meaning this 1+
and now we can directly evaluate
x
x
x
sin x
.
x
46
So, when x → ∞. So, if this x → ∞ here and this sin x something finite is sitting there. So,
something finite and divided by ∞ this will go to 0. So, the second part here
sin x
as x → ∞
x
will go to 0. So, we have 1+ 0 means this limit is 1. So, if we would have concluded here by
applying the L’Hospital’s rule, because this limit does not exist. And, we could have claimed
that the
lim
x →∞
x+sin x
x
does not exist, but that would have been a wrong conclusions. So, that is what in the rule
every time we have written provided the limit of the ratio of the derivatives exist. So, that is
very important.
(Refer Slide Time: 20:04)
Now, one example here so, let this α , β ∈ R . So, they are the real number and we have this
f ( x )=
{
α tan x + β sin x
,x≠0
x3
1 , x=0
So, in this case we want to find for what values of α and β the functions f is continuous. So,
the function f is continuous in the interval
(
−π π
, . So, now for the continuity what do we
2 2
)
47
need? So, for the continuity of this function this limit of this
α tan x+ β sin x
should be 1,
x3
because at x ≠ 0 the function is defined as 1. So, rest everywhere the function is continuous
the 1 the problem is at x=0.
So, here for x ≠ 0 we have this nice function defined over −π ¿ π the tan. So, it is a
continuous sin is continuous x3 is continuous. So, the function is continuous, the only
problem it could create at x=0. So, we are now setting here that if limit this
x→
α tan x+ β sin x
=1 then this function will become continuous. So, out of this condition
x3
we will compute α and β. So, for what values of α and β this expression here or this limit
here is equal to 1. So, now let us compute this limit here
α tan x+ β sin x
. So, when we take
x3
x → 0 the tan 0 is 0 sin 0 is 0 and x 3 is also 0 .
So, we have basically the 0/ 0 form. So, let us apply the L’Hospital’s rule to this expression.
So, if we apply L’Hospital’s rule α and tan x will become sec2 x+ β sin x will become cos x
and divided by 3 x2. So, the x2 x 3 when we take the derivative will become 3 x2 and we take
the limit here x → 0. So now, if you realize what is happening to this function now here. So,
we have the α and then x → 0 this sec x which is 1. So, here you have α + β cos 0 is also 1. So,
we have here α + β divided by 3 x2 and then x → 0. So, here we are getting this 3 x2 is going to
0.
So, we have α + β divided by something which is going to 0. Now, the only possibility to
move further or to have this limit as 1 will be when α + β is equal to 0. Because then we will
get 0/ 0 form and we can further apply the L’Hospital’s rule. But in this case so, what we can
2
α sec x+ β cos x
set to move further that this
to have this limit as 1, we can set that α + β is
2
3x
equal to 0 because when x → 0 then we can move further and apply the L’Hospital’s rule
again. So, applying this L’Hospital’s rule again so, we got already one condition on α + β
which is equal to 0 and now if we apply. So, again so here the derivative of α sin 2 so 2 α sin x.
So, 2 α sorry sec x and the derivative of sec x will be sec x tan x−β because cos x will give
you −sin x.
48
So, it is a −β sin x and divided by 6 x and now if we check what form we are getting now
here. So, this tan x will make this0 here sin xwill make the 0. So, we are getting 0 in the
numerator and divided by 6 x which is again 0. So, we are getting 0/ 0 form. So, we can apply
the L’Hospital’s rule once again to this expression. So, here limit x → 0. So, here we have this
2
2
sec x and tan x . So, the sec x will give 2 sec x sec x tan x . So, then it becomes 4 x 4 alpha and
sec x sec x tan x and the tan x remain as it is plus this 2 α sec 2 x and then tan x will become
again the derivative sec2 x−β sin x will become cos x and divided by the 6 here because this
was 6 x and derivative is 6.
So, now if we check again what is the value here? So, this sec x 1 tan x will be 0. So, this
expression will become 0 and then here when x → 0 this is like 2 α and then −β. So, this in
the numerator we are getting 2 α−β and divided by 6 and the limit x → 0. So,
2α −β
. So, to
6
have this value as 1 we need to set that 2 α−β=6. So, another condition we got that
2 α−β=6. So, if you solve these two equations α + β=0 and 2 α−β=6. So, we will get that
α =2 and β=−2.
So, for these values of α and β this function will become continuous or in other way this limit
here
α tan x+ βsin x
x3
will become as1. The function was defined as 1 at x=0.
49
(Refer Slide Time: 26:38)
So, these are the references which we have used to prepare these lectures. So, the integral a
Differential and Integral calculus by Piskunov and this is Volume 1; the Kreyszig Advanced
Engineering Mathematics and also the Thomas’ Calculus.
(Refer Slide Time: 26:55)
So, what did we learn today these indeterminate forms, they may take these several forms
like 0/ 0, ∞ / ∞, 0 ×∞, ∞−∞, 00, ∞ 0 1∞. So, what we have learn today how to compute such
limits when we have the 0/ 0 or ∞ / ∞ form. And, the L’Hospital’s rule which was useful to
compute this limit was that whether we have the 0/ 0 or ∞ / ∞form here for the ratio f / g(x).
50
We can apply this rule which says that this limit will be equal to the limit of the ratios and if
again this f ' and g ' they both becomes or takes the form 0/ 0 or ∞ / ∞ then we can again apply
the rule. And, then we will get this limit is equal to the limit of the ratio of the second
derivatives and so on we can continue further till we get the limit.
But that important point was that these rule is valid when, when those limits exist we cannot
conclude if those limits here of the derivatives do not exists then we cannot conclude that the
original limit does not exist. So, this rule is a very useful rule. In the next lecture we will
learn now how to deal the other forms; for example the 0 ∞, ∞ / ∞ and so on.
Thank you.
51
Engineering Mathematics – I
Prof. Jitendra Kumar
Department of Mathematics
Indian Institute of Technology, Kharagpur
Lecture - 04
Indeterminate Forms Part - 2
Hai, welcome to the lectures on Engineering Mathematics I and today we will be continuing
this Indeterminate Form Part 2 and this is lecture number 4.
(Refer Slide Time: 00:32)
So, we are discussing differential calculus and functions of single variable, and today this
indeterminate forms of the type infinity minus infinity 0 power 0 infinity power 0 and 1
power infinity will be discussed.
52
(Refer Slide Time: 00:47)
Just to recall from the previous lecture we have discussed already these two forms or rather
these two indeterminate forms of the type as 0 by 0 and infinity by infinity. There what we
have seen that this L’Hospital rule was very helpful which says that the ratio of this f ( x) and
g( x) where f ( x) and g( x) both either goes to 0 or they both go to infinity. So, in that case
the limit of this ratio will be equal to the limit of the derivatives provided this limit on the
right hand this exist.
And we have also seen that we can continue this process provided that all the conditions on
this f ' and g ' satisfies and again we have the situation that this f ' and g ' both they go to 0 or
they go to ∞. So, in that case the rules says that we can continue this process of these
differentiation and again we can differentiate the numerator and again the denominator. We
can take the limit till the time we achieve this limit.
The important point was that this limit of the ratio of these two functions exist when the ratio
of this derivative of the. So, this limit here exist otherwise for example, this limit does not
exist we cannot claim that the original limiter here of the ratio f ( x) over g( x) does not exist.
53
(Refer Slide Time: 02:28)
So, the today we will continue with the indeterminate form of the type 0 into ∞. So, we have
the situation suppose f ( x) goes to 0 and this g( x) goes to 0 goes to ∞ sorry. As x goes to a
then this product of f ( x) into this g( x) as x goes to is a is indeterminate, because this is
precisely 0 and into ∞ which we have to evaluate using the L’Hospital rule discussed in the
last lecture. So, in this case when we have such a product where one f ( x) this function goes
to 0 and the other one goes to ∞.
In that case we can rewrite this expression here
f ( x ) × g ( x)=
f (x)
1
g(x)
So, in this case here f ( x) goes to 0 and then this 1/ g(x ) will also go to 0. So, we have here
0/0 form we can also rewrite in this form that we keep this g( x) in the numerator and in the
denominator we take this f ( x) as 1/ f ( x). So, in this case this g( x) goes to ∞ and 1/ f ( x)
since f ( x) goes to 0. So, this term here 1/ f ( x) also goes to ∞. So, we have ∞ / ∞ form. And in
either case we know that we can apply the L’Hospital rule. So, applying the L’Hospital rule
we can get the limit because we have already learnt in the last lecture how to deal search
forms 0/0 or ∞ / ∞.
So, let us take a simple example here, we want to apply this L’Hospital rule to this problem
54
x sin
2
x
()
and x goes to ∞. So, here we have x goes to ∞ and sin
( 2x ) so, 2x goes to 0. So, this is ∞ ×0
form. So, in this case we will use this idea which is discussed above that we can make either
0/0 or ∞ / ∞ form. So, in this case we can bring this x to the denominator as 1/ x and then
sin
( 2x ).
So, in this case we have the 0/0 form. So, sin
( 2x )goes to 0 and 1/ x also goes to 0. So, in this
case we can now apply the L'Hospital rule because we have the 0/0 form and then the
derivative of the sin 2 xwill be the cos 2 x and the derivative of 2/ x will be
we have
(−2x ) and here also
2
−1
2
1
2 . So, this is the situation now this 2 and this 2 .
x
x
x
So, the x2 terms get cancelled and then we have here cos
( 2x )as x goes to ∞. So, x goes to ∞,
2 goes to 0 and then
cos 0 goes to 1 and we have this 2 here so, the limit is 2.
x
(Refer Slide Time: 05:54)
55
Now, as we have seen in the last slide that although this L’Hospital rule can be apply to 0/0
or ∞ / ∞ form, but one may be better in a particular case this we will see with the help of
example in a minute.
So, the point is that we can change between these form. So, for example, this f /¿0 is of the
form 0/0 then we can rewrite it as
1
1
divided by and in this case if thisf and g both goes to 0
g
f
here we have the ∞ / ∞ form. Or other way around if f and g both goes to ∞ in that case 1¿ g
divided by 1/ f we will have form of 0/0. So, we can interchange these two forms 0/0 and
∞ / ∞ depending on the convenience of the derivative there. So, for instance we consider this
example the limit x goes to 0 from the right side x n ln x the natural logarithmic n so, here n is
a natural number.
So, in this case it is much convenient to consider this is 0/0 form sorry ∞ / ∞ form. So, we
keep this ln x in the numerator and bring this x n as
1
in the denominator. So, in this case
xn
when x goes to 0 the numerator goes to infinity and also here the denominator goes to
infinity.
So, we have ∞ / ∞ form. And in this case the derivative of this ln x is simply
derivative of
1
and then the
x
1
−n
and then we can simplify this. So, we have the limit 0 because
n is minus
x
xn+1
here we have the x power n plus 1and n is a natural number. So, 12 3 so, here whatever n is
you will get some x power something there and x goes to 0 it will become 0.
But, in the same example if you would have consider x n and over ln x then the derivative of
this
1
would have created a problem there and it was certainly it would have not been such
ln x
a simple calculation. So, we have to observe that which form whether 0/ 0 or ∞ / ∞ is
convenient in a particular example.
56
(Refer Slide Time: 08:34)
So, now we will discuss the indeterminate form of the type ∞−∞. So, suppose f ( x) goes to
infinity and g( x) goes to infinity as x goes to a then this f ( x) minus g( x) this difference here
as x goes to a is indeterminate because, we have ∞−∞ form. And we have already discussed
in the last lecture that this ∞−∞ form is an indeterminate form and we have to be careful to
evaluate such limits.
So, we can again the rewrite this f ( x )−g ( x ) term; if we divide here f ( x )−g ( x )by f ( x ) g ( x )
and multiply by f ( x ) g ( x ) and then this term here we have the
(
1
1
−
g(x) f (x)
.
1
f ( x ) g( x)
)
So, in this case we have the situation and then since f ( x) and g( x) both goes to infinity. So,
we have this 0 and then again minus 0. So, 0 here and 1 over ∞ into ∞ will be ∞ it is not an
indeterminate form so, 1 over ∞ this is 0 again. So, we have here 0/0 form which we can
easily handle with the help of L’Hospital rule.
So, in this case let us take the simple example again if we have this
lim
x →0
(
1
1
− 2
2
x sin x
)
57
So, here since x goes to 0 so, we have this 1 over 0 this which is which is going to ∞ and
minus again here 1 by 0 it is also going to ∞. So, we have ∞−∞ form in this problem.
So, as discussed above here we can rewrite this as sin 2 x and then this −x2 divided by x2 sin 2 x .
So, in this case now the sin square x goes to 0 and minus the 0 so, we have the 0 divided by 0
. So, this ∞−∞ form changes to 0/ 0 form which using L’Hospital rule we can get the limit.
(Refer Slide Time: 11:04)
So, here moving further. So, when we take this when before we applying the L’Hospital rule
we just rewrite this expression in this form. So, we take 1 x2 there and divided by 1 x2 here.
So, this x2 and sin 2 term together and then the rest here sin 2 x−x 2 there was x2 already and 1
we have divided so, we have x 4. So, this limit sin 2 x−x 2 divided by x2 sin 2 x we have
rewritten in this form the limit of x2 sin 2 x and the limit of sin 2 x−x 2 over x 4.
And now let us consider the 2nd one 1st so,
2
2
sin x−x
.
4
x
So, in this case we have again this 0/0 form. So, sin 2 x goes to 0 minus 0 and then divided by
0. So, we can apply the L’Hospital rule here which says that the derivative of the numerator
will be 2 sin x and the derivative of sin x will be cos x and minus the derivative of x2 will be
58
2 x divided by the derivative of x 4 which is 4 x 3 and the limit x goes to 0. So, again here we
have the sin 2 x 2sin x cos x is sin 2 x −2 x divided by 4 x 3.
So, sin 2 xgoes to 0 minus this x goes to 0. So, we again end up with 0 0 0 by 0 form which
we have to differentiate again. So, here we have the sin 2 x which will become 2 cos2 x and
minus the derivative of this 2 x will be just 2 and the 4 x 3 which will become the 12 x2. So
now, here when we check this limit again so, 2 cos2 x and x goes to 0.
So, this is 1 so, 2−2 which is 0 so, again we have 0/0 form. But, what we can do now we can
simplify a little bit so, which is cos 2 x we can write down as so, let me just workout here. So,
this cos 2 x term we can write down as 1 minus 2 sin square x and then you have minus 1
there. So, the 2 we can take common and we will divide to this 12 here.
So, this is 1 minus 1 we will get cancel and we have minus 2 sin2 x and divided by 6 x2. So,
this 1/3 so, minus 1/3 and then here we have the limit x goes to 0 sin 2 x over x2 . And now this
limit here is x goes to 0 sin 2 x over x2 is similar to what we have their x2 over sin 2 x . And we
will see in a minute there these both limit can be handle in a similar fashion.
(Refer Slide Time: 14:45)
So, what we have let us consider this 1 we could have considered sin 2 x over x2 both will
result to the same limit. So, you just consider limit x goes to 0 x2 over sin 2 x . And in this case
we can apply again a L’Hospital rule because this is 0/0 form and then we have limit here.
So, the derivative of x square will become 2 x and then the sin 2 x will give us 2 sin x cos x. So,
59
here we have 0/0 form again and then so, the derivative of the numerator will give us 2 and
the derivative of the denominator which is sin 2 x will give us 2 cos2 x.
So, now if we take the limit here x goes to 0 so, the cos 0 is 1 so, 2/2 the limit is 1. The same
thing if we consider here sin 2 x over x it will result exactly in the same form and will lead to
the limit 1 because, we will have just the reverse the denominator will become numerator and
numerator will become denominator and we will end up with limit 1. So, here now getting
2
back to the original limit. So, this limit x goes to 0
sin x
2 is 1 and the limit here for the second
x
term which we have evaluated which was minus 3 and this one becomes 1.
So, the
lim
x →0
So,
(
1
1
−1
− 2 =
2
3
x sin x
)
−1
−1
−1
this is 1 and this is
so, the limit is
.
3
3
3
(Refer Slide Time: 16:36)
Now, the next example again of this nature is like
60
(
lim x+
x →∞
1
1
ln 1−
x
( )
)
.
So, in this case the question is what is the form of this expression when x goes to ∞. So, let
us just check so, here x goes to ∞. So, we have the ∞ here now this
lim 1
x→∞
1
ln 1−
x
.
( )
So, here we have to be careful when x goes to ∞. So, this is 1 minus something this
1
expression here is 1− . So, this is less than 1 and the logarithmic here ln .
x
So, if we just draw the graph so, here it is 1. So, less than 1 the logarithmic take the negative
( 1x )are taking all negative value. So, when x goes to
value. So, here in the denominator ln 1−
∞ this is going to 1, but 1 over and so, all these values were negative. So, 1/0 this is actually
tending to −∞. So, what we have there this form is −∞−∞. form.
(Refer Slide Time: 18:02)
And therefore, we can apply so, again I have written down. So, that this term here
61
lim 1
x→∞
1
ln 1−
x
=−∞
( )
so; we have ∞−∞ form.
( 1x )
And now we will deal exactly as done in the previous example. So, we have x over ln 1−
we can rewrite this
( 1x )+1
1
ln ( 1− )
x
x ln 1−
So, we need to check again what is the limit now here. So, when x goes to ∞ so, this 1 ln 1
( 1x ) as limit x goes to ∞.
goes to 0. So, here we have 0 now this 1 this x ln 1−
So, in this case we have the ∞ here and then ln 1 so 0. So, we have to rewrite this as limit x
( 1x ) divided by 1/ x. So, now, this is go going to ln1 that is 0 and here
goes to ∞ and ln 1−
also 0. So, we have 0/0 form and we can apply the L’Hospital rule. So, here ln 1−
1
will
x
become the derivative x over x−¿1 and the derivative of this term which is 1/ x2 and then we
have here also
1
with minus sign.
x2
So, this gets cancelled and then we have with minus sign. So, x and then minus 1 plus 1
divided by x−1. So, which we can write down as 1 plus 1 over x minus 1 and when limit x
goes to ∞. So, this goes to 0 and we get this limit as here -1. So, in this case we have 1 minus
1 0 and divided by 0. So, this is basically 0/0 form.
62
(Refer Slide Time: 20:19)
And now we can apply the L’Hospital rule. So, we have to get the derivative of this term and
the derivative of this term. So, the derivative of this term will be the product rule will be
( 1x )then plus
applicable here. So, derivative of x is 1 and then we have exactly this term ln 1−
x will remain the derivative of this which we have just evaluated before that was
1
x
and 2
x−1
x
x 1
and this x makes this x2 and again here this is
. So, in this case it is a 0 and then here it
x−1 x 2
is again 0.
Similarly, here also 0 so, we have again 0/0 form and we need to apply the L’Hospital rule to
this expression once again. So, the derivative here is
derivative of this term which is
1
as done before. So, and the
x−1
1
1
1
will become minus
. So,
2 here we have
x( x−1)
x−1
( x−1 )
the derivative will be −x2 ( x−1 )2 and the derivative of this x ( x−1 )which will 2 x−1.
So, the now again we need to simplify this term a little bit and what we will get. So, if I
multiply here in the numerator by x2 and x−1. So, what we will get; we will get x ( x−1 ) and
then −x2 here with the negative sign and then this will become 2 x−1. So, this x with x2 will
63
get cancel we have this 1−x and this minus will make it plus. So, x over 2 x−1 and then
again here the 2 so, we can divided by 2 and here −1+1. So, we have this 1+
1
and
2 x−1
when x goes to ∞. So, it will become 1 and the half is sitting there so, this value will be half.
(Refer Slide Time: 22:43)
So, this
(
lim x+
x →∞
1
( 1x )
ln 1−
)
=
1
2
(Refer Slide Time: 22:49)
64
Now, we will move to the indeterminate form of the type 0 0, ∞ 0 and 1∞. So, all these type of
limits we can handle all together. So, we consider this f ( x )g( x) as x goes to a and suppose this
f ( x) and g( x) have the following properties.
So, the 1st one f ( x) goes to 0 and g ( x ) goes to 0 so, we have this limit here as 00. In the 2nd
case we will consider that f ( x) goes to 0 and this g( x) goes to 0. So, in this case we have ∞ 0
form and in the 3rd case we take as f ( x) goes to 1 so, we have 1 here and this g( x) goes to
∞ so, 1 power infinity the 3rd form. In all these cases we consider a new function here y as
f ( x) power g( x) this one.
So, f ( x) power g( x) in either case when we take the limit as ln y ( x) it will become g( x)
ln f (x) and now consider for example, the 1st case when f ( x) goes to 0. So, this will go to
−∞ and g( x) goes to 0 so, 0 ×∞ form. In the 2nd case when f ( x) goes to ∞ . So, this will
become infinity and g( x) goes to 0. So, again 0 ×∞ form in the 3rd case when f ( x) goes to 1
when f ( x) goes to 1 and g( x) goes to ∞. So, ∞ and ln1 will become 0 so, we will have
again the 0 ×∞ form.
So, in all these cases whether f ( x) goes to 0 g( x) goes to 0 means 00form ∞ 0form 1∞ form in
all these cases we will have 0 ×∞ form once we take the logarithmic of this expression and
then we take the limit; so while taking the limit here this will become ln and since this is a
continuous
function.
So,
we
can
bring
this
limit
to
the
argument
here
ln lim ⁡y ( x ) =lim g ( x ) ln f ( x ) .And now here as we have already discussed that these are the
(
x→a
)
x →a
forms of 0 ×∞ form which we can deal as we have done before and suppose that this limit is
L. So, once we compute this limit and we can take this algorithm exponential both the sides
to get this limit x goes to a y ( x) because this was the desired limit here this was the y ( x).
So, limit x goes to a y ( x) will become the exponential of this number L. So, what we have
seen that all these types of form 00 ∞ 0 and 1∞ they can be handled by taking the logarithmic
of this function y which we have assumed here f ( x) power g( x).
65
(Refer Slide Time: 26:08)
Now, let us take the example here x x and x goes to 0. So, in this case as discussed before we
will take y ( x) is equal to x x we will assume this x x as y and then take the logarithmic here.
So, log y will become x ln x and in this case. So, we have the limit now of x ln x.
So, we have 0 ×∞ form which we have already discuss before so, we will bring into this ∞ / ∞
form. So, ln x and 1/ x and now apply the L’Hospital rule here. So,ln x will give 1/ x and here
2
−1/ x which is simply x here and x goes to 0. So, this will become 0 and then taking the
exponential of both the sides. So, we will get the limit y is equal to e 0here which is 1.
Another example if you consider here x →0+¿lim
(cos x )
1/ x
¿
¿which is ∞ so, cos x goes to 1 and 1/ x
1
goes to ∞.
So, we have 1∞ form and the similar process we assume y as (cos x)1/ x take the logarithmic.
So, log y will be
ln cos x
and now this is log 1 and here again goes to 0. So, 0/0 form once we
x
take the limit. So, we will apply the L’Hospital rule. So, we will take the derivative here of ln
cos x which will be
1
and this cos x the derivative of cos x will be −sin x divided by the
cos x
derivative of 1 which is 1. So, we have here minus 10 x x goes to 0 whether this goes to 0.
So, we have this limit is 0 and then taking this exponential both the sides we will get the limit
y which was the desired function here (cos x)1/ x it goes to 1.
66
(Refer Slide Time: 28:22)
Finally we consider one more example of this limit 1/sin x and 1/ln x; x goes to 0. So, we
have the ∞ 0form and again taking the limit here or taking the logarithmic we will get
ln y=
1
1
ln
ln x
sin x
( )
So, here it will become 1/ln x x goes to 0 so,∞. So, it is a 0 and then here ln ∞. So, 0 ×∞
form and then we can rewrite it as while taking the logarithmic I am we taking the limit now.
So, this is
1
1
1
ln
. So,
is ∞ here. So, ln ∞ ∞ and divided by the ln x. So, this is
ln x
sin x
sin x
( )
also going to ∞. So, we have ∞ / ∞ so, the numerator is ln
we can apply the L’Hospital rule. So, here ln
the derivative of
( sin1 x ) and denominator is ln x now
( sin1 x ) the derivative of this will be sin x and
−1
1
will be
and the derivative of sin x will be cos x.
sin x
sin 2 x
So, this is the derivative of the numerator term ln
( sin1 x ) and then ln x will give 1x . So, now,
we can simplify this so, we have here the cos x /sin x term and then the x there. So,
x cos x /sin x and now we know that this x sin x this limit as x goes to 0 becomes 1 and then
67
cos 0 is also 1. So, we have this −1 and then taking the exponential both the sides. So, we
will get ln x goes to 0 y as e−1. So, we have considered at least 1 example of each nature.
(Refer Slide Time: 30:26)
And these are the references used for preparation of this lecture.
(Refer Slide Time: 30:32)
So, what we have learnt today the indeterminate form of 0 ×∞ ∞ ×∞ and these three which
were handle using the logarithmic function. In all the cases we have used the L’Hospital rule
68
which says that the limit of the ratio of the two functions when they go to the 0/0 or ∞ / ∞
form will be equal to the limit of the derivatives ratio of the derivatives.
So, that was a very useful rule L’Hospital and we have handled with the help of the single
concept all these types of indeterminate form. So, that is the end of the lecture.
Thank you very much.
69
Engineering Mathematics - I
Prof. Jitendra Kumar
Department of Mathematics
Indian Institute of Technology, Kharagpur
Lecture – 05
Taylor Polynomial and Taylor Series
Welcome back to the lectures on Engineering Mathematics-I and today’s we will learn
Taylor’s Polynomial and Taylor Series.
(Refer Slide Time: 00:22)
So, these are the topics you will cover will start with the Taylor’s polynomial and then
coming back to this Taylor series from the Taylor’s polynomial and some worked out
examples.
70
(Refer Slide Time: 00:36)
So, this Taylor’s formula which is a generalization of the mean value theorem or the
Cauchy's mean value theorem which we have learned in the last lecture. So, we assume that
the function here f has all the derivatives up to the order n plus 1.
In some interval which contains the point x is equal to x 0. Having this we wish to find this
polynomial Pn ( x) of degree n, with the conditions that this polynomial satisfies that Pn ( x)
polynomial at x is equal to 0 is equal to the function value at 0. Second: that the first
derivative of this polynomial is equal to the first derivative of the function. The third
condition the second derivative of this polynomial at this point x 0 is equal to the second
derivative of the function at the x 0 and so on.
So, what we assume basically that all these derivatives, the function value itself the first
derivative, the second derivative, and n-th derivative they all are equal to this derivative of
the polynomial. So, having these conditions what do we expect from such a polynomial? We
expect such a polynomial if we construct then there should be close to the function f
naturally at x 0 function value is 0. So, at x 0 several so the derivatives of 2 order n are equal.
But in general also we expect that because of these conditions that this polynomial of degree
n and somehow in some form will represent the function f .
So, now the question is how to construct such a polynomial.
71
(Refer Slide Time: 02:33)
So, we assume here a general polynomial of degree n. So, we take that
2
3
P n ( x )=c0 +c 1 ( x−x0 ) +c 2 ( x−x0 ) +c3 ( x−x0 ) +…+ cn ( x−x 0 )
n
So, in this is special form we have taken this polynomial because of the convenience to
evaluate these unknown c 0 , c 1 and c 2, but one can also assume any general polynomial of
degree n. So, now we want to find these coefficients c i’s based on the conditions which we
have set or we wish to have that this up to n-th order derivative of this polynomial at the point
x 0 must be equal to the respective derivative of the function at the same point x 0.
So, having this condition first we know that the first derivative of this polynomial here is
equal to. So, once we this is a constant so this will be equal to 0 and then we take the second
term which will be having here x there so we will get the c 1 out of this. Similarly from here
we will get 2 c2 ( x−x 0 ) then we will get here 3 c3 and so on.
Then if we move further, than the second derivative, so out of this first derivative we can
again differentiate this you will get 2 c2 here 3 ⋅ 2c 3 and so on and then we can repeat this
process further to get the n-th order derivatives. In n-th order derivative because this was n-th
order polynomial we will get only a constant term which will see here the n ! and the
coefficient the c n will come in the expression here and there will be no x term present here
because, the polynomial was degree and then we have differentiated n times.
72
So, having this what we get out of the first condition when we substitute because, our
conditions is the polynomial value at x 0 must be equal to the function value at x 0 and further
derivatives upto order n. So, from the functional value we will get here P n x 0and is equal to
all these terms will vanish and we will get only c 1. So, P n; sorry from the from the
polynomial itself when we substitute x is equal to x 0 we will get c 0 as Pn ( x0 ) and Pn ( x0 ) is an
f n (x 0 ), so we get the c 0 as f ( x0 ) over here.
The second when we substitute in this first derivative so you will get c 1, as the first derivative
of the function the c 2 again from this condition so we will get the second derivative at x 0
divided by this factorial; sorry to factorial 2 here and then the
f ( n ) ( x0 )
c n=
.
n!
(Refer Slide Time: 06:00)
So, having these coefficients now what we have we have the polynomial, the polynomial says
that this Pn ( x) will be f ( x0 ) which was c 0 there and all these coefficients are substituted here,
so this is what we call the Taylor’s polynomial of order n.
Now if you go through the example here and construct the Taylor’s polynomial of the
exponential function for example, e x around x=0. So, the P0 ( x) the degree 0 polynomial
only the first term will be present there and e 0, so e 0 will be just 1. So, the polynomial of
73
degree 0 will be simply 1. And if we plot this, so this is the green plot here of the exponential
function and this polynomial of degree 0 is just a constant line; so the straight line going
through this 0 1 point. And if we compute the polynomial of degree once we will get this
f ( x0 ) as 1 and then we have the derivative here e x the derivative will be e x and then at x=0
this will again 1.
So, we have the polynomial of degree 1 as 1+ x which is plotted here. So, it is again the
straight line with slope 1. And now if we continue this process we can evaluate P 2 then P 3, so
P 3 again we have plotted here with this curve and then going further so we will get like P 5.
So, for example, if we plot here P 5 then this P 5 if you see it is pretty close to the exponential
function in a very wide range of or in a wide interval around this point 0 which was the point
of this expansion here. And if we move further for the polynomial of degree 6 or 7, then we
will be moving closer to the exponential function and that’s the idea of the Taylor’s
polynomial. So, by increasing the degree of the polynomial we can approximate our function
as good as we like and we will discuss on these further.
(Refer Slide Time: 08:22)
So, what is the relation of the Taylor’s polynomial and the function. That means, because this
Taylor’s polynomial representing the Taylor functions to some approximation. So now, we
will denote this Rn ( x) as the difference between the values of the given function and the
constructed polynomial Pn ( x). So, in this case what we mean this the Rn ( x) we define this
74
difference of the function and the polynomial Pn ( x).. And in this case now the Rn ( x) is called
the remainder, because this is the difference between the actual value of the function minus
the polynomial value at the point x.
So, now the question is the how to evaluate this R n ( x) and to go further for the evaluation we
first note that the R n at x 0, because f ( x 0 ) −P n ( x0 ). and by construction this polynomial at x 0 is
equal to the function value at x 0 so this is 0, and the first derivative the same thing. So, the
first derivative of a f at x 0 is equal to the first derivative of the polynomial at x 0 this was the
construction of this polynomial. So, all these n-th order derivative are equal, so in this case
therefore this R n at point x or the first derivative at point x 0 the n-th derivative at point x 0 all
are equal to 0.
(Refer Slide Time: 09:57)
So, we have these conditions Rn ( x), now we consider another function here which is
g ( x ) =( x−x 0 )
n+1
,∀ x ∈ I
So, this function has also the property if we notice this that g ( x 0 ) =0 in fact the first
derivative, because n plus 1 and x minus 0 power n that will also become 0 at the point x 0
and so on. So, we can continue this differentiation here up to the n-th order and all these
derivatives at point x 0 will be 0. So, what condition we have now that
75
( k)
(n+ 1)
g ( x0 ) =0 ,k =0 , 1, … , n∧g
( x0 ) =( n+1 ) !
So, we have two functions one is Rn ( x) and another one is g( x) they have similar properties
here that all these derivative upto order n they vanish.
So, in this case if it take to x point in the interval and suppose that x is bigger than x 0 we can
also assume that x is less than x 0. And now we apply the Cauchy mean value theorem for
these two functions the R n function and the g function in this interval x 0 to x and now
remember what was the Cauchy mean value theorem that you have this function
R n ( x )−Rn ( x0 )
¿¿
Now if he noticed that R n ( x0 ) =0 and also the g ( x 0 ) =0, so we have this result that are
'
⟹
Rn ( x ) R n (ξ1 )
=
, x0 <ξ 1 < x
g ( x ) g' ( ξ 1 )
we can continue this process.
(Refer Slide Time: 12:12)
76
So, further if we apply again the Cauchy mean value theorem for the derivatives R n ' and g '
in the interval [ x0 ,ξ 1 ]. So, what we will get we will get because this first derivatives are also
0. So, simply we will get the result that the
''
Rn ( x ) R n ( ξ2 )
=
g ( x ) g' ' ( ξ 2 )
Now the some point in between this interval which we have considered from x 0 to ξ 1. And
now we can continue this process further for the next derivative supplying this Cauchy's
mean value theorem further what you will get we can go up to the (n+1)th derivative
because, up to n-th derivative R n and g both are 0.
So, we will end up with this term and then here the ξ n+1lies between x 0 and the ξ n and there
was a continuity up to x there. So, what we get out of this
⟹
(n+1 )
R n ( x ) R n ( ξ n+1 )
= ( n+1 )
, x 0 <ξ n+ 1 <ξ n <…<ξ 1 < x .
g ( x ) g ( ξ n+ 1 )
So therefore, this factorial and plus 1 term came here and then we have this g( x) which was
the function ( x−x 0 )
n+ 1
and the ξ which is introduced here ξ n+1. Now we have replaced by ξ it
lies somewhere between this x 0 and the point x, so which is written here the ξ lies between x 0
and x.
(Refer Slide Time: 14:12)
77
So, now we note that these Rn ( x) which we have define that was the difference between the
function f ( x) and Pn ( x). So, if you take the n+1 and derivative here of this reminder term
the (n+1)th derivative of this a f minus the (n+1)th derivative of this p is equal to the (n+1)
th derivative of x because the (n+1)th derivative of n and P n was the polynomial of degree n.
So, if we take the (n+1)th derivative this term will become 0 and we have these
R (nn+1 ) ( x ) =f ( n+1 ) ( x ). And now we can substitute in our formula which was the R n
(n+1)
here, so we
have substituted now this value of this R (n+1)
as f ( n+1 ) ( ξ ).
n
Well now, we got the polynomial here which is the remainder term the Rn ( x) which is the
Lagrange form of the remainder, this term we can also rewrite this remainder form in this
form. So, this ξ which appear there we have just replaced by this x 0 +θ ( x−x 0 ) ,,0<θ<1. So,
here if θ is close to 0 then this argument here is moving to x 0when theta goes to one this
argument here goes to simply x. So, this argument of this f lies between x 0 and x, so it has
the same similar meaning what the other form has. So, we can we write this Lagrange form of
the remainder in this form as well.
(Refer Slide Time: 16:18)
The Taylor’s theorem or the Taylor’s formula now if you summarize we have the function
'
f ( x) which can be expanded in this form of f ( x0 ) f ( x 0 )( x−x0 ) up to the this n-th order term
and plus this reminder term R n ( x ) which we have just derived as this form which is called the
Lagrange form of the reminder.
78
So, in the special case when we take n=0 so that means, this up to the order one we have to
'
f (ξ )
write this reminder term. So, we get this
and then we have this x minus either way or x 0
1!
whatever we consider. So, then this x the ξ lies between this a and x. And in this case we get
this form of the Taylor’s theorem which was which was the Lagrange form mean value
theorem. So, this is a special case of the mean value theorem which we have seen just by
putting n=0 in this case.
(Refer Slide Time: 17:20)
Some remarks so if we set x 0=0 point of expansion in this Taylor’s formula then it is called
the Maclaurin’s formula and in the Taylor’s formula if it is reminder Rn ( x)→0 as n → ∞, so
this is an important remark here. If this Rn ( x)→0 as n → ∞, then we can write down that
Taylor’s formula in the form of the series so f ( x0 ) and so on you can continue for infinite
term and this is called the Taylor series. So, for x=0 if we said this x 0=0 , then this is called
the maclaurins series.
79
(Refer Slide Time: 18:03)
So, what we have seen this last remark again, so it is necessary and sufficient for the
convergence of the Taylor’s or the Maclaurin series that R n ( x)→0 as n → ∞. Because they
are examples of smooth function series. So, smooth means we have the derivatives of
whatever or we lie, but the Taylor series diverges everywhere rather than the point of
expansion, because a point of a expansion the series will have the value same as the function
as per the construction so. And there are the example (Refer Time: 18:35) of this smooth
functions whose Taylor series converges to some other function and for instance you take this
example
{
−1
x2
f ( x )= e , x ≠ 0
¿ 0 , x=0
So, x not equal to 0 and x is equal to 0 so in this case one can easily show that (Refer Time:
−1
19:01) f ( n ) (0). So, if we take the derivative of this function here e x which I am not doing this
2
calculations, but one can easily compute at all the derivations of any order of this function at
0 will be 0 and then we if we write the Taylor series or other Maclaurin series around x=0
then we will get because all the derivative are 0.
So, you will get
2
n
0+0⋅ x+0⋅ x +⋯+0⋅ x +…
80
And then what we see here whatever x we keep the maclaurin series is giving 0. That means,
the series converges whatever x the series is just 0, but it does not convert to the function of
−1
the function was for x ≠ 0 , e x . So, it is very simple example where we can see that these
2
Maclaurin or Taylor series they do not converge to the function.
(Refer Slide Time: 20:02)
Now, let us just take this example of the Maclaurin series of e x now again not that. So,
whatever derivative we take for this function exponential it is just the exponential x and at 0
we have the value 1. So, for all values of n all the derivatives of this function exponential
function is 1. So, we can easily write down this maclaurins series theorem that
2
n
n
x
x
x
e =1+ x+ + …+ + R n ( x ) ,all the derivatives are 1. And + R n ( x); so the R n ( x) is the
2!
n!
n!
x
remainder term which we have just seen the remainder term
R n ( x )=
( x−x 0 )
n+ 1
( n+1 ) !
f
(n+ 1 )
( x0 +θ ( x−x 0 ) )
So, here if we substitute this x 0=0 then we get this reminder
n+1
R n ( x )=
x
θx
e , 0<θ<1
( n+1 ) !
So, one now you will see what happens to this R n ( x) as n → ∞.
81
If this is the case if R n ( x ) → 0 as n → ∞, then we can write down the Maclaurin series of
exponential function x. So, here again we note that this is the Maclaurin’s theorem. So, if this
term goes to 0 as n → ∞, in this case we can write down this exponential function as a series.
So, we can remove this Rn ( x) and then we can continue with the further terms as a series.
So, let us just check.
(Refer Slide Time: 22:04)
So, this is the R n ( x) the remainder term
n+1
x
θx
R n ( x )=
e , 0<θ<1
( n+1 ) !
n+1
and if you take the absolute value of this remainder term as
x
θx
e , then we notice that
( n+1 ) !
this e θx for whatever given x this will be a finite quantity. So, this will not disturb because
there is no n term here. So, if we can now focus on this term that what will happen when
n → ∞.
We should notice that when x whenn → ∞. So, this becomes ∞ and here whenever this x is
for example, large number greater than 1, then this term is also going to ∞. So, we cannot
simply say that what will happen to this term when n goes to ∞. So, we have to carefully
check this limit that what will happen to this term when n goes to ∞. So, what we consider
82
for a fixed value of x, we can always whatever x as could be very large number, but we can
find a natural number n such that the |x|is greater than n.
So, whatever x we take here, then this N the big N we take greater than the |x|. Having this
we will also consider one more n the small n term which is appearing there in the formula.
So, which is a bigger than this number n as well. Now we consider this term
n+1
n+ 1
|x|
( n+1 ) !
=
|x|
¿¿
So, we can write down in the form of the product as. So, factorial x divided by 1 factorial x
again divided by 2 and so, on we can continue now just look at this term here which have
appear after this N−1 term. So,
|x|
|x|
N
N +1
here also this
out of this expression what we see? The
|x|
N
and so, on. So, this term here
|x|
N
. So,
<1and now if we assume if we assume this as a
number q. So, we have a q here and this is in fact, divided by N +1. So, this is less than q and
all other terms here less than q.
So, again you note that this n> N. So, we have gone up to this N +1 term. So, this with this N
will be somewhere in the middle and now this q what we have notice because the
|x|
N
<1. So,
what we see now we can replace this equality by the inequality. So, less than equal to because
here we have replace by q and this one is also replace by q<1, q this is also less than q all
these terms are less than q.
So, now how many q’s we have here. So, N−1 terms are already there and then the total
terms were n+1. So, if we can re write now the total term n+1 and already these terms are
N −1
n−1; so ( n+1 )−(N −1). So, this number here is n−N +2 and
|x|
( N−1 ) !
and now we can take a
limit here in this case and q<1 and when n → ∞. So, this n → ∞ this n here it goes to infinity
and q<1 then this goes to 0 and here some fixed number is appearing. So, this everything
goes to 0.
83
So, this absolute value of this remainder term which goes to 0; so what we have seen that this
reminder term which was a part of the remainder term is less than which term which goes to 0
lim R n =0.
as n → ∞ and we can conclude that the n→
∞
(Refer Slide Time: 27:04)
So, these are the references which were used for preparing these lectures and the conclusion.
(Refer Slide Time: 27:12)
So, what we have learnt today is the Taylor’s formula and very important topic in the
differential calculus. So, a function which is smooth enough we can write down as
84
f ( x0 ) f ' ( x0 )(x−x 0 ). So, on plus this Rn ( x) term which is called the remainder term, and this
is one form of the remainder term
f ( n+1 ) ( ξ )
n+1
x−x0 ) . And what we call this the polynomial
(
( n+1 ) !
term we call the Taylor’s polynomial, the whole result this is called the Taylor’s formula.
And then we have also observed that when R n ( x)→ 0 as n → ∞, then we can write down this
Taylor’s formula as in the terms of a series which is called the Taylor series. So, that is all.
Thank you for your attention.
85
Engineering Mathematics - I
Prof. Jitendra Kumar
Department of Mathematics
Indian Institute of Technology, Kharagpur
Lecture – 06
Limit of Functions of Two Variables
I welcome to the 6th lecture on Engineering Mathematics I and today, we will discuss Limit
of Functions of Two variables.
(Refer Slide Time: 00:31)
So, today in particular we are talking about now the Differential Calculus and Functions of
Several Variables and before I introduce the limits, I will also discuss what type of these
functions we mean for the several variables.
86
(Refer Slide Time: 00:49)
So, here first let me introduce you the Functions of Two Variables. So, a function z=f ( x , y )
so its a similar to what we have the function of one variable, but there we consider like y=x.
So, they are used to be only one variable. But now here the zdepends on two variables x and
y.
So, it’s a function of two variables and this is a real valued function of two variables x and y
if to each ( x , y ) of a certain part of x-y plane which is called the domain of the function and if
we assign this value using this rule f is equal to z=f ( x , y ) to a real value z according to some
given rule f ( x , y); then, we call this function as a Function of Two Variables. So, what is the
Domain? Domain is basically the set of points ( x , y ) the x-y plane for which f ( x , y )is defined
and the Range, Range is the collection of overall possible values of z corresponding to the
point ( x , y ).
So, here we will call this ( x , y ) as dependent independent variables and z as dependent
variable. So, in this situation let us consider the coordinate system of ( x , y ) and z axis and for
example, this is the point in the xy plane denoted by this point here and the coordinate of this
point in the xy plane are given by x and y. So, corresponding to this point, if we compute this
f ( x , y ), then for instance this is the point here the height this in along the z axis at this point
( x , y ) of this function is f ( x , y ). So, this is the point here in 3-dimensional system x
coordinate comma y coordinate and thez coordinate which is f ( x , y ).
87
So, now if we collect all these points corresponding to the point in the domain, we this
surface will be formed. So, this a function of two variables in 3-dimensional plane here
represents a surface. So, this is the function z is equal to f ( x , y )which represents the surface
in this 3-dimensional coordinate system.
(Refer Slide Time: 03:31)
So, now let us consider this example here
z=√ 1−x 2− y 2
So, since this z is real, then the argument here (1−x 2− y2 )≥ 0 and this will imply that
2
2
2
x + y ≤1 . And therefore, we get the domain D which is all contains all the points here in R
means both are the real. So, in xy plane, where the x2 + y2 ≤1.
So, this is a circular disc in the xy plane D={( x , y ) ∈ R 2 : x2 + y 2 ≤1 } which forms the domain,
for this function z=√ 1−x 2− y 2 and the range. So, the values of the possible values of z for
the points from this domain, is the Range. So, if you notice here that x2 + y2 ≤1, this value of
the square root here will belong to 0 and 1. So, that is the Range here R¿ {z ∈ R ,0 ≤ z ≤1}
So, this is the graph of this function which is the sphere basically the half sphere above part
of the sphere. So, here the z axis, the x axis and the y axis. So, this domain here which is the
circular disc xsquare plus y square less than equal to 0 and at each point of this circular disc,
88
we have computed the value of z and the surface is plotted. So, this is the interpretation the
geometrical interpretation of the functions of two variables.
(Refer Slide Time: 05:37)
Now, before we move to the limit and continuity part later on, we need the concept of the
distance between the two points in xy plane. So, if we have two points P and Q having two
coordinates here ( x0 , y0 ) and ( x1 , y 1 ); then, what is the distance between these two points?
So, which we can easily find out if we form this triangle, then this here the height will be like
y 1− y0 and this is going to be x1 −x0 and now, this distance P Q will be the
√ ( x −x ) + ( y − y ) .
2
1
0
2
1
0
89
(Refer Slide Time: 06:36)
So, this is the distance ( x1 −x0 )2 + ( y1 − y 0 ) 2. We will also introduce now the neighborhood of
√
a point P( x0 , y0 ). So, if a point has given in the xy plane as ( x0 , y0 ) which is noted by P here.
Then, we will introduce now what is the delta neighborhood of this point P which is usually
denoted by N δ ( P )or N ( P , δ ). So, here
{
2
2
N δ ( P ) ≔ ( x , y ) : ( x−x0 ) + ( y− y 0 ) <δ
√
}
So, naturally now this one here is again open circular disc which is represented here; that is
the point P and the radius here is δ and now all the points here in this disk satisfies this
relation here. So, this is the neighborhood of this point P. Note that this is a open
neighborhood because the boundary is not included. If we include here less than equal to δ,
then the boundary of this disc that means, the circle will be also included in the point here.
So, this is the open neighborhood of P or with radius this δ.
This is the most common definition for the neighborhood, but we can also introduce
neighborhood as the square around this point P introduced by the x which lies between x 0−δ
and the x 0 +δ ; y also lies between y 0 −δ and y 0 +δ and this is represented by this is square
here, with this half of the side is this δ and the P is the point.
90
(Refer Slide Time: 08:39)
So, Limit of a Function of One Variable. So, we recall now before we introduce the limit the
concept of the limit for the functions of two variables, it is important to first discuss the limit
of a function of one variable. And mainly, we will now focus on this delta epsilon definition
of the limit which is very important to discuss the limit for the functions of several variable.
f ( x ) =L, if for every ϵ >0. If for every
So, here we say for the one variable case that this xlim
→x
0
ϵ >0, there exists a δ >0 such that for all x; so, all those x which belongs to this neighborhood
of this x 0 implies that |f ( x ) −L|< L.
So, let me just explain you this concept with the help of this plot here. So, we have a function
f ( x) whose limit as x approaches to this x 0 is L. So, this is the point x 0 here. This is the point
x 0 and then corresponding to this, this value is L there and now.
So, what this definition says is that corresponding to every ϵ ; so this could be a very small
number or anything. So, this epsilon positive that means, which is denoted by this distance
here around this L. So, for every epsilon, however small, there always exist in neighborhood
of this x 0 which is the case here and now, if you take any x in this neighborhood.
So, this is the δ. So, if you take any point in this neighborhood now, the value will be or the
difference of the value of this f ( x) and minus this L will be less than the ϵ . Yeah, this is ϵ
here. So, let me just write it. This is ϵ . So, that difference will be less than ϵ and this function
91
may not be defined at x is equal to x 0. So, define the limit of function at point x 0, the function
may not be defined at this point x 0 and therefore, here that x is equal to x 0 is excluded.
So, in other words, if we can make the difference this f ( x )−L, this difference here f ( x )−L; if
we can make this difference as a small, as we like by considering a small neighborhood
f ( x ) =L.
around this x 0, then we say that this f ( x) is equal to the limit of this f ( x0 ); xlim
→x
0
(Refer Slide Time: 12:06)
So, here now let us take this example that what is the limit of this ( 3 x+1 ) as x goes to 1 and
its clearly visible here that the limit is 4 in this case and we will prove now using this ϵδ
approach that this indeed is the limit of this function ( 3 x+1 ). So, what we need to show? We
need to show that for a given ϵ that there exists a δ. So, that the |x−1|<δ. So, this is the
neighborhood around this one δ neighborhood around 1. If we take any point from this
neighborhood, this will imply that this |( 3 x +1) −4|<ϵ .
So, to prove this, we will start with this difference here |( 3 x +1) −4|=|3 x−3|=3|x−1|. So,
this modulus |x−1|; so, all x from this neighborhood satisfies that |x−1|<δ. So, here 3 was
there; so, less than 3 δ and what is our aim now? To make this difference is smaller than the
given epsilon. So, if we set here less than or equal to ϵ now, so we get a relation between δ
and ϵ . For given up δ, we can choose now the δ such that this difference will become less
than ϵ . So, what is the relation?
92
ϵ
Relation is 3δ ≤ ϵ. So, if we choose this δ ≤ ; for given ϵ if you choose this delta less than
3
equal to ϵ by 3. Then, for any given ϵ , we have this relation that the |( 3 x +1) −4|<ϵ which we
have just seen here, this difference is less than ϵ . Because of this relation we can choose this
δ≤
ϵ and then, this difference will become less than whenever for all
ϵ
|x−1|<δ .
3
We can see the situation in this plot. So, this is a for instance and at x=1 the value of the
function is 4 and now, you take any neighborhood. So, for example, we have chosen this
neighborhood here. So, there exist a δ neighborhood again correspond around this one in this
case. So, usually when the ϵ is a small, the δ is also small and when the ϵ is big, we can have
a big neighborhood around that point.
So, for instance here we have chosen a big neighborhood of around this point 4 and then,
correspondingly this δ is also big. So, all these x in this δ neighborhood satisfies this relation
that the difference of this function minus 4 is less than the ϵ .
(Refer Slide Time: 15:36)
So, what will happen for the Non-Existence of a Limit if the limit does not exist for? For
example, in this case as x approaches to x 0, the limit whether right or the left is not equal toL
and what will happen if we try to get a neighborhood around this Lhere and whether the
corresponding neighborhood around this x 0 will exist or not?
93
So, here for example, this is ϵ neighborhood, but what we see in this case that there is no
neighborhood of this x 0. Whatever small neighborhood you take and any x between this, the
difference between the function value and this limit will exceed than this ϵ here. So, it is
clearly visible that you take any point in this neighborhood here and then, the difference
between the f ( x) and this L will increase than the given ϵ here.
So, in this situation we do not have the possibility that for a given ϵ , there exists δ
neighborhood around this x 0. So, for a given ϵ in the situation, there does not exist any δ such
that the difference f ( x) minus L is less than ϵ wherever this x is from this δ neighborhood.
Recall again, the existence of the limit was that limit x goes to x naught f ( x) is equal to L
and this means that every neighborhood of L. So, every neighborhood which is represented
by N ϵ ( L) of L there exist some neighborhood of x 0 such that the f ( x) belongs to this
neighborhood of L whenever x belongs to the neighborhood of this is x 0, the δ neighborhood
of x 0.
(Refer Slide Time: 17:40)
So, Limit of Functions of Two Variables, we will discuss now. So, this is just the
continuation of this rather general definition here. So, this x maybe point in two dimensional
plane for example, in xy plane and again, this definition will be carried for generalization the
definition of the limit.
94
So, for example, in this case we take z as f ( x , y) function of two variables which is defined
domain D and let this P ¿, y 0) be a point D. So, if for a given real number epsilon however is
small, we can find a δ a real number δ such that every point ( x , y) in the δ neighborhood of
this point ( x0 , y0 ) it satisfies that |f ( x ) −L|<ϵ; then, we will call tell theL is limit.
So, what does that mean? That this f ( x , y) minus L is less than epsilon wherever these ( x , y)
fall in the delta neighborhood of this x naught y naught. So, if for a given epsilon, we can
find such a delta, then we will call that the cell is the limit of f ( x , y).
Again, the function may not be defined at ( x0 , y0 ) for the existence of this limit and therefore,
we may exclude this that ( x , y) is equal to ( x0 , y0 ) and then, this real number L is called the
limit of the function f ( x , y)
lim
as ( x , y ) → (x 0 , y 0 ). Symbolically, we denote again the
f ( x , y )=L .
( x , y ) → ( x0 , y0 )
(Refer Slide Time: 19:44)
So, if you take this problem now from these functions of two variable
lim
⁡
( x 2 + y2 ) sin
( x , y ) → (0 ,0 )
95
( x +1 y )=0.
2
2
We want to prove that this limit is 0 using ϵδ approach. So, how do we proceed? Now for
( x , y ) ≠(0 ,0) because at (0 , 0) this function is not defined and we consider this difference.
2
2
Difference of the function which is ( x + y ) sin
( x +1 y )less than the limiting value 0.
2
2
If we start with this difference now, this is ( x2 + y 2 ) because this is going to be positive and the
absolute value of sin
(
1
and we know that that this value of sin always lies between −1
x + y2
2
)
and 1. So, this is this quantity is bounded by 1. So, this is less than equal to 1; that means, this
difference is less than equal to x2 + y2 and what we know, the neighborhood of this (0 , 0)
point because taking this limit to (0 , 0).
So, what is the neighborhood of this (0 , 0) or rather δ neighborhood of (0 , 0)? It is ¿:
2
2
2
2
2
0< √ (x−0)2 +( y−0)2 <δ or ( x + y ) <δ . So, we can use this inequality there. So, ( x + y )in
this neighborhood is less than δ 2 and we want to set for a given ϵ , this difference less than ϵ
and we are looking for such a δ which we satisfy that inequality.
So, here this difference if you want to make less than ϵ ; then, we can get this relation between
δ and ϵ . So, for given ϵ , if you choose δ such that the δ 2 ≤ ϵ then, we are done with the limit
that this difference is less than ϵ for the δ which satisfies the δ 2 ≤ ϵ .
So, here if for a given ϵ if you choose the δ 2 ≤ ϵ , then this difference between the function
value and its limiting value will be less than ϵ . So, this here will be less than ϵ , will be less
than ϵ ; if we take ( x , y) from this neighborhood. So, again just to recall this so, we have
started with the difference between the function value at any point ( x , y)≠(0 , 0) and its
limiting value and somehow we have to write down this difference in terms of the so that we
can use this delta inequality from the neighborhood of the ( x0 , y0 ) point which is (0 ,0) in
this case.
And once we reach to this δ, then we can set that this must be equal to less than equal to ϵ
because we want to make this difference; this difference here of the function N minus this
limiting value less than ϵ . So, out of this difference, we can say that for a given ϵ there exist a
δ . The relation is that the δ 2 ≤ ϵ .
96
(Refer Slide Time: 23:45)
So, another example where we need to prove that this
lim
( x , y ) → ( 0, 0 )
(√
xy
=0 .
x 2+ y 2
)
So, in this case again the function is not defined when ( x , y ) → (0 ,0) because of this reason
2
2
x +y .
So, this will become 0, but what we will prove now that this limit is 0. We cannot just
directly substitute (0 , 0) here because this will be like a 0 and then 0 here and there is no
such an L'Hospital rule which we can apply in case of one variable we had the L'Hospital rule
and which says that the derivative this limit will be equal to the limit of the ratio of the
derivatives.
But here so we cannot substitute thus directly the value of x and y in this expression. So, we
will prove that this limit is 0 using again ϵ δ approach. So, we take this ( x , y)≠(0 , 0) and we
consider again as in the previous example, this difference between the function which is
|√
xy
x2 + y 2
|
−0 .
97
And now, so this is equal to the
¿ xy∨
xy
|√ |
x2 + y 2
and now this is positive. So, we have basically
¿
¿ and now, the aim is the same that you want to write down everything in terms
√ x + y2
2
of the neighborhood inequality which will be in this case because we have (0 , 0) point will be
x square plus y square and the square root. So, we will convert now this expression in the
form of the x2 + y2.
So, we notice here now that if you take this inequality ( x− y )2 which is always greater than or
equal to 0 because of the square here and then what do we get here? x2 + y2 −2 xy >0 which
will imply that this x2 + y2 >2 xy and then, so here you have x2 + y2 greater than 2 xy.
We can take the absolute value. So, here x2 + y2 is positive. So, we have the 2 and the xy is
greater than this 2 times the xy and this is greater than the xy. So, in this case, we can use this
inequality that x2 + y2 is greater than the absolute value of xy.
(Refer Slide Time: 27:01)
Having this relation, we can now replace this ¿ xy∨¿ by the bigger quantity which is x2 + y2
and divided by
√ x2 + y 2.
So, here now we got the
neighborhood of this delta which is
√ x2 + y 2
and we considered again this
√ (x−0)2 +( y−0)2 < δ .
So, we said that the
√ (x−0)2 +( y−0)2 < δ at least in the neighborhood of the (0 ,0) and now we want to said that
this difference is less than epsilon.
98
So, again we will use the same idea. So, here it is less than equal to ϵ . So, this expression the
difference between the function and the limiting value is less than ϵ . If we choose this
relation δ is less than or equal to ϵ . So, for given ϵ , if we choose δ equal to ϵ or anything less
than ϵ ; then, the difference between the function value and its limiting value will be less than
ϵ.
So, for given ϵ which use δ less than ϵ , then this difference between the function value and
the minus the limiting value will be less than ϵ .
(Refer Slide Time: 28:26)
So, now to conclude, so we have discussed the functions of two variables mainly Z=f ( x , y).
So, here x and y they are the independent variable. So, they can take any values and this that
depends on the value of the I mean this functional value here and we have also seen that such
functions represent surface in the xyz plane. We have also discussed the definition of the
limit in particular the ϵδ definition which is very useful to prove somehow that a given
number L is the limit.
With the help of this ϵδ approach, we cannot just get the limit; but if we have some idea
about the limit, then this ϵδ approach may be used to verify that L is the limit. Because what
we will observe now in case of these two variable when we have the functions of two
variables, then there are several parts which approaches to a particular point in the xy plane
and this is completely different what we have seen in case of one variable where there were
99
two parts; one from the right side, one from the left side and we used to get the limit from the
right side, from the left side and then, if they both are equal we say that the limit exist.
But now, in this case since you can approach to a particular point from the several direction,
it is not possible to get as the along some path because there are infinitely many parts
involved in this limiting process. So, this eϵδ approach will be very useful once we have
some idea about the limit and then, this value of the limit can be used to prove that this is
indeed the limit. So, in the next lecture, we will learn more on the Limits, the evaluation of
the limit in particular.
(Refer Slide Time: 30:35)
These are references which we have used to prepare these lectures and.
Thank you very much.
100
Engineering Mathematics - I
Prof. Jitendra Kumar
Department of Mathematics
Indian Institute of Technology, Kharagpur
Lecture - 07
Evaluation of Limit of Functions of Two Variables
So, welcome back to the lectures on Engineering Mathematics I and today, we will be talking
about Evaluation of Limit of Functions of two variables.
(Refer Slide Time: 00:26)
So, in the previous lecture what we have seen that this limit f ( x , y) as ( x , y) goes to ( x0 , y0 )
is equal to L. It was defined using this epsilon delta approach that means, if for a given real
number epsilon positive, we can find a real number delta positive such that f ( x) minus L less
than epsilon.
So, the difference between the two here f ( x , y) and minus its limiting value L is less than
epsilon for a given number for an arbitrary epsilon whenever this ( x , y) is in the delta
neighborhood of ( x0 , y0 ) point and we have seen that this approach is useful in verifying that
the given number L is the limit. But we have to guess that what is the limit L. So, in today’s
lecture we will look for more practical issues that how to get this number L and the epsilon
delta approach may be used to verify that this given number is L.
101
(Refer Slide Time: 01:39)
So, while talking the limit for a function of single variable, we had two paths to approach a
particular point here x 0 on x-axis like in this case. So, we can approach from the right side to
this point or we can approach from the left side to this point x 0 in case of a functions of single
variable.
(Refer Slide Time: 02:06)
But now, we have functions of two or more variables, in this case suppose here this is x- axis,
y- axis and z- axis. So, in the xy-plane, we have this domain of this function. Suppose we
want to approach to this point P here, then there are several paths to approach this point. We
102
can approach from any direction and using any path to this point to get the limit as ( x , y)
approaches to this particular point P.
While in the case of single variable, we had only two paths. So, we used to check the limit
from the right side and limit from the left side and if they are equal, then we say that the limit
exist and limit is equal to that value or if those two limits are different then, we call that the
limit does not exist. In this case we have infinitely many paths a ways to approach to a
particular point. So, in that case it is difficult to find a limit along some particular path and
saying that this is the limit.
So, as I have remarked here say note that this ( x , y) goes to ( x0 , y0 ) point in two dimensional
plane, there are infinite number of paths joining ( x , y) to this particular point ( x0 , y0 ). Since,
the limit if exist is unique the limit should be same along all the paths. That means, the limit
cannot be obtained by approaching the point P along a particular path and finding the limit of
the function f ( x , y) as ( x , y) goes to ( x0 , y0 ). However, if the limit is dependent on the path,
so if we find along two different paths that the limit is different; then, at least we can
conclude that limit does not exist.
But getting the limit along two or many different paths though that limit may be the same, but
we cannot conclude that that particular number is the limit. Because there may be some other
path where the limit may be different. So, concluding that the limit along some particular
path is not possible, but what we can conclude that if we find two different paths and along
those paths, the limit is different then we can say that the limit does not exist. So, now, we
will see some possibilities finding the limit of a function of f ( x , y )as ( x , y) approaches to
some particular point.
103
(Refer Slide Time: 04:59)
For example, consider this example x square y over x 4 plus y square and( x , y) goes to
(0 , 0) along y is equal to x path and second we will take again the same function and ( x , y)
goes to (0 , 0) along this path y is equal to x square. So, we have chosen two different paths
to approach this origin here. One the straight line y is equal to x and the second, we have
taken this y is equal to x square paths.
So, here how what will be the limit of this function as we approach to (0 ,0) point along this
y is equal to x square or y is equal to x, two different path. So, if we take along y is equal to
x here, we are approaching to the origin and in this case so when y is equal to x. So, we will
put here x square and then, y will be x again.
So, we have here x power 4 plus x square and now, we can approach as x goes to 0 because
along this path now we have restricted this y is equal to x or we have substituted here y is
equal to x. So, now, we can take that x goes to 0, what will happen to this limit. So, here the
limit x goes to 0 and this is x square we can cancel out. So, you have still x over x square
plus 1 and then, this limit will go to 0. So, along this path y is equal to x along this straight
line which is approaching to the origin, we are getting this limit as 0.
104
(Refer Slide Time: 06:54)
Now, if we take y is equal to x square. So, along this path we have this limit the given limit
as the limit x goes to 0. Again, this path y is equal to x square is also approaching to 0, but
with this parabolic path. So, here we have x is equal to 0 and x square the y is again x square.
So, we have x power 4 and y square is x power 4. So, in this case what we observed because
here x power 4 and then, you have 2 x power 4 here. So, this limit will be just half.
(Refer Slide Time: 07:36)
105
So, along this path we are getting the limit half; along the straight line, y is equal to x, the
limit is 0; along y is equal to x square, the limit is half. So, now, we can conclude that this
limit as ( x , y) goes to (0 ,0) x square y over x 4 plus y square does not exist.
(Refer Slide Time: 07:59)
The second example of this kind, we will take limit ( x , y) goes to (0 , 1) tan inverse y over
x . So, in this case we have again this xy - plane and we are approaching to the point (0 , 1). So,
x 0 y 1 so, somewhere here, we will take two paths again. We can fix this y and we will take
these two paths as x approaches to this point from the right side or from the left side. So,
along this path we will fix y as 1 and then, we will take or we will evaluate this limit along
these two particular paths. So, in this case if we fix y is equal to 1 and then, approach x to 0
from this two directions. So, if we go from the left hand side as x goes to 0 tan inverse 1 over
x.
So, you note that if x is negative. So, all these numbers here 1 over x will be negative. So, as
x approaches to infinity. This 1 over x will approach to minus infinity. So, so tan inverse
minus infinity will get minus pi by 2. Similarly, if we approach this point from the right side
of this (0 , 1) point, then in this case this will become infinity and the tan inverse infinity.
So, we will get limit as pi by 2. So, I will again in this example we have seen that along two
different paths, we have two different limits. So, what we can conclude that again the limit
106
the given limit here ( x , y) goes to (0 , 1) tan inverse y over x does not exist because this
limit depends on the path.
(Refer Slide Time: 09:59)
So, if you take this next example xy over x square plus y square and ( x , y) goes to (0 , 0), in
this case we are approaching to the origin and the function is xy over x square plus y square.
So, we take a general path y is equal to mx; that means, here y is equal to mx with varying m.
So, we are approaching to this point along different straight lines. So, the slope of these
straight lines are different. So, m can take any value and then, we are approaching to the point
here the (0 ,0) along these straight lines. So, in this case what will be this limit here?
So, y we can substitute as earlier, y is equal to mx. So, here we will have m x square and then
x square plus m square x square and x square will get cancelled and we will get m over 1 plus
m square. So, what we observe again that for different values of m, we are getting a different
limit. Therefore, the limit depends on the path and again, this limit does not exist.
107
(Refer Slide Time: 11:31)
An alternative approach to compute such limit could be using a different coordinate system
and we can use the change of coordinate system from Cartesian to Polar coordinates. So, in
this case we know the transformation that is x is equal to r cos theta and y is equal to
r sin theta.
So, in this case we are will be now working on different coordinate system and in polar
coordinate. So, this limit here ( x , y) goes to (0 ,0) xy over x square y square will be given as
sin theta cos theta because if we change here x, we substitute as r cos theta and y as r sin theta
. And then, x square plus y square will become r square and this limit here ( x , y) goes to
(0 , 0) will be the limit as r goes to 0. Because in the polar coordinate, now we have this r
and this is theta and this is r. So, independent of theta, we if we approach r to 0; we will
approach to infinite to this origin. So, here this limit ( x , y) goes to (0 ,0) will be equal to this
limit rgoes to 0.
And r cos theta r sin thetaover r square and this r gets cancelled. So, we get this cos thetaand
sin theta. So, what we observe again that this limit is depend on is depending on theta. So, if
we choose a different angle to approach to this limit or to this approach to this point here the
origin then the value will be different.
108
(Refer Slide Time: 13:36)
So, again the similar situation that the limit depends on the angle theta or it depends on the
path, we can say and hence this does not exist. But what we will realize in today’s lecture that
this change of coordinate system in many cases is very helpful.
(Refer Slide Time: 13:54)
For example, we consider this problem x square y over x square plus y square and if we
change the coordinate system to the Polar, then by the substitution as earlier, we can write
down this limit ( x , y) goes to (0 ,0); x square y over x square y square is equal to. So, the
109
limit again r goes to 0 and x square as r square cos square theta y is r sin theta and we have
here x square again r square cos square thetaplus r square sin square theta.
This will become r square. So, we have limit r goes to 0. So, this is r; the r square will get
cancelled. And then, we have cos square theta and sin theta and divided by r square which is
already cancelled. So, r square and then here r square. So, these cancel out and r goes to 0.
So, independent of theta, we are not fixing theta here because whatever the value of theta is
we have a finite number here and then, r goes to 0 then this will become 0. So, in this case
this limit here is 0.
So, by changing to the coordinate system, the evaluation was easy and we could conclude
now that the limit is 0. We have again note that we have not fixed the path because this limit
here, we got independently of theta. So, without fixing the value of the theta, we have
approach to this 0. So, this is this limit is not along a particular path. This is independent of
the path. So, that that is what I said that thus changing the coordinate system is very helpful.
For example, we have seen in this particular case, one might again get confused that
approaching to this point along y is equal to mx line will also give limit as 0, but here we
have fixed the path. So, y is equal to mx these are the straight lines approaching to (0 ,0)
point. So, by doing so y is equal to mx and letting this x goes to 0, we cannot conclude that
the given number or the given limit is the limit of the of the function.
But in this case by changing the coordinate system. So, now, we have a different coordinate
system equivalent coordinate system where we have not fixed anything. So, here if it is
independent of theta we are approaching to some number then that would be the limit we will
explore this in some more example.
110
(Refer Slide Time: 17:05)
So, no dependency on theta. Therefore, in this case the limit exists and the limit is equal to 0.
(Refer Slide Time: 17:11)
So, here sin inverse x plus 2 y and 3 x plus 6 y. So, this is another approach without
changing to the coordinate system. We can observe here that this is x plus 2 y and here also
the tan inverse the argument again it is 3 times x plus 2 y. So, we have the same number x
plus 2 y and here also 3 times x plus 2 y. So, if we substitute x plus 2 y as a new parameter,
as a new variable t. Then, we can convert this limit to the limit sin inverse. So, this is
substituted as t and then, tan inverse three times t and the limit t approaches to 0.
111
So, this is another convenient way without fixing the path. So, we are not fixing again the
path, but we have substituted this relation here x plus 2 y which appear everywhere in this
function. So, here it was x plus 2 y and again same functionality, but with the 3 times x plus
2 y which we have replaced by t. So, it becomes 3 times t and now, we have changed the
problem to the problem of limits of single variable which we can compute. So, in this case it
is like 0 by 0.
So, we can apply the L’Hospital’s rule which says this will be 1 over square root 1 minus t
square and then here this will be 3 and 1 plus 9 t square and then, the limit t goes to 0. So, in
this case t goes to 0, it will be 1 and then, we have 3 here. So, 1 by 3.
So, this limit will be 1 by 3.
(Refer Slide Time: 19:04)
So, again this is another approach which could be helpful in some cases when we can change
the functions of two variables to one variable.
112
(Refer Slide Time: 19:22)
So, now few remarks on working with the limits. So, what we have if the limit f ( x , y) goes
to L1 and ( x , y) goes to x 0 as L2; if these two limits are given, then we can compute this k
times f ( x , y) will be the k times the limit of the f ( x , y) as ( x , y) goes to ( x0 , y0 ). So, this k
is a constant. So, this limit will be k times L 1.
Again, when we add this f ( x , y) and plus or minus g( x , y), we have finite quantities there L 1
and L2, then this limit will be also as L1, L2. L 1 plus and minus if it is a plus there here plus
will appear; if it is a minus here, then minus L2 will come. Now, we have the product here of
the two functions f ( x , y) into g( x , y). So, this will be the product of L1 and L2. If we have
the quotient here f ( x , y) over g( x , y); in this case if this L2 is not zero because we cannot
divide by 0. So, if L2 is not zero then this limit will be also as L1 L2 provided L2 is not equal to
0.
113
(Refer Slide Time: 20:48)
Some more generalization of these working rules, we can also work when we have infinities
there. So, if this limit f ( x , y) as ( x , y) goes to ( x0 , y0 ) is infinity.
And here, the limit of this g( x , y) as ( x , y) goes to ( x0 , y0 ) is also infinity. In that case, we
can compute this limit of the product of these two functions f ( x , y) g(x , y) as ( x , y) goes to
( x0 , y0 ). So, infinity and here also we have plus infinity plus infinity. So, this limit will be
also plus infinity. We can also add the two functions and take the limit as ( x , y) goes to
( x0 , y0 )and this will be infinity plus infinity which we know it is the infinity again.
So, here we have the limit plus infinity and the g is approaching to minus infinity as ( x , y)
goes to ( x0 , y0 ). So, in this case since one is minus infinity another one is plus infinity. So,
the product will be infinity again with minus sign because one is negative here, one is plus.
So, plus and into minus one is minus one and then, here we have infinity. So, this will
approach to infinity.
114
(Refer Slide Time: 22:10)
If we have f ( x , y) as is going to infinity and g( x , y) is approaching to some finite real
number, in this case we can evaluate such limits f ( x , y) plus or minus g( x , y), since this is
infinity already and then we have here g( x , y) is equal to L. So, will does not matter what is
this number here as long as this is a finite number. Then, this limit will be just infinity.
Second, if we have infinity and then this limit L is positive, in this case we can go for the
product rule. So, limit ( x , y) goes to ( x0 , y0 ) f ( x , y) and g( x , y) and this product will be
equal to because this L is positive. So, sign will not change and it will be plus infinity.
(Refer Slide Time: 23:07)
115
Some more. So, here we have f ( x , y)
as approaching to infinity and this g( x , y)
approaching to L which is a negative number. In that case, this product since f ( x , y) is
approaching to infinity; the product will go to infinity. But because of this negative L, this
will approach to minus infinity.
Now, if f ( x , y) approaching to infinity and g( x , y) is a real number, the limit is a real
number; in that case the quotient here the g( x , y) over f ( x , y). So, here this f ( x , y) is going
to infinity and g( x , y) is some number L, in that case this limit will be 0.
(Refer Slide Time: 23:53)
So, what we have seen today that the limit, we can evaluate the limit in two in case of two
variables. But we have to be careful because in the xy plane now we can approach to a
particular point from several directions, while in case of one variable we had only two
directions from the right hand side from the left hand side, but now we have infinitely many
directions. So, computing limit along a particular direction will not help, but at least if we
find two directions where the limits are different, then we can conclude that limit is a limit
does not exist.
But we cannot conclude by evaluating the limit along two or many paths that this is the limit,
even though that limit may be same along several paths. So, computing the limit then what
we have seen two approaches the one was changing to the polar coordinate would be helpful.
In that case if r approaches to 0. For example, if the limit in the question is approaching to
116
the origin. So, in that case we can say if r approaching to 0 independently of theta. So, then
whatever number we get that does not depend on the path and that will be the limit.
Another approach we have seen that we can convert in many situation a number into one
variable problem and then, we can apply L’Hospital’ rule for example, to conclude the limit.
So, one again we will see more in the next lecture that changing to the Polar coordinate is
helpful for evaluating the limit. But we have to be again very careful that while taking the
limit as r goes to 0 that should not depend on the angle theta.
So, if the independently of theta we are approaching to some number, we can conclude that
that is the limit. But if that limit is depending on theta, then again it is a path dependent limit.
And therefore, the limit does not exist. So, we will talk more on these issues that what could
be the problem by while changing the coordinate system to polar coordinate and more
examples later.
(Refer Slide Time: 26:30)
So, these are the references used for preparing these lectures and.
Thank you very much for your attention.
117
Engineering Mathematics - I
Prof. Jitendra Kumar
Department of Mathematics
Indian Institute of Technology, Kharagpur
Lecture – 08
Continuity of Functions of 2 Variables
Welcome to engineering mathematics 1 and today we will be talking about a Continuity
of Functions of two Variables.
(Refer Slide Time: 00:25)
So, we have already seen the definition of a limit indeed the limit delta epsilon definition
of limit and this is motivated by that definitions. So, a function z=f ( x , y ) is said to be a
continuous at a point ( x0 , y0 ). If this f ( x , y) is defined at ( x0 , y0 ) second, this limit
( x , y) goes to ( x0 , y0 ) f ( x , y )exists. And the third one that this limit is equal to the
function value at that point.
If all these three conditions are met; so function is defined this limit exists and the limit
is equal to the function value at 0 ( x0 , y0 ), then we call that the function is continuous at
the point ( x0 , y0 ). And if a function f ( x , y) is continuous at every point in the domain D,
then we call that function is continuous in that domain D. So, eventually here the
continuity is basically finding the limit as ( x , y ) → (x 0 , y 0 ) and then observe that this limit
should be equal to the function value at ( x0 , y0 ).
118
(Refer Slide Time: 01:53)
So, we can define delta epsilon definition as for the limit. So, a function z=f ( x , y ) is
said to be a continuous at a point ( x0 , y0 ). If for a given epsilon there exist a real number
delta positive; such that this difference between f ( x , y) and this value of f at ( x0 , y0 )less
than epsilon whenever these x and y’s if we take from the delta neighborhood of ( x0 , y0 )
point.
So, this is the def this was eventually the definition of the limit as well, where we have
seen that this function has a limit l. So, this instead of f ( x0 , y 0 ) we used 1 there and now
since the function has to be defined at ( x0 , y0 ) and we are looking at whether the limit of
f ( x , y) as ( x , y ) → 0 is equal to f ( x0 , y0 ) or not. So, this epsilon delta definition is
exactly the same as earlier for the limit.
Indeed, here we have not used now that this must be a greater than 0 to avoid this
( x0 , y0 ) point. Because, now the function is defined at ( x0 , y0 ) point; earlier in the case of
limit function may not be defined at that point is still we can talk about the limit. But,
now in this case for the continuity the function has to be defined at ( x0 , y0 ) and therefore,
this inequality at this end is no more required. So, there is another term we used for
removable discontinuity. So, all the conditions we have discussed before like the
function is defined at ( x0 , y0 ), and the limit ( x , y ) → (x 0 , y 0 ) f ( x , y) exists.
119
And the third one when this limit is not equal to the function value at ( x0 , y0 ), then we
call that the function has removable discontinuity. If this is equal, then this is the
definition of the continuity we have just discussed.
(Refer Slide Time: 04:17)
Let us go to some examples now. So, discuss the continuity of this function
2 x 4 +3 y 4
, ( x , y ) ≠ (0 , 0)
f ( x , y )= x 2 + y 2
¿0 , (x , y)=( 0 , 0 )
{
And we will discuss the continuity of this function at the origin.
4
So, we need to check basically that this limit here of this function
4
2 x +3 y
is equal to 0.
2
2
x +y
If this limit is equal to 0, then the function will be continuous; otherwise, in case this
limit does not exists or this is not equal to 0 then we call the function is not continuous.
So, as discussed in the last lecture, this changing to polar coordinate is very helpful for
finding out the limit.
So, in this case also we will follow those steps. So, we will change now from Cartesian
to the polar coordinate, because whenever we see the x2 + y2 term in the denominator
than this changing to polar coordinate is very, very useful. So, in this case when we
120
change the coordinates from ( x , y) to (r , θ), then x will be a so we basically substitute
here. The x=r cos θ and y=r sin θ. So, by doing so we got here
4
lim
( x , y ) → ( 0, 0 )
4
4
4
4
4
2 x +3 y
2r cos θ+3r sin θ
2
2 =lim
2
r →0
x +y
r
So now this r square will get cancel with this, so here also we will get 2. And now we
have 2 or rather r 2 if we take common; so we have here cos 4 theta plus sin 4 theta times
2
2
r . So, let us just see and which, so this will become as the limit r → 0 and then here r ,
and then cos 4 θ+sin 4 θ and since this is a bounded function cos 4 θ+sin 4 θ when r → 0 this
limit will be 0. And the function value at (0 ,0) is also 0, so the function is continuous;
function is continuous.
(Refer Slide Time: 07:17)
Now, moving further to the next example, we will discuss now the continuity of the
function
2
( x− y )
f ( x , y )= x2 + y2 , ( x , y ) ≠ ( 0 ,0 )
¿0 , (x , y)=( 0 ,0 )
{
So, in this case we will choose the path y=mx because here this choosing path is again
conclusive, because the order here is 2 in y also it is 2 and there also it is 2. So, choosing
121
y=mx the x will get cancel and this limit will depend on m, in most of the cases this will
work.
So, when we do this substitution here for y is equal to mx we will get this limit as ( x , y)
goes to (0 ,0) and x and y is substituted as mx divided by x square m plus m square x
square. So, from there we take common x so, this x will be removed and we will get this
2
lim
( x , y ) → ( 0, 0 )
2
( x− y ) ( 1−m )
=
x2 + y 2 (1+m 2 )
2
free from x. So, this limit which is x → 0 will be nothing but
( 1−m )
.
(1+m 2 )
(Refer Slide Time: 08:57)
So, this limit depends on path because choosing different value of m we will get a
different value of this limit and hence this limit does not exist and then again we will call
that this function is not continuous.
122
(Refer Slide Time: 09:11)
The next example we will discuss the continuity of this function
f ( x , y )=
{
2
sin √ x + y
√ x2 + y 2
2
, ( x , y ) ≠(0 ,0)
¿0 ,(x , y)=( 0 ,0 )
And we will discuss again the continuity of this function at (0 , 0).
Because, when ( x , y)≠(0 , 0), the function can be easily shown that this is continuous.
So, at this point again because x square y square term is there, changing to polar
coordinate will be helpful and we will do so the limit ( x , y ) →(0 ,0) sin this given
function will be changed to as limit r → 0. So, here sin this x2 + y2 will be r 2 by that
transformation x=r cos θ, y=r sin θ is equal to r cosθ and y=r sin θ.
So, by this substitution we will get here r 2 and here also this r 2 and then this will be like
lim
r→ 0
sin r
sin r
. So, lim
we have only one variable limit now. So, this is using the for
r
r
r→ 0
example, L Hopital’s rule we can get the derivative of sin r. So, that will be cos r and this
derivative 1 and limit r → 0, so this will be 1.
123
(Refer Slide Time: 10:53)
So, this limit here is 1 and the function value defined at (0 ,0) point is given as 0. So, the
limit exist in this case, but the function is not continuous at (0 ,0), because this limiting
value is not equal to the function value which is prescribed at (0 , 0). So, this is the case
of the removable discontinuity where the limiting value.
So, the limit exists the function is defined, but this limiting value is not equal to the
functional value at that point. So, this is the example of the removable discontinuity.
(Refer Slide Time: 11:31)
124
Another example we will discuss the continuity of this function again at origin. So, we
have
{
x4 y 4
, ( x , y ) ≠(0 ,0)
f ( x , y )= ( x 2 + y 4 ) 3
¿0 ,(x , y)= ( 0 ,0 )
So now, in this case we will choose the special path y 2=mxbecause, y=mx will not be
helpful we have a different order here. So now, choosing this y 2=mxdue to this y power
4 here. So, if y 4 will be in then m 2 x2 and then we can have this x2 common. So, if we
take this we substitute this y 2 there, then in that case this limit x → 0 because this path
will take to the origin only.
So, now we will take the limitx → 0. We have their x 4 and then y 2 whole square. So, this
will become x2 and x2 . So, m 2 x2 and then divided here by x2 + y4 whole square again. So,
this is m 2 x2 and power this 3. So, this x6 , so from here also x2 and then power 3. So, x 6
3
will be cancelled and then we will get this limit m 2 over ( 1+m 2 ) ; which is given here.
3
So, this is m 2 over ( 1+m 2 ) . So, again we see that if we choose this path, then the limit
depends on the path again; for different values of m you will get a different value of this
limit and therefore, this limit does not exist and hence this function is not continuous.
(Refer Slide Time: 13:27)
125
Now, moving next to the next problem. Here we have the
2
2
x +y
f ( x , y )= tan xy , xy ≠ 0
¿ 0 ,elsewhere
{
So, looking at this function is a difficult to see that what will be the limiting value where
the limit will exist on what substitution we should make. So, again let us choose this
y=mx and see what is happening in this case. So, this given limit if we have this
2
( x , y ) →(0 ,0) and then this
2
x +y
. And if we substitute this y=mx, so we will take then
tan xy
the limit x → 0.
And here we will have x2 + m2 x2 and then this tan m x 2. So, y is mx, so x2 , so now, when
a x → 0 we are getting like a 0 in the numerator and also tan 0 in the denominator. So, 0
by 0 case this is a one variable limit so, we can use basically L’Hospital.
So, in that case we will get this limit equal to this limit 2 x +2 x m 2 and then here we will
have the sec2 m x 2 and then this is 2 mx. So, this 2 x 2 x and then also 2 x from here will
get cancel and when x → 0. So, this is 1 here, this is 1 and in this case it is a there as m 2
square.
So, we have m 2 over this is 1 and this is; so 1 plus m 2 this is
depending on the path.
126
2
1+m
. So, again this limit is
m
(Refer Slide Time: 15:25)
2
And this is
1+m
which we have just evaluated there. So, limit depends on path and
m
hence the limit does not exist and the function is not continuous again in this case.
(Refer Slide Time: 15:39)
So, some remarks which are very useful for finding out the continuity or basically the
limit. So, changing to polar coordinate as we have seen by substituting x=r cos θ and
¿ r sin θ . And it is investigating the limit of the resulting expression as r → 0 is very
useful.
127
As we have seen in many examples ah, like again if we take the simple example
3
x
2
2;
x +y
so we can just change the polar coordinate. So, our limit will be the r → 0and here we
3
3
r cos θ
will have r cos q sorry
. And then thisr 2 will be cancelled with this r here.
2
r
2
So, you will have r and then limit r → 0, so this cos2 θ is a bounded function. So, here
when r → 0 this limit will be 0 so limit is 0. In another example what we have seen this
changing to polar coordinate is also useful for determining that the limit does not exist.
So, this was the case when limit exist and now in this case when we substitute this again
x=r cos θ and y=r sin θ . In this so we will get here r 2 and here r 2 cos2 θ . So, this r
square will get cancelled with this r 2 square and in this case then there is no r left so, this
is cos2 θ .
So, this limit is cos2 θ and it depends on θ. So, when the limit depends on θ; that means,
the limit does not exist the limit should be independent of θ. So, here for different values
of theta we are getting the different real number. So, this limit is not unique and hence it
does not exist. So, we have seen other examples also that this choosing to polar
coordinate is very useful for determining the limit as well as for determining that the
limit does not exist.
128
(Refer Slide Time: 17:59)
The next remark that changing to this polar coordinate does not always helps, you know
this is very important now that it does not always help and the transformation may tempt
us to false conclusion we will see some supporting example in this case. So, if we take
2
this
x y
4
2 , this is example; then if we substitute x=r cos θ , y=r sin θ. So, here we will
x +y
get because one r 2 from here r from here. So, we will get r 3, and then cos2 θ here sin
theta and again here r 2 cos4 θ and r 2 sin2 θ .
So, in this case this r 2 we can cancel out. So, will get r cos2 θsin θ, and in the
denominator we will get r 2 cos4 θ+sin 2 θ. And now if someone just directly try to put the
limit here and think that r → 0. So, this term becomes 0 we have something a sin 2 θ. And
then here r co s2 θsin θ when r → 0, that this becomes 0; but that is the wrong conclusion.
Indeed, if we fix sin 2 θ; so if we fix some value of sin 2 θ, in that case this limit is 0
because, whatever sin 2 θ including 0 also when sin 2 θ is 0. So, this will become 0 and
then everything will become this 0, so limit will be 0 in that case also. If we take any
other theta so, some number will be sitting here and then r → 0. So, we can conclude that
this is 0, no doubt about it. But the point to be noted here that we have we have fixed θ.
So, for fix value of θ, this is 0, but as for the limit we have already discussed that we
should not fix θ to conclude that this is the limit.
129
Fixing theta means fixing the path. So, in some particular path this limit is 0; where we
are fixing the θ it is like the straight line in the case of the straight line when we are
fixing the theta, we are basically approaching towards. So, if we have this is our r θ
plane, and then by fixing θ; so if we have fixed θ and then approaching to origin here. In
that case we are basically approaching to origin by the straight line. So, again this fixing
θ will not conclude that the limit is 0.
So, for example, if we take another path where y=x 2 path so, this is again going to 0 to
origin. And in this case if you choose this particular path y=x 2 or in polar coordinate
r sin θ=r cos θ , then this limit will become r → 0 and here this x =r cos θ and then this
2
2
2
2
2
r sin θ again we have used this r 2 cos2 θ.
Same thing here r 2 cos4 θ and this r sin θ is replaced with r 4 cos 4 θ . So now, what are we
getting here it is a r 4 cos 2 θ and again cos2 θ and cos2 . So, we are getting here r 4 cos 4 θ and
here we are getting 2 r4 cos4 θ . So, this will get cancel and we will get this limit half or
directly one can see just substituting y=x 2 there taking this path.
So, we will get that limit the desired limit as x → 0 and then x2 . So, y=x 2 and then we
4
have x 4 + y 2 again here x 4. So, again this x 4 and so, limit x → 0 and we have this
x
4.
2x
So, x 4 get cancel and then we get this limit half. So, this is the limit in this case, along
this particular path while by not closely looking at this here we could have done this
mistake by putting taking this limit here as r → 0 and then this limit is 0.
But this is not the case. As we have seen that this is indeed 0, but in that case we have to
fix this θ and then the limit is 0. And we have any way seen here by taking this path the
limit is half. So, limit and does not exist in this case.
130
(Refer Slide Time: 23:13)
And we will discuss now the continuity following by that motivation the continuity of
2
2
x y
this 3 3 . And in this case if we change the coordinate to this polar coordinate what
x +y
will happen now? So, this is r 2so r 4 will come and cos2 θ sin 2 θ and from here r 3 will
come. So, we have one r in the numerator and then cos3 θ and sin 3 θ.
So, again when r → 0, we should not do that mistake that r → 0 and something is here; so
it is it becomes 0, no. We have to closely look at this function whether if it is a bounded
function of theta sitting here and then we are taking this limit r → 0 goes to 0, then this
will be 0. But in this case this function here for example, this cos3 θ−sin3 θ may become
unbounded because this may go to very close to 0 and this is actually unbounded
function so this is not a bounded function. So, we cannot use that fact that this limit as
r → 0 is 0.
So, if we fix θ, if we fix θ, then again like in the previous example this limit is 0. Thus,
we cannot conclude that the limit is 0. So, we will see in this in the next slide now again
that limit in this case does not exist.
131
(Refer Slide Time: 24:45)
So, if we look at the continuity of this function; so along this path y=mx or fixing θ it is
the same thing. So, you will get this limit as 0. So, we take a y=mx or changing to polar
coordinate and fixing θ as we discussed both are equivalent. So, here this limit is 0 along
this path, but if we take this special path y=−x ex , then we are getting here x2 and for y 2
again x2 e 2 xand then x3 and this y 3 will be −x3 e3 x.
So, now in this case this x 3 will get cancel. So, we have in the numerator x e 2x and
denominator 1−e3 x . So, what is the limit here? x → 0 so it is a 0 by 0 case. So, we can
use the L’Hospital rule. So, this limit will be equal to; so e2 x + x e 2x . And then 2; so this is
a differentiation there and then −3 e3 x.
So, when x → 0 this goes to 0 so, we have
1
−1
. So,
is the limit of this function along
3
3
this particular path again. This path is going to 0 so, we have taken a very special path to
show this is a very typical example and difficult to realize that a long this path we will
get a different limit. But this proves that the limit does not exist and hence that function
is not continuous at (0 ,0). The last example, we will discuss here the continuity of this
function at origin.
132
(Refer Slide Time: 26:49)
So, again we change the to the polar coordinate, it will simplify this x2 + y2 and we have
−¿
1
x 2 + y2
−1
( x , y ) → (0 ,0) e
. So, changing to polar coordinate it will be like e r .
4
4 ¿
x +y
2
And this here this will be r 4 and cos 4 θ+sin 4 θ because of this term. So now, here we can
do little bit manipulations. So, here cos 4 θ+sin 4 θ we can just a make 4 a square term here
2
so, the ( cos2 θ+sin 2 θ ) . So, the 2 cos2 θ will come and then minus we can subtract that
2
2
2
2
cos θ sin θ . So, this term is equal to this term. The whole square minus 2 cos θsin θ.
Because when we open this square we will get cos 4 θ+sin 4 θ and 2 times the product
which is it will be cancelled by this one, so we will get this term. So, that sin square theta
plus cos square theta became 1 here, and then we have this 1 minus this term.
133
(Refer Slide Time: 28:11)
Which again we can write down here 2 cosθ sin θ as sin 2θ. So, this is square because of
the square this square will appear here, and we have to make this 4. So, we multiplied
and divided by 2. We have this term now so we have to see what is this limit here of this
function.
(Refer Slide Time: 28:29)
134
1 2
So, what is the limit? So, in this case we have to note now, that this 1− sin 2 θ this
2
term here lies between
1
1
and 1. So, this is like 1 when θ is 0 and then this and the
2
2
2
sin θ is 1. So, this will become as
1
1
, so this lies between to a 1. So, in this case, then
2
2
4 r ≠ 0 we can estimate this term the whole term there
1
and 1 minus this one.
r2
So, this is a positive term, this is a positive term and here also we have this is greater
than half. So, certainly this is greater than 0 and also now because we know that this is
always greater than half. So, we can replace this half also and get the upper bound for
−1
r2
this term. So, replacing this by half we will get this 2e . So, this time here lies between
4
r
0 and this term. And now for this we can look at what is the limit of this as r → 0 of this
−1
r2
2e , which we have this used another variablet here, which goes to ∞ ; because 1 we
4
r
r
are substituting as t.
2
So, if → 0 , t → ∞. So, here we have 2 e−t and then this
2
can now handle easily. So, this is 2t 4 and e t . So,
1
will become t 4 and this we
r4
∞
case, and then we can use L’Hospital
∞
rule with to get this limit as 0. So, this limit is 0 so, what we see that this from here
which we want to get the limit as r → 0. This is between 0 and something here which
again goes to 0 as r → 0. So, by this squeeze theorem or sandwich theorem, we can get
now the limit this as because this is 0 and here also in the limiting scenario this is also
going to 0 so, this limit has to be 0.
135
(Refer Slide Time: 30:45)
So, in this case now we can conclude this. So, we have discuss the continuity; that
f ( x , y) is equal to the limiting value here is equal to the function value of the of that
function, then we call that the function is continuous. And naturally the function must be
defined at this point, then only we can talk about the continuity. And what we have also
observed that we need to be very careful while changing to polar coordinate, because that
may mislead to some wrong limit in min many cases.
(Refer Slide Time: 31:23)
So, these are references we have used to prepare these lectures.
136
And thank you very much.
137
Engineering Mathematics - I
Prof. Jitendra Kumar
Department of Mathematics
Indian Institute of Technology, Kharagpur
Lecture – 09
Partial Derivatives of Functions of two Variables
Hello, welcome to the lectures on Engineering Mathematics I and today’s, this is lecture
number 9 and we will be talking about Partial Derivatives of Functions of Several variables.
(Refer Slide Time: 00:26)
So, what is Partial Derivative? It is a usual derivative of a function of several variables with
respect to one of the independent variable while keeping all other independent variables as
constant. So, as written here it is a usual derivative, when we have to keep other independent
variable as variables as constant and taking the derivative of one particular with respect to
one particular variable and this is called the partial derivative with respect to that variable.
So, for example, we have a function z is equal to f ( x , y) and ( x , y) belongs to some domain
in R square and z ∈ R the real line. So, this partial derivative of f with respect to x at a point
( x0 , y0 ) which we also denote by f x ( x0 , y0 ). This is defined as we said before it is the usual
derivative. So, we are taking here derivative with respect to x. So, the fundamental definition
of the derivative that limit delta x goes to 0 and then, we will make an increment here in x 0.
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So, x 0 + Δx ; this is the increment in x because we are taking partial derivative with respect to
x.
So, here x 0 + Δx and then, y 0 we are not changing y 0. So, it is at the point y 0. So, y 0 will
remain as it is and minus the function value at ( x0 , y0 ) and divided by Δ x and if this limit
exists, then we call that this is a partial derivative with respect to x at this point ( x0 , y0 ) and
it is denoted by f x at ( x0 , y0 ) or delta f over Δ x at ( x0 , y0 ).
We can also understand this as so here in the function we have kept this y 0 as constant and
taking this usual derivatives. So, now, this is here the function of x because we have
substituted like y 0 in the function and then, this become a function of one variable x and then,
we can take the derivative with respect to x and then later on we can substitute x is equal to
x 0.
So, as written there it is a usual derivative with respect to x. Like here we have use the usual
derivative
d
of this function which is a function of x because we have substituted y= y 0, the
dx
constant value of this y. Similarly, the derivative of f with respect to y at the point ( x0 , y0 )
and now because we are taking the derivative of with respect to y. So, in that fundamental
definition now we will make an increment in y 0 arguments. So, here y 0 + Δ y and minus the
function value of f at ( x0 , y0 ) and here divided by Δ y.
And again, if this limit exist we call that the partial derivative exists and this is the notation
that f y (x 0 , y 0 ) over del f over del y at ( x0 , y0 ). And again, we can also take like fixing this x 0
, then we have this function as a function of one variable y and then we can take this usual
derivative with respect to y at y is equal to y 0 later on. So, this will be the partial derivative
with respect to y at the point ( x0 , y0 ).
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(Refer Slide Time: 04:10)
So, coming to Geometrical Interpretation of the Partial Derivative, we consider this plane
three dimensional. So, x axis, y axis and then, we have z axis there and we have some
surface where the function z=f ( x , y ). So, the surface is represented by this function as
z=f ( x , y ).
So, if we cut the surface by a plane here y= y 0. Now, this is the plane y= y 0, this is y axis.
So, here we have point on the y axis y= y 0 and this is the plane. So, here all the values on of
y as y 0 another are constant. So, here we have this plane y= y 0. If we cut this surface, then
we as we can see here there is a curve of intersection here by this plane and the surface. So,
now, here on this curve if we take a point for example, ( x0 , y0 ) and we are talking about the
partial derivative with respect to x.
Then, if we draw the tangent on this curve at this point ( x0 , y0 ) and then, we note here the
angle from the x axis and the tangent of this angle would be precisely the partial derivative of
f with respect to x at ( x0 , y0 ). So, this is the same definition same geometrical interpretation
as we have for the function of one variable. The only difference is that because we have fixed
this y. So, y= y 0 if we cut this plane over this surface or by this plane, then we will get a
curve and then we have we are again back to in one dimensional case and the geometrical
interpretation is same that the tangent of this theta is equal to the partial derivative of f with
respect to x at this point ( x0 , y0 ).
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(Refer Slide Time: 06:13)
Coming to the problem find the value of del f over del x at the point x 0 y. It is a general point
here of the function f ( x , y )= y e−x from the first principle. So, we will use the definition we
which we have discussed in the previous slide to compute this
∂f
∂f
and
at a general point
∂x
∂y
( x , y).
So, what was the definition?
∂f
at a point ( x , y). So, we have taken here a general point
∂x
( x , y). So, the definition says that we have to make an increment in x because we are talking
about the partial derivative with respect to x. So, x+ Δx and y minus the function value at
¿. So, then we
( x , y); so, f ( x , y) and divided by this increment Δx and then, taking the Δlim
x →0
have to take this function which is f ( x , y )= y e−x .
So, in this case this + Δ x , y. So, it will become y and instead of x we will have x+ Δ x. So,
e−¿ ¿ and for x , we have x+ Δ x minus the function value at ( x , y) which is y e−x the function
itself and divided by Δ x. So, now, if we see here y e−x and y e−x is common. So, we can take
this outside y e−x . So, what will remain here? e−Δ x and then, −1 because y e−x we have taken
common from these two terms and then, we have divided by Δ x there.
Now, if you compute this limit because as we see Δ x goes to 0. This e 0 it is one minus one
0/ 0 form. So, we can apply the (Refer Time: 08:11) rule which says that this limit will be like
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−e
−Δ x
, the derivative of this e−Δ x and the derivative here will be just 1. So, when Δ x goes to
0. So, this limit is going to −1. So, the derivative here in this case will become −1 and y e−x ;
so, − y e−x .
Now, for the partial derivative with respect to y, so our definition will change now. So, x will
remain as it is and they will be incrementing y. So, y + Δ y and −f ( x , y) divided by Δ y. So,
now, substituting this in the function; so, we have y + Δ y and e−x minus the function y e x and
divided by this Δ y.
So, again here y e power minus x is common term y e−x and here also we have y e−x . So,
what is left then e limit Δ y goes to 0 and e−x we. So, here we have y e−x which cancel out.
−x
So, y e−x and minus y e−x cancel out and then, we have
Δye
. So, the Δ y also cancels out
Δy
and we get only e−x. So, no need to take common because this is y e−x and here we have
minus y e−x is the same term.
So, it basically gets cancelled and then, we have here y e−x term divided by Δ y and this Δ y
and Δ y will be cancelled. So, we will get e−x and there is no Δ y term here. So, this is just e−x
. So, what we can also see? So, this was using the basic definition of or the fundamental
definition of partial derivatives. We got the derivatives here with respect to x was minus y e−x
and with respect to y it was e−x.
We can also compute this directly because this function is given as f ( x , y) is equal to y e−x
and the partial derivative with respect to x at the point ( x , y). So, keeping here y constant;
treating y constant and taking the usual derivative. So, y will remain as it is and the
derivative of e−x will be −e −x.
So, we have the partial derivative with respect to x which we have already seen here − y e−x .
Same we can do to get the partial derivative with respect to y. So, if we take the partial
derivative with respect to y here, then this is now x will be treated as constants. So, e power
minus x will be treated as constant and then, when we take the partial derivative with respect
to y. So, this will be 1, then we have e−x. So, again we get. So, we can directly compute using
the usual definition or usual a known reserves of the derivatives keeping other variable
constant ok.
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(Refer Slide Time: 11:46)
So, the next. So, what is the Relationship between the Partial Derivatives and the Continuity?
We have already discussed in the previous lecture Limits and Continuity. So, if a function
can have partial derivative that is a interesting result. So, if a function can have partial
derivative without or with respect to both x and y at a point without being continuous there.
So, we do not need continuity to for the existence of partial derivative in this case. When we
talk about the usual derivatives, the continuity is must for the differentiability, but that is so
far we called differentiability or getting the derivative there in case of single functions of
single variable, we need continuity of the function.
But in this case a function can have partial derivatives with respect to both x and y at a point
without being continuous there. On the other hand, a continuous function may not have
partial derivatives. So, other way around, we can have a continuous function, but it may not
have partial derivative. So, anything is possible and we will see now in this example which
shows that this function we will check now is continuous, but its partial derivatives do not
exist. So, that is a one example which shows that the function is continuous, but the partial
derivative both the partial derivatives do not exist at (0 ,0).
143
(Refer Slide Time: 13:17)
So, let us check the continuity at ( 0,0 ). So, we have this function here
1
( x+ y ) sin
, ( x+ y ) ≠0
(
)
f x,y =
x+ y
¿ 0 ,elsewhere
{
( )
and we want to show for the continuity that this limit if we take the limit ( x , y) goes to (0 , 0)
of this function f ( x , y), then this limit is 0.
If this limit is 0 because the function value we have seen a 0 here at (0 ,0). So, we will show
that this limit here is equal to the function value than the function is continuous. We can use
for example, the delta epsilon definition or we can also do directly as we have done in the
previous lecture. So, let us just go through the standard definition of epsilon delta.
So, here f ( x , y) minus this 0, the function value is equal to this difference. So, here that is in
the function itself ( x + y ) sin
( x+1 y ) and we know that the sin is bounded by 1. So, we can
write down this x plus less than |( x+ y )|. So, this is bounded by 1. So, this will be a big
quantity than this one and again, we can we know also this result the triangular inequality that
the absolute value of ( x + y ) will be less than equal to the |x|+| y|. And now, we will make a
calculation here to substitute this x+ y or to estimate this absolute value of x+ y in terms of
the neighborhood.
144
If you remember we need to set this inequality in terms of the neighborhood so that we can
get the relation with δ and ϵ . So, here if we consider this absolute value of x minus absolute
value of y whole square, this will be always positive because this is a whole square term and
then, we have here x square plus y square and this will be minus 2 times absolute value of x
minus absolute value of y which I can bring to the right hand side. So, the right hand side
will become 2 absolute value of x and absolute value of y.
Now, if I add both the sides this x2 + y2. So, here it will become 2(x 2 + y2 ) and the right side, it
will be ( x2 + y 2 ) and this plus 2 times x and y. This the right hand side, we can write down as
¿ x∨¿ , ¿ y∨¿ whole square and the left hand side is a 2(x 2 + y2 ). And now, we can take the
square root here. So, this will become as a √ 2 √ x 2 + y2 and less than equal to this is less than
less than equal to |x|+| y|.
This is other way round. So, this is greater than equal to because here it was greater than. So,
this is greater than, this is greater than equal to this |x|+| y|. So, now, we can replace this term
here the |x|+| y|≤ √ 2 √ x 2 + y2.
(Refer Slide Time: 17:14)
So, let us do that. A √ 2 √ x 2 + y2 and now here exactly we have the neighborhood of this (0 , 0)
point which we can bound by δ. So, here this term can be we can use that inequality for the
neighborhood and then, we have a
√ 2δ and now this difference the aim is to make this
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difference of f ( x , y) minus 0 less than ϵ . So, this is less than ϵ now, if you make then we will
get the relation between δ and ϵ .
So, here choosing this δ from here ϵ by a square root 2 because ϵ is given and then we can
find this δ by this relation anything is smaller than this ϵ divided by square root 2 and then,
we have that this difference is less than ϵ whenever we choose ( x , y) from this
neighborhood; from this neighborhood δ neighborhood of (0 , 0) point.
So, this is the ϵ δ approach to prove that the given number is the limit of a function which is
here ( x + y ) sin
1
. So, what we have seen that this 0 is the limit of this f ( x , y) function.
x+ y
( )
So, in that case we have proved the continuity. So, this implies that f ( x , y) is continuous.
(Refer Slide Time: 18:46)
Next moving to the partial derivatives at (0 ,0) points, so we have this function and we can
use now this definition Δ x goes to 0 and x 0 + Δ x and minus f ( x0 ) over Δ x. So, since x 0 and
y 0 (0 , 0) , so we are talking about the derivative at (0 ,0) point. So, we are taking this limit if
this limit exist, then we call the partial derivatives with respect to x exist. If it does not exist,
we will call the partial derivatives or partial derivative with respect to x does not exist. So,
here in this case we have as per the definition now and Δ x goes to 0.
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So, this f ( Δ x 0 ). So, here y is 0 and then, x will be replaced by Δ x. So, we have Δ x and
sin
( Δ1x ) and this Δ1x term is here. So, Δ x and Δ x gets cancelled and then, we get simply
this limit Δ x goes to 0 sin
( Δ1x ) and Δ x approaches to 0. So, in this case here we do not
know what is this limit because the sin is not definite when this Δ x goes to 0. We cannot
conclude the value of this one.
So, basically in this case this limit does not exist because we do not know what number is
this. So, the limit; that means, the partial derivative with respect to x does not exist in this
case and obviously, the similar calculation we can do for y or the partial derivative with
respect to y to get this limit and we will find that that the partial derivative with respect to y
also does not exist.
(Refer Slide Time: 20:30)
Moving to another problem, we have a now we will show that this function
xy
, ( x , y ) ≠(0 , 0)
f ( x , y )= x2 +2 y2
¿0 , elsewhere
{
is not continuous, but its partial derivatives exists at (0 ,0). In the earlier problem, we have
seen the function was continuous, but the partial derivatives do not exist. In this case, we will
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show that the function is not continuous, but its partial derivatives exist; both the partial
derivatives exist. So, why this is not continuous? If we choose the path y=mx; then, what we
find because y=mx and then here also y=mx.
So, here x2 here x2 and here also x2 will come and what we will get. So, here we have x and
then y=mx. So, we have x2 +2 x2 m 2 and then, the limit x goes to 0.
(Refer Slide Time: 21:39)
So, here x2 here x2 here x x 2. So, we will get this limit as
m
m
and hence, this
2 .
(1+2 m ) (1+2 m2 )
limit does not exist because it depends on m. For different values of m, we are getting
different value of this limit.
Hence, this limit does not exist and the function is not continuous. Now, we will show that
though the function is not continuous, but we can compute the partial derivatives with respect
to x and with respect to y.
148
(Refer Slide Time: 22:03)
So, the partial derivatives again with the definition we have this fundamental definition which
at the (0 , 0) points become Δ x 0 and f (0 , 0) over Δ x. Now, Δ x 0. So, one argument is 0, the
y is 0. So, this will become 0 because of this product. So, here it is 0 and then, f (0 , 0) this is
also 0. So, 0 minus 0 over Δ x and this is 0 whatever Δ x’s so, we have here the value of this
limit is 0; that means, the partial derivative with respect to x is 0.
Now, the partial derivative with respect to y. So, we have y 0 + Δy in this case and x 0 y over
Δ y . As per the definition, we have because this x 0 y0 ; you will put 0, we are talking about the
partial derivatives at (0 , 0) point. So,
f ( 0 , Δy ) −f (0, 0)
, again the same argument because
Δy
here the x is 0 and the product in the numerator is there x y term. So, here this will also
become 0. So, this will become 0 and this is 0.
So,
0−0
and this is 0 and we got again this limit as 0. So, both the partial derivatives with
Δy
respect to x and with respect to y exist the value of the partial derivatives in both the cases
are 0 and the function was are not continuous in this case.
149
(Refer Slide Time: 23:46)
Another problem, so here we have
3
3
2 x +3 y
f ( x , y )= x 2 + y 2 , ( x , y ) ≠(0 , 0)
¿ 0 , elsewhere
{
. So, we want to compute f x and f y at (0 , 0) point.
So, again as per the definition, we have f x (0 ,0)is equal to
f ( Δx , 0 )−f (0, 0)
and Δx goes to
Δx
0. This is the partial derivative with respect to x. So, we will substitute now f ( Δ x , 0). So, y
will be set here 0; there also 0 and we will get this 2 times. So, this when y will be 0. So, this
function will become 2 x; simply 2 x. So, we will get this one2 Δ x and divided by again Δ x
here and then, limit Δ x goes to 0. So, Δ x gets cancel and we will get this value as 0.
150
(Refer Slide Time: 25:00)
So, we will get here the value 2 ok. Now, the second when we compute f y (0, 0). So, we have
a similar situation again; the 0 Δ y. So, this time the Δ y will remain. So, we will have here 3
times and the Δ y minus. This f (0 , 0) is 0 and over Δ y and Δ y → 0. So, here this gets
cancelled and you will get this value as 3. So, this is 3 here. So, we have the partial derivative
with respect to x as to partial derivative of f with respect to y here as 3. The function is
continuous and also partial derivatives exist.
Because in this case we can easily prove that the partial derivatives exist and we have already
done this in previous lectures. So, here for example, to see the continuity, we can put
x=r cos θ and y=r sin θ and then, this will become simply here the r 2 will come and then, 2
3
3
3
3
r and cos θ+3 r and sin θ and limit r → 0.
So, in this case the 1 r will survive in the numerator and the rest term here is bounded. So,
this will go to 0. So, the function is continuous and the partial derivatives exist. So, anything
is possible. We have seen the example when the function was not continuous partial
derivatives exist or the function was continuous, partial derivative do not exist. In this case
function is also continuous and partial derivatives also exist.
151
(Refer Slide Time: 27:01)
So, this example again to compute the partial derivatives with respect to x and with respect to
y and also you will discuss the continuity of these partial derivatives. So, here the f x when
( x , y)≠(0 , 0). So, when ( x , y)≠(0 , 0), we have this nice function. The problem is at 0 where
we have to defined separately as 0. So, when ( x , y)≠(0 , 0), we have this function and we can
take the partial derivative x here treating this y as constant.
So, just doing this calculation for a partial derivative with respect to x, we can just follow the
usual calculations. So, here x2 + y2 and the square root; so this will be its whole square and
then, in the numerator when we take the derivative treating this y as constant.
So, here we will get the partial derivative with respect to x of x y will become y and then,
minus this x y as it is and the partial derivative here which will be 1/2 and then, here this is
√ x2 + y 2 and we are taking partial derivative with respect to x . So, this 2 x will come and this
2 will get cancel and then, we can multiply here this y term in x2 + y2 and then we have
minus. So, here again this term will go there because this quotient rule.
So, x2 + y2. So, this will become y ( x2 + y 2) and minus this x2 y and then, divided by this x2 + y2
term and this will also come. So, this will become 3 by 2; 1 and this half will make it 3/2.
2
3
So, this here x y will get canceled, we will get y . Exactly this term here
similarly, we can compute this f y with respect to y.
152
y
3
3/ 2
( x 2 + y2 )
and
(Refer Slide Time: 29:16)
So, we will get just instead of the y will be replaced by x because of the symmetry of the
functions. So, we got
{
{
y3
, ( x , y ) ≠(0, 0)
f x ( x , y ) = ( x2 + y2 ) 3/ 2
¿ 0 , elsewhere
y3
, ( x , y ) ≠(0, 0)
f x ( x , y ) = ( x2 + y2 ) 3/ 2
¿ 0 ,elsewhere
and here this partial derivative with respect to x, we have to also compute this with the
definition the fundamental definitions.
So, we have to use here what is the f x at (0 ,0). So, at non zero point that non zero point, we
can directly compute from here and at (0 , 0) we have to use the fundamental definitions. So,
the Δ x goes to 0 and we have
f ( Δx ,0 )−f (0, 0)
. So, f (0 , 0) is 0 and here if one of the
Δx
argument is 0.
So, this function will become 0 and so, we have
0−0
. So, this will become 0. Therefore, we
Δx
have used here 0, but we have to use this fundamental definition to get the partial derivative
153
at 0. So, here also we get 0 and now moving next. So, we want to discuss the continuity. So,
to discuss the continuity, we have to see what is the limit of this? Whether the limit of this
function is 0 or same thing here that the limit of this is 0; then, the function is continuous,
otherwise it is not continuous.
(Refer Slide Time: 30:34)
So, we have to check those limits here. In this case, when we substitute x=r cos θ y=r sin θ.
So, we will get here r 3 and here also r 3 sin 3 θ . So, r cube will get cancelled and we will get
3
sin θ. So, this limit depends on θ . So, therefore, this is not continuous. So, this f ( x , y) is not
continuous because this limit does not exist the limit depend on θ.
For continuity, the limit should exist and it should be equal to the value at (0 , 0). So, in this
case the limit does not exist. Therefore, these partial derivatives are not continuous; (Refer
Time: 31:18) did we can do the similar calculation for f y. So, both a partial derivatives f x and
f y, they are not continuous.
154
(Refer Slide Time: 31:26)
We have one result on the sufficient which is the sufficient condition for continuity in a
region r.
So, what it says if a partial if a function f ( x , y) has partial derivatives f x and f y everywhere
in a region R. So, the existence of the partial derivative here and moreover, these partial
derivatives are bounded by some number M here. So, f x ( x , y) in that region is bounded and
f x also the f y is bounded, where this M is independent of x and y . So, then the f ( x , y) is
continuous everywhere. So, that is the one sufficient condition for the continuity again.
So, if the partial derivatives exists and they are bounded, then the function will be continuous
and we can also have a sufficient condition to discuss this continuity at ( x0 , y0 ). So, here if
one of the partial derivatives exist and is bounded in the neighborhood; so, here we have to
talk about the neighborhood. So, if one of the partial derivatives exist and is bounded in this
neighborhood and the other one exist at this ( x0 , y0 ) point. Then, also we can say that the
function is continuous at ( x0 , y0 ) point ok.
155
(Refer Slide Time: 32:46)
So, what we have learnt today which will be used in following lectures is the Partial
Derivatives; so, it with respect to x this definition we are going to use in many other lectures.
So, here the increment with respect to x and then, this quotient we have to take the limit if
this limit exists, we will call it partial derivative with respect to x and same thing for the
partial derivative of with respect to y.
(Refer Slide Time: 33:15)
These are the references used for preparing these this lecture and.
Thank you very much.
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157
Engineering Mathematics - I
Prof. Jitendra Kumar
Department of Mathematics
Indian Institute of Technology, Kharagpur
Lecture – 10
Partial Derivatives of Higher Order
Welcome to the lectures on Engineering Mathematics-I.
(Refer Slide Time: 00:21)
And this is lecture number 10 we will be talking about Partial Derivatives of Higher
Order.
158
(Refer Slide Time: 00:27)
So, in the last lecture what we have seen the partial derivatives of f . So, basically the
first order partial derivatives we have done in the last lecture. So, what was? It was the
fundamental definition of the partial derivative with respect to x at the point ( x0 , y0 ). So,
the increment in x 0 because we are taking or we are talking about the partial derivatives
with respect to x and the function value divided by the delta x increment and taking this
limit if this limit exists. Then we have the partial derivative with respect to x the first
order partial derivative.
f ( x 0 + Δx , y0 ) −f ( x0 , y0 )
∂f
|( x , y )= lim
.
∂x
Δx
Δx →0
0
0
And similarly we have the first order partial derivative with respect to y at this point
( x0 , y0 ) when this limit exists.
f ( x0 , y0 + Δy ) −f (x 0 , y 0 )
∂f
|( x , y )= lim
.
∂y
Δy
Δy → 0
0
0
Now we will continue this for higher order derivatives. So, what will happen if we take
for example, the derivative of f with respect to x two times?
159
(Refer Slide Time: 01:23)
So, in this case this is the notation. So, we have the partial derivative with respect to x
and for example, we want to take again the partial derivative with respect to x. So, the
idea remains the same because we have this partial derivative with respect to x. So, this
is another function and then we are taking partial derivative with respect to x of this
function.
So, basically this is like
∂
. And then we have some other function which I am calling
∂x
as f x which is basically the notation of this a partial derivatives. So, what we are now
doing we are taking the partial derivatives of this function f x. And now the same
definition what we have learnt already that the Δ x → 0, we will be talking about this
limit so our function here f x. And for example, if we take at ( x0 , y0 ) points. So, here with
respect to x means we will make an increment here in x and then y 0will remain and then
we have the f x at ( x0 , y0 ) and divided by this Δ x.
So, that will be the definition now of the second order derivative here with respect to x.
160
(Refer Slide Time: 02:43)
So, here the notation again: so we instead of this we also use f xx the derivative of f two
2
times with respect to x or we also use this notation
∂ f
2 . So, again the second order
∂x
derivative a partial a derivative of f with respect to x. Similarly, we have the notation
when we take the partial derivative of x and then we are taking the partial derivative with
2
∂ f
respect to y. So, in this case we use this notation f yx or
. In some literature people
∂ y∂x
use this as f xy or some other way around ∂ x ∂ y.
So, we will be following this notation the partial derivative with respect to x and then we
take once again the partial derivative with respect to y. So, we will keep this order here
2
∂ f
. This means the first three partial derivative with respect to x and
yx and f yx and
∂ y∂x
then with respect to y. So, similar notation for the two times partial derivative with
2
respect to y which we will denote as here with respect to x here f yy or
∂ f
2 . Or we can
∂y
have like first the partial derivative with respect to y and then we take the partial
derivative with respect to x. So, this we will denote as f xy or
2
∂ f
. So, this is just the
∂x ∂ y
notation. And these derivatives here f xy or f yx are called the mixed derivatives, because
161
here we have used both the variables x and then with respect to y or first with respect to
y then with respect to x. So, these are called the mixed derivatives.
(Refer Slide Time: 04:39)
2
So, let us directly compute these mixed derivative
2
∂ f
∂ f
and
, at the origin of this
∂x ∂ y
∂ y∂x
function.
xy( x2 − y2 )
, ( x , y ) ≠(0 , 0)
f ( x , y )=
x2 + y 2
¿ 0 , elsewhere
{
2
∂ f
So, as we have discussed the
means that the partial derivative with respect to x of
∂x ∂ y
the partial derivative of with respect to y. So, in short we can denote this like f y.
2
∂f y
∂ f
∂ ∂f
=
=
∂x ∂ y ∂x ∂ y
∂x
( )
So, we have partial derivative f y and with respect to x. So, we want to get the partial
derivative of this f y with respect to x. So, note that this f y is also a function of x and y.
So, we can talk about this partial derivative with respect to x.
162
So, as per the definition again with respect to x, so a partial derivative with respect to x.
So, we will take the limit Δ x → 0, f y the function and we are talking about the derivative
at origin, so ( 0,0 ). So, you have
¿ lim
Δ x→0
f y ( Δ x , 0 )−f y ( 0 , 0 )
Δx
The same definition of the partial derivatives with respect to x: so instead of f we have
used here f y, because we are taking the derivative with respect to x of the function f y.
So now, in this case we have to know what is f y ( Δx ,0 )then only we can use that
f y ( Δx , 0 ) here. And also we should know what is f y (0, 0). Then we can use that value
here and compute this mixed order partial derivative. So, let us compute what is
f y ( Δx , 0 ). So, f y ( Δx , 0 ): now here it is a partial derivative with respect to y . So, we have
to use the definition now that Δy → 0 and of the function f . So now, we are using the
function f , but the partial derivative with respect to y. So, the increment has to be in y
argument. So, f the ( Δ x , 0 ): so the Δ x will remain as it is because we are not touching
Δ x. We are taking the derivative with respect to y. So, this Δ x will remain and here
0+ Δ y which I am writing just Δ y and minus f the function value at ( Δ x ,0 ); so ( Δ x , 0 )
and divided by then Δ y, because we are taking the derivative with respect to Δ y.
So now, if we compute this deltaΔy → 0, f ( Δ x , 0 )so we have to substitute there now.
Then we have
f y ( Δ x , 0 )= lim
Δ y→ 0
f ( Δ x , Δ y ) −f ( Δ x ,0 )
=Δx
Δy
(Refer Slide Time: 08:43)
163
So, this is Δ x. Now we have to compute f y (0, 0). So, what is f y (0, 0). ? So, f y (0, 0)
again we have to use the definition. So, Δ y → 0 and then we have the function f we are
taking the partial derivative of the function f so with respect to y. So, the 0 will remain
as it is so 0+ Δy. So, here we have Δ y minus this f (0 , 0) and divided by Δ y.
So, this limit Δ y → 0. And this one argument is 0 and we have the product. So, this will
become 0 and minus f ( 0 , 0 ) =0 that is given and Δ y. So, this will be 0. So, the partial
derivative with respect to y at (0 ,0) is 0. So, this is here 0.
f y ( 0 , 0 )= lim
Δ y →0
f ( 0 , Δ y ) −f ( 0 , 0 )
=0
Δy
164
(Refer Slide Time: 09:41)
So, we got these two.So, f y ( Δx , 0 )= Δx and here f y ( 0 , 0 )=0.So, we can substitute now:
lim
Δ x →0
f y ( Δ x ,0 )−f y ( 0 , 0 )
Δx−0
= lim
=1
Δx
Δx
Δx → 0
So, we have this limit as 1. So, we got this paths of mixed order partial derivative of this
2
∂ f
=1
∂x ∂ y
165
(Refer Slide Time: 10:09)
Similarly, now we will compute the second order partial derivative with respect to y and
then with respect to x.
So, in this case we have the order now
∂ ∂f
; that means, the partial derivative with
∂ y ∂x
( )
respect to y of the function f x. So, in this case again we use the definition. So, Δ y → 0
this Δ y and we have this function f x.
∂fx
f x ( 0 , Δy ) −f x ( 0, 0 )
= lim
∂ y Δy → 0
Δy
So now, as in the previous slide we have to compute this f x ( 0 , Δy )and f x ( 0 , 0 )instead of
f y earlier. So, what is this derivative here f x ( 0 , Δy ). So, again the limit has to be for Δx
because we are taking the derivative with respect to x and f here the increment.
f x ( 0 , Δy ) = lim
Δx → 0
f ( Δx , Δy ) −f ( 0 , Δy )
=− Δy
Δx
2
2
¿ and this will be again and this term ΔxΔy( Δ x − Δ y ) −0 and then we
So, here the Δxlim
2
2
→0
Δx +Δ y
¿.
have Δx here. So, in this case this Δx and Δx will be cancelled, and we are taking Δlim
x →0
166
So, this term will go to 0 and this term will go to 0 and when we will get here
3
2
−Δ y / Δ y , which will be −Δy .
(Refer Slide Time: 12:25)
So now this derivative here is −Δy− Δy and then f x (0 ,0). So, this f x (0 ,0). We have to
use again this definition. So, the limit Δ x → 0 and then we have
f ( Δ x , 0 )−f ( 0 , 0 )
. And
Δx
since one of the argument here is 0, because of this product this will become
0−0
and
Δx
the limit Δ x → 0. So, we have here again this 0. So, we got this 0. Now we will substitute
here.
167
(Refer Slide Time: 13:03)
So,
Δ y−0
and this limit this Δ y∧ Δ y gets cancelled and we will get minus -1. Note
Δy
2
that: the earlier we got in the previous slide
2
∂ f
∂ f
=1, and now we got
=−1. So,
∂x ∂ y
∂ y∂x
what we have observed that these two partial derivatives first with respect to y and then
with respect to x or first with respect to x and then with respect to y they may not be
equal. Like in this present example here it is value -1 and here the value is 1.
(Refer Slide Time: 13:47)
168
So, when the equality of this mixed partial derivatives happen. So, we have this (Refer
Time: 14:02) if f x f y and the f yx all exist in the neighborhood of this point. So, not only at
this point, but also in the neighborhood of this point all these three exist and this on the
top here f yx is also continuous at that point 0 0. In that case we have the result that f xy
will also exist at this point and it will be equal to the value of f yx. So, basically the
continuity plays here major role of this f yx or we can do this result for x; xy. Or there is
another parallel result here.
If the mixed derivatives this f yx and f xy it is easy to remember. So, when the equality
happens when these two derivatives f yx and f xy are continuous in an open domain D.
Then at any point in the domain we have the partial derivatives equal; this mixed order
partial derivatives equal. So, in the previous example definitely those partial derivatives
were not equal. Otherwise we would have got the equality there that f xy is equal to f yx,
because that is a sufficient condition. So, if these derivatives are continuous second order
partial derivatives are continuous then they will be equal.
So, what we can conclude from here, if they are not equal, if they are not equal; that
means, the derivatives were not continuous because if the derivatives were continuous
then they has to be equal. So, if these derivatives are not equal this mixed order
derivatives are not equal, then we can conclude that the partial derivatives these mixed
order partial derivatives are not continuous.
(Refer Slide Time: 15:51)
169
So, let us see in this example. So, we will compute first the f xy ( 0 , 0 ) and f yx ( 0 ,0 ) for this
problem
3
xy
2
f ( x , y )= x+ y 2 , x ≠− y
¿ 0 ,elsewhere .
{
So, we compute the second order partial derivative with respect to x of the function f y.
As per the definition we have
2
f y ( Δx , 0 ) −f y ( 0 , 0 )
∂f y
∂ f
=
= lim
∂ x ∂ y ∂ x Δx →0
Δx
And similar to earlier example we need to compute now this f y ( Δx ,0 )which is again this
will be the limit here and the Δy → 0. And f we are talking about Δy. So, the Δx will not
change and we have the 0+ Δy then we have f ( Δx , 0 ) and divided by Δy. So, what is this
limit? So, if we take this limit here Δy → 0 here it will become Δx and Δy cube from here
and then divided by Δx and Δ y2.
So, again this y is 0. So, this will become 0 and Δy. So, this Δy will be cancelled here,
and then Δy if we are taking to 0 because this is let us write down again Δy → 0. So, we
have
3
ΔxΔ y
2
Δx+ Δ y
So, when Δy → 0. So, this term goes to 0 and here this goes to 0. So, you have this limit
as 0. So, in this case this is 0.
170
(Refer Slide Time: 17:47)
And f y (0, 0) will be let us compute here. So, f y (0, 0) is the limit Δy → 0. And then we
have
f y ( 0 , 0 )= lim
Δy → 0
f ( 0, Δy )−f (0 , 0)
Δy
So, again in this case here it is 0, and this is also 0, so
0−0
.
Δy
So, this partial derivative at (0, 0) is 0.
171
(Refer Slide Time: 18:21)
So, in this case we got this 0 both are 0.
lim
Δ x →0
0−0
Δx
So, in this case we got this partial or a second order derivative as 0. Now you will
compute the other one f yx of this function.
(Refer Slide Time: 18:35)
172
So, here now the partial derivative of y with respect to the y of this function f x again the
definition. So, increment in y and then. So, we need to compute again this partial
derivatives here at f x ( 0 , Δ y ) and at f x ( 0 , 0 ). So, if you compute this at f x ( 0 , Δ y ). So, in
this case what will happen? So, we are computing now f x ( 0 , Δ y ).
Let us take the limit as Δ y → 0 sorry Δ x → 0, not Δ y → 0this will be Δ x → 0. And we
have f . So, the increment in x Δ y as it is the function as it is now divided by this
increment Δ x. So, this will be the partial derivative with respect to x. So, the limit delta
3
ΔxΔ y
−0
x goes to 0. So, this will be Δx+ Δ y2
.
Δx
So, this Δx will get canceled. And now we will let Δx → 0. So, this term will become 0
3
and we will get
Δy
2 . So, in this case we are getting Δy this derivative here, we are
Δy
getting this Δy.
(Refer Slide Time: 20:11)
So, in this case: and now f x ( 0 , 0 ), so this will become 0 because of this product there, so
no need to compute again. And now we have Δy for this
173
f x ( 0 , Δ y )−f x ( 0 , 0 )
. So, in this
Δy
case now
Δy
, so this will become as 1. So, this f xy ( 0 , 0 ) at these two derivatives again at
Δy
this origin are not equal. So that means, they are not continuous also at ( 0,0 ) because we
have seen if the derivatives these derivatives are continuous then we will get the equality
there.
So, there they are certainly not continuous which we can check.
(Refer Slide Time: 21:05)
So, it is a continuity check of these two. So, what to do to con to check the continuity we
have to get the these mixed derivatives when x ≠− y2 and also elsewhere the both the
places. So, here for ( x , y)≠(0 , 0) if we compute or rather we will say when x is not
equal to. So, here instead of this we will call as x ≠− y2. So, at all these points where
there is no problem. We can just compute this derivative by keeping y as constant.
So, from here if you want to take this derivative from this f xy with respect tox what will
happen; so here x+ y 2 the constant rule the whole square will come. And this term sorry
2
x+ y and the derivative of this with respect to x will be
3
5
( x+ y2 ) y −x y 3 ( 1 )
y
=
2
2
( x+ y 2 )
( x+ y2 )
174
So, the partial derivative of f with respect to x we are getting
y
5
( x+ y2 )
2
.
And now, if we compute now again the partial derivative of this with respect to y for
those points where we have we do not have any problem when x ≠− y2. So now we will
again differentiate this function with respect to y keeping x constant. So, doing exactly
what we have done there now the function is this one. We will get here
6
y +5 x y
4
3
( x+ y 2 )
When we take the derivative of this one with respect to y keeping x as constant we will
get this term here. So, we have this partial derivative f yx ( x , y) when x ≠− y2.
Now we have to also compute at other points. So, first of all we should note that if we
compute the partial derivative here first with respect to y, and then with respect to x we
will get the same quantity as before. The reason is clear, because when x ≠− y2 we have
this nice function and which is certainly continuous we can prove that. So, all these
derivatives, so they must be equal at these points where there is no problem the problem
will be when this is defined as 0. So, when x ≠− y2. So, at those points we have to be
careful. At all other points certainly the derivatives will be continuous and the value will
be equal. So, we do not have to compute this again one can check or one can verify this.
So, one has to first compute the partial derivative with respect to y and then with respect
to x. And then you will notice that we will get exactly the same expression.
So now for this to check the continuity of this, what we have to now take this limit of
this f yx. Now if we take this x=m y2 this special path then what will happen? Let me
clear this first. So, what will happen now, because x=m y2. So, if we put here x=m y2
what will happen.
175
(Refer Slide Time: 25:11)
So, we have
6
2
y +5m y y
4
3
( m y 2+ y 2)
and then we can take the limit as x → 0 to approach to the origin because this path will
take us to the origin.
And now when x → 0, so what is happening here sorry y → 0 because we have replaced
3
the x. So, y → 0. And now y 6, so here y 6 y 6 and ( y2 ) . So, everything other than this m
will cancel out. And we will get
1+5m
3
( 1+ m )
So, that is the limit now along this path which depends on m. So, it depends on the path.
So; that means, this limit does not exist and now we can just say that the function is not
continuous because for continuity this limit should exist and should approach to the
derivative at (0, 0) point.
So, in this case the partial derivatives either f xy or f yx they are not continuous which was
clear because those values were not same as we have observed. And we know the result
that if these two derivatives are continuous then the value should be also equal. So, the
176
value was not equal at (0, 0). So, naturally these functions are not continuous, which is
clear again from here we have taken this f yx and then along this path we have seen that
the limit depends on path and hence the limit does not exist and the function is not
continuous.
(Refer Slide Time: 27:05)
So, the limit depends on the path and hence these two are not continuous at (0, 0).
(Refer Slide Time: 27:11)
177
So now the next example it shows the existence of the second order partial derivative
though the function is not continuous.
3
3
x +y
f ( x , y )= x− y , x ≠ y
¿ 0 ,elsewhere
{
We have seen such results earlier for the first order partial derivatives that the function is
not continuous, the partial derivatives, exist and function is continuous partial derivative
does not exist. So, here also we have for example, this result that the second order partial
derivatives exists, but the function is not continuous. So how: if you take this path
y=cos x this special path, and try to get the limit for the continuity what will happen. So,
here x 3 and then y 3=x 3 cos3 x and again note that that this will take us to the origin x → 0
y → 0. So, this path is correct it is taking us to the origin and we are checking the limit
here as ( x , y )=(0 ,0) along this particular path y=x cos x. So, substituting this x cos x we
have x−x cos x.
And in this case what is this limit here now. So, this x will get cancelled we will have
2
2
3
x + x cos x
1−cos x
And then we will take the limit as x → 0. So, in this case, so 0/0 forms we have to use
the L’Hospital’s rule. So, this will be the limit now here. So,
3
2
2
2 x+2 x cos x+ x 3 cos x
sin x
but again we see it is a 0 by 0 form.
So, we have to take we have to apply the L’Hospital’s rule again. So, here we will have
3
2
2
2
2+2 cos x+6 x cos x+6 x cos x+6 x cos x
cos x
2 without x. And in this case also we will have 2 without x when we do this derivative
here the product rule cos3 x . And in the rest I see the x will appear. So, terms with terms
178
with x and then we have here cos x. So, in this case now if we take the limit this is 1 and
here
2+2=4
So, this limit is 4.
(Refer Slide Time: 30:05)
And now, we have seen that this limit as ( x , y) goes to (0,0) is 4 and not 0. So, here the
function value is 0, but along this particular path we have seen the value is 4. So,
certainly this function is not continuous. One can also take some other path and can show
that the limit is different.
For example, if we take y is equal to 2 x if we take this path here and then again we will
see that the limit is different. So, if x goes to 0 and we have taken this x cube along this
path. So, y is 2 x. So, 8 x is cube and then here we have minus x and minus 2 x. So,
minus x and this will become 0. So, we have 2 different path one is y is equal to x cos x
another one is y is equal to 2 x here the limit is 0 there the limit was 4. So, that
concludes that the limit does not exist. And in any case here the value of the function is 4
and we have already seen along this path the value is 4. So, there itself we can conclude
that the function is not continuous.
179
(Refer Slide Time: 31:23)
So, in this case the function is not continuous.
(Refer Slide Time: 31:27)
But what we can do we can evaluate f xx at (0, 0). How? So, f x at (0, 0) as per the
definition
lim
Δ x →0
f x ( Δ x ,0 ) −f x ( 0, 0 )
Δx
180
the usual definition we have used several times. And then again we have to see what is
this function here f x ( Δ x , 0 ). So, we have to see what is f x ( Δ x ,0 ). So, this will be the
limit Δ x → 0, because of the derivative here and f . So, better to use because delta x is
already there let us compute this f x at general point x 0.
(Refer Slide Time: 32:11)
So, instead of this we will choose now here f x ( x ,0) instead of f x ( Δ x ,0 ) later on we will
substitute for this point as ( Δ x , 0 ).
So, at any general point x 0 we are computing the derivative. So,
lim f ( x+ Δ x , 0 )−f ( x , 0)
Δx → 0
(
)
f x x,0 =
Δx
because increment has to be made in the first argument then 0, and minus f ( x , 0) sorry
¿ this will become because the
and then we have Δ x there. So, in this case this Δlim
x →0
second argument is 0. So, there is no y term we will have x square simply there; that
means, the
2
lim ( x+ Δ x ) −x 2
Δx → 0
Δx
So, this will become
181
lim 2 x Δ x+ Δ x
Δx → 0
Δx
2
=2 x
So, in this case we will get this result as 2 x.
(Refer Slide Time: 33:27)
So, this is 2 x and f x ( 0 , 0 ) it is much easier to compute that this will come as 0. So, in this
case we have this 2 times. Now the result was f x at point ( x , 0) is 2 x. So, here is Δ x. So,
lim
Δ x →0
f x ( Δ x , 0 ) −f x ( 0, 0 )
2 Δx−0
= lim
=2
Δx
Δx
Δx →0
So, what we have seen the second order derivative with respect to x of this function is 2.
182
(Refer Slide Time: 34:05)
Now, we can also compute the derivative with respect to y like we have done before.
f yy (0 , 0)= lim
Δy → 0
f y ( 0 , Δy )−f y ( 0 , 0 )
Δy
So, with now we have to compute this f y ( 0 , Δ y )as earlier. So, we will get −2 y because
of this sign here the 1 minus will appear. So, we will get −2 y in this case. And the (0, 0)
will be 0 so now this f yy. So, second order derivative with respect to y will be
lim
Δ y→0
f y ( 0 , Δ y ) −f y ( 0, 0 )
−2 Δy−0
= lim
=−2
Δy
Δy
Δy →0
So, in this particular case we have seen the function was not continuous, but we have a
second order derivative not only the first order we have the existence of second order
derivatives.
183
(Refer Slide Time: 34:55)
So, what is the conclusion for today’s lecture we have now learn the partial order
derivative of higher order. And what we have also seen that the continuity of the partial
derivative is sufficient to ensure that these two mixed order partial derivatives are equal.
Or if they are not equal that means these partial mixed partial order derivatives are not
continuous in that case.
(Refer Slide Time: 35:25)
So, these are the references used for preparing this lecture.
Thank you very much.
184
185
Engineering Mathematics - I
Prof. Jitendra Kumar
Department of Mathematics
Indian Institute of Technology, Kharagpur
Lecture – 11
Derivative & Differentiability
Welcome to the lectures on Engineering Mathematics - I. And today’s we are discussing
lecture number 11 on Derivatives and Differentiability of One Variable; actually it is very
important to discuss differentiability of one variable before we go for the several variable
case in the next lecture.
(Refer Slide Time: 00:37)
So, what is derivative for a function of single variable? So, let y=f (x) be a function of
single variable and if this ratio f ( x + Δ x ) −f ( x ); so this difference when we make an
increment Δ x in x and minus thef ( x ); the value at x divided by this increment. If this ratio
here has a limit a definite limit as Δ x → 0 then we call that this limit is the derivative of the
functionf ( x ) at the point x. So, that is the definition of the derivative usually we have. So,
that limit here when we take the limit Δ x → 0, if this limit exists we call that this value is the
derivative of the function f ( x ).
And, this is usually denoted by this f ' (x ) or sometimes y ' ( x) or
dy
that is a notation for the
dx
derivative which is the limit of this quotient here, when we make the increment in x by Δ x
186
and take this difference by the value of the function at x divided by the increment. So, if this
limit exists we call the function has the derivative and its value is exactly that limit. And, the
notation here we use for the derivative are f prime y prime over
dy
.
dx
(Refer Slide Time: 02:13)
Now, what is differentiability and differential usually? So, this is more formal definition and
the reason behind this we will we can easily extend it to the case of several variable. So, a
function f ( x) is said to be differentiable at the point x, if when x is given the increment Δ x
arbitrary increment and the increment Δ y can be expressed in the form of Δ y is equal to
some constant A into Δ x plus epsilon times Δ x; where this epsilon goes to 0 as Δ x goes to 0
and this number here A is independent of Δ x. So, the point is that this f ( x) is said to be
differentiable if we can write down the increment Δ y when we give the increment Δ x in x.
So, there will be an increment in y that is denoted by Δ y. So, if we can write down this
Δ y= A Δ x+ϵ Δ x, this is a linear term here because A is independent of Δ x and +ϵ Δ x and
this ϵ has the property that it goes to 0 as Δ x → 0. Then we call that this function is
differentiable and the first term on this right hand side; that means, this A times Δ x.
So, this first term on the right hand side A Δ x, this is called differential or the total
differential of y and this is usually denoted by the notation dy. Thus, what we have that dy
that is another notation for what we call differential. So, dy is denoted by this A Δ x those are
187
the linear term in this expression of this Δ y. So, again this Δ y was the increment. So, this
Δ y is as f because the function is y=f (x).
So, Δ y=f ( x+ Δ x )−f ( x ). So, if this increment here Δ y we can write down in this form. So,
this linear term A which is independent of Δ x times Δ x and plus ϵ Δ x, then we call that the
function is differentiable. And, this epsilon very important that it should go to 0 as Δ x goes
to 0 and this A should be independent of Δ x.
So, if we can do that we call the function is differentiable and now in the next slide we will
see that this definition of the differentiability; what we have here that we can express this Δ y
in terms of this A Δ x+ ϵ Δ x. And, the earlier definition of the derivative they are actually
equivalent.
(Refer Slide Time: 05:33)
So, let us move to the next slide here. So, here differentiability and the derivative what is the
relation or basically they are the same what we will observe now here. So, the necessary and
sufficient condition that the function y=f (x) is differentiable at the point x is that it
possesses a finite derivative at this point. So, as written here it is a necessary and sufficient;
that means if a function is differentiable we have given already the definition of the
differentiability. So, if a function is differentiable then it will have derivative at that point or
it if it has a derivative at that point then the function must be differentiable at that point.
188
So, the conclusion is that the both the definition what we have given earlier the derivative and
the differentiability they are basically the same in case of the single variable. So, the now we
will see the differentiability implies the existence of the derivative at a given point. So,
suppose the function y is equal to f ( x) is differentiable. So, we assume that the function is
differentiable and what this will imply as per the definition that we can express this Δ y
increment in Δ y when an increment in x is given by Δ x.
So, if we can write down this Δ y as A Δ x+ ϵ Δ x then this function is differentiable. So, we
have assumed that the function is differentiable; that means, we can express Δ y is equal to
A Δ x+ ϵ Δ x and now we can take the limit. So, before we take the limit we can divide by this
Δ x. So, this will be like
Δy
is equal to A+ϵ . So, A+ϵ and now we will take the limit as
Δx
Δ x → 0.
So, by taking this limit here so, this will be
Δy
. So, the limit is equal to here A there is no
Δx
Δ x and A is also independent of Δ x as per the definition of the differentiability plus this ϵ
and the limit Δ x goes to 0. And, we know that this ϵ has the property that it goes to 0 as
Δ x → 0. So, in this case we get now here this lim ⁡
Δ x →0
Δy
.
Δx
So, this is the constant in the derivative we have seen f . So, this term here is nothing, but
lim f ( x+ Δ x )−f (x)
f ( x+ Δ x )−f (x )
. So, this here is Δx → 0
. So, this term here becomes the
Δx
Δx
derivative which we have discussed in the previous slide so; that means, we get here let me
erase this ok.
So, we get this here f ' ( x )=¿ A because this will go to 0. So, what we have observed that if
the function is differentiable then the derivative f ' (x ) this is the derivative here when take
the limit. So, the derivative is nothing, but this A which is written here in the linear term A
which is free from Δ x. So, this A is nothing, but this f ' (x ) the derivative. So, what we have
seen that if the function is differentiable then the derivative exist and that derivative is equal
to A, this is given in this expression of Δ y.
189
(Refer Slide Time: 09:18)
So, if f ( x) is differentiable then f ' exist and has definite value A. The next we will see that if
the derivative exists that will imply that the function is differentiable.
(Refer Slide Time: 09:27)
So, we assume that f ' ( x ) has definite value A; that means, this f ' exist or the function
derivative exist. So, in this case we have that this is given here f ( x + Δ x ) −f ( x ) over Δ x has
the limit which is equal to A because we have assumed that the derivative exist. So; that
means, this limit exist which is equal to A this is what we have assumed and now since this
190
¿
limit is equal to A we can define or let me just explain you here. So, if for example, the Δlim
x →0
of some function is equal to let us say L then what we can write down out of it that this
f ( x )−L we can define this difference in the neighborhood of point x is equal to some ϵ for
example.
So, what this ϵ will have the property that this ϵ will go to 0 when Δ x → 0. So, if we take the
limit here. So, ϵ and then the limit Δ x → 0will be equal to this limit here Δ x → 0 and f ( x )−L.
So, since the limit f ( x ) → L so, this will go to 0. So, if we write down this expression here for
ϵ as f ( x )−L then this ϵ must have this property that ϵ will go to 0 as Δ x → 0 .
So, we have a similar situation here this limit Δ x → 0 of this expression of this quotient is
given as A. So, what we have written down that this quotient
f ( x+ Δx ) −f ( x )
we can write
Δx
down as A plus so, the value of this limit which was L there. So, A+ϵ and this epsilon will
have the property that ϵ → 0 , as Δ x → 0. So, this concept we will also use later on in many
lectures.
So, having this now out of this limit we have in introduce this epsilon which has this property
that it will go to 0 as Δ x → 0, because we know that this limit of this quotient is nothing but
A. So, this epsilon must be 0 as Δ x → 0.
(Refer Slide Time: 12:10)
191
So, in this case now we can rewrite this as f ( x + Δ x ) −f ( x ) and this Δ x we can take to the
right hand side. So, we get A Δ x+ ϵ Δ x and ϵ has this property that it goes to 0 as Δ x → 0.
And, now this implies because we got this expression here that this difference or this is we
can also call like Δ y. So, the Δ y we can write down as A Δ x+ ϵ Δ x and this A was
independent of Δ x because this was the limiting situation here of A. So, A was free from Δ x,
and now we got that this f is differentiable because that was the definition of the
differentiability. So, this implies that f is differentiable. So, what we have learnt now that the
differential of the function which we have introduced as dy which was this term A Δ x.
So, this differential of the function is the product of its derivative because A is nothing, but
the derivative of this f and the arbitrary increment Δ x. So, this term here is dy. So, the dy
which is called differential is nothing, but the product of its derivative and the increment Δ x
of this independent variable or we can write down simply that dy is nothing, but the f ' ( x ) Δ x .
What else we have learned that having the derivative or having the differentiability they are
basically the same concept when we are talking about function of one variable. Well so,
moving next to the geometrical interpretation of differentials.
(Refer Slide Time: 14:05)
So, we are talking about the differential. So, what do we mean here in geometry now? So,
what we have introduced already that the Δ y when the function is differentiable then Δ y we
can write down as A Δ x+ ϵ Δ x and we have also seen that when we take this limit so, we got
192
that this A is nothing, but the f ' ( x ). This is also we have observed and this we have
introduced that the differential y is nothing but the A dx or A is nothing, but the f ' so, f ' dx .
Now, we will see here what do we mean by this increment and this differential in terms of the
geometry. So, let us consider here we have the function y so, this is x axis and this is your y
axis. So, we have this curve y=f (x) or the function of one variable and let us take a point x
here so, the value on this curve. So, the point here is x and f ( x) so, this value here the height
is f ( x).
So, this point is x into f ( x) and then we make an increment in x by Δ x. So, we have another
point here x+ Δ x and we take the value of the function as f ( x+ Δ x). So, this point becomes
x+ Δ x because of this x coordinate and the y coordinate. So, it is the value of this function at
this point f ( x+ Δ x).
Now, we draw an tangent here because we are talking about the derivatives so, that we will
come into the pictures. So, we draw a tangent at this point x which is denoted by this dotted
line and then this is naturally Δ x the increment we have given in x. So, this is Δ x or dx term,
we will come to this point why we have written this dx and then this is the increment in Δ y.
So, when we have made an increment in Δ in x by Δ x then this is the change in Δ y.
So, that is a change in Δ y because earlier the y was this one and now the y has become this
one here at this point. So, now, the difference here is the Δ y or the increment in Δ y this one
and what is this dy, dy is this first derivative here f ' (x ) and dx. So, if this is the derivative so,
the tangent f ' (x ) is nothing, but the tangent of this angle. So, that is equal to this divided by
this one.
So, here this distance from here to here will be nothing, but the f ' into this Δ x because in this
triangle you can figure out that this tangent of this angle here is this f 'at this point x. And,
this will be equal to this distance which we will call as dy and is equal to this Δ x. So, this dy
is nothing, but delta x f x or this distance here is f ' (x ) Δ x. So, this distance is f ' (x ) Δ x this is
what we are calling as dy. So, let me erase this.
193
(Refer Slide Time: 17:50)
So, this is dy in our definition. Now, one can observe that this was Δ y which was the
increment in y increment in the function when we made an increment in x by Δ x and this is
dy the differential of y in our definition. So, what do we observe here that this Δ x and the
Δ y are the changes in the function when we changes x by Δ x then there was a change of Δ y
in the function and this dy was the change in the tangent.
So, along this is change along the tangent and this is changed along the along the function
itself when we make an increment in x by Δ x. So, we have also noted here that dy and dx dy
and this dx measure changes along the tangent line. So, this is here along the tangent line; we
have made the change here by Δ x or we can call it dx both are same and here that is the
change in the tangent line with dy. So, that is the notation heredx and dy; on the other hand
the Δ y and Δ x measure changes for the function f ( x).
So, here when we make an increment Δ x then there was a increment Δ y in the function y.
So, this is we have denoted Δ y and this Δ x and Δ or dx they are the same because change in
the x whether we denote by this Δ x or Δ y they are the same. But, along this y axis this is the
change in the function and this is the change in the tangent line of the function. So, there is a
difference or from the formula itself we can observe because we have this definition dy is
equal to the derivative of this f ' (x ) and dx and if we substitute for y is equal to x suppose
y=x.
194
So, y=x and then the derivative will be 1 into dx or Δ x; so, because this was originally taken
as the linear term which was A times the Δ x. So, originally this was defined as a f ' (x ) Δ x.
So, the dx will become just the Δ x or we can understood from this note that dy and dx we
take the notation the measure changes along the tangent line. So, we have the dx and we have
the del dy and then we have again this Δ x and then we have here Δ y ok.
(Refer Slide Time: 20:59)
So, now we have the geometrical interpretation for the differentiability again just though we
have rewritten that definition. So, a function y=f (x) is said to be differentiable at the point
( x0 , y0 ); if it can be approximated in the neighborhood of this point by a linear function. So,
this is another interpretation of the differentiability we have written earlier in terms of that
linear expression and then ϵ Δ x.
So, here also we have another interpretation that this function is said to be differentiable, if it
can be approximated in the neighborhood of this point by the linear function. So, what do we
mean by this mathematically that f ( x), if we can approximate this function in the
neighborhood of this ( x0 , y0 ) point by the linear function. So, this is the linear function here
f ( x0 )+(x−x 0 ) A +ϵ ( x−x 0 ). So, this is a similar definition what we have earlier.
So, if we bring this f ( x0 ) to the left hand side this will become like Δ y and we have here Δ x
here and A+ϵ and again Δ x the change in x. So, but here now we are talking about that if we
can express this f ( x) by a linear function in the neighborhood of ( x0 , y0 ) point. So, what does
195
that mean? This is the linear function and plus we have ϵ and this x−x0 note that this ϵ → 0 x
as x → x0 . So, this term is pretty smaller because this is also going to 0 and this is also going
to 0.
So, we have somehow the quadratic term in terms of x the difference x−x0 and we have the
linear term here in terms of x−x0. So that means, this function we can approximate by this
linear function at least in the neighborhood of this point x 0. So, this is a linear function of x
and the equation of the tangent of the curve this is nothing, but the equation of the tangent at
the point x 0 f ( x0 ) of this function y is equal to or curve y=f ( x ) .
So, again like this is x axis y x is here and then we have some curve y=f ( x ) and at this point
x 0this is a point x 0 f ( x0 ), if you draw the tangent then what we want to say here that if the
function is differentiable then we can approximate in the neighborhood by a linear function.
And, the accuracy of this linear approximation will depend that where we take x for example,
if x is this point this is very well approximated and if it is a far point from this x 0 then we are
not approximating well.
So, at least in the neighborhood of this point if we can approximate by a linear function then
we call this function as differentiable or equivalently or mathematically, if we can write down
this f ( x) in terms of the linear function and plus the epsilon and this is the difference x−x0
and this ϵ → 0. So, then this is not the linear term in terms of this difference it is a non-linear
term here. So, we have this linear term plus this non-linear term and this is the equation of the
tangent which we have written down in this case.
196
(Refer Slide Time: 24:41)
Now, the testing of the differentiability what we have learned so far. So, we have basically
the three concept one was the existence of this derivative which is evaluated by this quotient
and taking the limit. So, if this limit exists we call that the function has derivative or the
function is differentiable because that was equivalent to the differentiability of the function.
And, we have denoted this by f ' (x )or dy dx or y '. The second concept we have seen that the
Δ y=dy +ϵ Δ x that we can write down this expression in terms of dy and ϵ Δ x and the
dy = A Δx the increment and this is called the differential term.
So, this is another way of testing differentiability that we can write down this Δ y in this
form; the third one was which is which can be deducted from this term itself. So, the Δ y and
we take this dy to the left and divided by this Δ x and then take the limit since this epsilon
goes to 0 as Δ x goes to 0. So, this expression must be 0. So, we can use either this one to test
the differentiability that this quotient must be 0 or we can use this one or we can use the
derivative one.
197
(Refer Slide Time: 26:03)
We have the three concepts. So, the example 1 we will show now that the function f ( x)=x2
is differentiable its a pretty simple. So, we have y=x 2 and then the Δ y; that means,
f ( x + Δ x ) −f ( x ) this will be equal to so, here f ( x + Δ x ) ; that means, ( x + Δ x )2 −f ( x ). So, −x2 if
I expand this you will get x2 + Δ x2 +2 x Δ x− x2.
So, this term get cancelled and we will get a 2 x Δ x and Δ x 2. So, we get precisely here 2 x Δ x
and into Δ x. So, we have this form this is A Δ x and the ϵ Δ x. So, here this ϵ naturally it goes
to a 0 as Δ x → 0 here this term 2 x is A, the Δ x is given. So, we have that format which we
have used for the differentiability.
198
(Refer Slide Time: 27:11)
That means this function is differentiable and the derivative value which is this A the term
sitting with this Δ x in the linear part is f ' (x ) and this is ϵ . So, this implies that the function is
differentiable and its derivative is 2 x; alternatively we can show that this quotient here
lim f ( x+ Δ x )−f (x)
Δx → 0
.
Δx
So, again heref ( x + Δ x ) −f ( x ) is given already there. So, we have the 2 x Δ x+ Δ x 2 term there
and divided by Δ x and then we will take the limit as Δ x → 0. So, we have here this 2 x term
and plus the Δ x term and then we take the limit as Δ x → 0. So, this will go to 0 and we will
get only 2 x. So, we have this showing that these limits exist for whatever for any value of x.
199
(Refer Slide Time: 28:13)
So, the function is differentiable in the whole domain or we can also show that this here the
ratio. So, when Δ y−dy if we take so, this is Δ y−dy. So, we take this will be Δ x 2 and when
¿, this will go to 0.
we divide by Δ x. So, we will have Δ x only and taking this Δlim
x →0
So, that was another the third concept of showing the differentiability either by expressing
this Δ y in this format or alternatively taking this quotient here and limit if this exists or this
one.
(Refer Slide Time: 28:50)
200
So, here this example shows that y=x 2 we find Δ y and dy at x=2 when
Δ x=1 , Δ x=0.1, Δ x=0.01. So, the point here x=2 and these are the increment. So, we have
the Δ y=f ( x+ Δ x )−f (x) and the dy =f ' ( x ) dx.
So, if you make this table here. So, for example, this value we can evaluate here Δ y Δ y is f
the x is 2+ Δ x we take first as 1 so, 1−f (2). So, here this is f (3) and x x 2 so, 9 and f (2). So,
2
x this will become 4. So, this value is 5 and so, here it is 5 and now dy . So, simply from here
f ' is 2 x. So, 2 x into dx or dx is Δ x is the same. So, here 2 and this 2 x so, this will be 4 and
dx is 1. So, we will get 4.
Similarly, while taking Δ x we will get this 0.41 and dy will be 0.40 and while taking a
smaller Δ x we will get Δ y as 0.0401 and here we will get 0.0400. So, in this case what we
have observe when Δ x is large; obviously, these 2 the Δ y and dy they are not the same. So,
the difference is large between the 2 its it is not approximating this dy is not approximating
this total change in y, but when the when this Δ x is smaller. So, if you make a smaller
increment then this dy is well approximating this Δ y term and that was the meaning of that,
if we can approximate by the linear term at least in the neighborhood of the point.
We can also test the differentiability of this function of f ( x )=1+ √3 ( x−1 )2 at x=1.
(Refer Slide Time: 31:05)
201
So, this function so, if we take this Δ y as this difference because at x=1 so, f ( 1+ Δ x ) −f ( 1 ).
So, 1+ Δ x we will submit here. So, it will be like 1 plus and this will become Δ x 2 and the
cube root minus f (1) will be 1.
So, we will get only the √3 Δ x2 and now we will check whether it is possible to find a number
A such that this quotient here which we have just talked, that if we can get this quotient and
the limit is 0 then we call the function is differentiable. So, we will check whether we can
find such a A. So, that this quotient here the dy is A Δ x over Δ x, if this quotient is 0 then we
call the function differentiable and if we cannot find such a A then we will call the function is
not differentiable.
So, we will take this quotient here
A there. So, you have
Δ y− A Δ x
. So, we can divide by this Δ x so, we will get
Δx
1
Δy
Δy
what is
. So, Δ y is here ( Δ x2 ) 3 and when we divide by this Δ x
Δx
Δx
so, here its power 2 by 3rd and this will be minus. So, let me just compute this one, this term
here
2
−1
Δy
term Δ y was Δ x 2/3 and then we have Δ x here. So, this will be Δ x 3 −1 so, Δ x 3 − A .
Δx
(Refer Slide Time: 32:53)
Now, what we observe that this function here is approaching to plus ∞ when Δ x is
approaching to 0 from the right side for example, when Δ x is positive or this is approaching
to -∞ when this Δ x we are approaching from the left side when Δ x is negative and
202
approaching to 0. So, in that case this limit whatever A we choose this will be ∞ when Δ x
approaches from the right side when Δ x approaches from the left side this will become -∞
irrespective of this number A.
So, meaning that we cannot get such a constant so, that this quotient has a finite limit oh sorry
the 0 limit this should be 0 for the differentiability. So, we do not have such a A a finite
number A. So, that this quotient has the limit 0 as Δ x approaches to 0. So, this limit in fact,
does not exist. So, in this case the function is not differentiable at x is equal to A.
(Refer Slide Time: 33:59)
So, coming to the conclusion we have seen that this function is said to be differentiable, if we
can write down Δ y in terms of this expression A Δ x+ ϵ Δ x, where A is independent of Δ x
and epsilon is a function of Δ x such that epsilon goes to 0 Δ x → 0 this was one definition we
have discussed. And this term A Δ x is called the total differential of y at this point ( x , y) and
is denoted by delta y and the value of A is nothing but it is the derivative of f at x.
203
(Refer Slide Time: 34:38)
Again we call a function y=f (x) differentiable if this quotient exist which we call derivative,
but both the definitions were equivalent we have seen. And, what we also realized now
introducing that differential dy is equal to f ' dx that this is not just a notation for f ', but this
ratio of the 2 differential because dy was f ' dx makes sense.
So, writing this dy and dx separately makes sense and that we will also use now in the in the
in the next lecture which will be on the differentiability of the two functions. And, very
important that we first understood this differentiability of the function of one variable mainly
these three concept having the derivative and then this linear expression and also writing that
limit to 0 gives us differentiability.
So, there in case of the two variables what we will see that just having the derivatives, where
the partial derivative with respect to x and partial derivatives derivative with respect to y will
not help or will not be equivalent to two differentiability. The differentiability will be a
different concept than having just the partial derivatives, though in case of one variable if we
have the derivative of the function, then the function is differentiable or the other way
around.
So, that will be in the next lecture..
204
(Refer Slide Time: 36:13)
And these are the references we have used to prepare these lecture.
Thank you very much.
205
Engineering Mathematics – I
Prof. Jitendra Kumar
Department of Mathematics
Indian Institute of Technology, Kharagpur
Lecture – 12
Differentiability of Two Variables
Welcome to the lectures on Engineering Mathematics-I, and today we will be talking about
the Differentiability of Two Variables functions of two variables and this is lecture number
12.
(Refer Slide Time: 00:27)
So, we will introduce first the differentiability concept for the functions of two variables, and
then we will derive necessary conditions for differentiability, and also sufficient conditions
for differentiability.
206
(Refer Slide Time: 00:41)
So, just to recall the differentiability of single variable which we have learnt in the previous
lecture. So, there were three concepts we have used to define the differentiability of a
functions of single variable. So, one was the existence of the derivative. So, if this limit f x
plus delta x minus f x over delta x exists, which we call derivative of the function f at x and it
was denoted by f prime x. So, if this derivative exists we call the function as differentiable.
Existence of lim ⁡
Δx →0
f ( x+ Δx )−f ( x )
=f ' ( x)
Δx
The second concept which is more general and we will indeed now use this to derive
differentiability of functions of more than one variable. And in this case we have seen that if
we can express this delta y which is the change in delta y when we make a change of delta x
in x. And if we can express this change in terms of this dy plus epsilon delta x the dy which
we call differential was a linear term in delta x and this was A times delta x; here the A was
independent of delta x and this epsilon term must go to 0 as delta x goes to 0.
Δ y=dy +ϵ Δ x ,dy= A Δ x
207
So, the third one was useful for testing the differentiability and what we have seen that if this
quotient here delta y minus d y over delta x and when we take this limit delta x goes to 0 is 0
then we call the function differentiable. So, this was just a consequence of this relation when
we have taken this dy to the left side and divided by delta x taking this limit since epsilon
goes to 0. So, we get this relation out of this relation, but this was useful for testing the
differentiability in many cases.
lim Δ y−dy
Δx → 0
Δx
=0
But what was the point here that all these three definitions are equivalent for differentiability.
So, the existence of the derivative we have seen that imply implies the differentiability of a
function of single variable, but today what we will observe that the existence of derivatives
which we call the partial derivatives with respect to x or y that is not sufficient for the
differentiability of the functions of two variables for example.
(Refer Slide Time: 03:17)
So, here just to define the differentiability of functions of two variables, we take this function
z is equal to f x y and this is said to be differentiable at the point x y a general point xy, if at
this point we can express this change in z. So now, we have this z as an dependent variable
on x and y. So, xy here are independent variables. So, if this change in delta y is equal to a
times delta x plus b times delta y plus epsilon 1 times delta x and epsilon 2 times delta y.
208
So, this is just the extension of the definition which we have used for functions of one
variable. And now in this case again this a and b they are independent of delta x and delta y
and this epsilon 1 and epsilon 2 they must go to 0 as delta x delta y goes to 0. So, again we
have this linear term and then this quadratic or the non-linear rather non-linear term because
we do not know; what is the dependency of delta x one epsilon 1 and epsilon 2. So, we have
the linear term here and pass this rest.
lim ϵ 1 =0 ⁡
Δ x →0
Δ y→0
¿
lim ϵ 2 =0
Δ x →0
Δ y→0
So, again here the a and b are independent of delta x and delta y and epsilon and epsilon 2 are
functions of delta x and delta y such that, this epsilon 1 is equal to 0 when we take this limit
delta x goes to 0 and delta y goes to 0 also this epsilon 2 when we take this limit as delta x
and delta y both go to 0 epsilon 2 must be 0. So, this linear function again which is a delta x
plus b delta y is called the total differential we have also introduced in case of functions of
single variables. So, here this is called total differentiable of z at this point general point x y
and this is denoted by delta z.
dz=a Δ x+b Δ y=a dx +b dy
So, now we will denote this by d z which is the linear part of this expression and delta z that
is the change in z when we make change in x and y by delta x and delta y. So, this dz which
we call the differential term or the total differential this is equal to a delta x and b delta y, that
is the linear part of this change in z; which we can also denote because we have observed
yesterday that delta x or dx they are the same because they measure the change in x and in
this case because we have two variables.
a=f x ( x , y )=
∂z
∂z
b=f y ( x , y )=
∂x
∂y
So, the delta x and delta y or dx or dy they make they measure the change along the x and the
y axis, while this delta z is the change in the function value. Whereas, this dz is different
which is the change in a along the tangent plane in this case in case of one variable it was
along the tangent line.
f ( x + Δ x , y + Δ y )−f (x , y)=a Δ x+b Δ y +ϵ 1 Δ x+ϵ 2 Δ y
209
Taking limit as Δ x → 0 , Δ y → 0
lim f ( x+ Δ x , y+ Δ y ) =f ( x , y )
Δ x →0
Δ y→0
So, if this delta x and delta y are small enough then this dz which is the differential of z at this
point will be a close approximation or will be a good approximation of the change in delta
change in z that is delta z, because this non-linear term here for small values of delta x and
delta y that will go to 0. And we will have a very good approximation by this linear term or at
least in the neighborhood of the point x y.
(Refer Slide Time: 07:21)
So, the geometrical interpretation of the differentiability: so what we have seen in case of
function of single variable that at this point x 0, the value of the function is f x 0. So, this is
the tangent line; that means, if the function is differentiable we have observed that we can
draw a tangent line or in other words we can approximate this function in the neighborhood
of this x 0 point by this tangent line.
So, as we see here in the graph as well in the close neighborhood of this point x naught, we
can have a very good approximation by this tangent line at least in the close neighborhood of
this point x naught in case of the functions of two variables. So, we have a similar argument
or the extension of this concept.
210
So, if suppose there is a point here x naught y naught, and then in this case if the function is
differentiable. Then we can approximate in the close vicinity of this point by the tangent
plane; in case of single variable this was tangent line. And now we have the function of two
variables in this picture and then we can approximate by this tangent plane which is actually
the linear part which we have seen in the definition there.
(Refer Slide Time: 08:48)
So, what are the necessary conditions for differentiability we will derive now? So, if this z is
equal to fx y, this function is differentiable. That means, we can express this delta z is equal
to a delta x plus b delta y plus epsilon delta x plus epsilon delta y. Then we will show that the
function is continuous and has partial derivatives with respect to x and y, at that point where
we are talking about the differentiability. That means, here that discontinuity and the partial
derivatives the existence of the partial derivatives at x y point with respect to x and y these
two are the necessary conditions for the differentiability, because we will show that the
differentiability implies that f is continuous and the partial derivatives exist at the point x y.
a=f x ( x , y )=
∂z
∂z
, b=f y ( x , y ) =
∂x
∂y
Moreover we will also see that this a here which is the constant free from delta x and delta y
this is nothing, but the partial derivative of f with respect to x at that point and this b here is
nothing, but the partial derivative of f with respect to y at that point. So, let us prove this. So,
if we assume that this f is differentiable; that means, we can write down this delta z here
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which is we have made an increment here delta x and delta y in x and y and this difference is
nothing but the delta z. So, delta z we can express as a delta x plus b delta y plus epsilon delta
x plus epsilon delta y, the definition of the differentiability.
f ( x + Δ x , y + Δ y )−f (x , y)=a Δ x+b Δ y +ϵ 1 Δ x+ϵ 2 Δ y
So, having this now, if we take the limit here as delta x goes to 0 and delta y goes to 0 what
will happen? Since delta x goes to 0 and a is a finite number. So, this will disappear. And
here again this will go to 0 here also since the epsilon goes to 0 and also delta x goes to 0. So,
this term will go to 0 and also this term will go to 0. So, we will have simply when we take
the limit delta x goes to 0 and delta y goes to 0, we will have that the limit of this f x plus
delta x y plus delta y as delta x goes to 0 delta y goes to 0 is equal to f x y and this is nothing,
but the continuity of f.
Taking limit as Δ x → 0 , Δ y →0
lim f ( x+ Δ x , y+ Δ y ) =f ( x , y )
Δ x →0
Δ y→0
So, to show the continuity, we take a point in the neighborhood and then we take the limit
and this should approach to the function value x and y independent of this how we approach
to this xy point by taking this delta x to 0 delta y to 0 along any path. So, that was the
continuity here of f which we have proved. So, what we have assumed that if f is
differentiable then we can express it in this form and simply by taking this limit we have seen
that the function must be continuous if the function is differentiable.
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(Refer Slide Time: 12:10)
So, next we will continue the necessary condition the second one. So, we again assume that f
is differentiable then we can express this delta f or delta z in terms of a delta x b delta y and
this epsilon and epsilon 2. And now since delta x and delta y are arbitrary we can choose in
the neighborhood anything here. So, what we have set if we say delta y equal to 0 and then
divide by delta x then what we will get. So, we will get this here x plus delta x and then delta
y is 0 minus this f xy and divided by delta x. So, that will be equal to so here a will remain
and then here delta y is set to 0. So, this term is 0 and then epsilon 1 because we are dividing
with delta x and this term will go to 0 because we have set delta y to 0, because this is again
we are in the neighborhood of this point x y.
So, if we are talking about a point here x y, so we have set just delta y to 0. So, again we are
in the neighborhood along this x axis by having this delta x not equal to 0 at this moment
here. And so we can do that we can take any delta x and any delta y and this relation must
hold. So, what we have done we have taken delta y to 0 and then divided by delta x and now
we can we can take the limit of this delta x as it goes to 0. And what we will get this is the
partial derivative the definition of the partial derivative of f with respect to x and this will be
equal to.
So, we have taken the limit here both the side, so limit delta x goes to 0 and here also then in
this case the limit delta x goes to 0. So, a is independent of delta x delta y, so this will remain
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as a and this epsilon term will go to 0 as delta x go to 0. So, we will get this relation that fx is
equal to at x y is equal to a.
Setting Δ y=0 and dividing by Δ x yield the relation
f ( x+ Δ x , y )−f ( x , y )
=a+ϵ 1 ⟹f x (x , y)=a
Δx
(Refer Slide Time: 14:29)
And now similarly what we can do we can set delta x to 0 now and then we can divide by
delta y. And again by taking the limit as delta y goes to 0, so we will get that f xy sorry f y xy
is equal to b. So, here if we take this limit now delta y goes to 0, so the right hand side will be
just b because epsilon 2 will go to 0 as delta y goes to 0. So, in this case you will get that the
partial derivative of y at x y is equal to b.
So, what we have observed now that if function is differentiable, in that case the partial
derivative of x at this point x y will exist, and that will be nothing but this number a here
which is appearing in the definition of this differentiability. And also the partial derivative
with respect to y at that point will exist and the value will be b which is appearing here in the
definition of the differentiability.
Similarly, setting Δ x=0 and dividing by Δ y yield the relation
f ( x , y+ Δ y ) −f ( x , y )
=b+ ϵ 2 ⟹f y (x , y)=b
Δy
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So, what we have seen now that if a function is differentiable then it must be continuous and
not only continuous, but the partial derivatives must exist at that point where the function is
differentiable. So these are the necessary conditions for differentiability. If one of the
conditions like the function is not continuous or one of these two partial derivatives does not
exist. In that case we can immediately claim that the function is not differentiable because,
these two are the necessary conditions the continuity of the function and the existence of
partial derivatives.
So, if one of these conditions is violated then we can immediately conclude that the function
is not differentiable, but if the function is continuous and also the partial derivatives exist at
the point x y, where we are talking about the differentiability. In that case we have to go for
the further test, because these are the necessary conditions based on these 2 conditions we
cannot say that the function is differentiable or it is not differentiable because these are the
necessary conditions for differentiability.
(Refer Slide Time: 16:57)
So, now we will move to the sufficient conditions for differentiability and in this case if one
of the partial derivatives of z exists and the other is continuous at a point x y then the function
is differentiable at x y. So, what we have now for the sufficient conditions that if one partial
derivative exists and that is anyway necessary condition for the differentiability. So, we must
have the existence of both the partial derivatives for discussing the differencability.
f ( x + Δ x , y + Δ y )−f ( x , y + Δ y )= Δx f x ( x+θ 1 Δx , y+ Δy ) , 0<θ 1 <1
215
So, this is natural now because this is a necessary condition that both the partial derivatives
exist. So, basically for the sufficient condition we have to check that if one of the partial
derivative is continuous, x then the function is differentiable at x y. So, we need continuity of
one partial derivatives and the existence of both because the existence is necessary condition
for the differentiability. So, we suppose here that fy exists and fx is continuous.
f ( x + Δ x , y + Δ y )−f ( x , y + Δ y )= Δ x f x ( x , y ) +ϵ 1 Δ x
f ( x , y+ Δ y ) −f ( x , y )=Δ y f y ( x , y ) +ϵ 2 Δ y
So, in this case we will now observe what is this delta z the variation in this z when we vary x
by delta x and y by delta y. So, we will consider this difference now and we will see whether
we can express this difference in terms of that linear function plus this epsilon delta x plus
epsilon to delta y. So, here now moving further we have f x plus this term and then we have
subtracted this f x y plus delta y and we have added the same here f x y plus delta y minus f x
y which was already there in the difference and then since we have taken here the existence
of fy and continuity of f x. So, we will use these two conditions to prove that the function is
differentiable.
So, the existence of f y implies that this limit here f x y plus, so here this is the partial
derivative with respect to y. So that means, f x y plus delta y we will make an increment in y
and minus the function value at that point divided by this increment delta y, and taking this
limit delta y that this exist. That is the meaning of the existence of f y that this partial
derivative exists.
So, this is equal to f y and now we can use that idea which we have used in the previous
lecture, that once the limit is given we can define or we can introduce some epsilon. That
means, that f this x and y plus this delta y and minus this f xy over this delta y is equal to or
with minus sign here. So, minus this f x y we can set as epsilon in this case we will take this
epsilon 2 and then later on we will introduce epsilon 1 as well.
So, if we set this difference to epsilon 2 then we know that when we take the limit as delta x
or in this case only this delta y goes to 0 then this epsilon 2 must go to 0, because when delta
y goes to 0 this is nothing but exactly the f y at x y. So, this we will introduce now and then
we will multiply this whole expression by delta y, and we will get that this difference is equal
to f y delta y plus epsilon 2 delta y.
216
(Refer Slide Time: 20:55)
So, having this we will get this f x y plus delta y minus this f x y is equal to this delta y will
be multiplied to the right side. So, we will have f y x y and plus this epsilon 2 which we have
introduced here and this delta y and note that this epsilon 2 which appeared here now must go
to 0 as delta y goes to 0, this is the condition because of this limit we have observed.
(Refer Slide Time: 21:26)
Next now we will use the LaGrange mean value theorem. So, just to recall; what was the
LaGrange mean value theorem we have function which is continuous and differentiable in the
open interval a b. So, here f b minus f a over b minus a is equal to this quotient will be equal
217
to the derivative at some point in the interval a b, that was the LaGrange mean value theorem
we have studied before and now so using this LaGrange mean value theorem to this
difference here. So, notice that here f x plus delta x and f x so y is not changed.
So, why we are not changing y plus delta y here also y plus delta y, so this is with respect to x
only the change is in x and if we divide. So, this is like f b minus this f a here a is x and this b
is x plus delta x and divided by the difference which is b minus a, so in this case this will be
delta x but we will multiply to the right side will be delta x and the derivative naturally with
respect to x because we are talking about with respect to x the y is unchanged.
So, f x the partial derivative and at which point here the xi was between b and a, so here also
this argument will vary between this x and x plus delta x. So, we have introduced this theta 1
so that this will be precisely between x and x plus delta x when theta runs between 0 and 1.
So, this theta 1 here is between 0 and 1 when this is close to 0 this argument is close to x and
when this is close to 1 this will go to and go to x plus delta x. So, exactly which xi was doing
here in the open interval a b, we have this argument x plus theta 1 delta x for this theta 1 0
and 1, so we have this LaGrange mean value theorem now.
(Refer Slide Time: 23:50)
And now the continuity of f x because we have assumed the continuity of f x and the
existence of f y. So, this continuity of the partial derivative f x we will apply here. So, if f x is
continuous and we take the limit delta x go to 0 and this delta y goes to 0, then this f x x plus
218
theta one delta x y plus delta y will simply go to f x at the point x y this is what we have used
now.
Δ z=f ( x + Δ x , y + Δ y )−f ( x , y + Δ y ) +f ( x , y+ Δ y ) −f ( x , y)
¿ Δ x f x ( x , y ) + Δ y f y ( x , y ) +ϵ 1 Δ x+ϵ 2 Δ y
ϵ 1 , ϵ 2 → 0 as Δ x , Δ y → 0
Existence of f y and continuity of f x ⟹ Differentiability of f
So, the continuity of f x will give us that this derivative with respect to x will be f x at x y
point and then again we will introduce another epsilon as we have done before, so that is f x
and this x plus theta one delta x y plus delta y is equal to f x at x y plus this epsilon and this
epsilon will have property that this will go to 0 as delta x and delta y goes to 0. So, now out
of this we get this relation now that f x plus delta x and y plus delta y minus f x y plus delta y
is equal to delta x is multiplied and f x x y plus epsilon 1 delta x and now so we are here now
f x plus delta x and y plus delta y minus f x y is equal to we can write down as delta x times f
x at x y plus epsilon times delta x no well.
(Refer Slide Time: 25:44)
And now what we have we have these two relations we have proved that this difference when
we have variation here is delta x and y is unchanged y plus delta y. This was because of the
continuity we got this relation, and because of the existence of f y we got this relation and
219
having these 2. So, remember the delta z we had written by subtracting and adding to this
difference here. And now we will replace this f x plus delta x y plus delta y minus this by this
expression and this f x y plus delta y by this expression here.
So, we will get now by substituting these that delta x f x delta y f y plus epsilon delta x plus
epsilon 2 delta y and epsilon and epsilon 2 goes to 0 as delta x delta y go to 0. That means,
the existence of f y and the continuity of f x these 2 conditions we have proved that the
function is differentiable because, this is the definition of the differentiability. And we have
used to arrive to this expression, we have used the fact that the we have used that the function
derivative with respect to y exists and the function derivative with respect to x is continuous.
(Refer Slide Time: 27:14)
So, just a remark that the function may not be differentiable at a point even if the partial
derivatives f x and f y exist at p, why because the existence of partial derivatives at the point
p is necessary condition this is not sufficient for differentiability which is written here
existence of partial derivatives necessary condition, so they cannot guaranty differentiability.
On the other hand a function may be differentiable if f x and f y are not continuous because,
what we have observed that the continuity of f x over f y or both that is the sufficient
condition not the necessary condition. So, the function may be differentiable if f x or f y or f
x and f y are not continuous because, this existence of partial derivatives and continuity of the
other the existence of 1 partial derivative and the continuity of other these 2 are the sufficient
conditions.
220
(Refer Slide Time: 28:17)
So, let us go through quickly 2 problems, so here find the total differential and total
increment of the function z is equal to xy at the point 2 3 for delta x is equal to 0.1 and delta y
is equal to 0.2. So, the total increment is defined as f x plus delta xy plus delta y minus f x y
which is equal to. So, this is x plus delta x and y plus delta y we can use this z now z is equal
to x y, so this is our function f x y is xy f x y is x y. So, in this case we have here the product
of this x plus delta x y plus delta y minus this product here x y and then we can open this so x
y and then x delta y delta x y delta x delta y so x y will get cancelled and we will have this
expression. And then we can compute this delta z at this point two three and the delta x is
given point one delta y is given 0.2 and this will come 0.72.
Δ z=f ( x + Δ x , y + Δ y )−f ( x , y )= ( x+ Δx ) ( y + Δy )−xy = y Δx+ x Δy + Δx Δy
Δz=3 ×0.1+2 ×0.2+0.1 ×0.2=0.72
dz=
∂z
∂z
dx +
dy= y dx + x dy= y Δx +x Δy
∂x
∂y
dz =3× 0.1+2× 0.2=0.7
Now, coming to the differential which is dz del z over del x dx plus del z over del y dy and in
this case del z over del x is y and del z over del y is x. So, we have this expression now to be
computed at this two three point with delta x 0.1 delta y 0.2. So, note that is dx dy is nothing
but delta x and delta y because, these are independent variable for independent variables
221
these notations are the same dx is equal to delta x dy is equal to delta y. But for dependent
there is a difference which we are going to observe now here and this dz the differential of z
now at this point will be 0.7. So, there is a difference naturally on delta z and the dz, but
when this delta x and delta y are small numbers, then they will be a good approximation to
each other here or dz will approximate well this delta z term.
(Refer Slide Time: 30:39)
Another problem we will show now that this is differentiable and write down this total
differential. So, to show the differentiability we will use this basic definition of expressing
delta z in terms of this a b and epsilon-epsilon 2. So, this delta z is nothing but this difference
here and then we can compute this, because this is our f x y. So, here we have substituted this
x plus delta x and y plus delta y and this is f x y which is given as x square minus x y and
minus x y square.
Δ z=f ( x + Δ x , y + Δ y )−f (x , y)
2
2
¿ ( x+ Δ x ) + ( x + Δ x ) ( y+ Δ y ) + ( x + Δ x ) ( y+ Δ y ) −x2 −xy−x y2
2
2
2
¿ Δ x +2 x Δ x+ x Δ y+ y Δ x+ Δ x Δ y+2 xy Δ y+2 y Δ x Δ y+ x Δ y + Δ x Δ y + Δ x y
2
¿ Δ x ( 2 x+ y + y ) + Δ y ( x+2 xy ) + ( Δ x+ Δ y+2 y Δ y ) Δ x+ ( x Δ y + Δ x Δ y ) Δy
2
dz=( 2 x + y+ y ) dx+ ( x+2 xy ) dy
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2
So, we can open now here the square so x squared term will be 3 2 times x delta x term will
be there delta x square term here also 1 term will be xy here also there will be a term xy
square. So, these terms will be canceled and the other we will collect together. So we will
have these many terms this is coming from x plus delta x square other than x square because
that was canceled out.
Now, these terms are because of this product here there will be 4 terms, but 1 will be
cancelled here and there will be 6 terms. So, we will have 5 terms here because this xy square
will be cancelled. And now we will try to get the coefficient of delta x and the delta y and
plus in terms of epsilon delta x and epsilon delta 2. So, here if we take common delta x, so
this is two x from here and then y from here and then we will get delta x from here also y
square. So, this is with the delta x term and from delta y you will get x from here and delta y
is appearing here too. So, del 2 x delta 2 x y with delta y, so this is the linear term which we
have collected and the rest we have kept in this format that with delta x whatever the
coefficients here and we delta y also we got this 1.
So, what we observe that we can introduce now that this is our epsilon 1 and this is our
epsilon 2. And they have the property that they will go to 0 as delta x and delta y will go to 0
and this 1 is the total differential term and since this a and this is b in our notation and they
are independent of delta x delta y.
So, we have a into delta x b into delta y epsilon 1 delta x epsilon 2 delta y, so we can express
this z this function in this form. So, the function is differentiable and its total differential will
be given by this linear term which is written here 2 x plus y plus y square dx and x plus 2 x y
dy. So, this is the differential.
223
(Refer Slide Time: 33:27)
So, coming to the conclusion now: so we have discussed that this function z is said to be
differentiable at the point xy; if at this point we can express this delta z as a delta x b delta y
plus epsilon delta x epsilon to delta y, where this epsilon 1 and epsilon 2 they go to 0 as delta
x delta y go to 0 and also this number a and b they are independent of delta x and delta y. So,
the necessary conditions we have a studied they were continuity of f is necessary for
differentiability and the existence of partial derivatives of f x and f y at that point where we
are talking about the differentiability.
So, these two are the necessary conditions and for the sufficient conditions we have seen that
the continuity of the partial derivatives of f x and f y are important or in other words we can
say the existence of 1 and the continuity of the other. Meaning the continuity of one is
sufficient here, because the existence anyway we have to have to discuss the differentiability
because, the existence of partial derivative is important is necessary then only we can talk
about the about the differentiability. So, basically the continuity of one partial derivative is
sufficient for differentiability.
224
(Refer Slide Time: 34:56)
These are the references we have used to prepare these lectures.
Thank you very much.
225
Engineering Mathematics - I
Prof. Jitendra Kumar
Department of Mathematics
Indian Institute of Technology, Kharagpur
Lecture – 13
Differentiability Functions of Two Variables
Welcome back to the lectures on Engineering Mathematics I, and this is lecture number 13.
(Refer Slide Time: 00:21)
And today we will talk about again Differentiability of Functions of Two Variables. And
basically we will go through the limit test for differentiability, and that is very useful to test
differentiability of a function of more than one variable. And then some worked problems
will be done in this lecture.
226
(Refer Slide Time: 00:41)
So, just to recall from the previous lecture we have discussed already the differentiability of
functions of two variables. And what we have learnt that a function z is said to be
differentiable at the point x y, at any point x y; if at this point we can express this delta z
which is the variation in z when we when we vary x by delta x and y by delta y. So, if we can
express this delta z in terms of delta x and delta y; a times delta x plus b times delta y, that is
a linear term a and b are independent of delta x and delta y. Plus, this epsilon times delta x
plus epsilon 2 times delta y, and here epsilon and epsilon 2 must go to 0 as delta x and delta y
go to 0.
Δ z=a Δ x+ b Δ y+ϵ 1 Δ x+ϵ 2 Δ y
So, the necessary conditions we have learned that the continuity of f is necessary for
differentiability and also the existence of partial derivatives f x and f y is necessary for the
existence of or for the differentiability. And we have also seen the sufficient conditions where
we observe that the continuity of one derivative or continuity of both partial derivatives is
sufficient for defining differentiability of functions of two variables. So, what we have seen
that to prove the differentiability either we can use the sufficient condition. So, if you observe
that the function is or the partial derivative is continuous, then we can claim that the function
is differentiable. Or we can test directly this definition, so we have to express this delta z in
227
terms of this delta and delta y, in this form and then we can claim that the function is
differentiable.
Today we will learn another way now which is equivalent to this expression here that can be
used to prove differentiability easily and that is a limit test.
(Refer Slide Time: 02:47)
So, here we will now show that this differentiability is equivalent to saying that this limit here
delta rho which is square root delta x square plus delta y square. If we take this limit here of
this expression delta z minus dz over delta rho is equal to 0, then we can prove that the
function is differentiable. Or if we have differentiability it will imply that this limit is 0, and
this limit 0 will imply differentiability. So, the both are equivalent definition or testing for
differentiability.
Differentiability ⟺ lim Δ z−dz =0 , Δ ρ= √ Δ x2 + Δ y2
Δ ρ →0
Δρ
So we now, we will show that if a function is differentiable, then this limit must be 0. So, if f
is differentiable; that means, we can express this delta z as a delta x plus b delta y and plus
epsilon 1 delta x plus epsilon 2 delta y. And now this is the dz term which we call
differential. And if we take this differential term to the left hand side, and divide by this delta
rho, delta rho is given as square root of delta x square plus delta y square. So, if we divide
this, then we will get the right hand side as epsilon delta x over delta rho a plus epsilon delta
y over delta rho term.
228
Δ z−dz
Δx
Δy
=ϵ 1
+ϵ 2
Δρ
Δρ
Δρ
And now we will take the limit here as delta x and delta y goes to 0 and observe what is the
value right hand side when we take the limit delta x, and delta y goes to 0. We already know
that this epsilon 1 and epsilon 2 they go to 0 as delta x goes and delta y go to 0. But we have
to make sure that this term here sitting with epsilon 1 that is delta x over delta rho and delta y
over delta rho, they both are bounded. But this we can easily see because this delta rho is the
square root of delta x square plus delta y square. So, this term here because this is bigger than
delta x. Here we have delta y square as well and then the square root.
lim
Δ ρ →0
Δ z−dz
Δx
Δy
= lim ϵ 1
+ lim ϵ 2
=0
Δρ
Δ ρ Δρ →0 Δ ρ
Δ ρ →0
So, this term is certainly bigger than this delta x. So, the modulus of this term so, absolute
value of this delta x over delta rho is bounded by 1. And similarly the absolute value of this
delta y over delta rho is also bounded by 1. And then when we take the limit as delta rho goes
to 0. So, this side will go to 0, because epsilon 1 will go to 0 and epsilon 2 will go to 0. So,
when taking the limit, so we observe because of the boundedness of these 2 terms and we
know the properties of these epsilon 1 and epsilon 2 that they go to 0 as delta rho goes to 0
means delta x go to 0 delta y go to 0.
Note that
Δ x 1∧ Δ x
≤
≤1∧ϵ 1 , ϵ 2 tend ¿ zero as Δ ρ → 0
Δρ
Δρ
So, in this case we can prove that this limit is equal to 0 because of the boundedness of these
terms there. So, this is this limit which we want to show that if f is differentiable this must be
equal to 0.
229
(Refer Slide Time: 06:03)
So now we will go the other way round; that means, if this limit is equal to 0 we will show
that the function is differentiable. So, in this case we now let that this limit equal to 0. And
now we will use the fact which we have also used in the previous lecture, that this limit equal
to 0 then we can introduce one epsilon here. So, this term minus 0 which is 0 here so this
term is equal to epsilon. And this epsilon will have property that when we take the limit delta
rho go to 0 goes to 0 or delta x, and delta y go to 0 then this epsilon must go to 0, because the
limit of this is precisely 0. So, the epsilon must go to 0 and we take the limit here as delta rho
go to 0 goes to 0.
lim
Δ ρ →0
Δ z−dz
=0 , Δ ρ= √ Δ x 2 + Δ y 2
Δρ
Let lim
Δ ρ →0
Δ z−dz
Δz−dz
=0⟹
=ϵ , ϵ → 0 as Δρ → 0
Δρ
Δρ
So, in this case we have this property of the epsilon that this must go to 0 when delta rho goes
to 0. And if we take now this term to the right hand side. So, we have delta z minus this dz is
equal to epsilon times this delta rho which implies, so the epsilon and this delta rho we have
substituted a square root delta x square plus delta y square. Now we can divide by the square
root delta x square plus delta y square and multiply it by the same term.
230
2
2
⟹ Δ z−dz=ϵ Δ ρ=ϵ √ Δ x + Δ y =ϵ
2
2
Δ x +Δ y
√ Δ x 2+ Δ y 2
To get this following expression here, so epsilon delta x square plus delta y square divided by
this square root here. And now we can break into 2 parts, so here the epsilon and one delta x
divided by this term, and the other delta x here we have written down in the product plus,
again the same concept here this epsilon together with 1 delta y, and divided by this term
square root delta x square plus delta y square and delta y.
¿
(
ϵ Δx
Δx+
√ Δ x 2+ Δ y 2
) (
ϵ Δy
Δy
√ Δ x2 + Δ y 2
)
So, what we observe now that this delta z variation in z we can write down as dz plus epsilon
1. So, this is our epsilon 1 delta x plus this epsilon 2 delta y. And which implies, but this was
the dz that goes to the right hand side, and then this is like epsilon 1 term, and this is epsilon
2 term, and they have the property that they will go to 0 again which we have just learned
before.
⟹ Δ z=dz+ϵ 1 Δ x+ϵ 2 Δ y ⟹Differentiability of f
So, this we will go to 0, this also go to 0, and we have written this delta z in terms of this dz
plus epsilon delta x plus epsilon delta y. And that is precisely the definition of the
differentiability. So, we have observed now that for the differentiability we have the
equivalent definition here. That the differentiability imply this that this limit must be equal to
0, or we have also seen that if this limit equal to 0 then the function must be differentiable.
So, we can use this limit definition here for testing the differentiability, because getting this
limit is easier than expressing this delta z in terms of this dz and epsilon delta x delta y term.
So, we most of the time you will use this definition to prove the differentiability of the
function, because this is easier then that the other definition of the differentiability.
231
(Refer Slide Time: 09:57)
So, here we take this problem number 1, and we will show that in this case the function is
continuous and the partial derivatives exists, but the function is not differentiable. And this is
because of the reason that the continuity and the partial derivatives the existence of partial
derivatives. These two are necessary conditions for the differentiability. So, we cannot claim
based on these two conditions that the function is differentiable.
So, this is one example, we will show that this function is continuous and the partial
derivatives exist both f x and f y, but the function is not differentiable. So, first the existence
of partial derivatives; so, we know the definition of f x at 0 0. And naturally we will show
this existence continuity at 0 0 and also that the function is not differentiable at the origin. So,
here the f x at 0 0 is as per the definition the limit delta x goes to 0 and f delta x 0 minus f 0 0
over this delta x term.
f ( x , y )=
{
xy
, ( x , y ) =(0, 0)
√ x 2+ y 2
¿ 0 ( x , y ) ≠(0 , 0)
So, this will be 0, because here we can see this product of x y. So, when we have this
argument here in f as 0. So, this will make the function 0 and f 0 0 is defined at 0. So, we
have this 0 minus 0. So, this is here 0 and this is also 0. So, 0 minus 0 by delta x and we get
this 0. So, similarly for f y; f y 0 0 we have f 0 delta y minus f 0 0, and again this because of
232
this x argument here it is 0. So, this product will make this again 0, and this is 0, so 0 minus 0
and we will get again this value as 0.
f x ( 0 , 0 ) = lim
f ( Δ x , 0 ) −f ( 0 , 0 )
=0
Δx
f y ( 0 , 0 )= lim
f ( 0 , Δ y ) −f ( 0 , 0 )
=0
Δy
Δ x→0
Δ y →0
So, we have the existence of the partial derivatives at 0 0, and the value of the partial
derivative with respect to f x 0 and also the value of the partial derivative at 0 0 with respect
to y is also 0. Now for the continuity again it is a simple function we have already tested
before. So, we can change this to polar coordinate that is easier. So, x is equal to r cos theta
and y is equal to r sin theta we can substitute there, and take the limit as r goes to 0. So, when
we substitute here x is equal to r cos theta y is equal to r sin theta, and here we will get simply
r the square root of r square, and then one r will get canceled, and we have here r and cos
theta sin theta; as r goes to 0 this cos theta sin theta they are bounded. So, here this limit will
go to 0.
lim f ( x , y ) =lim r cos θsin θ=0=f (0 ,0)
x →0
y→0
r→ 0
So, we have and the function is also 0 at 0 0. So, the function is continuous at the point 0 0.
So, we have the existence of partial derivatives, we have the continuity of the function and
now we will test for the differentiability of this function.
233
(Refer Slide Time: 13:23)
So, for the differentiability of this function we need to now go to this limit definition.
Because as I said this is much easier so, we will find out that what is limit here delta z at
minus dz at over delta rho.
So, the delta z is this increment in x and y so, at point 0 0; so 0 plus delta x and 0 plus delta y
and minus the f 0 0. So, this f 0 0 is 0; so here we have f delta x delta y. So, here this x y will
be replaced by delta x delta y, we have delta x delta y and the square root delta x square plus
delta y square. And, now this dz the differential of z is partial derivative with respect to x
delta x plus the partial derivative of z with respect to y and then we have delta y.
Δ z=f ( 0+ Δ x , 0+ Δ y )−f ( 0, 0 )=
Δx Δy
√ Δx2 + Δy 2
So, these two here we have just seen before that they are 0. So, we will get this dz term as 0,
and then this limit delta z over dz over delta rho. So, this delta z is delta x delta y over delta x
square plus delta y square and d z is 0. So, we will get now this delta x delta y and this delta
rho was also the square root delta x square plus delta y square and we have one square root
here. So, we will get this term without square root.
dz=
∂z
∂z
Δ x+
Δy=0
∂x
∂y
234
So, delta x delta y over delta x square plus delta y square and now we will see this limit. So,
if we take this path here delta y is equal to m into delta x. One can clearly see because we
have this quadratic term there and each of them is also quadratic. So, we will easily realize
that, when we take this special part delta y is equal to m delta x the linear path to go to delta x
0 delta y 0. So, we will get here simply when we put delta y is equal to m delta x here also m
square delta x square delta x square will get cancel and we will get this m over 1 plus m
square.
lim
Δ ρ →0
Δ z−dz
Δx Δy
= lim
2
2
Δρ
Δρ→0 Δx + Δy
(
)
Along the path Δ y=m Δ x
lim
Δ ρ →0
Δ z−dz
m
=
Δρ
1+m 2
So, this function is not differentiable, because this limit does not exist the limit depend on this
path, so the function is not differentiable. For differentiability this limit must be equal to 0,
but what we have seen that this limit does not exist. And hence the given function is not
differentiable. So, what we have observed in this example, that the function is continuous and
it is partial derivatives exist, but the function is not differentiable.
(Refer Slide Time: 16:09)
235
Another example of similar kind here also we will see that the function is continuous partial
derivatives exist, but it is not differentiable. So, the existence of partial derivatives again it is
easier. So, we have the definition of the partial derivative with respect to x so, f delta x 0, so
here if you put y 0. So, we will get delta x cube over this delta x square, and this is f 0 0. So,
and this delta x square is also there, so we will get basically this limit as 1. Because f delta x
0 will be delta x cube over delta x square, and then 1 delta x from this definition. So, this will
be as 1.
x3 +2 y3
, ( x , y ) =(0 , 0)
f ( x , y )= x 2 + y 2
¿ 0 ( x , y ) =(0 ,0)
{
(Refer Slide Time: 16:59)
So, this limit is 1, and similarly the f y at 0 0 when we compute. So, in this case we will get
this x 0 here 0, and it will get 2 delta y cube and then here also y delta y square and this delta
y will make delta y cube. So, this delta y cube will get cancel and we will get this limit as 2.
So, the partial derivative with respect to x is 1 and the partial derivative with respect to y is 2.
f x (0 , 0)= lim
f ( Δ x , 0 ) −f (0 , 0)
=1
Δx
f y ( 0 , 0 )= lim
f ( 0 , Δ y ) −f ( 0 , 0 )
=2
Δy
Δx → 0
Δ y →0
236
Now, coming to the continuity of this function again it is simple, and we can show by
changing it to polar coordinate as x is equal to r cos theta y is equal to r sin theta and it is easy
to see that this will be r square term here. And we will get r cube from there also r cube from
there. So, we will get one r in the numerator and together with some bounded function of this
cos cube theta plus 2 sin cube theta. So, when r goes to 0 this limit will be 0.
3
lim
x →0
y→0
3
3
3
3
3
x +2 y
r cos θ+2r sin θ
=0=f (0 ,0)
2
2 =lim
2
r
→0
x +y
r
So, we have this limit is equal to 0 the function value is 0. So, hence this function is
continuous the partial derivatives exist and the function is continuous. So, we will now move
to show that the function is not differentiable at this point.
(Refer Slide Time: 18:13)
For that we need to show this limit again, where this delta z is this difference, which is again
this is 0 and delta x delta y. So, this x y will be replaced simply by delta x and delta y terms.
The dz term, so del z over del x at 0 0 is 1 and this was 2 there. So, we have delta x plus 2
delta y, and if we take this limit again here 1 over this is rho and then this delta z which is
delta x square plus 2 delta y square, and this delta y square plus delta x square.
3
3
Δx +2 Δy
Δ z=f ( 0+ Δ x , 0+ Δ y )−f ( 0, 0 )=
2
2
Δx + Δy
237
dz=
∂z
∂z
Δ x+
Δy=Δ x+2 Δy
∂x
∂y
Minus this dz term delta x plus 2 delta y, and this one can simply simplify that delta x square
delta y square in this denominator. So, we have then delta x square and 2 times delta y square
minus this product; which will give us delta x cube as one term, and minus 2 times delta x
square and delta y term. Then we will also get this delta y square, delta x term and then minus
2 times delta y cube term.
So, and this is 2 times delta y cube. So, here also this cube, so this delta x cube will get
cancelled 2 times delta y cube will get cancel. And you will get only these 2 terms there with
this delta y square and x square. So, we get precisely this minus 2 times, this delta x square
and this delta y there, and minus delta x and delta y square, and this delta x square plus delta
y square together with this you will get this power 3 by 2.
2
lim
Δ ρ →0
1
√ Δ x 2+ Δ y 2
(
2
lim −Δx Δ y −2 Δ x Δy
Δ x 3 +2 Δ y3
Δρ →0
2
2 −( Δ x+2 Δy ) =
3 /2
Δ x +Δ y
( Δx 2 + Δy 2 )
)
(Refer Slide Time: 20:21)
And now if we take this path delta y is equal to m delta x. So, again the same situation, so
here we have then delta x cube common here also we will get delta x cube common and this
delta y square will become m square delta x square.
Along the path Δ y=m Δ x
238
2
¿
−m −2m
( 1+m 2 )
3/2
So, this is power 3 by 2. So, here also we can take common delta x cube. So, delta x cube will
get cancelled everywhere, and we will get the limit minus this is m square minus here m and
here 1 plus m square 3 by 2. So, this will be the path dependent limit. Depending on m we
have a different number for the limit, and hence this limit does not exist, and again this
function is not differentiable.
(Refer Slide Time: 21:07)
Next problem, here we will show that the function is differentiable, but f x and f y the partial
derivatives are not continuous. Remember that the continuity of partial derivatives is
sufficient for differentiability. So, the function may be differentiable, but the f x and f y may
not be continuous. If we can prove that f x and f y are continuous, that will simply imply that
the function is differentiable, but if we if we cannot prove the continuity of f x and f y, we
cannot conclude anything about the differentiability of f.
f ( x , y )=
{
( x2 + y 2 ) cos
(√
1
)
, ( x , y ) =(0 ,0)
2
x +y
¿ 0 ( x , y ) =(0 , 0)
2
So, this example precisely shows that the function is differentiable, but f x and f y are not
continuous in this case. So, the existence of partial derivative now because we have to show
239
that the necessary conditions are satisfied for differentiability, if one of the necessary
conditions violated then we can immediately claim that the function is not differentiable.
So, here the existence of partial derivative again we have the definition f delta x 0 minus f 0 0
over delta x. And in this case we can show that this is 0, because here f delta x; so this delta y
0 and we have the delta x. So, we will get this delta x square, and then we will get here the
cost term with 1 over square root delta x square, and divided by this delta x term here, and
this is 0. So, this will get cancelled. So, here is something bounded and then delta x goes to 0,
so this limit will be a 0.
f x (0 , 0)= lim
f ( Δ x , 0 ) −f (0 , 0)
=0
Δx
f y ( 0 , 0 )= lim
f ( 0 , Δ y ) −f ( 0 , 0 )
=0
Δy
Δx → 0
Δ y →0
(Refer Slide Time: 22:55)
And similarly we can show that f y at 0 0 is also 0, because this function is symmetric
anyway. So, we will get this again as 0. Now checking the continuity of the function is again
e 0, because when x y goes to 0 0 this is something bounded sitting here and x y goes to 0.
So, naturally this f x y will go to 0 as x y goes to 0 0. So, again we can see by changing to the
polar coordinate also. So, we have here we can get this like r square, there and then rest
240
everything will be bounded and as r goes to 0. So, this will be 0, whether showing by
changing to polar coordinate or directly here when x y goes to 0.
lim ( x2 + y2 ) cos
x →0
y→0
(
1
=0=f (0, 0)
√ x + y2
)
2
So this term go to 0 and this is something bounded something finite here. So, we will get this
limit as 0 directly; which is the function value at 0 0 point. So, we have seen the other
function is continuous.
(Refer Slide Time: 24:05)
And now coming to the differentiability, so of this function, so we will take this delta z direct
definition here, I mean for the delta z, the variance in z. So, f 0 plus delta x 0 plus delta y
minus f 0 0 so, this will be because f 0 0 is 0 we have delta x delta y. So, x y will be replaced
by delta x delta y term is there. And this dz the partial derivative with respect to x at 0 partial
derivative of z with respect to y at 0, and we have seen those values were 0. So, this dz is 0
and now this quotient here, and then we will take the limit.
Δ z=f ( 0+ Δ x , 0+ Δ y )−f ( 0, 0 )
¿ ( Δ x 2 + Δ y 2 ) cos
(√
1
Δ x + Δ y2
2
)
241
So, the delta z which is given here delta x square and delta y square cos term and minus this
0; so, and this delta rho is a square root delta x square plus delta y square. So, we will see that
what is this limit here, and this can get cancelled. So, we will get in the numerator this is
square root term with this cos 1 over this term. And now this term we can easily see that here
the limit as delta x goes to 0 delta y goes to 0. Because this term will go to 0 and something
bound it is sitting here. So, directly we can show that this value is 0.
( Δx2 + Δy2 )
Δ z−dz
lim
= lim
cos
2
2
Δρ
Δ ρ →0
Δx →0 √ Δ x + Δy
Δy →0
Δ x2 + Δ y2 cos
√
Δ x →0
¿ lim
Δ y →0
(
(√
1
2
Δ x + Δy
2
)
1
=0
√ Δ x + Δ y2
2
)
That means, if this limit is 0 then the function is differentiable. So, in this case we have
observed that this function is differentiable. So, what is now to show, that we will go to the
continuity of the partial derivatives and we will observe that the partial derivatives are not
continuous in this case.
(Refer Slide Time: 25:59)
To prove the continuity of f x and f y, we need to get the f x and f y at non origin point. So,
here it is non equal to 0, and when equal to 0 we have the 0 0 may be the same here also the
same mistake. So, we can hey this was like at 0 0 the function was defined as 0 and then we
have 0 when x y equal to 0. So, this is also not equal to 0 and it is 0 when x y is 0.
242
Well, so now, we will show the continuity of the partial derivatives at 0 0 point. So, we need
to compute the continuity of the partial derivative at non 0-point. We have to compute the
partial derivatives at non origin here and then the f x and f y at 0 0 which we have already
computed the value was 0. So, to get the partial derivative of this function at this non origin
point, then we have to we can just directly get the derivative of this function with respect to x
treating this y as constant.
So, at x y not equal to 0 0 point, we can get this derivative by the direct differentiation of this
treating this y constant. So, this is a product tool. So, this x square y square term, this cos will
be with minus sign and 1 over the square root x square plus y square, and the derivative of
this term which will be minus 1 by 2 and this 2 x over x square plus y square 3 by 2, and then
this will remain unchanged here the cause and the derivative of this term will be 2 x.
At ( x , y ) ≠(0 ,0)
f x ( x , y ) =−( x 2 + y2 ) sin
¿ sin
(√
1
x + y2
2
)( √ (
(√
x
2
x +y
2
1
2
x +y
))
2
)(
−1
2
+2 x cos
(√
2x
3
2 2
( x 2+ y )
1
x + y2
2
)
+2 x cos
(√
1
2
x +y
2
)
)
So, this is the partial derivative of the function at with respect to x at the point x y which is
not equal to 0 0. We can simplify little bit this term here. So, we will get sign of this 1 over
square root x square plus y square, and this term will become x over the x square root x
square plus y square plus this 2 x cos 1 over a square root x square plus y square. And now if
we take path here along the x axis; that means, the delta y this y will be set to 0. So, we are
approaching to 0 0, we want to see whether f x as x y goes to 0 is equal to the partial
derivative of f at 0 0 point which was 0 there. So, along x axis if we move towards the origin,
then what will happen?
The y is 0, now, so we have here x and divided by the square root of x square which is
absolute value of x, and then we have this sign here, 1 over again absolute value of x plus;
this 2 x and the cos again this here 1 over square more absolute value of x. So, this is along x
axis; that means, the y is 0, so we have kept here y 0. So, we are taking a particular path
243
along this x axis. And now if we realize here for example, this one when we go x to 0 from
the right side this is plus 1, when x goes to 0 from the negative side this will become as minus
1. And in any case this is also not definite what will be the value here a same situation. At
this point here anyway this x goes to 0. So, this is something bounded, so this will vanish this
will go to 0.
Along x -axis
lim f x ( x , y )=lim ¿
x →0
x→0
But at this point here this is like plus 1, and this is undetermined in that case and this is can be
minus 1 also and here we do not know what is the value when x goes to 0. So, this limit does
not exist or certainly this is not equal to 0 which we were looking for that this limit. If this
limit is equal to 0, we can claim that the function is the derivative f x is continuous, because
this was the derivative value at 0 0 point, this was f x at 0 0 point. So, in any case this is not
equal to this one. In fact, the limit does not exist. So, there is no question about the continuity
of this partial derivative hence this f x is not continuous.
(Refer Slide Time: 30:51)
And similarly we can show because this function is just now symmetric. So, we can show
also that f y is not continuous. So, in this example we have seen that the partial derivatives
are not continuous though the function is differentiable.
244
(Refer Slide Time: 31:07)
So, this remark the above example shows that the continuity of partial derivatives is not in
necessary condition for differentiability. A function can be differentiable without having
continuous first order partial derivatives. Another example of similar kind one can show
again here that the function is differentiable and f x and f y they are not continuous. So, this is
left to the participants we are not going to show that this function is differentiable, but f x and
f y are not continuous, but the working steps are similar to the earlier problem. And one can
easily show that f x and f y are not continuous for this problem as well.
(Refer Slide Time: 31:55)
245
So, the conclusion here, we have what we have seen, we have seen the differentiability an
equivalent definition which is the limit here, showing to 0 is equivalent to saying that the
function is differentiable. So, this is useful in testing the differentiability of the function. And
here the function may not be differentiable at a point even if partial derivative is exist,
because the existence of partial derivatives is a sufficient condition for differentiability it is
not necessary condition. And, we have also seen that function may be differentiable even if f
x and f y are not continuous this is what we have also; so because this is sufficient condition,
ok.
(Refer Slide Time: 32:39)
These are the references we have used to prepare these lectures.
Thank you very much.
246
Engineering Mathematics - I
Prof. Jitendra Kumar
Department of Mathematics
Indian Institute of Technology, Kharagpur
Lecture – 14
Differentiability Functions of Two Variables (Contd.)
So, welcome back to the lectures on Engineering Mathematics-I, and this is lecture number
14. And, again we will continue here the Differentiability of Functions of Two or Several
Variables and there will be many worked problems today.
(Refer Slide Time: 00:31)
So, what we have observed in the previous lecture that the equivalent definition of the
differentiability here instead of expressing this delta z in terms of a delta x be delta y and
epsilon delta x plus epsilon delta 2 y we can also find out this limit. And, if this limit equal to
0 then the function is differentiable and we have observed that this limit definition was quite
useful for proving the differentiability of various functions.
We have also observed that the necessary conditions are also important because if we check
these necessary conditions first and we observe that for example, the function is not
continuous or the partial derivatives do not exist. Or one partial derivative does not exist in
that case we can immediately conclude there that the function is not differentiable, because
these are the necessary conditions for differentiability. And, we have also observed the
sufficient the importance of sufficient conditions because if we show that the partial the
247
continuity of one of the partial derivatives. And, then we can conclude that the function is
differentiable without proving the differentiability using the definitions here given above.
(Refer Slide Time: 01:58)
So, let us discuss the problem here the differentiability at the origin of this function f xy is
equal to xy cube over x square plus y square and this is 0 when x y equal to 0 so, at origin this
is 0. First we check the necessary conditions and we should do basically in all the problems
because, these are the necessary these are the conditions which the function must satisfy
before going to the differentiability. So, here we will check the continuity of the functions.
So, when f x y when x y goes to 0 and in this case without doing much so, we can either
change again to polar coordinate then r square term will be coming in the denominator and
here in numerator there will be a 4 terms.
x y3
, ( x , y ) ≠(0 ,0)
f ( x , y )= x 2 + y 2
¿ 0 , ( x , y ) =(0 , 0)
{
lim
( x , y ) →(0, 0)
f ( x , y )=0 , f x ( 0 , 0 ) =0 ,f y ( 0, 0 )=0 ⁡
248
And so, we will get r square still in the numerator and with this cos and sin together so, that
will make this limit to 0. So, we have the limit of this function f x y as x y goes to 0 is equal
to 0 also the partial derivative f x at 0 0, we can easily show the others is like limit delta x
goes to 0 and f 0 plus delta x into the y is unchanged here. So, f 0 0 and divided by this delta
x so, since this 0 is here and you have we have the product there. So, this will be 0 and here
also this will be 0 so, 0 minus 0 we will get this partial derivative as 0. So, this is 0 and
similarly the f y will be also 0.
(Refer Slide Time: 03:32)
So, we have the existence of the partial derivatives and we have the continuity of the function
at 0 0 point, now for the sufficient conditions. So, we need to check the continuity of these
partial derivatives at 0 0. So, we need to get these f x the partial derivative of x at all the
points here at least in the neighborhood of this 0 0 point. So, to get the partial derivative at x
y with not equal to 0 0 we have to differentiate this function with respect to x. So, treating
this y constant so, we will just differentiate this with respect to x keeping this y as constant
because of these partial derivatives.
So, this will be equal to this is square here whole square and this will be x square plus y
square. So, the derivative of this with respect to x will be y cube then minus this x y cube and
the partial derivative here with respect to x this will be 2 x. So, this x square y cube here also
we have 2 times x square y cube. So, we will get minus this x square y cube and then here y
for 5, this is y power 5 here and x square plus y square whole square this is here. So, at this
249
non-zero point x y not equal to 0 0 we can get the derivative partial derivative with respect to
x or partial derivatives with respect to y directly from directly taking derivative of this
function with respect to that variable and keeping the other variable as constant.
f x ( x , y)=
{
−x2 y 3 + y 5
∧0 , ( x , y )=(0 ,0)
2
( x ¿ ¿2+ y2 ) , ( x , y ) ≠(0 , 0)¿
And to get this at 0 0 we have to use this fundamental definition of the differentiability which
we have just got here this f x at 0 0 was 0 so, which is 0 here. So, we have now evaluated f x
at x y point in the neighborhood and also at 0 0. And, now for continuity we have to take the
limit as x y goes to 0 0 of this function whether it is equal to 0 or not. If it is equal to 0 then
we can claim that the function is differentiable because of the sufficient condition otherwise
we have to go further to test the differentiability.
lim ¿
¿¿
(Refer Slide Time: 05:58)
So, here if we take this limit as x y goes to 0 0 f x x y. So, in this case we will observe that
like if we substitute here if we change to the polar coordinate now. So, we will get here r
square so r 4, but there we will have r power 5 and so, we will get 1 r there in the numerator
250
and since r goes to 0 to go to xy to 0 0 this value will be 0. So, here the f x is continuous
because this is equal to 0 and this is the value of the f x at 0 0 point.
So, we have the continuity of the partial derivative and that we can use now to prove the
differentiability of this function because the partial derivatives exist, function is continuous.
So, we have the necessary conditions for the sufficient condition we have to show that one of
the partial derivatives is continuous. So, here we have shown that this f x is continuous and
that is enough to prove that the function f is differentiable at 0 0 point.
(Refer Slide Time: 07:12)
The next problem here we will discuss the differentiability again at the origin of this function
y cube sin 1 over y square when x is not equal to 0 and x is equal to 0 this is 0. So, we will
test the continuity of f and the existence of the partial derivative which is easy to see again.
So, for the continuity we have to take this limit as x y goes to 0 0 of this function x y; and
which is trivial here because this is a bounded function sitting and then x y go to 0. So, this y
goes to 0 and this will be 0 and this is the function value also at 0 0 point.
f ( x , y )=
{
3
1
2 , x ≠0
x
¿0 , x=0
y sin
( )
So, we have the continuity of this function and also the existence of the partial derivative for
that we have to consider for partial derivative with respect to x we need to show that this limit
251
0 plus delta x. So, 0 and minus this f 0 0 divided by delta x. So, here this y is 0. So, this will
be 0 and this is again here 0. So, we have the 0 minus 0 so, this is also 0. So, we have the
existence of the partial derivative with respect to x. Similarly we can show the existence of
partial derivative with respect to y for that we have to consider the delta y goes to 0 and f 0
delta y minus this f 0 comma 0 and then here delta y.
So, since x is 0 the value of this function is again 0 and here also the value of this function is
0 so, we have again 0. So, the continuity of the function and the existence of the partial
derivative we have seen it is a easy.
−2 y 3
1
cos 2 , x ≠0
f x ( x , y ) = x3
x
¿ 0, x=0
{
( )
(Refer Slide Time: 09:15)
Now, and next we will show the continuity of this f x we will try to get the continuity of the
partial derivative with respect to x. So, for that again we need to compute the partial
derivative when x is not equal to 0 and we have to also get this we have already seen that this
value is 0 and x is equal to 0. So, at least at 0 0 point we have seen it as 0. Now, here to get
this when x is not equal to 0 we have to directly differentiate this with respect to x. So, the y
cube will be as taken as constant here.
252
For sin we will get the cos term 1 over x square and the derivative of 1 over x square will be
minus 2 over x cube. So, this is the derivative when x is not equal to 0 and we can get this
derivative as 0 when x is equal to 0 by the fundamental definition of the derivative. So, in this
case and what we will show now that what happens when we take the limit here as xy or
basically this x goes to 0, we are approaching to the to the to the origin here what will happen
to this limit. So, when we take this limit here f x at x y goes to 0 along this path y is equal to
x this is easy to see. So, if we take this path y is equal to x here then what will happen to this
limit, this limit will be minus 2 and this cos 1 over x square.
And the limit x goes to 0. So, in this case this limit does not exist if clearly one can see this
cos 1 over x square this is not a definite value. So, here this limit does not exist along this
particular path itself. So, we can say that the limit of f x as x y goes to 0 0 does not exist and
hence this f x is not continuous function in this case. So, it does not help us to prove the
differentiability of this function at this point of time.
Along the path y=x :
lim ¿
¿¿
What we can do we can further test the continuity of the other partial derivative f y with
respect to. So, the partial derivative of f with respect to y and if that comes to be continuous
again we can claim based on the sufficient condition of differentiability, because we need to
have the continuity of one of the partial derivatives to claim that the function is differentiable.
So, in this case this f x is not continuous. So, we will check now the continuity of the other.
253
(Refer Slide Time: 11:53)
So, in this case again f y we can compute here by treating this x constant. So, this will be just
3 y square when x is not equal to 0 and 0 when x is equal to 0. So, we will take again this
limit as xy goes to 0 0 of this f y at the point xy. So, here one now the situation is different.
So, here when x y goes to 0 0 so, this term will go to 0 and something finite is sitting here.
So, in this case there is no problem to get this limit as 0. So, the limit of this function 3 y
square cos 1 over x square when we take xy to 0 0 from any direction. So, this is come this is
to be 0.
f y ( x , y )=
{
2
1
2 ,x≠0
x
¿0 , x=0
3 y cos
( )
lim ¿
¿¿
And this is the value at the origin as well of this function f y. So, hence this f y is continuous,
this is a special example where f x was not continuous and now f y is continuous. So, in this
case the f x and f y obviously, both exist there and we have on the top this continuity of f y;
continuity of one of the partial first order partial derivatives. And, then we can claim that the
function is basically differentiable or we can prove that this function is differentiable based
on these sufficient conditions of differentiability.
254
(Refer Slide Time: 13:22)
Another problem here we take this f xy is equal to square root x y when we have this xy, this
product positive greater than equal to 0 otherwise there will be problem here and the
definition and 0 elsewhere. So, we will determine whether the function is differentiable at the
origin or not. So, the continuity of this function is again simple because when x y go goes to
0 0 from any direction this will go to 0 and the function is also defined as 0 at the origin.
f ( x , y )=
xy , xy ≥0
{¿ 0√elsewhere
.
So, we have the continuity of the function and the existence of partial derivatives also we can
use this definition again. So, here f delta x and y is 0 so, this product is 0. So, we will get this
again here 0 and this f 0 0 is 0 so, 0 minus 0. So, this f x at 0 0 point is 0 and similarly f y is
also 0 at 0 0. So, this function is continuous and the partial order first order derivatives exist.
lim
( x , y)→(0 ,0)
√ xy =0
lim f ( Δ x ,0 )−f ( 0, 0 )
f x=
Δ x →0
f y=
Δ y→0
Δx
=0
lim f ( 0 , Δ y )−f ( 0, 0 )
Δy
=0
255
(Refer Slide Time: 14:30)
Now, to check the differentiability we will get this limit at what is the value of this limit, if
this limit comes to be 0 then the function is differentiable. If we do not get this limit as 0 then
the function is not differentiable. So, for that we need to get this dz which is the partial
derivative of z with respect to x partial derivative of z with respect to y at the origin which
was and both of them was 0. So, we have this dz is equal to 0 and this delta z which is f the 0
plus delta x 0 plus delta y minus this f 0 0. So, this will become as the square root of delta x
delta y and this is minus 0 here assuming that this product is 0. So, we get this delta set as
square root delta x and delta y.
ty f ( x , y )=
xy , xy ≥0
{¿ 0√elsewhere
.
Now, this constant here delta z minus d z over delta rho will be so, this delta z was square
root x and square root x delta x delta y and this data rho was the square root delta x square
plus delta y square here. And, in this case when we take this limit as delta rho goes to 0 we
will also that this limit does not exist because if we take this path delta y is equal to delta x
for example, here. So, what will happen you have delta x there and delta x square here. So,
this delta x square terms will vanish there so, we will have 1 over square root 2.
dz=
∂z
∂z
Δ x+
Δy=0
∂x
∂y
Δ z=f ( 0+ Δ x , 0+ Δ y )−f ( 0, 0 )=√ Δx Δy
256
So, along this path delta y is equal to delta x what we observe that delta rho goes to 0 this
delta z minus dz over delta rho is 1 over square root 2. Because we will get here 1 when delta
x square get cancelled here 1 plus here 1 against the square root 2 will come. So, along this
path linear part delta y is equal to delta x this limit is equal to 1 by 2. So, certainly it is the
limit cannot be equal to equal to 0. So, here itself by taking one path and showing the limit is
different from 0 though the limit may exist or limit does not exist, it does not matter to prove
that the function is not differentiable. Because, for differentiability this limit must be equal to
0 and what we have observed that along this particular path this limit is equal to 1 over square
root 2.
lim
Δ ρ →0
Δ z−dz
= lim
Δρ
Δρ→0
√ Δx Δy
√ Δ x2 + Δ y 2
Along the path Δ y= Δ x
lim
Δ ρ →0
Δ z−dz 1
= ≠0
Δρ
√2
And hence, we can now say that the function is not differentiable because this limit cannot be
equal to 0. So, the function is not differentiable in this case.
(Refer Slide Time: 17:16)
Another example which discuss the differentiability again at the origin of this function: so, x
power 5 by 2 and the sin 1 over square root x plus y 5 by 2 and then we have cos 1 over
257
square root y and this is defined for x not equal to 0 and y not equal to 0 and this is 0
elsewhere. So, again this we can directly use this definition of the differentiability which is
much easier in this case because of this special structure. We can easily express this function
into this form because if you observe that this delta z is f delta x delta y minus this f 0 0 at 0 0
point.
{
5
2
f ( x , y )= x sin
5
1
1
+ y 2 cos
, x ≠0 , y ≠0
√x
√y
¿ 0elsewhere
( )
( )
Δ z=a Δ x+ b Δ y+ϵ 1 Δ x+ϵ 2 Δ y
This is equal to just the x and y will be replaced by delta x delta y, this is 0. So, we have here
delta x power 5 by 2 then the sin term 1 over square root delta x plus here we have delta y
power 5 by 2 and then we have 1 sin and then here we have the cos term this is cos here; so,
the cos 1 over square root y. So, in this case we can now write down the 0 into delta x plus 0
into delta y that linear term because there is non-linear term. So, we have just introduced here
0 times delta x and 0 times delta y. Here we have taken this 1 delta x outside because, we
need that format here epsilon 1 delta x.
5
2
5
1
f ( Δ x , Δ y ) −f ( 0 , 0 ) =Δ x sin
+ Δ y 2 cos
√Δ x
( )
( √ Δ1 y )
This here epsilon 1 delta x so, this will become as epsilon 1 here and then plus from here also
we have taken this delta y term away, then we have delta y 5 by 2 and this is again this cos
term; so, the cos 1 over square root delta y. So, in this case we have expressed this delta z in
this form where this is your epsilon and which has this property that this goes to 0, when f
delta x delta y go to 0. Because, of this term here delta x and the positive power 3 by 2 and
this is bounded function. So, this will go to 0 and here again the same argument. So, this with
cos function also this will go to 0.
3
2
3
1
¿ 0⋅ Δ x+0⋅ Δ y+ Δ x Δ x sin
+ Δ y Δy 2 cos
√Δ x
(
( )) (
258
( √ Δ1 y ))
(Refer Slide Time: 19:52)
So, in that case this function is differentiable because we can easily express this function in
this form a times. So, this is going to be the partial derivative of that function at 0 0 with
respect to x and this is going to be the partial derivative with respect to y at 0 0 point and
delta y plus this delta x this is epsilon 1 term delta by epsilon 2 term. So, here it was very
easy directly because of this function to express in this form as delta z is equal to a delta x
plus b delta y plus epsilon delta x plus epsilon delta y. And, then we have proved that this
function is differentiable. Another example here that f x y is defined at 0 when x y not equal
to 0 and when x y the product is 0 then this function is 1.
f ( x , y )= 0, xy ≠ 0
¿ 1, xy =0
{
And we want to check the existence of f x and f y at the origin and we also want to know
whether the function is differentiable at origin or not. So, to get the existence of f x and f y at
origin we use this definition, fundamental definition of the derivative. So, we have limit delta
x to 0 f delta x 0 minus f 0 0 over delta x and now you note that this f delta x 0. So, this
product here of the argument is 0. So, the value will be coming from this 1 so, this is 1 and f
0 0 so, again this is 1 here. So, 1 minus 1 this is 0 that limit is 0, similarly for f y when we
have delta y to 0; now this is delta y to 0 and f 0 delta y minus f 0 0 over delta y. So, in this
case again this is 1 and this is also 1. So, we have 1 minus 1 again 0 as the partial derivative
of f with respect to y at a origin.
259
lim f ( Δ x , 0 )−f ( 0, 0 )
f x=
Δ x →0
Δx
=¿
So, now we will check the continuity of f at origin. So, for that if we observe that what is the
value of this f xy of this limit when x y goes to 0 0 along x is equal to y. So, along this x is
equal to y, what we have along x is equal to y line if we approach to this origin point here
what will happen. So, here at all these points the x y the product is not equal to 0 at any other
point then origin the product is not equal to 0. So, the function will take the value 0. So, when
we are approaching along this x is equal to y line towards this 0 0 we have the function value
0 0 0 0 whatever neighborhood we take close to 0 the value of this function is 0.
lim f ( 0 , Δ y )−f ( 0 , 0 )
f y=
Δx → 0
Δy
=0
Because the function is defined as the value is 0 when x y not equal to 0. So, when we are
approaching to origin along this line here, the function is taking value as 0 whatever close we
are to the origin the function will take the value 0 only at the origin it is 1. So, there is a point
of discontinuity here. So, this limit along this x is equal to y is0 whereas, the value of this
function at origin is 1. So, we can clearly see then the function is not differentiable or is not
continuous.
lim
f ( x , y ) along (x= y )=0 ≠ f (0, 0)
( x , y)→(0 ,0)
(Refer Slide Time: 23:51)
260
So, this is not equal to the function value at 0 0. Hence, this function is not continuous at
origin and since the function is not continuous we can decide about the differentiability.
Because the continuity is the necessary condition for differentiability, if the function is not
continuous definitely the function is not differentiable. So, this function f is not differentiable
at origin. We do not have to go for any other test for differentiability because the function is
not continuous though the partial derivatives exist in this case.
So, one of the necessary conditions here is fulfilled, but the continuity is also the necessary
condition for differentiability which is not fulfilled here. So, the function is not continuous
and hence the function is not differentiable without any further test.
(Refer Slide Time: 24:47)
The last example here we take this fx y is equal to x square plus y square divided by absolute
value of x plus absolute value of y when x y not equal to 0 0. And the value is 0 when x y
equal to 0 0. So, this is a mistake here we have we have this equal to here not this is equal to
0 0. So, we will check whether this function is differentiable at origin. So, the continuity
check for this one is can be done by various ways we can change to the polar coordinate or
we can also use the epsilon delta approach to show that this function is continuous.
f ( x , y )=
2
2
x +y
∧0, ( x , y ) =(0, 0)
|x|+¿ y∨¿ ,( x , y)≠(0 , 0) ¿
{
261
So, here let us take the epsilon delta approach. So, we take the difference of the function f xy
xy not equal to 0 0 and minus this f 0 0 which is given as 0. So, we will show that this limit of
this f xy is equal to 0 or not by taking this difference and expressing this in terms of the delta
and epsilon. So, here this is x square plus y square over modulus x plus modulus y.
2
|f ( x , y ) −0|=
2
x +y
|x|+¿ y∨¿ ¿
And then we can add here plus 2 times absolute value of x plus absolute value of y to make
this square here. So, when we are adding in the numerator the quantity will be bigger. So, we
have this bigger quantity and that is whole square because we have added this term here plus
2 times absolute value x absolute value of y. And, then this term here becomes this whole
square and now this is the same term so, one will get cancelled. So, we have absolute value of
x plus absolute value of y. And, in one of the earlier lectures so, we have seen that this here is
less than equal to square root 2 and square root x square plus y square which is again we have
converted into the delta neighborhood of this 0 0 point. So, this is less than square root 2 and
in terms of delta and this everything we want to make less than the arbitrary number epsilon.
2
(|x|+| y|)
¿
|x|+¿ y∨¿=|x|+| y|< √ 2 √ x 2 + y2 < √ 2δ < ϵ ¿
So, by this relation here we have found that for given epsilon we can set this delta so, that this
f xy minus this 0 0 term is less than the epsilon. So, by choosing this delta here less than
epsilon by square root 2 for given epsilon we have this delta now. So, that this difference is
less than epsilon and that is the epsilon delta approach for showing the continuity. So, we
have shown that if we choose this delta here then this difference will be less than epsilon.
And, whenever we take any point xy from this delta neighborhood of this 0 0 point and this
implies that the function is continuous. So, moving next now to show the differentiability of
this function.
δ<
ϵ
,|f ( x , y )−f ( 0 , 0 )|<ϵ whenever 0< √ x 2 + y2 < δ
√2
262
Refer Slide Time: 27:52)
So, now we will show the existence of the partial derivatives again as per the definition here.
We have f delta x 0 minus f 0 0 delta x. So, what is this term here the y will be set to 0. So,
you have this delta x square and divided by this absolute value of delta x and delta x. So, this
will be this term and now so, this delta x will get cancelled we have delta x over more
absolute value of delta x. So, if this delta x goes to 0 from the right side then this will be 1
and when it goes to 0 from the left side then this value will become minus 1. So, hence this
limit does not exist because these we have here either plus 1 or minus 1 as the limit, when
this delta x approaches to 0 from the positive side, when delta x approaches to 0 from the
negative side.
lim f ( Δ x , 0 )−f ( 0 , 0 )
lim Δ x
Δx → 0
Δx →0
Δx
=
2
¿ Δx∨Δx
lim Δx
=
Δx →0
¿ Δx∨¿¿
So, hence this limit does not exist and in this case we can now we have shown just for the one
derivative over the f x here, but since the nature of this function it is a symmetric here x
square plus y square over this one; we can also show that f y at 0 0 does not exists. And, in
this case since in fact, showing the existence of one of the derivatives is enough to tell that
the function is not differentiable because the necessary condition for differentiability is the
existence of both the partial order derivatives. So, in this case if this f x does not exist there
itself we can tell that a function is not differentiable though in this case both the derivatives
do not exist. And hence, the function is not differentiable.
263
(Refer Slide Time: 29:54)
So, coming to the conclusion now; so, what we have observed that the investigation of this
differentiability we can use the necessary conditions because the continuity and the existence
of partial derivative; if one of them is violated then we can prove that the function is not
differentiable. Or using the sufficient conditions also we can discuss the differentiability of a
function. So, the continuity of one of the partial order derivatives is sufficient for the
differentiability of the function. And the final check if we have observed that the function the
necessary conditions are fulfilled. And we are not meeting with the sufficient conditions for
example, that the partial derivatives are not continuous In that case we have to go for the final
check that one can use the limit test which is often easier to prove that this limit is 0 or it is
not equal to 0 or we can directly use the definition of the differentiability in this case.
264
(Refer Slide Time: 31:02)
So, these are the references used for preparing the lectures.
Thank you very much.
265
Engineering Mathematics - I
Prof. Jitendra Kumar
Department of Mathematics
Indian Institute of Technology, Kharagpur
Lecture – 15
Composite and Homogeneous Functions
So, welcome back to the lectures on Engineering Mathematics-I. And this is lecture number
15. And, today we will be discussing about a Composite Functions and Homogeneous
Functions.
(Refer Slide Time: 00:28)
So, what are the composite functions? So, if we consider here function z is equal to f x y a
function of 2 variables x and y and we let that this x depends on t. So, x is a function of a phi
t and here y is also a function of t which we are denoting by psi t. Or this could be that this
function x is a function of 2 variables again u and v and y is a function again of 2 variables u
and v.
z=f ( x , y )
x=ϕ ( t ) , y=ψ (t)
x=ϕ ( u, v ) , y=ψ (u , v)
266
Let us call this equation number 1, here the equation number 2 and this is equation number 2
prime. Then the equations here 1 and this 2 together or 1 with this 2 prime together are said to
be to define z as composite function of t or in this case composite function of u and v. So,
here what basically we will be discussing here, if we want to get for example, the derivative
of this z function with respect to 2 t. Because in this first case when we are taking equation
number 1 and equation number 2 then this z depends on x and y, but the x and y again
depends on t.
So, basically this z is a function of t and then we can define here the derivative of z with
respect to t directly. So, we will derive a formula how to get the derivative of this z without
substituting this x and y here in this function. And, then making this as a function of t and
then taking the derivative. But, we will find a direct formula which will use the partial
derivatives of this function f which is a function of 2 variables x and y.
And also here the x and y are the functions of phi and psi both are the functions of t and we
will also make use of the derivative of these functions to get the derivative of this z directly
without substituting this x and y in terms of t in this function f x y. And, the same scenario we
have when we have this equation 1 z is equal to f x y with equations x is equal to phi u v and
y is equal to psi u v.
(Refer Slide Time: 03:06)
So, let us take the first case lets z posses continuous partial derivatives or we can also take
this assumption that this z is equal to f x y is differentiable. So, this f x y is a differentiable
267
function and then we also let that x is equal to phi t and y is equal to psi t, they posses a
continuous derivatives or again we can assume that they are differentiable functions.
dz ∂ z dx ∂ z dy
=
+
dt ∂ x dt ∂ y dt
So, in that case we will derive this formula which says that we can get this dz over dt. So, the
ordinary derivative of z with respect to t which makes sense now because x and y are
functions of t. So, basically this z is a function of t alone. So, in this case this formula says
that we can get this ordinary derivative dz over dt is equal to the partial derivative of z with
respect to x. So, we are making use of that this f is a function of 2 variables x and y. So, we
will take here the partial derivatives of this f with respect to x and multiplied by the
derivative of x with respect to t.
z=f ( x , y ), x=ϕ ( t ) , y=ψ (t)
Δ z=z x Δ x+ z y Δ y +ϵ 1 Δ x +ϵ 2 Δ y
So, it is again an ordinary derivative because x is a function of 1 variable t and then this is a
chain rule against of plus this because z is a function of y as well. So, we will have here the
partial derivative with respect to y and multiplied by dy over dt. So, we will derive this
formula now. So, we take that this z is equal to f x y and x is a function of phi t and y is a
function of t as denoted by psi t and this is a composite function which we call as for the
definition.
268
(Refer Slide Time: 04:56)
And since we have assumed already the differentiability of all these functions z phi and psi,
then differentiability of z will imply that we can write down this delta z is equal to del x delta
x plus del y delta y and psi 1 delta x plus psi 2 delta y. And, if we divide this expression both
the sides by delta t then what we will get; here the delta z over delta t and this delta x divided
by delta t. Here again we have delta y divided by delta t and it is dx and dy will be also
divided by delta t.
Δz ∂z Δx ∂z Δ y
Δx
Δy
=
+
+ϵ 1
+ϵ 2
Δt ∂ x Δt ∂ y Δt
Δt
Δt
And now if we take the limit as delta t goes to 0 then what will happen. So, delta t goes to 0
means delta x goes to 0 and delta y goes to 0 because this x and y they are functions of
function of t and if there is no variation in t. So, naturally the x and y variation in x and y will
be also 0. So, as delta t goes to 0 meaning delta x goes to 0 and delta y goes to 0.
d z ∂ z d x ∂z d y
=
+
dt ∂x dt ∂ y dt
So, when we take the limit now as delta t goes to 0 in this expression here. So, this delta z
over delta t as per the definition of the of the derivative of z will be like dz over dt and this
will remain as it is; we are not touching this del z over del x and then we have here delta x
over delta t. So, again this is the derivative of x with respect to t. So, we get here dx over dt;
269
similarly here again we have delta y over delta t which will become here dy over dt and these
terms.
So, assuming again that these partial these derivatives exist and they are finite numbers. So,
this when delta x and delta y goes to 0. So, this epsilon will go to 0 and this epsilon 2 will go
to 0 and then these expressions here epsilon delta x over delta t and this epsilon delta y over
delta t they both will go to 0. So, we have only this these three terms here. So, dz over dt we
can find out directly in terms of the partial derivatives of z and the ordinary derivatives of x
and y.
(Refer Slide Time: 07:26)
So, the differentiation of the composite functions when we have this more general case that x
and y they also depend on u and v a functions of 2 variables. In that case the formula will
become that so, we can prove similarly as above. So, this del z over del u so now, the partial
derivatives here because the x and y they are the functions of 2 variables u and v.
z=f ( x , y ) , x=ϕ ( u , v ) , y=ψ (u ,v )
∂ z ∂z ∂ x ∂z ∂ y
=
+
∂u ∂ x ∂u ∂ y ∂u
∂ z ∂z ∂x ∂z ∂ y
=
+
∂v ∂ x ∂v ∂ y ∂v
270
So, we will have here the partial derivative of z with respect to u without substituting all these
into this function and then getting this partial derivatives. We can use this formula to get the
partial derivative of this z with respect to u is equal to the partial derivative of z with respect
to x and partial derivative of x with respect to u plus partial derivative of z with respect to y
and partial derivative of y with respect to u again because we are differentiating partially z
with respect to u. So, here the u will come and here also the u will come.
Similarly, for del z over del v now the similar formula, but instead of this u it will be replaced
by v. So, what is important if you are differentiating here with respect to u then this u will
appear both the places and if we are differentiating with respect to v here. So, then this v will
appear at these places here when we are taking the partial derivative of x with respect to u
and here with respect to v well; so, moving next.
(Refer Slide Time: 09:14)
Let us solve some problem based on this. So, for example, we have this function z is equal to
xy x is a function of t. So, which is defined here as x is equal to cos t and then y is a function
of t again which is denoted by y is equal to sin t and then we want to find here what is dz over
dt. So, one direct way would be that we substitute this x and y here in this function and then
we will have z as a function of t which we can differentiate to get dz over dt. But, we will
follow the idea of the composite functions that z is a function of x y and x is a function of t
and y is a function of t with the formula derived above.
z=xy ; x=cos t , y=sin t .
271
So, that formula says that dx over dt will be the we can have here with partial derivative with
respect to x of the function z and dx dt and similarly here the partial derivative of y and then
dy dt. So, what do we get here? Del z over del x. So, z was x y and when we differentiate
here with respect to x then this will become only y and dx over dt so, x was cos t. So, we will
get here minus sin t. Similarly, here del z over del y will become x and dy over dt which is
sin t it will become cos t.
dz ∂ z dx ∂ z dy
=
+
dt ∂ x dt ∂ y dt
¿ y (−sin t ) + x cos t
2
2
¿−sin t +cos t
¿ cos 2t
So, we get here the y is sin t so, we have minus sin square t and x was cos t. So, we have cos
square t. So, this is the derivative of z with respect to t without substituting into this function
and then getting the derivative. So, this we can again simplify to have this cos 2 t. As I said
they alternatively we can just substitute here z is equal to x x is the cos t and then y is sin t.
So, now we have z as a function of t and we can get dz over dt there. So, this we can write
first as half of 2 cos t sin t which is sin 2 t and then when we take the differentiation here this
will be a sin t will be cos 2 t and this 2 will get cancelled.
So, we will have this cos 2 t directly as well, but we used here the idea of this composite
differentiation or the differentiation of composite function. And, with this formula we do not
have to substitute the values of these x and y into the function, but directly with the help of
the partial derivatives of this z we can get the derivative of z directly with respect to the
function t.
272
(Refer Slide Time: 12:14)
So, again next problem here z is a function of x and y and further it is given that x is a
function of 2 variables now. So, u and v and then we have here y is a function of again of 2
variables u and v. And, then we will show that this partial derivative the difference of this
partial derivative here with respect to u and v can be written as x partial derivative of z with
respect to x minus the partial derivative of z with respect to y. So, z is a function of x and y.
So, with the formula derived above we have a partial derivative of z with respect to u in terms
of the partial derivatives of z with respect to x and y and the partial derivative of x with
respect to u partial derivative of y with respect to u again.
u
−v
−u
x=e + e , y=e + e
v
So, if it is u here it will be u there in both the terms here with respect to u of the derivatives x
and y. So, in this case so, del z over del x because this is not given function here z is a just
function of x y. So, this partial derivatives z with respect to x will come and del x over del u.
So, what was x here? e power u plus e power minus v. So, if we take that partial derivative
with respect to u; that means, treating v as constant so, this term will be treated as constant.
So, the partial derivative of x with respect to u will become simply e power u which is written
here and then del z over del y and del y over del u.
∂ z ∂z ∂ x ∂z ∂ y ∂z u ∂z
=
+
=
e+
(−e−u )
∂u ∂ x ∂u ∂ y ∂u ∂ x
∂y
273
∂ z ∂ z ∂ x ∂ z ∂ y −∂ z −v ∂ z v
=
+
=
e +
e
∂v ∂ x ∂v ∂ y ∂v
∂x
∂y
∂ z ∂z ∂ z
∂ z −u v
∂z
∂z
− = (e ¿ ¿u+e−v)−
(e +e )=x
−y
¿
∂u ∂ v ∂ x
∂y
∂x
∂y
So, the y is here e power minus u plus e power v and when we take the partial derivative here
with respect to u. So, it we will get here minus e power minus u which is the term here and
now the partial derivative of z with respect to v again the similar formula, but instead of u we
have v here and then when we take the partial derivative of x with respect to v. So, it is e
power minus v. So, we will get minus e power minus v and then we have here y with respect
to v.
So, this will become e power v which is written here and now since we want to get the
difference of these 2 partial derivatives. So, we will take this difference here and then so, del
z over del x here this will become plus. So, del z over del x when we take common it will be
e power u and plus e power minus v; similarly here when we take these 2 terms common with
minus sign.
So, again e power minus u plus e power v will come as written here and then this e power u
plus e power minus v is x and here e power minus u plus e power v is y. So, we get precisely
the desired term here. So, this is x del z over del x and then minus this is y and del z over del
y.
(Refer Slide Time: 15:25)
274
So, another part of this lecture is to define the homogeneous functions. So, what are the
homogeneous functions we will learn now? So, an expression in xy is homogeneous of order
of order n there. So, any expression here in x and y of order n, if it can be expressed in this
form that x power n n some function of y over x then we call such a function of homogeneous
function of order n.
xn f
y
x
()
So, this is important this n here. So, if we can bring this x power n and then the rest term can
be written as a function of y over x, then we call such expression as a a homogeneous
expression of order n or in terms of the functions we can define. So, a function f x y is said to
be homogeneous of order n if it satisfies so, we replace here the argument x and y by t x and t
y. And, if we can bring this t power n out of the function; that means, this t power n. And
again the function of x y remain then we again have this the concept of the homogeneous
function of order n. So, we will call this such a function a function a homogeneous function
of order n.
n
f ( tx , ty )=t f ( x , y )
(Refer Slide Time: 16:55)
So, these are the examples of the homogeneous functions. So, we consider for example, this f
x y is equal to a 0 x power n a 0 x power n minus 1 into y again here x power n minus 2 y
275
square and so on a n y power n. So, what we observe that this is a function this is a
homogeneous function of order n. Why order n, because if we take this x power n from all
these terms outside then what will we will get here this will be a 0 because x power n we
have taken out.
n
f ( x , y)=a0 x + a1 x
n
n−1
y+a2 x
n−2
y
y 2
y
+a2
+…+an
x
x
x
( () ()
¿ x a0 +a1
2
y +…+ an y
n
n
( ))
And here we have again x power n taken out so, 1 x we have to divide. So, a 1 and then y
over x it will become. So, here we have taken x power n out again and then this x power
minus 2 will remain; that means, a 2 and then y over x power 2 and so on. Here a n and x n
we have taken out. So, this x power n will be in the denominator term and in this case again
we have here y over x power n.
So, all these terms here or this function we can denote as the function of y power x, because
this y power x appears together in all the terms. So, we can consider this function here as a
function of y over x and then x power n is sitting outside. Therefore, this is a function of
homogeneous function of order n. Similarly, if we consider this 1 f x y is equal to square root
y plus square root x over y plus x.
√
y
√ y+ √ x = √ x x
f ( x , y )=
y+ x
x
+1
y
+1
x
=x
−1
2
g
( yx )
In this case also we can write down this as x power something and the rest we can consider as
the function of y over x. And, this is simple because we can take here square root x out and
here we will take x out. And in this case so, here square root x over x and then inside here we
have a square root y over x plus this 1 and here also y over x plus 1.
So, now this one here, this expression we can consider as the function of y over x and then
what is together here x power minus half. So, this we can write down as x power minus half
and some function of y over x because in all these terms y over x comes together. So, we can
say that this is some function of y over x and in this case as we see here x power minus half.
276
So, this is a homogeneous function of order minus half. So, what is the importance of these
homogeneous functions: we will see in the next slide.
So, based on these order for example, here it was n or here minus half we can have some
formula for the derivative term in terms of this n.
(Refer Slide Time: 20:04)
So, precisely what it is called the Euler’s theorem on homogeneous function and it says if z is
a homogeneous function of x and y of order n and these partial derivatives exist and so on.
So, then we have this expression here x partial derivative of z with respect to x y the partial
derivative of z with respect to y will be equal to nz.
x
∂z
∂z
+y
=nz , ∀ x , y ∈ D
∂x
∂y
So, we do not have to compute this separately, if we know that it is a homogeneous function
then x z x y z y will be simply n z and for all x y point in the domain of the order domain of
the function here. So, given that z is equal to f x y is a homogeneous function.
277
(Refer Slide Time: 20:49)
So, if it is a homogeneous function in that case we can write down wait a minute. So, here if
it is a homogeneous function of x and y of order n; so, we can write down that this x power n
and g y over x and then we know already we have to differentiate with respect to x. So, we
can do that.
z=f ( x , y ) =xn g
( xy )
So, here when we take the partial derivative of this with respect to x so, what we will get here
we have to differentiate. So, n x power n minus 1 and this is remain as it is here product rule.
So, and then x power n as it is and we differentiate this g of y over x which is g prime y over
x. So, the derivative of g with respect to its argument g prime y over x and then here the y
over x will be differentiated with respect to x that will give us here minus y over x square.
∂z
y
−y
y
y
y
=n xn−1 g
+ x n 2 g'
=n xn−1 g
− y x n−2 g'
∂x
x
x
x
x
x
() ( )()
()
( )
So, in this case now if we simplify this a bit. So, here we have n x power n minus 1 and g y
over x and then here we can have this minus sign there. So, y and n x power n minus 2 and
the g prime y over x and the partial derivative of z with respect to y we can get again we have
to differentiate now with respect to y. So, x power n will be treated as constant and we have
to just differentiate this term.
278
So, we have g prime y over x and then this y over x the derivative with respect to y will be
just 1 over x the partial derivative of y over x with respect to y will be 1 over x. And, then if
we add these 2 terms with the product of x here and y there what we will get. So, we have
multiply it here by x so, we will get here x power n and here x power n minus 1 and here we
have multiply it by y.
∂z
1 ' y
=x n
g
∂y
x
x
()( )
So, this is exactly here n x power n and minus y x power n minus 1 and in this case we have x
power n and then we have multiply it here by y term. So, this is x power n minus 1 and we
have multiply it by y. And, then these 2 terms will get cancelled and we have here n x power
n g y over x and what was x power n g y over x this was z or the function f x y. So, here these
terms cancel out and then we get simply n into z term.
x
∂z
∂z
y
y
y
+y
=n xn g
− y xn−1 g'
+ y xn−1 g'
=nz
∂x
∂y
x
x
x
( )
()
()
So, n and this is z here and these 2 gets cancelled. So, the problem number 3; so, here we
have a u which is given as tan inverse x cube plus y cube over x minus y x is not equal to y
because this is now defined as x is equal to y. And, then we show that x del u over del x plus
y del u over del y is equal to sin 2 x. And in this case so, again, but we should note that this
given function tan inverse y cube plus x cube over x minus y is not a homogeneous function.
Because of this tan inverse we cannot bring x power something and the rest we cannot write
in terms of y over x. But what we notice that this argument of this tan inverse; that means, x
cube plus y cube over x minus y that is a homogeneous function. And, we will make use of
this because we can define here z as tan u because, if we take this tan to the left hand side we
will get tan u is equal to x cube plus y cube over x minus y. And, now this z here is a function
of is a homogeneous function of order 2 because here we can write down as x square.
So, x cube we have taken common in the numerator and x here from the denominator. So, we
will have x square and this will become 1 plus y over x cube. So, this is 1 here. So, it is 1 plus
y over x whole cube and then we have 1 minus y over x. So, this is a homogeneous function
of order 2 and then we can apply the Euler’s theorem on this function z.
279
So, applying the Euler’s theorem on z is equal to tan u. So, this will give us 2 times the z and
z was tan u. So, we get here the x x and then del z over del u. So, we need to see what is this
1 here del z over del u. So, we have z is equal to tan u. So, del z over del x here will be the
sec square u and del u over del x, the partial derivative of z with respect to x here the we will
take exactly the partial derivative of z with respect to x, but it is given in terms of u. So, we
will get the sec square u and then del u over del x.
So, here then y and similarly we have del z over del y. So, we have your tan will become a
sec square u and del u over del y is equal to 2 times z. So, z is tan u and then we have here x
the sec u it is common here, we can bring to the right hand side. So, we get x and del u over
del x plus this y del u over del y and this sec u goes there as a cos square u. So, this is 2 tan u
and then here sec square u will be in the denominator which will become cos square.
So, here a sin u over cos u for tan and this is cos is square u. So, this gets cancelled we have 2
sin u cos u which is a sin 2 u. So, we get this the z quantity as x del u over del x plus y del u
over del y is equal to sin u. So, here the important point is that this u was not a homogeneous
function, but by defining this as the z is equal to tan u which is x cube plus y cube over x
minus y it becomes homogeneous. And, then we have applied the Euler’s result on z and
noting this z is equal to tan u, we have computed here partial derivatives of z with respect to x
and y and then we get the desired result.
(Refer Slide Time: 28:04)
280
So, generalization of this Euler results says that so, this was in case of the first order partial
derivatives, but we do have results in case of the second order partial derivatives as well.
And, which is the extension of this which says that x square the second order partial
derivative 2 x y the mixed from y square again second order derivative with respect to y is
equal to n n minus 1 z for all xy in the domain of the function.
x
2
2
2
2
∂ z
∂ z
2 ∂ z
(
)
+y
2 +2 xy
2 =n n−1 z , ∀ x , y ∈ D
∂
x
∂
y
∂x
∂y
(Refer Slide Time: 28:38)
So, this is the generalization of this result and now we will have this one problem on this. So,
suppose we have z is equal to x y a function of y over x and plus g y over x and where this f
and g are 2 times differentiable function and then we will evaluate this expression here for z.
So, again the same problem the z is not a homogeneous function of x and y, but here this is a
homogeneous function and the second g is also a homogeneous function because we can
write down as it is a directly given in terms of y over x.
z=xy f
x
2
2
( xy )+g ( yx ),
2
2
∂ z
∂ z
2 ∂ z
+y
2 +2 xy
2
∂
x
∂
y
∂x
∂y
281
So, what we take here we take z is equal to u 1 plus u 2 the first term here we take as u 1 and
here we will take as u 2 ; that means, this u 1 is x y and f y over x and u 2 is g y over x. So,
this is a function of homogeneous function of order 2 because this we can again write down
as x square and y over x and this f y over x. So, this is a function of y over x and then we
have x square there. So therefore, the order is 2 and in this case this g is a function of y over x
and here there is no term of x so, x power 0. So, this is a homogeneous function of order 0.
u1=xy f
u2=g
( yx )(Hom. function of order 2)
( xy )(Hom. function of order 0)
(Refer Slide Time: 30:07)
So, well we have u 1 homogeneous function of order 2 u 2 a homogeneous function of order
0 and z was as u 1 plus u 2 u 1 is given by this, u 2 is given by this. So, the Euler’s theorem
we can apply on u 1 and u 2 because they are the homogeneous functions of order 2 and 0.
So, on u 1 for second derivative theorem it says 2 times so, n is 2; so 2 times and n minus 1
and then the u 1.
u1: Hom. function of order 2
2
∂ u1
2
2
∂ u1
∂ u1
x
+2 xy
+ y2
=2u1
2
∂ x∂ y
∂x
∂ y2
2
282
u2: Hom. function of order 0
2
x2
∂ u2
∂x
+2 xy
2
2
2
∂ u2
∂ u2
+ y2
=0
∂ x∂ y
∂ y2
So, n n minus 1 u 1 so this order was 2; so, we get 2 into 1 into u 1 and this u 1 now we know
already is x y f y over x, we will substitute later. And, then for u 2 because that is also a
homogeneous function of order 0 and because of that order 0 here right hand side we will get
0 term because here its n n minus 1 and because n is 0 the order is 0. So, this will become 0
there and now we have these 2 expressions directly without computing these derivatives here,
we have used this Euler’s theorem and we can add them because we want to get this result in
terms of z there is u 1 plus u 2.
So, if we add these two so, we have x square and this will become as the partial derivative
with respect to 2, the second order u 1 plus u 2 here u 1 plus u 2 here also u 1 plus u 2 and
that is z. And, now equal to here 2 u 1 plus 0 so, 2 and u 1 so, plus 0. So, we get only 2 u 1
and u 1 is x y f y u over x. So, we get this result 2 times xy f function of y over x of this
expression here.
x
2
2
2
2
∂ z
∂ z
y
2 ∂ z
+y
2 +2 xy
2 =2 xy f
∂
x
∂
y
x
∂x
∂y
()
(Refer Slide Time: 31:51)
283
So, coming to the conclusion we have learn today the differentiation of composite functions
which was very useful when z is a given function of x y and x is and y they are the function
of t. So, in that case we can directly compute the derivative of z with respect to t; by this
formula and then this is generalization of this one that x and y. They may be functions of u v.
And in that case also we can compute the partial derivatives now of z with respect to u and
the partial derivative of z with respect to v by these formulas.
And, then the Euler’s theorem for homogeneous functions we have learnt that if z is a
homogeneous function, then we can have this for homogeneous function of order n then it
will be here the n z. So, x partial derivative with respect to x plus y partial derivative with
respect to y will be equal to n z or there was a generalization that we can also care these
second order derivatives as n n minus 1 into z when this z is a homogeneous function of x
and y of order n.
z=f ( x , y ), x=ϕ ( t ) , y=ψ ( t ) ,
d z ∂ z d x ∂z d y
=
+
dt ∂x dt ∂ y dt
z=f ( x , y ) , x=ϕ ( u ,v ) , y=ψ (u ,v)
∂ z ∂z ∂ x ∂z ∂ y ∂ z ∂z ∂x ∂z ∂ y
=
+
, =
+
∂u ∂ x ∂u ∂ y ∂u ∂ v ∂ x ∂ v ∂ y ∂ v
'
Eule r sTheroem : x
x
2
2
2
∂z
∂z
+y
=nz
∂x
∂y
2
∂ z
∂ z
2 ∂ z
(
)
+y
2 +2 xy
2 =n n−1 z
∂x ∂ y
∂x
∂y
284
(Refer Slide Time: 33:11)
So, these are the references used for preparing these lectures.
Thank you very much.
285
Engineering Mathematics - I
Prof. Jitendra Kumar
Department of Mathematics
Indian Institute of Technology, Kharagpur
Lecture – 16
Taylor’s Theorem for Functions of Two Variables
So welcome back to the lectures on Engineering Mathematics I and today this is lecture
number 16 and we will learn the Taylor’s Theorem for Functions of Two Variables.
(Refer Slide Time: 00:25)
So, what is the Taylors theorem of function of single variables we need to just recall. So,
assume that f has all derivatives up to order n plus 1 in some interval containing the point x is
equal to x 0. And in that case we can write down this f x 0 plus h. So, this is a point in the
neighborhood of 10 x 0 and we can express this as f x 0 plus the h the first order derivative at
x 0 h square by factorial 2 the second order derivative and so on.
'
f ( x 0 +h ) =f ( x 0 ) + h f ( x 0 ) +
2
n
h ''
h ( n)
f ( x0 ) +…+ f ( x0 ) + R n
2!
n!
So, in terms of the these higher order derivatives plus the remainder term because we have
the equality here. So, this is this is called the Taylor’s polynomial and then this is the
remainder the error term in this Taylor’s polynomial which is denoted by the h power. So, it
is a continuation again so h power n plus 1 over n plus 1 factorial and then we have the n plus
286
1 and derivative at some point xi which lies between these two points. So, x 0 where we have
expanded this around and the point which we are considering here x 0 plus h.
n+1
h
(n+1 )
( ξ ) , x0 <ξ< x
Rn =
f
( n+1 ) !
So, or x yeah this is x 0 plus h means the x the point in the neighborhood of this x 0. So, the
this point here xi where we have evaluated this n plus 1 a derivative it is not known, but it
exists and it is somewhere between x 0 and x 0 plus h. So, precisely we can also write here x
0 plus h. So, this is the point here between x 0 and x 0 plus h.
(Refer Slide Time: 02:05)
So, now the Taylor’s theorem for a function of two variables so we have again that function
is defined in some domain D and we have continuous partial order derivative up to n plus 1th
order in some neighborhood of a point here x 0 y 0 in the domain, then we can express here
this f x 0 plus h y 0 plus k. So, again this is a point in the neighborhood of this x 0 y 0 and we
can get this function value at this some other point in the neighborhood by the by this
expression here.
(
f ( x 0 +h , y 0 +k ) =f ( x 0 , y 0 ) + h
2
n
∂
∂
1
∂
∂
1
∂
∂
+k
f ( x 0 , y 0) +
h
+k
f ( x0 , y 0 ) +…+
h
+k
f ( x 0 , y 0)+
∂x
∂y
2! ∂x
∂y
n! ∂x
∂y
)
(
287
)
(
)
So, f x 0 plus y 0 x 0 y 0 and then these are the first order terms, now we have the partial
derivative. So, these are the first order partial derivatives. So, h or the partial derivative with
respect to x and with k the partial derivative with respect to y and then we have one over
factorial terms similar to the single variable case and then here we have this higher order
term. So, we will get in the proof now what do we mean by this h, the partial derivative with
respect to x plus k the partial derivative y and then whole square. So, do we need to wait a
little bit here.
And then this will be continued up to this n and one over factorial n and f x 0 y 0. So, these
partial derivatives will be evaluated at x 0 y 0 all the partial derivatives here and then the rest
term here the remainder term which is again the continuation of these terms and the only
difference is that this point here the argument of the n plus 1th derivative. Which will be
appearing because of this term, will be evaluated here at x 0 plus theta h y 0 plus theta k and
this theta is between 0 and 1.
Rn =
1
∂
∂
h
+k
∂y
(n+1)! ∂ x
(
n+ 1
)
f ( x0 +θh , y 0 +θk ) , 0<θ<1
So; that means, again this is the point here between this x 0 y 0 and this point here x 0 plus h
y 0 plus k. So, this first argument will vary from x 0 to x 0 plus h when theta varies from 0 to
1. Similarly here y 0 will vary from y 0 to or can vary from y 0 to y 0 plus k when this theta
varies from 0 to 1. (Refer Slide Time: 04:25)
So, we will just go through the proof of this result. So, this is the Taylor’s theorem for two
variables and we will take just for simplicity the n is equal to 2; that means, you will consider
288
the terms in the expansion up to order 3. So, we take this x is equal to x 0 plus th and y as y 0
plus th and t is some parameter from this interval 0 and 1, including 0 and 1.
So, having this we define a function now phi t, the same the function f x 0 plus th y 0 plus tk.
So, this h and k were given already in the in the expansion which is somehow fixed, x 0 y 0
that point is also fixed and then this t varies here in this interval 0 and 1. So, this function
here f x 0 plus th and y 0 plus tk having this x 0 and y 0 fixed h and k fixed we have a
variable t here. So, therefore, we have defined this as function of t, function of one variable.
And for function of 1 variable we can get the derivative here with respect to t. So, this is
derivative with respect to t and remember this is like composite function. So, here we have
like x and here we have y where the x is x 0 plus this th and this y is y 0 plus tk. So, by that
differentiation which we have learnt before we can compute this phi prime t the derivative of
phi with respect to t or the derivative of this f with respect to t because this is a function of 1
variable when x and y are given in terms of t. So, that composite formula we have written del
f over del x dx over dt and so on.
And now we will compute this; so, here this is dx over dt. So, x was x 0 plus th so dx over dt
means only the h will remain here. So, this is h, this is partial derivative with respect to x we
have written as it is then here we will get k the derivative of y with respect to t will be k and
then del over del y. So, del over del y and we have just because this was del over del x on f
and this del over del y on f. So, that f at x y point we have just written outside. So, let me
erase this. So, we have written just this f. So, the meaning is here h and del f over del x at the
point x y or this x 0 plus th y 0 t k point plus k and again del f over del y here.
289
(Refer Slide Time: 07:13)
Similarly, we can compute the second order derivative. So, we have the first order terms here
with this h. So, this was h here and this was k here. So, we will compute now the second
order derivative of this term. So, h will remain as it is and the derivative of del f over del x,
again the same this chain formula will be used so the partial derivative of this with respect to
x again. So, the double derivative and here again dx dt will term come then h then plus, the
partial derivative of f with respect to y so it was already with respect to x. So, we have the
second order partial derivative with respect to y and x and then this k will come because we
have dy over dt term again.
So, we have used this chain rule again from this del f over del x and also similarly for del f
over del y which is the term given here. So, again to make it more clear. So, this was this is
the term when we take the derivative of del f over del x and then this h was already here, here
the k was there and this is the term which we have computed here d over dx of del f over del f
over del y. So, with respect to t sorry with respect to t
So, here we have used this chain rule again so that the partial derivative with respect to x
again of this second order derivative and then dx over dt which becomes h then the partial
derivative of this with respect to y here and then dy over dt which is k.
290
(Refer Slide Time: 09:05)
Well and then we have written again this little more simplified form so here h is square and
then this hk and here also k h assuming that this second order partial derivatives are equal.
So, we have the two term 2 times this h k and the second order mixed order derivative and
plus this k square and the second order term here, and then we can compute the third order
term as well. So, this is before we compute the third order term we have written this little
more a convenient form that this is like h square two h k plus k square.
So, h plus k whole square term, and for the notational convenience we have also because here
we have this like del to the second order term here also it is mix term like hk, here also the yy
term. So, the second order term so that also we have written inside this these parentheses.
So, the meaning of this square will be h square the second order partial derivative is not a
whole square of this, but rather second order derivative of this with respect to x, then the k
square and the second order derivative with respect to y and then the two times hk and the
mixed order derivatives. So, that we have to understand that this whole square is not literally
the a square of this two terms, but the meaning of this here is in terms of the second order
partial derivatives.
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(Refer Slide Time: 10:31)
Similarly, we can compute the third order derivative. So, in I am not going to explain this
now. So, here for h square term we have again here we have to use this chain rule then again
here and here so, all these three terms. So, with h square this is from this first term then 2 hk
and k square and so on and again the simplified form we have written this.
So, 3 times h square k is appearing here we have h square k and then here also we have h
square k and assuming the continuity of these third order derivatives. So, we can assume that
this partial derivative with respect to x 2 times and then partial derivative with respect to y or
first partial derivative y and then 2 times with respect to x they are equal.
So, treating them equal we have these terms and again we have written for simplicity because
this is like a cubic term in terms of h cubed 3 h square k 3 h k square and k is cube. So, we
have combined this again these third order derivatives terms as well in this cubed term and
this f at this point.
Now, we will use the Taylor’s theorem for phi t phi is a function of one variable. So, it is a 1
variable case again for the Taylor’s theorem which says that phi t will be phi 0 t phi prime 0
and so on. And up to the third order term so t cube or factorial 3 so, we can extend this to any
order term. So, phi the third order derivative and theta t and now we will substitute this t is
equal to 1 here because t varies from 0 to 1 it is this is true for any value of this t. So, we have
taken the t is equal to one and then we get here just only theta. So, now we will substitute this
phi which we have computed.
292
(Refer Slide Time: 12:25)
So, the phi t was this phi prime phi double prime and so on and this was because of the
Taylor’s theorem of one variable and then we can substitute now. So, the phi at 1, so phi at 1
means the t will be substituted as 1 here. So, this t will be 1 1 so we will get phi f x 0 plus h
and y 0 plus k this term, phi 0. So, when t is 0 so we will have x 0 y 0 term and then here phi
prime at 0. So, phi prime is here and at 0 means x 0 and y 0 point. So, this parenthesis here h
and the partial derivatives the first order partial derivatives and this evaluated at x 0 y 0 point.
Similarly, for the second order term again t will be set to 0. So, we will have this x 0 y 0 point
and for this one the here the third derivative will be computed as theta point. So, here the t
will be replaced by this theta, here also by theta and then we have this third order terms. So, x
0 plus theta h and y 0 plus theta k and this is the result which we want to, we want to prove
for this Taylor’s theorem for the functions of two variables.
293
(Refer Slide Time: 13:49)
And in general also we can extend that expansion in this Taylor’s theorem of one variable
and then we can have for the two variables as well. So, in general we have this result that x 0
plus h y 0 plus k and we can keep on this continuing. So, a squared from the cubic term and
then we have this nth order partial derivatives terms and then this is the remainder term where
we have this general variable this x 0 plus theta h y 0 plus theta k where theta is between 0
and 1. So, this argument here varies from x 0 to x 0 to this x 0 plus h and here also this varies
now from y 0 to y 0 plus k when theta varies from 0 to 1. And alternatively we can also write
down this expression in terms of like f x, y is equal to f x 0, the only difference is here it was
x 0 plus h and now we have x y point here. So, this difference again. So, this h was nothing,
but the difference here x 0 plus h and x 0.
So, here also now the h will become as the difference of x and x 0; that means, x minus x 0.
So, exactly we have this is like h here. So, here also we have this h and because this was just
the difference of this x 0 plus h and this x 0 trump. So, that h was appearing there, but in this
case when we have taken x y point in the neighborhood of x 0 y 0 then this institute of h we
will write down the difference here x and this x 0. So, x minus x 0, here again this y minus y
0 from here and so on so, the rest everything will be the same. Only this h will be like here
the it was theta h so theta x minus x 0 and y 0 plus theta k so this is y minus y 0 term, the rest
everything will remain the same.
294
(Refer Slide Time: 15:59)
So, we have now some problems based on these Taylor’s expansion the first one is find the
quadratic polynomial approximation of the function this fx y is equal to x minus y over x plus
y. And we want to expand this only the quadratic polynomial around this point 1 1. So, we
are not writing here the whole Taylor’s theorem, but only the quadratic polynomial term we
want to approximate that.
So, when we take the in our formula if we just go back. So, up to this one if we write down at
this x 0 y 0 point. So, this is like a Taylor’s polynomial up to this point here and if we include
this one then we call this the Taylor’s theorem or Taylor’s result including the remainder
term. But if we leave this remainder term then this will be the approximation of this f at this
point in which is in the neighborhood of this x 0 y 0 point. So, that is the Taylor’s polynomial
if we fix this n and do not write this r n term.
So, here we are interested in a quadratic polynomial of this function and about this point x 0 y
0. So, we need to get the partial derivative of this with respect to x; that means, we will be
differentiating this with respect to x treating y is constant. So, this quotient rule here the
whole square term there and this term as it is and the partial derivative of this numerator term
with respect to x which is 1 and then we have the minus term x minus y and this partial
derivative of this term with respect to x which is again 1. And when we simplify this so this is
a 2 y term here; so, 2 y over x plus y a whole square and then at this point 1 1 we can
evaluate this, this is 2 over 2 squares this is 1 over 2. So, we have the one over 2 term here.
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Then the partial derivative similarly with respect to y; so, again we will get here minus 2 x
over x plus y square and we can compute this at 1 1 point which will give us minus half.
Similarly, we need because for the quadratic term we need at least we need to go to the
second order term so f xx. So, we have to take the derivative again with respect to x here
treating y is constant. So, the 2 y will be treated as constant.
So, we have 1 over this x plus y whole square and when we take the derivative with respect to
x. So, this will be minus 2 over x plus y cube. So, minus 2 and this 2 y will become minus 4
and x plus y power 3 and again at this point 1 1 this will become minus half similarly the f yy
when we differentiate this with respect to y. So, we will get here 4 x over x plus y cube
whose value at 1 1 is is again half, then we substitute x and y as 1 1. So, we will get half
there.
And the mixed order term so, whether we can differentiate this f x is equal to 2 y over x plus
y square term with respect to y or we can differentiate this term here with respect to x to get
this term 2 x minus 2 y and x plus y power 3 and again we need to compute this at the point 1
1. So, we will get as 0 because when we substitute 1 1 there 2 minus 2 that will become 0.
(Refer Slide Time: 19:35)
So, here we have now the f x we have evaluated at 1 1 which is 1 by 2 f y at 1 1, it is minus 1
by 2 f x x the second order term its minus 1 by 2 and so on. So, the second order polynomial
as I said we will just go up to the second order term. So, we have f 1 1, the partial derivative
with respect to x at 1 1 and then x minus 1 from here to here these are the first order terms.
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So, the f x at 1 1 x minus 1 so, this difference again in terms of h f y 1 1 and then we have y
minus 1 again the difference here.
Then we have the second order term with this 1 over factorial 2, which is 1 over 2 there is a
higher order term f x 6 and then this a h square term. So, x minus 1 is square and this is 2
times, but with that half it will become 1 now. So, x minus 1 x minus y minus 1 term and then
here the second order term with respect to y and the way we have y minus 1 whole squared
term.
So, we can substitute these values here f 1 1 will be 0, because of that function and here f x at
1 1 was half. So, we have half x minus 1 here again we have minus half and y minus 1 1 by 2
and f xx minus half. So, will become minus 1 by 4 x minus 1 whole square, this will be 0
because this mixed order term is 0. So, this will become 0 and then we have 1 over 2, again
here we have 1 over 2 so 1 over 5 and y minus 1 square.
So, this is the quadratic polynomial which is approximating the function in the neighborhood
of this 1 1 point and the accuracy of this polynomial will depend on how far we are from the
point 1 1 if we are in a very close neighborhood of 1 1 this will give us a very good
approximation of the function.
(Refer Slide Time: 21:49)
So, another problem when we have the function here x square plus xy and plus this y square
be linearly approximated by the tailors polynomial. So, it is given here that this function is
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linearly approximated by the Taylor’s polynomial about this point 1 1. And now we want to
find out the maximum error in this approximation at a point in this square here x minus 1 less
than 0.1 and y minus 1 less than 0.1.
So, what you want to discuss here that if we have this 1 1 point for example, and we are
expanding this function around this point by a linear approximation. So, only the linear terms
are considered in the approximation and we want to find out the maximum error in that linear
approximation, if we take any point here in this square around this point 1 1 by this point. So,
this x so, this is 0.1 here this is 0.1, 0.1 or the whole h here is 0.2 from here to here.
So, if we take any point in this neighborhood square neighborhood around this point and what
will be the maximum error in that linear approximation of this function we want to evaluate.
So, in that case we will use make use of the remainder term directly. So, we have f x x y as
we can compute from here it is 2 x the partial derivative of this f with respect to x this would
be 2 x plus y and f y x plus 2 y again the second order terms. So, here will be just 2 and there
with respect to y it will be again two there with the mix term we will have 1 and then the
remainder which is written after this linear term.
So, the quadratic term will be coming the remainder. So, this is exactly the remainder which
we have discussed before and which we can write down in this form. So, x minus 1 whole
square and this will be the second order term f xx 2 times f x minus 1 y minus 1, the mixed
order term and y minus 1 is square and then we will have here again the second order term
with respect to y.
So, all these second order derivatives we have already computed and they are actually
constant in this case. So, we have here 2 and here we have 1 and then 2 again. So, after
substituting this we have this remainder term and now that is given that this x minus 1 is less
than 0.1 and y minus 1 is less than 0.1. So, we can write down those terms here now to get
this approximation that R 1 will be less than equal to we have written down the maximum
error here which is 0.1 this x minus 1. So, 0.1 square here again this product will come and
0.1; so, 3 times this 0.1 square which will be 0.01 03.
298
(Refer Slide Time: 25:07)
The last problems here so we have Taylor’s formula about this 0.00 involving derivatives up
to these third order terms of this function cos x plus y. So, again this is the Taylor’s theorem
including this remainder term and we have considered these third order terms here. So, we
need to compute this f 0, 0 which is in this case 1.
So, a cos 0 will be 1 and then the first order derivatives we have to compute for this f x
simply it is a minus because this cos will give minus sin x plus y and then again at 0 0 this
will be 0. Partial derivative with respect to y will be again sin and this will become 0 there.
So, similarly the second order derivatives here the sin will becomes the cos and then we have
minus cos x plus y which we can evaluate at this 0 0 point and this will give us minus 1.
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(Refer Slide Time: 26:05)
So, the third order derivatives similarly we will get back to again the sin x plus y and for the
third order derivative we need to compute at theta x point. So, at theta x point this will
become as theta x plus theta y and now we can substitute here in this expansion. So, we will
get 1, this was the 0 the first order derivative terms then again here one was the value and
then we have the sin the third order derivatives in this term.
So, we have the sin theta x plus theta y and after the simplification. So, we will get 1 minus
this x plus y whole square this is x plus y whole cube and then this is sin because it was
common in also sin theta x plus theta y.
300
(Refer Slide Time: 26:51)
Well so, we have learned this Taylor’s theorem for a function of two variables and in this
case what we have what we have observed that it is just the extension of these functions of 1
variable. So, this is the point in the neighborhood of this 1 and we can expand this function or
get this value or write down this value in terms of this expansion here.
(Refer Slide Time: 27:35)
So, these are the first order terms here we have the second order terms with h square k square
and 2 hk and similarly we will have the higher order derivatives there. Up to this n and this is
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the remainder term which is often useful to get the error or the estimation of the error in the
approximation. So, these are the references used for the preparation of this lecture.
Thank you very much.
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Engineering Mathematics - I
Prof. Jitendra Kumar
Department of Mathematics
Indian Institute of Technology, Kharagpur
Lecture - 17
Maxima & Minima of Functions of Two Variables
So welcome back to the lectures on Engineering Mathematics I and this is lecture number 17
and today we will be talking about the Maxima and Minima of Functions of Two Variables.
(Refer Slide Time: 00:25)
And in particular we will be talking about the necessary conditions that are required to find
the maximum and minimum values of the function in a certain domain.
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(Refer Slide Time: 00:37)
So, what is local maximum and minimum we will define here. So, a functions z is equal to
f ( x , y) has a maximum at the point ( x0 , y0 ) if at every point in a neighborhood of this point
the function assumes a smaller values then the point itself. So, meaning here we are calling
that the function is taking maximum value at this ( x0 , y0 ) point if at all other points in the
neighborhood of this point function is taking smaller values. So, naturally the function has
attained local maximum at this point and similarly we can define for the minimum also. So, if
a function has a minimum at this point in that case all the points in the neighborhood function
will take larger values than the point itself, then the value of the function at that point itself.
And such maximum or minimum is called relative or local maximum because we are not
talking about or we are not searching the local and maxima, the maximum and the minimum
of the function in the entire domain. But we are talking about at a certain point and checking
its behavior locally. So, if at all other points in the neighborhood of this function of this point
the function is taking smaller values then we call that this is a local maximum and if the
function is taking larger values in the neighborhood of a certain point then we call that this is
a point of local minimum. And these maximum and minimum values together these are called
the extreme values.
2
2
So, here the a plot or the surface of this z=( 4 x2 + y 2 ) e−x −4 y and we can see here clearly. So,
at this point for example, at this point here the function in the neighborhood of this point is
taking a smaller values at that point itself. So, here this is a maximum locally. So, we call this
304
local maximum here also at this point the same situation is happening that all the points in the
neighborhood of this point the function is taking smaller values. So, here also there is a local
maximum of this function.
Similarly, at this point here, but this is other way around that all the points in the
neighborhood function is taking more values than this point itself then the value of the
function at this point itself. So, this is a local minimum.
(Refer Slide Time: 03:31)
So, what is the absolute or the global maximum minimum? So, the smallest and the largest
values attained by a function over entire domain including the boundary of the domain are
called absolute or global minimum or absolute or global maximum respectively.
So, here we are not talking about what is happening locally around a certain point, but we are
talking about the smallest and the largest values attained by the function over the entire
domain. So, in that case if such a value we will call that we will call that this is the absolute
minimum value of the function or the absolute maximum value of the function. And we have
to include naturally the boundaries as well or the boundary of the domain to find out such a
local such absolute maximum or absolute minimum.
So, for example, if we consider this function f ( x , y )=x 2 +2 y 2 and we have this bounded
domain your x2 + y2 ≤1. So, this disk of this radius one and including this 1 so, we have the
305
boundaries there. So, what we have to search now we have to search inside the domain so at
all these points.
So, this is a surface here x2 +2 y 2 so, as we can see somewhere here there will be a local
minimum at this (0,0) that is local minimum, but that will be also a global minimum because
we do not see any other point where the function will take a smaller value than this point and
to find the maximum for example. So, at these points here it seems that the function is taking
the maximum values so at though this point here or that point here.
So, we have to later on mathematically find out what is the where what is the point in the
domain where the function is taking the maximum value or it is taking the minimum value so,
but in this case when we consider the boundaries and find among all these local maximum
local minimum and also at the boundaries we have to check whether the function is taking
somewhere again the maximum value. So, among all these local maximas a local maxima and
minima we have to find the largest the smallest values and then we can claim that this is the
absolute maximum or the absolute minimum or the a function.
(Refer Slide Time: 06:05)
So, there will be another terminology used for used in this lecture there will be critical point
and the saddle points. So, the point ( x0 , y0 ) is called critical point or we also call like a
stationary point of f ( x , y) if the partial derivative of the function f x at this ( x0 , y0 ) point is 0
and f the partial derivative of the function f y at that point ( x0 , y0 ) is 0 or simply they fail to
exists. So, that possibility is also included. So, that is the definition of the critical point or the
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stationary point, and a critical point so among these all the critical points where f x is 0 and f y
is 0 where the functions a critical point where the function has no minimum and maximum is
called a saddle point.
So, basically what we will observe now in the next slides that if the local maximum or
minimum exists. So, that will exist only at these critical points, but there will be some critical
points, where the function is not taking the maximum the local maximum or the local
minimum and those critical points will be called as saddle point. So, just to realize this
situation here so we have many critical points which we will identify again mathematically
little later.
So, this here for example, will be the point where the partial derivative will be 0 with respect
to x and also with respect to y there will be a point here where both the partial derivatives
would be 0, there will be a point here where the partial derivatives will be 0. So, basically the
tangent plane will be parallel to the xy-plane, here also there will be a point where f x and f y
will be 0 and there will be a point here as well where f x and f y will be 0.
So, all these points for example, the critical points and for this will be like the local minimum
here also local minimum there will be local maximum here again this is local maximum. But
at this point if we see that in the neighborhood it is not showing the same behavior because if
we go in the direction of y. So, here the function is taking lower values, but if we take in go
in the direction ofx then the function is taking the more values or the larger values then this
point itself.
So, this point we cannot identify as the local maximum or local minimum though the partial
derivatives here at this point was 0 with respect to x and with respect to y but. So, this is a
stationary point, but this is not a local maximum or local minimum and such a point, such a
point will be called as the saddle point. So, that terminology we will use in today’s lecture.
307
(Refer Slide Time: 09:19)
And then we come to the necessary conditions or condition for a function to have extremum.
So, what are these conditions let f ( x , y) be a continuous and have partial order derivatives at
a point P(a ,b)., then necessary conditions for the existence of an extreme value. So, extreme
value means the maximum and the minimum value or rather the local we are talking about
local maximum and local minimum values of it at this point P(a , b)..
So, these are the necessary conditions what are the necessary conditions that the partial both
the first order partial derivatives. So, with respect to x and with respect to y at that point they
should be 0. So, this is what we are calling necessary conditions; that means, without these
conditions we cannot have the local maximum and local minimum at this point. So, if a point
P(a , b) is a point of local maximum or local minimum, then definitely these partial
derivatives must vanish at that point.
But this is not sufficient to say that yes definitely there will be a if we have a point where f x
and f y both are 0 then we cannot claim that at this point certainly there will be a local
maximum or local minimum. Because this is a necessary condition, necessary conditions
means that this is the first condition we have to check where and or in other words these
points will be the candidates for the local maximum or minimum. They may be there may be
local maximum minimum there may not be a local maximum or minimum. So, if there is a
local maximum minimum we will identify those points local maximum and minimum and
among these which are the critical points as per the definition we have defined.
308
So, at these critical points we will identify if there is no local maximum and minimum that
point will be called as the saddle point. So, this point P is a critical point as per the definition
we have discussed or in other words if a point P(a ,b) is a relative extremum of the function
f ( x , y) then this is a critical point of f ( x , y) because yes as we said that this is the necessary
condition to have the extremum minimum. So, if we have a point a b where the relative
extremum exists of this function then definitely that has to be a critical point because those x,
local extremum and local minimum will exist only on the critical points.
(Refer Slide Time: 12:01)
So, here the necessary conditions again to just to have a simple proof that how do we care
that these partial derivatives must be 0 at these at the points a b if a b is a critical point. So,
we consider this a+hand b+kthis is a point in the neighborhood of the point P(a ,b). So, this
is a neighborhood because h and k can take any value. So, then this a+hand b+k will be the
neighborhood of this point P(a , b). So, there is no restriction on h and k if they can take any
value to have a point in the neighborhood.
Then P this point P(a , b). will be a point of maximum, when it will be a point of maximum?
When the values of the function in the neighborhood of this point are lower; so, it takes the or
the function attains lower values in the point of the neighborhood of this P(a ,b) point. So,
for example, if we define this difference delta f; that means
Δ f =f ( a+h , b+k ) −f ( a , b )
309
So, if it is a point of local maximum; that means, this f ( a, b ) will be bigger will be the large
value then this these values in the neighborhood of this function. So, this value will be less
than equal to 0 if this is a point of maximum for all sufficiently small h and k. So, whatever h
andk we take in the neighborhood of this ( a, b ) point. So, this value will be less than equal to
0 or in other words there will be definitely a neighborhood of this point ( a, b ) where this Δ f
will be less than equal to 0 for all values of h and k.
And this point will be called a point of minimum if this Δ f will be greater than equal to 0.
So, in other way round that this f ( a, b ) will be a lower value than this f ( a+h ,b+k ) for all
sufficient values of h and k.
(Refer Slide Time: 14:19)
So, if we use this Taylor series expansion about this point (a , b). So, what will happen here
f ( a+h , b+k ) will be equal to f ( a, b ). So, we have already discussed this in the last lectures.
So, h f xand this is the first order derivative here the second order derivatives at (a , b) and so
on we can write down this expansion about this (a , b) point as long as these derivatives exist.
And this noting that this Δ f which we have defined in the previous slide the difference of this
f ( a+h , b+k ) −f ( a , b ).
Δ f =h f x ( a , b ) +k f y ( a , b ) +
1 2
h f xx +2 hk f xy + k 2 f kk )( a, b ) +…
(
2
310
So, for sufficiently small h∧k , the sign of Δ f will depend on the sign h f x ( a , b ) +k f y ( a , b ).
So, what we will also see in the detail in the next slide that the sign of this expansion here in
the right hand side the sign of this Δ f basically we will depend on the sign of this first term,
these are the leading terms here in terms of h and k. So, therefore, sufficiently small h and k
the sign will be determined by the sign of this term, if this is a positive term Δ f will be
positive. If this is a negative term Δ f will be negative irrespective of whatever we have here
because these are the higher order terms in terms of h square there is a product h and k. So, at
least we can find small h and k here sufficiently small so that the sign will be determined by
this factor only in the close neighborhood of this point (a , b).
(Refer Slide Time: 16:11)
Let us explore this little bit more. So, we have this
Δ f =h f x ( a , b ) +k f y ( a , b ) +
1 2
( h f xx +2 hk f xy+ k2 f kk )( a, b ) +…
2
Letting h → 0 , here because we are a still neighborhood we are we are just Letting h → 0 , and
then they will be k in the direction of y. So, if we let this h tends to 0 we are still in the
neighborhood of this (a ,b) point, then what we are getting here
1
Δ f =k f y ( a , b ) + k 2 f kk (a , b)+…
2
311
So, let me just again repeat the argument we had in the previous slide. So, note that the sign
of this Δ f will depend on again this leading term here of this k, the rest all these terms are k
square k cube and so on for very small k these terms will be like negligible and the sign will
be determined again by this term just for example, we take an example here that you have h
and this expression here h minus 1000 h square minus this 2000 h cube.
So, what we will notice here the sign will be determined by this h only not by these terms
here, but we have to go to a sufficiently small h close to 0 to see that this will be determined
by the sign of this one. For example, we take h is equal to 0.1 this is too large value and we
are getting this negative number because these are the dominating term for this h 0.1. But if
you go pretty close to h to 0 for example, h is equal to 0.1. So, this is a still negative we have
taken a much smaller h now. So, this is a still negative because these are the last term, but
what we see here now we have taken this point 0001 and in that case now we can see that the,
since h is positive.
So, this value of this expression is also positive and that is a point here. So, for sufficiently
small h the sign of this term here; however, large these numbers are the sign will be
determined by the first term which is which one can see here. So, same argument we are
making here that there will be a sufficiently small k for which the sign of this expression will
be determined by this first term which is k f y ( a ,b )and that is the point here. So, if we assume
this f y ( a , b )is positive because either this is a derivative, partial derivative with respect to y at
( a, b ). So, it has it must have some determined sign. So, if it is positive in that case this delta f
will be positive for k positive and this delta f will be negative when we have k negative. So,
this k we are moving in the direction of.
So, for example, this is h is equal to 0 and k is equal to 0. So, this we already let this h to 0
here. So, we are in only the neighborhoods points are in this direction of y. So, either when k
positives, we are in the neighborhood of upper side when the h k is negative we are in the
neighborhood somewhere here. So, then we have that for k positive. So, in the neighborhood
a point where Δ f is greater than 0 there is a point in the neighborhood where Δ f is less than 0
and if you assume again if Δ f is less than 0 because we do not know what is f y at this point
(a , b). But if it is negative then we have this Δ f less than 0 and delta f greater than 0 again in
the points in the neighborhood. So, what we observe here that whether the f y is positive here
or f y is negative this Δ f is changing its sign, the Δ f was the difference between the value at
312
the point (a , b) and the value in the neighborhood. So, this is changing time, but changing its
sign, but to have the local maximum or local minimum it should not change its sign, because
then only we can determine this is a point of local maximum or local minimum when all the
points in the neighborhood it behaves us with the same sign.
So, here this Δ f is positive Δ f is negative for some points in the neighborhood and that will
conclude that the function cannot have an extreme value extremum here, unless this f y is 0
because k is we are moving in the neighborhood of this point. So, if this f y is not 0, if f y is
positive it is changing sign if f y is negative it is again changing sign, but if this is a point of
local maximum then this should not change its sign and for that the f y must be 0, because if
f y is not 0 then we have observed that it is changing sign. So, this is necessary that f y has to
be 0 otherwise it will not be a point of local extremum. So, moving next now similarly what
we can do.
(Refer Slide Time: 21:31)
So, we had this expression already we have seen; now similarly we will add that k tends to 0
and then we will find that Δ f changes sign for h. So, here for example, if we take f x positive
and then we will realize that this Δ f is positive and then when h is negative Δ f is negative
when we let this (h , k) tends to 0. So, this term is anyway is removed here and this all these
terms were the k is there we will go to 0. So, we have h f x ( a , b ) and so on.
313
Δ f =h f x ( a , b ) +k f y ( a , b ) +
1 2
h f xx +2 hk f xy + k 2 f kk )( a, b ) +…
(
2
So, here assuming this f x is less than 0. So, the sign will depend on now this h. So, if h is
positive with theΔ f is negative h is negative then Δ f is positive. So, again the same
argument what we have observed that in the neighborhood of again of this point (a , b) point
this is changing its sign, whether if Δ f is positive it is changing sign if Δ f is negative it is
again changing sign.
So, again to have that this (a ,b) is a point of local extremum we must have that this f x is
equal to 0. So, this f x must be equal to 0 earlier we have observed that f y must be equal to 0.
So, we can conclude that the necessary conditions so here the necessary conditions for the
existence of an extremum at this point a b is that f x should be 0 and also the f y should be 0.
These two derivatives must be 0 this is what we have observed, if these are not 0 this Δ f was
changing sign and then that point cannot be a point of local maximum or local minimum.
(Refer Slide Time: 23:31)
So, now let us just go through this problem number 1. So, we here we will find all the critical
points of this function f ( x , y)=x 3 + y 3−3 x−12 y+20. So, what we have to do now to find
the critical points, we have to compute the partial derivatives with respect to x and with
respect to y and set them equal to 0 and out of these two conditions we will find out how
there are how many points which satisfy f x ( x , y)=0∧f y ( x , y )=0.
314
So, this f x ( x , y) is equal to 0 and f y (x , y) is equal to 0 will give us; so, here the 3 x2 −3=0 to
find the critical points similarly when we take f y (x , y) here. So, we will have 3 y2 −12=0.
So, out of these what we get so from this first condition we are getting like x2 is equal to 1.
So, x is ±1 and from the second condition we are getting y 2 is equal to 4 and this implies y is
equal to ± 2. So, we have all these points when x is equal to ±1 and y is equal to ± 2these f x at
( x , y) all these points this is 0 and f y is also 0.
(Refer Slide Time: 25:13)
So, these are the critical points now here we can take one and then 2 then we have ( ± 1, ± 2 ).
So, all these four possibilities are there and if we can see here in the surface of this function.
So, there are the points for example, 1 and 2. So, here in x we have 1 and then we have 2
there. So, here there is a point where there will be a critical point, again here the +1 and then
we have a −2.
So, there will be a point here and there will be −1 and +2. So, there will be a point here and
(-1, -2) there will be a point here. So, all these 4 points are there which are the critical points.
So, here the function derivative with respect to x and with respect to y both are 0 at these
points.
315
(Refer Slide Time: 26:15)
So, now moving to the next problem here again we will find all the critical points of this
−x 2− y 2
function, f ( x , y )=( x2 − y2 ) e 2 . So, again proceeding with this f x=0 . So, we will get what
is f x=0 in this case. So, we are differentiating this. So, we will get this x square minus y
square and the derivative of this term again the same x square minus y square by 2 and then
there will be a term minus 2 x divided by minus x plus the derivative of this with respect to x
and then we have e power minus x square minus y square by 2 term.
So, if we take this e power minus x square and minus y square by 2 term common and also
we can take this x common. So, here we will have 2, then here we will have minus x square
minus y square yeah. So, and this cannot be 0 this exponential of minus x square minus y
square. So, we will have that this is equal to 0 this is the condition precisely here that this x
multiplied by 2 minus x square minus y square must be 0
¿
316
(Refer Slide Time: 27:33)
So, out of this condition and now for f y also similarly we will get ¿. So, now, we can solve
this. So, for example, we said here we take here x is equal to 0. So, from this first equation if
we take x is equal to 0 irrespective of y there this f x will be 0. And if we take x is equal to 0
and in this case we will get minus 2 and then plus this y square. So, this x is 0 and then we
have y is equal to 0.
So, from here we got y is equal to 0 and y is equal to ± √ 2. So, corresponding to x is equal to
0 we have the point x 0; so, here ( 0,0 ) and 0 ± √ 2. Now, similarly what we can do we can set
like here y=0 first and then because this for y is equal to 0, this will be also 0 and then
setting here y is equal to 0 you will find the values possible values of x. So, in this way we
can find all the critical points by solving these equations.
317
(Refer Slide Time: 28:49)
And what do we get? So, we have the critical points is ( 0, 0 ) , ( ± √ 2, 0 ) , ( 0 , ± √ 2 ). So, all these
are the critical points which we can see here in this problem.
So, these are the critical points. So, we can see like this one this one this one this one and
again here there is a critical point. So, these are the critical points of this problem.
(Refer Slide Time: 29:23)
So, what is the conclusion now, we have found the necessary conditions for the extrema
f x ( a , b ) =0 and f y ( a , b ) =0 should be 0 if this point a b is a point of local maximum or local
318
minimum and these critical points are the candidates for the local extrema and the saddle
points. So, because the saddle points are those critical points where the local extrema do not
exists. So, these are the points called saddle points.
So, we have already seen that this is for example, the point of local extrema here, we have a
local maximum. Here also local maximum these are the points here with local minimum and
there is a point at this place here where we do not have a local maximum or minimum and
then this will be called a saddle point.
In the next lecture we will learn now, how to identify whether it is a local maximum or local
minimum because necessary conditions will give us the candidates for the local extrema and
also for the saddle points, but then we have to identify whether a given point a given critical
point the function is taking local maximum local minimum or it is a saddle point. So, that will
be the content of the next lecture.
(Refer Slide Time: 30:45)
These are the references.
Thank you.
319
Engineering Mathematics - I
Prof. Jitendra Kumar
Department of Mathematics
Indian Institute of Technology, Kharagpur
Lecture – 18
Maxima & Minima of Functions of Two Variables (Contd. )
So welcome back to the lectures on Engineering Mathematics 1 and this is lecture number 18
will be talking about the Maxima and Minima Functions of two Variables.
(Refer Slide Time: 00:25)
In particular today we will be talking about the sufficient conditions and that will be used for
getting the or characterizing the point, the critical points whether it is a point of local
maximum or local minimum or it is a saddle point.
320
(Refer Slide Time: 00:43)
So, in the previous lecture we have seen that a point ab will be a point of local extrema, if this
delta f which is the difference between the function value at the neighborhood points of this
ab and minus this the function value at ab point. So, if this does not change its sign for
sufficiently a small h and k then, we call that then this has a local maxima or local minima,
but if it changes its sign then; that means, the point is a saddle point.
So, in that case we try to get the behavior of this delta f or rather the sign of this Δ f by using
the Taylor series expansion of this function f ( a+h ,b+k ) −f (a ,b) around this ( a, b ) point and
from where we got the necessary conditions that to have that this point ab is a point of local
maxima or minima, f x ( a , b ) has to be 0 and f y ( a ,b ) has to be 0. So, which we have observe
that, there are for example, here 5 points, which satisfy these conditions, so there is a point
here and there is a point there and also there are a total 5 points here.
And now in the this lecture, we will identify whether for example, here we can see that this
looks like a point of local minimum, here also its a local minimum, these points are local
maximum, but at this point here in the neighborhood of we see if we go in this direction, in
the direction of this x, then the function is increasing if we go in the direction of y the
function is decreasing. So, this point is a point of it is not a point of local x, local minimum or
it is not a point of local maximum, but it is a saddle point.
So, now mathematically we will identify all these points with the help of the second order
derivatives.
321
(Refer Slide Time: 02:59)
So, here this is the sufficient condition. So, we will just for simplicity we use these notations :
r=f xx ( a , b ) , s=f xy ( a , b ) ,t =f yy (a , b).
And then if this function f ( x , y)
let us assume that is a continuous and have continuous
second order partial derivatives at this point P(a , b) and if this point P(a , b) is a critical
point then at this point P(a ,b) there will be local maximum, if rt−s2 >0 and r <0. This
point will be a point of local minimum, if rt−s2 >0 and r >0, saddle point if this rt−s2 <0 and
the test will fail if thisrt−s2=0 and then we have to find some other ways to characterize the
behavior of this function at this point P(a , b) .
So, here as written there if P(a , b) is a critical point so definitely, because this is a necessary
condition to discuss that, this point is a local point of local maximum minimum or a saddle
point. So, definitely this is a critical point and then we discuss with the help of the second
order test here, mainly this rt−s2 the value of this expression we can find out whether it is a
point of local maximum minimum saddle point and in this situation when rt−s2 is 0 this just
fails.
So, let us prove this result, how do we get these expressions.
322
(Refer Slide Time: 04:47)
So, we consider again this Δ f because, finally we need to get based on these sign of this Δ f ,
the behavior of this f in the neighborhood of a point. So, which we have already seen that the
Taylor series expansion of this will lead to this h, the first order terms and then the second
order terms there.
Δ f =h f x ( a , b ) +k f y ( a , b ) +
1 2
h f xx +2 hk f xy + k 2 f kk )( a, b ) +…
(
2
Since this (a ,b) is a point of is a critical point, so the f x ( a , b ) and f y ( a ,b )will will vanish and
then we have this Δ f is equal to these second order terms, so h2 f xx and so on.
Δf=
1 2
( h f xx +2hk f xy + k 2 f kk )( a, b ) +…
2
So, now, the behavior of this Δ f , we have to get based on these second order derivatives
terms or the first this leading term here, the rest all the terms in the expansion will be higher
order terms.
So, with our notation we have used this f xx is r and this f xy is s and this f kk is t at this point
( a, b ). So,
1
Δ f = ( h2 r +2hk s+k 2 t ) +…
2
323
(Refer Slide Time: 05:55)
So, now we will consider that or we will assume that r ≠ 0, here we can also assume that if
this r=0in case then we can assume that t ≠0. So, at least one of them is 0 then we can
proceed in this way. In the situation when both are 0. So, suppose we have a r=0 and we
have r=0, so in that case naturally we cannot proceed in this way, but in that case directly
from here we see that if these two are 0 r=0 and r=0.
So, s may be 0, s may not be 0, so if s is 0 again the second order test will not give anything
and we have to go for the higher order derivatives. So, suppose this s is not equal to 0 then
what do we get here? We get this two hk and s term. So, s may be positive, s may be
negative, let us just assume that s is positive the same thing we can argue when s is negative.
So, we have this two hk terms, when this product h and k whether if both are positive for
example, of both are negative this hk term is positive. So, we have this Δ f , the leading term
is positive and we have already discussed in the last lecture that this leading term will decide
the sign of this Δ f . If it is positive then the sign of this Δ f have in the neighborhood of this
(a , b) point will be positive, if this leading term is negative then the sign will be negative.
So, here this hk if this product is positive, we have this Δ f positive, if this h is negative and k
is positive or k is negative h is positive, we have Δ f negative. So, means this Δ f is changing
its sign in that case and this will be a saddle point definitely, which will be also concluded
from this result, which I have shown you before. So, we do not have to discuss this
separately.
324
So, let us assume that either r is 0 or t is sorry, r is not equal to 0 or t is not equal to 0. So,
here we assume r is not equal to 0 and now, proceed so same thing we can do when t is not
equal to 0.
(Refer Slide Time: 08:17)
So, let us assume this r not equal to 0 we can divide by r and multiplied by r. So, we got this
expression here and this Δ f 1 over 2 r. So, in this case we have added here k 2 s2 term and also
subtract to this k 2 s2 term plus this k 2 rt plus the higher order terms.
Δf=
1 2 2
( h r +2 hk rs+k 2 rt ) +…
2r
Δf=
1 2 2
( h r +2 hk rs+k 2 s2 −k 2 s2 + k 2 rt ) +…
2r
Now, the first 3 terms will make a whole square; that means,
Δf=
1
( ( hr+ ks ) 2−k 2 s2 +k 2 rt )+…
2r
So, here from the last 2 terms also we take k 2 common.
Δf=
1
( ( hr+ ks ) 2 +k 2 (rt −s2 ) )+…
2r
(Refer Slide Time: 09:11)
325
So, this is the expression Δ f =
1
( ( hr + ks ) 2 +k 2 (rt −s2 ) )+…. So, we will consider this case 1
2r
now, where we assume that this rt−s2 >0, that is the case 1. So, in this case what will happen,
when this is positive here then we have this whole square term and we have this k 2 term. So,
we have to now see how this Δ f > 0if r >0.
So, r >0 we further assume that r <0, r >0 it is the second order a derivative at this (a , b)
points. So, depending on that ab and the function but it will be a definite sign whether it will
be positive or negative. So, if this r >0, we will see now here that Δ f will be positive why?
So, if we have this expression here this is positive, so this one is strictly positive. Now for the
neighborhood either h will be 0 then k will be non-zero or if k is 0 then h has to be non-zero,
both cannot be 0 because both 0 means we are at the point ab and we are looking at the
neighborhood point. So, here for example, we have this (a , b) point. Now in this
neighborhood of this point, either in this increment in this direction is denoted by h increment
in the y direction was denoted by k.
So, to have a point in the neighborhood one of them has to be non-zero both of them have to
be non-zero. So, if both are non-zero means this k is non zero, then we have a positive term
here this strictly positive term; does not matter what is this term here, we have the overall
positive number here. So, in that case it is fine. Suppose this k=0, so if if k=0 then h has to
be non-zero, h has to be non-zero to have in the neighborhood.
326
So, in that case this term is 0 and here the k is 0. So, we have h2 r 2. So, again this h2 r 2., we
have the positive term, because h cannot be 0. So, in any case whatever the situation is, if this
2
rt−s is positive then, this Δ f will be positive for r, positive and when r is negative just the
sign will change because this r is sitting here; otherwise the rest everywhere we have the s
square there.
So, what do we get now that this Δ f is positive if r is positive 0 and this Δ f is negative when
r is negative.
(Refer Slide Time: 12:03)
So, we can conclude now that we have the same sign because r will have either positive or
negative. So, if r is positive Δ f is positive and then we have this point is a local minimum in
this case and when r is negative this point will be a point of local maximum when this rt−s2
is positive and r is negative because having this Δ f a negative means that, this (a , b) point is
taking more larger values than the points in the neighborhood at; that means, this point is a
point of local maximum.
327
(Refer Slide Time: 12:49)
Moving further to the case number 2, when we take this rt−s2 <0what will happen this case,
when this rt−s2 <0is less than 0? So, we have to again carefully look at, so we take this
possibility that is let this k → 0 & h ≠0 as I discussed before that one has to be non-zero. So,
you are letting at k to 0 and the h non zero. So, k to 0 means this is 0 and h is non-zero. So,
from here, we will conclude that this Δ f > 0if r >0. So, let us assume here r >0, r <0. So, this
Δ f will be negative in that case nothing else will change.
So, here r >0, so this Δ f is positive because, h is non zero and k is 0. So, we will have here
2 2
h k term, which is positive and then the second observation we will take let us take k is not
zero. So, k is not zero and this rt−s2 <0.
So, we have something negative sitting here now and we choose this h now such that ( hr +ks )
is equal to 0. So, we choose our h, so that this relation holds for given k whatever is small
you can take, you can we can choose this h here as
−ks
−ks
. So, we choose h as
for any
r
r
value of k.
So, on this point here, when h is chosen from this
−ks
for whatever k, this first term this
r
hr+ ks=0 because, we have chosen our h in such a way that, this hr + ks=0. So, this term is 0
and here this k 2 is positive, but this is negative this is less than 0. So, the overall, this first
328
term for r positive, which we are considering at this moment, this will be less than 0, Δ f < 0
because of this term, there is no first term if we are in the neighborhood which satisfies all
these points.
Then this Δ f < 0 if r >0 and we are not restricting for the neighborhood that we have to be a
far at some point where this relation holds, we you can be as close as to this (a,b) point by
choosing this hr + ks is equal to 0 for any small value of this k and not equal to 0.
So, what we have realized here that the Δ f > 0if r >0, Δ f < 0if r >0. So, we assume if we take
r <0 then naturally, this Δ f > 0. So, the point is that this Δ f is changing its sign in the
neighborhood and that is exactly the case of the critical point. So, this is the condition if this
2
rt−s <0 then we can conclude immediately that this will be a point of this will be a point of
a saddle point.
So, this is a saddle point, it is not a point of maximum or minimum because this Δ f , the sign
of this Δ f depends on h and k. So, there are points in the neighborhood where the Δ f is
positive and there are points in the neighborhood where this Δ f is negative. So, it is changing
sign in the neighborhood of this (a,b) point and; that means, this is a saddle point.
(Refer Slide Time: 16:35)
The third situation we will take when this rt−s2=0. So, in this case there is no the second
term here rt−s2=0. So, the behavior will be discussed with the help of the first term ( hr +ks )2
329
. At a first glance, we will see that this ok, this is a positive greater than equal to 0 term
hr + ks is equal to 0.
Δf=
1
2
( hr+ks ) +…
2r
It ( hr+ks )2 so this is definitely greater than equal to 0, but there is a possibility of having
equal to 0 because if this term the second order term becomes 0 then we cannot identify the
behavior of this Δ f because, then the behavior will be determined from the other higher order
terms which can make this Δ f to negative as well because, if this term is 0 the next term will
decide the sign because that might be the case that Δ f is negative. So, we cannot conclude
out of this immediately by having this that this is here greater than equal to 0.
Now, we have to have a strict sign here then only we can say this will be the whole Δ f will
be of that sign. So, here if we can prove that this is strictly greater than 0 then it is fine, we
can conclude about the point but here this is now greater than equal to 0 why? Because if we
choose our hr and ks in the neighborhood such that this hr + ks is 0. So, in that case at all
those points in the neighborhood, this term will become 0 and the behavior will be
determined from the next higher order terms.
So, let us write down this here. So, if we take this h and k such that; this hr + ks is 0, that
means, hr =−ks. So, all these are the points in the neighborhood where Δ f is 0 plus this third
order terms, then the second order terms of the right hand side vanish and then therefore, the
conclusion will depend on the higher order terms. So, we cannot conclude anything in this
situation about the sign of this Δ f , it may be negative, it may be positive. So, one has to find
some other ways to investigate such points.
330
(Refer Slide Time: 19:05)
So, what is the working rule now for investigation of local extrema
•
Find all critical points f x=0∧f y =0
•
For each critical point, evaluate r=f xx ,s=f xy ,t =f yy
•
Identification
 If rt−s2 >0∧r <0
maximum
 If rt−s2 >0∧r >0
minimum
 If rt−s2 <0 Saddle point
 If rt−s2=0 Test Fails
331
(Refer Slide Time: 20:09)
So, we will consider this example
Example: Find all critical points of f ( x , y )=x 3−6 x2 −8 y2 and investigate their nature for
local maximum/minimum and saddle point.
So, we will compute first all the critical points and for each critical point we will investigate
that rt minus s square term to discuss the behavior of those critical points in the
neighborhood. So, here we have the critical points now the f x=0 and f y=0, so what do we
get, f x=0 is what is f x=3 x2 −12 x.
So, in this case this will be set to 0 and this f y will be −16 y is equal to 0. So, from here we
will get 0 and from the first equation we will get 3 x( x−4), so is equal to 0. So, we will get x
is equal to 0 and 4 from the first equation, from the second equation we will get y is equal to
0. So, we have the (0 ,0), we have the (4,0) these are the critical points.
332
(Refer Slide Time: 21:27)
(0 , 0) and (4 ,0) these are the 2 critical points of the problem and in this case, now we will
compute the behavior of the of this rt−s2, the sign of rt−s2 at all these points. So, (0 ,0)
point this second derivative which we have to be computed here. So, this f x=3 x2 −12 xx and
then f x x. So, f x x will be 6 x−12 and this f x y will be 0 because there is no term of y here. So,
f x y will be become 0 and this f y was−16 y , so, this f yy will become −16.
So, based on these 3 now, we will compute these r. So, f x x at 0 0, so will be −12 will be −12
and then s is 0, so this is 0 and t here this is t. So, t is −16. And at this ( 4 ,0 ) point, so r here,
so 6 × 4, so 24−12. So, this will be 12, s is 0 and fyy is constant here minus 16.
So, if rt−s2 this is a product of these two negative number we have 192 and in this case we
have −192. So, here when we have this rt−s2 >0∧r <0. So, this is a point of local maximum.
So, (0 , 0) is a point of local maximum and in this case this is rt−s2 is negative less than 0.
So, it will be a saddle point, which we have discussed in the in the sufficient conditions.
333
(Refer Slide Time: 23:39)
So, coming to the conclusion here, so what are the necessary conditions we need to know the
critical points because, these critical points are the candidates for the local maximum and
minimum. So, we have f x=0∧f y =0 that will gives give us the necessary conditions for the
extrema. So, these points will be the candidates, which we need to investigate for the
behavior the local behavior of the function at that point.
So, the sufficient conditions help us to identify these points for local maxima, minima or the
saddle point. So, for that we need to compute If rt−s2, if it is positive and r is negative in
that case, this comes to be a local maximum. If this If rt−s2is positive and r is also positive
then, such point will be a point of a local minimum and if If
2
rt−s is less than 0 then this
will be a saddle point. So, it is not a point of maximum, it is not a point of minimum and it is
a saddle point as per the definition we use.
And here If rt−s2 square will be 0 in that case we cannot identify the behavior of this point
and then this test at least the second order test fails and we have to find some other ways;
either they the higher order terms we have to investigate or we have to directly investigate the
behavior of this function at that point, but it is certainly needs further investigation.
334
(Refer Slide Time: 25:15)
So, these are the references we have used to prepare these lectures and in the next lecture now
we will see more such examples to identify the local the behavior for the local maxima
minima and the saddle point.
So, thank you very much for your attention.
335
Engineering Mathematics – I
Prof. Jitendra Kumar
Department of Mathematics
Indian Institute of Technology, Kharagpur
Lecture – 19
Maxima & Minima of Functions of Two Variables ( Contd. )
So, welcome back to the lectures on Engineering Mathematics I and this is lecture number
19. And, today we will continue our discussion on Maxima and Minima of Functions of Two
Variables.
(Refer Slide Time: 00:25)
And in particular today we will see some typical problems where, we will apply the idea
which was discussed already in previous lectures.
336
(Refer Slide Time: 00:37)
So, just to recall in the last lecture, we have investigated the local extrema and that was the
sufficient conditions and necessary conditions. So, we need to find first all the critical points
and those critical points we will get by solving these equations. So, the partial derivative of f
with respect to x will be set to 0 and partial derivative of f with respect to y will be set to 0
and then we will solve these two equations to get all the points which satisfy these equations.
So, those will be the critical points and then for each critical point, we will evaluate this is the
notation r we have used for the second derivative with respect to x at each critical point and
then s here the mixed derivative and twhich is the 2× y derivative of this a function f .
And then for the identification we have realized that if this rt−s2, so rt−s2, this is positive
and this r is negative, then we have the point of a maximum. And if at a point if we have
again this rt−s2 positive and ris positive, then this is a point of minimum. Similarly when
this is negative rt−s2 then, this comes out to be a saddle point and if this rt−s2 is 0 then, this
test of the second derivatives this fails and we need to go for further investigation by some
other ways.
337
(Refer Slide Time: 02:19)
So, let us discuss the problem here. So, we want to find the local extrema of the function
f ( x , y )=( 4 x2 + y2 ) e−x −4 y . So, we need to compute the f x. So, f x is simply if we differentiate
2
2
this with respect to x keeping y constant, so we will have this first term let us keep it as it is it
2
2
is a product rule and then here we will differentiate. So, e−x −4 y and then the derivative of this
2
2
−x −4 y , which will be −2 x with respect to x and then plus.
2
2
So, here this is 8 x the derivative of this first term 8 x and then here e−x −4 y . And then this
2
2
term we can take common with this x here; in fact, this 2 x we can take common and e−x −4 y ,
then from this first term, we will get this 4 x 2− y2. So, with minus sign because there was
minus there and then here we have taken already 2 x and this exponential function, so we will
get simply 4 here. So, that is the first derivative of this function with respect to x which is
written here. So, 2 x and this exponential function 4−4 x2− y 2.
2
2
f x ( x , y ) =2 x e−x −4 y ( 4−4 x2 − y 2 )
338
(Refer Slide Time: 04:05)
Similarly, we can get the first order derivative with respect to y. So, in now we will
differentiate here with respect to y and again the product rule will be applicable. So, we will
get this 2 ynow and here the same exponential function and that the extra term 1−16 x 2−4 y2.
And the critical points we can now get by solving these two equations. So, if we set these
equations to 0, in that case we will get from this first equation because exponential cannot be
0. So, either x has to be 0 or this term in this bracket has to be 0. So, let us assume that x is 0.
So, when x is 0 f x at whatever point along this x is equal to 0, for whatever y this will be 0.
So, we have x is equal to 0there which can set to this f x and now when x is 0, so from here
we get two possibilities to make this term 0; either y will be 0, in that case also this term will
be 0 or here when we set x to 0 we are getting this 1−16 x 2−4 y2 is equal to 0 because x is
said to be 0 here.
So, we have 1−4 y2 0, where we will get 1 is equal to ±1/2. So, in this case, when we set
x=0 there we have two possibilities from the second equation to set to 0; either y is equal to
0, we will make this 0 or along with this x=0, if we take y is equal to ±1/2 that will also be
0 . So, basically we are getting the three points in fact, so was a ( 0,0 ) is 1 point and then
1
1
point and 0 ,− .
2
2
( )
0,
(
)
339
So, there are three points here which can make this f x and f yboth 0 with this possibilities and
now we have another one. So, if y is 0 from the second equation which is making this
derivative 0. From the first now we can get when we set y to 0 here, so we will have 4−4 x2
is equal to 0, so we will get x is equal to ±1. So, these are the other points with y=0, so we
have (1, 0) and (−1, 0). So, on other points will be ( 1 , 0 ) and then we will have (−1 ,0 ).
And now if we try to make this and this term 0, so what we will get from the first equation,
we are getting 4 is equal to 4(x 2 + y 2), well from the second equation we are getting
2
2
2
2
16 x +4 y correct, 16 x +4 y . So, if we take a close look, so here if we take this 4 common,
so will be 1 by 4 there in the right hand side, this 4 x 2 + y 2. So, we cannot get any solution out
of these 2 equations because first equation says 4 x 2 + y 2 is equal to 4 while the second says
that 4 x 2 + y 2 will be 1/4.
So, we cannot get a solution out of this system. So, we got all these points one was (0 , 0) and
(0 , 12 ) ,(0 ,− 12 ), (1, 0 ) , (−1, 0 ). So, at all these points we need to further investigate. So, as
discussed, so we have these 5 critical points or the points where both the derivatives vanished
and now we have to discuss for each critical point that whether it is a point of local minimum
or it is a point of local maximum or it is a critical point. So, that there we have to use the
sufficient conditions that were discussed before so, moving a next now.
(Refer Slide Time: 08:31)
340
So, we have this f x and we need to get the second order derivative to further investigate these
points for the extrema. So, to again we have to get the derivative of this function, so once
again we have to differentiate this with respect to x and then we can get this term because
there will be now many terms. So, if we can I will show you just for this one the next we will
directly write. So, here again the product rule will be applicable.
2
2
So, we will do here 2 e−x −4 y , we will take as the first function and then the derivative of the
second with respect to x. So, that will be 4 and then we will have minus here a 12 x2 and then
here − y 2, so with respect to x and then plus, so here this 2 times. Again this term will remain
2
3
4 x−4 x and −( xy ) and we will differentiate this first term, so which will be e−x −4 y and
2
2
then −2 x term when we differentiate this −x2 −4 y2.
2
2
So, in this case we will take this common 2 e−x −4 y , from here we have 4−12 x2 − y 2, here we
have taken this common here 2 x. So, we will get this −2−2. So, here this is 4, so −8 x then,
we will have here 8 x3 and then we will have this +2 x 2 y2and this is already taken care well.
So, here the x was there, so we have x2 term and then here since this is x, so x 4 will be there
and this x will make this x2 y2. So, this is the term here, we have the 4, then we have a x2
term. So, 8 x2 and then this is 12 x2, so which makes 20 x 2 then we have x 4. So, 8 x4 , we have
2
2 2
− y and then we have this 2 x y term along with this 2 e−x −4 y . So, this is the derivative, the
2
2
second order derivative with respect to x, which is already written there. And then we need to
compute the other derivatives as well, so the mixed order derivative; so, with respect to x and
y and then also the derivative with respect to y 2 times.
341
(Refer Slide Time: 11:41)
So, here we have the f y from the previous slide and then we can differentiate this with respect
to x, here with the idea which we have just discussed above. So, we will get this term here
and then we need to differentiate this once again with respect to yto get the f yy the second
order derivative with respect to y. So, this is f y and when we differentiate this to get this t we
will
get
this
term
now
with
exponential
function
and
this
expression
2
2
2 2
4
1−20 y −16 x −128 x y +32 y .
So, we have three equations now, this is for three expressions, so here for the r and then we
have the s the second order derivative the mixed derivative and then we have the second
order derivative with respect to t. So, with the help of these derivatives we will investigate
further whether this point is a point of maximum or it is a point of minimum or it is a saddle
point.
2
2
r=f xx ( x , y )=2e−x −4 y ( 4−20 x 2 +8 x 4 − y 2 +2 x2 y 2 )
2
2
s=f xy ( x , y )=4 xy e−x −4 y ( −17+16 x 2 +4 y 2 )
2
2
t=f yy ( x , y ) =2e−x −4 y ( 1−20 y 2−16 x2 −128 x 2 y2 +32 y 4 )
342
(Refer Slide Time: 12:53)
So, now as discussed we have this r , s and t these three expressions, so we will now identify
these points. So, the first point remember it was ( 0,0 ). So, at this ( 0,0 ) point, we will compute
r , s and t. So for r, this is when x and y both have 0. So, we have 0 term there, so we have 4,
this e 0 this becomes 1. So, we have 4× 2=8, so r is 8 and this when ( 0,0 ), we substitute here
in s, so this these terms are 0 now, so we have −17, but here the x y is sitting in the product,
so the s will become 0 and in this case again you will have some non-zero numbers. So, all
these terms will be 0, we have 1 there and 2 multiplied by 1. So, this will be 2.
So, we have at this point r is 8, s is 0 and t is 2 at the ( 0,0 ), this we have taken the first
critical point among all these critical because, for each we have to identify whether it is a
point of a maximum, minimum or a saddle point. So, this point we have computed all these
higher order derivatives and then we have to compute rt−s2. So, r and t this product is 16.
So, we have the 16, which is positive. And remember in the sufficient conditions we have
seen that if rt−s2 is positive, when r is positive then this is a point of local minimum. So, we
got this point of local minimum. So, ( 0, 0 )the function has a local minimum and now we will
consider the other points.
343
(Refer Slide Time: 14:45)
So, for the second point this was second and third points, so 0 plus half and 0 minus half.
Since this y appears in all these expressions in the even power. So, we can deal this together
because, so whether we take the plus sign there or the minus sign the y power even will have
the same number.
So, we can consider this to these two points together, so we are now discussing point number
2 and point number 3, so 0 and ±1/2. So, in this case we need to again compute this r, so x is
0, so these terms, so what we will get here 2 times. So, r will be 2 times and then we have
exponential function x is 0 and y 2 is 1/4.
So, here you will get e−1 and then here we have 4 then minus this is 0, here also 0 x is there
only this y 2 term will survive. So, here we will have 1/4 and then here again x that will
become 0. So, what do we get here, e−1 and then here 15 /4. So, this two also gets cancelled,
so we get
15
15
, it is
that is a value of r at this point.
2e
2e
344
(Refer Slide Time: 16:13)
So, that is a
15
similarly, we can compute now this s and also t just by substituting this 0 and
2e
±1/2. So, when we substitute in this s, since there is a term x and y here in the product when
x is 0 the whole term will become 0. So, the s is the straightforward 0 and the t again we need
to substitute this y 2 here and all these terms these 2 terms have x, so they will vanish and here
we have this y 4 , so these terms, when we solve this we will get this
−4
and this rt−s2.
e
So, rt when we make this product, so since the minus sign is there with the t we will get this
−30 . So, this time now this
2
rt−s is negative. So, negative means, as per the sufficient
e2
conditions now, this will be a saddle point. So, these point number 2 and point 3, (0 , 1/ 2) and
(0 ,− 21 ) they both are the saddle points, with our sufficient conditions we are able to identify
that these two points are the saddle points.
Now, the last here the 4/5 we have ( ± 1, 0 ) these two points again we can deal together
because the x also appears in the power either square or we have the 4. So, this value whether
we take plus and minus will remain the same. So, when we computer, so we need to
substitute y to 0 here. So, these two terms will go to 0 and then we have here again this e
345
power this is 0 and then you have 1 there, so e−1, again the same, so which will come in the
denominator and then after simplifying this we will get 16.
So, here r is
−16
and the s is 0 because of here y is there in the expression. So, we will get 0
e
and t will be
−30
from this expression here. So, again this rt−s2 we have to compute and in
e
this case, so rt here we get this positive again. So,
480
and now based on this sign of r, we
e2
can decide whether this is a point of maximum or minimum. So, in this case the r is negative
and remember when rt−s2 is positive and r is negative then we have a point of maximum.
So, in this case these two points (1, 0) and (−1, 0) these are the points of local maximum.
(Refer Slide Time: 19:11)
And now we move to the next problem which is the function here f x y is equal to y square
plus x square y and plus x 4 we want to discuss four local extrema. So, in this case we will
again compute the first order partial derivative which is f x is equal to here 2 xy and 4 x 3, will
come and then we have the partial derivative with respect to y. So, here 2 y+ x2, so we have
the partial derivative and to get the stationary points, so we need to solve these 2 equations;
that means, the f x is equal to 0 and this f y is equal to 0.
346
So, from here we have this 2 xy + 4 x3 and f y is equal to 0, we will get when we set to zero;
that means, 2 y+ x2 is equal to 0. So, out of these two equations we need to get all the points
which satisfy these two equations. So, from the first immediately we see that x equal to 0
satisfies at least there is a point here x is equal to 0 which satisfies this one. So,
corresponding to this when x is equal to 0 here what will be the value of y from the second
equation because we are looking for all the points which satisfy both the equations together.
So, here x is equal to 0 makes this f xx ( 0, 0 )=¿0 and then when x is 0, so y has to be also 0
from the second equation. So, we got this 1 point, now x is equal to 0 and y is equal to 0.
Now we have to also look for any other possibility which can make these two terms 0, so
here either x is 0 or the second term here is 0; that means, that 2 y+4 x 2 is 0 and together with
this 2 y+ x2 has to be 0. So, again out of these two equations we see that the only number
which satisfy only point which satisfy these two equations 2 y+4 x 2 is equal to 0 and this
2
2 y+ x is equal to 0, the only point is (0 ,0) again. So, we are not getting any other point then
(0 , 0), so a stationary point in this case is (0 , 0).
(Refer Slide Time: 21:41)
So, at this (0 , 0) point, we have to now discuss for the identification whether this is a point
of local maximum or it is a point of local minimum. So, for that we need to compute now the
second order derivatives. So, f xx ( 0, 0 ), so what will be f xxnow, so this was f x, so this f xxwill
become with respect to x again. So, we will get 2 y and we will get here 12 x2. And at (0 , 0)
347
points, so again here we have the x and y term, so this will become 0 and if we compute here
f xy which also needed, so with respect to y here we will get 2 x term and then f yy we will get 2
because here when we differentiate with respect to y we will get 2 there.
So now, if we compute r=f xx ( 0 , 0 ) this will be 0, the s at this the mixed order partial
derivative at ( 0,0 ) point, this will be also 0 because x is sitting here and then at this point
again this t=f yy ( 0 , 0 )=2 there is no x and y term, so this is 2.
(Refer Slide Time: 23:05)
And then so what we have rt−s2we need to compute rt−s2, so this is r and t here, so, this
will be 0 and minus s2 0. So, we have this rt−s2and in this case the test fails. So, we cannot
conclude anything based on this rt−s2 or the second order derivative test and we have to find
some other ways to conclude if we can that what is this point (0 ,0). So, this test fails, but we
can easily identify this point whether it is a point of maximum minimum or a saddle point, if
we compute now directly the idea was this Δ f . So, based on the sign of this Δ f , we can
decide whether this is a point of maximum point of minimum.
So, if it is negative, if it is negative; that means, this f (0 , 0) is larger than the points in the
neighborhood, so we have the local maximum at the in this case. If this Δ f is positive in the
neighborhood then this is a point of local minimum naturally and if we changes sign in the
neighborhood of this (0 , 0) point, then we will conclude that this is a saddle point. So, we
will try here now to observe the sign of this Δ f directly without going through the second
348
order derivative test. So, in many cases this works, but sometimes this is difficult to realize
the sign of Δ f directly from the function.
So, in this case perhaps it is possible. So, when we have this f ( 0+h ,0+k ) −f (0, 0) is 0. So,
this f ( 0+h , 0+k )will be just simply this y will be replaced by k and this is h2 k and then here
h will be replaced by this h , so here we have h4 for this x 4. So, we have this value of this Δ f
in the neighborhood of this point (0 , 0) and then we can make we can rewrite this term as so
the first term if we look at, so we have
(
2
k 2
+h .
2
)
2
So, what we are getting out of this whole square,
So, the
k
. So, there is a term here, let us discuss.
4
2
3 2
k
and we have h 4 term and then we have a k h square term and then this is k . So,
2
4
this k 2 /4 and then here we have this k 2 was
3 2
k . So, these will be, so here we can combine
2
these 2 terms to get this 4 and then you have 1 there and this will become, so this would be a
3
3
. So, this is here , instead of 2 we have the 4 there.
4
4
So, 3/ 4 and this 1/4 will make exactly this k 2 there and we have h4 and we have k into h
2
square. So, this is just
3
k there. So now, we will 3/ 4. So, now, we can try to get the
2
behavior based on this term here, which is clear because when h and k both are non-zero.
349
(Refer Slide Time: 26:29)
So, 1 can be 0, we will be still in the neighborhood, but we cannot set both to 0 together. So,
if 1 of them is non-zero for example, k is non-zero. So, here we have something positive and
here also we will have positive this is a square there. If h is non-zero again this can be 0. So,
h is the again this will be a positive term. So, for as long as both h and k are not 0, they can
be negative positive does not matter but this term here for h either h is non-zero or k is nonzero or both are non-zero, this term will remain positive and what does that mean that in the
neighborhood whatever point we take and; however, small this neighborhood is, this Δ f is
positive, Δ f is positive.
So, in this case if this is positive; that means, in the neighborhood the function is taking more
values, the larger values and then this (0 , 0) has to be a point of a local minimum. So, this
(0 , 0) is a point of local minimum. So, it was easier in this case because the test fails, so but it
was a sufficient condition. So, we could not get any sufficient condition, which can tell us
about this local minimum, but this function was very easy to discuss directly, the behavior
and we realized that whatever point be taking the neighborhood the function this Δ f will be
positive and hence, this is a point of local minimum.
350
(Refer Slide Time: 28:19)
So, we will discuss one more example here ( x , y )=2 x4 −3 x2 y + y2 . It is a similar to the
earlier example, so we will compute f x f y and then the stationary point again and we will find
that there is only one stationary point that is a ( 0,0 ) in this case also similar to the earlier
problem and when we compute this f x x from here we will at 0 zero this will become 0
because the term will x or y term will survive there and when we compute this f xy.
So, again here this 6 x will come and at ( 0,0 ) point this will be 0 and when we compute f yy,
so we differentiate this with respect to y again, so we will get 2. And similar to the previous
case we have rt−s2, so this test fails again and we will use the same idea a similar idea what
we have done before. So, we will compute this Δ f now, the point in the neighborhood minus
this f (0 , 0) which will come exactly that function 2 h4 −3h2 k +k 2and in this case we will do
again little manipulation here.
So, −3 h2 k , we have written this −2 h2 k and −h2 k and then we take common from the first 2
terms the 2 h2. So, here we will get h2 and −k and then −k from here we will get h2 −k . So,
we have the product of h2 −k into 2 h2 −k and now we will discuss how to identify the sign of
this delta phi whether we have a definite sign either positive or negative at all the points in
the neighborhood of this point or the sign changes.
So, first possibility we will take that if this k is negative for example, So, if k is negative
irrespective of whatever our h is, so as long as this k is negative, this term will be positive
351
here −k term, here also −k term will be positive. So, whatever h is h may be 0 or h may be
non-zero, but when k is negative this product is going to be positive. So, we have this Δ f
positive, in this situation when k is negative. Now we will see the other situation actually in
this case that the Δ f can be also or Δ f is negative in the neighborhood, how whatever close
you go to the h is equal to 0, k is equal to 0 and this Δ f will be negative.
So, here we have seen that for k less than 0 Δ f is positive and in the other possibilities, we
will choose our h and k such that; the k is bigger than h2 and less than 2 h2. So, by choosing
this, so we are not restricting our neighborhood, we can go as close as to this (0 , 0) point
meaning h 0 and k 0 point having this path here h2 this k should be greater than h2 and less
than 2 h2. So, if we choose our h and k point in the neighborhood which satisfies this
inequality, what will happen? This h square is smaller than the k.
So, k is larger this is a negative number and then 2 h2 is bigger than k, so this is a positive
number. So, this negative positive will make it negative now. So, in their neighborhood when
this h and k satisfies these points then we have this Δ f negative. So, what we have observed
this is interesting that in the neighborhood of this (0 ,0) point, we realize that this Δ f is
positive and also this Δ f is negative. So, it is changing sign in the neighborhood of this (0 , 0)
point and that means, this (0 , 0) is a saddle point.
So, again we were able to identify this point here the (0 , 0) point and this comes to be the
saddle point and which was not possible with the second order test, but the direct observation
we can find that this is a saddle point.
352
(Refer Slide Time: 32:43)
Well so, this is again a simple problem of this maximum and minimum of this function
2
2
f ( x , y )=2+2 x+2 y−x − y over a triangular plate in the first quadrant which is bounded by
this 0, x is equal to 0, y is equal to 0 and y is equal to 9−x line.
So, this is the domain of the problem the bounded domain here because these x is equal to 0
y is equal to 0 and y is equal to 9−x , they are the boundaries here this is (0 ,9) point this is
(9 , 0) point. So, we have to now in this case we have to consider the interior and also the
boundary. So, when we have the bounded domain we have to consider because this maximum
minimum may exist at the boundaries. So, this interior points, so that means, leaving the
boundary now.
So, at interior points we will search for the stationary point and separately we will handle or
we will look for the extrema at the boundary. So, the stationary point for this problem we can
easily compute f x and f y and they come to be 1 and 1. So, this here is the stationary point in
the domain or in the interior of this domain excluding the boundary. So, we have to consider
this point because we can have local extrema at this interior point, but in this problem we our
interest is to find absolute maximum and minimum. So, definitely we will test at this point
what is the value of the function later on.
353
(Refer Slide Time: 34:19)
So, we have one point and none the boundary points we have to look. So, along this OA line
here, so y is 0. So, we can set in our f x as y 0. So, f will be 2+2 x−x2 and x varies from this
0 to 9. So, this is a 1 variable problem and we can look again for the critical point were just
by differentiating this function with respect to x so which comes to be at x is equal to 1 there
might be a point of local maximum or minimum.
We do not have to actually go for the identification, so there will be few points where the
maxima minima can appear and then we will get the function value at those points. And, from
there we can conclude which is the maximum value and which is the minimum value.
So, here this is a stationary point in this one dimensional problem. So, the possible candidates
for this extrema because again for this problem now; we have the boundary points 0 and 9.
So, we have to also consider (0 , 0) points in (9 ,0) point and ( 1 , 0 ), this at x is equal to 1,
which is a stationary point for this problem. So, we have three points, we have the (0 ,0)
points, which is the boundary of this one dimensional problem and here we have the (9 ,0)
point and we have the ( 1 , 0 ) point. So, these are the three points; similarly along this OB line,
we have to search now here along this line x is equal to 0, we can set in f ( x , y )function.
So, again this is a problem of one variable we have to look for the stationary point and along
this boundary here we have the (0 , 0), this boundary point which was already taken earlier we
354
have (0 , 9)
point and from here by differentiating this we will get the stationary point as
(0 , 1). So, we have these three stationary points I mean these three candidates for extrema.
(Refer Slide Time: 36:19)
Similarly, we have to do again this third boundary y is equal to 9−x. So, we will substitute y
is equal to 9−x intof ( x , y ). So, we will have again a problem of 1 variable, we have to look
for the interior point there. So, f x is equal to 0, we will get only 1 point which is 9/ 2 and 9/ 2.
So, this is the point there and obviously, these boundary points which are already being taken
care by the earlier problem. So, we have so many points here the 7 points where the function
may take the maximum or the minimum value. So, we have to consider the interior and we
have to also consider the boundaries when we have a closed and the bounded domain.
So, here these points we will evaluate simply the function because our interest is to find the
global maximum minimum, we do not have to identify each of this point for local maximum
or minimum. So, at (1, 1) the function f ( x , y ) is taking value 4 and (0 ,0) is taking 2 and so
on. So, here we have computed all these values at these points and then we realize this −16,
which is the which the function takes at (9 ,0) and (0 ,9) point is the minimum among all and
this 4 is the maximum 1 at (1,1) point.
So, the function takes maximum at the interior here at (1,1) point and it takes the minimum
at (9 , 0) or (0 ,9) point which is the value is −61. So, this was a problem with bounded
domain the earlier problems were in the open domain. So, where we have not considered the
355
boundary because, there were no boundaries in the problem, but this here we have to if there
is a boundary the domain is closed. Then we have to also look for the boundaries because
maximum and minimum may occur on the boundaries which is the case here like minimum is
at(9 , 0) or (0 ,9).
(Refer Slide Time: 38:17)
So, the conclusion here that maximum and minimum can occur only at the boundary points of
the domain; that is a one and the second the critical points which we have to look for the
possible candidates for maximum and minimum.
(Refer Slide Time: 38:29)
356
So, these are the references.
Thank you very much.
357
Engineering Mathematics – I
Prof. Jitendra Kumar
Department of Mathematics
Indian Institute of Technology, Kharagpur
Lecture – 20
Constrained Maxima & Minima
Welcome back to the lectures on Engineering Mathematics-I and this is lecture number 20.
So, today we will discuss the Constrained Maxima and Minima.
(Refer Slide Time: 00:27)
So, in particular we will be talking about the method of a Lagrange’s multiplier which is used
to solve such problems, where we have along with the function some constraints.
358
(Refer Slide Time: 00:39)
So, what is the method of Lagrange multiplier? So, let me just discuss the problem. So, if you
want to find the maxima and minima of the function, let us say u=f ( x , y ) where we have
some other constraint that ϕ ( x , y ) =0. So, we have some relations some conditions on x y is
given. So, basically this is similar to the problem we have discussed in the previous lecture
where, we had taken the boundaries into the consideration.
So, those boundaries were actually the additional constraint on the function that x y satisfies
either x is equal to 0 or y is equal to 0 or there was a line there y is equal to 9−x. So, there
was an additional condition along with the function given. So, there we have substituted the
value of the y into this f ( x , y )and then the function was converted into a function of one
variable problem which we have discussed for along each of these conditions.
So, today we will have this method of Lagrange multiplier where we do not have to substitute
the value of y in this one and then getting a function of one variable and proceeding further
for extrema. In this case we have some direct method which without substituting we can
actually get all these critical points where, the function may take maximum minimum or it
may be a saddle point. So, let me just explain that what is the method of Lagrange multiplier.
So, using this chain rule here u is equal to f ( x , y )we can get the
du
directly because, here the
dx
function y is a function of x out of this we have the relation here between x and y.
359
So, by removing this y from here we have basically this function of x. So, this
du
make sense
dx
here, but with the rule with the idea of this chain rule we can get that derivative in terms of
the partial derivatives here. So, the partial derivative of f with respect to x and then
a 1; so, plus this partial derivative with respect to y and then
point of extrema because we know that the
du
that is
dx
dy
dy
. So, with this
and at the
dx
dx
du
must be 0, it is a 1 variable problem where we
dx
have to get this derivative of u with respect to x and that has to be 0 at the point of extrema.
So, we at this point here we get that this derivative must be 0. That means, this
∂ f ∂ f dy
dy must be 0.
+
=0 into
∂ x ∂ y dx
dx
(Refer Slide Time: 03:27)
And at the point of this extrema we have this
∂ f ∂ f dy
+
=0. So, that is a one condition one
∂ x ∂ y dx
equation we are getting which has to be satisfied at the point of extrema, and then this
equation which is the constraint that a ϕ ( x , y ) =0. So, whatever point is the point of extrema
this condition has to be satisfied because, that is given in the problem that the relation x and
y must be satisfied with this relation.
360
So, in that case whatever point we have the point of extrema this equation must be satisfied at
any point; so, naturally at the point of extrema as well. So, we have won this equation out of
this equation if we get the derivative with respect to x, if we differentiate this one. So, we
will get a partial derivative with respect to x and the chain rule plus this
∂ f dy
=0. So, we
∂ y dx
got this another equation which is satisfied at the point of a extrema. And now we will try to
eliminate out of these two equations this
dy
term and we will have everything in terms of
dx
partial derivatives.
So, here we have this
∂ ϕ ∂ ϕ dy
+
this second equation the left hand side of this equation here
∂ x ∂ y dx
and we have multiplied by
−f y
. The idea is because this ϕ y and this ϕ y will get canceled and
ϕy
we have this with minus sign
dy
∂f
∂f
dy
and this
and here we have
and . So, these two will
dx
∂y
∂y
dx
get cancelled and we have the term free from this
dy
. So, in this case when we added here;
dx
so, these terms will get canceled and we will get simply assuming that this here is λ just for
simplicity now.
So, here we get
∂f
∂ϕ
+λ
=0. So, this is the equation we got that that has to be satisfied at
∂x
∂x
the point of extrema that is a one equation, another one since λ has appeared here. So, we
have this relation from here that
∂f
is equal to this ϕ y λ which is the relation here. So, we
∂y
have another relation which has to be satisfied, and we have this equation ϕ ( x , y ) =0. So, we
have these three equations which has to be satisfied at the point of extrema there.
361
(Refer Slide Time: 06:23)
Now, moving further so, we have considered we have discussed that these are the equations
which has to be satisfied at the point of a extrema and we have here 3 parameters. So, 1 will
go 3 unknowns basically the x y will be the points we are looking for and also this λ we have
introduced. So, there is a x y and λ. There are three points we have to we have to get by
solving these 3 equations and that those points will be the points of extrema in terms of the
x y.
So, what is the method of the Lagrange multiplier which we have just discussed we can there
is a way to remember easily how to set up these equations. One way which we have just seen
bit along bit long derivation, but now we will here write down in a more precise form which
is very easy to remember and how to get these equations directly. So, we have a problem that
we want to minimize or maximize this u is equal to f ( x , y ) under this condition that x y
satisfies this relation.
And we just define in an auxiliary function. So, we define that F ( x , y , λ )=f ( x , y ) + λ ϕ (x , y).
So, we take this function f ( x , y ) which we want to minimize or maximize and plus this
lambda and this constraint this ϕ ( x , y). So, we need to define such an auxiliary function. In
fact, this idea can be extended for when we have many constraints like phi 1 phi 2 and so on.
So, again we have to introduce the λ 1 ϕ1 + λ 2 ϕ 2. So, we have to introduce more lambdas. So,
we are discussing just for under one constraint. So, this is the auxiliary function we define
362
and then we find the necessary conditions on for the extrema of this F. That means, the
derivative of this F x=0. F y=0 , the F λ =0 to find the a critical points of this F here.
So, by doing this we will precisely get those equations which we have derived that at the
point of extrema these equations must be satisfied. So, when we derive when we get the
derivative with respect to x here we will get f x + λ ϕ x . That means, this F x=0 is nothing but
the f x + λ ϕ x=0 which was the first equation there; the F y=0 . So, here again this f y + λ ϕ y =0
which is the second equation listed there. And, the third one is the F λ =0 and that will be just
the ϕ=0.
So, we have these three equations which can be obtained just by defining this auxiliary
function here F, do as that you we substitute here this f ( x , y ) as it is, introduce one λ and we
have just one constraints here. So, λ ϕ (x , y), but we can extend this idea for many constraints
as well. So, if we have for example, two constraint ϕ 1 ( x , y ) is equal to 0 ϕ 2 ( x , y ) is equal to 0
then we will just introduce more lambdas there λ 1 ϕ1 ( x , y ) + λ2 ϕ2 ( x , y ). And again then we
have to discuss these necessary conditions. So, there will be two λ ' s there so, two one more
equation will be coming with respect to λ 1 here or with respect to λ 2 there. So, at another
constraint will appear here and then some more terms there.
(Refer Slide Time: 10:35)
So, anyway let us discuss for this one very one constraint and the remark here that using this
method of Lagrange multiplier, we will obtain the stationary point or rather we call the
363
candidates for the extrema. This will be the candidates for the extrema; we will not compute
the behavior of this point whether it is a point of local minimum local maximum.
We will we do not determine the nature of the stationary point in practice we in many
problems we are usually interested finding the maximum and the minimum; so, the global
maximum and minimum value of the function under given constraint. So, here in this lecture
today or by this Lagrange multiplier we basically find all the candidates. So, called the
critical points of the problem where the maximum or the minimum may take place. So, we
will find all these critical points and find the values of the function at all these points.
And then, we can identify that which one is the point of maximum the global maximum or
which one is the point of a global minimum. So, usually there are a few candidates as to the
critical points. So, we can evaluate f at all of them and choose the largest and the smallest
values. And, hence no further test is required if we wish to find only absolute maximum and
minimum. So, our aim is now to find only the absolute maximum and minimum. And
therefore, we do not need any other test to compute whether this point is a point of a local
maximum minimum or a saddle point. But rather we will look for just the global maximum
minimum of the problem under that given constraint.
(Refer Slide Time: 12:27)
So, let us move to the example here. So, we have the maximum minimum of this function
2
2
2
2
x − y −2 x in the region here, x + y ≤1 . So, this is the constraint is given that we want to
364
find the maximum minimum of this function x2 − y 2−2 x under this constraint that the x y
satisfies only x2 + y2 <1.
So, we are talking about this disc including the boundaries including this circle here. So, x y’s
are restricted only within this a domain here. So, there is a restriction on x y. So, this is a
problem of a constraint and now we will deal in two ways. So, in all these problems when we
have such a reason is given; so, this boundary is the boundary is here x2 + y2 =1. So, we will
break into two problems. So, we will find the maximum or the critical points of the problem
here x2 − y 2−2 x inside this domain.
And then on the boundaries; that means, with the restriction x2 − y 2 is equal to 1 exactly and
that will be the problem which we have discussed that using the Lagrange multiplier we will
get; when we are talking inside this domain that means, this x2 + y2 <1. So, our domain is open
here and we will apply the idea which is discussed already in the previous lecture that we will
find directly, the critical point and then we will see that if some of the critical points fall in
this region.
So, we will take them for further evaluation of f at those points, if the critical points does not
fall in the domain they are outside the domain then we can leave them because that is not of
our interest. So, this problem is taken into two parts: one is finding the critical points or the
candidates for local extrema minima inside the domain. So, x2 + y2 <1 and the one problem
will be that finding these maxima minima the candidates for maxima minima on the
boundary. So, let us move to inside the domain first.
365
(Refer Slide Time: 15:05)
So, for the interior; that means, x2 + y2 <1. So, here we do not have to use any other idea then
the discussed in already in the previous lecture. So, we will have the f ( x , y )=x 2− y 2−2 x . So,
we will get f x here which will give us thex=1. So, f y=0 so, we have y=0. So, it means y is
equal to 0.
So, the only critical point which we are finding for this f ( x , y )=0 is 1 and 0, but this 1 and 0
point does not fall in our interior. Here indeed this is the point on the boundary, but that will
be automatically taken care when we will consider the boundary in the next slide. So, here
there is no critical point inside this interior here. This is outside the interior or in this case
rather it is sitting exactly on the boundary, this point may be absolutely outside the domain
which we are considering.
So, in any case you will not consider this point now in this sub problem where we have we
are looking for this extrema in the interior of the domain. So, this is a point certainly because
falling on the boundary. But, this will automatically come in the problem and we discuss
when we find the critical points on the boundaries.
366
(Refer Slide Time: 16:41)
So, the next problem will be that we will look now for the local extrema on the boundary the
2
2
x + y =1. So, the problem is to find the maxima minima subject to this condition and that is
the problem which we will solve using the Lagrange multiplier. Because here, we have the
exactly this constraint that x2 + y2 =1 and for that we have to define the auxiliary function, so
that will be this function itself ( x2− y2 −2 x ) + λ( x2 + y 2−1).
So, you will find the critical points f x=0 which will be 2 x−2+2 x. So, then we can simplify;
so, x and 1+ λ is equal to 1. So, just to look again here so, what we get with respect to x we
got 2 x and we got 2 here plus this λ and we got 2 x here is equal to 0 we want to set this. So,
this 2 will be cancelled from everywhere we have x and plus this λ x. So, x we can take
common here. So, we have 1+ λ and then this 1 this minus 1 the other side will go plus 1. So,
this isx ( 1+ λ ) =1.
367
(Refer Slide Time: 18:11)
And then F y=0 . So, we have to differentiate now this with respect to y treating λ and x as
constant. So, we will get y and ( λ−1 ) is equal to 0 another point. So, with respect to λ when
we differentiate we will get exactly our constraint. So, it is x2 + y2 is equal to 0. So now, we
have to solve these three equations to get the points to get the critical points and those points
will be the candidates for the local for the maximum or the minimum of the problem.
(Refer Slide Time: 18:47)
So, we have these three conditions. So, these three equations which we want to solve now
which let us start with this middle one. So, this is satisfied when λ is equal to 1 or this y is
368
equal to 0 or both. So, here if we consider the y is equal to 0 first. So, let us consider that y is
equal to 0 so, this equation is satisfied. Now, for this y is equal to 0 from this third equation
when we set y is equal to 0 here we have x2 is equal to 1; that means, x is equal to ±1 this
equation is also satisfied.
So, if we have y is equal to 0 and x is equal to ±1 these two equations are satisfied, now from
the left one having this x is equal to ±1. So, if x is x is +1 and we have 1+ λis equal to 1 so,
this λ will be 0 in that case and when we have x is equal to. So, this was x is equal to 1 and
then if we have x is equal to −1 in that case we have minus and the 1+ λ is equal to 1. So, this
λ=−2 in this case. So, we have λ=0 ,−2 from this first equation.
(Refer Slide Time: 20:13)
Lambda is equal to 0, lambda is equal to 2. So, what are our points here? Our points are like
x is equal to 1 y is equal to0 and λ is equal to 0. This is the point which is which satisfies all
these three equations. The second point we have here −1, we have 0 there and we have −2
there. This is another point which satisfies all these equations.
Now, from here we have another possibility is that we get λ is equal to 1. So, if we take λ is
equal to 1 from here; when λ is equal to 1 we can get x from this first equation when λ is
equal to 1 we can get x has to be a
1
1
to satisfy this equation. So, we have x= λ=1 and
2
2
369
since x is
1
1
3
2
so, from here we can get y. So, y =1− . So, we get y=± √ . So, we got
2
4
2
another points here. So, our points are now, so x=
1
3
and then y let us take the ± √ , λ=1,
2
2
this is 1 point which satisfies all these three equations.
1
− 3
And then we have x= we have √ and then we have 1 again here, this is another point
2
2
which satisfies all these three equations and these are all possibilities; we do not have any
more possibilities which can satisfy all these equations. So, the candidates for this extrema I
am just writing here in term the x y points because, we will compute the function value at the
x y point λ is not interesting for us there. So, we have the candidates now the plus and the −1
with the 0. So, the first two points there and these two points so,
( 12 , ± √23 ). So, these are the
candidates now for the extrema.
(Refer Slide Time: 22:29)
So, now the function values we will compute at this ( ± 1, 0 ) and
( 12 , ± √23 ), this was our
3
function now. And, these are the points (1,0) (−1 ,0 )and then here the y is taken ± √ ,
2
because it is a y 2 there. So, does not matter if we take plus sign there or minus sign the value
370
will be the same. So, the function value at these points so, when (1,0) so, 1 here the 0 and
then 1 so, −2+1−1.
So, the function value at this (1,0) point is −1, here we take (−1, 0) so, −1 for x. So, we
have 1 there and then this is 0 and then here we have 2. So, we have 3 and at this point
similarly, if we compute we will get
−3
. So, here we see that the maximum value is achieved
2
at this point (−1, 0) which is the value of the function is 3 at this point.
So, the maximum value of the function is 3 and the minimum value is at these two points
here or the value of the function is
−3
. So, in this problem again to conclude that we have
2
computed all the critical points of the problem. And, then we have evaluated the function
value at all these critical points and then we have selected well this is 3 is the maximum value
which is attained at (−1,0),
this point
−3
is the minimum value of the function which is attained at
2
( 12 , ± √23 ).
(Refer Slide Time: 24:13)
Next problem so, we want to find the maximum and the minimum value of this function
2
2
2
2
f ( x , y )=x + y and in the region ( x−2) +( y−1) ≤ 20. So, again we have the similar
371
problem, we have the region given here and this function x square plus y square. So, we will
deal exactly in a similar fashion. We will first look for the extrema in the interior point; that
means, when strictly less than 20. So, we have this open domain no boundaries and we will
compute the f x=0 and which is just 2 x=0.
So, x=0
we will compute f y=0 so, we see here y=0. So, our critical point of the problem here at 0
the only point which falls inside the domain. So, we have this point (0 , 0) in the domain
itself. So, we will consider, now for this we will find the value of the function and then that
will tell us whether the function has the local minimum or the rather minimum or the global
minimum at this point.
So, the local extremum on the boundary so now we will consider the boundary points that
( x−2) +( y−1) =20. So, this with equality and now we have to use the method of Lagrange
2
2
multiplier. So, the problem is that we want to minimize maximize this function subject to this
constraint. So, we will define this auxiliary function as usual by putting the λ in front of this
constraint there and find all the critical points by differentiating F with respect to x y and the
λ partially.
(Refer Slide Time: 26:07)
So, the critical points by differentiating F x=0we will get this equation F y=0 , we will get this
equation and F y z is equal to 0 we will get this equation. So, out of these three equations we
372
have to find all the critical points, from this equation first we will write down this ( x−2 ) in
terms of λ. From the second equation also we write ( y−1 )in terms of λ we will substitute
here. So, we will get possible values of λ. So, in this way we will solve this let us from the let
us look at the first problem here. We have x+ λ ( x−2 )=0.
So( x−2)=
−2
. So, if we want to get x=−2, 6 there. So, in this case x=−2, 6.
1+ λ
(Refer Slide Time: 27:27)
So, here we have ( x−2 )=
−2
−1
. Similarly, we can get out of this equation ( y−1 ) =
. And,
1+ λ
1+ λ
from the third equation now we can substitute this ( x−2 ) and ( y−1 ) here and we will get a
1
equation in λ which can be easily solved. So, we have ( 1+ λ )=± . So, from here we got
2
basically λ; once we have the λ here we can get x and then corresponding y from this first
and the second equations.
So, from here we get λ=
−1 3
,− , these are the possibilities which will finally satisfy this
2
2
equation. From this equation when this λ=
−1
−3
we will get −1, when λ=
we will get 3
2
2
373
and from the first equation we will get x=−2, 6. So, we have the solution of this problem.
(
So, −2,−1,−
1
3
is one point another one is 6 , 3 ,− .
2
2
)
(
)
So, naturally we will take just the x y points, because we want to compute the value of the
function at the x y point. So, we have 3 points 1 was in the interior that is a (0 , 0)point and 2
points we got here (−2,−1) and (6 ,3). These are the three points we will evaluate the
function value. So, at the ( 0,0 )the function value is 0, (−2,−1)it is 5 and the (6 ,3) if we
evaluate that x2 + y2. So, the function was the function was f ( x , y )=x 2 + y 2.
So, here it is 5 and then here 62 =36+9=45. So, we have these functions values and we
realize now here this is the minimum value of the function takes which is naturally true and
directly we can get from the function itself where, the function is x2 + y2. So, the minimum
value will be a 0, because it has to be greater than equal to 0. So, the minimum at ( 0,0 ) will
be 0 and the maximum value here is attained at the point (6 ,3) and the value of the function
is 45. So, we have the minimum value 0 and the maximum value of this problem is 45.
(Refer Slide Time: 30:05)
Conclusion now: so, the method of Lagrange multiplier is that we want to find the maximum
or minimum of a function here f ( x , y ) with the sum given constraint. The idea is that to
formulate an auxiliary function here by introducing this λ in front of this phi. And, then the
374
necessary condition for extrema of this F will be the setting F x=0 and then which gives us
this equation F y=0 we get another equation and F λ =0 we get the ϕ ( x , y ) =0.
So, out of these three equations we have to find all the points meaning this x , y and λ. So, all
possible values of x , y and λ we have to see which satisfy all these three equations and then
at all those points we have to compute the function value f ( x , y )and see where the function is
attaining its maximum and its minimum. So, in this way we can find the global maximum and
minimum of the problem.
But, we have to be careful that the problem when we have that close and the boundary
domain we have to look inside the domain. Because, there may be some candidates where the
function may take maximum or minimum value and there may be the points on the
boundaries where the function may take its extrema.
(Refer Slide Time: 31:33)
So, these are the references we have used for preparing these lectures.
Thank you very much.
375
Engineering Mathematics – I
Prof. Jitendra Kumar
Department of Mathematics
Indian Institute of Technology, Kharagpur
Lecture – 21
Improper Integrals
So, welcome to the lectures on Engineering Mathematics – I and this is lecture number 21
and today, we will talk about the integral calculus and in particular Improper Integrals.
(Refer Slide Time: 00:26)
So, we will introduce first the improper integrals and then how to evaluate such integrals that
will be the topic of today’s lecture.
376
(Refer Slide Time: 00:37)
So, here there are proper integrals which we usually studies so, those integrals. So, for
b
example, the
∫ f ( x ) dx
is proper. If the range here a ¿ b of this integral is finite and the
a
integrand so, the f x is the integrand here and that is bounded in this range a ¿ b for x. So, if
this f ( x ) is bounded and a∧b both are finite so, the range of the integral is finite in that case
we call that this integral is proper. But, now we will discuss today what are the improper
integrals.
b
So, improper integral the integral ∫ f ( x ) dx is improper. If this a is for example, −∞ and or b
a
is ∞ so, one of them is basically infinity that a=−∞ or b=∞ or both, a=−∞ and b=∞ and
this f ( x )the integrand here is unbounded at some point in the range of this x from a ¿ b. In that
case we call this integral improper integral of first kind.
The second, if this f ( x ) is unbounded at one or more points in this range a ¿ b then we call
improper integral of second kind. So, in the first kind the limits either a∨b is ∞ or both are ∞
−∞ to +∞, but here the f x is bounded in the integral of first kind whereas, in the integral of
the second kind we assume that this f ( x ) is unbounded between this a∧b and these limits
here the range of the integral is finite from this a ¿ b.
377
The second type or the third kind of improper integral is when we mix these two types of
integrals. So, the range one of the range here either a or this b is ∞ or both are ∞ as well as
the f ( x ) the integrand here is also unbounded at some point in the within the range of this x
then we call that this is the third kind of improper integral.
(Refer Slide Time: 03:11)
2
So, the example for proper integral: so, for example, we have here
∫ √( x¿¿ 2+1)d x ¿. So,
0
here the integrand which is √ (x ¿¿2+1)¿ is finite or bounded in the range of this x from 0 to
2 and also the range the 0 and the 2 that is also finite. So, we do not see here ∞ and also this
f ( x ), the integrand is also bounded. So, in this case we call such integrals proper integral or
1
for example, we have
∫ sinx x d x. So, in this case one might think that here the
x is in the
0
denominator and the range of this x is from 0 to ∞ then the integrand may become
unbounded.
But, this is not the case because if we consider the limit here so, the limit as x goes to x goes
to 0 and this
sin x
. So, this limit is finite, the limit is 1, we know that this limit is 1. So, here
x
the limiting value of this integrand as this x goes to 0 is ∞, in that case also we will call this
integral as a proper integral because this is not unbounded at any point in the range of this 0
to 1 and the limits are also finite the range is finite. So, this integral is also proper integral.
378
∞
Coming to the improper integrals: so, here we have for example, this
∫ cos x dx. So, this is
0
the improper integral of a kind one or the first kind because here in the limit we do see a
infinity, but the integrand at any point is bounded. So, this is the integral of first kind.
1
Second, this
1
dx;
∫ x−1
so, in this case the range is finite, but the integrand here as x
0
approaches to 1, this is becoming unbounded. So, in this case the integrand is becoming
unbounded as x approaching to this upper limit a and then this is the improper integral of a
second kind. This is here the integral of third kind or the mixed kind because the limit is also
unbounded. So, here we have this ∞ in the upper range and then this integrand which is
1
is also going to ∞ as x is approaching to 1 and 1 is in the range of the integral. So, in
2
( 1−x )
this case the integrand is also becoming ∞ and the range is also not finite. So, this is the
integrand of improper integral of third kind.
(Refer Slide Time: 06:13)
So, how to evaluate such improper integrals, for first we will discuss for the first kind. So, for
∞
example, we have this ∫ f ( x ) dx and suppose that this b is going to ∞. So, this is the integral
a
of first kind and we assume here that this b is approaching to ∞. So, in this case what we will
379
consider, we will consider the integral a to R. So, this b which is ∞ we have replaced by this
R and then we are taking this R to ∞.
So, in this case here we have like this infinity. So, this is the integral of first kind and to
evaluate this what we will do we will replace that ∞ by R this number and then later on after
R
evaluating this integral here
∫ f ( x ) dx and then we will take this limit and if this limit exists
a
we call that this is the value of this integral or the integral converges and if this limit does not
exist then we call the limit that the integral does not exist or the integral diverges.
Well, so, now in this case we may have the situation that this instead of showing the earlier
integral here this b was ∞ and now we have considered that this the lower range is −∞ so,
again we have the improper integral of first kind. So, in this case we will use the same idea
here the limit we will take over R to ∞ and this −∞ will be replaced by −R. So, now, we will
b
evaluate this integral first from
∫ f ( x ) dx and then we will take the limit as R tending to ∞
−R
after the evaluation of this integral.
So, if we have this again the first kind integral, but we have the both the limits here −∞ and
this +∞ so, in this case what we will do we will break this integral into two parts. So, one we
⁡ . So, here we have taken this −R for this ∞ we have replaced by R
will take this lim R →∞
1
1
1
and for this R 2 we have replaced by this is replaced by this +∞ there.
∞
And, we have taken this R 1 → ∞ and R 2 → ∞ so, again we have this integral here
∫
and if
−∞
these limits here on the right hand side these exists then we call that this limit is convergent
or this integral is convergent otherwise this is a divergent integral or the integral does not
exist. Or, we can combine these two like
lim
⁡
R 1 →∞
R 2 →∞
same.
380
R2
and we take from
∫ f ( x ) dx this is the
−R 1
(Refer Slide Time: 09:23)
So, now we will do how to evaluate improper integrals of say first kind. So, this is the
∞
example here we have the ∫
2
2 x2
dx. So, in this case the three case that we will this we will
x 4 −1
remove this ∞ and we will replace that with the R. So, we have this integral equal to the limit
2
2x
R to ∞ and this 2 to R and 4 . This is equal to limit R to∞ this we will do the partial
x −1
1
1
+ 2 . So, this will be exactly
x −1 x −1
fractions now. So, this will be x2 −1 and x2 +1; so,
2
2
2x .
4
x −1
⁡
And, then again we will do this partial fractions there. So, we have the lim R →∞
R
1
1
∫ x 21−1 + x21−1 dx . So, this is the partial fraction of x2 −1 + x 2−1 and everything over this dx
2
term.
So, here we have the
lim ⁡
R →∞
R
∫
2
( x 1−1 + x 1−1 )dx
2
381
2
R
¿ lim
∫
R→ ∞
⁡
2
R
¿ lim
⁡
R→ ∞
1 1
1
1
+
+ 2
dx
2 x−1 x−1 x +1
[(
]
)
R
R
[∫[ ( )] [ ( )] [( )] ]
2
1 1
1 1
1
dx −∫
dx +∫ 2
dx
2 x−1
x +1
2 2 x+1
2
¿ lim
lim ⁡
R →∞
⁡
R→ ∞
[
1
x−1
−1
ln
+ tan x
2
x+1
( )]
]
R
2
1
R−1 1 1
− ln +tan−1 R−tan−1 2
ln
2
R+1 2 3
[ ( )
1
1
π
¿ × 0− ln 3+ −tan−1 2
2
2
2
1
π
¿− ln 3+ −tan−1 2
2
2
(Refer Slide Time: 14:06)
382
]
So, this integral exists and this is the value of this integral improper integral. The second for
∞
example, we have
∫ sin x dx.
So, in this case also using the same tricks so we have the
0
R
lim ⁡
∫ sin x dx will be sin x will become −cos x and then we have the limit 0 to R.
R →∞ 0
⁡ [1−cos R] and when we take the lim ⁡ this for this cos R this limit
So, this will be lim R →∞
R →∞
does not exist. So, therefore, in this case the second case the limit does not exist and we have
⁡ [1−cos R]
precisely this lim R →∞
and the limit does not exist or the integral does not
converge in this case. So, the limit does not exist. So, in the first case this improper integral
we call it is convergent and the values given by this expression here. In the second case this
integral the improper integral does not converge because we cannot evaluate this integral, ok.
(Refer Slide Time: 15:33)
So, the evaluation of the improper integral of the second kind we will learn are the integral of
b
the kind
∫ f ( x ) dxwhere
f ( x ) is unbounded at some point in the range a ¿ b. So, here this
a
suppose this f ( x ) is ∞ as x tending to b; so, the upper limit when x is approaching to upper
limit this f ( x ) is converging to is going to 0 is becoming unbounded and for such type of
integrals when this f ( x ) is approaching to infinity as extending to b.
383
We will evaluate by taking by introducing this epsilon here and so, here we have introduced
this epsilon and subtracted from the b. So, the problem was at b. So, we have voided now by
b−ε because the f ( x ) was tending to infinity when x was tending to b . So, we have avoided
now this b making this b−ε and now this is nice integral, the proper integral which we will
evaluate using the techniques developed for the evaluation of the integral and later on we will
take the limit epsilon tending to 0 from the right side.
So, here we have this integral and in the second case when f ( x ) goes to infinity as x goes to a
you will use the same idea that now we will use here a+ϵ . So, again this a where the
problem was there in the integral is avoided and later on we will take this epsilon approaches
to 0 from the positive side to evaluate this improper integral.
(Refer Slide Time: 17:22)
So, evaluation of improper integral of second kind when we have a some other point as x
approaching to c c is a point in a b and this f ( x ) is becoming unbounded so, this is the
improper integral of the second kind and in this case again we will use the trick that the
integral a ¿ c and c ¿ b we will break and since the problem was at c so, we have avoided by
c−ε
subtracting ϵ from c. So, we have a to
∫ f ( x ) dx and in the second case the again the integral
a
was c ¿ b, but we have the problem at c, so, we have removed this by taking ε →0
lim
+¿
b
∫
c +ε
384
¿
.
f ( x ) dx ¿
So, if you have this integral f ( x)→ ∞ at both the ends like x → a and also x → b. So, this is
b
another for the second kind of integral. So, in this case we can evaluate this
∫ f ( x ) dxby
a
introducing this ε →0
1
lim
¿
b −ε2
+¿¿
+¿
ε 2 →0
∫
f ( x ) dx ¿
.
a+ε 1
So, now, we will take some example here where we can evaluate just again a remarks. So, in
this case when we have this f ( x) is approaching to infinity at both the ends. We can also
break this integral at some middle at some other point here like a ¿ c and then c ¿ b and then
we can handle exactly the integrals we have handled before. So, we will avoid we have two
different integrals and c dot this one and the first one we will we will handle with this a+ε 1
and the second one we will handle with that point to this b−ε 2 similarly as we have done
above.
(Refer Slide Time: 19:26)
So, here how to evaluate such improper integrals of the second kind; so, for example, we
1
have this
∫
0
dx
. So, here when x approaching to 1 this is becoming unbounded our
√ ( 1−x )
385
integrand is becoming unbounded. So, in this case what you will do we will take the
lim
1−ϵ
+¿
ϵ→0
∫
0
¿
dx
¿
√ ( 1−x )
.
So, the point where the function was becoming unbounded we have replaced by 1−ϵ . So, we
are not going from 0 ¿1, but we are going from 0 to 1−ϵ where we have the nice integral and
later on we will substitute we will take that limit there. So, what is this integral value here?
−1
So, we have this integral 0 to 1−ϵ and this is ( 1−x ) 2 which if we integrate we will have
1 1−ϵ
2
[ ]
−( 1−x )
1
2
1−ϵ
=−2 [ √ 1−x ]0
0
¿−2[ √(1−(1−ϵ ))−√ 1]=−2[ √ ϵ −1] .
lim
¿
ϵ → 0+¿−2 √ ϵ+2=2¿
(Refer Slide Time: 21:46)
So, that is the value of this integral this integral convergence and the value is 2.
386
(Refer Slide Time: 21:54)
2
∫
The second example we take
0
dx
=
lim
( 2 x−x2 ) ϵ → 0 ∫ dx +
( 2 x−x )
¿
1
+¿
1
ϵ1
2
ϵ 2→ 0+ ¿
lim
2−ϵ 2
∫
1
¿¿
.
dx
¿
( 2 x− x 2)
So, we have avoided also this a point where the function was becoming unbounded. Now, in
both the cases we have to integrate this and then later on we will supply this limit as ϵ 1 → 0+¿ ¿
and this ϵ 2 → 0+¿ ¿. So, let us look at this integral here we have basically this
1
dx
= ∫
∫ x ( 2−x
) 2
[
1
1
1
1
x .
+ dx= [−ln(2−x)+ ln x ]= ln
2−x x
2
2 2−x
]
¿ from the positive side and this values we will write
For example, here where the ϵ lim
→0 ¿
+¿
1
down first so, above here 1.
lim
+¿
ϵ1→ 0
[
ϵ
1
ln 1 +
2 2−ϵ 1
]
¿
lim
ϵ 2→ 0+ ¿
[
¿¿
2−ϵ 2
1
ln
¿
2 2 −( 2−ϵ 2)
]
So, ϵ goes to 0 this will be becoming again the ∞. So, this integral does not converge because
we are getting the ∞ here.
Well, so that is true. So, we have 2 minus x and this is ∞. So, here going to ∞ this is also
going to ∞. So, our integral is becoming ∞ and this ∞ +∞. So, this is not convergent.
387
(Refer Slide Time: 25:29)
So, in this case we will we get this value I mean this integral does not converge because, this
is converging to going to ∞ the value and this integral diverges.
(Refer Slide Time: 25:39)
So, here we also use in further lectures about this test integrals because in the next lectures
we will learn about how to how to test whether this converge this integral converges or it
diverges without evaluating them. So, there we will be using some this test integral. This is
for test integral – I, this will be used there.
388
R
¿. So, we have this integral ∫ 1 dx
So, we have a to R this later on we will as take the Rlim
p
→∞
a x
which we can easily evaluate, this a is positive number; so, here
this. So, we have
1
. So, when we integrate
x−b
1
and this −1 will come. So, x−p+1 and then so, − p+1, so, 1− p will also
xp
comes. So, this is the integral then a ¿ R, but this is like taking p≠ 1 because when we take
p=1 we will get this
1
which is the logarithmic here of x and then we will take the limit.
x
So, there will be two possibilities when p=1, then this will be the integral and then p=1, this
will be the integral and then we can substitute the upper limit here R power p minus 1 and we
can substitute the lower limit as a and we will get this integral here. So, for p=1 we have
ln
1
1
1
R
− p−1 .
and in the second case when p≠ 1 we will get
p−1
1−p R
a
a
[
]
(Refer Slide Time: 27:31)
So, these are the integrals integral value for this a to R and now we will see here what will
¿. So, basically we have a interest about test integral is
happen to when we take this Rlim
→∞
∞
∫ x1p dx. We have already evaluated this now we will put the limits.
a
389
R
∞ , p ≤1
1
1
dx=
lim
⁡
dx
=
1
1
∫ x p R→ ∞ ∫ x p
, p>1
a
a
p−1
p−1 a
∞
{
(Refer Slide Time: 29:08)
We have one more integral which will be used for the test integral. This is called test integral
– II
b
∫(
a+ϵ
1
dx
p
x−a )
and in this case we consider this integral a+ϵ to b. So, here we have that this as x → a this is
becoming infinite so, unbounded so, we have the second kind integral. So, a+ϵ we will
evaluate this and again here the same idea. We have
1
p which we can integrate. So, there
( x−a )
will be two cases again, when p=1 we will have this ln x−a which will give us this is
ln ( b−a ) −ln ϵ
and when
p≠ 1 then you will integrate this with this will be
1
1
1
− p−1 .
p−1
1−p ( b−a )
ϵ
[
]
¿.
Now, we will again discuss the case that what will happen when ϵ lim
→0 ¿
+¿
390
b
∫(
a
1
dx=
p
x−a )
lim
b
+¿
ϵ → 0 ⁡∫
a+ϵ
¿
∞, p≥ 1
1
dx= 1
¿
1
p
, p< 1
( x −a )
1− p (b−a) p −1
{
So, again this integral will be used as a test integral in further lectures and we will use this
result that p ≥1this is ∞ and this integral converges when p<1.
(Refer Slide Time: 31:33)
b
So, today we have discussed the improper integral of this kind
∫ f ( x ) dx and there were two
a
are possibilities that a=−∞and/or b=∞ and f ( x ) is bounded. We have also seen that there
might be a situation that f ( x ) is unbounded at one or more points of a ≤ x ≤ b. . So, in that case
also this is a improper integral of a second kind and they we have seen how to evaluate those
integrals basically using that limit. So, if it is a of first kind then this infinity whether it is in
the lower range or the upper one you will replace by a finite number R and then we will
evaluate the integral later on we will take the limit as R tending to ∞.
In the second case at the point where the function is unbounded that we have to avoid by
introducing a small number epsilon which later on we will move to 0 will go to 0 in the
limiting scenario. So, we can evaluate those integrals, but in the next lecture we will see that
how to how to find without evaluating whether the integral converges or it diverges.
391
(Refer Slide Time: 32:57)
So, here these are the references we have used to prepare these lectures and.
Thank you very much.
392
Engineering Mathematics - I
Prof. Jitendra Kumar
Department of Mathematics
Indian Institute of Technology, Kharagpur
Lecture – 22
Improper Integrals (Contd.)
So, welcome back to the lectures on the Engineering Mathematics-1, and this is lecture
number-22. And we will continue our discussion on integral calculus, so in particular
improper integrals of type-1. And today we will be talking about the convergence of type-1
integrals. In the last lecture we have seen how to evaluate those integrals or the integrals of a
type-1. And with the basis of the evaluation indeed we can conclude whether the integral
converges or it diverges. But today we will be talking about, some test based on which we
can conclude that the integral converges or diverges without evaluating the integral.
(Refer Slide Time: 01:02)
So, just to recall in the previous lecture we have a discuss this test case, which was
∞
∫ x1p dx converges for p>1 & divergesif p ≤1. So, this test this test integral will be very useful
a
to compare the integrals with other improper integrals and to conclude the convergence of the
underline integral. So, here again this convergence of this depends on p. So, for all values of
p>1, this integral convergence; and p ≤1, this integral diverges.
393
(Refer Slide Time: 01:42)
So, here we have we will be talking about type-1 integrals again to recall, we have this type-1
b
integral ∫ f ( x ) dx when our f ( x ) is bounded for any value of in this range a to b, a=−∞and/or
a
b=∞. So, both are so like −∞ to +∞ that is also type-1 integral. So, we will be talking about
the convergence of such integrals where infinity appears in the in the limit.
So, this is a very useful test comparison test it is called comparison test-1. So, here we
suppose that f and g are integrable over [ a , c ] ,∀ c ≥ a and that 0 ≤ f ≤ g , ∀ x>a , then
∞
i.
∞
∫ f ( x ) dx converges if ∫ g ( x ) dx converges
a
∞
ii.
a
∞
∫ g ( x ) dx diverges if ∫ f ( x ) dx diverges
a
a
So, here what is this integral, integral g so g is taking larger values then f . So, in that case if
this integral, which is the bigger integral in that sense, because g is taking larger values. So,
∞
we have indeed this relation when we have this g bigger that this integral this
∫ g ( x ) dx will
a
∞
be having the larger value then this integral
∫ f ( x ) dx, whatever range we are talking about
a
394
here like a to ∞ and a to ∞ if these integral exist naturally in that case, we have such a result
here.
So, if this integral converges if this integral converges, because and this is the larger value,
because g is taking a larger value at any point x greater than a so, in that case naturally that
this integral also converges, because if this value exists and this is a smaller than that so
naturally this also exist.
∞
And there is another conclusion which we can conclude out of it that
∫ g ( x ) dxdiverges if this
a
∞
∞
integral ∫ f ( x ) dx diverges. So, if ∫ f ( x ) dx the smaller integral this diverges. So, naturally we
a
a
can conclude that the integral over a to ∞ of this gwill also diverge, because this f is taking
the smaller values at any point x here greater than a; and if this integral exist, so naturally if it
does not exist this f with a smaller values, then naturally this will also not exist this will also
diverge.
(Refer Slide Time: 04:40)
So, here we have another test which is called the limit comparison test or the comparison test2, so this is again a useful test here. So, we have this f and g they are integrable over the
range a ¿ c for any value of c greater than a. And we have again this that f is taking nonnegative values and g is strictly positive for all values of xgreater than a.
395
And then we will be talking about, when we take this limit x approaches to infinity. So, if we
take this lim
x →∞
f ( x)
, we have two functions we are actually getting this behavior that when
g( x)
x → ∞ of this ratio of the two functions, and if it comes to be just k a non-zero number; in that
case we can conclude from here that the behaviour of this f and g is similar when x → ∞,
because we are getting some constant there.
∞
So, regarding to this integrals where we have
∫ f ( x ) dx the type-1 integral. And this table can
a
converge or diverge together because of this reason here; we have the problem when x → ∞,.
And what we have observed that this f ( x ) and g ( x ) , they behave a similarly as x goes to
∞
infinity and in that case we can easily conclude that so this integral
∫ f ( x ) dx or this integral
a
∞
∫ g ( x ) dx ; either both will converge or both will diverge when we get this limit as k non-zero
a
number.
So, we are not proving all these results, but by intuition we can see all these results that they
hold. Here for example, so again as I said here this behavior of both the functions is same as
x → ∞. And in that case, we can conclude easily that if this converges the other one will also
converge and if this converge this was also converge. And this is going to be very useful,
because for a given integrals suppose this f ( x )is given we will construct a g whose a
behaviour is known or that the integral over this g is known whether it converges or diverges.
And based on the behaviour of this integral, we can easily conclude the convergence of the
other integral.
Further, what will happen if this k is 0 for example; k is 0 means that this is converging faster
to infinity than this one than this function, then only this will dominate here the g will
dominate and then x → ∞ this will go to 0. So, in this case here g is a dominating function,
therefore we will get this k=0. And in this case if this g integral converges, so we have this
function which is a dominating as x → ∞ than thisf ( x ).
And if this converges then naturally the other integral will also converge, so that is again a
simple so, because this limit is coming to be 0 as x → ∞. So, here the g is a growing faster
396
thanf as x → ∞, so this is approaching to infinity faster. Then this one, and therefore we are
getting this 0 there. So, this is a and that is a reason here if this g converges, so the f will also
converge.
And again in the other way around if this k is ∞ here. So, if this is approaching to infinity in
the limiting case this ratio, so in that case this f is dominating and then if we can conclude
that if g diverges. So, if we have the result that this g diverges, so we can tell something
about the integral f which will also diverge, because this gthe f is dominating function. So,
this is taking its growing much faster than this one and that is the reason we are getting
infinity there. So, in that case this integralf will also diverge.
∞
If k=∞ and
∫ g ( x ) dx diverges then
a
∞
∫ f ( x ) dx diverges. So, these are the simple comparison
a
test which we have discussed and now we will go for some problems sample problems.
(Refer Slide Time: 09:23)
∞
So, for example this we will test now the convergence of this integral
∫
1
dx
2
x √ x +1
. So, we
have to here now actually find another function which we can relate or which is comparable
to a similar to this as x approaches to infinity, again we have the problem at infinity only that
is what we are talking about these improper integrals. So, here we will see that how this
function is behaving as x approaches to infinity.
397
So, basically what usually we do here, so we take x and here also we will take xcommon. So,
we have then the
1
2
x √ x +1
. So, this term will just approach to 1 as x approaches to ∞, so we
will not create any trouble. And now we have here x2 , so
function the integrand is behaving as
1
behaves as x goes to ∞ and that will help us to set the
x2
functions for comparison.
Note that
1
1
2
x √ x +1 x
2
(Refer Slide Time: 10:37)
Let f ( x )=
1
2
x √ x +1
and g ( x ) =
1
. So, this function here the given
x2
1
x2
lim x
f ( x) x→∞
lim
= 2 =1(≠0)
x →∞ g ( x )
√ x +1
398
∞
∞
As ∫ 1 dx converges ⟹ ∫
2
1
x
1
dx
2
x √ x +1
converges
(Refer Slide Time: 12:09)
So, here the main point is that we have to look for the behavior of this integrand as x
approaches to infinity, how this behaves in terms of the simple function 1 over x power p
type function. And then we can easily get this limit, in this case it is like non-zero, so 1 and
then we can conclude that this converges. So, this will also converge.
(Refer Slide Time: 13:09)
399
∞
Well. So, the next problem we will test the convergence of this integral
∫
1
x2
√ x5 +1
dx. So, in
2
this case also you will also approach same as before. So, we have this f ( x )=
x
, and then
√ x5 +1
the behavior of this function we will test here. So, we have x2 in the numerator and here if
you take x5/ 2 So, x 5 and there is square root there are so 1 plus 1 over x power 5 and then
there
1
here.
√x
So, this term will not create any problem well as x → ∞, because this will be just one here.
So, the behavior will be influenced by these two terms. So, we have 1 over here x power 5 by
2 and then minus 2, and this term square root which is not creating any trouble as x → ∞,, and
this one here. So, this is behaving like 1 over x power 1 by it is a 4 5 minus 4 by 2, so 1 by 2.
So, square root x, so
1
1
. So, this is behaving like
as x → ∞. So, based on this now we can
√x
√x
select the function, we can choose the function g.
(Refer Slide Time: 14:35)
2
lim f ( x ) lim x √ x
1
So, g ( x ) = . And now in this case, we will take again this x → ∞
= x→∞ 5
=1. So, this
√x
g(x)
√ x +1
limit will be coming as 1.
400
So, again we are getting a non-zero limit here as the ratio of these two functions. So, both the
integrals will behave the same. So, in this case we know about this g ( x ) integral that this
1
.
√x
So, here the power of xpower is half which is less than 1. So, this basically diverges this
integral as this was a test integral and we have the divergence whenever this power of this x
is less than or equal to 1.
So, this integral diverges and since both the integrals will behave the same. So, the integral
given in the problem will also diverge by the comparison test, so this will also diverge. So,
we have proved the divergence of this integral again using these comparison test ok.
(Refer Slide Time: 16:01)
∞
So, this is again another problem where we will show that this integral
2
∫ e−x dx converges.
0
1
So, in this case we will break this integral as
∫e
−x
0
2
∞
2
dx +∫ e−x dx . So, this is the trick very
1
1
2
often we used to remove this part here 0 to 1, because ∫ e−x dx is a proper integral. So, this is
0
not bothering as now, so this is proper integral that means we have the value of this finite
value of this integral.
401
Now, we will discuss only this integral here. So, if it is a convergent integral, then everything
will converge; if it diverges naturally, the given integral will diverge. So, here now to apply
2
the comparison test what we observe, now that this e x if we expand this one. So, we have
4
x
1+ x + +…, this is the expansion. And this will be greater than always greater than x2 term.
2!
2
So, we have all these positive term whether x is strictly positive. So, for all x>0 or x<0 this
term is going to be larger than this x2 term, because x2 term already sets here. In fact, x=0
also, so for all x this is this is true now. So, 1+ x 2 this will be larger than this one, because this
is exactly the same term we have taken here and all other are positive terms.
So, in any case this for all x this equality this inequality will hold. And now what we will do
2
here. So, this we have the result now that e x > x2 . So, what we can invert it, so we have
2
e−x <
1 . So, we are not taking here
x=0 indeed in our problem now x>1. So, this is naturally
x2
true for x>1.
2
−x
So, we have this result that e <
∞
integral
2
∫ e−x dx
1
term. And we also know the test integral that this test
x2
that this converges. So, this converges and then by this comparison test,
1
because this function here
1
is taking larger values then e−x , and this integral which is
x2
2
dominating integral this converges. So, naturally this one will also so this will imply that this
∞
2
integral ∫ e−x dx also converges.
1
1
And now this integral converges and also this is proper integral
2
∫ e−x dx
which converges,
0
so the given integral here it will converge, so this converges which we want to show. So, this
was easy by handling in this way that we have taken this two integrals into consideration the
idea was because if x is greater than 1, we can easily in fact this not equal to 0; we can just
402
divert it here
1
and by breaking into two integrals that help, because the first one was the
x2
proper one.
And now we have x greater than 1 anyway. So, it convergence and then both the sum
converges and the original the given integral converges. So, now you will take one more
∞
example. So, so that this integral ∫
0
sin 2 x
dx converges.
x2
(Refer Slide Time: 20:02)
So, in this case we will also deal this because at x=0, we have this problem here with x. So,
1
∞
sin 2 x
sin 2 x
let us just break again this ∫ 2 dx +∫ 2 dx. And this the first integral here is a proper
x
x
0
1
integral, we have discussed already in the last lecture, because this limit as x → 0is finite; so
this is a proper integral where that was the only problematic point as x → 0, but this limit is
finite. And in that case, we do not have any problem in the integrand even when x → 0, so
this is a proper integral.
The second one, now we will discuss that this also convergence and the reason is clear,
2
because the integrand which is given here
sin x 1
2 ≤ 2 . And we know that the
x
x
403
∞
∫ x12 dx
1
that
also converges. So, here this is a dominating integral, because this
1
is taking larger values
x2
2
then the
sin x
2 .
x
∞
And this integral converges, so by comparison test we know that this
2
∫ sinx 2 x dx will also
0
∞
converge. So, this will also converge based on this test integral which was
∫ x12 dx
this
1
converges. So, here this was a dominating integral. So, naturally this will also converge. So,
we will take another example now.
(Refer Slide Time: 22:16)
∞
x tan
Here we will show that this ∫
1
−1
x
1
4 3
d x diverges. So, now we will show here that this
( 1+ x )
integral diverges. So, let us take this f ( x) the integrand term, so let us assume that this is
−1
f ( x). So, over f ( x) is
x tan x
1
4 3
( 1+ x )
. So, this x we can bring to the denominators. So, tan−1 x and
we have x here and this x 4 also we will take out of this, because we want to see the behavior
as x → 0. So, in that case we take this x 4/ 3 here out, this is x−1 this x.
404
And then we have again here x−4+1. So, this is 1/3. So, as x approaches to infinity this will be
1 and here was x → 0, this will be also a constant term
can see by this term here which is a x power, so
1
x
1
3
π
. So, mainly the behaviour of this we
2
or this is x
−1
3
as x → ∞.
So, this function here the integrand behaves as x1−1/ 3 when x → ∞. And now we got this
function g( x). So, if we take this function as g ( x ) is equal to
1
1
x3
, then we can gather limit. So,
−1
x tan x
f ( x)
1 ,. So, I can
the limit as x → ∞ of this
function. So, the limit x → ∞ the f ( x) was
4 3
g( x)
( 1+ x )
1
take to the numerator x 3 .
So, now what do we have here. So, this is limit x → ∞ and there we have
as x → ∞ this will be
x
4/ 3
−1
tan x
1
x4 /3 ( 1+ x−4 ) 3
. So,
π
π
, and this will go to 1, so we got this limit.
2
2
So, now since this limit is same this limit is a non-zero number. So, now the conclusion is
that both integrals integral over f and integral over g, they will behave the same. So, in this
∞
case we know that this integral here 0 to so ∫
1
1
x
1
3
d x . So, here the p the power is less than 1.
So, this integral will diverge so diverge. So, this integral will diverge and so the integral over
f will also diverge, which is given here in the problem.
∞
So, this integral diverges because we have a compared this with
∫ 11 d x integral, and they
1
x3
both has two different has to be behave the same because of the reason that they behave their
behaviour is same as x → ∞, well.
405
(Refer Slide Time: 26:26)
So, having this now we go to the conclusion now. So, we have seen two comparison tests test
number 1 was when we have a this relation in f that 0 ≤ f ( x ) ≤ g ( x ). And in that case if we
know that this g converges, this g converges the larger integral which is taking the function
taking large values. If this converges, the other one will also converge; and if that diverge the
smaller one if that diverges, the other one will also diverge.
(Refer Slide Time: 27:03)
So, the comparison test-2 was again we have these relations. So, it can be strictly equal also
because that x → ∞ at least this g( x) should not be 0. So, we have if this value is k, and in
406
that case if k ≠ 0, they both will behave the same and if k comes to be equal to 0 that means,
this is growing much faster; in that case if this converges, the other one will also converge.
And we have also the result if this k is come infinity, and this diverges the g integral which is
the smaller one now this f dominates. So, in that case if the smaller one diverges naturally,
the bigger one will also diverge so that was the conclusion of this lecture these comparison
test-1 and the comparison test-2.
(Refer Slide Time: 28:03)
So, these are the references we have used to prepare these lectures.
And, thank you very much.
407
Engineering Mathematics - I
Prof. Jitendra Kumar
Department of Mathematics
Indian Institute of Technology, Kharagpur
Lecture - 23
Improper Integrals (Contd.)
(Refer Slide Time: 00:25)
So, welcome to the lectures on Engineering Mathematics-1, and this is lecture number-23.
We are talking about the Integral Calculus in particular Improper Integrals. So, we will
continue on the discussion on the convergence of type-1 integrals. Now, we have one more
test, which was not discussed in the previous lecture that is called the Dirichlet’s test. And
this is again a very important test for testing the convergence of improper integrals of type-1.
408
(Refer Slide Time: 00:31)
So, here we suppose that this f and g, these are the two functions defined from this a to ∞,
and they are taking the real value. And they are such that, that f is integrable on each interval,
so a, b and b can take any value greater than a. So, this f is integrable on each integral [ a ,b ],
b>a.
b
And the other one the integral
∫ f ( x ) dx, this is the
second condition, these integrals are
a
uniformly bounded. So, integrals means that this b can take any value, so we are not
restricting on b. So, these integrals here are uniformly bounded. So, what does that mean that
there exists number C here, a constant C >0 such that the value of this integral, the absolute
b
value of this
|∫ |
f ( x ) dx ≤C and for all b>a.
a
Naturally, b is some finite number we are talking about. So, here for any value b here, which
is greater than a, if this integral is bounded by a constant C, it is this C is not depending on
the number b here. So, in that case we call that this is uniformly bounded. So, this bound is
independent of this b. Then we call that this integral is uniformly, these integrals are
uniformly bounded. So, these are the two conditions on f , f is integrable. And this is uniform,
these are these integrals are uniformly bounded.
409
And the function g is monotone and bounded. So, again the g is either monotonically
increasing or monotonically decreasing, and the bounded so the values of these functions are
g( x)=0 . So, this is a g is a monotonic
also finite the bounded And then we have this lim
x →∞
function which is approaching to 0 as x → ∞, so that is another condition.
∞
And in that case this improper integral
∫ f ( x ) g ( x ) dx. So, we have now the product here in
a
the integrant f ( x ) g ( x ), and this converges that is the result. So, we will not go through the
proof of this result. So, what is important here the important is to check that these integrals
here, they are uniformly bounded. So, meaning we have to compute this integral, and see
whether we can bound that integral by some constant C, which is independent of this number
b here for any finite number b if we can do that, so we have this condition uniformly
boundedness.
And the second one then g is monotone, and so monotonically it is going to 0 as x → ∞. So,
in that case so these are the two main conditions here. And that those conditions we have that
this integral the product here converges. So, this is a very useful integral, because if we know
that one of them, so for example g is here this bounded and monotone going to 0 here. And
then these integral over f converges, so then we can conclude about this product as well that
what will happen to this integral. And it will converge, if these conditions are satisfied.
(Refer Slide Time: 04:09)
410
So, here let us go through this example, where we can use this result easily to prove such
∞
integrals
∫ sinx px dx is convergent for
p greater than equal to 0. So, for any p here greater
1
∞
than 0. This integral
∫ sinx px dx is convergent, we will show with the help of this is Dirichlet
1
test.
So, in this case we let that f ( x )=sin x, so we have this function here f ( x )=sin x. And the
other one g ( x ) =
1
. So, how this is useful now, because we know about this g that this is
xp
monotonically decreasing to 0 as x → ∞ for this p positive. And this integral the sine x, which
b
we can easily show that
|∫
1
|
sin x dx .
So, this integral will be the −cos x, and then we have this limit 1¿ b, so which will be this
|cos 1−cos b|this integral value, and then the absolute value of these we have to consider. So,
this cos is bounded by 1 always, and we can use this inequality here that this is
≤|cos 1|+|cos b|, and in that case this will be bounded by 2.
(Refer Slide Time: 05:45)
411
So, we can easily show that this integral here is bounded by by 2 for any value of b we take,
which is greater than 1 less than ∞. So, what we have seen that there is a uniform bound here.
So, whatever values of b we take, the value of this integral will be bounded by 2. And this is
what we want for uniform boundedness of this integral now.
The other function g, which is monotonically decreasing function and naturally this tends to
0 as x → ∞ for any value of p positive. So, using that Dirichlet test now, so we can talk about
this product the integral of this product sin x and product with
1
that means, this integral
xp
∞
∫ sinx px dx this converges for any value of p>0.
1
So, this Dirichlet test is very useful now that we have just consider as a ratio of the two
function again one was f ( x ), and another was g ( x ) . And this function f ( x ) has this nice
property above this uniform integrability. And the second one was monotonically decreasing
to 0, therefore we could conclude with the Dirichlet test that this converges.
(Refer Slide Time: 07:05)
∞
Another of similar kind we can test the convergence of this ∫
0
∞
1
∞
∫ sinx x e−x dx =∫ sinx x e−x dx +∫ sinx x e−x dx
0
0
1
412
sin x −x
e dx this integral here.
x
∞
∞
∫ sinx x e−x dx is convergent and ∫ sinx x e−x dx is proper integral. Note that
1
0
|
b
e
−x
||
∞
|
∫ sinx x e−x dx ≤ ∫ sin x dx =|−cos x|=|cos 1|+|cos b|≤ 2
1
1
−x
is monotone and lim e =0 .
x →∞
(Refer Slide Time: 07:16)
So, for any value of this a positive. So, what we will do we can prove this again easily, so
∞
this is e power minus x cos x wait so this was the first problem ∫
0
sin x −x
e dx.
x
So, in this case we will again this use the idea that we can break this easily
1
∞
∫ sinx x e−x dx +∫ sinx x e−x dx this function. So, the limit here of the integrant exist, when
0
1
x → ∞ and at all other point there is no singularity is there is no unboundedness. So, this is a
proper integral the first one. The second one now because of this x → ∞, we have to test that
what type of result we get from here.
413
So, now we take this integral here a ¿ b, so note that the integral one and then above one we
can take this b is
sin x
sin x
dx. So, if we take this integral here
dx, this x is going from 1 to
x
x
any number this b. So, here the x we can replace, because this is the minimum value if you
replace this one, so that the integral will be the larger one.
So, we have 1¿ b and this x is replaced by 1 the lowest possible value, so that the integrant is
the larger one. So, we have this integral bigger than this one as sin x dx. And this we know
now that we can integrate this with −cos x and then 1¿ b, so this value will be cos 1 and
−b cos b, and we can take that absolute value there. So, this absolute value of these functions
will be again the absolute value here, which is bounded by 2.
sin x
So, we can prove that that this integral here
the
x
b
∫ sinx x dx for any value of b>1 any
1
finite value. This is uniformly bounded, and this value is 2. So, this is uniformly this integral
is uniformly bounded by 2. And this e−x function, so e−x function is is monotone, and it
¿ this e−x that approaches to 0. So, we have that
approaches to to 0 if we take the lim
x →∞
property also satisfy.
So, then we can use Dirichlet test there, because the one integral with the
sin x
we have seen
x
that there is monotone that is uniformly bounded, and the other one here the g( x), which is
−x
e , it is going to 0 as x → ∞ ,. So, in this case we can apply now Dirichlet results Dirichlet
test, which says that this integral converges. So, this integral converges and the other one is
∞
proper integral. So, we have the original integral here ∫
0
sin x −x
e dx that converges well.
x
So, we do one more problem here that is problem number-2. So, here we will it is a similar
∞
kind of problem, we have
∫
( 1−e−x ) cos x
x2
a
dx in this case. And the idea again we can break
∞
∞
cos x
cos x e−x
into two parts, so I will take here ∫ 2 dx as 1 integral and the other ∫
dx.
x
x2
a
a
414
So, this first integral this is converges absolutely this converges converges absolutely. And
the reason is clear, because by the comparison test we can easily conclude, because this
cos x 1
≤ 2 indeed this absolutely a concept we will explain again.
x2
x
| |
So, here the absolutely means that the if we take even the absolute value of the integrant this
is this converges, so this
have
cos x 1
≤ 2 , because this cos x the values are bounded by 1. So, we
x2
x
| |
1
and we know, this integral as
x2
∞
∫ x12 dx that converges. And in that case this integral
a
will also converge, because this is dominating function
1
and the whose integral converges.
x2
So, naturally that integral will converge.
−x
The second one here we can again use the same argument, because again this
cos x e
. So,
2
x
this value here will be bounded by by 1, because e−x is also decreasing function. So, it will
take the highest value, and we have then x → a. So, in any case this will b will also bounded
by1, so this second integrand is also bounded by
1
. And by the comparison test, we can
x2
again conclude that this also converges absolutely.
We can also use here the Dirichlet test, because we can take this e−x. So, if you want to use
−x
this Dirichlet test here, we can take this function
e
2 . And this function approaches to to 0 as
x
x → ∞ as x → ∞ is this is monotone function e−x is decreasing, and
1
is also decreasing. So,
x2
both the functions, so here we have the decreasing function for whatever values of x. And
∞
this cos x the integral of the ∫ cos x dx, we can again show that this is bounded by 2.
a
415
So, we can use that Dirichlet test like we have done in the earlier problems, so this is similar
to the problem number-1. So, we can here also show that this converges, so this also
∞
converges. And the given integral the ∫
( 1−e−x ) cos x
a
x2
dx converges.
(Refer Slide Time: 15:02)
b
So, now just in note here that this integral of this type
∫ f ( x ) dx, we are not you normally
−∞
discussing here, we are just discussing the integrals where we have infinity in place of this b.
So, this a to infinity type of integrals we are discussing, but this also we can easily b, because
we can substitute for example here x=−t.
∞
And then what will happen that this integral will be just minus
∫ f ( −t ) dt. So, again we have
−b
a similar type integral, which we are discussing for the convergence where the infinity
appears above. So, if we have this type of integrals, we can just by simple substitution, we
can converge into this type of integral, and then discuss the convergence the way we have
discussed earlier.
416
(Refer Slide Time: 15:55)
So, this is the point here, we were talking about the absolute convergence, another important
factor here that what is the absolute convergence. So, as I said before, the absolute
convergence means that this integral converges absolutely meaning that if with the absolute
value over the integrant, if this converges. So, if this integral converges, we call that this
integral converges absolutely, because we have taken the positive values of this integrant for
the range here 0 to ∞. So, if this converges, then we call that this integral converges
absolutely.
And there is a one more term, which is used here that this integral converges conditionally
meaning that this integral converges, but does not converge absolutely. So, here meaning is
that this converges, but not absolutely. In that case, we call usually that this integral
converges conditionally.
417
(Refer Slide Time: 17:00)
∞
So, here like coming to this example here
∫ sinx px dx, and we can show now that this integral
1
converges absolutely for p>1. For any value of p>1, this integral converges absolutely. The
idea is clear which was also mentioned in previous example a bit. So, here we have the finite;
if you take the absolute value of this integrant that means, the
sin x
, so here x taking positive
xp
values, and p>1.
¿
So, here we have the ¿ sin x∨ p ¿, because this is positive. So, we do not have to use the
x
absolute value here. And this is bounded by
greater than 1. So, we can bound this by
∞
the integral ∫
1
1
, because the sin x will never take value
xp
1
, and we know now by the comparison test that
xp
1
dx.
xp
This integral converges. So, what we have shown here that this integral also converges,
because this we have by the comparison test. So, we have taken the integrant here that the the
418
sin x
1
, which was less than equal to p . And we have we know the result that this
p
x
x
∞
convergence the integral of ∫
1
1
dx, this converges.
xp
And therefore, by the comparison test we have shown that this converges meaning that
integral in other words converges absolutely yeah. So, we can take (Refer Time: 18:47) of
because at present this is taking positive, negative values in the range. So, if we take indeed
all the positive values, in that case also this integral converges for this is important that we
have shown here for p>1. This integral converges absolutely, when p>1. We will see later
on that when p=1, this integral does not converge absolutely indeed it converges, but not
absolutely.
(Refer Slide Time: 19:18)
So, going to the next problem. So, this is there is a result also that this is such integral
converges if this converges meaning with the absolute value, because this is going to be a
larger value than this value of the integral. So, if this converges naturally this converges,
because here we have positive negative and many this cancelation will come and this value
will be less than the value here with while taking the absolute value of the integrant.
So, this converge so if this converges, but the converse is not true meaning that so this will
converge if this converges, but if this converges this may not converge. So, if the integral
converges, the integral may not converge absolutely. And that is the way standard example,
419
∞
which we will discuss now here that ∫
1
sin x
dx this integral converges conditionally meaning
xp
that this integral converges, but when we take the absolute value here over the integrant, then
this integral does not converge.
∞
So, first we will show that this integral converges, which is a trivial task now. So,
∫ f ( x ) dx,
0
1
sin x
we have written as the sum of these two integrals ∫
dx and
x
0
∞
∫ sinx x dx. And note that
1
the first integral is a proper integral, so naturally it converges.
And the second one, we have this infinity this we have to discuss. So, this is a proper integral.
And here we have seen in this example-1 that here given x p and p is greater than 1, so that
integral converges. So, naturally this integral also converges
sin x
, we have seen already in
x
∞
example-1, so that means, that this integral converges. So, the integral ∫
0
sin x
dx converges.
x
And next, we will show that when we take the absolute value over the integrant that integrant
does not converge. So, we have this example, which says that the converse is not true,
because here we have the convergence of the integral, but the integral does not converge
absolutely. The other way around this is obvious, because if we taking all the positive value is
still we can get this integral, so naturally when we take the when the function takes positive
and negative values, it will certainly converge.
420
(Refer Slide Time: 22:04)
So, here we will show now that this integral does not converge, and this is a bit involved
∞
|sinx x |dx. And we write down this as a sum of
now. So, we take this absolute value∫
0
∞ ( n+ 1 ) π
∑
∫
n=0
nπ
|sin x|
x
dx.So, here this integral is exactly this integral. So, we have broken here like
n=0, so we are going from 0 to π. And then this integrant and the value, then we are going
from n=1, so that means π to 2 π, and this integrant and so on. So, we have taken the 0 to π,
π to 2 π, and then 2 π to 3 π , and so on. So, this integral 0 to ∞, we have broken into several
this is small segments over the range. So, 0 to π, π to 2 π, and then 2 π to 3 π, and so on,
which we have written in this summation form.
421
(Refer Slide Time: 23:14)
So, this is the integral. And now, we will make a substitution here that x=nπ + y. So, we
make the substitution x=nπ + y in this integral, so by substituting this, so dx will become this
dy. And then this nπ will be replaced accordingly and (n+1) y as well. So, when we
substitute this in this integral, so when the x was taking nπ, the y will take 0 values.
So, again to see here, because this relation was x=nπ +1, so x was taking the lower value as
nπ , so nπ + y. So, this y will take as the value 0. So, this goes from 0 now. And when x was
taking the value nπ +π, so in that case we have this nπ + y, so the y will take the value π. So,
our integral is now 0 to π. And this absolute value ¿ sin nπ + y∨¿, so we have nπ + y. And
here also we have substituted for xthat is nπ + y.
422
(Refer Slide Time: 24:24)
So, we will make another use of this inequality a of this equality that sin ( nπ + y ) =(−1 )n sin y .
For example, n=1. So, we have sin ( π+ y ), which is −sin y, and similarly for other n. So, we
have this equality the technomatric equality, which we will use here now in this numerator,
because we have sin ( nπ + y ).
So, y using this we have now in the numerator minus 1 power n, and then we have this
∞
π
∑∫
n=0 0
|(−1 )n sin y|
(nπ + y )
dy the same integral, which was here. So, this sin ( nπ + y ) is replaced by
∞
n
π
(−1 ) sin y . So, now this summation as it is we have ∑ ∫
n=0 0
|(−1 )n sin y|
(nπ+ y )
dy .
So, here we should note that we have we are not using now the absolute value, because the
this sin yis will take positive values from 0 ¿ π takes value 0 there, and then at
π
1, and then
2
gradually goes to 0. So, this never takes negative values in 0 ¿ π, therefore we have remove
∞
π
this absolute value here. So, ∑ ∫
n=0 0
sin y
dy.
(nπ + y )
So, in this case now this denominator if you take a look here, so y in this case for this integral
as taking 0 value at 2 the π value. So, if we want to make this integral larger, then some other
423
integral here, so we can replace this y by the larger value or this denominator will be the
taking the largest value here at pi here. So, we have nπ +π. So, this y is replaced by this π, so
that this is this integral here becomes smaller integral for this integral, and therefore we have
this inequality that this integral is larger, because this is taking now the integrant is taking
lower value, because this denominator was replaced by the lowest value the highest value
here which is pi there. So, nπ +π, so this is a smaller integral.
∞
π
∞
π
sin y
sin y
dy ≥ ∑ ∫
dy
∑ ∫ (nπ
+y)
n=0 0
n=0 0 (nπ +π )
And now here this is nothing but because there is no y here. So, we can take this out, so we
have n sorry this π we can take out. So, we have pi there, and the integral also we can
evaluate. So, we have this summation n goes 0 to ∞ and
also this π here. So,
1
so will remain as it is so we have
n
1
1
and this n+1, so
or again let me write down.
π (n+1)
π
So, we have this summation n goes from 0 to ∞, and this is
1
, and integral 0 ¿ π and
π (n+1)
we have this sin y. So, this will be −cos y and then 0 ¿ π y. So, when we substitute π, we will
have again −1 there. So, minus minus will be plus, and then minus minus will be plus again,
so say 1 we have to value 2 here. So, this is replaced by 2. So, we have
∞
π
∞
sin y
2
1
dy = ∑
∑ ∫ (nπ
+ π)
π n=0 n+1
n=0 0
So, this integral is equal to this 1, and this is a well known series here, which is the divergent
series, so that means, the sum is ∞ here. So, what we have seen that this integral, which we
have started with the absolute value. This is greater than this sum of the series, which is ∞, so
that means this integral is a divergent integral, because the value of this interval will b larger
than the value of the sum, which is coming already to ∞.
424
(Refer Slide Time: 28:46)
∞
So, in that case we have that this improper integral
|sinx x |dx, the given integral this
∫
0
diverges.
(Refer Slide Time: 28:53)
So, the conclusion we have the Dirichlet test, which says that if we can show this uniform
boundedness of this integral for all b>a. And g is a monotone decreasing to 0 as x → ∞. Then
∞
we have this product in the integrant
∫ f ( x ) g ( x ) dx converges. And we have also discussed
a
425
about the absolute convergence there. And we have seen this nice example which does not
converge absolutely, but it converges, if we do not take this absolute value for the integrant.
(Refer Slide Time: 29:34)
So, these are the references used for preparing these lecturers.
And thank you very much.
426
Engineering Mathematics - I
Prof. Jitendra Kumar
Department of Mathematics
Indian Institute of Technology, Kharagpur
Lecture – 24
Improper Integrals (Contd.)
So, welcome to the lectures on Engineering Mathematics-1, and this is lecture number-24.
And we will be talking about this improper integrals. So, in particular we will discuss today
the convergence of improper integrals of type-2.
(Refer Slide Time: 00:32)
b
And in the last lecture, we have seen this test integral, which was
∫(
a
1
dx. So, this is the
p
x−a )
improper integral of type-2, because this integrand is unbounded, when x goes to this a. So,
this integral which is the test integral, and today we will use this integral to prove the
convergence of other integrals. And this integral converges when p<1, they strictly less than
1. And this integral diverges, if this p ≥1.
427
(Refer Slide Time: 01:11)
So, to start with so we have again to remind we have this type-2 integrals the improper
integral, and we will be using this notation which says that this a plus this notation means,
this f ( x) becomes unbounded, when x goes to a. So, this will be the notation use, when we
used here b−¿ ¿ for example that means this f becomes unbounded, when x goes to b.
So, for the case when we have this integral, where f ( x) becomes unbounded as x goes to b.
In that case, we will de ligand this integral similar to this type of integral by just this
substitution. So, if you substitute here x=b−t, so if you substitute x=b−t, so in that case
this integral will become this
∫
0+¿ ¿
b−a f ( b−t ) dt , so with minus sign.
And then when x is a, so this t will become b−a, and this when x=b−¿¿ , so this will become
0. So, this we will change the limit and we will get this
∫
0+¿ ¿
b−a f ( x ) dx. So, this will be
integral, which will be considering again that f x is unbounded, when x goes to the 0.
So, in this situation so we will be considering this integral, which is again of this nature,
which we take as these standard for the discussion that f your integrand is bounded, when the
argument is going to the lower limit. So, in this case here the when t is approaching to 0, so
this is becoming unbounded. So, the same situation like we have in the above integral that
this f ( x) is unbounded, when x goes to a from the right hand side.
428
(Refer Slide Time: 03:29)
So, moving now to the next, we have the comparison test the similar to the one which we
have already discussed in previous lecture for integral type-1. And in this case, again we
suppose that we have this inequality for the values of the function between this a< x ≤ b. So,
this f is taking non-negative values, and g is taking larger values thenf ,0 ≤ f ≤ g.
And in that case, we have that if this g the bigger one here that theg is taking large values, so
if this integral
∫ b g ( x ) dx
a+¿ ¿
this converges, then naturally the integral, which where this
integrand is taking the lower values, then the
∫ b g ( x ) dx this will also converge. Second
a+¿ ¿
term, we can also conclude out of this relation that if
will diverge if the integral the smaller one,
∫ b g ( x ) dxdiverges, so the ∫ b g ( x ) dx
a+¿ ¿
a+¿ ¿
∫ b f ( x ) dx if it diverges.
a+¿ ¿
So, since this f is taking the lower values than the g. So, if this integral diverges, so naturally
this integral the value is suppose to be bigger than this integral, so definitely this will also
diverge. So, it is a simple comparison test, but very useful for deciding the behavior of such
improper integrals.
429
(Refer Slide Time: 04:58)
There is another test here, it is we call comparison test-2 or it is called the limit test also limit
comparison test. And in this case, again we consider this in same inequality that f is taking
non-negative values, and g is taking larger values then f , 0 ≤ f ≤ g
So, in that case we compute this
lim
x →a+¿
¿
f ( x)
¿
g( x )
. And we have to find for what g, we are getting
this conclusion here. So, we take another function based on the given function f this will be
rather clear, when we discuss the examples.
So, in this case we will choose some gfor given f for the other way around, and compute this
limit. Suppose, this limit
lim
f ( x)
x →a
=k ¿
g( x )
+¿
¿
, then based on the value of k we can decide the nature
of the integral for the given nature of the other integral. So, here for example if k ≠ 0, so if
you are getting this non-zero number as the limit.
So, in that case both the integrals meaning this
∫ b f ( x ) dx , and this ∫ b g ( x ) dx will behave
a+¿ ¿
a+¿ ¿
the same. And the reason is clear that when x → a, this ratio is the non-zero number that
means, they behave the same as x → a. And then there integral will also either both will
diverge or both will converge in this case, and this is very useful.
And most of the examples exactly we use this conclusion that this k, which we got after this
limit of this ratio. And then either depending on these stress integral whether that that
430
converges or the diverges, the other integral the behavior of the other integral we can
conclude.
Further if we get for example this k=0, so in this case what is happening that means, this this
g is growing much faster and perhaps going to ∞ than this f , so we got this 0. So, it this
seems that g is having larger values as x → a. So, if this integral converges in this case this g
integral, then we can tell that this integral f will also converge because of this limit here. This
g will be growing faster, and that is a reason here we are getting this 0. And in that case if
this
∫ b g ( x ) dx converges, the ∫ b f ( x ) dx will also converge. So, without the proof we can
a+¿ ¿
a+¿ ¿
just by intuition, we can feel that these results hold.
Again this is the other way around that if k=∞, so in this case this x → ∞ is getting
unbounded faster than this g function. And if this g function diverges here, so if this g
diverges we can conclude that the other integral, which is
∫ b f ( x ) dx that also diverges. So,
a+¿ ¿
with this simple test we can discuss many examples, where based on the integral which we
have just discuss this test integral. So, this is going to be very useful.
(Refer Slide Time: 08:35)
There was another Dirichlet’s test for type-1 interval we have discussed, and again we have a
parallel test here also, which is more or less the same that if this f integral is uniformly
431
b
bounded that means for any b>a,
|∫ |
f ( x ) dx <C ,∀ b>a , we can bound this by some
a+ϵ
constant for any epsilon positive. In that case, further if we assume that another function g is
monotone and bounded, and approaching to 0 as x → a,
lim
x →a+¿ g ( x )=0 ¿
test concludes that this integral a to b, taking this product
¿, and then this Dirichlet’s
∫ b f ( x ) g ( x ) dx also converges.
a+¿ ¿
So, this Dirichlet’s test is also useful, when we have this kind of product of two integrals.
And one we know that this is monotone, and this is bounded also approaching to 0. And the
other one, here this integral is uniformly bounded. So, in that case we can use this Dirichlet’s
test to prove the convergence of this, when we have in the integrand this product.
(Refer Slide Time: 09:43)
3
For example, we want to test the convergence of this integral ∫
0
dx
2
( 3−x ) √ x +1
. So, if you want
to test the convergence of this integral, then we should not first that the integrand is
unbounded at the upper end.
So, we are not as per the our earlier discussion that we are talking about I mean though
similarly we can deal, when we have this integral is becoming unbounded at the upper bound.
But, the other way we can just converted by a suitable substitution to again, so that our
432
integrand is going to unbounded, when we are approaching to this lower boundary of the
integral.
So, by substituting this 3−x=t. So, if you substitute this one 3−x=t, and the that will
implies that dx =−dt.
3
∫
0
3
dx
2
( 3−x ) √ x +1
=∫
0
dt
2
t √ ( 3−t ) +1
(Refer Slide Time: 11:30)
We can now talk about the convergence of this integral. And now our integrand here is
getting unbounded, when this t is going to 0, so at the lower end of the integral. So, we take
1
1
now for the comparison this function g(t) is equal to . And the reason, why we are taking
t
t
is clear.
In this case when we have this function, so our integrand was f (t ) here
1
2
t √ ( 3−t ) +1
, so this
was the integrand. And as t approaching to 0, so as t approaching to 0, whereas the problem
this
1
is creating problem here there is absolutely no issues this square root, when t
t
approaching to 0 is bounded. So, this function is getting unbounded, because of this
433
1
so that
t
means this behavior when t approaching to 0 is said by this
1
. So, we will choose therefore
t
1
1
this function g(t)= , so that is the reason we have taken here, this function .
t
t
By taking this one now we can compute this lim
t →0
1
f (t )
, so this f ( t )=
. And then
2
g (t )
t √ ( 3−t ) +1
1
g(t)= , so this t will get cancelled, and then in that case when
t
3
so that case this integral
∫
0
lim 1
t →0
2
√ ( 3−t ) +1
, we will get
1
,
√ 10
3
dx
2
( 3−x ) √ x +1
this diverges, because this integral here
∫ 1t dt,
0
when we take this integral, because we are going to now compare with this integral g(t),
because this tests says when limit is a non-zero number, this is a non-zero number here.
3
Then both the integrals both means the one is we have this
∫ 1t dt, so this integral. And this
0
integral which we are testing for the convergence, and the originally this was in the form of x
3
given by this function here. So, since this integral which we are comparing with
∫ 1t dt , this
0
integral diverges. So, if this interval diverges, then this integral will also a diverge. And
3
naturally the given integral this
∫
0
dx
2
( 3−x ) √ x +1
will also diverge. So, by simple this
comparison test, we are able to tell without explicitly computing the value of the integral that
this integral diverges.
434
(Refer Slide Time: 14:24)
4π
Now, we will test the convergence of this integral, where we have
∫ 3 sin x
π
√ x−π
dx. So, now we
have this problem again, when x is approaching to ∞, this integrand is approaching to is
getting unbounded. And now we will discuss the convergence here. So, here it is rather easy
to see that if we take given the absolute value of the integrand, and we have discussed already
in the last lecture, which is called the absolute convergence.
So, if you if we take the absolute value of this integrand, and even though that integral
converges we call that the integral converges absolutely. So, in this case even we take this
absolute value of the integrand, and this is bounded by 1 over because this sin x is bounded
by 1. So, this integral this absolute value is bounded by
|
1
.
√ x−π
3
sin x
1
≤3
√ x−π √ x−π
3
|
And in this case when we have this so and we know about from the test integral, how this
π
integral behaves. So, this integral
1
1
dx, this converges because the power is which
3
√ x−π
∫3
0
is less than 1. So, this integral converges for any power here x minus p, x−π power any
435
power here less than 1this integral converges. So, here we have this power
1
, so this
3
converges.
4π
And in that case we have that this integral ∫
π
sin x
dx over this converges absolutely. So, we
√ x−π
3
have the absolute convergence here, and this is that means, definitely this integral converges,
because we have the absolute convergence. So, absolute convergence implies the
convergence of the integral.
(Refer Slide Time: 16:21)
Now, if we have the improper integrals of 3rd kind, so you just to recall what was the
improper integral of third kind, it was the mixture of the integral of kind 1 and kind 2. So, we
∞
have for example we have this integral ∫
1
1
dx. So, we have both the cases here, we do
x √ x−1
see this infinity in the limit of the integral. And also when x is approaching to 1, this
integrand is getting unbounded. So, we have the integral of type-1 as well as type-2.
So, such improper integrals of we can express in terms improper integrals of the first and the
second kind, and then we can basically evaluate such integrals or we can talk about the
∞
convergence of such integrals. So, in this case, we have an integral is
∫ x √ 1x−1 dx, and this
1
436
we can break into two integrals. So, in the first one we have taken the
2
∞
1
dx .
∫ x √ 1x−1 dx +∫ x √ x−1
1
2
And in the second case, then we have taken from 2 to ∞. Now, what happen? In the first
integral, we have no ∞. So, in that case, but this integrand is getting unbounded, when x is
approaching to 1. So, this is the, a improper integral of improper integral of type of type-2,
which we have discussed.
And here this is the infinity, other than this there is no problem the function is bounded. So,
in whole range from 2 to ∞, and this is the improper integral of so improper integral of type1. So, we have break this integer into two parts. The first one is the improper integral of type2, the second one is then improper integral of type-1, and then we can discuss separately the
convergence of each.
(Refer Slide Time: 18:41)
So, for the convergence, we will use this comparison test and for the first integral here, if we
discussed first. So, we will take this function
1
, and the reason is clear. If we take a look
√ x−1
at this integral, here the first integral in this case, we have this integrand
437
1
, which is
x √ x−1
getting unbounded, when x is approaching to 1. So, this factor here
1
is creating this
√ x−1
unboundedness in the function as x approaching to 1.
So, we will choose we will compare with the function, which is we have taken this here
g1=
1 . If we take the ratio of the here, so f was our integrand. So, here the
2
f ( x) in the
√ x−1
first integral was
lim
x →1
1
. In the second integral, the g( x) is
x √ x−1
1
and if we take this
√ x−1
f ( x ) , and then the take the limit approaching to .
x
1
g(x)
In this case what will happen now, so limit x approaching to 1. And this
survive, the other one will cancel out
f ( x)
1
, so this will
x
g( x)
1
, and as approaches to 1 this is 1. So, we are
√ x−1
getting a non-zero number here as the limit of this ratio that means, this integral the given
integral and the integral of this g.
2
So, integral
∫ x √ 1x−1 dx,
they both will behave the same. And we know from the test
1
2
integral that this integral ∫
1
1
dx this integral converges. So, hence the given integral this
√ x−1
also converges. So, this integral converges, we have seen from the comparison tests that this
integral of type-2 converges.
438
(Refer Slide Time: 20:55)
Now, we will look at the second integral. So, for the second integral, we will choose the
function
1
the reason is again clear, because here the function the problem is because of this
x2
∞. So, we will check the behavior of this function as x approaching to ∞, and from there we
will take the function g for the comparison.
So, if we take this f ( x) here, which was the integral. So,
1
, so it behaves when x
x √ x−1
1
approaching to ∞ so, this is like first I will rewrite this 2 , and here I will take
x
1
√
1−
1 , so I
x
have taken this x 1 to this one.
And now if we look at the behavior as approaching to ∞, so this term here is not creating
trouble, because x goes to ∞ this is 1. And so we have this x2 , so the behavior of this function
is like
1
1
1
2 as x approaching to ∞. So, we got the function here g as 2 , so taking this g as 2 .
x
x
x
439
And if we get this limit for the comparison test, so the limit xgoes to ∞. And this
f ( x) was
f ( x)
, so the
g( x)
1
1
, and then this g( x) again 2 , so we have x square there. So, this is limit x
x √ x−1
x
goes to ∞, and again we will take this common x, so x2 we will get cancel. And we will have
1
√
1−
1 . So, as x goes to infinity, this will approach to 1.
x
So, we have this limit as 1, and in this case again for the same reason both the integrals will
behave the same. So, one this given integral here, and the another one our test integral, what
∞
is the test integral? Test in a
∫ x12 dx. And we know already that the test integral converges,
2
so this converges so this integral converges. And hence the given integral the second integral,
which was the type-1 integral that also converges.
(Refer Slide Time: 23:24)
So, what we have observed now here that both the integrals on the right hand side, they both
converges by this comparison test.
440
(Refer Slide Time: 23:28)
∞
Indeed now in this case, we can also evaluate this function
∫ x √ 1x−1 dx exactly using the
1
same idea by breaking this integral into two parts, where the first integral will be a integral of
type-2, the second will be of integral type-1. And then we can evaluate each one or we can
consider now this integral the same idea, that we have avoided here this one by taking this
1+ ϵ, and the upper limit which was ∞, we have replaced by a number R a large number R,
R
and taking this ∫
1
1
dx .
x √ x−1
Now, we will evaluate first this integral, because it is a proper integral we have no problem at
the lower end and also at the upper end. So, here we will substitute this
√ x−1=t , and then
we can easily evaluate this integral. So, by doing this substitution, so we have a substituted
this x−1=t 2 that means, we have this dx =2t dt.
So, in this case this integral so if we talk about this in definite integral, so x is replaced by
t +1 now, and this is √ x−1 is replaced by t , and this dx =2t dt . So, this t gets cancelled, and
2
R
this integral becomes just
2
dt , which we can integrate. So, this is 2 tan−1 t , so that is a
∫ 1+t
2
1
value of the integral.
441
And now we can put that limit back. So, first this x, so here 2 tan−1 t was a
√ x−1. So, and
now the limit from 1+ ϵ to R, so putting these limits here now, we got 2 and this
2 tan
−1
√ R−1, and then minus this tan inverse, and this x is replaced by 1+ ϵ , so 1 gets
cancels and we get a tan−1 √ ϵ . So, this is the value of the integral the given integral, which is
given here the 2 ( tan−1 √ R−1−tan−1 √ ϵ ) .
(Refer Slide Time: 26:23)
So, with this now we can approach to the given integral. So, in this case, this integral here
∞
∫ x √ 1x−1 dx will get when we take the limit this ϵ to 0 and R to ∞ . So, in this integral the
1
value of this integral, we will pass the limit as R approaching to ∞ and ϵ approaches to 0.
So, when R approaches to ∞, you note that here R approaches to ∞ tan−1 ∞ , which will
become
π
. And minus this tan−¿❑¿, when this ϵ approaches to 0, so tan−1 0 that will become 0
2
π
. So, we have because of this 2 , so this value will be just π.
2
442
So, again not, so this is this is here tan inverse tan inverse pi by tan inverse infinity, so that
will be
π
and this is −0 and then we have here 2 times. So, therefore, we got this pi here. So,
2
the value of this integral is π, and we have seen before as well without evaluating the value
that this is a convergent integral.
(Refer Slide Time: 27:27)
And just a remark here that one needs to be careful to evaluate the improper integrals where
the integrand is not defined or it is not bounded at an interior point of the range of the
integral. So, if we have a point not the end point, if you have a point in the middle where the
integral is getting is integrand is unbounded or it is not defined, we have to be very careful
for the evaluation.
b
So, suppose we have this integral a to ∫ f ( x ) dx. And further we let that f ( x) is unbounded at
a
a point c, where this a<c< b. So, in this case this f ( x) is getting unbounded somewhere at the
point c, which is in the range of this integral. And then we have to break this at that point, so
c
b
a
c
∫ f ( x ) dx +∫ f ( x ) dx .
Now, we have these two integrals of type-2 integral. So, here this is getting unbounded, when
x → c. And this is also getting unbounded, when x approaching to c. So, there are two ways
443
¿. And we
now, one can think of getting this limit one is like we introduce this ϵ , and ϵ lim
→0 ¿
+¿
c−ϵ
have
∫
a
b
f ( x ) dx + ∫ f ( x ) dx. So, at these two places, we have avoided that point c−ϵ and
c +ϵ
c+ ϵ, we have taken the same ϵ at both the places.
In the second case, which is the correct one for the evaluation basically, we take the epsilon
one here for the first integral, and because these two are the separate improper integrals, so
we should deal differently. So, here we have the
c−ϵ 1
lim
ϵ 1 → 0 ⁡ ∫ f ( x ) dx+
lim
a
ϵ 2→ 0+ ¿ ⁡ ∫ f ( x ) dx ¿
+¿
b
¿
¿¿ . So, we have now
c+ ϵ
2
considered, these two integrals differently here we have introduced one epsilon, and then
avoided these point there.
So, in the case when we have convergence integrals, so there is no problem whether you take
this one or this one, this will come of the same. So, but in other cases we have the problem
here, because this might be a divergent integral and this might be the divergent integral and in
that case the value will differ.
(Refer Slide Time: 29:59)
444
1
So, for instance we consider the simple problem here
∫ x13 dx,
and this is equal to
−1
0
∫
−1
1
1
1
dx +∫ 3 dx . So, if we take these two integrals, in that case if we use this first approach
3
x
0 x
by introducing only one ϵ here. So, what will happen, when we compute this
lim
−ϵ
+¿
ϵ→0
[∫
−1
¿
1
]
1
1
dx +∫ 3 dx ¿
x3
ϵ x
which will be ϵ → 0
+¿
lim
¿
−1 1
−1
1
−1 +
1− 2 ¿
2 ϵ2
2
ϵ
[ ( ) ( ) ( ) ( )]
.
So, now in these will cancel out, and this limit will be 0 basically. But, what will happen if
−ϵ 1
evaluate separately the first integral from lim ⁡∫
ϵ 1 → 0 −1
1
1
1
in that case none of the
3 dx + lim ⁡∫ 3 dx
ϵ →0 ϵ x
x
2
2
integrals here, because we will get exactly this here with ϵ 1. And when ϵ 1 goes to 0, this will
be unbounded this will go to ∞. Similarly, this is also not in convergent integral, this also
diverges and that is the reason, we cannot evaluate this here, because both the integrals
appearing at this place they diverges and they do not exist. But, if we do this mistake here by
taking the epsilon at both the places, then the value is coming to be 0.
(Refer Slide Time: 31:33)
So, coming to the conclusion. So, we have these two important comparison test, so which
was one here, where we take this 0 ≤ f ( x ) ≤ g ( x ) , a< x ≤b. And then, we notice that if this g
converges the bigger one the natural, this is smaller one will also converge. The other way
445
around if this smaller one diverges, we can conclude that this the larger one will also diverge,
so we have this comparison test.
∫ b g ( x ) dx conveges⟹ ∫ b f ( x ) dx conveges
a+¿ ¿
a+¿ ¿
∫ b f ( x ) dx diverges ⟹ ∫ b g ( x ) dx diverges
a+¿ ¿
a+¿ ¿
(Refer Slide Time: 32:03)
And another this comparison test-2, which was also very important and very useful where we
of take a another function g and compute this limit, and they based on this number k, we can
conclude the convergence. Let 0 ≤ f ( x ) ≤ g ( x ) , a< x ≤b and
integrals
∫ b f ( x ) dx∧ ∫ b g ( x ) dx behave the same .
a+¿ ¿
a+¿ ¿
And
f ( x)
=k ¿
g( x )
And
a+¿ ¿
∫ b g ( x ) dx conveges⟹ ∫ b f ( x ) dx conveges .
a+¿ ¿
lim
x →a+¿
if
k=∞,
¿
So, if k ≠ 0, both the
if
which
k=0,
implies
∫ b g ( x ) dx diverges ⟹ ∫ b f ( x ) dx diverges . So, this was the comparison test-2.
a+¿ ¿
a+¿ ¿
446
then
that
(Refer Slide Time: 32:38)
And these are the references we have used for preparing the lectures.
And thank you very much.
447
Engineering Mathematics - I
Prof. Jitendra Kumar
Department of Mathematics
Indian Institute of Technology, Kharagpur
Lecture – 25
Beta & Gamma Functions
So, welcome back to the lectures on Engineering Mathematics-1, and this is lecture number
25. And today we will be talking about some special functions beta and gamma which fall
into the class of these improper integrals. So, mainly we will be discussing again the
convergence of such a special functions which are improper integrals.
(Refer Slide Time: 00:38)
So, in the previous lecture we have seen very useful comparison test. So, for example here we
have the function f andg which satisfy this inequality that f ( x ) is a non-negative. So, all the
values of f at any point here a< x ≤ b(∞) in the case of improper integral of type-1. So, we
have this inequality that f ( x ) is taking in non-negative values. And also this g ( x ) is taking
even larger values then f ( x ) for all the given range. And in that case, we compute this ratio of
his f and g and based on the limit of this ratio you will conclude, whether the f converges for
given the convergence of the integral of g or other way round.
So, here we are considering both the cases the integral of type-1 as well as integral of type-2.
So, for integral of type-1 we have the problem because of the infinity where the integrand is
448
bounded, in type-2 integral we have a unbounded function somewhere between this a and b.
So, we considering both together here. So, in case of the improper integral of type-1 you will
take this limit as x → ∞, in case of integral of type 2 we will consider this limit as x → a+ ¿¿.
So, based on this limit here if this k ≠ 0 in that case the both the integrals, the integral
b (∞ )
∫
b (∞ )
f ( x ) dx ; and the other one
a
∫
g ( x ) dx , they both will behave the same this was the
a
b (∞ )
conclusion. And there was another inclusion form there that if k=0, and
∫
g ( x ) dx converges
a
b (∞ )
then
∫ f ( x ) dx converges.
a
So, in that case if that integral converges the other one will also converge and we had also the
b (∞ )
result that if this k=∞, in that case if the other one the
∫
g ( x ) dx integral this diverges in that
a
b (∞ )
case we can also conclude that this
∫ f ( x ) dx
will also diverge. So, based on this these
a
resources some on convergence we will prove the convergence of two special functions
which are the beta and gamma functions. So, first I will define what are those functions.
(Refer Slide Time: 03:29)
449
So, before that we also need these test integrals again from the previous lecture. And there we
have seen these two types of integrals; this was used when we were talking about type-2
b
integrals. So, here the convergence this integral ∫
a
1
dx converges for p<1 and diverges
( x−a ) p
when this p ≥1.
In this case, when we were discussing the integral of improper integrals of type-1; in that case
∞
this integral
∫ x1p dx converges when we have
p>1, and this integral diverges for p ≤1. So,
a
these two integrals will be used again for the comparison test of those special functions.
(Refer Slide Time: 04:21)
So, coming to these functions we have beta and gamma functions. And this beta function is
1
defined by this integral the notation we used here B ( m, n )=∫ x m−1 ( 1−x ) n−1 dxwhere this
0
m>0 , n> 0.
So, we will see that this integral converges only for these values when m and n both are
strictly positive. And there is another function it is called gamma function. So, gamma n is
450
∞
defined by this improper integral. So,
∫ e−x x n−1 dx.
And we will also observed that this
0
integral as well converges when n>0.
1
So, coming to this again this first integral here on B ( m, n )=∫ x m−1 ( 1−x ) n−1 dx , ,m> 0, n>0.
0
So, this integral is integral of type-2 improper integral of type-2, because this integrand
integrand may become unbounded when x → 0 or x → 1. In case this x m−1, so the power here
is negative, then this will become unbounded when x → 0 or when x → 1 in case this n−1 is
negative. So, this will be improper integral of type-2.
In the second case, you have 0 ¿ ∞. So, this infinity appears in the limit. So, this is a improper
in this is an improper integral of type-1 as well as it can be in proper integral of type-2,
because this x n−1 and if this n−1 is negative in that case when x → 0 the integrand will
become unbounded. So, this is a mixed kind integral or the type 3 integrals as per our
notations used in previous lectures. So, we will be discussing today the convergence of these
integrals based on the knowledge we have received from earlier lectures.
(Refer Slide Time: 06:40)
So, here the convergence of the beta function and we will consider the first case when
m , n ≥1. So, in this case what is happening here the m and n? So, this power here for x is
positive and also for ( 1−x )n−1, because n is also greater than equal to 1 that power is also
positive.
451
So, in both the cases this when this m and n are strictly greater than 1, we see that the
integrand is not unbounded function. So, this integral is indeed a proper integral and we do
not have to discuss about the convergence of improper integrals. So, for this case this is
special case and this integral is proper and hence it is convergent.
So, the interesting case here is when m ,n<1. So, in this case when the both m , n<1. So, here
this is m−1 and here n−1; so these powers will become negative, and in that case we have
the improper integral of type-2, because the integrand is becoming unbounded. So, we will
consider now this one. So, we will break into two, because we have the problem at both the
ends when x → 0 as well as when x → 1 because of this term.
So, we will break this integral into two parts,
1
∫x
0
c
m−1
( 1−x )
n−1
dx=∫ x
1
m−1
( 1−x )
n−1
dx +∫ x m−1 ( 1−x )
n−1
dx
⏟ ⏟
0
c
I1
I2
So, let us call this I 1 and I 2, because you will discuss now separately the convergence of
these two integrals.
And in this first case because of this x m−1term, this integral is becoming improper. And in the
second case because of this ( 1−x )n−1 as x → 1, this integral is becoming improper.
(Refer Slide Time: 09:00)
452
c
So, let us discuss one by one. So, in the first case we will take this I 1 =∫ xm−1 ( 1−x )n−1 dx.
0
And we take this f ( x ), the integrand of this improper integral as x m−1 ( 1−x ) n−1. And now for
the comparison we have to take another function.
So, we consider here g( x), because the problem is clear we have in this integrand as x → 0.
So, the problem here is when x → 0. This is not creating any unboundedness as x → 0. So, this
is the function here the part of the function, which is actually making this integrand
unbounded. So, the behavior of this function as x → 0 will be taken from this x m−1.
And that is the reason we have chosen here this g ( x ) =
1
x
1−m
. So, we have taken this g, and
now you will take the ratio, so that this which is making it unbounded will cancel out, and
you will get some finite limit in that case.
¿ , and this f ( x ) , so
So, if we take that limit f x is equal to so we are taking now the xlim
→0 ¿
g( x)
+¿
taking this limit here. This is equal to this
lim
x →0+¿ x 1−m × x m−1 ( 1−x )
n −1
=1 ¿
¿.
And when taking the limits, so here this is cancel out and when taking the limit x → 0, so we
will get this as 1. So, this limit here the ratio of this
f ( x)
is getting 1. And then from the
g( x)
comparison test, we have just reviewed in previous slides, we know that the behavior of this
integral this improper integral will be the same as the behavior of the integral over this g( x)
function.
453
(Refer Slide Time: 11:32)
c
So, what is our integral of g( x) function, this is a ∫
0
1
x
1−m
dx, this is the test integral we have
discussed. And this integral converges, when this 1−m<1, so that means the m>0. So, this is
this integral converges, and m>0.
(Refer Slide Time: 11:59)
And here we will conclude that again the test integral converges, when this 1−m<1 meaning
this m>0, and also we know about the divergence as well. So, this integral diverges, when
this 1−m≥ 1 meaning this m ≤0. So, we know the convergence and the divergence behavior
454
of this test integral here. So, this was the test integral. And behavior of this test integral, and
the given integral is the same, because this ratio is coming to be constant.
So, as per the comparison test our integral I 1 will also converge, when m>0; and it will
diverge, when m ≤0. So, what is the conclusion now here about the integral I 1. So, when we
have this m>0, and we are considering a case when m<1. So, if this 0<m<1, the integral test
integral convergence; and when m ≤0, this integral diverges.
(Refer Slide Time: 13:20)
Now, coming to the second integral, because we have break the original integral into two
1
parts. So, this was the second integral I 2 =∫ xm−1 ( 1−x )
n−1
dx. And in this case again, we take
c
this integrand as this f ( x) function.
And now in this case remember; now the behavior is governed by this ( 1−x ) as x → 1,
because this n−1 is negative. And this part of this function is making this integral divergent
this unbounded. So, this is (1−x )1−n, when x → 1. So, this is the term here in this function, so
accordingly we will choose now again our g( x).
455
So, our g ( x ) =
1
1−n . So, having this g, now we will compute again this limit as before, so
( 1−x )
will compute this
lim
x →1−¿
f (x)
¿
g( x)
¿
. So, this limit here
f (x)
is precisely this one, so ( 1−x )1−n will
g(x)
go to the numerator, and this is the f ( x) here. So, again this term will cancel out, and this we
have x m−1, so as x → 1. So, again we are getting the similar result, which was earlier one.
(Refer Slide Time: 15:02)
¿, and now the behavior of this integral as well. We can conclude from the
So, this x lim
→1 ¿
−¿
behavior of this improper integral this test integral. So, this was our test integral here
c
1
dx . And this converges, when 1−n<1 that means, n>0.
∫ (1−x)
1−n
0
And this diverges, when 1−n ≥1 here that means, the n ≤0. So, this part of the integral also
have this behavior that when 0<n<1, the integral converges; and when n ≤ 0, this integral
diverges. We had the similar results for the other one, but that was the condition was on m.
456
(Refer Slide Time: 15:53)
1
So, combining these two what we get that our beta function, which was
∫ x m−1 ( 1−x )n−1 dx.
0
And this was taken into two parts, and we have discussed each separately for the
convergence. So, this integral will converge for m∧n>0. And any value they can take,
because we have already discussed the convergence.
And for the divergence, if m∧n ≤0 in that case this integral diverges. So, whenever we will
be taking these beta this beta function, we will assume this m∧n>0, because the integral
diverges when these 2 m and n are less than equal to 0.
457
(Refer Slide Time: 16:40)
∞
Now, we will discuss the convergence of gamma function, which is Γ ( n ) =∫ e −x x n−1 dx, and
0
here we will conclude that this exist when n>0.
So, we consider the case that n ≥ 1. And in this case, the integrand is bounded. So, we will
take into two parts again this integral one is 0< x ≤ a, and then the other one is a → ∞. So,
when we take 0 to a, the first integral so meaning this
∞
a
∫e
−x
0
x
n−1
∞
dx=¿∫ e x
−x
0
n−1
dx +∫ e−x xn−1 dx ¿
a
So, this part here, when n is greater than equal to 0. So, n is greater than equal to 0 means this
is there is no singularity, there is no unboundedness coming here of due to this part. And this
a
is nice here e−x, so that means, this integrand is bounded now in ∫ e−x x n−1 dx.
0
458
(Refer Slide Time: 18:11)
And the other parts, so we will check the convergence of the second integral, which is
∞
∫ e−x x n−1 dx, because here we have now this infinity problem even in this case, when n>1.
a
So, we are considering the case n>1 and because of this infinity now, we have this improper
integral of type-1, and we will go through the convergence.
So, here this is have a integrand f ( x )=e−x xn−1. And based on this behavior of this function,
which is as x → ∞ because of this x here, we will consider this g(x) or let us considered this
g ( x)=
1 1
1 1
∨ p , p>1. Anything we can take either 2 ∨ p , p>1 what is the reason taking this.
2
x x
x x
n+1
f ( x)
x
= lim x =0, so what we will get here, we have the x power
So, if we take this lim
x →∞ g ( x )
x →∞ e
suppose we have taken this x2 . So, this x square will be multiplied here with x n−1, so we have
n+ 1
lim
x →∞
x . The reason whether we take here p, so in that case this will be
n−1+ p does not
x
x
e
matter, and n ≥ 1. So, here we have in that case also some x power positive there.
So, now this limit we know already the behavior of these exponential function is this is a is a
fast growing function, then x power whatever positive power we have there. So, this limit
459
here, because this e x will go will grow faster to x → ∞ than this one, so this limit is going to
be 0, this limit is going to be 0 whatever positive power we have.
So, for example if you take x p, so this limit will become the lim
x →∞
x
n−1+p
e
x
. So, this is again, so
we have here n ≥ 1. So, this is greater than equal to 0 . And here we have the p>1.
So, in this case also we have this power greater than 1, so this power x here is greater than 1
¿, the limit will be
in any case, and the same scenario will happen that when we take the lim
x →∞
0. And now we have the comparison test, when this x then the limit is going to 0 that means
this g( x) is growing faster than this f ( x) function, and this is our g( x) here
1
.
x2
So, if this g( x) integral g( x) converges, then we can conclude that the f will also converge,
because from this ratio if we are getting this 0 and that was one case in the comparison test. If
this g is the meaning here is that g is growing faster or getting unbounded faster here or to
when x approaching to ∞, now going growing faster than the f function. And if the integral
of this g converges, then naturally the integral of this f will also converge.
(Refer Slide Time: 21:47)
460
So, here the Γ ( n ) converges for n ≥1, because this integral the g the integral over the g the
∞
integral
∫ x1p dx this converges, when p>1. So, we have seen by taking this g that this ratio
a
here is 0.
And then if this g converges, then the integral over the f will also converge that means, the
integral or the Γ ( n ) will converge. So, here the Γ ( n ) converges n ≥1, because this Γ ( n ) we
have the two integral. One was 0 to a which converges for any arbitrary a, and the second one
we have just tested that this integral will also converge, whenever we have this n ≥1. So, you
need a case we got this convergence for n ≥ 1 case.
(Refer Slide Time: 22:49)
Now, we will consider the second case, which is 0<n<1, when 0<n<1. So, in this case we
have this integral. So, again we have taken 0 ¿ a, and a ¿ ∞.
∞
a
∫e
−x
0
x
n−1
∞
dx=¿∫ e x
−x
0
n−1
dx +∫ e−x xn−1 dx ¿
a
So, based on the earlier study again this converges the same reason, which we have use the
same argument which we have used here, we will choose this function and again here when
0<n<1, the same limit you will get 0. And since this integral g converges, the other one will
also converge.
461
So, in that case with the second integral converges. And now we will test for the second
a
integeral
∫ e−x x n−1 dx, because when 0<n<1, we have this x power negative, so because of
0
this when x approaches into 0, we are getting the unbounded function meaning this is a type2 improper integral.
So, consider the convergence of this we take this f ( x )=e−x xn−1 the integrand again the same
procedure, and the g because this function is getting unbounded because of this g ( x ) =xn−1.
So, we will take this g ( x ) =xn−1 and if we take this limit as x approaching to a , when x
approaching to 0.
In fact, not to a because the problem here is when is approaching to 0. So, e 0, so when x
approaching to 0. So, we will take x approaching to 0, because the integrand is getting
unbounded, when x approaching to 0. So, when we take x approaching to 0, so e 0and this
will become 1, which is not equal to 0. So, this is this is 1 here. So, this limit is coming to be
1, and which is not equal to 0. So, the behavior of this function here will be the same as the
behaviour of this function.
lim
x →0
f (x)
=lim e −x=1 ≠0
g ( x ) x →0
So, now we know already that this function here
1
x
1−n
. So, the integral over this g 0 to 1 and
this converges whenever this 0<n<1, this integral converges we know this test integral. And
based on this we can conclude that our gamma integral will also converge, because they will
behave the same this integral. If this integral of converges, the other one will also converge; if
this diverges, other one will also diverge because of this limit is one here. So, in this case this
integral converges again, when this is 0<n<1.
a
∫ x 11−n dx conveges for 0<n<1⟹ Γ ( n ) converges for 0<n<1.
0
462
(Refer Slide Time: 25:48)
So, now the 3rd case, we will discuss when n ≤0. So, so n ≤0, so that is the case left case-3.
∞
So, in this case,
a
∞
∫ e−x x n−1 dx=¿∫ e−x xn−1 dx +∫ e−x xn−1 dx ¿
0
0
a
and we will separately discuss the convergence.
∞
So, this integral converges which we have already seen before, when we have
∫ e−x x n−1 dx,
a
because we take this now for this one f ( x )=e x
−x
g ( x ) =x
n−1
n−1
, we take this function f ( x). And
, so when n is negative. So, this is getting unbounded. And again we have to take
this limit as x approaching to 0, so when we take this limit x approaches to x approaches to 0.
So, in this case again let me point out here, this is x approaching to 0 from the right side. And
in this case also this e 0 will be 1 which is not equal to 0. So, we have again the same situation
a
∫ x 11−n dx, they will behave
that if this integral if this integral over g, and this integral here
0
a
exactly the same. And that can be seen here, so we have
∫ x 11−n dx.
And this integral
0
diverges, when n ≤ 0. Because, when n is less than equal to 0, we have this integral 1 plus
463
something. So, this integral will diverge, because this integral converges when 1−n is less
than 1.
But, in this case this power here, when n is negative this will be a greater than equal to 1. And
a
in that case this integral will diverge, and therefore the integral ∫ e−x x n−1 dxwill also diverge.
0
So, we have this Γ ( n ) that it diverges, when n ≤ 0.
So, we have considered all the possibilities for n, and less than equal to 0 and between 0 and
1 and greater than equal to 1. So, in those cases other than this one that integral converges,
but when n ≤0, this integral diverges because of this reason, which we have explained here.
And now that is the conclusion now, so both the integrals this Γ ( n ) the beta, they have the
convergence issues and based on this power here. So, in case of this Γ ( n ) whenever n is
positive, this converges. And the other one also for m ,n positive, it converges.
(Refer Slide Time: 28:42)
1
So, coming to the conclusion we have this beta function
∫ x m−1 ( 1−x )n−1 dx, which converges
0
∞
if m∧n>0 and it diverges when m∧n ≤0. And this Gamma function
∫ e−x x n−1 dx, And we
0
have the same situation here that it converges, when n>0or diverges when n ≤0. So, in the
464
next lecture, we will also see some nice properties of this beta and gamma function also the
evaluation of these such special functions.
(Refer Slide Time: 29:28)
So, these are the references we have used to prepare these lectures.
And thank you very much.
465
Engineering Mathematics – I
Prof. Jitendra Kumar
Department of Mathematics
Indian Institution of Technology, Kharagpur
Lecture – 26
Beta & Gamma Function (Contd.)
So, welcome back to the lectures on Engineering Mathematics – I, and this is lecture number
26. We will continue the discussion on Beta and Gamma Function.
(Refer Slide Time: 00:41)
So, these are the special type of examples for improper integrals and today we will be talking
about mainly their properties and how to evaluate those integrals. And, in the last lecture we
have already seen the convergence of those integrals.
466
(Refer Slide Time: 00:45)
1
And, if you recall that the beta function was this improper integral
∫ x m−1 ( 1−x )n−1 dx which
0
converges when this m∧n>0, this number here in the power exponent so, m and n are strictly
positive and the diverges this integral diverges when m∧n ≤0. And, we have also seen for the
∞
gamma function that this integral , this mixed type of improper integral
∫ e−x x n−1 dx,
0
converges if this n>0and divergence when n ≤ 0.
(Refer Slide Time: 01:31)
467
it
First we will discuss the property one nice property of this beta function which is the
1
symmetry property and this is the improper integral for B ( m, n )=∫ x m−1 ( 1−x ) n−1 dx. And, if
0
we substitute here for 1−x= y then we can easily see that the B ( m, n )=B(n, m).
So, let us do it here. So, if we substitute here 1−x= y; that means, minus this −dx =dy and
then this integral; so, then x=0, so, y=1 and then we have here 0. So, x m−1; so, from here the
x becomes 1− y . So, x m−1; so, we have (1− y)m−1 and 1 minus x is y. So, y n−1 and this
1
dx =−dy. So, this we can revert now. So, ∫ yn−1 ( 1− y )
m−1
dy.
0
So, this is the again the beta function and which as per our notation this will be denoted by
beta. So, this is here ( 1− y )m−1. So, this is y n−1. So, per our notation this is B(n, m). So, this
B ( m, n )=B(n, m). So, we have the symmetry property here. So, does not matter whether we
compute this B ( m, n )∨B(n, m).. So, that now you will keep in mind now in future discussion.
(Refer Slide Time: 03:41)
And so, how to evaluate this beta function; the question is the evaluation because these are
the special integrals, improper integrals and special care has to be taken to evaluate this beta
function. So, suppose here n is a positive number. So, in many cases we can discuss that or
468
we can compute that integral in a very simple fashion. So, first space we are considering here
when n is a positive number.
1
So, in this case we consider this B ( m, n )=∫ x m−1 ( 1−x ) n−1 dx and now, if we integrate this by
0
parts and taking that this n is taken as a positive integer so, we will differentiate this part of
the function and then this will be for the integration when we do this integration by parts. So,
this will be here the integral of this
m
1
1
m
[
x
x
( 1−x )n−1 +∫ ( n−1 )( 1−x )n−2 dx .
m
0 0 m
¿
( n−1 )
∫ x m ( 1−x )n−2 dx
m 0
¿
( n−1 ) ( n−2 ) ⋯ ( n− ( n−1 ) )
xm+ n−2 dx
∫
m ( m+1 ) ⋯ ( m+n−2 ) 0
¿
( n−1 ) !
m ( m+1) ⋯ ( m+n−1 )
]
1
1
So, same thing we can do when we assume this m to be positive and noting that this beta is a
symmetric function which we have just seen before. So, there is absolutely no issues, you will
get the similar formula, this m will be replaced by n.
469
(Refer Slide Time: 11:33)
So, in this case we have just seen when we have taken any positive number. So, when we
have taken any positive integer we got this B ( m, n )=
( n−1 ) !
. And, we can do
m ( m+1 ) ⋯ ( m+ n−1)
the same similar calculations when m is a positive integer and in that case also we will get a
similar result. And, the result will be just we can replace this m by n we will get this
expression here with
( m−1 ) !
.
n ( n+1 ) ⋯ ( m+n−1 )
If both are integers m∧n both are integers so, in this case we have you can either take this
case for example. So, when here m is also integer, so; that means, we have m(m+1) and so
on this product again appearing of these integers with the difference of 1.
B(m, n)=
( m−1) ! ( n−1 ) !
( m+n−1 ) !
470
(Refer Slide Time: 14:25)
∞
Now, evaluation of the gamma function: so, can we have this integral Γ ( n+1 )=∫ e−x xn dx
0
and you will use the same idea that just integrating by parts we will get at nice relation of this
∞
Γ ( n+1 )=−x e | +∫ n xn−1 e−x dx ⟹ Γ ( n+1 ) =n Γn..
n −x ∞
0
0
(Refer Slide Time: 16:25)
So, we got this relation that Γ ( n+1 )=n Γ n. and that is a very useful relation which will be
used to evaluate this gamma functions. If n is a positive integer if n is a positive integer, what
471
will happen? So, when n is a positive integer this we can easily valuate in this case because as
per the definition or as per this reference relation what do we get Γ n, we can write down in
terms of like Γ ( n ) =(n−1) Γ (n−1).
So, here if I if I use here Γ ( n ) and we use this relation so, this will be (n−1) Γ (n−1). Again
for Γ ( n−1) we can use that relation. So, we have Γ ( n−1) =(n−2) Γ (n−2) and we can keep
on repeating this. So, n minus n minus 2 gamma n minus 2; so, Γ ( n−2) =(n−3) Γ (n−3). So,
this we will continue until we get this 2, 1 and then Γ (1). So, at some point we will get like
Γ ( 2 ) there then which we can write Γ ( 1 ) and so on.
So, we will get this when n is a positive integer the simplification we will get that the Γ ( 1 ) we
can easily compute now and now we will end up with because this was a positive integer and
we are just reducing by 1, so, we will end up with this Γ ( 1 ) naturally. So, this Γ ( 1 ) if we see
from this integral so, when we substitute there n=0 if we substitute here n=0 in the integral
we have Γ ( 1 ). So, n=0 means this term will disappear.
(Refer Slide Time: 18:31)
∞
So, we have simply the integral
∫ e−x dx =Γ ( 1 )
and this integral we can evaluate easily
0
because e−x integral is −e −x and we have the limit here 0 to ∞. So, when x approaches to ∞
we will get 1. So, this will be 1. So, the value of this Γ ( 1 ) is the 1.
472
So, having this value now, so, we have Γ ( n ) =( n−1) ( n−2 ) ⋯ ( 2) (1) Γ (1); that means,
Γ ( n ) =( n−1) !. So, that is a nice relation we got that such integral which was improper
integral, but we have a nice relation in terms of the this factorial when n is integer at least in
this simple case. So, the Γ ( n ) =( n−1) !.
(Refer Slide Time: 19:37)
Now, moving next we have another nice of relation not only for the interior, but also for this
half at least Γ
( 12 ) which is going to be again useful. So, if we have that reference relation
Γ ( n+1 )=n Γ n. So, for some non-integer number as well we can compute using that reference
relation and at the end if we end up with this gamma half then certainly we can get the value.
So, this will be again helpful. So, Γ
1
= √ π which we can observe by the definitions of
2
()
∞
Γ ( n ) =∫ e−x x n−1 dx . So, if we substitute this x= y 2 in this integral if we substitute x= y 2 this
0
will have another form of this gamma integral.
So, what will happen for the substitution the dx will become the 2 y dy the limits will remain
∞
2
the same. So, we will have Γ n=2∫ y2 n−1 e− y dy. So, this is another form of this gamma
0
integral Γ nwhich we will use now.
473
∞
2
So, this is 2∫ y2n−1 e− y dy and now, you will substitute for this n as n=
0
∞
1
So, having this we
2
2
have 2∫ e− y dy and this is well known integral which we will also evaluate in when we will
0
∞
2
be talking about the double integral later on, but this one the value of this integral 2∫ e− y dy.
0
∞
2
∫ e− y dy =√ π.
In fact, we will see later that value of this
−∞
So, at present we will assume this, but it is very useful integral and better to also remember
this value
√ π and having this one now we can evaluate this 2 times and the value of this
interval is coming to be
√ π because this is in the half range not from minus into infinity, but
2
it is 0 to ∞. So, this is 2 times and the
√ π and this to get cancel and then we have the value
2
1
√ π . So, Γ 2 = √ π what provided we have just without evaluation we have use this value, but
()
after few lectures we will evaluate this when discussing the double integrals. So, this for later
and.
(Refer Slide Time: 23:21)
474
Now, we have also several different forms like we have seen just for the gamma function by
some substitution we got another form, same thing happened for beta function B ( m, n ). So, let
us just go through some forms that could be useful to evaluate the integrals or to recognize
integrals in a different format. So, we have this standard definition of the B ( m, n )and if you
substitute for example, here x=
−1
1
dy . So, this
; so, in that case your dx will become
2
1+ y
( 1+ y )
will be the relation and then this B ( m, n ) will become. So, let us talk about the limits. So, x,
when x was approaching to 0 here, the y is approaching to ∞. So, this we will get this
integral as this minus sign and this will be approaching to ∞ when x is approaching to 1 when
x is approaching to 1 the y will approach to 0.
So, 1=
1
;; that means, the 1+ y=1. So, y is 0. So, y will approach to 0 and then we have
1+ y
∞
y n−1
dy.
this minus sign so, which will revert the limits. ∫
m+n
0 ( 1+ y )
∞
So, combining those powers now so, here we have this
∫(
0
y m−1
dy. So, this is another
m+n
1+ y )
form of the beta function given here y power n minus 1 and 1 plus y power m plus 1 and we
should note that with this substitution the integral limits change completely.
(Refer Slide Time: 26:21)
475
So, in the original integrate was 0 to 1, but this is also the same function, but having this 0 to
∞ now, into the picture or we can because this is a symmetric function. So, m , n we can
∞
replace m by n and n by m the value will not differ you have ∫
0
y m−1
dy. So, this integral
( 1+ y )m+n
or this integral they are the same having the same value as B ( m, n ).
(Refer Slide Time: 26:51)
We can get another form when we substitute this x=sin 2 θ . So, having this
2
2m−1
2n−1
x=sin θ , dx=2sin θ cosθ so, again this is this will become sin
θ cos
θ dθ. So, limits
when x=0, so, that will be also 0 when 1, so, this will be
π
2
2∫ sin
2m−1
θ cos
2 n−1
θ dθ
π
. So, we are getting here
2
. So, this is the one another form of this B ( m, n ) in terms of the
0
π
2
trigonometric functions. So, B ( m, n ) is coming to be 2 sin 2n−1 θ cos2m−1 θ dθ . And, now we
∫
0
∞
can also look for some different forms of the gamma function Γ ( n ) which is ∫ e−x x n−1 dx.
0
476
(Refer Slide Time: 28:47)
And, in this case if you substitute this ¿ λy . So, x=λy the limits will remain the same and we
∞
have a slightly different form. So, we have this integral, ∫ e−λy λ n−1 yn−1 λ dy.
0
∞
∫ e−λy y n−1 dy= Γn
λn
0
(Refer Slide Time: 29:13)
477
Also if we substitute this e−x=t so, we will have again very difficult form because from here
now we get
1 x
1
1
=e so, we get x=ln , e−x=t and then x=ln ,. So, this −e−x dx=dt . So,
t
t
t
0
there will be no extra term will be coming here and then we have Γ n=−∫
1
(Refer Slide Time: 30:01)
1
0
478
1
t
[ ( )]
So, this is another form of this gamma function ∫ ln
n−1
dt =Γ (n).
1
ln
t
[ ( )]
n−1
dt.
(Refer Slide Time: 30:09)
So, the relation between this gamma and beta function we know that when m∧n both are
integers we have seen this relation that B ( m, n ) is can be written as the
(m-1 ) !(n-1) !
. In
( m+n-1 )!
terms of the gamma function because we have seen that relation that this Γ m is nothing,
B ( m, n )=
Γm Γn .
Γ ( m+ n)
And, what is interesting which we are not showing here that this result with which at the end
here this result also holds when this m and n are not integer, so, they are just the positive
number in that case also this results holds because this gamma is defined for non-integer
values as well naturally. So, this beta m, n is defined as gamma m gamma n over gamma m
plus n.
479
(Refer Slide Time: 31:11)
1
Just an example to show, if you have this integral which we want to evaluate
∫ x 4 ( 1−√ x )5 dx
0
we know the definition of the beta function. So, sometimes difficult to recognize, but in this
case it is not so difficult we by some substitution we can easily convert into this form and
then we can evaluate.
So, for instance in this case if we take this
√ x=t so, if we have taken here x=t 2. So, this will
be 1 minus t which we want to have to see this beta form. So, that means, this x=t 2 we can
1
1
write. So, this would become
∫t
0
8
5
5
( 1−t ) 2 t dt=¿. So, this 2∫ t 9 ( 1−t ) dt. So, this is exactly
0
the integral we are talking about in this beta. So, we can write down this as the beta.
¿ 2 B ( 10 , 6 ) =2
Γ 10 Γ 6
9! 5 !
1
=2
=
Γ 16
15 ! 15015
So, we can you evaluate several such integral with the help of this beta function.
480
(Refer Slide Time: 32:59)
So,
the
conclusion
1
B ( m, n )=∫ x m−1 ( 1−x )
0
is
that
we
have
seen
these
two
definitions
∞
n−1
dx , Γ ( n+1 ) =∫ e−x x n dx and how to evaluate these functions and
0
mainly the interesting was that the gamma half we have a computed Γ
1
= √ π and also this
2
()
relation is very useful between beta and the gamma functions. So, B ( m, n )=
(Refer Slide Time: 33:23)
481
Γm Γn
.
Γ ( m+ n)
So, these are the references used and thank you very much.
482
Engineering Mathematics – I
Prof. Jitendra Kumar
Department of Mathematics
Indian Institute of Technology, Kharagpur
Lecture – 27
Differentiation Under Integral Sign
So, welcome back to the lectures on Engineering Mathematics-1, and this is lecture number27. And we will be continuing our discussion on integral calculus. And today’s topic will be
Differentiation Under Integral Sign. So, in particular we will be talking about Leibnitz rule,
so that is rule where we apply for differentiation under integral sign. And we will go through
also the derivation of this Leibnitz rule and some worked problems.
(Refer Slide Time: 00:43)
So, before we go to define this rule discuss this rule, we need to recall the mean value
theorems. So, we already discuss in previous lectures in differential calculus. This Lagrange
mean value theorem, which says that this ratio here
f ( b ) −f ( a ) '
=f ( ξ ) , where ξ ∈(a ,b). And
b−a
here we had this assumptions that f is continuous on the [a ,b] and differentiable on the (a , b)
.
483
So, having this mean value theorem, we can also include little more based on this Lagrange
b
mean value theorem. So, we have this
1
f ' ( x ) dx=f ' ( ξ ). Because, this integral if we
∫
b−a a
solve, so this is nothing but the f ( x), and then we have limit here a to b, so which will be
f ( b )−f ( a ). So, this f ( b )−f ( a )exactly the same term here, and divided by this (b−a)f ' ( ξ ).
(Refer Slide Time: 02:11)
So, now we will make another assumption. We assume that this f ' ( x )=g( x). So, having this
now we get another mean value theorem for integral calculus. So, here the f ' ( x )=g( x), and
b
we get this rule here so ∫ g ( x ) dx=( b−a ) g ( ξ ) so what is this; what does that mean here.
a
b
So, if you want to integrate this
∫ g ( x ) dx, this will be equal to this difference here of the
a
limit so ( b−a ). And then g ( ξ ), so g the integrand is taken at some point in the interval a to b.
So, we do not know exactly where is this point, but this mean value theorem says that this
integral will be equal to g at some point in the interval, and multiplied by this difference of
the limit ( b−a ).
484
(Refer Slide Time: 03:25)
So, with this with these two mean value theorems, we can now proceed for the Leibnitz rule.
So, this rule says if we have this integral u1 ( α ), so these are the limits. So, it could be some
function of alpha to u2 (α ) and the integrand is function of two variables, so
u2 ( α )
Φ ( α ) = ∫ f ( x , α ) dx . So, this we assume that this is a function of α , because we are
u1 ( α )
integrating over this x. So, what is this integral? This integral depends on alpha. So, we
assume that this is a function of α.
And this rule says that if these u1 ( α ) & u2 (α ) the function sitting in the limits here as a
function of α, they have continuous first order derivatives with respect to α. So, if this is the
case, we can differentiate this Φ with respect to α or in other words we can take the
differentiation of this integral, and the rule says that this
dϕ
will go into the integral.
dα
And we will have the derivative or the differential of this f ( x , α ) . So, the derivative of this
f ( x , α ) will be the partial derivative with respect to α , because there are two variables here.
u2 (α )
So, the partial derivative with respect to α of this
∫
f α (x , α )dx, and then there will be some
u1 ( α )
more terms. So, this rule says that we can take the derivative inside here; we will get partial
derivative of f with respect to α.
485
Then plus the derivative of the upper limit here, so f ( u2 (α) , α )
d u2
. And then minus now for
dα
the lower limit you will have again the derivative of the lower limit f ( u1 (α) ,α )
d u1
. And the
dα
function here, the integrand will be now evaluated at this u1 ( α ), so that is the rule that is easy
to remember.
dϕ
of this integral here
So, when we want to take this, like
dα
u2 (α )
∫ f α (x ,α )dx, so this will be
u1 ( α )
equal to the this derivative will go inside the integral so over the integrand. And this will
become like partial derivative of f with respect to α, so that is the one term.
The second term will have the derivative of the upper limit, and the integrand will be
evaluated again at this upper limit. So, this x will be replaced by u2 ( α ) minus the same
scenario here again. So, the derivative of u1 ( α ) the lower limit with respect to α, and then the
integrand will be evaluated again at this u1 ( α ). So, this is the Leibnitz rule and we will go
though this proof also, which is very simple with the knowledge what we have already
discussed in differential calculus, we can prove that this
u2 ( α )
du
du
dϕ
= ∫ f α (x , α )dx +f ( u2 (α ) , α ) 2 −f ( u1 (α ), α ) 1
d α u (α)
dα
dα
1
(Refer Slide Time: 06:59)
486
So, here the Leibnitz rule, so we check this integral as a function of alpha
u2 ( α )
Φ ( α ) = ∫ f ( x , α ) dx . And then consider this ΔΦ , the increment in Φ as by taking the
u1 ( α )
increment Δ α in α, so we have Φ ( α + Δ α )−Φ ( α ) . So, with this we have here α + Δ α for α.
u2 ( α+ Δα )
So, here α will be replaced by α + Δ α, so we will have ¿
∫
f ( x , α + Δ α ) dx. So, wherever
u1 ( α+ Δα )
we have α that will be replaced by α + Δ α and −Φ ( α ), so again the same integral
u2 (α )
− ∫ f ( x , α ) dx .
u1 ( α )
¿ Φ ( α + Δ α ) −Φ ( α )
u2 ( α+ Δα )
¿
∫
u1 ( α+ Δα )
u 2 ( α)
f ( x , α + Δ α ) dx− ∫ f ( x , α ) dx
u 1 ( α)
Now, we will do little more manipulation here. So, u1 ( α + Δ α ) , and we are going to some
number this which is actually appearing here u1 ( α ), so up to u1 ( α ) this integral, we are going
to break now. So, u1 ( α + Δ α ) to this number u1 ( α ) plus then u1 ( α ) to some other number this
u2 ( α )or the function here u2 ( α ), and then from u2 ( α ) to the end of the limit.
¿
u 1 (α )
u2 ( α )
u2 ( α+ Δα )
∫
f ( x , α + Δ α ) dx + ∫ f ( x , α + Δα ) dx +
∫
u1 ( α+ Δα )
u1 ( α )
u2 ( α )
u2 ( α )
f ( x , α + Δα ) dx− ∫ f ( x , α ) dx
u1 ( α )
So, we have taken this integral as a sum of these three integrals. So, u1 ( α + Δ α ) to u1 ( α ), then
u1 ( α ) to u2 ( α ) and then u1 ( α ) to the end limit there u2 ( α + Δα ), so that is a property of the limit.
We can take many intermediate points and we can break the integral into several integrals.
Then we have here minus exactly that term. So, now we will combine the similar terms, so
here we have the same limits u1 ( α ) to u2 ( α ) here alsou1 ( α ) to u2 ( α ). So, these two terms you
will combine, and we will have then integrand f ( x , α + Δ α ) and minus f ( x , α ) . So, these two
terms will be combined, and then we will have these two terms as well.
487
(Refer Slide Time: 09:19)
So,
combining
u2 (α )
these
two
terms,
u 2 (α +Δα )
∫ [ f ( x , α + Δ α ) −f ( x , α ) ] dx+ ∫
u1 ( α )
u1 ( α+ Δα )
f ( x ,α + Δα ) dx −
u 2 (α )
∫
we
will
get
the
f ( x , α + Δα ) dx.
u1 ( α )
So, now we have these three integrals. And now we will apply the mean value theorem. So,
u2 (α )
here we have this difference of
∫ [ f ( x ,α + Δ α ) −f ( x , α ) ] dx=¿. So, the difference is with the
u1 ( α )
second argument α + Δ α. So, here we will apply the mean value theorem from differential
calculus or the Lagrange mean value theorem
u2 (α )
∫
u 2 ( α)
[ f ( x , α + Δ α ) −f ( x , α ) ] dx=Δ α
u1 ( α )
∫ f α ( x , ξ 1 )dx , ξ1 ∈ ( α , α + Δα )
u 1 ( α)
And in these two integrals, we will apply the mean value theorem from integral calculus,
which was reviewed in the previous slide.
u2 (α +Δα )
∫
f ( x , α + Δ α ) dx=f ( ξ 2 ,α + Δα ) [ u2 ( α + Δα )−u2 ( α ) ] ,
u2 ( α )
ξ 2 ∈ ( u2 ( α ) , u2 ( α + Δα ) )
488
u1 ( α + Δα )
∫
f ( x , α + Δα ) dx=f ( ξ 3 , α + Δα ) [ u1 ( α + Δα ) −u1 ( α ) ] ,
u1 ( α )
ξ 3 ∈ ( u1 ( α ) , u1 ( α + Δα ) )
(Refer Slide Time: 10:34)
So, using mean value theorem for the first integral, we have this difference. So, we will apply
the mean value theorem here, which says that if we divide here by Δ α and multiply by Δ α,
so we can do so, so divide by Δ α and then multiply by Δ α here.
So, this one by Lagrange mean value theorem will be the derivative with respect to α,
because the increment is in α, so derivative with respect to α. And f ( x) and some point here
ξ , which we are calling here ξ 1. So, the ξ 1 belongs to exactly the interval here ( α , α + Δ α ).
And note that this ξ 1 will go to 0 and Δ α goes to 0. So, later on we will be going taking on
the limit. So, this ξ 1 will approach to α as Δ α goes to 0.
u2 (α )
u 2 ( α)
∫ [ f ( x , α + Δ α ) −f ( x , α ) ] dx=Δ α ∫ f α ( x , ξ 1 )dx , ξ1 ∈ ( α , α + Δα )
u1 ( α )
u 1 ( α)
u2 (α +Δα )
So, with this now we have the second integral, which was
∫
f ( x ,α + Δ α ) dx. And now we
u2 ( α )
will use the again the mean value theorem from integral calculus, which says that this here
because the integral is over x, so this like a constant in this reference, so integral is over x.
489
So, there will be a point somewhere in this range or the limits of this integral, where the
integral value will be equal to f . So, some point here, so this x will be replaced by this ξ 2.
And α + Δ α as it is, and then the difference of the limits so u2 ( α + Δ α ) −u2 (α ), where now this
ξ 2 ∈ ( u2 ( α ) , u2 ( α + Δ α ) ) belongs to this limits somewhere. so that was the mean value theorem
for integral calculus.
u2 (α + Δα )
∫
f ( x , α + Δ α ) dx=f ( ξ 2 , α + Δα ) [ u2 ( α + Δα )−u2 ( α ) ] ,
u2 ( α )
ξ 2 ∈ ( u2 ( α ) , u2 ( α + Δα ) )
And now for the second integral the same theorem we will apply, so this x will be replaced
by some ξ 3. So, this function evaluation at the point ξ 3 and the difference this of the limits, so
u1 ( α + Δ α ) −u1 ( α )and again this ξ 3 ∈ ( u1 ( α ) ,u1 ( α + Δα ) ). So, these three integrals will be
replaced by these right hand side terms. So, again we will have one integral, and these two
terms.
u1 ( α + Δα )
∫
f ( x , α + Δα ) dx=f ( ξ 3 , α + Δα ) [ u1 ( α + Δα ) −u1 ( α ) ] ,
u1 ( α )
ξ 3 ∈ ( u1 ( α ) , u1 ( α + Δα ) )
(Refer Slide Time: 13:05)
490
So, this was the original the ΔΦ in the form of integrals. And now we will replace all here we
have applied this Lagrange mean value theorem, then the mean value theorem for integral
here also mean value theorem for integral. So, we will get from the first term, because we
have applied the Lagrange mean value theorem here. So, we get this f alpha and this d delta
alpha will be there.
u 2 (α )
u 2 ( α+ Δα )
ΔΦ= ∫ [ f ( x , α + Δα ) −f ( x , α ) ] dx+
u 1 (α )
∫
u1 ( α+ Δα )
f ( x , α + Δα ) dx −
u 2 ( α)
∫
f ( x , α + Δα ) dx
u1 ( α )
From the second term, we have applied the mean value theorem from integral calculus and
again here as well the mean value theorem from integral calculus. So, these three integrals
now, we have new terms here. And we will divide by Δα to each of these terms. So, here Δα,
and here also this Δα, here we will divide by Δα, and here also we will divide by Δα.
u (α)
f ( ξ 2 , α + Δα ) [ u2 ( α + Δα ) −u2 ( α ) ] f ( ξ 3 , α + Δα ) [ u1 ( α + Δα )−u1 ( α ) ]
ΔΦ Δα
=
f α ( x ,ξ 1 ) dx +
+
∫
Δ α Δ α u (α)
Δα
Δα
2
1
And now with this Δα, so this is Δ u 2. So, this is the increment in u2, when we make
increment in α by Δα. So, this is Δ u 2and this is Δ u1.
(Refer Slide Time: 14:21)
So, with this notation we move now. So, here
Δ u2 Δ u1
, and the Δ α gets cancelled
Δα Δα
491
u2 ( α )
Δ u2
Δ u1
Δϕ
= ∫ f α ( x , ξ 1 )dx +f ( ξ 2 , α + Δα )
−f ( ξ 3 , α + Δα )
Δ α u ( α)
Δα
Δα
1
here. So, in this equation now, we can pass the limit as Δα goes to 0. So, passing this limit
Δα goes to 0, this will become
dϕ
the derivative.
dα
u2 (α )
And this first term, so the ξ 1 will go to α, and we have
∫
f α (x , α )dx. So, this ξ 1 → α , when
u1 ( α )
Δ α → 0. Here we have the ξ 2 term, and the ξ 2 lies between u2 (α ) and u2 (α + Δ α). So, when
Δ α → 0, this ξ 2 will go to u2 (α ). And then here also when Δ α → 0, this will go to α .
And here this will become the derivative again the
d u2
minus the same thing here the ξ 3 lies
dα
between u1 (α ) and u1 (α + Δ α), and when Δ α → 0. So, this term ξ 3 will go to u1 (α ), so we
have here f ( u1 , α + Δ α ) will be alpha as delta alpha goes to 0. And then this delta u 1 over
delta alpha will become the derivative again of u 1 with respect to alpha.
So, we are done with the rules, so which says that this differentiation under integral sign, so
when you want to take the differentiation. It will go to the integrand, so the partial derivative
of f with respect to α, and there will be a term
d u2
so the upper limit, the derivative of the
dα
upper limit. And that same limit will be substituted in the integrand, and minus the lower the
derivative of the lower limit and the same limit will be substituted into the function. So, a
particular case which we often use, so we assume that u1 (α ) and u2 (α ), they do not depend
on α, so they are constant. So, in that case the second and the third term here will be
cancelled, because
d u2 d u1
will become 0.
d α dα
u2 ( α )
du
du
dϕ
= ∫ f α (x , α )dx +f ( u2 (α ) , α ) 2 −f ( u1 (α ), α ) 1
d α u (α)
dα
dα
1
And then we have only the first term. So, if we have the constant limits, we will have only the
one term, when we take the derivative or we differentiate under integral sign. So, we will
492
have only the one term, because these two terms will cancel out or usually we write like
d u1
dα
b
of this integral. So, this
d
∫ f ( x , α ) dx, because these limits are constant, we will go to the
dα a
integrand. And then we have this differentiation under integral sign. So, we can take this
derivative into the integral, but this will be like partial derivative, because f depends on two
b
variables x and , ∫
a
∂f (x, α)
dx.
∂α
b
b
b
d Φ (α )
∂ f ( x ,α ) OR d
∂ f ( x ,α )
=∫
dx
f ( x , α ) dx=∫
dx
∫
dα
∂α
dα a
∂α
a
a
(Refer Slide Time: 17:33)
∞
So, now in this example-1, we will show that the integral
∫ tan
0
ax
π
dx= ln (1+ a),if a≥ 0.
2
2
x ( 1+ x )
−1
So, this idea of this differentiation under integral is very often used to compute such
complicated integrals, and they become very simple with the idea of this differentiation under
integration sign, because we consider actually in this case this integral the given integral as a
function of a, because the a is there as the tan inverse.
493
∞
So, this is the Φ ( a )=∫
0
tan−1 ax
dx. So, the given integral, we have taken as a function of a.
x ( 1+ x2 )
And now we will take the derivative or we will differentiate this with respect to a. So, with
the differentiation if we can see some a simplification or the resulting integral, we can easily
solve then this going to be useful, because the integral given in this form is its a little bit
complicated to evaluate.
But, if we differentiate here with respect to a, then what will happen? So, the limits will be
treated as a constant here though one should note that in general this Leibnitz rule or the
differentiation under integral sign, which the rule we have proved that is not valid for
improper integrals. We need some extra conditions in general, but all these examples that is
applicable and we are not going much into the details about the applicability of this Leibnitz
rule for improper integrals. So, all the examples considered here that rule is applicable.
So, here treating these constant limits, we can just differentiate with respect to a. So,
∞
Φ' ( a )=∫
0
∞
1
1
1
a2
dx=
−
∫ 1−a2 1+ x 2 1+a2 x 2 dx
( 1+a2 x2 ) ( 1+ x 2 )
0
[
]
(Refer Slide Time: 22:01)
And now we can easily integrate this. So,
¿
1
[ tan−1 x−a tan−1 ax ] |∞0 = 1 2 π (1−a)
2
1−a
1−a 2
494
(Refer Slide Time: 23:05)
'
So, we got this derivative here Φ ( a )=
1 π
, and that is the always the process this which
1+ a 2
makes it easier, because now we can easily integrate this differential equation. So, when we
π
integrate this, we will get ϕ ( a ) = ln (1+a)+c, constant of integration an arbitrary constant.
2
But, that constant also is easy to evaluate, because we should know that this
∞
Φ ( a )=∫
0
tan−1 ax and the
dx
ϕ ( a ) , so the value ϕ ( 0 ) =0.
x ( 1+ x2 )
∞
So, we will take the point where we know the integral value easily, so this is
∫ tan 0dx =0.
0
So, here we have v, this information is given which we can use here to evaluate this constant
c.
So, with this ϕ ( 0 ) =0 so here 0. And then this is 0, so ln1=0 again, so we get the c is equal to
π
0. So, we got this constant as0, and then ϕ ( a ) = log (1+ a). So, this log is just the natural
2
logarithmic log(1+a), and this what we want to prove. So, we got this value here by using
this Leibnitz rule or the differentiation under integral sign very quite easily.
495
(Refer Slide Time: 25:00)
∞
−α 2
π
So, another example, where we will be calculating evaluating ∫ e cos αx dx= √ e 4 , so
−x 2
0
∞
2
2
again the same process. So, we will take this as the ϕ ( α )=∫ e−x cos αx dx, and then get the
0
∞
2
derivative of this ϕ ( α ) ,ϕ ' ( α ) =−∫ e−x x sin αx dx. And this αx with respect to α will give us x,
0
and that is exactly the point, because this was not easy to differentiate partially.
2
But, now since x has appeared here with e−x , and now we can easily differentiate. So, we can
easy integrate this term, and differentiate this term, so we can apply the rule of partial of
integral by parts. So, by doing so, we will get this
2
∞
2
e−x
−e−x
ϕ (α )=
sin αx |∞0 +∫
cos αx α dx
2
2
0
'
(
)
So, we have now this here when x goes to infinity, this will go to 0. And when x goes to 0,
because of the sin this will go to 0. So, first term will vanish. And then the second one, we
have this minus sin with half and this e power minus x square cos alpha x the same integral,
which we have started with phi a. So, this will be like minus alpha by 2 and this phi a. So,
minus alpha by 2 and this phi a and this same integral.
496
(Refer Slide Time: 27:25)
'
So, we get this differential equation ϕ ( α ) =
−α
ϕ (α ), so which is very simple to integrate. So,
2
we can take this ϕ (α ) to the left hand side, and here the d α will go to the right hand side and
2
integrating this. So, we will get this ln ϕ(α )=
So, we will get c e
1
c1 e
−α 2
4
−α 2
4
2
−α
−α
+c and this ϕ ( α )=c e 4 .
1
4
, and that e power c we can assume another constant. So, we have
∞
. And now we note that this ( 0 )=∫ e
−x
2
∞
dx. And that
0
2
∫ e−x dx, this integral we have
0
seen in previous lectures as well.
And in next lectures, we will also prove that this is equal to
standard integral is
√ π . So, this value of this
2
√ π . So, we know the ϕ ( 0 ), so by this we can compute now the c , so the
1
2
∞
−α 2
c 1 will be √ π . Because, when we put ∫ e−x cos αx dx= √ π e 4 , and that is the value of the
2
2
2
0
integral we want to evaluate.
497
(Refer Slide Time: 29:18)
α2
So, the third example is again very simple. So, we have this ϕ ( α )=∫ sin αx dx, and we want
x
α
to get this ϕ ' ( α ), where α ≠ 0.. So, it is a direct application of this differentiation under
integral sin.
So, if we differentiate this under integral
α2
3
cos αx
sin α sin α
ϕ ( α ) =∫
x dx +2 α
2 −
x
α
α
α
'
3
sin αx α 2 sin α sin α
| +
−
α α
α
α
2
¿
3
3 sin α −2sin α
¿
α
2
So, this was the direct application of the Leibnitz rule.
498
2
2
(Refer Slide Time: 32:07)
And coming to the conclusion so, we have learned this Leibnitz rule, which is useful for
u2 ( α )
differentiating the integrals. So, if we have the integral Φ ( α ) = ∫ f ( x , α ) dx, we want to take
u1 ( α )
derivative with respect to α there. So, this rule says that
u2 ( α )
du
du
dϕ
= ∫ f α (x , α )dx +f ( u2 (α ) , α ) 2 −f ( u1 (α ), α ) 1
d α u (α)
dα
dα
1
so that is the Leibnitz rule.
499
(Refer Slide Time: 32:53)
These are the references we have used to prepare these lectures.
And thank you very much.
500
Engineering Mathematics - I
Prof. Jitendra Kumar
Department of Mathematics
Indian Institute of Technology, Kharagpur
Lecture - 28
Double Integrals
(Refer Slide Time: 00:22)
So, welcome back, this is lecture number-28. And today, we will be talking about Double
Integrals. In particular, we will go through some simple cases of evaluation and before that
we will introduce the double integrals and their its properties.
501
(Refer Slide Time: 00:35)
So, the integral of functions of single variable, so before we define for the two variables. Let
us just recall the definition, what we had for the single integrals. So, integral
b
n
∫ f ( x ) dx =lim ∑ f ( ck ) Δ x k ⁡. So, what was this sum, if we just take this function f ( x) which
a
n→ ∞ k=1
is defined from a ¿ b, this is x axis and we have your y axis.
So, if we divide this domain from a ¿ b into several pieces, so let us denote here the first point
a by x 0, then we have a x 1 here, x2 , x3 and so on, x n−2, x n−1 and x n. So, we have divided the
whole range a ¿ b into n intervals. And then in each interval we have taken a point, it could be
a middle point or it can be any other point here. So, this c k point and then so c k is between x 0
and x 1 in this interval.
And we take the value at this point of the function, which is here. And so f ( ck ) Δ x k. So, this is
Δ x 1 in our notation. And this is c 1, because this is the first interval and this will be the
f ( x ) f (c1 ). And we are now integrating this summing this product. So, what is this product,
this product will give the area of this rectangle here.
So, this width x1 −x0 , which is Δ x 1 and multiplied by the height, which is f (c 1). So, and then
we are summing all these cases, so all these rectangles. So, we will get finally the area of
these rectangles in the finite sum. But, when we are taking the limits, so in the limiting case
when n is approaching to ∞. So, this error for in each rectangle there is a error part here,
502
which we are for example here we are taking more area then the area under the curve perhaps
or the less we do not know. Here also the same situation, so they could be in error, because
we are taking the area of these rectangles.
So, in the limiting case, when these intervals the length of these intervals are going to 0. So,
in that case this limit here will approach to the area under the curve. So, this integral
b
∫ f ( x ) dx represents the area under this curve here and this is defined as the limit of this sum.
a
So, similar concept we have for the integral of two variables or the double integrals.
(Refer Slide Time: 03:49)
So, let us just go through. So, in this case let f ( x ) be defined in a closed region D of the x , y
plane. So, this is the region here we have define a closed region D, it can be of any shape in
general. So, here we have x axis and the y axis and then we divide again this region into sub
regions so into a smaller regions of area Δ A j. So, the each the sub region will have area Δ A j
and j varies from 1 ,2,… ,n.
So, such rectangles or the squares are actually n and each of them has the area Δ A j. So, if we
take a point ( x j , y j ) are some point in the area Δ A ja general point, we are calling this a
general square or the rectangle, we are calling as Δ A jand we take a point there at which is
denoted by this ( x j , y j ).
503
And then we consider this sum, so we are summing here again the value of the function at
that point. So, here we have taken some point ( x j , y j ) and then there will be a function
defined as a surface in the third dimension, which is usually taken as z axis. So, there will be
a surface. So, on the surface you will have this point there f ( x j , y j ) corresponding to this
point ( x j , y j ) and we have multiplied by this area, this Δ A j.
And if this limit exist, then we denote this integral this value of this limit by this integral over
D, so the double integral D. And f ( x , y) and this d A or sometimes you also used the
notation sheet of dA, which is representing this area of this infinite symbol sub region. So, we
can also denote by dx , dy∨dy , dx, so that is the definition here again its the limit of this sum,
which where we are adding actually this f ( x j , y j ) and multiplied by the area of the smaller
region in the x , y plane.
❑
❑
❑
∬ f ( x , y ) dA∨∬ f ( x , y ) dx dy∨∬ f ( x , y ) dy dx
D
D
D
And the there is a point ( x j , y j ) in each sub region in corresponding to this point, we have the
function value f ( x j , y j ). Adding all these and then taking the limit as n approaches to the
number of these rectangles goes to infinity. And in that case if this limit exist, we define this
by such integral notation.
(Refer Slide Time: 06:38)
504
So, it can easily be proved that this above limit exist, so or this integral exist. If this function
is continuous or piecewise continuous in D, so we have this closed boundary domain and if
the function is piecewise continuous or continuous, so this is sufficient for the existence of
such integrals.
And coming to the geometrical interpretation for these double integrals. So, this is also clear
from the limit definition, which we have seen. So, we have taken the small region in the xy
plane or the whole domain in the xy plane was divided into many sub regions. And this
consider for example, one sub-region here and these corresponding to this we have the
surface here, which is actually the projection given by this one.
So, the double integral. So, we will take a point in this sub region at some point we will take
which we can denote by usually ( x j , y j ). And then corresponding to this point, we will have a
point there in the on the surface here. So, we have this point like ( x j , y j ), and then f ( x j , y j ).
And there we take this product of the area of this rectangle or the sub-region in the xy plane
and multiplied by this height. So, what do we get, when we have this area here and then we
have multiplied by some height. So, we will get again this volume of this sub-region, which
is define here. And then we add such sub regions and take the limits. So, we will get just the
area under this the volume of under the surface in this case.
n
lim ∑ f ( x j , y j ) Δ x Δ y
n→ ∞ j=1
So, this limit here, because this Δ x Δ yfor example this was the area of this triangle or the sub
region defined here, which is multiplied by this f . So, this gives the volume of this smaller
505
sub region and such sub regions we are adding here and that will be denoted by this integral,
so that will represent the volume.
❑
∬ f ( x , y ) dx dy represents volume
D
And in case for example we said this f ( x , y ) is equal to 1, so what we will get, so if we set
here f ( x j , y j ) is equal to 1, so we take the function 1. So, the same value will also give the
area, because we are basically adding Δ x Δ y. So, we are adding all these areas, and we will
get the total area of the domain D. So, in that case the area of D if we for example take this f
to be 1, so why with this double integral, we can get the area of the domain in the xy plane or
basically this represents the volume under that surface over the xy plane well.
(Refer Slide Time: 10:02)
So, coming further now for the evaluation part, before we go to the evaluation let us discuss
some properties of the double integrals. So, similar to the single integral, we have more or
less the same properties for the double integral as well. So, for example we have here the k
multiplied by this function f ( x , y ) and in that case this is a constant k, so that can just come
❑
❑
out the integral, and then we have this integral ∬ k f ( x , y ) dA=k ∬ f ( x , y ) dA
D
D
Similarly, we have the addition here or the difference of the two functions f and g. So, in that
case this integral will be over this f and plus or minus the integral over D.
506
❑
❑
❑
∬ [ f ( x , y ) ± g(x , y)] dA=∬ f ( x , y ) dA ±∬ g ( x , y ) dA
D
D
D
So, another property of this double integral. So, here if we take the function as a non-negative
on the domain D, and we are integrating here this over D, then this value of this integral will
be also greater than or equal to 0.
❑
∬ f ( x , y ) dA ≥0 if f ( x , y ) ≥ 0 on D
D
Similarly, if we have two functions f ( x , y ), which is greater than the function g ( x , y )or its
this f is taking move the larger values then the g on the domain D. In that case, the integral of
this g, f will be also greater than the integral of this g here.
❑
❑
∬ f ( x , y ) dA ≥∬ g ( x , y ) dA if f ( x , y ) ≥ g( x , y) on D
D
D
So, the last one which we are discussing here, so that is the additive property. So, we have
this integral over D and this domain if we break into two parts.
❑
❑
❑
∬ f ( x , y ) dA=∬ f ( x , y ) dA +∬ f ( x , y ) dA if D=D1 ∪ D2
D
D1
D2
So, we had for example the some domain here, which was D and we have defined I mean
divided this into the domain D 1 and then D 2. So, in that case this integral over D, we can
write the integral over this D 1 and then the integral over D 2 the same integrant, same integral
only this region has changed. So, this is over D 1 and this is over D 2, so that is also common
property of the of this single integral. So, we have more or less the similar properties for the
double integral.
507
(Refer Slide Time: 12:18)
Now, coming to the evaluation. So, for the evaluation, we have we will consider first the a
simple scenario. So, for example, if this f ( x , y ) is continuous or defined and bounded in that
case also this integral will exist on this rectangular region. So, we have taken we have
restricted our self to the rectangular region. So, x varies from a ¿ b and yvaries fromc ¿ d.
So, for this rectangular region, which is also given in this figure. So, this rectangular region is
from a ¿ b for x and in the in the y axis, it varies from c ¿ d. So, we have these constant
numbers a ¿ b, and there also c ¿ d . D:a ≤ x ≤ b ,c ≤ y ≤ d . So, this is the simplest case, where we
can define this integral over the such a nice domain, because the limits are clear that for x, we
are wearing from a ¿ b. And in the direction of y we are wearing from c ¿ d and the limits are
constant they are not depending on each other.
So, here this integral over this. So, over such a rectangular region dA will be defined as so
first we will compute the integral with respect to x. For example, we could do with respect to
y first does not matter in this case. So, let us just discuss this one first. So, we take this
❑
integral
∬ f ( x , y ) dA over this dx treating this
y as a constant. So, taking this integral here
D
from a ¿ b, because x varies from a ¿ b. So, we have taken this integral a ¿ b. And now this
will become a function of y, so because we have integrated over this x from a ¿ b. So, the y
will remain as it is so this will be a function of y and then we can so this is a repeated integral.
508
So, now we can integrate with respect to y, so y will vary from c ¿ d. And since the integral
limits here, because the region was nicely define from a ¿ b and the c ¿ d in the direction of y.
We can also compute first the integral with respect to y so this dy. And then y varies from c
to d or which we can call like phi of x, so because this will be a function of x after the
integration. And then this is a function of one variable and we can now integrate over the x,
so the x will vary from a ¿ b.
❑
b
b
∬ f ( x , y ) dA=∫ f ( x , y ) dx=∫
D
a
a
d
{∫
c
f ( x , y ) dy
}
So, in this case it does not matter whether we first compute with respect to x and then with
respect to y or first with respect to y and then with espect to x for such domain. And here the
getting the limits are also much easier, but this is not always the case, because we do not have
a such nice domain all the time.
(Refer Slide Time: 15:23)
So, for non-rectangular regions. Again we will consider a some simple cases and then later on
we will come to the more general cases. So, here for example have a domain is of this kind.
So, this is the domain here. So, our domain of the function so, our function is defined this
domain. So, correspondingly we have the function values there.
So, for this domain here and this domain is bounded by this v1 ( x). So, here we have now
function v1 ( x), which depends on x. So, as the x varies, we have the curve here. And the
509
same thing, we have some other v2 ( x )function with respect to x, so that is boundary here also
changes. In the direction of y, we have this constant. So, at this from x is equal to a to this x
is equal to b, this is the cut. But, in this direction here in the direction of y we have the
functions v1 ( x) and their v2 ( x).
❑
b
{
v 2 (x )
∬ f ( x , y ) dA=∫ ∫
D
a
v 1 (x )
}
f ( x , y ) dy dx
So, for such domains also it is a not difficult to compute the integral over such over this
domain D. And first we assume that this f ( x) naturally is as defined and bounded or
discontinuous or piecewise continuous in this domain. And this v1and v2, they are the
continuous functions. So, we have this domain, which is depicted in this figure.
❑
And then this integral the double integral of this
∬ f ( x , y ) dA , we will take now so we have
D
to now follow this sequencing. So, first we have to integrate with respect to y, because now
with respect to y we have so for getting this limit this is the easiest way to understand. So, of
with respect to y draw a line this parallel to y and then we see that this line here enters at this
v1 ( x )and go out at v2 ( x). So, these are the limits now in the direction of y and then such lines
they goes from here x is equal to a to and everywhere up to x is equal to b.
So, this is the first integral, which is taken over y. So, for ydirection we have the limits here
v1 ( x ) to this v2 ( x ). So, v1 ( x ) to v2 ( x )in the direction of y . So, having this integral first this
again a single integral, because this x we are not touching here, we are integrating with
respect to y. And then taking these limits also from v1 ( x )to v2 ( x ).
And then once we are done with this integral, we will integrate with respect to x. So, for x
now the limits are from a ¿ b. So, for the x limits are constant here a ¿ b, but for the inter inner
integral y the limits were depending on x. So, v1 ( x ) to v2 ( x ). So, for such domain naturally
510
we will follow the sequencing because if we go the other way round that first we integrate
with respect to x, then there will be a problem, because we do not have a such a nice
representation of this whole integral so or whole domain here. So, in this case the y we have
the nice limits here v1 ( x )to v2 ( x ). And for then x we have the constant limits from a ¿ b, so
that is the sequence, we should follow for such domain.
(Refer Slide Time: 19:23)
And if we have for example the domain given in this figure, so here there is a curve which
depends on this y. So, we have v1 ( y ) to this v2 ( y ) and then in the y we are going here from
c ¿ d. So, this is just the other way round, then the earlier one. And for this case also, so we
have this f ( x , y ) access why is defined and bounded in D and v1 and v2are continuous.
So, in this case this integral of this f ( x , y )over such domain will be now we will integrate
first with respect to x. So, in this case we will now discuss, now we will take first with
respect to x. So, here therefore the limit we are going from this v1 ( y ) to v2 ( y ) . And such lines
now will vary from this c, they will go from c ¿ d, so that is the way we can compute the
easily put the limits for x and y.
And in this case, so here the first integral as I said, we will compute with respect to x now.
So, for x we have v1 ( y ) to v2 ( y ) , we have v1 ( y ) and to v2 ( y ) . And once we have this integral,
which is will be a function of y. So, we will integrate with respect to y then and y the limits
will be c ¿ d. So, again in this case we have to see which one is convenient now.
511
So, we have this first we need to get the integral with respect to x, because getting this limits
are as much easier now and for the while we will put the limits fromc ¿ d. So, in this case also
we have seen that we can easily evaluate the integrals, because once we have these limits we
are going back to the single integrals, which we know how to evaluate.
(Refer Slide Time: 21:24)
❑
So, let us go to the example here. So, the first example we have this
∬ xy ( x+ y ) dA=¿. And
R
we will compute what is the value here. And this region is R is bounded by the line y=x and
the curve y=x 2. So, we have this line here and the parabola y=x 2.
So, if we draw I mean this is the first step that we have to draw the region of integration, and
sometimes that is the difficult part. So, here we have y=x line, this y=x line and then we
have y=x 2 this parabola. So, they intersect because y=x and y=x 2, if we solve here. So,
y=x we put, So, here we have x and ( x−1) is equal to 0. So, x is equal to 0, and x is equal to
1 at these two points they intersect.
512
So, at x is equal to 0 we have the y also 0 naturally, so this is the point of intersection. And
the other one when x is equal to 1, so y is also 1 so, this is the another point. So, this is the
area which we call the area of integration. So, over this region or the region of integration, we
❑
will now integrate ∬ xy ( x+ y ) dA . So, having this now we have the possibilities that we first
R
take the integral with respect to y. So, while taking the integral with respect to y, we will
take so with respect to y.
So, we will draw a line here parallel to the y, y is equal to y axis, so this is the line. So, now
you will see that this enters here at x2 and it leaves the domain at this y=x. And that that is
true everywhere in the domain at whatever point you draw this line parallel to the y axis, it
will enter through the y=x 2, and it will leave the domain at y=x line. So, and then such lines
will vary from x is equal to 0 to x is equal to 1.
So, first we will put the limit here for y, because we are in going to integrate first with
respect to y for x for instance in this case now. And so here the limit for y will be x2 , because
at this point when it enters the domain it is x2 . So, here we have this y is equal to x2 . And
then when we leave the domain, this is y is equal to x, so we have y is equal to x. So, these
1
are the limits we have the integrant here
x
∫ ∫
xy ( x + y ) dy dx.
x=0 y =x 2
So, once we integrate this one, then we have to integrate then with respect to x. And the x
limits are clear, because these lines which we have drawn here. They are going from x is
equal to 0 to x is equal to 1 and then we are swapping the whole domain. So, here from x is
equal to 0 to x is equal to 1, these are the limits for x. So, x is equal to 0 to x is equal to 1 or
what we can do we can take the limits other way round as well.
513
(Refer Slide Time: 25:12)
So, if we take if we consider for example first with respect to x, so what we have to do? Let
us draw a line parallel to thisx axis. And this line will is entering here at y=x and living this
domain this region at y=x 2 this parabola. So, while entering here, so we are putting the limit
for x. So, here the x is equal to y that is the entry point, so x is equal to y that is the limit.
And it is leaving at, so here from this y is equal to x square, we can take y is x is equal to √ y .
So, we are leaving the domain here at x=√ y , so these are the limits now for x. And now such
lines will vary from this y is equal to 0 to y is equal to 1, so that will be the limit for y. So,
for y limits here we have y=0 ¿1 or we have for this x limit we have y to
√ y . So, in this
cases we have either the possibility of taking first with respect to y, then with respect to x or
the other way round.
1
√y
∫∫
xy ( x+ y ) dx dy
y =0 x =y
514
(Refer Slide Time: 26:30)
So, let us take the first one and then compute this integral. So, we will take the first one. And
then taking this one we will integrate now. So, after this integration of the inner integral, we
will get this. So, this is how we are doing. So, this integral of the inner one, so we have this
integral here 0 ¿1 and then x2 y+ x y 2 with respect to dy and then dx.
1
x
∫ ∫
(x 2 y+ x y 2 )dy dx
x=0 y =x 2
2
2
3
x y xy
So, this is 0 ¿1 and this we will integrate now with respect to y. So, we will have
,
+
2
3
that is the integral and then we will put the limit from x2 ¿ x . So, x2 ¿ x and then we have the
4
outer integral dx. So, this will be 0 ¿1 the one we putting x there. So, we will have
4
x x
+ ,
2 3
because y is substituted this x is substituted for y. Then3 and then minus for this x, so we
have here x2 . So, you will be substitute x2 , so you will have
7
4
6
6
x
and we will also get here then
2
7
x . So, now this one it is 5 x that is a term here x , we have here and x , we have the term
3
6
2
3
there.
515
(Refer Slide Time: 28:14)
So, this is after the first integration. And now again this is a single integral, so with respect to
x, so we can integrate again this. And we will get these number, because from here we will
get
5
7
1
x
x
will get cancel, we will get . Here we will get . So, when we put the limit here
6
5
7
7
only this x is equal to 1 will contribute, because at x=0 will make this 0. So, here
x
so,
7
7 ×2=14 and then that x7 when we put the limit that will become 1.
8
x
Here also we have , so this is 8 ×3=24 and that is the one there. So, this is a value we can
8
simplify this. So, it will be 3 by 56, so that is the value of this double integral by taking the
integral first with respect to y and then with respect to x what we can do the other way round
as well.
516
(Refer Slide Time: 29:05)
So, one more example here, we want to evaluate now this ∫∫ e
2 x+ 3 y
R
dxdy over this region R.
And R is the region bounded by the x axis, so x is equal to0that means, the y axis y is equal
to 0 we have the x axis and then there is a line x+ y=1. So, the first step always it should be
the sketching the domain. So, if you put the domains, so we have x axis, we have y axis and
then we have the line here x+ y=1. So, this is a triangular domain and again very simple to
evaluate the limits whether we take first in the direction of x or we take first in the direction
of y.
So, suppose we are taking the direction of y first. So, now draw the line parallel to the y axis.
And then we will see that this line enters here at y is equal to 0 and this leaves the domain at
y is equal to so from here we will have y is equal to 1−x. So, the limits will be y is equal to
0 ¿ y is equal to 1−x. And then the x limits will be from 0 ¿1, so then x from 0 ¿1 and then
the dx, so that is the one way of computing the integral. But, in this domain again it is a same
similar limits we can get, when we integrate first with respect to x.
517
(Refer Slide Time: 30:40)
So, for integrating first with respect to x, we will have the limits for the x, so we will draw a
line parallel to this x axis and that enters here x is equal to 0. And leaves here at this line,
where the value of x will be now 1− y. So, x from here will be 1− y. And then we can
integrate with respect to y, so the y will go from 0 ¿1, so y will be from 0 ¿1. And then either
we can take the first integral or the second one to compute.
(Refer Slide Time: 31:13)
So, let us take the first one. So, taking this one here 0 to 1−x, and this xgoes from 0 to 1. So,
we will integrate this the inner one with respect to y. So, we have here e2 x +3 y that is our
518
integrant. So, we are integrating with respect to y that means, after this integral of this
1−x dy the inner one. So, we will get here e2 x is a as a constant with respect to this y and e3 y
3
you will become
e
and then this limit 0 to 1−x. So, when we put the upper limit here, we
3
will get ( e3−3 x−1 ) here and e2 x is sitting here.
1
1
¿ ∫ e2 x ( e3−3 x −1) dx
3 0
(Refer Slide Time: 32:10)
So, this is the first integral the inner one. And now we can integrate this as well because the
single simple integral. So, we will
2
1 −3 e
3 1
+e + .
3 2
2
[
]
So, this when we have the simple domain like the rectangular triangular or when we have
seen with which was bounded by the line and the parabola. So, in these cases finding the
limits whether first with respect to x and then with respect to y or with respect to y first, and
then with respect to x, in either way it is simple to get the limits. And we can evaluate the
integrals by this repeated a single integrals.
519
(Refer Slide Time: 33:07)
So, the conclusion is we have seen the the definition also for the double integral and which
represents the volume or if we said this function to be 1.
n
❑
lim ∑ f ( x j , y j ) Δ A j=∬ f ( x , y ) dA
n→ ∞ j=1
D
It represents volume (or area if f ( x , y )=1).
It can also represent the area of the domain D. The hardest past hardest passed usually is the
evaluating multiple integral is the finding the limits of integration, but in cases which we
have considered today there was simple a geometry and we can easily find the limits of
integration. And the sketch of region of this integration is important, because without
sketching the region we cannot get the idea of the limits of the domain.
520
(Refer Slide Time: 33:55)
So, these are the references which were used to prepare these lectures.
And thank you very much.
521
Engineering Mathematics - I
Prof. Jitendra Kumar
Department of Mathematics
Indian Institute of Technology, Kharagpur
Lecture - 29
Double Integrals (Contd.)
So, welcome back this is lecture number 29 on integral calculus Double Integrals.
(Refer Slide Time: 00:21)
And in particular we will be focusing on the change of order of integration in this lecture.
522
(Refer Slide Time: 00:27)
So, just to recall from the previous lecture, we have gone through some simple geometry and
for example, if you take the rectangular domain of the function f ( x) then this integral the
double integral, we can evaluate by repeating the single integrals 2 time. So, either we can
integrate this function over this dx and the limit for x will go from a ¿ b. So, first evaluate the
singer integral with respect to x and this will be function of y, which can be integrated and
second the step with respect to y. So, the limits of y will be from c ¿ d.
We can change the order; here meaning that we can first evaluate the integral with respect to
y, where the limits of y will go from c ¿ d. And then, at the end we can evaluate the single
integral with respect to x and the limit will be a ¿ b in that case.
❑
d
∬ f ( x , y ) dA=∫
D
c
{
b
b
a
d
} {∫
∫ f ( x , y ) dx dy =∫
a
}
f ( x , y ) dy dx
c
523
(Refer Slide Time: 01:33)
We have also seen some other domain. So, for example, this domain is bounded by the
function v1 and v2 in the direction of y; and in the direction of y we have these constant
limits from x is equal to a ¿ x is equal tob. So, in this situation we have the possibility now
that we should first integrate or it is convenient or it is easier to integrate first in the direction
of y, and the limit of y will be from v1 ( x) to this v2 ( x) and then for the limit of x we will
have a ¿ b.
❑
b
∬ f ( x , y ) dA=∫
D
a
v 2 (x )
{∫
v 1 (x )
}
f ( x , y ) dy dx
So, in this case this double integral will be first the inner one. So, with respect to y the limits
as I said will be from v1 ( x )to this v2 ( x )and later on or at the end we can integrate this with
respect to x the limits will be from a ¿ b. Or, we have this domain for instance. So, in this case
we have to consider first the integration with respect to x because it is easier to set this
variable limits for x. So, the limit of x will be from v1 ( y) to v1 ( y) and then in the second
524
step we will do the integration with respect to y. So, here we will have this integral the inner
one with respect to x, the limits will be v1 ( y) to v1 ( y) and the outer one will have the limits
from c ¿ d.
❑
d
{
v 2 ( y)
∬ f ( x , y ) dA=∫ ∫
D
c
}
f ( x , y ) dx dy
v 1 ( y)
So, for these simple cases we have seen when we have the rectangular domain it is much
easier that we can interchange the limits whether we compute first with respect to x or with
respect to y it does not matter. But, when we have such a domains for example, in this figure
or in the other one then we should see that which one is convenience. So, in the first one we
have realize that, we should first integrate with respect to x and then with respect to y while,
in the second case it is much more convenient to first integrate with respect to x and then
with respect to y. So, this is one a situation where we can see the convenience of the order
here, but there will be other reasons depending on the integrand that which integral whether
with respect to y is easier or possible or convenient or with respect to x first. So, that will be
the discussion now here.
525
(Refer Slide Time: 04:10)
So, for instance we have. So, the question is here that why do we change the order. So, the
a
one reason which will be very clear from this example
a
∫∫
y =0 x =y
have this integral the inner
x
dx dy that suppose we
x + y2
2
x
with respect to x and the limits for x goes from y to a. So,
x + y2
2
x is equal to y to x is equal to a that is the limit of the inner integral and for the outer one we
have y is equal to 0 ¿ y is equal to a.
The question is now if we differentiate if we integrate here with respect to x first in the given
order, then what will happen? We should consider here like 1/2 in the integrand and then we
have
2x
this is our integrand; and then we have the integral here with respect to xfrom y
x + y2
2
to a and with respect to y from 0 to a we have dx dy. So, with respect to xthis will be the
logarithmic functions. So, we will have the ln(x 2 + y2 ) the integral of this with the half and
then we have to also put the limits now for x from y ¿a and then the outer one 0 ¿ a and we
will have dy.
So, this will become the function of y, now which will be a ln(a2 + y2 ) and then here we will
have also minus ln(2 y2 ). So, the point is now integrating this function the ln(a2 + y2 ) or this
ln (2 y ) is very difficult, but what we will realize here? If you change the order of integration
2
526
then this will be much easier to compute. So, if you change the order of integration the first
step is going to be always that we have to the sketch the domain of integration.
(Refer Slide Time: 06:23)
So, in this situation in this case what we have? We have this let us say this is xaxis and we
have here this y axis. So, this limit of the inner one goes from x is equal to y. So, this is the
line here xis equal to y and this goes the inner integral from x is equal to yto the constant
limit a. So, x goes from y to a. So, we starts from this line and it goes up to a. So, let us say
this is the point a here.
So, this is x is equal to a and naturally this is also then y is equal to a point. So, this limit
here x goes from y to. So, this line to this point x is equal to a; so, this is the domain of
integration in our case now, and if you want to change the order of integration. So, this figure
will be very helpful now. So, what we are going to have? We are going to have the same
integrand naturally and then here dy and dx. So, the order will be change.
So, first we have to fix the limit for y here. So, for the limit of y, now we will go through a
line in the y direction and see where it enters and where it leaves the domain. So, it is
entering here always for covering the whole domain, it is entering at this y is equal to 0. So,
the limit here will be like y is equal to0 and it is leaving the domain exactly at this line where
y is equal to x.
527
So, from y is equal to 0 to this y is equal to x and the limit for the x will be from this is xis
equal to 0 to x is equal toa; this is x is equal to a. So, x is equal to 0 to a. So, in this way we
have changed the limit now fromx=0 ¿a and y goes from this 0 ¿ x. And now we can
integrate this because with respect to y when we integrate, this is going to be easier we have
this is x is constant. So, we have
1
term.
x + y2
2
(Refer Slide Time: 08:51)
So, let us just go through here. So, what do we have now? We have this after changing the
order of integration, we have this y goes from 0 ¿ x and x goes from 0 ¿ a. So, while
integrating the inner one first. So, we fix the outer one. So, x from 0 ¿ a and for the inner one
we can integrate now. So,
1
1 −1
tan ( y/ x); with respect to y and then we have
2 will give
x
x +y
2
the limit for yfrom 0 ¿ x and the outer integral x. So, this x will get cancel and we have
π
−1
−1
tan ( y / x), which will be 0 ¿ a. So, tan inverse the upper limit when we put tan (1)= −0.
4
So, tan−1 0=0 and then this dx.
So, what do we have here
π
and then this integral 0 ¿ a dx will be just the x and the upper
4
limit will give us a. So, that is the value of the integral. So, this change of order was very
useful if you would have not change the order of integration, the direct integral was difficult
528
to compute purchase by changing the order of integration it has become trivial and we got the
value as this
π
a.
4
(Refer Slide Time: 10:19)
So, the value of this integral is
π
a.
4
(Refer Slide Time: 10:24)
529
1 2−x
So, we consider few more problems. So, let us discuss this
∫∫
xy dy dx .. And we want to
0 y=x 2
change the order of integration and then we want to evaluate though in this case our integrand
is just xy. So, it will not matter whether we first differentiate with respect to y or with respect
to x the complicacy level would be more or less the same.
So again to change the order of integration we have to sketch the domain here. So, in this
case what do we have? We have this x axis and we have the y axis there and we want to now
draw this. So, this y goes from x2 and up to y is equal to 2−x. So, y is equal to x2 is this
parabola here. So, we have y is equal to x2 and then we have the line. We have the line y is
equal to 2−x. So, whenx is 0 , y is 2 somewhere and when this y is 0, x will be 2 again. So,
this is the line y is equal to 2−x.
So, the region here of the interest is going to be because y goes from x2 ; so, this parabola
2−x. So, this is our region of integration and now we can easily change the order, but before
we need to get what is this point of intersection. So, here y if we put x2 is equal to 2−x and
then we have x2 + x−2; that means, we have x−2 and. So, ( x+2) and ( x−1) is equal to 0. So,
we have the point here x−1,−2. So, x is equal to 1 this is x is equal to 1 and y will be also 1.
So, this point of intersection here is (1,1) the value of x is 1. And also the value of y is 1 and
at this point here the value of yis2. So, y is 2 and here thex will be 2 and y is 0.
So, we have all the coordinates we can change the order now. So, this was 0 to 1and now we
want to have the order xy and dx dy; so, first with respect to x. So, for with respect to x we
have to draw the line parallel to this x axis, and see where it enters and leave the domain. So,
in this case it always enters through this x is equal to zero line. So, here we have x is equal to
0 as the limit and then we have the upper limit here which is up to this x yis equal to 1 we
have the limit of x always enters through this x is equal to 0 and leaves through this y is
equal to x2 ; that means, x is equal to
√ y . So, here x goes from 0 to this parabola which is a
√ y , and this is up to the point y is equal to 1. So, y goes from 0to 1.
So, up to this point we have this and then we will have one more part because when ygoes
from 1¿ 2, then the limits of x will be from 0to this line; that means, we will have another
part here when y goes from 1¿ 2 the limits of x will be again from 0 to 2− y, but now it exists
exit from the line; that means, there xis 2− y. So, this 2− y and the integrand dx dy. So, we
530
have the two parts now; one where y is going from 0 ¿1; that means, this domain here this
domain and the other one the upper one where y is from 1¿ y is from 1¿ 2 where the x is
going from 0 ¿2− y.
1
√y
∫∫
y =0 x =0
2
xy dx dy + ∫
2− y
∫
xy dx dy
y=1 x =0
(Refer Slide Time: 15:02)
So, having change this order of integration, we have this situation of this two integrals. So,
we will evaluate these integral separately and then we can find the value of the integral.
(Refer Slide Time: 15:10)
531
So, considering the first integral here y goes from 0 ¿1 and x goes from 0 to √ y so, the first
integrals so, 0 ¿1 and then here with respect to x. So, y and with respect to x when we
integrate x this will be x2 /2, and we have to put the limits for x is 0 to √ y . And this is dyhere.
So, 0 ¿1and we have y by 2 x 2 will become y again when we put
this dy. So, we have
get the value
√ y−0. So, we have y and
1 2
1
y this will be y 3 and when we put the limit 1 it will be 1−¿. So, we
2
3
1
here.
6
(Refer Slide Time: 16:04)
So, the value of the first integral is
2
2− y
∫∫
1
and then we can evaluate the second integral again
6
xy dx dy . So, here we have the y from 1¿ 2and the second integral x is 0 to 2− y . So,
y =1 x=0
2
x
we are integrating now for x will be
and the limits will be 0 ¿2− y and then we have dy.
2
y
So, this is equal to 1¿ 2 we have , and then x2 . So, 2− y 2 and then minus 0; so, we have this
2
dy. So,
1
and this 1¿ 2. So, y we have has to be multiplied here.
2
532
2
1
So, we will get this ∫ (4 y−4 y 2 + y3 )dy . So, now, we can easily integrate this 4 y and this
21
is −4 y 2 and then we have this y 3 term and with this
is coming to be
1
. So, this evaluation when we do, this
2
5
5
the integral of this. So,
we need to add here.
24
24
(Refer Slide Time: 17:29)
So, this
5
3
3
if we add than this will become as . So, the value of this integral is and we
24
8
8
have used again the idea of change of order of integration which was not necessary in this
case indeed it has the integral has become more complicated by changing the order of
integration, because we have to now considered two domains one was from y ¿1 and the
other one was from 1¿ 2.
So, in this particular case it was absolutely not necessary to change the order in fact, the
direct evaluation would have been easier; but, here the inverse to learn how to change the
order of integration.
533
(Refer Slide Time: 18:13)
Now, this is problem number 2, where we will again change the order of integration and
evaluate
a
∫
y =0
a−√ (a¿¿ 2− y 2)
xy ln ( x+a )
dx dy
( x−a)2
∫
¿¿
and indeed in this case change of order of integration is
x =0
necessary because if we just considered first with respect to x, this integrand is pretty
complicated for integrating with respect to x, but if you change the order. So, you will have
here dy first we will be integrating with respect to dy and in that case we have only this y
there. So, by changing the order we can see directly looking at the integrand, that we can
easily integrate with respect to y this given integrand.
So, for changing the order we have to now draw this order of integration or the domain of this
integration here. So, we have the x axis we have here y axis and xgoes from 0. So, xgoes
from 0 ¿this point. So, what is this? x is equal to a−√(a ¿ ¿2− y2 )¿. So, if we bring this
( x−a) and take whole square, then this will be (a ¿¿ 2− y 2 )¿. And this implies this is
2
( x−a ) +( y2 −a2 ). So, this is the circle which we can draw now first. So, x is equal to a . So,
this is the center a and0 is the center and ais the radius again. So, this is going to be
somewhere here. So, we have the circle of radius a and now if we look at the integral limits.
So, for x it goes from0 to the circle. So, from 0 to the circle and it in the direction of y it is
exactly going to y is equal toathat is the radius of this circle. So, this is small region here that
is the order of that is the region of this integration. So, this region is the order is the region of
534
the integration this d y, which we have to now see when we change the order of integration
what will happen. So, if we change the order. So, first we have to fix the limits for dy.
So, the integrand will remain as it is and then later on with dx. So, changing for y we have to
now draw a line parallel to the y axis and see where it enters the domain and where it exit the
domain. So, for y it is now entering through the circle at any point here to cover the whole
domain; so, entering through the circle. So, what is the question here?
So, that will be y is y=√ a2−( x−a )2 that is the lower limit of y, and it exist from y is equal to
a. So, this is y is equal to a the upper point. And now for the x direction it is fixed now from
xis equal to 0 and to x is equal to a, because these lines move from x is equal to 0 to xis
equal to a. So, that is the outer limit of the integral. So, we have change the order here easily
and now the inner one goes from y is equal to y=√ a2−( x−a )2 to y is equal to a and the inner
one goes from x is equal to 0to x is equal to a.
a
I=∫
a
∫
x=0 y= √ a2 −( x−a )2
xy ln ( x +a )
dy dx
( x−a )2
(Refer Slide Time: 22:05)
So, we have this for the inner one with respect to y and then we have also for with respect to
x. So now, we can easily integrate the inner one. So, integrating this one with respect to y we
will have just the y squared term there. So, the outer integral will be 0 ¿ a, and then the
535
integrand which is if I leave y. So, we have
1 x ln ( x+ a )
. So, now, putting the limit for y 2; so,
2 ( x−a )2
we have [a2 −(a2−(x−a)2 )].
So,a2 −a2 will cancel out then we have ( x−a)2 and that whole will go out with this one. So,
a
I=
1
∫ x ln ⁡( x+a)dx
20
(Refer Slide Time: 23:11)
And this is exactly the result of the integral the inner integral, and now we can evaluate this
outer integral as well.
536
(Refer Slide Time: 23:18)
a
1
I = ∫ x ln ⁡( x+a)dx
20
¿
1
2
[{
2
a
2
] ]
1
a
a
dx
ln ( 2 a ) − ∫ ( x−a ) +
2
20
x+a
} [
2
a
¿ [1+2 ln a]
8
2
a
So, this will be [1+2 ln a] that is the value of this integral.
8
537
(Refer Slide Time: 26:25)
1
√x
Now, this problem we want to change the order of integration ∫ ∫ f ( x , y ) dy dx. So, changing
0 x
the order of integration in this case again let us quickly draw the region, and we have this x
here and we have y. So, ygoes from x ¿ y is equal to
√ x which is y 2=x . So, we have the
parabola we have this line here. So, y is equal to x line; y is equal to x line and we have this
y is equal to y 2 is equal to x the parabola. So, they will intersect first of all at x2 −x is equal
to 0. So, x and ( x−1). So, x=1. So, this parabola will have this. So, this is y 2=x and now if
you look at. So, this y goes from the line to the parabola. So, this was the given order of the
integration, and now we want to change the order. So, what we will have now the f ( x , y) and
we will have dx and dy there.
So, the we will fix first the limit of x. So, we need to draw a line here in the direction of this a
parallel to x, and we was see there where it enters the domain and where it exit the domain.
So, here the entry point is again the parabola. So, thexis the limit will be from y 2 to and this is
leaving the domain exactly at x= y. So, here the x= y and in the direction of. So, this was the
point the (1, 1) point and in the direction of this y now. So, such lines are moving from y is
equal to 0 ¿ y is equal to1. So, that is the change of order of integration. So, for the limits of y
will be 0 ¿1 and limit of x will be y 2 ¿ y and then f ( x , y)dxdy.
1 y
∫∫ f (x , y)dxdy
0 y2
538
(Refer Slide Time: 28:47)
So, one more example that is the last example for this lecture; so, we have now again we
want to change the order of integration and our limit for the y goes from x square by this 4 a
to 3 a minus x. So, again we need to draw the region here. So, x axis and then we have here
the y axis. So, this y goes from x square over 4 a or this x square is equal to 4 a y, and then
we have y there at the upper limit 3 a minus x. So, this is the line here. So, let us draw first
this line and that will be x 0. So, we have a 3 a point here y is equal to 3 a point.
And when the y is 0 so, x is 3 a. So, here x is equal to 3 a. So, this is the line and now we
have this x square is equal to 4 ay. So, that is the parabola x2 =4 ay . So now, here if we look
at the limits for y; so, y goes from the parabola to the line. So, this is the direction for the y
from the parabola to the line, and now we want to change the order of integration; that means,
in the direction of x we have to draw the draw a line parallel to the x axis and again we have
that situation which we have discussed earlier, that there will be two domains because of this
problem here.
Before this line so, in this domain in the lower domain here, we have the entry point exactly
at x is equal to 0, but the exit is the parabola. And in the upper part of this domain here, we
have the entry point this x is equal to 0 the same, but the exit point here is again this line. So,
there are two different regions we have to consider now if you want to change the order of
integration.
539
So, for this first part we will have we want to have now the dx dy. So, for this x here x goes
from0 xgoes from0 and to this parabola which is a √ 4 ay or square to 2 √ ay and then in this
case we have to also see what is this point here which will come as 2 a into 0. So, 2 a into a.
So, up to this xy is equal to a. So, from 0 to y is equal to a, and then the plus from a to 3 a.
So, y is equal to a to 3 a now for the range of this x is this line to that line. So, here xgoes
from 0 again and to this line which was y is equal to 3 a−x. So that means, x here is 3 a− y
and then we have this f ( x , y) and dx and dy.
So, this is the order the change of order of integration we got this two integrals because we
need to divide now the domain into two parts because the line this parallel to this x axis was
not following from one to the another one, but in this lower domain we had from this x is
equal to 0 to this parabola while in the upper domain here when y is greater than a, then we
have the situation that it is entering again at x is equal to 0, but it will exit now from this line
y is equal to 3 a−x and that is the reason we have to break the integral into two parts. So, this
is the result which we have written already here from a to 3 a and 0 to 3 a− y.
a 2 √ ay
∫∫
0
0
3 a 3 a− y
f ( x , y ) dx dy +∫
a
∫
f ( x , y ) dx dy
0
(Refer Slide Time: 33:19)
So, coming to the conclusion what is important when we change the order of integration?
First of all this is very useful very convenient in many situations, when we observe that the
given integrand is not possible to integrate in that given order. So, by changing the order of
540
integration it becomes easier to integrate, but for that the sketching of the region of
integration is very important and then putting the limit of the integration after sketching the
domain it becomes much easier.
(Refer Slide Time: 33:55)
So, these are the references we have used for preparing the lectures.
Thank you very much.
541
Engineering Mathematics - I
Prof. Jitendra Kumar
Department of Mathematics
Indian Institute of Technology, Kharagpur
Lecture - 30
Double Integrals (Contd.)
So, welcome back to the lectures on Engineering Mathematics I and this is lecture number 30
and you will continue our discussion on Double Integrals.
(Refer Slide Time: 00:27)
And today in particular we will be talking about some applications of double integral.
542
(Refer Slide Time: 00:31)
So, let us start with the problem number 1 Using a double integral find the area of the region
enclosed by the parabola y=x 2 and the line y=x .. So, here using a double integral find the area
of the region enclosed by the parabola y is equal to x2 and y is equal to x. So, we have two
curves here: one is the parabola y is equal to x2 , and the other one is the straight line y is
equal to x. So, the area here enclosed by these two curves, we want to find using the double
integral. And, the idea was which has already been discussed in the previous lecture, that we
can simply integrate the function one or the constant function 1 over this region and then we
can get the area of the region bounded by these two curves.
So, precisely we have these two curves y is equal to x this line, and this y is equal to x2 this
parabola. And then we have to get this point of intersection which is pretty simple in this
case. So, we have y is equal to x2 and y is equal to x these two curves. So, y is equal to x is
equal to x2 and this will be give us the point of intersection. So, x2 −x is equal to 0. So, we
will get x is equal to 0 and x is equal to 1 these two points and corresponding y of course, we
can easily get. So, here x is equal to 0 is this point, where we have x 0 and 0, the another
point we have here where the x is 0 and x is 1 and y is1.
So, these are the two points and now we will compute the area of this region enclosed by
these two curves. So, we need to find the limits now. So, the limits of this integral; here we
have the possibility whether we take first in the direction of x and then in the direction of y it
does not matter. So, we will take now for example, in the direction of yfirst and then in the
543
direction of x. So, in the direction of y we have the limits from this curve to that curve, and
these lines will run from x is equal to 0 to x is equal to 1 to go through all this enclosed area.
So, here this curve is y is equal to x2 and the above one here is y is equal to x. So, in the
direction of y we are entering through this curve x is equal to yis equal to x2 and leaving this
area, by this curve y is equal to x the straight line. So, here the limits of y will be y is equal
to x2 to yis equal to x and such lines are moving from x is equal to 0 to x is equal to 1. So,
this integral will give us the area of the integral of the region bounded by these two curves.
(Refer Slide Time: 04:01)
So, we need to compute simply this integral now, which will give us the area of the region
enclosed by these two curves. So, here the outer one we keep as 0 ¿1 and the inner one with
respect to y. So, we will get y and then the limit x2 ¿ x and then we have here dx. So, this will
be0 ¿1 and ythe upper limit is x and the lower limit is x2 . So, x−x2 and then we have dx here
2
and now we can integrate the single integral again. So, here we will get
3
x
x
and
and the
2
3
limits will be 0 ¿1. So, substituting this upper limit here, the lower limit will make anyway
1 1
1
this 0. So, when putting at a limit we have the − . So, this will give us . This is the area
2 3
6
of the region bounded by this parabola and the straight line y is equal tox. So, that is the
answer of this integral.
544
1 x
1
1
2
3 1
∫∫ d y d x=∫ [ y ]x d x=∫ (x−x ) d x= x2 − x3 = 61
0
0 x
0
0
x
2
2
2
[
]
So, we will take another example of little more slightly more complicated. Using a double
integral find the area of the region enclosed by parabola y=x 2 and the line y=x+2. So, here we
have instead of y is equal to x for example, y is equal to x+2 line. So, in this case again we
have one parabola and then the line instead of y is equal to x we have y is equal to x+2.
(Refer Slide Time: 05:46)
So, this is the situation we have this y is equal to x+2 line. So, this intercept here when x is 0
, the y value is2 and when x is −1 this is 1 here. So, this point of intersection again; so, y is
equal to x2 and y is equal tox+2. So, we can get just by this solving this equation x2 is equal
to x+2 or x2 −x−2 is equal to 0. So, this will give us x−2 and x+1 is equal to 0.
So, x is equal to −1 and x is equal to 2. So, we will have these two points now where these
two curves intersects. So, one is precisely this one, the other one at this point here. So, here
when x is 2 so, ywill be 4. So,(2, 4) is this point and then −1 and 1 is this point here and we
are computing this area bounded by these two curves. Now again the same idea that in the
direction of y we are always moving from this curve to this line and in the direction of x will
be moving than. So, this is in the direction of y which is always from this curve y is equal to
545
2
x to y is equal to x+2 and then this line will move from this x is equal to −1 to x is equal to
2.
So, this area bounded by these two curves we will compute this integral, the integrand will be
1 and then we have dy dx. The limits of the inner integral y with respect to y will be from
this yis equal to x2 to y is equal to x+2the parabola to this to this line from this parabola to
that line. And now for the x as discussed we will take from −1¿ 2. So, which we can easily
integrate and then we will get the area enclosed by these two curves.
2 x+2
∫ ∫ d y d x= 92
−1 x 2
(Refer Slide Time: 08:13)
546
(Refer Slide Time: 10:15)
So, now we will take one more example for the area, Using a double integral, determine the area
2
bounded by the curves xy =2, y= x and y=4 ,where we want to again use the double integral
4
and determine the area bounded by the curves xy is equal to 2. So, now, we have 3 curves
2
here xy =2 and y=
x
and y is equal to 4 this line parallel to the xaxis. So, let us look at the
4
graphs of all these curves. So, we have here y is equal to4 this straight line, we have this
2
parabola again y is equal to
x
and then we have xy =2.
4
So, here as x is approaching to0 this will go to infinity and same thing when x will go this y
will go to 0and x is going to ∞. So, we have this curve here xy =2 and then we have here y is
2
equal to
x
and this straight line is yis equal to 4. So, these three curves and the area bounded
4
by these three curves will be precisely this one here.
So, this is the area we are going to compute now with the help of double integral. So, in this
we have the possibility though going first through the x axis and then the y axis or the other
way round, but in this case if we go first to watch the x axis. So, in this direction if we take
the integral first in the direction of x what will happen? We are going from this curve which
547
is xy is equal to 2 to this curve which is y is equal to
x
, and in this direction we will go from
4
y from this point which we will compute now what is the point of intersection here to yis
equal to 4 which is already given.
But, if you first fix in the direction of x; so, if we first fix in the direction ofx, then the
problem will be that going from this curve to the line always, but up to this point; next to this
2
we will be going from this parabola yis equal to
x
to that line. So, we have to break this
4
integral into two regions one would be this one and the other one would be this one.
So, instead of this if we take first with respect to x that will be easier, because we do not have
to break the domain in that case we will go from this curve to this curve always and in the
direction of y from this point to that point. So, we need to compute this point of intersection
2
of this xy =2. So, xy is equal to 2 and the other one is yis equal to
the value of y from there, its a
x
. So, here we substitute
4
2
2
x
is equal to
and from here we can get that x will be just 2.
x
4
So, this point here is a precisely x is equal to 2 and the y will be given by. So, this is x is
equal to2 and y is
2
2
. So, y is so; that means, 1. So, here y is 1. So, this point here y is 1
x
x
and that line here y is 4.
So, in the direction of y we will go from 1¿ 4, and in the direction of x we will go from this
2
x
xy is equal to 2¿ y is equal to . So, let us write down this integral. So, inner one with
4
respect to x and the outer one with respect to y. So, for the inner one we are moving from
this xy =2; that means, xis
2
and their the upper boundary will be given by this x2 is equal to
y
2
x is equal to 4 y ; that means, x is 2 √ y . So, here 2 √ y and for y we will go from 1¿ 4 . So, that
is the area bounded by these three curves, we can compute this integral. So, let us do that.
548
(Refer Slide Time: 15:11)
So, here this is the integral we want to compute now. So, for the inner one so, we fix the outer
one y is equal to 1¿ 4, for inner one with respect to x first.
4
2√ y
∫ ∫
y =1
[
x=
2
y
1
(
dx dy= ∫ 2 √ y−
y=1
3
2
dy
y
)
4
]
2
4
¿ 2 × y 2 −2 ln y = ( 8−1 )−2 ln 4
3
3
1
¿
4 ×7
28
−2 ln 4= −2 ln 4
3
3
So, we have seen these 3 examples, where we have computed the area bounded by the curves
or sometimes these two curves or the three curves. So, in this case it was enclosed by a three
curves and we had another application which we have already discussed that is the volume of
the solid. Using double integrals find the volume of the solid below the z=xy over the region
enclosed by y=4−x 2 , x=1, x=2 and the x -axis.
549
(Refer Slide Time: 17:13)
So, here the using the double integrals, we want to find the volume of the solid below this z is
equal to xy. So, we have this zis equal to x y surface over this region which is enclosed by
this parabola, again yis equal to 4−x 2 and x is equal to 1 and x is equal to 2. So, these are the
three curves again and the xaxis the forth one is the x axis. So, we have to first see what is
the region here in the domain, and then we have the function surface here z is equal to xy. So,
we want to find the volume below the surface and over this enclosed area in the xy plane. So,
we have to first sketch the region as usual. So, we have the parabola and then these three
lines.
So, we have this parabola which is other direction now y is equal to 4−x 2. So, when x is 0
the y is 4. So, the vertex is here. So, we have this y is equal to 4−x 2 that is the parabola, we
have x is equal to one this is the line x is equal to 1 and we have the line x is equal to 2. So,
this is the line here and then we have the x axis. So, the region bounded by all these four
curves is this one. So, this is the region bounded by the four curves though this y is equal to.
So, x is equal to 2 does not play a role here because this is precisely passing through this
point of intersection of this parabola and this line.
So, even if we remove also this xis equal to 2 the region enclosed will be the same. So, now,
we also need to get this point here. So, that is pretty clear when x is 1. So, when x is 1 the y
is 3. So, this x is 1 and then y will be 3 at this point by this y is equal to 4−x 2 and that this
point here the x is 2 and the y is 0. So, now, we need to just put the limits and the function
550
will be xy. So, this automatically this double integral with the integrand this xy will give us
the volume of the solid below by the surface and above this xy plane. So, to get this these
limits here again we have both the possibilities we can go first in the direction of xor we can
go in the direction of y first. So, let us go in the direction of x and then for the inner integral
and then this line will move from this to that point. So, fordx and the dywill be the outer one
the integrand now will be xy because we are going to get the solid if we put this instead of xy
once again we will get the area of this region, which is shown in the figure.
So, now, for the limits of x, we are now moving from x is equal to1 this is the line x is equal
to 1. So, always we are entering here to the domain from xis equal to 1 and the exit point is
this a parabola yis equal to 4−x 2 or from there we can get this y is equal to 4−x 2. So, we
can also write that x is equal to √ 4− y . So, for the inner one x we are moving from 1 to the
√ 4− y and for the outer one for the y , we are moving from this line to that line; that means, y
is equal to 0 ¿ y is equal to 3. So, this is the integral we want to now evaluate and that will
give us the volume of the solid below this that surface z is equal to xywhich is the integrand
of this integral.
(Refer Slide Time: 21:47)
So, now let us evaluate this integral. So, here this was the limit we have just seen. So, xfrom
1 to this is √ 4− y and y is 0 to 3and we want to integrate this with respect to x first. So, this v
2
the volume with respect to x we have to integrate first so; that means, this will be
551
x
. So, this
2
is y and by2 and we have x2 and these limits from 1 to 4− y and the outer integral is y. So,
this is 0 ¿3 and
y
1
. So, here , the 4− y−1 which will be 0 to 3. So, here this is 3− y. So,
2
2
outside and then 3 y− y 2 that is the integrand and dy.
So, we have 3 y− y 2 dy and the integral over y from 0 ¿3 and this vector half there so, half the
integral 3 y. So,
2
3
1
3y y
the limits 0 ¿3. So, and then that is post the upper one here 3
−
2
2
3
lower one will make anyway0. So,
again so, here
3
and this y 2, here 9−3 and then 27. So, which is a 9
2
27
1
27
and then −9. So, this is and
and this minus 9 there. So, this is 18 and
2
2
2
then we have a 27 there. So, we get 9 and then this is by4. So, we get the value
9
of this
4
volume which is below the surface y is equal to xy and above the region close by the
parabola and these three lines. So, that is the answer of this integral as
9
.
4
(Refer Slide Time: 24:15)
So, we move further to discuss another example where again we will compute the volume of
the solid whose base is in the xy plane, and it is bounded by the parabola. So, this is not the
parabola. So, bounded by the curves here xy is equal to 1, y is equal to x and this y is equal
552
to 0and y is equal to 8. So, in this case again we need to draw these curves here we have the
4. So, among them three are the lines and then this xy is equal to 16 with similar kind of a
graph we have in the earlier figure. So, to have this now this is the situation we have this xy is
equal to 16 and then this is the line y is equal to x, we have y is equal to 0 and we have x is
equal to 8. So, these are the four curves. So, again let me just mention. So, this is y is equal
to x and we have xy is equal to 16 and we have in this case x is equal to 8 and the region
enclosed by these four curves.
So, this is not the region here this the region by this 4; so, 1 the2 and the third and the 4. So,
the enclosed by this we have precisely this region here and now we will compute the volume
of the solid which is above this region here and below the bounded by the xy is equal to 1.
So, whose base is this and the plane here z is equal to xyeah. So, this is the function this is a
surface z is equal to x. So, our integrand is going to be x now. So, again we have the situation
that we can either go in the direction of y first or we go in the direction of x first, but in either
case we have to break the domain because even if we go first in the direction of x. So, up to
this point here we have a different situation and after that it goes from y is equal to x and exit
from x is equal to a.
Or, if we set in the direction of a y first; so, we have again these two regions one is this one
which is from the entries from y is equal to 0 to this curve which is given by y is equal tox
and then from 4 ¿8 we have this different curve. So, what is this point here? The point of
intersection that is also important. This is y is equal to x and this is xy is equal to 16and y is
equal to x. So, out of this we will get x2 =16. So, x is 4. So, this is exactly x is equal to 4 and
naturally this y is equal to 4. So, that is the point here. So, we can let us take now first in the
direction of y, and then in the direction of x. So, we need to have two integrals now; so, in
the direction of y first and then in the direction of x.
So, and integrand is going to be x. So, in the direction of y; however, ranges for this first
region which is this triangle y goes from 0 to this curve which is x and then for x we will
move from this point to this point; that means, 0 ¿ 4 the another one then 4 ¿8 our x will vary
from 4 ¿8 and now for the y its again the same curve y is equal to 0here, but now there it is
xy is equal to 16. So, this y is equal to 0 ¿
16
and then we have again dy and dx.
x
553
(Refer Slide Time: 28:39)
So, this integral we will compute which is given here. So, again this is a simple each of the
integral is simplest the first one is 0 ¿ 4 and then with respect to y. So, x remain as it is and
with respect to ywe will get x there. So, we have dx here we have 4 ¿8 and then thisx and
again y there. So, upper limit will give us
16
and minus the 0. So, dx. So, these are the two
x
integral the first one is x2 . So, we will get
x
and the limits 0 ¿ 4 and then here we have the 16
3
3
and then x; so, 8−4. So, we will get 4 the integral value there. So, here
this is 64 if I take common this is
So,
64
and plus 64. So,
3
1
4
256
+1; so, 64 × so,
.
3
3
3
256
is the volume of the solid below this plane zis equal to x and bounded by that region
3
xy is equal to 16 and yis equal to 6, y is equal to xy is equal to 0 and x is equal to 8. So, that
is a value
256
.
3
4 x
8 16/ x
0 0
4
∫∫ x dy dx +∫ ∫ x dy dx= 256
3
0
554
(Refer Slide Time: 30:14)
And now the last example of today’s lecture; so, we want to compute the volume of the solid
whose base is again the xy plane, but now we have the parabola here y is equal to 4−x 2, we
have the straight line y is equal to 3 xand on the top we have the plane zis equal to x+ 4. So,
again the similar problem; so, the integrand will be x+ 4, and the region will be computed
from this y is −x2 and y is equal to 3 x. So, these are the lines here. So, the curve so, y is
equal to 3 x. So, we have y is equal to 3 x and this parabola is y=4−x 2 we have seen earlier
also similar regions.
So, now this is the region where we want to get the volume below the z=x+4 plane. So, on
this we need to compute now what is this intersection points. So, for this we can now have y
is equal to 4−x 2 and y is 3 x. So, I will put 3 x there and then 4−x 2. So, we have
2
x +3 x−4=0 or x minus 4 and so, ( x +4 ) ( x−1 ) is equal to 0. So, x is equal to 1 and x is
equal to −4. So, these are the two points. So, here we have x is equal to 1 and
correspondingly why we can get from this y is equal to 3 x and here we have x is equal to −4
. So, the corresponding to this y will be −12 and here the y will be 3 x so, 3.
So, these are the points and now we want to integrate. So, we have to write the integrals and
again the question that we either we can integrate first in the direction of x or in the direction
of y . So, let us integrate first in the direction of y because that will be easier. So, in the
direction of y it always goes from it enters through this line and exit through this parabola so;
555
that means, this y and then dx. So, the y goes from 3 x to this parabola, y is equal to 4−x 2
and the limits of x will be from −4 to this x is equal to 1.
So, this integral we need to compute and the integrand will be x+ 4 and plus we have to yeah
that is the only integral because we have cover the whole region here, this line is always
entering through this line and exit from that parabola and then for x we have also taken from
−4to 1. So, we need to just integrate the simple integral to get the values.
1 4−x 2
V =∫
∫
( x+ 4 ) d y d x
−4 3 x
(Refer Slide Time: 33:31)
So, in this case this is the volume in terms of the double integral, which we can again this is
simple integrand we can integrate. And, after the calculations here the values are coming as a
625 it is a simple integral one can easily compute.
12
556
(Refer Slide Time: 33:53)
So, what we have seen today that applications of this double integrals mainly the computation
of area and also the computation of volume. In later lectures we will also see another
application where we can compute the surface area as well. So now, we have computed only
the area of the domain of the function and then, later on we can also compute the surface area
of the given surface using the double integrals. But this was a straight forward application;
the computation of the area and the computation of the volume which we have cover today.
(Refer Slide Time: 34:34)
These are the references which are used to prepare these lectures.
557
Thank you very much.
558
Engineering Mathematics – I
Prof. Jitendra Kumar
Department of Mathematics
Indian Institute of Technology, Kharagpur
Lecture – 31
Integral Calculus – Double Integrals in Polar Form
Welcome back. So, this is lecture number 31. And today we will again continue with this
double integrals, but in particular this polar form. So, the double integrals in polar forms are
very useful, and we will see with the help of many examples that if we convert in to the polar
forms the integrand, and also the corresponding limits, then this integral becomes much
easier to evaluate.
(Refer Slide Time: 00:46)
So, the situations we will be considering for the polar form when we have for example the
domain which can be described in polar form. So, we have for example, the circular domain
or we have some other domain where the boundaries are prescribed in terms of this variable
here r and θ. In those cases, we can easily think of converting the integral to polar form or the
integrand itself.
559
So, when we do see in the integral like x2 + y2form and then again since we know in the polar
form the relation from the Cartesian is x=r cos θ, and y=r sin θ. So, this x2 + y2 becomes r 2.
So, this also simplifies in several cases. So, based on the integrand of the integral as well as
the domain of integration we will decide whether it is better to go for the polar form. So, let
us discuss now.
(Refer Slide Time: 02:00)
So, suppose we have situation here, the domain is bounded by this curve this is r 2 (θ), it is a
function ofθ; and the another curve which is a function of θ here. So, we are talking about the
polar coordinate. So, the distance from the origin to this point will be r, and the angle from
this θ is equal to 0 x s will be θ. So, in that situation, whenever domain is bounded by these
two curves here r 1 (θ) and r 2 (θ), then we can actually do the integration easily with the help
of the polar coordinates, because the limits for this domain here r easy to evaluate because in
560
the direction of r here entering the domain from this r 1 (θ) and it existing this domain from
r 2 (θ).
In that case and also for the angle θ, we have that theta is going from this alpha angle. So, this
is θ is equal to alpha angle, and 2θ is equal to beta angle. So, we can curve domain with the
help of polar coordinating easily, but if we discretize in the a Cartesian form, then they will
be many difficulties to cover this domain, because it has this curve which is described nicely
by r 2 (θ) and here r 1 (θ).
So, now, how the integral will take the form when we have this polar coordinates. So,
suppose we discretize this domain again as similar to the Cartesian one into smaller cells, and
let us consider for instance here this cell which we are calling has Δ A i . So, for this cell we
have the radius from here to this point or any point on this line here it is r i So, this distance
from the origin to this is r i, and then we take and increment here in r i by r i + Δ r i to reach to
the other radius here from thus to this point.
Δ A i =( r i + Δ r i )
¿ (2 ri Δ ri + Δri
(
¿ ri +
2
2
)
Δθ i 2 Δ θ i
−ri
2
2
Δ θi
2
Δ ri
Δ r i Δ θi
2
)
¿ r¿i Δ r i Δ θ i
So, this is let us different than what we have in the Cartesian case. So, when we have done
same thing for the Cartesian one, so we were getting this increment in this direction by Δ x
and in this direction by Δ y, and then this area becomes just Δx or Δy.
But in this case when we are talking about the circular or the polar coordinate in this case the
area of this region in this domain here ¿ r¿i Δ r i Δ θi. While in the Cartesian coordinate, it was
simply the product of the increments we have made in the direction of x and in the direction
of y. So, here one additional factor this r i has appear. And now because of this, we will get
we will define now the integral this i over this domain here which is described in the polar
form by this limit.
561
So, we will do the same. So, the function if a function is given. So, we will compute that
function value at a point r i θ j so somewhere in this between. Now, this region r ij point and θ ¿j
point multiplied by this Δ A j, similar to the Cartesian one. But now this one since this
n
lim ∑ f (r j ,θ j ) Δ A j , so now this limit will take the or we will write in the integral form now
¿
¿
n→ ∞ j=1
β
r=r2 (θ )
∫ ∫
f ( r , θ ) r dr dθ .
θ=α r=r1 (θ )
Another now the limits we have to now consider for this domain, but these are given in the
polar form. So, it is very easy for the r. We are entering the domain from r 1 (θ) and existing
the domain from r 2 (θ), and the theta varies from the angle α ¿ β. So, this is now these are the
limits for r and θ and one we should know that this r factor has come as an extra in
comparison to the polar coordinate, where we use to have just simply dx and dy. So, here also
we have the function and then that will be evaluated r at this coordinate here (r , θ), and they
will be additional factor r and then dr dθ. And to cover the region we have to consider this
rangers for θ and r.
β
r=r2 (θ )
∫ ∫
f ( r , θ ) r dr dθ
θ=α r=r1 (θ )
(Refer Slide Time: 12:21)
562
Well, so now we will see here that how to change the Cartesian integral to polar coordinate,
though we have already discuss that that how to write the integral in the polar coordinate. So,
❑
suppose integral is given in Cartesian coordinate
∬ f ( x , y ) dx dy form over some region R.
R
So, what you will do now to convert into the polar form we will replace this x=r cos θ we
will replace this y=r sin θ, and this dx dy will be replaced by r dr dθ, and then the
corresponding to rin θ we have to place the limits that covers the region R.
❑
So, here
∬ f ( r cos θ ,r sin θ ) r dr dθ, because the area of the this is small region which we
G
have just seen in case of the polar coordinate it comes to be r dr dθ while in the case of
Cartesian it was dx dy. So, having this what is important again the steps are that we substitute
x=r cos θ , y=r sin θ in the integral, and this dx dy will be replaced by r dr dθ. And this Gis
basically the same the same region naturally, but now it is described in terms of the polar
coordinates. So, we will see with the help of examples now how to evaluate integrals in polar
form.
(Refer Slide Time: 14:08)
So, now, we want to compute, we want to start with a very simple example compute area of
the first quadrant of a circle of radius r, of radius a. So, we have a circle of radius a, and we
want to compute the area. So, again for the area we have to just assume that this function f is
1 and then we will get this integral over r dr dθ that will give us the area as we have done in
563
the Cartesian coordinates. So, here this area of the circle will be given by θ=0 ¿
π
because
2
our first quadrant. So, here this is the circle and the radius is a. So, the limits for θ for
instance first. So, theta vary from this 2, it will go up to
π
.
2
π
π
So, for theta from 0 ¿ , at this line the θ is 0 and here at this line when we reach θ is . And
2
2
r is varying from 0, it is starts from 0 up to this circle here which is r is equal to a. So, the
limits of r will be from 0to the circle r is equal to a, and theta varies from this 0 to θ is equal
to
2
π
a
. So, we need to just evaluate this integral, which will be here r is . So, r the upper
2
2
2
a
limit is a. So, . And then from the outer integral we will get when we integrate with respect
2
to θ so the θ will appear which will gives us
π
, so that is that is the area of this part of the
2
circular region.
π
2
A= ∫
a
∫ rdrdθ
θ=0 r=0
2
¿
a π
2 2
¿
πa
4
2
2
So, here
πa
which we have easily evaluated with the help of the polar or the integral in the
4
polar form. But if we do in the Cartesian form, then there will be little it will be little bit
complicated because we have to now draw the limits of the x∧ y and that will contain the
circle in the in the Cartesian form. So, the equation of the circle in the polar form is simply
r=a. So, we are just to cover this region, we are going from r=0 ¿ r=a, but in the Cartesian
coordinate this will be x2 + y2 =a2. So, we have to whether first compute the limit of x∨ y
accordingly we have to get this
√ a2−x 2∧√ a2− y2,
complicated ok.
564
so that will make the integral bit
(Refer Slide Time: 17:25)
❑
So, moving next, we have this problem now you want to evaluate this integral
2
2
∬ ex + y dy dx .
R
And this R is the semi circular region bounded by the x−¿axis and the curve y=√ 1−x2. So,
this curve is nothing but y 2=1−x 2meaning this x2 + y2 =1. So, this is the circle. So, we have a
circle here. And then the x−axis and we are talking about the semi circular region. So,
bounded by this x−axis and this upper part of the circle.
So, if you draw the region here that will be this half circle with radius 1 and then we have
here the x−axis. So, this is the region which we want to integrate over this region. So, again
since we have the circular region it would be much convenient to use the polar coordinate.
Secondly, our integrand here has this term like x2 + y2, so that will or some make it easier for
the evaluation of the integral.
❑
So, if we now compute the integral, so that will be
2
2
∬ ex + y dy dx . So, here if you write down
R
in the polar form first of all the integrand, because we are substituting now x=r cos θ and
y=r sin θ in the integrand, which will gives us which will give us this x2 + y2 =r 2. So, we will
π
get here
1
∫ ∫e
θ=0 r=0
π
r2
r dr dθ= ∫
θ=0
er
2
2
[ ]
1
1
dθ= ( e−1 ) π .
2
0
So, can using this polar coordinate, this is much simpler to evaluate this.
565
(Refer Slide Time: 21:23)
So, let us take another problem here, problem number 2 which will give us now again little
more inside. So, here calculate the area which is inside the cardioid r=2(1+cos θ) and the
outside the region r=1, the circle r=1. So, there is a circle here of radius this 2, centre 0. So,
this is the circle. And then we have this cardioid r=1. So, we have to first draw the region as
usual.
So, in this case we have this cardioid and then we have the circle here of radius 2, and then
what is ask here which is inside the cardioid and outside the circle r is equal to 2, so that
means, we are talking about this here outside the circle and inside this cardioid. So, this is the
region we want to find the area now, and naturally we will use the polar coordinates because
this is again much easier.
π/ 2 2(1+ cos θ)
∫ ∫
π
θ=
2
π/ 2
∫
θ=
π
2
r dr dθ
r=2
1
2
[ 4 ( 1+cos θ ) −4 ]dθ
2
π/ 2
∫
θ=
2[2cos θ+cos2 θ]dθ
π
2
566
1
sin 2 θ
4 2 sin θ+ θ+
2
2
[
[
4 2+
(
)]
π /2
0
1π
π
=4 2+ =8+ π
22
4
] ( )
So that is the area here of this region which is drawn in the figure. And with the help of again
the polar coordinate this was very easy to evaluate.
(Refer Slide Time: 26:29)
So, coming to the next problem; problem number 3. So, using the polar coordinate will find
the area of the region R in the xy-plane enclosed by the circle x2 + y2 =4 above the line y=1
and below the line y= √ 3 x . So, we have two lines y=1 and we have y= √ 3 x , and then the
circle here x2 + y2 =4 . So, that is the situation. The circle and then we have this line here
y= √ 3 x and then we have the line y=1, y=√ 3 x , and then this is the circle which is given by
x + y =4 .
2
2
Now, enclosed by the circle above by this line and below by this line; so, this is the region
which we want to now find the area using the polar coordinate. So, in this case, now we have
to see the limits for R and also for theta. So, we are moving in the direction of r. So, from this
point or for θ from this point and we have going up to that point. So, we need this here and
we need that, so these two angles.
567
So, first we need to compute these points where these lines are meeting. So, this is y is equal
to one line, and then we have the circle here x2 + y2 =4 . So, y if we put 1, x you will get as a
√ 3 and 1 that is the point here. And in this case since y is equal to √ 3 x if we put there, we
will well get x as 1 and y as a √ 3 . So, with this now we can compute easily this angle
because this is the vertical distance is 1, and here is a √ 3 for the first angle this one. So, tan of
this angle, tanof θ will be
π /3
r=2
∫ ∫
π / 6 r=cosec θ
r dr dθ=
1
π
π
, and we can get that theta s this . So, this is θ= .
6
6
√3
π− √ 3
3
So, the important is drawing the region and putting the limits of the integration.
(Refer Slide Time: 30:49)
Moving now to the next problem, you want to find the volume of the solid region bounded
above by the parabola, paraboloid z=9−x 2− y 2and below by the unit circle. So, above this
unit circle, we want to get the volume of this paraboloid. So, here can we have the unit circle.
So, we know the region now and we can easily draw the limits.
So, the r will go from 0 ¿1, we have unit circle for the limits of the theta it will be from 0 ¿2 π
, and we have the integrand 9−x 2− y2 which will be become r 2 in the polar coordinate and
then we have r dr dθ, so that is the integrand here, and then this r dr dθ. So, this x2 + y2 in
polar coordinate that becomes r 2. So, we got this 9−r2and r dr dθ which again it is a very
568
17
π the evaluation of this integral for these
2
simple to evaluate. And this will be coming up
limits for these limits ok.
(Refer Slide Time: 32:13)
∞ ∞
The last example another important example
2
2
∫∫ e−( x + y ) dx dy which with the help of the
0 0
polar coordinates we can evaluate, so 0 to ∞, and then we have so that means we are talking
2
2
about this first quadrant here in the whole up to ∞, and then e−(x + y ) dxdy . So, for this region
here now in terms of the r and in terms of the θ, so the θ varies from 0 to
2
π
, and r varies
2
2
from 0 to ∞ and e−(x + y ) square will become r 2 and then dr d θ.
π
2
∞
∫ ∫e
θ=0 r=0
−r
2
π
2
r dr dθ= ∫
θ=0
[( ) ]
−1 −r
e
2
So, the value of this integral is
∞
2
dθ=
0
1π π
=
22 4
π
, which is very simple when we use the polar coordinate, but
4
it is very difficult if you try to do on the Cartesian coordinates.
569
(Refer Slide Time: 33:45)
∞
2
Here just not based on the above example if you consider this I =∫ e−x dx which is equal to
0
another integral I am considering terms of y. So, just the integral variable changes name to y
now, but does not matter the value will be the same. And we consider the product of these
∞
two integrals, that means, ∫ e
−x
0
∞
I =∫ e
−x
2
0
∞
I =∫ e
−x
0
∞
2
∞
2
dx =∫ e− y dy.
0
2
dx=∫ e−y dy
0
2
∞
2
dx ∫ e− y dy
0
∞ ∞
2
¿∫ ∫ e−( x +y
¿ ¿ 2) dxdy
¿
0 0
¿
π
4
∞
2
⟹∫ e−x dx=
0
√π
2
570
(Refer Slide Time: 35:10)
So, the conclusion that is double integrals in polar forms are very important, and mainly the
some integrals like we have seen just before becomes very easier. We can evaluate easily by
change into the polar coordinate. And that based on the integral integrants and also the
domain, we can easily decide that which form is better whether the Cartesian or the polar
form we have seen through some examples.
(Refer Slide Time: 35:41)
And these are the references which we have used to prepare these lectures.
571
And thank you very much.
572
Engineering Mathematics - I
Prof. Jitendra Kumar
Department of Mathematics
Indian Institute of Technology, Kharagpur
Lecture - 32
Integral Calculus – Double Integrals: Change of Variables
Welcome back. And this is lecture number 32 and we will continue discussion on the
double integrals and in particular today we will discuss an very very important topic that
is Change of Variables in Double Integrals.
(Refer Slide Time: 00:33)
So, in the last lecture we have seen the double integrals in the polar forms and the idea was
to again break this integral the domain of integral into smaller parts of region, this delta i
which we have denoted here for this small portion of the area and which was coming to be
the r times the delta r and delta theta. Delta theta was the angle here and the r was the
radius from this to this circular arc. And in that case the whole idea was that if we have
the integral here in the Cartesian coordinate, we can convert into the polar coordinate by
just substituting this 𝑥 and 𝑦 to 𝑟 cos 𝜃 and 𝑟 sin 𝜃 and this 𝑑𝑥 𝑑𝑦 will become the
𝑟 𝑑𝑟 𝑑𝜃.
573
So, here also we have done eventually the substitution from (𝑥, 𝑦) to the (𝑟, 𝜃) coordinates
by this relation that 𝑥 was 𝑟 cos 𝜃 and 𝑦 was 𝑟 sin 𝜃. So, this was awesome substitution,
but directly with the help of the basic the fundamental definition we have seen that how
this 𝑟 is coming to the picture. And so, it is not just the substitution of the variables here
𝑥 𝑎𝑛𝑑 𝑦 to 𝑟 cos 𝜃 and 𝑟 sin 𝜃, but also this vector 𝑟 had appeared there and then we have
𝑟 𝑑𝑟 𝑑𝜃 and corresponding to this 𝑟 we have to also substitute now the limits of the 𝑟 and
𝜃.
So, now, we will go for a more general case and this will be a particular situation when we
go from Cartesian to polar coordinate. But in general, if we have any other type of change
of variables then what how to deal with this factor and we will see now in today’s lecture.
(Refer Slide Time: 02:37)
𝑏
Just to recall if we have a single integral ∫𝑎 𝑓(𝑥) 𝑑𝑥 and we want to substitute now to a
new variable. So, 𝑥 to let us say the 𝑡 here by, this relation 𝑥 = 𝑔(𝑡). So, what we do? We
immediately get that what would be our 𝑑𝑥 from here, so 𝑑𝑥 would be 𝑔′(𝑡)𝑑𝑡 in this form
and then we substitute this 𝑥 as 𝑔(𝑡) and dx will be the 𝑔′(𝑡)𝑑𝑡 and then the corresponding
the limits will appear there.
So, the idea used to be that we have the corresponding limits now of this 𝑡 and this 𝑥 was
substituted by the new function this 𝑔(𝑡) and for dx we have got here 𝑔′(𝑡)𝑑𝑡. So, that is
again here to recall that we got now there is something else here which is in case of a single
574
integral it is g prime the derivative of this 𝑔 and then 𝑑𝑡. And these corresponding limits
will be computed again from this substitution only. So, this a is going to be this 𝑔(𝑐) and
this b is nothing but 𝑔(𝑑). So, in that way we have also computed these new limits and we
have substituted the variable and instead of this 𝑑𝑥 𝑑𝑡 has come, but within another factor
which was in terms of the derivatives of this 𝑔. So, a similar kind of derivative we will
also get when we have a integral in two variables or the double integral.
𝑑
∫ 𝑓(𝑔(𝑡)) 𝑔′(𝑡) 𝑑𝑡
𝑐
(Refer Slide Time: 04:29)
So, if we have a double integral here ∬𝑅 𝑓(𝑥, 𝑦) 𝑑𝑥 𝑑𝑦, now if you want to make some
change of variables. So, let us say 𝑥 = Φ(𝑢, 𝑣) and 𝑦 = 𝜓(𝑢, 𝑣) . So, our new variables
are 𝑢 and 𝑣, the older one here were 𝑥 and 𝑦. So, we want to change the variables from 𝑥
𝑦 to 𝑢 𝑣 with this relation 𝑥 = Φ(𝑢, 𝑣), 𝑦 = 𝜓(𝑢, 𝑣) .
So, having this we have this integral here over the new region I mean the same region, but
for the new variables 𝑢 and 𝑣, so that is what we have used here different notation 𝑅′. And
then we straightaway substitute here for 𝑥 this Φ and for 𝑦, we will substitute the 𝜓 as a
function of this 𝑢 𝑣. Naturally, for the 𝑑𝑥 𝑑𝑦 we will get this 𝑑𝑢 𝑑𝑣, but with this another
factor which we have just discussed before also. So, in case of one variable or in case of
that polar coordinate you are getting just this as 𝑟 𝑑𝑟 𝑑𝜃, so there is some factor here this
575
𝐽 and we have taken this absolute value also; we are not going through the proof of this
because that is a bit involved a topic.
So, now what is this 𝐽 here? So, the way this 𝐽 is again kind of derivative which we call
Jacobian. So, Jacobian here 𝑥 𝑦 with respect to this 𝑢 and 𝑣 which is computed as in the
form of this determinant; so the partial derivative of this 𝑥 with respect to 𝑢 the first term,
here the partial derivative of 𝑥 with respect to 𝑣, now for the 𝑦 with respect to 𝑢 and this
partial derivative 𝑦 with respect to 𝑣.
𝜕𝑥
𝐽=
𝜕(𝑥, 𝑦)
𝜕(𝑢, 𝑣)
= |𝜕𝑢
𝜕𝑦
𝜕𝑢
𝜕𝑥
𝜕𝑣|
𝜕𝑦
𝜕𝑣
And that is this 𝐽 term here we will compute this one and absolute value will go to the
integral with this 𝑑𝑢 𝑑𝑣; so that is the important point here. And then these limits of this
𝑅′, it is nothing but the same region, but now we have to compute over 𝑢 and 𝑣. So, the
limits will now follow according to the new variables 𝑢 and 𝑣.
So, with this now we can compute or we can substitute the old variables here 𝑥 𝑦 to some
suitable because depending on again the integrand or the region r we may have some other
suitable substitution, which makes the integral evaluation easier.
(Refer Slide Time: 07:34)
576
So, if we want to see a particular case that what will happen if we substitute into the polar
form; that means, 𝑥 is equal to 𝑟 cos 𝜃 and y is equal to 𝑟 sin 𝜃. So, in that case 𝑥 is equal
to 𝑟 cos 𝜃 y is equal to 𝑟 sin 𝜃. If we compute this Jacobian here, so
𝜕𝑥
𝐽 = | 𝜕𝑟
𝜕𝑦
𝜕𝑟
𝜕𝑥
𝜕𝜃 | = |𝑐𝑜𝑠 𝜃 −𝑟 𝑠𝑖𝑛 𝜃| = 𝑟
𝜕𝑦
𝑠𝑖𝑛 𝜃 𝑟 𝑐𝑜𝑠 𝜃
𝜕𝜃
So, we have this determinant value as r, and now we can change this integral which was
given in the Cartesian coordinate to the polar coordinate and which we have already
discussed in the previous lecture that we need to substitute for 𝑥 = 𝑟 cos 𝜃 , and 𝑦 = 𝑟 sin 𝜃
and then 𝑑𝑥 𝑑𝑦 will become 𝑟 𝑑𝑟 𝑑𝜃. And this region now this 𝑅 will be 𝑅′ will be
described in terms of the polar coordinates 𝑟 and 𝜃.
∬ 𝑓(𝑥, 𝑦) 𝑑𝑥 𝑑𝑦 = ∬ 𝑓(𝑟 𝑐𝑜𝑠 𝜃 , 𝑟 𝑠𝑖𝑛 𝜃) 𝑟 𝑑𝑟 𝑑𝜃
𝑅
𝑅′
(Refer Slide Time: 09:08)
So, with this now we can go through the examples which we will use now the first one to
find the volume of one octant of a sphere of radius 𝑎; so, another very simple example
where we are using the sphere here. So, we know the equation of the sphere. So, its z
square in terms of the 𝑥 2 + 𝑦 2 we will get and then here the volume in one octant, so we
577
are considering basically the first octant here. So, this the lift the limit we can easily now
compute for this one and our surface here is given by the sphere.
So, we have the integral in terms of this volume that this is coming from the sphere because
the equation of the sphere was 𝑧 2 + 𝑥 2 + 𝑦 2 = 𝑎2 . So, from here you can compute 𝑧 2 is
equal to 𝑎2 − 𝑥 2 − 𝑦 2 . So, 𝑧 will be plus here because we are talking about just one octane
there. So, this will be the surface or this will be the integrand here √𝑎2 − 𝑥 2 − 𝑦 2 and
then 𝑑𝑥 𝑑𝑦 over this 𝑆 region.
∫ ∫√𝑎2 − 𝑥 2 − 𝑦 2 𝑑𝑥 𝑑𝑦 =
𝑆
So, that the radius 𝑅 the circular disk. So, this is in the first quadrant of the circular disk
𝑥 2 + 𝑦 2 ≤ 𝑎2 . So, that is the circular disk in the first quadrant and where we have this
domain for the integral.
And if we change now because direct evaluation over this Cartesian coordinate will be
definitely difficult because of already the integrand is difficult then the limits again this
square roots will appear because of the circular disk and overall the computation will be
will be difficult. But if we change to the polar coordinates by this 𝑥 = 𝑟 cos 𝜃 and 𝑦 =
𝑟 sin 𝜃 , in that case, we have already seen that this |𝐽| = 𝑟.
So, this additional factor which comes with 𝑑𝑟 𝑑𝜃 and then we have the simple idea that
we will change it to the √𝑎2 − (𝑥 2 + 𝑦 2 ) term will become √𝑎2 − 𝑟 2 and then we have
this Jacobian term here with this 𝑟 𝑑𝑟 𝑑𝜃. So, we have the integral now converted into
(𝑟, 𝜃) form and this additional factor 𝑟 has come and then we have 𝑑𝑟 𝑑𝜃 term.
∫ ∫ √𝑎2 − 𝑟 2 𝑟 𝑑𝑟 𝑑𝜃 = ∫ ∫ √𝑎2 − 𝑟 2 𝑟 𝑑𝑟 𝑑𝜃
𝑅
𝑅
𝜋
2
=∫
𝑎
∫ √𝑎2 − 𝑟 2 𝑟 𝑑𝑟 𝑑𝜃
𝜃=0 𝑟=0
=
3
𝜋
1 2
𝜋
(− ) ( ) (𝑎2 − 𝑟 2 )2 |𝑎0 = 𝑎3
2
2 3
6
578
(Refer Slide Time: 14:47)
𝑦−𝑥
Another example ∫ ∫𝑆 𝑒 𝑦+𝑥 𝑑𝑥 𝑑𝑦 where we can make use of this substitution like here
𝑦−𝑥
𝑒 𝑦+𝑥 is given and the region is this s here which is bounded by these two coordinate axis,
so 𝑥 axis and the 𝑦 axis here. And then we have the line 𝑥 + 𝑦 = 2or; so this is the region
𝑦−𝑥
where we are integrating this function 𝑒 𝑦+𝑥 . So, the suitable substitution looking at the
integrand seems to be 𝑦 − 𝑥, and this +𝑥 . So, if we substitute, we give a new name to 𝑦 −
𝑥 and also to 𝑦 + 𝑥 then perhaps this integral will become easier to compute.
So, we changed the variable from 𝑦 − 𝑥 = 𝑢 and 𝑦 + 𝑥 = 𝑣, and then out of these two we
can compute also 𝑥 and 𝑦 in terms of 𝑢 𝑎𝑛𝑑 𝑣 for example, 𝑥 =
𝑣−𝑢
2
,𝑦=
𝑣+𝑢
2
So, we have
𝑥 𝑎𝑛𝑑 𝑦 in terms of 𝑢 𝑎𝑛𝑑 𝑣 from the relations and we can get the Jacobean term because
that is needed for these transformation in the new variable in the new integral.
𝜕(𝑥,𝑦)
So, here the 𝜕(𝑢,𝑣) we have to compute. So,
−
1
𝜕(𝑥, 𝑦)
= | 12
𝜕(𝑢, 𝑣)
2
1
2| = − 1 − 1 = − 1
1
4 4
2
2
1
So, the Jacobian value in this case is− 2.
579
(Refer Slide Time: 17:15)
So, now this change of variable; we have to see now the domain the new domain here
corresponding to the new variables 𝑢 𝑎𝑛𝑑 𝑣.
So, the domain for 𝑢 𝑎𝑛𝑑 𝑣 plain would be because line 𝑥 is equal to 0 here. So, we have
this line 𝑥 is equal to 0, this is 𝑥 this is 𝑥 is equal to 0 and we have this 𝑦 is equal to 0 and
we have this line already 𝑎𝑥 + 𝑦 = 2. So, having this we have to see the new domain now
in terms of variables 𝑢 𝑎𝑛𝑑 𝑣. So, this line 𝑥 is equal to 0 what it implies here? 𝑥 is equal
to 0 implies 𝑣 𝑖s equal to 𝑢.
So, this region now is the new one for our integration, ok. So, now we can evaluate the
new limits based on this new region in terms of 𝑢 𝑎𝑛𝑑 𝑣.
580
(Refer Slide Time: 19:15)
1
So, remember the Jacobean was − 2and we are going to compute this integral
𝑦−𝑥
∫ ∫𝑆 𝑒 𝑦+𝑥 𝑑𝑥 𝑑𝑦, and our substitution was 𝑦 − 𝑥 = 𝑢, and 𝑦 + 𝑥 = 𝑣 and this is the new
domain given in terms of 𝑣 𝑎𝑛𝑑 𝑢.
So, having this integral we can now substitute exponential power this was 𝑢 and this was
𝑣. So,
𝑦−𝑥
𝑢
∫ ∫𝑒 𝑦+𝑥 𝑑𝑥 𝑑𝑦 = ∫ ∫ 𝑒𝑣
𝑆
𝑇
1
2
𝑣
𝑢
1 2
= ∫ ∫
𝑒 𝑣 𝑑𝑢 𝑑𝑣
2 𝑣=0 𝑢=−𝑣
1 2
1
= ∫ 𝑣 (𝑒 − ) 𝑑𝑣
2 0
𝑒
=𝑒−
581
1
𝑒
𝑑𝑢 𝑑𝑣
(Refer Slide Time: 22:15)
1
√2𝑥−𝑥 2
So, another problem where we want to integrate this ∫0 ∫𝑥
(𝑥 2 + 𝑦 2 ) 𝑑𝑦 𝑑𝑥 by
changing two polar coordinates and this is natural because the integrand is also supporting
that changing to polar coordinate will help and perhaps here as well, because it is a circle
here.
So, this region of this integration the given integral is given by 𝑦 = 𝑥, 𝑦 =
√2𝑥 − 𝑥 2 ,
𝑥 = 0 and 𝑥 = 1.
The region enclosed by this one, so from x to the circle, so this is the region of integration
which we are using here and we have to now cover with the help of the polar coordinate.
582
So, the polar equation of the circle. So, (𝑟 cos 𝜃 − 1)2 + 𝑟 2 sin2 𝜃 = 1, . So, from here if we
simplify this, we will get 𝑟 2 − 2𝑟 cos 𝜃 = 0. So, here the circle is 𝑟 = 2 cos 𝜃. So, the limits
of 𝑟 is also clear now, 𝑟 is going from 0 to 2 cos 𝜃.
(Refer Slide Time: 26:44)
1
√2𝑥−𝑥 2
So, this integral which was originally given ∫0 ∫𝑥
𝜋
2 cos 𝜃 2
2
this r square here, for = ∫𝜃=
𝜋∫
𝑟=0
(𝑥 2 + 𝑦 2 )𝑑𝑦𝑑𝑥 will be converted to
𝑟 𝑟 𝑑𝑟 𝑑𝜃 , so which we can easily compute this one
4
now.
583
(Refer Slide Time: 27:22)
∫
𝜋
2
2
∫
𝜋
𝜃= 4 𝑟=0
𝜋
2
𝜋
2
2 cos 𝜃
2
2
𝑟4
2 𝑐𝑜𝑠 𝜃
𝑑𝜃 = ∫ 4 𝑐𝑜𝑠4 𝜃 𝑑𝜃
𝑟 𝑟 𝑑𝑟 𝑑𝜃 = ∫ [ ]
4
𝜋
4
𝜋
2
𝜋
4
0
𝜋
2
= ∫ (2 𝑐𝑜𝑠 𝜃) 𝑑𝜃 = ∫ (1 + 𝑐𝑜𝑠 2𝜃) 𝑑𝜃 = ∫ (1 + 𝑐𝑜𝑠 2 2𝜃 + 2 𝑐𝑜𝑠 2𝜃)𝑑𝜃
𝜋
4
2
𝜋
2
𝜋
4
𝜋
4
𝜋
2
1
1
= ∫ (1 + (1 + 𝑐𝑜𝑠 4𝜃) + 2 𝑐𝑜𝑠 2𝜃) 𝑑𝜃 = (3𝜋 − 8)
𝜋
2
8
4
584
(Refer Slide Time: 29:22)
The last example which we want to evaluate here this integral over this
∬𝑅 √𝑥 2 + 𝑦 2 𝑑𝑥 𝑑𝑦 by changing to polar coordinates and this R is the region in the xy
plane bounded by the circles here 𝑥 2 + 𝑦 2 = 4 and 𝑥 2 + 𝑦 2 = 9. So, we have two circles,
with these two radius; so one is 𝑥 2 + 𝑦 2 = 4, 𝑥 2 + 𝑦 2 = 9. So, with radius 2 and this with
radius 3.
And we want to evaluate this given region here. So, again certainly in this case are
changing to polar coordinate will help us because of the integrand and as well as because
of the region here which is bounded by these two circle. So, by changing to polar
coordinate this is much simpler now to get the limits because putting 𝑥 = 𝑟 𝑐𝑜𝑠 𝜃 , , 𝑦 =
𝑟 sin 𝜃 , , |𝐽| = 𝑟 and then this integral will
585
2𝜋
𝐼=∫
0
3
2𝜋
𝑟3
3
∫ 𝑟 𝑟 𝑑𝑟 𝑑𝜃 = ∫ [ ] 𝑑𝜃
2
=(
0
3
2
27 − 8
38
) 2𝜋 = 𝜋
3
3
(Refer Slide Time: 31:18)
So, what we have seen now this that this change of variables is another important aspect
of evaluation of double integrals and changing to polar coordinate was a particular case.
But we have seen the other cases as well where by substituting two different variables
makes the domain of the integral easier and also the integrand easier.
586
(Refer Slide Time: 31:46)
So, these are the references used for preparing these lectures.
Thank you.
587
Engineering Mathematics – I
Prof. Jitendra Kumar
Department of Mathematics
Indian Institute of Technology, Kharagpur
Lecture – 33
Integral Calculus – Double Integrals: Surface Area
Welcome back this is lecture number 33 and today again we will continue with this Double
Integrals, but with a special application to surface computation of the surface area.
(Refer Slide Time: 00:27)
So, we have already seen the applications of double integrals for computing volume for
example, and also the area of the domain of integration. And today we will continue our
discussion for computation of surface area.
588
(Refer Slide Time: 00:47)
So, just to recall how do we compute the curve length in case of single integral. So, for
instance we have this curve here or the function which is given by 𝑓(𝑥). So, this is 𝑥 axis
and then here we have the y axis and this is the function 𝑓(𝑥) with respect to the values of
𝑥. And so, what we do we divide the whole domain from 𝑎 𝑡𝑜 𝑏 into small pieces of
intervals from 𝑥𝑖 to 𝑥𝑖+1 that is the one interval shown here. But, we have discretized this
the whole domain from 𝑎 𝑡𝑜 𝑏 into 𝑛 such intervals and let us consider this point for
instance this 𝜉𝑖 within the interval this 𝑥𝑖 to 𝑥𝑖+1 . And, corresponding to this point 𝑥𝑖 we
draw tangent at this point here which is given by 𝜉𝑖 and 𝑓(𝜉𝑖 ).
And this tangent the length of this tangent, if we add for all the intervals and later on we
take the limit as the width of these intervals go to 0 in that case we will end up with getting
the curve length. So, the idea is that here this is just shown this part of the tangent. So, this
589
is the tangent line at this 𝜉𝑖 point and this is the interval length here Δ𝑥𝑖 and this is suppose
the angle which tangent makes from the 𝑥 axis. So, that is we have denoted by theta.
So, we have this length here Δ𝑙𝑖 , this length here in the base here is Δ𝑥𝑖 and this is the
angle theta, now the length of the curve. So, length of this curve l which is so, this is our
curve l here and the length of this curve l will be given as when we sum all these parts of
the tangents and then take the limit. Because, if we do not take the limit we have the error
because this tangent is not representing the curve in this interval Δ𝑥𝑖 . But, when this Δ𝑥𝑖
will go to 0 eventually we will end up with calculating the length of this curve from 𝑎 𝑡𝑜 𝑏.
Δ𝑥𝑖
1
= cos 𝜃 ⟹ 𝛥𝑙𝑖 =
𝛥𝑥𝑖
Δ𝑙𝑖
𝑐𝑜𝑠 𝜃
cos 𝜃 =
1
√1 + tan2 𝜃
⟹
1
= √1 + (𝑓 ′ (𝜉𝑖 ))2
𝑐𝑜𝑠 𝜃
So, we want to get this limit from 𝑖 = 1 𝑡𝑜 𝑛 and then this summed up over this delta l i
and then the limit 𝑛 goes to ∞..
Length of the curve 𝐿 = lim ∑𝑛𝑖=1 Δ𝑙𝑖
𝑛→∞
(Refer Slide Time: 05:07)
𝑛
𝐿 = lim ∑ √1 + (𝑓 ′ (𝜉𝑖 ))2 Δ𝑥𝑖
𝑛→∞
𝑖=1
590
𝑏
= ∫ √1 + 𝑓 ′ (𝑥)2 𝑑𝑥
𝑎
(Refer Slide Time: 06:23)
So, what we have learnt that in case of the curve length the formula is given by the square
root of 1 plus and this derivative whole square. The extension of this we will have for the
computation of the surface where instead of the tangent line we will have the concept of
the tangent plane. And, and here we have taken the pieces of the tangent and then we have
added them after taking the limit we got the curve length. In case of the two dimensional
so, we have the surface there and we will take the tangent plane or the pieces of the tangent
plane. And, then we will sum those pieces and after taking the limit basically we will get
the surface area of the of the surface.
So, that is the trivial extension of this we will not go through the formal prove. But, if you
understood the concept here in case of this one variable the integral which computes the
𝑏
curve length is given by ∫𝑎 √1 + 𝑓 ′ (𝑥)2 𝑑𝑥, we will just make an extension here now, for
the computation of the surface area.
591
(Refer Slide Time: 08:35)
So, for the curve length we have just seen the formula was 𝑎 𝑡𝑜 𝑏. So, over the domain we
are integrating this integrand which was √1 + 𝑓 ′ (𝑥)2 . In case of the surface when we have
a function 𝑧 = 𝑓(𝑥, 𝑦) so, our here the surface is given by 𝑧 = 𝑓(𝑥, 𝑦).
So, this is the surface here and we can get the surface area again similar to what we have
for the 1 dimensional case. We have the double integral, we will discuss this what is the
domain here now and the square root we will take 1 plus like similar to here we have the
derivative here also we have the first order partial derivative. So, the first partial derivative
with respect to 𝑥 whole square plus the partial derivative with respect to 𝑦 of the given
function whole square. And then we integrate this integrand over the domain which is in
the 𝑥𝑦 plane and this is nothing, but the projection of the of the surface in the 𝑥𝑦 plane.
So, we have the some surface given or the 𝑧 axis and if we project that surface into this 𝑥𝑦
plane then 𝐷 will be exactly that domain in the 𝑥𝑦 plane. So, 𝐷 is the projection of the
surface in the 𝑥𝑦 plane and similarly because this equation we have formulated when the
function was given as this 𝑧 is equal to a function of 𝑥𝑦.
𝑆 = ∬ √1 + (
𝐷
𝜕𝑧 2
𝜕𝑧 2
) + ( ) 𝑑𝑥 𝑑𝑦
𝜕𝑥
𝜕𝑦
where 𝐷 is the projection of the surface in the 𝑥𝑦-plane.
592
And, here we should note there the partial derivatives were taken with respect to 𝑥 and 𝑦
which was natural. And, for instance the function is given like 𝑥 = 𝜇(𝑦, 𝑧) or it is given
𝑦 = 𝜓(𝑥, 𝑧) in that case the formula will change obviously. So, in this case for 𝑥 is equal
to the function of 𝑦 and 𝑧 then the partial derivatives with respect to 𝑦 and 𝑧 will appear.
And, this domain will be the projection of the surface in the 𝑦 𝑧 plane and in the case when
the function is given by 𝑦 = 𝜓(𝑥, 𝑧); the integrand will have the derivatives with respect
to x and z. And, again the domain of the integration will be in the 𝑥𝑧 plane where we are
integrating the will be integrating this integrand. So, here in the case first when 𝑥 = 𝜇(𝑦, 𝑧)
̂ and now the integral will become
we will have the domain here which is denoted by this 𝐷
𝑆 = ∬ √1 + (
̂
𝐷
𝜕𝑥 2
𝜕𝑥 2
) + ( ) 𝑑𝑦 𝑑𝑧
𝜕𝑦
𝜕𝑧
̂ will be the projection in the projection of that surface in the 𝑦 𝑧 plane or
And, now this 𝐷
in case when the function is given by 𝑦 = 𝜓(𝑥, 𝑧) . So, we will have the partial derivatives
of 𝑦 with respect to 𝑥 and 𝑧 and here this D double hat will be the projection of the surface
̂ are the domains in the 𝑦 𝑧; so, in this first case and then
in xz plane. So, this D hat and 𝐷
in the second case in 𝑥𝑧 planes in which the given surface is projected.
So, with the help of this double integral which is given by in most of the cases S is equal
to this domain
𝑆 = ∬ √1 + (
̂
𝐷
𝜕𝑦 2
𝜕𝑦 2
) + ( ) 𝑑𝑥 𝑑𝑧
𝜕𝑥
𝜕𝑧
And, if you want to compute the area of the domain of integration so, the area of D in this
particular case then we will just set this integrand as 1; so we have three applications: the
one was the area of this domain 𝐷, another one was the volume under this surface over the
𝑥𝑦 plane which was given by again double integral where, the integrand was simply
𝑓(𝑥, 𝑦). And, now we have for the computation of the surface area where the integrand is
𝜕𝑧 2
𝜕𝑧 2
𝑆 = ∬ √1 + ( ) + ( ) 𝑑𝑥 𝑑𝑦
𝜕𝑥
𝜕𝑦
𝐷
593
(Refer Slide Time: 13:33)
So, now we will take the problem here compute the surface area of the sphere
𝑥 2 + 𝑦 2 + 𝑧 2 = 𝑎2 . So, we have a sphere which is centered at (0, 0, 0) and given by this
equation 𝑥 2 + 𝑦 2 + 𝑧 2 = 𝑎2 . So, the radius of the a sphere is 𝑎; so, we want to compute
the surface area now. So, we will take for instance the upper half part of the sphere and if
we project that upper half part of the sphere; this will give us the circle of radius a in the
𝑥𝑦 plane. So, though we can we can project this to any one of these planes we either
𝑦𝑧 𝑜𝑟 𝑧𝑥 𝑜𝑟 𝑥𝑦. So, let us project into the 𝑥𝑦 plane. So, in the 𝑥𝑦 plane this will be a circle
of radius a center at 0.
So, we have the domain now that is the circular disk in the 𝑥𝑦 plane and then the surface
will be given by simply this 𝑧 = √𝑎2 − 𝑥 2 − 𝑦 2 . Because, we are considering the upper
part of this sphere and we can make it the double to compute the surface area of the whole
sphere. So, we are considering only the positive part; the upper half of the sphere and when
594
we project into the 𝑥𝑦 plane we will get this circle of radius a here and the center at (0, 0)
ok.
(Refer Slide Time: 15:19)
So, equation which will give us the surface here which is which I have just discussed for
the positive for the upper half we will take 𝑧 = √𝑎2 − 𝑥 2 − 𝑦 2 for the upper half. And,
now we need to compute because in the formula we need the partial derivative of 𝑧 with
respect to 𝑥 and also the partial derivative of 𝑧 with respect to 𝑦. So, for the partial
derivative of that with respect to 𝑥
𝜕𝑧
𝑥
=−
𝜕𝑥
√𝑎2 − 𝑥 2 − 𝑦 2
595
(Refer Slide Time: 16:45)
And then we have the partial derivative, similarly with respect to 𝑦;
𝜕𝑧
𝑦
=−
𝜕𝑦
√𝑎2 − 𝑥 2 − 𝑦 2
So, that is the surface we want to now compute this double integral which will give us the
surface area of the sphere.
𝑎
+√𝑎2 −𝑥 2
𝑆 = 2∫ ∫
√1 + (
−𝑎 −√𝑎2 −𝑥 2
596
𝜕𝑧 2
𝜕𝑧 2
) + ( ) 𝑑𝑦𝑑𝑥
𝜕𝑥
𝜕𝑥
(Refer Slide Time: 19:05)
So, that is what we have now the partial derivative with respect to 𝑥 is given by this partial
derivative of 𝑧 with respect to 𝑦 is given by this and that was the formula; here this was
with respect to 𝑦. So, if we if we substitute this now so, these are the whole square of these
terms. What we will get?
𝜕𝑧
𝑥
=−
𝜕𝑥
√𝑎2 − 𝑥 2 − 𝑦 2
𝜕𝑧
𝑦
=−
𝜕𝑦
√𝑎2 − 𝑥 2 − 𝑦 2
+√𝑎2 −𝑥 2
𝑎
𝑆 = 2∫ ∫
√1 + (
−𝑎 −√𝑎2 −𝑥 2
𝑎
+√𝑎2 −𝑥 2
= 2∫ ∫
−𝑎 −√𝑎2 −𝑥 2
2𝜋
𝑎
√𝑎2 − 𝑥 2 − 𝑦 2
𝑎
= 2∫
∫
0
0
𝜕𝑧 2
𝜕𝑧 2
) + ( ) 𝑑𝑦𝑑𝑥
𝜕𝑥
𝜕𝑥
𝑎
√𝑎2 − 𝑟 2
𝑟 𝑑𝑟 𝑑𝜃
= −2𝑎 2𝜋 √𝑎2 − 𝑟 2 |𝑎0
597
𝑑𝑦𝑑𝑥
= 4𝜋𝑎2
and that is exactly the surface area of this sphere which we already know. So, but with the
help of this double integral and with the help of the polar coordinates we could very easily
compute this surface area.
(Refer Slide Time: 23:55)
So, moving to the next problem we will find now the area of that part of the sphere. So,
we have again a sphere 𝑥 2 + 𝑦 2 + 𝑧 2 = 𝑎2 that is cut off by this cylinder
𝑥 2 + 𝑦 2 = 𝑎𝑥. So, this is a equation of the circle.
So, here 𝑥 2 + 𝑦 2 = 𝑎𝑥 is x square. So, this we can make it whole square so, a by two
whole square.
𝜕𝑧
𝑥
=−
𝜕𝑥
√𝑎2 − 𝑥 2 − 𝑦 2
𝜕𝑧
𝑦
=−
𝜕𝑦
√𝑎2 − 𝑥 2 − 𝑦 2
598
(Refer Slide Time: 25:07)
𝑆 = 2 ⋅ 2 ∬ √1 + (
𝐷
= 4∬
𝐷
𝜕𝑧 2
𝜕𝑧 2
) + ( ) 𝑑𝑥 𝑑𝑦
𝜕𝑥
𝜕𝑥
𝑎
√𝑎2 − 𝑥 2 − 𝑦 2
(Refer Slide Time: 27:25)
599
𝑑𝑥 𝑑𝑦
4∬
𝐷
𝜋
2
𝑎
√𝑎2 − 𝑥 2 − 𝑦 2
𝑎 𝑐𝑜𝑠 𝜃
𝑑𝑥 𝑑𝑦 = 4 ∫ ∫
0
𝑎
√𝑎2 − 𝑟 2
0
𝜋/2
𝑎 cos 𝜃
(−√𝑎2 − 𝑟 2 )
= 4𝑎∫
0
0
𝑟 𝑑𝑟 𝑑𝜃
𝑑𝜃
𝜋/2
= 4𝑎 ∫
[−𝑎 sin 𝜃 + 𝑎] 𝑑𝜃
0
𝜋
𝜋
𝜋
2
{𝑎
= 4𝑎 [ cos 𝜃}0 + 𝑎{𝜃}02 ] = 4𝑎 [−𝑎 + 𝑎 ] = 2𝑎2 (𝜋 − 2)
2
So, the only the difficult part here is locating the limits for the integral and that we have to
be careful. So, in this case the sphere was cut by the cylinder. So, this projection on the 𝑥𝑦
plane is nothing, but the circle because it was a circular cylinder and the projection is given
as here by this circle. And, then once we have figured out this projection on the 𝑥𝑦 plane
then we can easily put the limits of the integration.
600
(Refer Slide Time: 29:19)
So, the last example where we will determine the surface area of the part
𝑧 = 𝑥𝑦 that lies in the cylinder 𝑥 2 + 𝑦 2 = 1. So, again we have the cylinder, but it is it is
a cylinder with center (0, 0) and radius 1 and we have this 𝑧 is equal to 𝑥 𝑦 we want to get
the surface area. So, this is much simpler than the earlier examples.
So, again the projection of this cylinder because the cylinder cut that surface; so the
projection of that surface over the 𝑥𝑦 plane will be nothing, but the circle because the
cylinder is this circular cylinder and the cylinder is cutting the surface there. So, that will
be the projection here and we can get out of this 𝑧 = 𝑓(𝑥, 𝑦) = 𝑥𝑦 this partial derivative
with respect to 𝑥, partial derivative with respect to y which are required in the formula for
the computation here.
𝑧𝑥 = 𝑦, 𝑧𝑦 = 𝑥
601
𝑆 = ∬ √1 + 𝑥 2 + 𝑦 2 𝑑𝑥 𝑑𝑦
𝐷
2𝜋
𝑆=∫
0
2𝜋
= ∫
0
=
1
∫ √1 + 𝑟 2 𝑟 𝑑𝑟𝑑𝜃
𝑟=0
1 2
1
[(1 + 𝑟 2 )3/2 ]0 𝑑𝜃
2 3
2𝜋 3/2
(2 − 1)
3
(Refer Slide Time: 31:39)
So, what so, we have seen this so, another application of the double integral for the
computation of the surface area and, with the help of this simple formula over a
2
2
𝜕𝑧
𝜕𝑧
∬𝐷 √1 + (𝜕𝑥) + (𝜕𝑦)
𝑑𝑥 𝑑𝑦 only we have to be careful that this domain of this
integral is the projection of that surface in the 𝑥𝑦 plane or we have or it could be in the in
the 𝑦𝑧 or 𝑧𝑥 plane as well. Once we have that projection, if we identify that projection
putting the limits will be will be much easier and we can compute the integral.
602
(Refer Slide Time: 32:21)
So, these are the references we have used for the computation for this preparation of the
lecture and.
Thank you very much.
603
Engineering Mathematics – I
Prof. Jitendra Kumar
Department of Mathematics
Indian Institute of Technology, Kharagpur
Lecture – 34
Integral Calculus – Triple Integrals
So, welcome back and we will continue now this Triple Integral. So, we have already
finished evaluation of double integrals and many other applications of double integral.
(Refer Slide Time: 00:27)
Today we will learn about the triple integrals.
604
(Refer Slide Time: 00:31)
So, basically the evaluation part we will look into in this lecture. So, first how to define
this triple integral? So, we need to similar to what we have done for the double integrals
we divide the given region so now, our region will be a three dimensional in the three
dimensional space. So, we will divide that region into n sub regions and we call the volume
of these sub regions by this 𝛿𝑉1 , 𝛿𝑉2 , … , 𝛿𝑉𝑛 . So, we have a three dimensional a space now
and we have divided that into small regions and for each region we are denoting its volume
by this 𝛿𝑉1 , 𝛿𝑉2 , … , 𝛿𝑉𝑛 .
And, now we let a point (𝑥𝑗 , 𝑦𝑗 , 𝑧𝑗 ) an arbitrary point in 𝑗’th sub region. So, it is very similar
to what we have done in case of two dimensional; we have divided the domain the two
dimensional domain into a small sections, small cells. And, then in each cell we had taken
a point and the corresponding value of the function at that point was multiplied by the area
and that was summed up and taking the limit of that sum we introduce the double integral.
So, here as well so, we will consider this sum now. So, sum over these all the volumes
𝛿𝑉𝑗 multiplied by the function value.
So, the function value at this point (𝑥𝑗 , 𝑦𝑗 , 𝑧𝑗 ) which we have considered in j’th sub region.
And we are adding all these regions or this multiplication over these all regions; that
means, this 𝑗 is varying from 0 𝑡𝑜 𝑛. And, now if we take the limit here as n goes to infinity
or other way around that this 𝛿𝑉𝑗 we will go to 0 because the domain is fixed now. So, if
605
we are letting this n goes to infinity then the sub regions will go to we ∞; that means, the
volume of each sub region will go to 0.
So, taking if this limit exists as n approaches to ∞ in that case we call this as integral and
the notation for this integral in the three dimensional case or for the triple integral we use
this notation. So, we have now the three symbols for this integral over that volume here
which is a space in region in the three dimensional plane. Then we have the function here
𝑓(𝑥, 𝑦, 𝑧) integrating over that volume in space and then the definition is coming again
from the sum when we have added this 𝑓 with this 𝛿𝑉𝑗 and at the end taking this limit as n
approaches to ∞. So, again the physical interpretation if we want to take a quick look so,
if for example, this 𝑓 is said to 𝑉1 .
So; that means, there is no 𝑓 here so we are basically adding this 𝛿𝑉𝑗 the volume of each
sub region and then taking the limit. So, by considering this 𝑓 as 1 we will get nothing, but
the volume of that sub region again. So, using the double integral also we have computed
the volume that was one of the applications of the double integral, but in that case we
considered that 𝑓 in the integrand. So, the double integral with that integrand that surface
and was giving us the volume but, now in this case when we set this 𝑓 to 1 then we are
getting actually the volume of this domain there; the domain of the integration which is
denoted by 𝑉 in this case. And, we had also the application they are finding the area of the
domain in case of the second in case of the double integral and there also precisely we
have set this function to 1.
And, just integrating without this function we were getting the area of the domain of
integrations. Or, in this case when we said this 𝑓 to 1 you will get the volume of the domain
because now, we have a three then we have a volume in this three dimensional space.
𝑛
∫ ∫ ∫ 𝑓(𝑥, 𝑦, 𝑧)𝑑𝑉 = 𝑙𝑖𝑚 ∑ 𝑓 (𝑥𝑗 , 𝑦𝑗 , 𝑧𝑗 ) 𝛿𝑉𝑗
𝑛→∞
𝑉
606
𝑗=1
(Refer Slide Time: 05:29)
So, it represents the volume if we say at f is equal to 1. So, that also we can look into as
one of the applications of the triple integral.
(Refer Slide Time: 05:43)
So, how to evaluate the triple integral? So, having the knowledge of the evaluation of the
double integral, this will be also easier. So, this is the idea of this iterative computation of
single integrals and then so, like for instance if we consider that is the most general form
is given. So, this is the inner most inner one into integral this is taken with respect to 𝑥.
So, the limits for this 𝑥 in a very general case can be a function of this 𝑦 and 𝑧 to the
607
function 𝑓 2 their 𝑦 and 𝑧. So, as a result when we integrate the inner one we will get some
function because, over 𝑥 we have integrated. So, we will not see 𝑥 anymore, but the limits
can be here the function of 𝑦 𝑧 here can be the function of 𝑦 𝑧 and then we will have also
the 𝑦 𝑧 in the integrand.
So, in general the after evaluation of the inner integral we will get a function of 𝑦 and 𝑧.
So, then once having done the evaluation of the inner integral we will go to the outer one
now with respect to 𝑦. So, we are integrating now with respect to 𝑦 and the limits of the 𝑦
can be the function of 𝑧 can be the function of 𝑧. So, after the integration of this second
inner integral, we will get a some function of z and now at the end we will integrate this
𝑥𝑧 with respect to 𝑧. And, now the limit has to be constant because always remember the
value of this integral is a constant and you will not see anything of 𝑥 𝑦 and 𝑧 because, we
are integrating over the volume this 𝑉 or over all the points this 𝑥𝑦𝑧 in our volume. So, at
the end this will be free this integral value will not have anything of 𝑥𝑦𝑧.
∫ ∫ ∫ 𝑓(𝑥, 𝑦, 𝑧) 𝑑𝑉
𝑉
So, once we have the evaluation of the inner two integral that we are getting like a function
of 𝑧 and now we will integrate with respect to 𝑧. So, for the outer integral now the limits
must be constant. So, this is a general consideration which we should always keep in mind
while putting the limits of the of the integral. So, the inner one when we are integrating
with respect to x the limits can be the function of 𝑦 and 𝑧 both the limits are lower and the
upper one. After this integration the x is removed we have the function of variable 𝑦 and
𝑧 and then we are integrating with respect to y the limits still can be the function of 𝑧 like
here 𝜓1 (𝑧) and 𝜓2 (𝑧). So, these limits can be the function of 𝑧 and the outer one then that
has to be the constant limits for 𝑧 and now we will get a value of this integral which is free
from this 𝑥 𝑦 and 𝑧. So, this structure we must keep in mind.
𝑏
∫ ∫ ∫ 𝑓(𝑥, 𝑦, 𝑧) 𝑑𝑉 = ∫ {∫
𝑉
𝑧=𝑎
𝜓2 (𝑧)
𝑦=𝜓1 (𝑧)
608
𝑓2 (𝑦,𝑧)
{∫
𝑥=𝑓1 (𝑦,𝑧)
𝑓(𝑥, 𝑦, 𝑧) 𝑑𝑥} 𝑑𝑦} 𝑑𝑧
(Refer Slide Time: 09:09)
And then similar to the double integrals the order of integration is immaterial, if the limits
of the integration are constant. So, similar to double integral; similar to double integrals
the order of the integration is immaterial if the limits of the integration are constants.
𝑏
𝑑
𝑓
𝑓
𝑑
𝑏
∫ ∫ ∫ 𝑓(𝑥, 𝑦, 𝑧) 𝑑𝑥 𝑑𝑦 𝑑𝑧 = ∫ ∫ ∫ 𝑓(𝑥, 𝑦, 𝑧) 𝑑𝑧 𝑑𝑦 𝑑𝑥
𝑎
𝑐
𝑒
𝑒
𝑑
𝑓
𝑐
𝑎
𝑏
= ∫ ∫ ∫ 𝑓(𝑥, 𝑦, 𝑧) 𝑑𝑧 𝑑𝑥 𝑑𝑦
𝑐
𝑒
𝑎
So, here the 𝑒 𝑡𝑜 𝑓 so, here also we have 𝑒 𝑡𝑜 𝑓. So, simply we can change the order if the
if the if the range here of integration or the limits of integration is or are constant. Also we
can further change like the order 𝑑𝑥 𝑑𝑧 𝑑𝑥 𝑑𝑦 and that corresponding limits will against it
over the integrals. Once we have the functions here as above in a very general case then
we have to be or it is more difficult to change the order of integration. And we have to be
very careful now, once we change the order similar to what we have done in case of two
variables. Once we change the order again we need to be very careful about the finding
the limits of the integration, if these limits are not constant; if the limits are constant one
can simply change the order.
609
(Refer Slide Time: 11:19)
So, here now we will go for the evaluation so, that is the first example 𝐼 =
𝑎
𝑥
𝑥+𝑦
∫0 ∫0 ∫0
𝑒 𝑥+𝑦+𝑧 𝑑𝑧 𝑑𝑦 𝑑𝑥 we will consider the exponential function 𝑒 𝑥+𝑦+𝑧 the very
simple function and we want to integrate with respect to 𝑧 𝑦 and then 𝑥.
𝑎
𝑥
𝐼=∫ ∫
0
𝑎
=∫
0
0
𝑥+𝑦
𝑒𝑥+𝑦+𝑧 |0
𝑎
𝑥
𝑑𝑦𝑑𝑥 = ∫ ∫ (𝑒2(𝑥+𝑦) − 𝑒𝑥+𝑦 ) 𝑑𝑦𝑑𝑥
0
0
𝑎
𝑒 2(𝑥+𝑦) 𝑥
1 𝑎
|0 𝑑𝑥 − ∫ 𝑒 𝑥+𝑦 |0𝑥 𝑑𝑥 = ∫ {(𝑒 4𝑥 − 𝑒 2𝑥 ) − 2(𝑒 2𝑥 − 𝑒 𝑥 )}𝑑𝑥
2
2 0
0
=
1 𝑎 4𝑥
𝑒 4𝑎 3 2𝑎
3
∫ (𝑒 − 3 𝑒 2𝑥 + 2𝑒 𝑥 ) 𝑑𝑥 =
− 𝑒 + 𝑒𝑎 −
2 0
8
4
8
So, that is the evaluation of this simple integral. So, we have to again iteratively solve the
integrals like and then these becomes single integrals. So, first with respect to 𝑧, then with
respect to 𝑦 and at the end with respect to 𝑥.
610
(Refer Slide Time: 18:09)
So, next we have this problem 2 where we will evaluate this integral ∭𝑅
𝑑𝑥 𝑑𝑦 𝑑𝑧
(𝑥+𝑦+𝑧+1)3
and
this R the region R is given bounded by 𝑥 = 0, 𝑦 = 0, 𝑧 = 0 & 𝑥 + 𝑦 + 𝑧 = 1. these are the
planes now. So, we are talking about the three dimensional space. So, this 𝑥 = 0 is the 𝑦 𝑧
plane here this 𝑦 = 0is the 𝑥𝑧 plane and 𝑧 = 0 is 𝑥𝑦 plane and we have another plane here
𝑥 + 𝑦 + 𝑧 = 1.
1
1−𝑥
𝐼=∫
∫
∫
𝑥=0 𝑦=0
1
1−𝑥
=∫ ∫
0
0
1−𝑥−𝑦
𝑧=0
1
(𝑥 + 𝑦 + 𝑧 + 1)3
𝑑𝑧 𝑑𝑦 𝑑𝑥
1−𝑥−𝑦
1
[− (𝑥 + 𝑦 + 𝑧 + 1)−2 ]
𝑑𝑦𝑑𝑥
2
0
611
(Refer Slide Time: 25:41)
1
1−𝑥
𝐼=∫ ∫
0
0
1−𝑥−𝑦
1
−2
[− (𝑥 + 𝑦 + 𝑧 + 1) ]
𝑑𝑦𝑑𝑥
2
0
1
5
= [log 2 − ]
2
8
(Refer Slide Time: 28:27)
The last problem so, using this triple integral we want to find the volume which is common
to this is sphere. We have this sphere 𝑥 2 + 𝑦 2 + 𝑧 2 = 𝑎2 and this circular cylinder which is
given by 𝑥 2 + 𝑦 2 = 𝑎𝑥. So, this is again the circle which we have discussed a couple of
612
times before. This is a sifted circle where radius is
𝑎
2
and the center is
𝑎
( , 0) and I shown here in the in this figure.
2
So, the cylinder here whose objection on the 𝑥𝑦 plane is given by the circle and the center
𝑎
here is (2 , 0). So, the volume if you want to find using this triple integral so; obviously,
there will be no integrand now. And, we have this integral over this 𝑉 which is given by
the cylinder and then we have the sphere.
So, the sphere is the cylinder is cutting the sphere and that portion we want to find the
volume. So, the most important part again here is finding this limit for this 𝑉 and what we
do now first with respect to 𝑧 again we are putting.
𝑉 = ∫ ∫ ∫ 𝑑𝑥 𝑑𝑦 𝑑𝑧 = ∫ ∫ ∫ 𝑑𝑧 𝑑𝑦 𝑑𝑥
𝑉
𝑉
√𝑎𝑥−𝑥 2
𝑎
= 4∫ ∫
∫
0
𝑦=0
𝑎
√𝑎𝑥−𝑥 2
= 4∫ ∫
0
√𝑎2 −𝑥 2 −𝑦 2
𝑑𝑧 𝑑𝑦 𝑑𝑥
𝑧=0
√𝑎2 − 𝑥 2 − 𝑦 2 𝑑𝑦 𝑑𝑥
𝑦=0
613
(Refer Slide Time: 34:09)
√𝑎𝑥−𝑥 2
𝑎
√𝑎2 − 𝑥 2 − 𝑦 2 𝑑𝑦 𝑑𝑥
= 4∫ ∫
0
𝑦=0
𝜋
2
𝑎 cos 𝜃
√𝑎2 − 𝑟 2 𝑟 𝑑𝑟 𝑑𝜃
= 4∫ ∫
0
𝑟=0
𝜋
3 𝑎 cos 𝜃
2
1 2
= 4 (− ⋅ ) ∫
2 3 0
[(𝑎2 − 𝑟 2 )2 ]
0
𝜋
2
4
= − 𝑎3 ∫ (sin3 𝜃 − 1) 𝑑𝜃
3
0
2
4
= 𝑎3 (𝜋 − )
3
3
614
𝑑𝜃
(Refer Slide Time: 35:41)
So, that will be the limit for this direction the z for instance and once we fix this with
respect to 1 we can project the whole volume to the other plane. So, if we have fixed
already with respect to z then we can project the geometry, the volume to the 𝑥𝑦 plane.
And, once we project to the 𝑥𝑦 plane we are in the two dimensions and we know how to
fix the limit for two dimensional case.
(Refer Slide Time: 36:45)
So, these are the references we have used for preparing the lectures and.
615
Thank you very much.
616
Engineering Mathematics – I
Prof. Jitendra Kumar
Department of Mathematics
Indian Institute of Technology, Kharagpur
Lecture – 35
Integral Calculus – Triple Integrals (Contd.)
So, welcome back and this is lecture number 35 today we will continue again with Triple
Integral.
(Refer Slide Time: 00:25)
And we will look for the change of variables in case of a triple integral.
617
(Refer Slide Time: 00:29)
So, first you will go through the example where we change from Cartesian to spherical
polar coordinates; so a very important the change of coordinates. So, here what we have
first let us take these Cartesian coordinates and suppose there is a point here 𝑥𝑦𝑧 and when
we are talking about the Cartesian coordinate; so, this distance now from here to the 𝑥𝑦
plane. So, if we draw a perpendicular here to this 𝑥𝑦 plane then this distance is the 𝑧 coordinate from this 𝑥𝑦 0 point to this 𝑥𝑦𝑧 point. And, from the 𝑥 axis along this 𝑦 axis the
distance of this point which was the perpendicular from this 𝑥𝑦𝑧 point to the 𝑥𝑦 point is
given by the 𝑦 and then this 𝑥 coordinate that is the distance from the origin.
So, here we have in this point as the origin so, from the origin to this point that is the
distance given by 𝑥. So, this is 𝑥 here and then along this 𝑦 direction we have this distance
𝑦 and then this is the perpendicular distance along the 𝑧 axis, this is the 𝑧. Now, our
618
question is when we change into the; a spherical polar coordinates then what would be
(𝑟, 𝜃, 𝜙). So, the coordinates in a spherical polar coordinates we denote by (𝑟, 𝜃, 𝜙). So, 𝑟
is precisely the distance from the origin to this point. So now, in spherical polar coordinate
this distance from the origin to this point to the given point will be denoted by 𝑟 and then
these two are the angles; these two are the angles 𝜃 and 𝜙.
So, here the angle 𝜃 will be the angle to this point which was the perpendicular from this
𝑥 𝑦 𝑧 point to the xy plane. So, this is the angle phi from the 𝑥 axis. So, this here from the
𝑥 axis to this line which is in the 𝑥𝑦 plane to this point 𝑥𝑦 that is the angle 𝜙. And, now
this 𝜃 angle the 𝜃 is the angle with this z axis which this line which is joining this origin
to this 𝑥 𝑦 𝑧 point in the space that angle is denoted by 𝜃. So, again to represent this point
which was 𝑥 𝑦 𝑧 in a spherical coordinate and now in Cartesian coordinate and, now in
spherical coordinates this is denoted by 𝑟, 𝜃 and 𝜙 and now r is the distance from the origin
to this point theta is the angle from the 𝑧 axis.
And, now this 𝜙 is the angle from the 𝑥 axis to the point which was the perpendicular from
this given point to the 𝑥𝑦 plane. So, with this notation of the spherical polar coordinates
we will now introduce the change of variables in triple integral. So, the relation between
these (𝑥, 𝑦, 𝑧) and this (𝑟, 𝜃, 𝜙) is given by 𝑥 = 𝑟 sin 𝜃 cos 𝜙, and 𝑦 = 𝑟 sin 𝜃 sin 𝜙 , and
𝑧 = 𝑟 cos 𝜃. So, that is the standard relation between this spherical polar coordinates and
the Cartesian co-ordinate.
𝑥 = 𝑟 sin 𝜃 cos 𝜙,
𝑦 = 𝑟 sin 𝜃 sin 𝜙 ,
𝑧 = 𝑟 cos 𝜃
And, now we note that if we make this square here for 𝑥 2 + 𝑦 2 + 𝑧 2 because, this term
will be 𝑟 2 .
619
(Refer Slide Time: 05:51)
So, here if you have a triple integral given as ∫ ∫ ∫𝐷 𝑓(𝑥, 𝑦, 𝑧) 𝑑𝑥𝑑𝑦𝑑𝑧 over some domain
d and we want to change to the spherical co-ordinate; meaning that 𝑥 = 𝑟 sin 𝜃 cos 𝜙, 𝑦 =
𝑟 sin 𝜃 sin 𝜙 , and 𝑧 = 𝑟 cos 𝜃. So, with this change of variables now, the idea is that this
is the this simple idea which we have also explained in previous lecture; that we need to
substitute we need to substitute 𝑥 = 𝑟 sin 𝜃 cos 𝜙. So, this is 𝑦 = 𝑟 sin 𝜃 sin 𝜙 and for 𝑧 =
𝑟 cos 𝜃 which was the relation there 𝑧 = 𝑟 cos 𝜃.
∫ ∫ ∫ 𝑓(𝑥, 𝑦, 𝑧) 𝑑𝑥𝑑𝑦𝑑𝑧
𝐷
= ∫ ∫ ∫ 𝑓(𝑟 𝑠𝑖𝑛 𝜃 𝑐𝑜𝑠 𝜙 , 𝑟 𝑠𝑖𝑛 𝜃 𝑠𝑖𝑛 𝜙 , 𝑟 𝑐𝑜𝑠 𝜃) |𝐽| 𝑑𝑟 𝑑𝜃 𝑑𝜙
̂
𝐷
So, with this we have now this Jacobian term that is the extra term which comes when we
do this change of variables. So, that Jacobian we need to compute and then our dx dy dz
will become 𝑑𝑟 𝑑𝜃 𝑑𝜙. So, and then correspondingly the limits will change or the domain
in terms of now, r theta phi we have to represent. So, what is this Jacobian here? That is a
given definition which we have already discussed that
620
𝜕𝑥
𝜕𝑟
|
𝜕𝑦
𝐽=
𝜕𝑟
| 𝜕𝑧
𝜕𝑟
𝜕𝑥
𝜕𝜃
𝜕𝑦
𝜕𝜃
𝜕𝑧
𝜕𝜃
𝜕𝑥
𝜕𝜙
|
𝑠𝑖𝑛 𝜃 𝑐𝑜𝑠 𝜙
𝜕𝑦
= | 𝑠𝑖𝑛 𝜃 𝑠𝑖𝑛 𝜙
𝜕𝜙
𝑐𝑜𝑠 𝜃
𝜕𝑧 |
𝜕𝜙
𝑟 𝑐𝑜𝑠 𝜃 𝑐𝑜𝑠 𝜙
𝑟 𝑐𝑜𝑠 𝜃 𝑠𝑖𝑛 𝜙
−𝑟 𝑠𝑖𝑛 𝜃
−𝑟 𝑠𝑖𝑛 𝜃 𝑠𝑖𝑛 𝜙
𝑟 𝑠𝑖𝑛 𝜃 𝑐𝑜𝑠 𝜙 | = 𝑟 2 𝑠𝑖𝑛 𝜃
0
(Refer Slide Time: 09:49)
So, now we will remember that this Jacobian term which is coming because of this change
of variables in this triple integral the value of this Jacobian term is nothing, but r square
sin theta. So now, the Cartesian co-ordinate to cylindrical coordinates. So, again the
representation for this point 𝑥𝑦𝑧 in the Cartesian co-ordinate and it will be represented
now in these cylindrical coordinate by (𝑟, 𝜙, 𝑧); so again this triplet with (𝑟, 𝜙, 𝑧). So, 𝑧 is
precisely the same which was in Cartesian coordinate. So, the distance from this point to
this perpendicular drawn to the 𝑥𝑦 plane. So, that distance the third member of this point
here is the same as this set in the Cartesian coordinate.
621
And, now this phi is similar to what we have in the spherical coordinate that is the angle
precisely this one which we have already discussed. That is the phi which is from the 𝑥
axis to this line which is joining the origin to this 𝑥𝑦 point on the 𝑥𝑦 plane. And now this
𝑟 is again the distance here from this point to this. Now, that is the difference here in the
spherical coordinate we have this 𝑟 from this point to the given point, but now this 𝑟 is the
distance in the 𝑥𝑦 plane.
So, basically in the 𝑥𝑦 plane we are represented by the polar coordinate 𝑟 is the distance
to this point and 𝜙 is the angle from this 𝑥 axis and this 𝑧 here; it is the same as the given
in the Cartesian coordinate. So, there is no change in the 𝑧 and this 𝑥𝑦 simply they are
given by this (𝑟, 𝜙) and that is nothing, but the polar coordinate in this xy plane.
So, 𝑥 the relation again since it is the polar coordinate in this 𝑥𝑦 plane. So, 𝑥 = 𝑟 cos 𝜙,
and 𝑦 = 𝑟 sin 𝜙. So, in polar coordinate we used to call this angle 𝜃, but now it is different
representation we are calling it 𝜙. So, the relation remains exactly the same 𝑥 = 𝑟 cos 𝜙
and 𝑦 = 𝑟 sin 𝜙 and this 𝑧 is same as the 𝑧 in the Cartesian co-ordinate. So, this is the
simple relation when we talk about the cylindrical coordinate and here this relation again
coming from the polar coordinate that this 𝑥 2 + 𝑦 2 = 𝑟 2 in this case.
(Refer Slide Time: 12:31)
So, here now if we talk about the triple integral and the change of variables; so, we have
this integral which is written in the Cartesian coordinates and if you want to make this
change of variables here 𝑥 = 𝑟 cos 𝜙 and 𝑦 = 𝑟 sin 𝜙 and 𝑧 is equal to 𝑧. In that case this
622
integral will be written as so, the direct substitution for 𝑥 = 𝑟 cos 𝜙 for y we have replaced
by 𝑟 sin 𝜙 and 𝑧 because it was same there.
∫ ∫ ∫ 𝑓(𝑥, 𝑦, 𝑧) 𝑑𝑥𝑑𝑦𝑑𝑧 = ∫ ∫ ∫ 𝑓(𝑟 𝑐𝑜𝑠 𝜙 , 𝑟 𝑠𝑖𝑛 𝜙 , 𝑧) |𝐽| 𝑑𝑟 𝑑𝜙 𝑑𝑧
̂
𝐷
𝐷
So, this is the cylindrical not the spherical one, but the cylindrical coordinates; so,
cylindrical coordinates. So, in this case we have this Jacobian now which we can compute
by this determinant.
𝜕𝑥
𝜕𝑟
|
𝜕𝑦
𝐽=
𝜕𝑟
| 𝜕𝑧
𝜕𝑟
𝜕𝑥
𝜕𝜙
𝜕𝑦
𝜕𝜙
𝜕𝑧
𝜕𝜙
𝜕𝑥
𝜕𝑧
|
𝑐𝑜𝑠 𝜙
𝜕𝑦
= | 𝑠𝑖𝑛 𝜙
𝜕𝑧
0
𝜕𝑧 |
𝜕𝑧
−𝑟 𝑠𝑖𝑛 𝜙
𝑟 𝑐𝑜𝑠 𝜙
0
0
0| = 𝑟
1
So, again because there was no change in 𝑧; so, basically this is similar to what we do in
polar coordinate. So, we are changing from Cartesian to polar coordinates because the 𝑥
and 𝑦 are changed not the 𝑧; 𝑧 remains as it is in the polar in this is the cylindrical
coordinates. So, we have again the Jacobian which was also in polar coordinate as 𝑟. So,
the only change when we do this change in this triple integral this 𝑥 will be replace by
𝑟 cos 𝜙, 𝑦 will be replace by 𝑟 sin 𝜙 and 𝑧 will remain as it is. And, this factor here this
Jacobian will be 𝑟 and 𝑑𝑥 𝑑𝑦 𝑑𝑧 will become 𝑑𝑟 𝑑𝜙 𝑑𝑧.
(Refer Slide Time: 15:23)
623
So, let us go through some problems where we can use the idea of this changing to
cylindrical coordinates in this problem number 1. So, changing into cylindrical coordinates we evaluate this integral.
∭ 𝑧(𝑥 2 + 𝑦 2 ) 𝑑𝑥 𝑑𝑦 𝑑𝑧 , 𝐷: 𝑥 2 + 𝑦 2 ≤ 1, 2 ≤ 𝑧 ≤ 3
𝐷
So, the integrand is 𝑧(𝑥 2 + 𝑦 2 ) 𝑑𝑥 𝑑𝑦 𝑑𝑧 and this domain D is given by 𝑥 2 + 𝑦 2 ≤ 1. So,
that is the disc circular disc and then we have in the direction of 𝑧 as well; that means, it
varies from 2 𝑡𝑜 3. So, the 𝑧 coordinates are directly given that 2 ≤ 𝑧 ≤ 3 and then we
have this circular disc here in the 𝑥𝑦-plane 𝑥 2 + 𝑦 2 ≤ 1.
So, this is a cylinder here with radius 1 circular cylinder with radius 1 and it varies in the
direction of this 𝑧 from 2 𝑡𝑜 3. So, this is the picture here the 𝑥 coordinate 𝑦 coordinate
and this 𝑧 coordinate. And, we have from 𝑧 is equal to 2 to 𝑧 is equal to 3 that is the height
of the cylinder along the 𝑧 axis and that is the axis of the cylinder as well. And, then we
have the circular disc here which is 𝑥 2 + 𝑦 2 < 1. So, the radius of the cylinder is 1 and
the height here is from 2 𝑡𝑜 3. So, it is natural to use for instance the cylindrical coordinates because, this is exactly the cylinder is given and it will be much more convenient,
if we use cylindrical coordinate in such problems.
So, we have the relation that 𝑥 = 𝑟 and 𝑦 = 𝑟 sin 𝜙 , and 𝑧 = 𝑧 as usual. We also know
that this x square plus y square in this case will become this r square. And, also the Jacobian
term which will be requiring for this change of variable of this triple integral and that is in
624
cylindrical co-ordinate that is also 𝑟 we have just evaluated in the previous slide. So now,
this integral the given interval will be represented in this polar coordinate. So, this 𝑟 here
that is for the Jacobian. So, we have this term because of the Jacobian and then this dr d𝜙
and d𝜃 which will be replaced for this dx dy dz with this extra Jacobian term and then the
first let us discuss these integrands. So, the 𝑧 as it is and 𝑥 2 + 𝑦 2 has become now 𝑟 2 ; now
we come to the limits here for 𝑟.
So, it is very clear or let us first put the limits of the 𝑧. So, the 𝑧 limits trivially given there
that 𝑧 varies from 2 𝑡𝑜 3; so, the limits of 𝑧 2 𝑡𝑜 3. So, once we have covered the limits of
𝑧 then what is left it is a projection to the 𝑥𝑦 plane and that is nothing, but the circle. So,
here once we have fixed these limits of that z what is left now is a circle in the 𝑥𝑦 plane
which we know already in polar coordinates what will be the limits for the circle.
The 𝑟 is going from 0 𝑡𝑜 1 because, that is the radius here is 1 of the circle. So, 𝑟 is moving
from 0 𝑡𝑜 1 and then the theta is moving from 0 𝑡𝑜 2𝜋 the whole circle. So, or the 𝜙 in
this case we take a in cylindrical polar coordinates here 𝜙. So, phi varies from 0 𝑡𝑜 2𝜋
and 𝑟 goes from 0 to 1. And, then we have change the integral we have taken care for the
Jacobian term and now we can evaluate this integral.
(Refer Slide Time: 19:45)
3
𝐼= ∫
2𝜋
∫
1
∫ 𝑧 𝑟 2 𝑟 𝑑𝑟 𝑑𝜙 𝑑𝑧
𝑧=2 𝜙=0 𝑟=0
625
3
2𝜋
𝐼= ∫
1
∫ 𝑧 𝑟 2 𝑟 𝑑𝑟 𝑑𝜙 𝑑𝑧
∫
𝑧=2 𝜙=0 𝑟=0
3
= ∫
2𝜋
1
𝑧 𝑑𝜙 𝑑𝑧
4
∫
𝑧=2 𝜙=0
=
2𝜋 3
∫ 𝑧 𝑑𝑧
4 𝑧=2
=
5𝜋
4
(Refer Slide Time: 21:37)
So, we check another problem here changing to his spherical coordinates.
√1−𝑥 2
1
∫ ∫
0
0
1
1
∫
√𝑥 2 +𝑦 2 √𝑥 2
+ 𝑦2 + 𝑧2
𝑑𝑧 𝑑𝑦 𝑑𝑥
So now, we will change to his spherical coordinate to evaluate this integral. So, here we
need to be again very careful that what are the limits here?
626
𝑥 = 𝑟 sin 𝜃 cos 𝜙 ,
𝑦 = 𝑟 𝑠𝑖𝑛 𝜃 𝑠𝑖𝑛 𝜙 , 𝑧 = 𝑟 𝑐𝑜𝑠 𝜃
𝐽 = 𝑟 2 𝑠𝑖𝑛 𝜃 , 𝑥 2 + 𝑦 2 + 𝑧 2 = 𝑟 2
(Refer Slide Time: 25:17)
√1−𝑥 2
1
∫ ∫
0
1
1
∫
√𝑥 2 +𝑦 2 √𝑥 2
0
𝜋
2
𝜋
4
sec 𝜃
=∫ ∫ ∫
0
0
𝑟=0
+ 𝑦2 + 𝑧2
𝑑𝑧 𝑑𝑦 𝑑𝑥
1 2
𝑟 sin 𝜃 𝑑𝑟 𝑑𝜃 𝑑𝜙
𝑟
627
(Refer Slide Time: 29:29)
𝜋
2
𝜋
41
=∫ ∫
0
0
2
sec 2 𝜃 sin 𝜃 𝑑𝜃 𝑑𝜙
𝜋
𝜋
1 2
= ∫ sec 𝜃 |04 𝑑𝜙
2 0
=
(√2 − 1)𝜋
4
628
(Refer Slide Time: 31:35)
The last example where we will be changing again to its spherical coordinate to evaluate
this integral.
√1−𝑥 2
1
∫ ∫
0
0
√1−𝑥 2 −𝑦2
∫
0
1
√1 − 𝑥 2 + 𝑦 2 + 𝑧 2
𝑑𝑧 𝑑𝑦 𝑑𝑥
And, again the change into a spherical coordinate we have to look for the geometry what
we are having in this integral limits. So, for the inner one z goes from 0 to
√1 − 𝑥 2 − 𝑦 2 ; that means, 𝑥 2 + 𝑦 2 + 𝑧 2 is equal to 1. So, we are moving from in the
direction of z from 0 to that is sphere of radius 1, that is the equation z is equal to
√1 − 𝑥 2 − 𝑦 2 that is the sphere.
So, we are moving from 0 in the direction of z to that sphere. So, we have to now look a
little bit more careful. And then in the direction of this y and this x this is nothing, but that
circle only when this is sphere is projected to this 𝑥𝑦 plane. And, then we have x from 0
to 1 and 𝑦 from 0 to √1 − 𝑥 2 . So, that is precisely in the 𝑥𝑦 plane what we have that is the
values for this y and x that is a circle circular part of these limits and then here in the
direction of this z we are moving from 0 to the sphere. So now, again this is natural to
consider a spherical coordinate which will convert this.
629
(Refer Slide Time: 33:13)
𝑥 = 𝑟 sin 𝜃 cos 𝜙 , 𝑦 = 𝑟 𝑠𝑖𝑛 𝜃 𝑠𝑖𝑛 𝜙 , 𝑧 = 𝑟 𝑐𝑜𝑠 𝜃
𝐽 = 𝑟 2 𝑠𝑖𝑛 𝜃 , 𝑥 2 + 𝑦 2 + 𝑧 2 = 𝑟 2
𝐼=∫
𝜋
2
𝜋
2
∫
1
∫
𝑟 2 sin 𝜃
𝜃=0 𝜙=0 𝑟=0 √1
− 𝑟2
𝑑𝑟 𝑑𝜙 𝑑𝜃
(Refer Slide Time: 34:43)
𝐼=∫
𝜋
2
𝜋
2
∫
1
∫
𝑟 2 sin 𝜃
𝜃=0 𝜙=0 𝑟=0 √1
630
− 𝑟2
𝑑𝑟 𝑑𝜙 𝑑𝜃
1
First evaluate ∫
𝑟2
𝑟=0 √1
− 𝑟2
𝑑𝑟
𝜋
= ∫02 sin2 𝑡 𝑑𝑡 (sub. 𝑟 = 𝑠𝑖𝑛 𝑡)
𝜋
1 2
= ∫ (1 − cos 2𝑡)𝑑𝑡
2 0
=
𝐼=∫
𝜋
2
𝜋
2
∫
1
𝜋
4
𝑟 2 sin 𝜃
∫
𝜃=0 𝜙=0 𝑟=0 √1
𝜋
− 𝑟2
𝜋
𝑑𝑟 𝑑𝜙 𝑑𝜃
𝜋 2 2
= ∫ ∫ sin 𝜃 𝑑𝜙 𝑑𝜃
4 0 0
=
𝜋
𝜋𝜋
[− cos 𝜃]02
42
=
(Refer Slide Time: 35:29)
631
𝜋2
8
So, what we have seen now at least two special cases where the we have considered this
spherical coordinate and also the cylindrical coordinate; in this triple integral the concept
was this change of variables where the Jacobian we need to convert. And, the most difficult
part again in this triple integral is finding the limits corresponding to the given space here.
In spherical we have to represent the coordinates in spherical coordinate when we are
changing or in cylindrical coordinates, but the evaluation become much easier if we have
suitable change there in variables.
(Refer Slide Time: 36:11)
So, these are the reference we have used.
Thank you very much for your attention.
632
Engineering Mathematics - I
Prof. Jitendra Kumar
Department of Mathematics
Indian Institute of Technology, Kharagpur
Lecture – 36
System of Linear Equations – Introduction
Welcome back. So, this is a lecture number 36 and today we will continue with a new
topic that is a linear algebra a very important topic in these series in this series of lecture.
And, we will start with the system of linear equations and again this is a very important to
understand many concepts in linear algebra. So, we will start with a very basic Introduction
to the System of Linear Equations.
(Refer Slide Time: 00:43)
So, we will be covering the introduction to the system and also the solution of the system
of linear equations and their geometrical interpretation.
633
(Refer Slide Time: 00:57)
So, here the system of linear equations; so, this is a general system written in terms of the
equations. So, this is the first equation where we have the coefficients here 𝑎11 𝑥1 and this
coefficient of 𝑥2 is 𝑎12 and so on; 𝑎11 𝑥1 + 𝑎12 𝑥2 + ⋯ + 𝑎1𝑛 𝑥𝑛 = 𝑏1 . So, this is first
equation where these are 𝑎11 𝑎12 and 𝑎1𝑛 these are the coefficients. So, they are the real
numbers and the 𝑥1 𝑥2 and 𝑥𝑛 they are the unknowns and the right hand side 𝑏1 again
some real number.
So, similarly we have the second equation which we have denoted by 𝑎21 𝑎22 and 𝑎2𝑛
these are the coefficients of 𝑥1 𝑥2 and 𝑥𝑛 respectively and the right hand side is denoted
by 𝑏2 . So, here we have 𝑎𝑚1 the mth equations; so, we have considered 𝑚 equations and 𝑛
unknowns. So, here we have 𝑎𝑚1 𝑎𝑚2 and 𝑎𝑚𝑛 and these are the coefficients of 𝑥1 𝑥2 and
𝑥𝑛 and the right hand side is 𝑏𝑚 . So, we have 1, 2 and so, on 𝑚 equations and 𝑥1 𝑥2 and
𝑥𝑛 these are the unknowns.
𝑎11 𝑥1 + 𝑎12 𝑥2 + ⋯ + 𝑎1𝑛 𝑥𝑛 = 𝑏1
𝑎21 𝑥1 + 𝑎22 𝑥2 + ⋯ + 𝑎2𝑛 𝑥𝑛 = 𝑏2
𝑎𝑚1 𝑥1 + 𝑎𝑚2 𝑥2 + ⋯ + 𝑎𝑚𝑛 𝑥𝑛 = 𝑏𝑚
So, in the matrix form we can write these equations as follows 𝐴 𝑥 is equal to 𝑏. So,
634
𝐴𝑥 =𝑏
𝑎11
𝑎21
𝐴=[ ⋮
𝑎𝑚1
𝑎12
𝑎22
⋮
𝑎𝑚2
… 𝑎1𝑛
… 𝑎2𝑛
⋱
⋮ ]
⋯ 𝑎𝑚𝑛
𝑥1
𝑏1
𝑥2
𝑏
𝑥 = [ ⋮ ],𝑏=[ 2]
⋮
𝑏𝑚
𝑥𝑛
So, a system of equation is consistent that is the definition we will use the system of
equation is consistent; if it has at least one solution and inconsistent if it has no solution.
So, because we will observe now in today’s lecture that there are several possibilities for
the solutions. One that the system does not have a solution at all, the second it has a
solution and it is unique, the third one it has solutions and they are in finite in number; so,
infinitely many solutions. So, these are the three possibilities for system of linear equations
we will also and we will also look for the interpretation; so again getting back to this matrix
form. So, this is easy to understand so, we have this matrix 𝐴 and this 𝑥. So, if you multiply
this 𝐴 and this 𝑥.
So, we will get another vector whose first component will be this left hand side of the
equation 1; the second component of that vector when we do multiplication of this 𝐴𝑥 𝐴
and 𝑥. The second component of this vector 𝐴𝑥 will be the left hand side of the second
equation and so on. And then we have the right hand side vector here 𝑏 whose components
are these 𝑏1 𝑏2 𝑏3 𝑎𝑛𝑑 𝑏𝑚 . So, comparing the each component we will get basically we
will get back to these equations. So, this system writing in this 𝐴𝑥 is equal to 𝑏 is equivalent
to the given system of equations where, we have 𝑚 equations and 𝑛 unknowns in general
we have considered here.
635
(Refer Slide Time: 05:15)
And then let us take a simple example. So, we have this𝑥+2𝑦 = 4 one equation and then
we have one more equation that is given by 𝑥 − 𝑦 = 1. So, considering this these two
equations because, the geometrical interpretation will be very easy and then we can
generalize the concept for 4 more variables. So, let us call this equation number 1 because,
this is a line the equation of the line. So, we are calling this 𝐿 1 and here this is 𝐿 2 the
equation of the second a line which is given by 𝑥 − 𝑦 = 1.
So, in this case when we are considering only two variables 𝑥 𝑎𝑛𝑑 𝑦 these equations the
equations of the system are nothing, but the equation of the lines. So, we can write down
the system in the matrix form as well. So, this will be the coefficient matrix where we will
collect the coefficient from the first equation that is 1 𝑎𝑛𝑑 2 there.
1 2 𝑥
4
[
] [𝑦] = [ ]
1
1 −1
636
the vector of unknowns that is 𝑥 𝑎𝑛𝑑 𝑦. So, we can write down this equation these
equations in the form of the matrix vectors. And, now we look for the geometrical
interpretation of these equations and the solution basically.
So, here this is the first equation 𝑥 + 2𝑦 = 4. So, if we look at here we can easily plot this.
So, when x is 0 the y is 2. So, this is actually the point (0,2) and then we have when y is
0 the x is 4 so, this is the point 4 and 0. So, this is the equation number 1 and then we have
the equation number 2 as well this 𝑥 − 𝑦 = 1 and that is given by this red line. So, here
when 𝑥 is 0 𝑦 is −1 so, this is the point (−1,0) and then we can have another point for
instance here 𝑦 is 0 𝑥 is 1. So, this is the point (1,0) and then we can draw this line given
by 𝑥 − 𝑦 = 1.
Now, the question is that as we see in this picture that they intersect at some point here.
So, let us compute the point of this intersection. So, if we add if we subtract equation
number 2 from the equation number 1. So, we will get here the 𝑥 will get cancelled and
then we will get 3 𝑦 is equal to 3 which gives us the 𝑦 is equal to 1. So, that is the
coordinate of 𝑦 and then we can put in any of these equations to get for example, 𝑥 𝑖s equal
to 1 + 𝑦 ; that means, 2. So, here these two lines intersect at this point here are 2 and 1;
so, we can write down here 2 and 1.
Now, talking about the solution of the system of equations; so solution means a point 𝑥 𝑦
which satisfy satisfies both the equations. So, here clearly this point here lies on the red
line and this point also lies on the blue line. So, naturally this point where 𝑥 is 2 and 𝑦 is
1 it satisfies both the equation; it must satisfy both the equations because, it lies on both
lines and what else we see here. So, the solution is 𝑥 is equal to 2 and 𝑦 is equal to 1 of
this system. And, the other point here to be noticed that this is the only point, this is the
only point where these two lines intersect or this is the only common point on both the
lines.
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So, that this has become the solution now. So, precisely in this case what we got we got
the unique solution because these two lines intersect exactly at one point. So, there is no
other possibility to have a solution for this given system of equations and this is what we
call the case of a unique solution. So, in this particular situation when we have considered
these two simple equations, we are getting unique solution. If we consider another case
here for are not the another case.
(Refer Slide Time: 09:59)
So, we will just look for one more interpretation of the same equation in the in terms of
the vectors. So, we have seen that we had these two equations which we have also written
in the form of this matrix a vector multiplication. But, there is another way of writing this
equation because, if I consider here this 𝑥 + 2 𝑦 and the second equation is 𝑥 − 𝑦 . So, we
can also write down the system as like the first vector I will take x from this and also x
from here and in the second equation the rest the y term.
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1 2 𝑥
4
[
] [𝑦] = [ ]
1
1 −1
So, 2 to y and then the minus y here and then the right hand side is 4 and 1 or we can write
down this as x and the vector 1 1 plus this y and this vector 2 minus 1 is equal to this vector
4 and 1. So, we have written the given system of equations in this form where we see the
vector scalar multiplication vector scalar multiplication and here the vector again the right
hand side the scalar means this the number x here.
(Refer Slide Time: 11:23)
So, now so, this is exactly what is written here that the system of equations whether it is
given in the matrix form or we can also write down in this vector form where, x is
1
2
4
multiplied to this vector [ ] y is multiplied to this vector [ ] is equal to [ ]. So, this is
1
1
−1
another way of looking at this product here because, in this case we had the matrix vector
product. So, we can also look into in this form that this x is multiplied to this first column,
𝑦 is multiplied to this second column that is a way we also do the product of this matrix
4
and the vector. The right hand side we have this vector [ ].
1
1
2
4
𝑥[ ]+𝑦[ ] = [ ]
1
1
−1
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(Refer Slide Time: 15:13)
So, now coming to another example. So, we have now 𝑥 − 𝑦 = 1 and the second equation
we are considering −𝑥 + 𝑦 = 2. So, this is equation number 1 and then we have an
equation number 2 here. So, if we plot now these two equations as earlier; so, we have this
first equation here where x is 0 and then y is minus 1. So, this is the point here minus 1
and x at 0 and the second line −𝑥 + 𝑦 = 2.
So, in this case when 𝑥 is 0 here we are getting the point 𝑦 𝑥 2. So, this is the point
(2, 0) here and then we can draw this line which has slope 1 and this line also has slope 1.
So, basically what we realize now because both the equations this equation has slope 1 and
this equation also have has slope 1. So, both the equations are basically parallel to each
other.
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So, if we talk about the solution of this system; so, it does not exist in this case because
there is no common point on the lines where, we can we can say that this point will satisfy
both the equations equation number 1 and equation number 2 because they do not intersect.
So, they are the parallel lines and in this case we do not have a solution of this given system
of equations. So, with the help of this very simple example the geometrical interpretation
itself says that we do not have a solution of this system of equations.
In terms of the equations if we want to see directly because, the first equation is 𝑥 − 𝑦 = 1
and if I multiply here the equation number 2 by minus 1 we will get 𝑥 − 𝑦 = − 2. So, we
have these two equations 𝑥 − 𝑦 = 1 and 𝑥 − 𝑦 = − 2. So, this is not possible because in
equation number 1 we are telling that 𝑥 − 1 the difference of −𝑥 and 𝑦 will be 1 whereas,
in the second equation we are telling that the difference of 𝑥 and 𝑦 will be −2.
So, this is not possible to have a point which satisfy both the equations. And, the
geometrical interpretation also says the same that these two lines they do not intersect and
hence, we cannot have a solution for this given system of equations.
(Refer Slide Time: 18:03)
Well so, these lines do not intersect and this is the case of no solution.
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(Refer Slide Time: 18:11)
So, now coming to the next interpretation, which we call the vectors interpretation. So, we
can again rewrite these equation or the system of equation in this form in the vectors form.
So, x and multiplied by 1 and minus 1 and y will be multiplied by minus 1 and this 1 the
right hand side vector is 1 and 2. So, similar to the previous case let us draw this vector
here 1 minus 1. So, this is the vector 1 minus 1 and then the other vector here we have
minus 1 and n 1. So, these two vectors so, these two vectors and the third vector is 1 and
2. So, the third vector is this 1 and 2; this is the third vector 1 and 2.
1
−1
1
]+𝑦[ ] = [ ]
−1
1
2
𝑥[
−1
] And,
1
So, now the question is can we add by multiplying some number to this vector [
1
]. The question is can we
−1
1
add these two by multiplying some number to get this vector [ ]and which is clearly visible
2
then the multi we can multiply some number to this vector [
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here that this is not possible because, these two vectors are parallel. So, we can get anything
in this in this parallel to this these two vectors, but we cannot get for instance here 1 and 2
which is not parallel to these two vectors the given vectors. So, certainly by any
combination of these two vectors or rather we usually call it linear combination. And, later
on we will come up with more or a better definition of this.
So, here any linear combination of these two vectors will not give us the vector 1 in 2 and
hence this is the case of no solution. So, it is not possible to produce this vector 1 and 2 by
any linear combination of this column 1 and this column 2 and hence, this is the case of
no solution.
(Refer Slide Time: 20:37)
So, coming to the to the next and the last case here where we consider the system as 𝑥 −
𝑦 = 2 and −𝑥 + 𝑦 = −2. So, this is our system now and again if we look for this
geometrical interpretation we need to plot these two lines. So, the first 𝑥 − 𝑦 = 2 which is
given by this blue line here 𝑥 − 𝑦 = 2. And, now if we draw this minus x plus y is equal
to minus 2 again we get the same line because, this is nothing, but the same equation. Here
if you multiply by minus 1 to the second equation you will get the same equation.
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So, basically these are not two different equations this is the same equation we have only
one equations. And therefore, we get this line the red line is sitting over the blue line here.
So, they say these two are the same lines and now naturally any point we take on this line
I mean they are the same line. So, we can take a point here on the line and that will satisfy
both the equations. So, here we have infinitely many possibilities for the solution of the
given system of equations.
Because, when these two equations are given it was easy to see that these two equations
are the same equation. But, when we have more equations and this is not always the case
that we can identify that which equation is a combination of the others for example,
example. So, that we will deal little later all those cases, but here for this very simple case
we can see this geometrically also that now, in this case any point on the line is the solution
of the given system of equations.
So, now this is the case of this infinitely many solutions which we call. So, we have see in
these 3 possibilities of the solution. In one case we have the unique solution, in another
one we have no solution here we have infinitely many solutions.
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(Refer Slide Time: 22:55)
So, let us look for the vector interpretation for this case as well.
So, we had these two equations and now we can write down in terms of the in terms of the
vectors. So, we have x multiplied by 1 and minus 1 again these coefficients when y
multiplied by minus 1 and 1 the another vector and then the right hand side vector we have
2 and minus 2. So, now if we plot these vectors the one here 1 minus 1 then we have minus
1 1 and the other one is minus 2 and minus 2. So, we have these 3 vectors the 1 minus 1
and then we have this 1 minus 1 and the other vector here we have this 2 and minus 2.
1
−1
2
]+𝑦[ ] = [ ]
−1
1
−2
𝑥[
So, it is very clearly visible now that this vector 2 minus 1 we can get by one of these
vectors or by the linear combinations we can multiplies. There are so, many infinitely
many possibilities to have this linear combination of these two vectors to get this vector 2
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and minus 1 because, this is parallel to the given vectors. And, and there are infinitely
many possibilities now to combine these two vectors to get this vector 2 and minus 2. So,
from the vector interpretation as well we can see that there are infinitely many solutions
in such cases.
(Refer Slide Time: 24:35)
So, and now coming to the conclusion. So, what we have seen? We have seen the 3 cases
for the system of linear equations that can happen to a system that the system will have a
unique solution and that was the case in terms of the these two equations which we have
consider or with two unknowns that there are 2 lines and they intersect exactly at one point.
And, this is what we call the case of unique solution that we can also interpret when we
have the three variables. So, in this case we will have instead of a line we will have a
planes. So, there are three planes now and they intersect exactly at one point; these three
planes and that is the solution that unique solution of that system.
And then we have also seen that if the lines are parallel for example, that they never
intersect and the same thing we can interpret in case of the three variables also that we
have the parallel planes. So, here we have seen the in the second case that the lines are
parallel. And hence we do not have any solution to the given system whereas, the third
case we have seen that that the both lines were basically the same and then any point on
the line was the solution of the system of equation. So, we have seen the case of the unique
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solution, we have seen the case of the infinitely many solutions and we have seen the case
of no solution.
And, these all are the possibilities which can happen for a system of linear equations when
we consider a large system of 𝑚 equations with n unknowns. But, we have seen here with
the help of very simple examples where we have considered only two unknowns. So, in
the next lecture we will be talking about this now the solution techniques to find the
solution when we have a large system; not just 2 by 2 which we have considered here for
geometrical and the vector interpretation.
(Refer Slide Time: 26:45)
So, these are the references we have used to prepare these lectures and.
Thank you for your attention.
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Engineering Mathematics – I
Prof. Jitendra Kumar
Department of Mathematics
Indian Institute of Technology, Kharagpur
Lecture – 37
System of Linear Equations – Gauss Elimination
So, welcome back and this is lecture number 37. We will be continuing our discussion on
System of Linear Equations and in particular, today we will look for the solution
techniques that is the Gauss Elimination Method.
So, we will be talking about this Gauss elimination method and there we will see that how
to get the solution using this very systematic elimination technique and with the help of
the same we will also talk about the consistency of the solution; that means, about the
existence and non-existence of the solution with the help of this technique we can identify
whether the system has a solution and it is unique or it does not have a solution, all these
identification we can also check with this Gauss elimination method.
(Refer Slide Time: 01:10)
So, the system of linear equations, the solution methods we have basically the other
methods to like the method of determinants you must be familiar with or we call this
Cramer’s rule. So, with the help of this determinant we get the solution and this is possible
when we have a system where a unique solution exists, the same equation the number of
equations the same as the number of unknowns and the system has a unique solution in
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that case this is possible to get the solution of a rather smaller system of equations using
this method of determinants.
We have also the matrix inversion method where this 𝐴𝑥 is equal to 𝑏 this system we can
solve once we have the inverse of the matrix. So, out of this 𝐴𝑥 is equal to 𝑏 we can write
down like 𝑥 = 𝐴−1 𝑏 and once we have the 𝐴−1 . So, we multiplied by this 𝑏 and then we
will get the 𝑥. So, if we have the inverse of a matrix we can easily write down the solution
of the system.
Now, this is a little more general case which we will be talking about this Gauss
elimination method. So, here we will be also dealing the cases when we have non unique
solutions meaning infinitely many solutions or the system does not have a solution. So, all
these we can figure it out with this Gauss elimination technique. And, we have iterative
methods that is the Jacobi and the Gauss-Seidel method for solving when we have a system
of equations which is large in number so, really a very large system. So, these all these
methods which we have this method of determinant Gauss elimination, inversion method
they actually fails or rather they take longer time, so, we take these iterative methods.
So, the first three methods falls into the category of the direct methods. The direct methods
meaning that we get actually the exact solution of the system using these techniques
whereas, here the iterative methods the they give us the approximate solution of the given
system of equation. So, we will be talking about in this lecture about this Gauss elimination
method, which is a very general one in the category of this direct methods.
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(Refer Slide Time: 03:44)
And, we will be also talking about in this technique some elementary row operations. So,
what do you mean by elementary row operations? So, we can interchange the 𝑖 − 𝑡ℎ and
the 𝑗 − 𝑡ℎ row of a matrix and if going back to the system of equations because for the
system of equations we are writing the matrix 𝐴.
So, changing this row is nothing, but just the writing the second equation for example, at
the first place and the first equation at the second place. So, it has nothing to do with the
solution and that exactly in terms of the element row operations we will be doing with the
matrix that we can change we can interchange any two rows of the matrix.
The second one we can multiply the 𝑖 − 𝑡ℎ row means any row by a nonzero number
lambda and this is precisely equivalent to saying that we have those equations and we are
multiplying any equation by some number and again that system the solution of the system
will remain we remain the same 𝑅𝑖 with that operation. So, in terms of the matrices we
talked about this row operation that 𝑅𝑖 now the 𝑖 − 𝑡ℎ row has become like lambda times
𝑅𝑖 So, we have just multiplied lambda a non-zero number to the equation 𝑅𝑖 .
Here addition to the lambda times the 𝑗 − 𝑡ℎ row to the 𝑖 − 𝑡ℎ row. So, what we can do
that we will multiply the 𝑗 − 𝑡ℎ row by a number lambda again non-zero number lambda
because if lambda is 0, then we are not adding anything to this 𝑅𝑖 . So, this lambda times
the 𝑗 − 𝑡ℎ row and we will add this to the 𝑖 − 𝑡ℎ row and in terms of the operation we
write like now the 𝑅𝑖 has become as 𝑅𝑖 + 𝜆𝑅𝑗 . So, we have added 𝜆𝑅𝑗 to 𝑅𝑖 . So, the row.
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So, 𝑅 means the row. So, the i-th row we have added this lambda times this j-th row and
the new 𝑅𝑖 will come up.
So, here these are the elementary row operations which we will be using for this Gauss
elimination method.
(Refer Slide Time: 06:12)
Let us explain this with the help of simple example and then perhaps in the next lecture
we will go for more general problems. So, here we take a very simple example. So, we
have three equation
𝑥1 + 𝑥2 + 𝑥3 = 4
2𝑥1 + 3𝑥2 + 𝑥3 = 7
𝑥1 + 2𝑥2 + 3𝑥3 = 9
First we to write down this equation in this matrix form where we also augment this right
hand side of the equations 4, 7, 9. So, the right hand side here I will be basically doing
precisely what we do in Gauss elimination method and the left hand side we will be doing
the same thing with the equations directly. So, it is easy to understand that this Gauss
elimination is nothing very very complicated, but it is like eliminating the variables in a
systematic fashion.
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So, here we have first that will be the first step that we have to write this combined matrix
or so called the augmented matrix in this form. So, we have the coefficient matrix that this
I will matrix A. So, if we write down the system of equations into Ax is equal to b form.
1
So, this [𝐴|𝑏] is going to be in this case [2
1
1
3
2
1 4
1 | 7].
3 9
So, these two the matrix A and the matrix b basically have all the information of the given
system of equations and what we do in the augmented matrix we basically collect this a
1
here. So, this is precisely the A [2
1
1 1
3 1] and this b we have augmented here with the
2 3
4
same matrix as extra column. So, the [7]. So, this matrix we call as the augmented matrix.
9
So, here what we do now that the left hand side with the equations as I said also everything
with the equation and then we will translate into the form of this matrix and so called the
Gauss elimination technique. So, here what we are doing now with the help of this equation
number 1, we are trying to eliminate this 𝑥1 variable from the equation number 2. So, what
we are supposed to do actually that if you multiply this equation 1 by 2 and then subtract
this equation from this equation number 2.
𝑅2 → 𝑅2 − 2𝑅1 , 𝑅3 → 𝑅3 − 𝑅1
𝑥1 + 𝑥2 + 𝑥3 = 4
𝑥2 − 𝑥3 = −1
𝑥2 + 2𝑥3 = 5
1
[𝐴|𝑏]~ [0
0
1
1
1
1
4
−1 | −1]
2
5
𝑅3 → 𝑅3 − 𝑅2
𝑥1 + 𝑥2 + 𝑥3 = 4
𝑥2 − 𝑥3 = −1
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3𝑥3 = 6
1
1
[𝐴|𝑏]~ [0 1
0
0
1
4
−1 | −1]
3
6
Solution:
𝑥3 = 2
𝑥2 = −1 + 2 = 1
𝑥1 = 4 − 1 − 2 = 1
(Refer Slide Time: 14:09)
And, now the next one; so, we have this reduced system here or corresponding augmented
matrix and now, this is the one step which we have to do in this Gauss elimination to start
with this augmented matrix and to reduce into this form, which we call as echelon form.
So, this is we will be continuing this discussion in the next lecture what is exactly echelon
form in general.
1
[𝐴|𝑏] ~ [0
0
1
1
0
1
4
−1 | −1]
3
6
So, in this case roughly speaking now, we have to see whether we got this echelon form.
So, always we will have this structure for the echelon this step like a structure for this
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echelon form. So, here something non-zero, everything zero then here also then everything
zero and then everything zero here and the left hand side naturally. So, we have this
echelon form of the system and once we reach to this echelon form we can get the solution
very easily. So, first let us see with the equations how to get the solution now.
So, the solution will be from the last equation we can directly write down our 𝑥3 .
𝑥3 = 2
𝑥2 = −1 + 𝑥3 = 1
𝑥1 = 4 − 𝑥2 − 𝑥3 = 1
What we call in terms of the Gauss elimination we call it this back substitution that is the
actually the third step. So, the first step was putting into this your matrix and the right hand
side vector into this augmented matrix. The second one is to do reducing to this echelon
form by doing these row operations, the last step of the Gauss elimination is going to be
back substitution where we will get the solution of the system.
So, back substitution we will start from the bottom here where the 3𝑥3 is equal to 6. So,
the last equation yeah this is the last equation which says 3𝑥3 is equal to 6. We should
always keep in mind that corresponding to this we have actually the system of equations
you are given in the left. So, here we have actually 3𝑥3 is equal to 6; that means, 𝑥3 is 2
here then this 𝑥3 is known then from this we will go step by step to up now.
So, here from this equation we will get x 2 variable. So, 𝑥2 = −1 + 𝑥3 = 1. So, we get the
solution out of this back substitution once we have this echelon form of the equation.
There are a few more points to be noted here and we will explore them in the next lecture
again. So, what we have observed that we have the we have got a unique solution we have
got a solution of the system as 1, 1, 2 and indeed this is the unique solution of the system
we do not see any other possibility directly getting the solution from the equations. But,
what is now we are going to introduce this that these numbers here in this first row or in
this first column this is called the pivot elements. This is called the pivot element and pivot
element is always non-zero element and below this everything should be zero.
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And, now, here when we come to the next row or next column this number is again pivot
because everything here to zero and left to also zero and also in this sub matrix everything
is zero. Going to the next this is also a pivot element because this is nonzero and everything
3
left to zero and there is nothing here because the matrix was this . So, these such elements
3
the such non-zero elements will be called pivot because they play a very important role
now discussing about the about the consistency of the solution about the existence non
existence of the solution.
So, if we have like written here the number of pivots, so, here how many pivots we have?
We have one pivot, we have two pivots, we have three pivots. We have three pivots in our
in our system here in our matrix in our this augmented matrix. So, the number of pivots
here is equal to number of unknowns. How many unknowns do we have here? Three
unknowns 𝑥1 , 𝑥2 , 𝑥3 .
So, this number of pivots is equal to the number of unknowns, then we have exactly the
unique solution this is the case of the unique solution because our number of pivots is
equal to number of unknowns or in other words we say because each column is parallel
each column is associated with one variable. So, the first one with 𝑥1 , this is associated
with the 𝑥2 , this is associated with the 𝑥3 because if you remember we can write down this
𝐴𝑥 is equal to 𝑏, I mean this product here in terms of this column here 1 0 0 multiplied by
𝑥1 and then the second one will be multiplied by 𝑥2 , the third column will be multiplied
by 𝑥3 .
So, this first column is associated with 𝑥1 here, this is with 𝑥2 , this is with 𝑥3 . So, if the
number of I mean each column has a pivot we can in other way we can put it as that if each
column as a pivot in that case the solution will be a unique solution. So, every column has
a pivot and this is exactly the case when we have the unique solution. So, we will be
exploring a little more on the pivots and this is just a very very basic introduction to this
concept here and we are considering only this 3 by 3 matrix. So, we will be talking about
in the next lecture more about these pivots and echelon form.
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(Refer Slide Time: 21:15)
So, now going to the example number 2, we have changed the example now.
1
[𝐴|𝑏] = [2
1
1 1 4
3 1 | 7]
2 0 9
So, this augmented matrix is taking this form and by doing those elementary
transformation which we have done earlier the same transformation because there is only
little change in the matrix will be again trying to make this element 0 out of the first row;
that means, we have to do we have to subtract from this row 2 times the row 1. So, this is
exactly written here 𝑅2 → 𝑅2 − 2𝑅1 and 𝑅3 → 𝑅3 − 𝑅1 .
1
~ [0
0
4
1 1
1 −1 | −1]
1 −1 5
𝑅3 → 𝑅3 − 𝑅2
1
~ [0
0
4
1 1
1 −1 | −1]
6
0 0
So, by reducing to this echelon form we have also realized that the given system is
inconsistent in consistence mean again that it has no solution it cannot have a solution
here. So, by just reducing it to this echelon form we observe that the solution has knows
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that the system has no solution. Because from the last equation we are getting 0 is equal to
minus 6.
So, by observing this we conclude that this system the given system has no solution and
in other words in terms of the pivots we call that the right the rightmost column has a pivot
because now this is the pivot here and this is the pivot in the second row or in the second
column; the third column has no pivot because this cannot be pivot because this is a 0
element. The pivot has to be a non-zero element, but looking at this number here this is a
pivot. So, it satisfied that everything left to the 0 and there is nothing to go to the bottom.
So, we have this right here right most column has a pivot. So, which is which is the case
exactly of the inconsistency or the system has no solution because this rightmost column
has a pivot, this should not happen that. The pivot should be there in this main matrix here
not in this rightmost column which corresponds to this right hand side vector b ok.
So, this is the case of no solution and the equations are inconsistent and hence the solution
does not exist.
(Refer Slide Time: 25:02)
Now, we will go to the third example which is again slightly changed.
1
[𝐴|𝑏] = [2
1
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1 1 4
3 1 | 7]
2 0 3
So, here, but we will be doing the same operations this 𝑅2 → 𝑅2 − 2𝑅1 and 𝑅3 → 𝑅3 − 𝑅1 ,
the same operation because the aim is to eliminate 𝑥1 from equation number 2 and equation
number 3.
1
~ [0
0
1
1
1
1
4
−1 | −1]
−1 −1
So, by doing so we got this and then the further again the same operation because this
equation I mean this row 2 or we want to get rid from equation from row 3 this element 1
here. So, we will just subtract. So, 𝑅3 → 𝑅3 − 𝑅2 and in that way we will get this 0 here
also 0 and 0.
1
~ [0
0
4
1 1
|
1 −1 −1]
0
0 0
So, now this is different than the previous case we are not getting anything like 0 is equal
to 6 we were getting the last equation is completely 0. So, 0 is equal to 0, there is no
problem. Or in other words we do not have pivot in the last column here because what are
our pivot this is a pivot elements and as I said they play away play very important role in
discussing several properties of the vector and mattresses. So, here this is pivot everything
0 here and there is nothing left.
Now, coming to the right one this is pivot see this is not the pivot here because the left you
have a non-zero element. So, for the pivot it has to be 0 to the left and also to the bottom
and everything to the left has to be 0. So, what we have we have this pivot here, we have
this element as pivot and now going to the column number 3 or the row number 3 we do
not have any pivot. So, no problem. So, but, but at least the last the rightmost column also
does not have a pivot. So, the system is consistent.
So, the system is consistent means we will have solutions and in the indeed in this case
when we see that the equations are consistent and this number of pivots are less than
actually the number of unknowns. So, in this case we got in only two pivots and there were
three unknowns. So, this is the case where we have infinitely many solutions because in
the case of unique solutions the number of pivots will be equal to the number of unknowns.
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So, in this case we will have infinitely many solutions and in terms again a very systematic
approach to solve this, when we have this case of infinitely many solutions. So, this column
has the pivot this column has a pivot the third column does not have a pivot and as I said
this we say this corresponding to 𝑥1 , this is corresponding to 𝑥2 and this is corresponding
to 𝑥3 .
So, here since this 𝑥3 corresponding to the 𝑥3 the third column does not have a pivot, we
call that this 𝑥3 is a free variable because corresponding to this 𝑥3 the column does not
have a pivot and we call this like a free variable; free means we can choose this as we
want. So, this is exactly the point where we are entering into the discussion of the infinitely
many solutions and since this is free variable so, we can choose anything we want and that
is what we will do now.
So, here the number of pivots in a less than is equal to number of unknowns because there
are two pivots and there are three unknowns and the number of free variables will be
precisely the left one because this is occupied here we have pivot 𝑥2 has a pivot, but 𝑥3
does not have a pivot. So, this is free variable. So, number of free variables will be n minus
r. So, n is the number of the columns there in this main matrix here a or the number of
unknowns and minus this r, that is the number of pivots we give this name to the number
of pivots.
(Refer Slide Time: 29:12)
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So, now what we have? We have our system which is reduced in this form and where 𝑥3
we have denoted as marked as free variable and then we can choose this 𝑥3 some 𝛼 as a
real number and then once we have 𝑥3 then from equation number 2, we will get x 2
because these are the dependent variables now.
𝑥1
[𝑥2 ]
𝑥3
=
5
[−1]
0
−2
+ 𝛼[ 1 ]
1
𝑥3 = 𝛼
𝑥2 = −1 + 𝛼
𝑥1 = 5 − 2𝛼
So, this is the solution of 𝐴𝑥 = 0 and this is null space which we will call as null space and
more discussion on this we will have a little later, but the solution of this 𝐴𝑥 = 0and that
is a very particular structure. Here we do see once we have written this solution here. It
has two components; one without this alpha and the other one with alpha and this part here
2 minus, 1, 1 this exactly satisfy 𝐴𝑥 = 0 and if it satisfy x is equal to 0 it the alpha times
this also satisfy 𝐴𝑥 = 0 because it is alpha is just a constant and if x satisfy this 𝐴𝑥 = 0, A
into alpha x will also satisfy 𝐴𝑥 = 0.
And, this the solution the solutions of this 𝐴𝑥 = 0 because there are so many solutions here
for each alpha, alpha you take 1, you take 2, you take 1.5, anything. So, that is exactly the
solution of 𝐴𝑥 = 0 and this later on we will come to this, this is called the null space of a.
So, null space will be nothing, but the solutions all solutions of this 𝐴𝑥 = 0 together will
be called as null space.
Here this constant part without alpha this is a one particular solution of given system 𝐴𝑥 =
−2
5
0 or sorry Ax is equal to b. So, this [−1] will satisfy our equations and this [ 1 ] or
1
0
multiplied by any number this will satisfy 𝐴𝑥 = 0 that the so called the homogeneous
system of equations, a corresponding to our given system. So, homogeneous means the
right hand side we have set to 0.
660
So, this a very very special structure we will be talking about more later. So, always you
will have that this our x we can write down 𝑥𝑝 . So, one particular solution of the of the
given system plus this of the solution of the homogeneous system when we set the right
−2
hand side to 0, and one can easily check that this to [ 1 ] actually satisfy when we have
1
−2
−2
this homogeneous one ; that means, the right hand side 0. So, this [ 1 ] so, here [ 1 ], so,
1
1
this is equal to 0, it satisfies.
So, it will satisfy all these three equations this part and any number also we can multiply
it to this, it will not affect because the right hand side is0. So, this part is exactly the solution
of the 𝐴𝑥 = 0 corresponding to this, Ax is equal to b equation. So, this part here is
corresponding to 𝐴𝑥 = 0 and this is one particular solution which exactly satisfies this Ax
is equal to b equation.
(Refer Slide Time: 33:24)
So, more on this we will continue in our next lecture. So, what we have seen a very basic
introduction to this elimination technique and in particular very important is this echelon
form which we have seen at least for this 3 by 3 system and based on this echelon form we
can talk about the solution we can talk about the consistency of the solution or
inconsistency of the solution. So, all these cases of unique solution, infinitely many
solution can be discussed with the help of this echelon form.
661
(Refer Slide Time: 33:56)
So, these are the references I used for preparing the lectures and.
Thank you very much.
662
Engineering Mathematics – I
Prof. Jitendra Kumar
Department of Mathematics
Indian Institute of Technology, Kharagpur
Lecture – 38
System of Linear Equations – Gauss Elimination (Contd.)
Welcome, back. So, this is lecture number 38, on System of Linear Equations and we will
continue our discussion on this Gauss Elimination technique which was introduced in
previous lecture. So, today we will go for more general case where we will see these
echelon form, a very important reduced form where we can identify about the consistency
of the solution or the system of equations.
(Refer Slide Time: 00:45)
So, let us just go back to this what we have done in the previous lecture with the help of
simple example now we will consider a rather general example where we have matrix A
which has m rows and n columns and then we have this 𝑥𝑛×1 and the right hand side vector
𝑏𝑚×1 .
𝐴𝑚×𝑛 𝑥𝑛×1 = 𝑏𝑚×1
So, the echelon form what is this echelon form which we have introduced already in the
previous lecture in general this will have this form. So, what exactly this let us discuss
here. So, these elements denoted by this symbol here. These are the; these are the pivot the
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pivotal elements and these are nonzero elements. So, what is the property of this pivot
element?
So, this number we will call pivot in this matrix when everything below this is 0 and also
the left this is the first element. So, we will not be talking about the left to this one for
instance this one here this is the pivot element because everything to the left all the
elements are 0 and here also all the elements are 0 and in this sub matrix everything is 0;
so this is we call the pivot element.
Again, in this reduced form in this echelon form which we call this matrix will be called
or this element will be called a pivot because everything here is 0 everything here is 0 and
in this sub matrix everything is 0 here. Similarly, this one is a pivot if it is a nonzero
element and this everything to this one is 0 and everything to this one is 0. So, that is that
the that is the property of the pivot it is a nonzero number and all the elements below these
are 0 all the elements left to this are 0.
And, we have this special structure which of the matrix which we call the echelon form;
we have these the first few rows are the nonzero rows this non nonzero rows and the last
few rows here which we see are the 0s rows; so, this here 0s. So, all these rows on the
bottom we have set these with 0s here the right hand side this symbol can. So, this can be
0 these elements or they may not be 0 that is the symbol we are using here just to represent
this echelon form in general.
Here we have these pivot elements which are always nonzero elements and, now what
exactly the same identification. So, here these are the pivot elements and they are nonzero
and with this symbol or this star here these are the other elements they may be 0 and they
may not be 0. But, what is important here to put into this echelon form we have bring into
this structure which this matrix has.
⊠
0
0
0
[𝐴̃|𝑏̃] = ⋮
0
0
⋮
(0
∗
0
0
0
⋮
⋯
⋯
∗ ∗
⊠ ∗
0 ⊠
0 0
⋮
⋮
0 0
⋯ ⋯
∗
∗
∗
⋮
⋮
0
⋯
⋯
⋯
⋯
⋮
⋮
⊠
⋯
⋯
⋯
⋯
⋮
⋮
∗
⋯
⋯
⋯
⋯
⋮
⋮
⋯
⋯
⋯
⋯
⋯
⋯
⋯
⋯
⋯
664
∗
∗
∗
⋮
⋮
∗
0
⋮
0
| ∗
| ∗
| ∗
| ∗
| ∗
| ∗
| ⊗
| ⋮
| ⊗)
So, here then this is step like structure and we need to this is the aim to put into this form
so that we have these 0s on the bottom of this of this structure here this stair like structure.
And if there are the zero rows they are they are taken to this bottom of this matrix and then
from this augmented part which was correspond to this b here these elements may be 0
and may not be 0.
So, once we reduce and we will see in one of the examples a little more general example
how to exactly get this echelon form. So, if you remember there this Gauss elimination
was basically having three steps – the first one putting your system into this augmented
matrix then reducing the augmented matrix exactly into this echelon form which we call
and then we need to identify these pivot elements. The pivot element are the important
elements of this matrix.
And, then we can characterize about the solution also whether we are going to have a
unique solution or the solution does not exist or we have infinitely many solutions based
on once we identify these pivot elements we can characterize the system in the form of
this consistency. So, this putting into echelon form that is the most I mean the difficult part
here. So, once we have the echelon form getting the solution from the echelon form was
just through this back substitution which we have observed in simple examples in previous
lecture.
So, we have these so called the pivot rows. So, here we have taken these pivot rows to the
to the up and then if something is 0 here in our matrix A that we have brought down to this
bottom of the matrix. So, these are the pivot rows because the each rows has a pivot here
now, each rows has a pivot. These are the pivotal row pivot rows and the rest here these
are called the zero rows.
So, if these are the r in number which usually the notation we use for pivot rows. So, if
these are the r rows and we have total m rows, so, naturally these zero rows will be m
minus r this is the structure which we will be using.
665
(Refer Slide Time: 06:36)
So, having this let me just introduce here one definition though I can you will be talking
more about this rank because this is not the only definition for rank. There are many other
definitions for this rank, but this is one of them that this rank of a matrix is defined as the
number of the pivots because later on you will observe in an alternate definition that will
again lead to this that Rank (A) = r (number of pivots) and that is the reason this number of
pivots or the identification of these pivots a pivot elements are very important.
So, here the rank of the matrix another very important concept of a matrix is defined as
this number r which is the number of the pivots. So, with this definition, we go ahead and
later on we will be talking more about this rank. So, if we say that the Rank (A) = r (number
of pivots) in our this augmented matrix then what we can observe from this augmented
matrix we can identify so many things.
The one here is that if these elements here if these elements yeah if these elements here in
this zero rows; if they are not equal to 0 and if they are not equal to 0 then the equations
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become inconsistent and why inconsistent because we have the situation that 0 is equal to
something nonzero which we have already discussed in the previous lecture.
So, if in our echelon form we got these as nonzero number, then the system is inconsistent
and meaning that the system has no solution. So, that is the first identification we are going
to have from this echelon form or in terms of the rank we can define now because we have
defined already the rank of the matrix by this number r.
So, what in terms of the r we will be talking always about now the this is corresponding to
the matrix A this first part of this augmented matrix and the second one here this column
was this b and the whole matrix we are calling this [𝐴|𝑏]. So, this augmented matrix. So,
the point here is now the rank in this situation what is the rank of A? Rank (𝐴) will be
from this matrix how many pivots are there? There are r pivots; so, the rank of a will be r.
But, if these are the nonzero elements if these are the nonzero elements then there will be
also a pivot here. So, we can just reduce it further to make all these 0 and they will be
nonzero number here at least in one of the equations. The system is inconsistent, that is
clear whenever we have one of them is a nonzero number here corresponding to this zero
rows then the system is inconsistent, but in terms of the rank what we can call that the rank
of A matrix which is this one here the Rank (𝐴) matrix this one it is this r, the number of
pivot elements.
And, it is not equal to the rank of the whole this augmented matrix because in this situation
what will happen when these are the nonzero numbers then this rank of this whole matrix;
that means, the number of pivots in the whole matrix will be equal to I will be more than
the Rank (𝐴) and in that case then we have the inconsistency.
So, if the rank of these two matrices Rank (𝐴) and rank of this augmented matrix these are
different then we have inconsistency of the system and this will exactly happen when we
have these here corresponding to zero rows we have something nonzero, then there will
be a pivot element here in this column as well in the last column or in other words we call
that we have the pivot in the last column and this is exactly the case of this inconsistency.
So, in many ways we can discuss this the one was already here that because this is not
equal to 0. So, inconsistent or we call this rightmost column has a pivot element and this
will exactly happen when we have this here nonzero or sitting in one of one of these
667
number is nonzero. So, we have either the rightmost pivot has rightmost column has a
pivot or we call rank a is not equal to the rank of the augmented matrix or we call simply
by looking at this that if this is nonzero then we have a case of no solution, clear.
(Refer Slide Time: 11:53)
So, now moving further the next possibility will be when this is 0. So, if these elements
are 0 here then we have that the system is consistent because there is nothing like 0 is equal
to nonzero in that case. So, if these are 0 then we have two possibilities and you have one
more point to be noted because 𝐴𝑥 is equal to 0 is always consistent that is the inclusion
coming from here because when we have this 𝑏 is 0, so, this right this column everything
is 0.
So, naturally these numbers will be 0 everything is 0 and we are doing only the row
operations from the beginning. So, nothing will change. So, they will be definitely 0 when
we have 𝐴𝑥 is equal to 0. So, in that case the system is always consistent meaning this 𝐴𝑥
is equal to 0 will have always a solution.
So, anyway so, we will be talking little more later and now here when we take this case
that these are zeros meaning the system is consistent then there are two cases the first here
the number of pivot elements is equal to the number of unknowns and this case also we
have seen in the last lecture and this is the case when we have exactly or in other words
that each column has a pivot because if the column columns are exactly the number of
unknowns.
668
So, if each column has a pivot then the system has a unique solution because the column
cannot have more than one pivot that because of this property that below this everything
should be 0. So, the each column can have only one pivot and this number of pivot equal
number unknowns meaning that each column has a pivot because a number of columns
equal to the number of unknowns.
So, in that case the system will have unique solution we will discuss all these with the help
of example today itself. So, in terms of the rank now if we talk about the Rank (𝐴) and
Rank([𝐴|𝑏]), so, they will be the same now and that is equal to this 𝑟 or equal to this 𝑛.
So, that is the case of in terms of the rank which we will later on discuss more in details.
(Refer Slide Time: 14:05)
And, now coming to the another possibility that if this is 0 means we have the consistency
and the number of pivot elements is less than the number of unknowns.
So, in this case what will happen, the system will have infinitely many solutions when this
is the case and there we will introduce the concept of this free variables and dependent
variables and then we can get the solution as well I mean some we can generate those
infinitely many solutions or in terms of the rank what we will say again the Rank([𝐴])is
equal to Rank([𝐴|𝑏]) for the consistency, but this rank is less than equal to the number of
unknowns. This is what we talked about this consistency in terms of the rank as well.
669
(Refer Slide Time: 14:57)
So, coming to the problem here we will discuss now this solution of this system Ax is
equal to b where the augmented matrix corresponding to the system is written as here.
1
[𝐴|𝑏] = [ 2
−1
3
2
4
−2
6
−2
−4
3
−7
−1
0
3
1
1 1
3 2
| ];
4 3
1 𝛽
𝛽∈ℝ
So, we have how many do we have 4 rows and we have 5 columns. So, this is a case where
we have 5 unknowns actually. So, the number of columns means the number of unknowns
because here this is 𝐴 and we have our system this 𝐴𝑥 is equal to is equal to 𝑏.
So, this x here must be to make it consistent with this a will have like 𝑥1 , 𝑥2 , 𝑥3 , 𝑥4 𝑎𝑛𝑑 𝑥5 .
So, there are 5 unknowns and therefore, these 5 columns each column this will correspond
to 𝑥1 , this is𝑥2 , this is 𝑥3 , this is 𝑥4 , this is 𝑥5 . So, number of columns always represent the
I mean of A here yeah not the whole augmented matrix this is corresponding to a this is
corresponding to b, that is what we have here.
So, the number of the columns in this a will represent the number of elements in our
system. So, these are their 5 unknowns and there is another beta here is sitting which
belongs to this real number. So, this beta is a real number and we will discuss that for what
values of beta we have the solution of the system or we do not have the solution all these
consistency part will also depend on beta in this particular case, ok.
670
So, what is the idea of this Gauss elimination or the reduction technique to this echelon
form we want to have echelon form out of this augmented matrix and the first step is to
make these numbers here 2 minus 1 3, 0 out of this row 1.
So, if we subtract from this row 2, 2 times the row 1 then this will be 0, our focus is just
to set this 0 rest everything will change accordingly, but we will set we will eliminate this
x 1 from equation number 2, we will eliminate this corresponding to x 1, this coefficient
from equation number 3 and also from equation number 4. So, for that we need to do this
elementary operation.
𝑅2 → 𝑅2 − 2𝑅1, 𝑅3 → 𝑅3 + 𝑅1 , 𝑅4 → 𝑅4 − 3𝑅1
1
[𝐴|𝑏] = [0
0
0
2
0
0
0
−2 −1
0
2
1
2
−1 4
1
1
0
1
|
]
4
5
−2 𝛽 − 3
𝑅2 ↔ 𝑅3
1
[𝐴|𝑏] = [0
0
0
2
0
0
0
−2 −1
1
2
0
2
−1 4
1
1
4
5
|
]
0
1
−2 𝛽 − 3
𝑅4 → 𝑅4 + 𝑅2
1
[𝐴|𝑏] = [0
0
0
2
0
0
0
−2
1
0
0
671
−1
2
2
6
1
1
4
5
|
]
0
1
3 𝛽+1
(Refer Slide Time: 21:17)
𝑅4 → 𝑅4 − 3𝑅3
1
[𝐴|𝑏] = [0
0
0
2
0
0
0
−2
1
0
0
−1
2
2
0
1
1
4
5
|
]
0
1
0 𝛽+1
So, this is the reduced form we do not have anything now further to reduce in this case.
(Refer Slide Time: 24:25)
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So, this is the reduced form and now we will identify we will take the first case when
𝛽 ≠ −1. So, if we take this case when 𝛽 ≠ −1; that means, something nonzero is sitting at
this place here when 𝛽 ≠ −1 something nonzero, nonzero will sit here. And, if something
nonzero is sitting here; that means, the last column has pivot last column has a pivot
meaning that we have an inconsistent system and or in other words we have 0 is equal to
something nonzero. So, in either case when this 𝛽 ≠ −1, the system is inconsistent and we
do not have a solution for the system.
So, that means, this is the case of exactly no solution and the case 2 we will consider when
𝛽 = −1. So, when 𝛽 = −1, this is exactly the case where we will discuss about the solution
because the system becomes consistent now; the last row has become 0. So, now, this
system is consistent, we do not have pivot in the last column now. So, let us identify the
pivots because that is another important step.
This is the pivot here in the first column, this is not pivot. So, each row and each column
will have at most one pivot that is also true this is the fact here because if this is the pivot
then nothing else can be the pivot because 2 cannot be pivot it is left to this something
nonzero is sitting.
Take 𝑥2 = 𝛼1 ,
𝑥5 = 𝛼2 , then
1
𝑥4 = −2𝛼2 , 𝑥3 = 4 − 4𝛼2 , 𝑥1 = 9 − 2𝛼1 − 9.5𝛼2
𝑥1
9
−2
−9.5
𝑥2
0
1
0
𝑥3 = 4 + 𝛼1 0 + 𝛼2 −4
𝑥4
0
0
−0.5
[𝑥5 ] [0 ]
[0 ]
[ 1 ]
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So, this 𝑥2 , we have marked here as free variables and this column has a pivot so, this is
corresponding to 𝑥3 you are not taking as free variable. So, this is; so, called the dependent
variables we can call. So, this x 3 is dependent, this is not free we have the pivot there. So,
wherever we have a pivot we marked them as dependent variables and wherever we do
not have a pivot we will mark them free variable that is a that is a simple algorithm we
will follow here though it is not necessary that we have to do in this way, but this is a very
systematic approach.
So, go to the column number 1, if it has a pivot element mark this as a dependent variable;
go to the next column 2, it does not have a pivot element. There is no pivot element in
column 2 we will mark as a free variable; column number 3 has a pivot then this is marked
as a dependent variable; again column number 4 has a pivot. So, this we have marked as a
dependent variable, column number 5 again we have marked as a free variable because it
does not have a pivot.
By doing this now, we have so many things in hand because free variables means we can
choose, we can give the values whatever we like and whenever we have free variables in
the system ah; that means, we have infinitely many solutions. So, this is the case of
infinitely many solutions since we have the free variables; free variables again means they
are free, you can choose whatever you like.
(Refer Slide Time: 29:22)
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So, in this case we have two in fact, free variables. So, we have more possibilities to
actually choose the choose the variables now here. So, this 𝑥2 and 𝑥5 , these are the two
free variables. So, we will take like x 2 alpha 1 and x 5 as alpha 2. In that case now from
this equation number 3 we can write down x 4. So, from here 2 times x 4 is equal to minus
this x 5 and x 5 we have taken x 2. So, this x 4 we can write down in terms of alpha 2.
Similarly, the x 3 from this equation number 2 we can write down in terms of this the
dependent where these free variables and also the x 1 from equation number 1, we can
write down in terms of these free variables alpha 1 and alpha 2 in the vector form we can
place them. So, x 1, x 2, x 3, x 4 the constant from each; so, from x 1 we have this 9 as
constant from x 2 was alpha 1 only. So, there is nothing constant from this x 3 we have 4
as constant and from x 4 there is nothing exactly and from x 5 also we do not have anything
to take as a constant.
And, then this alpha 1 and then the rest from x 1, for example, we have 9 minus 2. So,
minus 2 was with alpha 1. So, minus 2 here and with alpha 2 we have with alpha 2 here in
x 1 we have minus 9.5 this is taken here in this vector. So, we can write down now these
x 1, x 2, x 3, x 4 and x 5 in this particular form where we have first vector free from these
free variables and then this is with alpha 1, the one free variable the vector again coming
here and x 2 again this free vector with this three variable again this vector is coming up.
(Refer Slide Time: 33:16)
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(Refer Slide Time: 34:05)
So, just to remark, that these free variables are the responsible for infinitely many
solutions, so, in conclusion whenever we have free variables in our system; that means,
some columns do not have a pivot and in that case there will be a case of this infinitely
many solutions. An invertible matrix will have no free variable the reason is clear because
once the matrix invertible we get actually unique solution and if we have a unique solution
definitely we will not have a free variables or other way around if we observe that there is
no free variable; that means, this A is also invertible.
So, this vector that generates solution of Ax is equal to 0, what we have just seen that for
Ax is equal to 0 all the solutions we can generate with the help of those two vectors
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[−2, 1, 0, 0, 0]𝑇 & [−9.5, 0, − 4, − 0.5, 1]𝑇 . This was appearing with alpha 1 and this
was appearing with alpha 2 we have just used here this transpose because they were given
in this column form. So, these vectors they are so called the generators of the solution of
this Ax is equal to 0, that is what we have observed.
And, these generators are the BASIS are the BASIS of; so, this term we will introduce
little later. So, some technical terms I am just mentioning here, but they will be discussed
in the next lecture. So, these generators which we have just seen before these are called
the BASIS because these are the main component that generator of the solution. So, these
are called BASIS of the solution space again one more term has come; the solution space
of this Ax is equal to 0 because this Ax is equal to 0 has infinitely many solutions in that
case for instance. And, these vectors are called basis of the space where we have so many
solutions there and that space is also called the null space which again we will be discussed
in the next lecture yeah exactly here the null space is a vector space.
So, again one more term here the vector space has come. So, this null space which is the
solution space here is a vector space and there are some properties of a set here, the solution
set which we are talking about which will be later called as a vector space like for instance
here we see that if we add two solutions the new solution will be also solution of this Ax
is equal to 0 equation. Because A and if x 1 satisfy this and A and the x 2 also will satisfy
this and then the A and this x 1 plus x 2 will also satisfy that equation because Ax 1 plus
Ax 2 both are the 0, 0 so, that will also satisfy.
So, this solution set here of Ax is equal to 0, has some nice properties like you add any
two solution that will be also solution or you multiply by n number to this solution that
will be also a solution. So, there are so many other properties which are vector space has;
so, that we will discuss exactly in the next lecture.
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(Refer Slide Time: 37:16)
And, the conclusion here; so for the system of equations this echelon form was very
important and we have introduced this concept of the free variable and that lead to basically
infinitely many solutions and the solution of non homogeneous system of equation has this
special structure that we get this x p which satisfy the given Ax is equal to b and we have
a part here x h we satisfy Ax is equal to 0 and as a whole also this x means this satisfy Ax
is equal to b.
(Refer Slide Time: 37:48)
So, these are the references used.
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Thank you.
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Engineering Mathematics – I
Prof. Jitendra Kumar
Department of Mathematics
Indian Institute of Technology, Kharagpur
Lecture – 39
Linear Algebra – Vector Spaces
So, welcome back and this is lecture number 39 and we will be talking about a very
important topic in Linear Algebra that is a Vector Spaces.
(Refer Slide Time: 00:25)
And, just to recall from the previous lecture what we have seen for instance this solution
set of the homogeneous system of equations Ax is equal to 0, that is a very special set. So,
the solution set of solution set of this homogeneous system of equation 𝐴𝑥 = 0. Why that
is very special? Because in this set; so, let us call this as the set V which contains all the
solutions of this equation 𝐴𝑥 = 0, then if we take any two elements of this set let us say
𝑥1 and also we take 𝑥2 from this set; that means, this 𝐴𝑥1 = 0 and 𝐴𝑥2 = 0 because the
𝑥1 this vector is also a solution and 𝑥2 is also solution.
So, if we take two elements of this set and if we add them; so if we consider now 𝑥1+ 𝑥2
and that will be also the solution of this system of equations 𝐴𝑥 = 0, because of this linear
property here we have 𝐴𝑥1 and then 𝐴𝑥2 that is a matrix vector product and this is 0 and
this is 0. So, we have this 𝑥1+ 𝑥2 is also a solution and not only this if we take any element
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of this set and if we multiply by a real number, so, let us say 𝜆. So, this 𝜆𝒙 will also belong
to V because this is again a solution if 𝑥 is a solution and then this 𝜆𝒙 because 𝐴𝜆𝒙 will be
𝜆𝐴𝒙 and this Ax is will give 0. So, lambda into 0 and then you will get the zero vector.
So, here what we have seen in this set which is a kind of a special set if we take any two
elements of the set and add them that is also an element of the set. And if you multiply by
a real number to this element of this set this new element is also an element of this set 𝑉.
So, we are going into this direction that what do we call these such special sets and that is
exactly the vector spaces coming into the picture.
(Refer Slide Time: 02:49)
So, here this vector spaces they are the special set here. So, a vector space again here we
need to define the vector space over ℝ for instance and this ℝ can be more general like
here we have considered this set of real numbers this can be a set of complex numbers and
in more general cases what we call the field. So, it can be any field, but we are not going
to discuss that into details. So, most of the time we will take this vector space over this ℝ
because with this set 𝑉1 another set must be associated which we have taken here this ℝ
set of real numbers. So, this vector space is a set 𝑉 of elements and called vectors.
So, that is another point here I would like to mention that the elements of this vector space
𝑉 will be called vectors. So, they can be matrices they can be functions depending on what
type of elements this 𝑉 contain, but the elements of this 𝑉 we will call vector, so, a little
more general definition here and together with two binary operations. So, we need other
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than this set here we which we will call vector space we need this set ℝ which we have
taken and then two binary operations which we call this vector addition and this is scalar
multiplication.
So, what do we have? We have this set of 𝑉, we have another set which we have taken
here the set of real numbers and then we will have this vector addition which we have to
define with this set 𝑉 and we have scalar multiplication. So, we need to define these four
four things to define this vector space. So, the mean set 𝑉 here one another set of real
numbers or the set of complex number and two algebraic operations which we call vector
addition and scalar multiplication.
So, having all these here; now when do we call this set 𝑉 a vector space when the following
axioms hold. So, what are these axioms? The first one is that if we take two elements u
and v from this set 𝑉 and if we add them the new elements should also belong to this set
𝑉. So, that is the additive closure property of this set that we take any two elements and
do this vector addition which has to be defined for this given set 𝑉 then this addition of
this u and v must belong to the set 𝑉.
Another property that this lambda times u, so, the 𝜆 which is called is scalar because 𝜆
belongs to this set of real numbers or little more general this set of complex numbers we
can talk about, but we are restricting to the set of real numbers here. So, this 𝜆𝒖 that is the
scalar multiplication. So, here this scalar 𝜆 is multiplied to this vector u. So, this 𝜆𝒖 must
also belong to V that is another closure property which we call with respect to this is scalar
multiplication. So, for any 𝑢, the 𝜆𝒖 must also be there in the set for V and these are called
the closure properties of the set.
𝒖 + 𝒗 ∈ 𝑉, ∀ 𝒖, 𝒗 ∈ 𝑉
𝜆𝒖 ∈ 𝑉, ∀ 𝜆 ∈ ℝ, 𝒖 ∈ 𝑉
And, there are many more properties which are not so important in our discussion, but we
need to just state them for completeness because most of our examples all these properties
will trivially a follow, but these two properties are the most important properties which we
call the closure properties. So, here the third one that 𝒖 + 𝒗 = 𝒗 + 𝒖, for 𝒖, 𝒗 ∈ 𝑉 and this
is the commutativity property of this these vectors.
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𝒖 + 𝒗 = 𝒗 + 𝒖, ∀ 𝒖, 𝒗 ∈ 𝑉
The another one that here u plus first we add 𝒗 + 𝒘 or first we add this 𝒖 + 𝒗 and then w
the result must be the same and this is called the associativity property. Again, so, we have
this existence of the 0 vector. So, what how do we define the zero vector. So, there must
be a zero vector in this set V which we are concerned now. So, here this 0 must be there
and what is the property of 0, that if we add this 0 to any element of this set V then that
element will remain as it is. So, this is called the zero vector and this must be there in the
set V.
𝒖 + (𝒗 + 𝒘) = (𝒖 + 𝒗) + 𝒘, ∀ 𝒖, 𝒗, 𝒘 ∈ 𝑉, Associativity
∃ 𝟎 (zero vector) in 𝑉, s.t., 𝒖 + 𝟎 = 𝒖,
∀𝒖∈𝑉
And, another one; so, for each 𝑢 so, for each element of this set V; there must exist a vector
V which is denoted by − 𝑢. So, this is just a notation here − 𝑢 the negative of u such that
when we add this 𝑢 and this − 𝑢 we must get the zero vector which already exists there in
the set.
For each 𝒖 ∈ 𝑉, ∃ a vector in 𝑉, denoted by −𝒖 (negative of 𝒖), s.t., 𝒖 + (−𝒖) = 𝟎
The another one again this is a property with respect to this scalar multiplication that we
are multiplying here 𝜇 with u or we first multiply 𝜆 and mu in this set of real numbers and
then we multiply to this 𝑢 the result must be same for all and 𝜇 from the set of real numbers
and this 𝑢 from the set 𝑉 and this is again this associativity property with respect to this
scalar multiplication.
𝜆(𝜇𝒖) = (𝜆𝜇)𝒖, ∀ 𝜆, 𝜇 ∈ ℝ, 𝒖 ∈ 𝑉
And, further we have this 𝜆(𝒖 + 𝒗) must be equal to 𝜆𝒖 + 𝜆𝒗 and for all 𝜆 ∈ ℝ and all
𝒖, 𝒗 ∈ 𝑉 . This is called the distributive property of this of these scalars and the vectors
𝜆(𝒖 + 𝒗) = 𝜆𝒖 + 𝜆𝒗, ∀ 𝜆 ∈ ℝ, and 𝒖, 𝒗 ∈ 𝑉
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And, then we have this (𝜆 + 𝜇)𝒖, then this should be equal to 𝜆𝒖 + 𝜇𝒖 and this should
hold for all lambda mu from ℝ and u from V. So, this is again this distributivity property
of these scalars and vectors.
(𝜆 + 𝜇)𝒖 = 𝜆𝒖 + 𝜇𝒖, ∀ 𝜆, 𝜇 ∈ ℝ, 𝒖 ∈ 𝑉
And, the last ones that for each u from this set V we should have that 1 into 𝑢; 1 is the
identity element in ℝ, so, which is the number ℝ in case of this real number set of real
numbers. So, this if you multiply u to this 1. So, this is again here the scalar multiplication
which must be defined for each this set here and it must be equal to u and this is again kind
of a property which all the elements of this u must satisfy.
For each 𝒖 ∈ 𝑉, 1𝒖 = 𝒖, 1 being the identity element in ℝ.
So, we have so many properties, but as I said at the beginning that these two properties are
most important here; these are the closure properties that 𝑢 + 𝑣 must be there in the set
and 𝜆𝒖 times u must be there in the set, for all u and for all 𝜆 from ℝ. So, with these two
properties and there are other properties as well which we have to keep in mind like the
existence of the zero vector, existence of this negativity and all other these distributivity
property associativity property etcetera must hold for this set V then we call this set V as
a vector space.
(Refer Slide Time: 11:14)
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So, now we go through some of the examples where we will understand now better what
is this vector space. So, here the first example we are considering that is the vector space
R n over R. So, what is the R n? How the elements of this R n looks like? So, this R n is
this set of all this ordered n-tuples means we have the elements of this type a 1, a 2, a 3, a
n and all these as are from the real number. So, this R and the elements of R n will have
these n-elements the n-components in each element of this of this set.
And, here we need to define again this vector addition and scalar multiplication then only
we will say that this is a vector space with respect to this scalar multiplication, this vector
addition and this set of real number ℝ which we are considering. So, here the vector
addition and this is scalar multiplication is defined as usual. So, when we take these two
elements here (𝑎1 , 𝑎2 , … , 𝑎𝑛 ), (𝑏1 , 𝑏2 , … , 𝑏𝑛 ) from this set 𝑉 and when we add them, so,
the new element will be just the component wise addition here.
(𝑎1 , 𝑎2 , … , 𝑎𝑛 ) + (𝑏1 , 𝑏2 , … , 𝑏𝑛 ) = (𝑎1 + 𝑏1 , 𝑎2 + 𝑏2 , … , 𝑎𝑛 + 𝑏𝑛 )
So, that is the how the addition works in this particular case. And, for the multiplication
the scalar multiplication so, when we multiply by this scalar 𝜆(𝑎1 , 𝑎2 , … , 𝑎𝑛 ) in that case
we should get here now the new element will be just the multiplication of this lambda to
each of the element of this element or this member of this set V and for all lambda from
R.
𝜆(𝑎1 , 𝑎2 , … , 𝑎𝑛 ) = (𝜆𝑎1 , 𝜆𝑎2 , … , 𝜆𝑎𝑛 ), 𝜆 ∈ ℝ.
So, this is how these vector addition and the scalar multiplication is defined for this set
and what we can easily now observe those two closure properties at least because all others
are trivial in this case again like for instance the zero element will be when all these
components are zero and so on, the negative elements will be when we put the minus sign
in front of each so, that will be the negative element of a given element. So, all those
existence all this commutativity property etcetera one can easily verify.
So, again what is also important the closure properties which are again trivial here. So,
when we add two elements here we are getting a new element and that also belongs to this
ℝ𝑛 because this is again a element of element in this ℝ𝑛 . So, this closure property with
respect to vector addition is satisfied and also for the scalar multiplication. So, this new
element here 𝜆𝑎1 , 𝜆𝑎2 , … , 𝜆𝑎𝑛 this also belongs to this set here ℝ𝑛 .
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So, the both the closure properties are satisfied, all other properties one can easily check
that they are also satisfied. So, this ℝ𝑛 this set here ℝ𝑛 is a vector space and we can talk
about and we can take any integers here. So, the ℝ over ℝ is a vector space ℝ2 over ℝ is
a vector space etcetera it is it is clear now, whatever and we take your ℝ3 for instance it is
also a vector space over ℝ.
(Refer Slide Time: 15:03)
Well, so, now going to the next example here we will consider the polynomial space. So,
here we are considering this 𝑃𝑛 (𝑡) again over ℝ as I said we will be restricting to a set of
real number here. So, this 𝑃𝑛 (𝑡); so, how the elements of 𝑃𝑛 (𝑡) look like? So, 𝑃𝑛 (𝑡) denotes
the set of all polynomials of degree less than or equal to n. So, all polynomial this is a set
of all polynomials of degree less than or equal to n and how these polynomials look like
they are of this kind 𝑎0 + 𝑎1 𝑡 + ⋯ + 𝑎𝑠 𝑡 𝑠 , and this s is less than or equal to n because we
are talking about the set of all such polynomials.
𝑝(𝑡) = 𝑎0 + 𝑎1 𝑡 + ⋯ + 𝑎𝑠 𝑡 𝑠 ,where 𝑠 ≤ 𝑛 and 𝑎𝑖 ∈ ℝ.
So, here the constants also belong to this, the linear polynomial belongs to this, quadratic
at so and so on depending on this n here. So, this is a set of all polynomials of degree less
than or equal to n for given n and all these a’s here are just the real numbers.
So, in this case when we have polynomials now so, our vectors are these polynomials and
now, again if we look at the closer properties here, but the closure properties are satisfied
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because if we take any two elements from this set. So, for example, we have taken like one
element here which I am calling 𝑝1 (𝑡) and which we can denote again here this with a’s.
So, this is one element so, let me take this 𝑛. So, t power n this is one element of this set
here the polynomial space which you are talking about. Now let me take the another ones
which I can denote by this 𝑏0 . So, 𝑏1 𝑡 and here the degree can be n or it can be less than
𝑛, so, 𝑏 and 𝑡 𝑛 . So, these are the two elements from the same set and when we add these
two so, 𝑝1 (𝑡) plus this 𝑝2 (𝑡) how the addition works in this set it is just we have to add
the corresponding coefficients here.
So, 𝑎0 + 𝑏0 these were the constant terms and with the t we have this 𝑎1 + 𝑏1 with this
𝑡 2 we will have this 𝑎2 + 𝑏2 and with this 𝑡 𝑛 we will have 𝑎𝑛 + 𝑏𝑛 . So, this is the new
polynomial now after this addition, but naturally when 𝑎0 and this 𝑏0 in ℝ the sum is also
ℝ similarly this 𝑎1 + 𝑏1 will be in ℝ and 𝑎𝑛 + 𝑏𝑛 will be also in ℝ.
So, here this is a new polynomial which belongs to again this set this 𝑃𝑛 (𝑡). So, how this
is I mean now one can see this closure property with respect to the addition here. Similarly
when we take this 𝑝(𝑡) from this from this set 𝑃𝑛 (𝑡) and if you multiply by any scalar here
the lambda times this 𝑝(𝑡). So, what will happen? The lambda will be multiplied to this
𝑎0 , lambda will be multiplied to 𝑎1 the lambda will be multiplied to this to this 𝑎𝑠 here and
this new polynomial will be again an element of this 𝑃𝑛 (𝑡).
So, again we have the closure property with respect to this scalar multiplication, we have
the closure property with respect to the addition and this addition is defined in this way
which we have explained the multiplication is defined in this way for this set. So, this
polynomial space 𝑃𝑛 (𝑡) over ℝ is a vector space which we have seen in this example.
The next one we are talking about the matrix space; here the 𝑀𝑚⋅𝑛 we are talking about.
So, why in this case it is a matrix space again this is a vector space because if we take two
matrices of what are m and n. So, let us consider this 2 by 3 matrices for instance just for
simplicity.
So, for 2 by 3 matrices so, we have taken one element like a b c and then the d e and f this
is the one element we have taken from this set. We have taken another element of this set
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which we are denoting by a 2 b 2 c 2 and a 3 d 3 d 2 and e 2 and f 2 these are the two
elements for example, from this set here 𝑀2×3 are all matrices of order a 2 × 3.
So, when we add these two matrix a and b and we can take any two matrices the result
would be again this matrix of order 2 × 3 and we will be just adding these components
here a 1 plus b 2 this b 1 a 1 plus a 2 b 1 plus b 2 and so on. So, this will be again a matrix
of order 2 × 3 and this is also over this ℝ set of real numbers we are talking about. So,
here these are all real numbers when we add them they would be also real number and the
new matrix will also belongs to this set here m 𝑀2×3 .
And, also the scalar multiplication; so, when we multiply any member of this set here by
𝜆 the new matrix will be just multiplied by 𝜆 each component will be multiplied by 𝜆 and
again, this new matrix will be also a member of this set here 𝑀2×3 . So, here what we have
seen that this matrix spaces of order this 𝑚 × 𝑛 is also a vector space.
(Refer Slide Time: 21:25)
So, another example which is also a very important example from the system of linear
equations; so, the solution of this 𝐴𝑥 = 0 this form a vector space and how it forms a
vector space? So, we take this set 𝑉 here 𝑥 ∈ ℝ𝑛 and it satisfy this Ax is equal to 0; we
have discussed at the very beginning of today’s lecture about this set that the special set
only.
𝑉 = {𝑥 ∈ ℝ𝑛 : 𝐴𝑚𝑛 𝑥𝑛 = 0𝑚 }
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So, this is called null space and has discussed before because here also those closure
properties are satisfied and the elements are from ℝ𝑛 ; ℝ𝑛 is a vector space already we
have seen. So, this elements of this 𝑉 belongs to ℝ𝑛 and ℝ𝑛 is a vector space. So, naturally
all the properties which we discussed, they are satisfied automatically about this
distributivity, commutativity all are valid here because these elements actually belong to
ℝ𝑛 ; ℝ𝑛 is a vector space.
And, about the closure property; so, whenever we have taken the two elements here 𝑥1 and
𝑥2 from this set a set 𝑉. So, naturally this the sum 𝑥1 + 𝑥2 that is the property of this
solution here that this a this 𝑥1 , 𝑥2 will be also the solution because 𝐴(𝑥1 + 𝑥2 ) will be 0
and also A times the 𝜆𝑥1 or 𝜆𝑥2 with any element you can multiply by 𝜆 that will be also
0.
So, that is special for this homogeneous system when we have the right hand side 0, if we
do not have this right hand side 0; if it is a non homogeneous system we do not have this
property these two properties for example, valid for the solution of non-homogeneous
system of equations. So, for the homogeneous system of equations the solution set is a
vector space and this vector space has a special name which we call as null space that is
what this space coming into the picture. So, this is called the null space of the matrix 𝐴.
Note that this 𝑉 the set 𝑉 is a subset of is a subset of this ℝ𝑛 and forms a vector space and
therefore, this is called also vector subspace. So, we will be talking about that soon that
what are these vector subspaces. So, this is for instance vector subspaces because this is a
vector space and this is a subset of another vector space ℝ𝑛 and therefore, we call this as
a vector subspace.
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(Refer Slide Time: 24:17)
So, here again this what are the subspaces here. So, let 𝑉 be a vector space and let 𝑊 be a
subset of 𝑉. So, 𝑉 is a vector space and 𝑊 is another subset of 𝑉 and if 𝑉𝑊 is also a sub
also a vector space in that case we call that 𝑊 is a subspace. If this 𝑊 itself is a vector
space over the same the set of these real numbers with respect to the same operations what
we are done in the vector space 𝑉 then this 𝑊 is called a vector subspace.
For every 𝑢, 𝑣 ∈ 𝑊 and 𝜆 ∈ ℝ, the following closure properties should hold
1. 𝒖 + 𝒗 ∈ 𝑊
2. 𝜆𝒖 ∈ 𝑊
The criteria for identifying the subspaces is much easier now and we do not have to worry
about all the properties which we have discussed for defining the vector space and here we
would be talking about only the closure properties because those are important now all
others will be trivially satisfied. So, basically when we check whether a given subset is a
vector space or not what we have to check we have to check these closure properties. We
take the two elements any two elements the sum the addition of these two must belong to
this subset and also the 𝜆𝒖 must belong to this subset and that is enough to check that the
given space given vector space is a is a subspace.
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(Refer Slide Time: 25:43)
\
So, what are the trivial subspaces? So, here is a examples now of the subspaces here the
set {0} itself. So, if we take just the 0 element of a vector space that will form also a vector
space because of the closure properties. For example, you take the element the 0 and add
to the 0 you will get 0 and we multiply by any scalar to the 0 the element will remain as a
0. So, this is a vector space because all these closure properties are satisfied and the whole
set 𝑉 itself we can call as a vector subspace of the vector space 𝑉.
Another example here when we take this set 𝑈 = {(𝑎, 𝑏, 𝑐): 𝑎 = 𝑏 = 𝑐} ⊂ ℝ3 is a
subspace of ℝ3 . So, basically we have taken the elements when all these components are
equal and this naturally the elements belong to this ℝ3 ; ℝ3 is a is a vector space and this
is also a subspace of ℝ3 the reason is again clear if we take two elements from this set 𝑈.
So, here this belongs to 𝑈 because all three are equal and if we take 𝑏 this is also an element
from this 𝑈.
And, now this closure properties we need to basically focus on if we add the two here. So,
this new element here all the components are equal again this will also belong to the same
set 𝑈 or we multiply by the 𝜆 to any element here, the new element will be also 𝜆𝑎. So, all
three components are equal and that that is the reason it must belong to this 𝑈 again. So,
these closer properties are satisfied for this set 𝑈 which is a subspace of now of ℝ3 .
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Example 2: So, if we take here the 𝑉 = {(𝑥1 , 𝑥2 ) ∈ ℝ2 : 𝑥1 ≥ 0, 𝑥2 ≥ 0}. So, here we are
talking about this positive so, both the components are non-negative here; 𝑥1 ≥ 0, 𝑥2 ≥ 0.
And the question is whether this 𝑉 forms a vector space? And the answer is no, because if
we take one element here and the closure property says that the 𝜆 times this we must be
there and if we take lambda minus 1, for example, so, the −1 into the this element from 𝑉
which are having this non-negative components, but when we multiply with the − sign it
will become −(𝑥1 , 𝑥2 ) and this will not belongs to this set 𝑉.
So, here this closure property is not satisfied like here I have stated again the −1 when you
multiply to any element here that will not belong to 𝑉 because the 𝑉 has the property that
both are non negative. So, this is one example where we can see that this does not form a
vector space though here the; this is a subset of this R square.
(Refer Slide Time: 29:06)
The last example here for the subspace we consider the vectors of this form
[𝑠 + 4𝑡, 3𝑠 − 𝑡, 5𝑠 + 𝑡, 2𝑡]𝑇 this type of vectors and this 𝑠 and 𝑡 they can take any real
number. So, now, the question is whether this set the set of these vectors form a subspace
in ℝ4 . So, ℝ4 because there are four components here of this vector so, they are they
belongs to ℝ4 . Now, the question is whether this is a subspace of the ℝ4 or not?
So, here now this it is easy to consider now in this way
692
𝑠 + 4𝑡
1
4
3𝑠 − 𝑡
3
−1
[
] = 𝑠[ ] + t [ ]
5𝑠 + 𝑡
5
1
2𝑡
0
2
So, this, the new vector, will also belong to the same set.
So, the both the closure properties are satisfied for the set and hence this will also form a
vector space of ℝ4 . So, this is a vector space of ℝ4 what is left within in the following
lectures we will learn that this is a smaller set naturally of this ℝ4 and now how to how to
quantify this in terms of what in terms of kind of dimension or some other criteria; so, that
we will be also talking about in following lectures.
(Refer Slide Time: 31:58)
And, now coming to the conclusion, so, for the vector space the most important properties
were these additive closer and this the scalar closer here. So, 𝒖 + 𝒗 ∈ 𝑉 and 𝜆𝒖 ∈ 𝑉; these
were the closure properties the most important properties of the vector spaces and we have
seen several examples including the space vector spaces vector subspaces; so, these are
the references used for preparing the lectures
693
(Refer Slide Time: 32:29)
And thank you for your attention.
694
Engineering Mathematics – I
Prof. Jitendra Kumar
Department of Mathematics
Indian Institute of Technology, Kharagpur
Lecture – 40
Linear Algebra – Linear Independence of Vector
So, welcome back and this is lecture number 40, and we will be talking about the Linear
Independence of Vectors. In particular, we will cover today the linear combinations of the
vectors and also this linear independence of vectors.
(Refer Slide Time: 00:31)
So, what is linear combination? So, an expression of this type 𝜆1 𝒗𝟏 + 𝜆2 𝒗𝟐 + ⋯ + 𝜆𝑛 𝒗𝒏 ,
where these 𝜆′𝑠 belongs to the set of real numbers it is called linear combination of vectors
𝒗𝟏 , 𝒗𝟐 , … , 𝒗𝒏 .
So, here for given vectors this 𝒗𝟏 , 𝒗𝟐 , … , 𝒗𝒏 this expression here 𝑣 = 𝜆1 𝒗𝟏 + 𝜆2 𝒗𝟐 + ⋯ +
𝜆𝑛 𝒗𝒏 this is called the linear combination of these vectors where these lambdas are belongs
to the set of real numbers and just a remark short remarks here. So, 𝑣 belongs to a vector
space over R which is 𝑉 is a vector space and 𝑣 is an event of this vector space 𝑉 this is a
linear combination of 𝒗𝟏 , 𝒗𝟐 , … , 𝒗𝒏 in V. So, these all are vectors in 𝑉 and we call that this
𝑣 is a linear combination of 𝒗𝟏 , 𝒗𝟐 , … , 𝒗𝒏 if there exists scalars 𝜆1 , 𝜆2 , … , 𝜆𝑛 and they
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belongs to ℝ because 𝑉 these as a as the scalars as such that we can write down this 𝑣 here
the given 𝑣 as a linear combination of the others.
Then we call that this 𝑣 is a linear combination of the vectors 𝒗𝟏 , 𝒗𝟐 , … , 𝒗𝒏 . If we can write
down this 𝒗 = 𝜆1 𝒗𝟏 + 𝜆2 𝒗𝟐 + ⋯ + 𝜆𝑛 𝒗𝒏 then we call this vector 𝑣 which is again this
vector is a more general term we are talking about from the vector space they can be
matrices, they can be polynomial etcetera. So, here if this given 𝑣 here we can write down
as a linear combination of these other vectors v then we call that this is a linear combination
of the vectors 𝒗𝟏 , 𝒗𝟐 , … , 𝒗𝒏 .
(Refer Slide Time: 02:45)
So, the example here we take these four vectors 𝛼 = [4, 3, 5]𝑇 ,
𝛽 = [0, 1, 3]𝑇 ,
𝛾=
[2, 1, 1]𝑇 , 𝛿 = [4, 2, 2]𝑇 . Now, we have the following questions now we want to
examine whether this 𝛼 is a linear combination of 𝛽 and 𝛾 we will also see that if this beta
is a linear combination of 𝛾 and this 𝛿 and this 𝛾 is a linear combination of 𝛼 and 𝛽.
So, we will answer these questions here. To answer the first one if this 𝛼 is a linear
combination of 𝛽 and 𝛾. So, if we can write down this alpha as 𝜆1 𝛽 + 𝜆2 𝛾 for some 𝜆′𝑠
then we will call yes, this alpha is a linear combination of 𝛽 and 𝛾. So, now, to do so, we
will check whether it is possible or not and eventually we end up usually with the system
of linear equations to answer all these questions. And we have to solve this system of linear
equations which we can write down like again we can simplify this little bit.
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0
2
4
[3] = 𝜆1 [1] + 𝜆2 [1]
3
1
5
⇒ 𝝀𝟏 = 𝟏, 𝝀𝟐 = 𝟐
So, naturally this alpha is a linear combination of is a linear combination of beta and
gamma which we have seen and in this example.
Now, we will check for the second one that if this 𝛽 is a linear combination of 𝛾 & 𝛿.
0
4
2
[1] = 𝜆1 [1] + 𝜆2 [2]
3
1
2
So, it is not possible to find such 𝜆′𝑠 which satisfy these two equations here, forget about
the first one.
So, here definitely then we cannot find such 𝜆′𝑠 which can satisfy this relation here that
beta is equal to this 𝜆1 𝛾 + 𝜆2 𝛿. So, what do we get now, so; that means, this beta is not a
linear combination of these two elements here gamma and delta. So, this is a inconsistent
system and it has no solution and therefore, beta is not a linear combination of 𝛾 & 𝛿.
Now, coming to the third one here 𝛾 is a linear combination of 𝛼 & 𝛽 we have just seen
in this case 1 here that this was 𝛼 here and then we had a 𝛽 and we have 𝛾 that 𝛼 is a linear
combination of this 𝛽 and 𝛾 and this was the relation 𝛼 is equal to 𝛽 plus 2 𝛾 and now, we
are asking whether 𝛾 is a linear combination of 𝛼 and 𝛽 . So, naturally gamma is a linear
combination of 𝛼 and 𝛽 and we do not have to prove anything because in the first part we
have already shown this relation that 𝛼 is equal to 𝛽 plus this 2 𝛾 and then we have naturally
that 𝛾 is equal to one half of 𝛼 minus 𝛽 .
So, this 𝛾 is a linear combination of 𝛼 and 𝛽 which is trivial from this part 1 of this problem,
ok.
697
(Refer Slide Time: 09:07)
Well, so, now going to the next topic here we will be talking about linear independence of
vectors and what do we have here. So, 𝑉 is a vector space we take and then a finite set of
vectors. So, 𝑣1 , 𝑣2 , … , 𝑣𝑛 of this vector space 𝑉 this is said to be linearly independent if
𝜆1 𝑣1 + 𝜆2 𝑣2 + ⋯ + 𝜆𝑛 𝑣𝑛 = 0 ⟹ 𝜆1 = 𝜆2 = ⋯ = 𝜆𝑛 = 0, 𝜆𝑖 : 𝑠𝑐𝑎𝑙𝑎𝑟𝑠
So, now we can investigate again here the linear independence of these given vectors
which are given here. So, what do we have to check for this linear independence that if
this linear combination which is given 𝜆1 𝑣1 + 𝜆2 𝑣2 + ⋯ + 𝜆𝑛 𝑣𝑛 = 0, after solving these
equations if we get the only solution as 𝜆1 = 𝜆2 = ⋯ = 𝜆𝑛 = 0, then we call that these this
set is linearly independent, but if we get some other solution of the 𝜆′𝑠 nonzero solution
then they will be dependent here. So, they are not independent in that case.
So, let us consider this example. So, here we have taken these three vectors 𝑣1 =
(1, −1,0), 𝑣2 = (0,1, −1), 𝑣3 = (0,0,1). So, we will consider now this expression
𝜆1 𝑣1 + 𝜆2 𝑣2 + 𝜆3 𝑣3 = 𝟎
⇒ (𝜆1 , −𝜆1 + 𝜆2 , −𝜆2 + 𝜆3 ) = (0,0,0)
⟹ 𝜆1 = 𝜆2 = 𝜆3 = 0
The given set of vectors is linearly independent.
698
So, naturally all the trivial solution what we call here is (0, 0, 0) for lambdas, but you will
see whether we have some non-trivial solution or not if you have only the trivial solution
that is the zero solution then we call that they are linearly independent.
So, that is the only possibility we are getting here that is a unique solution we can also in
general we will come right down in the form of this augmented matrix and then reducing
to this row echelon form from there also we can observe whether we are getting a unique
solution or we are getting infinitely many solutions or this is the this cannot be the case of
no solution naturally because this is always a system of this homogeneous equation which
always has at least a trivial solution.
So, here we have seen that these vectors are linearly independent.
(Refer Slide Time: 13:41)
And, we will check now the second example which says that 𝑣1 = (0, 0), 𝑣2 = (1, 2) are
linearly independent. And this is a trivial task to check whenever we have this zero vector
is there in the set because we can always consider this equality 𝜆1 (0, 0) + 𝜆2 (1, 2) = (0, 0)
and since the zero vector is there so, I can take any lambda here it does not where matter I
can take any lambda with this and then the 0 value to 𝜆2 and the equation will be satisfied
because with this 0 we have now the zero vector here and I can choose any 𝜆 still this will
be a zero vector and that 𝜆1 (0, 0) + 0(1, 2) = (0, 0).
699
So, for all 𝜆1 from this set of real numbers this equation is satisfied and therefore, we are
getting a nonzero solution that our 𝜆′𝑠 are not zero in this case, but they are we are getting
infinitely many possibilities it is which are adding to this zero here and therefore, this set
is linearly dependent.
So, whenever we have a zero vector included in the set of these vectors then these vectors
are going to be linearly dependent. So, this 0 is one of the vectors in the set and then the
set must be linearly dependent.
(Refer Slide Time: 15:19)
And, this problem 3 we examine these vectors 𝑣1 = (1,1,1)𝑇 , 𝑣2 = (1,1,0)𝑇 , 𝑣3 =
(1,0,0)𝑇 , 𝑣4 = (1,0,1)𝑇 are linearly independent. So, you have taken these 4 vectors now
and we want to check whether they are linearly independent or they are linearly dependent
the same process we will continue now. So, we will consider this linear combination here
and that will be set to 0. So, we 𝜆1 𝑣1 + 𝜆2 𝑣2 + 𝜆3 𝑣3 + 𝜆4 𝑣4 = 0.
⇒ 𝜆1 + 𝜆2 + 𝜆3 + 𝜆4 = 0,
𝜆1 + 𝜆2 = 0,
𝜆1 + 𝜆4 = 0
So, we have these three equations coming out of this linear combination because there
were three components in each vectors. So, having this now, this system of equations
which we want to solve for lambdas the general way would be always to write down in the
form of this augmented matrix and then do this elimination or reduce to this row echelon
700
form and from there the situation will be clear whether we are getting unique solution or
we are getting non unique solution.
So, here for this system of equations we have written here this augmented matrix.
1 1
[𝐴|𝑏] = [1 1
1 0
1
0
0
1 0
0 | 0]
1 0
The right hand side the zero vector which we have kept here the zero vector and then we
can reduce it to the echelon form which is very simple in this case because this first row
will remain as it is.
1
[𝐴|𝑏]~ [0
0
1
−1
0
1
−1
−1
1 0
0 | 0]
−1 0
So, here that is what we call and if you remember already from our previous lectures this
is corresponding to 𝜆4 and which we call the free variables; free variable in this case only
one and these are the dependent variables 𝜆1 , 𝜆2 , 𝜆3 , 𝜆4 . So, this having a free variable in
the solution in the system here we are naturally getting infinitely many possibilities of the
solution now. And once we have infinitely many possibilities of the solution that system
cannot be linearly independent because we do not have on the unique solution which is 𝜆1
all these 𝜆′𝑠 to be equal to 0.
So, in this case we have now the so many solutions infinitely many solutions because this
𝜆4 is a free variable and we can choose anything we want. So, 𝜆4 for instance we have
taken as 1, and then from this equation we can get this 𝜆3 = −1, 𝜆2 = 1, 𝜆1 = −1.
So, we are getting for this choice of this 𝜆4 all other lambdas here 𝜆3 = −1, 𝜆2 = 1, 𝜆1 =
−1. So, that means, we are getting this relation here. So, naturally they are dependent, they
are not independent and they depend on each other with this relation. So, this given set of
vectors here is linearly dependent.
701
(Refer Slide Time: 21:03)
So, the another problem here we will examine now this set {𝑣1 = (2,1,1)𝑇 , 𝑣2 =
(1,2,2)𝑇 , 𝑣3 = (1,1,1)𝑇 } whether it is linearly independent or it is linearly dependent. So,
just getting back to the previous one, what we have it is a similar problem. So, we had the
four vectors here and our augmented matrix was having this columns here all these vectors
were the columns here
2
[𝐴|𝑏] = [1
1
1 1 0
2 1 | 0]
2 1 0
So, we kept all these in the columns here and the right hand side this b is always 0, we can
0
we do not have to write also this [0]always because that is not going to change. So, we
0
can just work with the 𝐴 itself and get the row reduced this echelon form from there we
can conclude the existence of the solution.
So, here now getting back to this one the new problem we have these three vectors and
you want to check the linear independence of these vectors. So, what do we consider? We
consider directly the augmented matrix which we have earlier formed through the system
of linear equations.
702
So, now, we will reduce this to this reduced echelon form which is much easier in this
case.
2
1
[𝐴|𝑏]~ [0 3/2
0
0
1 0
1/2 | 0]
0 0
So, we will get finally, this row reduced echelon form and where now we have this is the
pivot element and this is the pivot element and this third column will not have a pivot
element. So, we have these two pivot elements here. This is the 0 rows and what we
conclude now?
So, we have the free variable. We have the free variable this 𝜆3 corresponding to the
column 3 it is a free variable and which we can choose whatever we like. So, just for the
sake of simplicity of the calculation we have taken 𝜆3 = −3.. And then this 𝜆2 and 𝜆1 from
this equation number 2 and from the equation number 1 we will get the other 2 once we
fix this 𝜆3 . We can take any other number also here for 𝜆3 we have free to choose 𝜆3 .
So, with this combination we are getting for instance 𝑣1 + 𝑣2 − 3𝑣3 = 0 and there are there
can be so many possibilities here if we choose this 𝜆3 as 1 we will get different 𝜆2 and 𝜆1
or any other number for this 𝜆3 correspondingly this 𝜆1 and 𝜆2 will change.
So, this is definitely not a unique representation we can have as many representations as
possible here, but what it says that these vectors are linearly dependent and one of the
relations here for their dependencies already here 𝑣1 + 𝑣2 = 3𝑣3 that is 1 dependency we
have shown in this case and this set is linearly dependent.
703
(Refer Slide Time: 24:48)
So, what we have learnt today, it is about the linear independence of the vectors. This is
again a very important topic when while considering the vector spaces which we will
continue our discussion in the next lecture and what was important that this linear
combination of the vectors l𝜆1 𝑣1 + 𝜆2 𝑣2 + ⋯ + 𝜆𝑛 𝑣𝑛 = 0.
If we out of this equation if we get the unique solution as the 𝜆1 = 𝜆2 = ⋯ = 𝜆𝑛 = 0, then
we call that this set is linearly independent. Because we can there is no dependency on the
vectors on each other because that is the only possibility coming out of this relation, so we
cannot write any dependency among these vectors and that is the reason we call this linear
independence that they are independent.
Whenever we are getting a nonzero solution and that will be the case of infinitely many
solutions. So, there we can have now many ways to write down these relations. So, 1 can
be written in terms of the others and that is the case where what we call a linear
dependence. So, in that case those vectors will be linearly dependent; so, these are the
references we have used in this lecture.
704
(Refer Slide Time: 26:13)
And thank you for your attention.
705
Engineering Mathematics – I
Prof. Jitendra Kumar
Department of Mathematics
Indian Institute of Technology, Kharagpur
Lecture – 41
Vector Spaces – Spanning Set
So, welcome back and this is lecture number 41 will be talking about this Spanning Set.
So, to define the Vector Space we need to prepare for many concepts and this is one of
them so, called the spanning set.
(Refer Slide Time: 00:33)
And here we will be talking about the linear span first and then will be talking about
spanning set.
706
(Refer Slide Time: 00:40)
So, what is the linear span? So, linear span of vectors of given vectors 𝑣1 , 𝑣2 , … , 𝑣𝑛 this is
defined as and usually we denote as a span of this 𝑣1 , 𝑣2 , … , 𝑣𝑛 and we define as the set of
all these linear combinations of these vectors 𝑣1 , 𝑣2 , … , 𝑣𝑛 . So, here exactly this is the
linear combination given 𝜆1 𝑣1 + 𝜆2 𝑣2 + ⋯ + 𝜆𝑛 𝑣𝑛 . So, this the linear combination of the
vectors and this lambda belongs to the set of real number.
So, this is a set of all linear combinations of the given vectors we call the span of
𝑣1 , 𝑣2 , … , 𝑣𝑛 . Exactly so, the collection of all linear combinations is called the linear span
of v vectors 𝑣1 , 𝑣2 , … , 𝑣𝑛 and the problem here we will discuss now, is this vector
[1, 0, 0]𝑇 just to tell that this is a column vector, but we can do also with the row vectors.
So, here is the vector here in the span of the vectors [4, 2, 7]𝑇 and [3, 1, 4]𝑇 .
So, what do we need to check basically can we write this vector as a linear combination of
these two vectors because this is the question here that is this vector in the span of the
vectors of this. So, what is the span of these vectors? All linear combinations of these two
vectors and to check whether this belongs to this a span of this these vectors on we will
just check whether this vector is a linear combination of these two vectors or not.
So, we need to basically check we need to find the 𝜆1 and 𝜆2 such that the
4
1
3
𝜆1 [2] + 𝜆2 [1] = [0]
7
0
4
707
whether this is possible to find such lambdas or this is not possible the system here is
inconsistent or it is a consistent system which can give us the 𝜆1 and 𝜆2 .
So, to answer this again we have to get back to this the idea of the augmented matrix and
the row reduced or echelon form again just note that this row reduced echelon form is very
important almost in our lectures we will be utilizing the idea of this solving the system of
equations, using gauss elimination or rather saying this reducing to this row reduced
echelon form because once we have this row reduced echelon form, we can tell exactly
that what type of solution we are getting out of this given system.
Now we need to reduce it to the row reduced echelon form and that idea is also we have
explained several times we need to make this elements 0, then this element 0 with the help
of this equation number 1 and then we need to make this element 0 with the help of the
equation number 2. So, that doing so, I am writing directly here what will be the reduced
echelon form.
4
[2
7
3 1
4 3 1
1 | 0] ~ [0 1 | 1]
4 0
0 0 2
this is the row reduce echelon form and from here now we can conclude because this is
now our the first column has a pivot element the second column has also the pivot element
and, but the problem here is with the consistency of the equation. The last equation says
that 0 is equal to 2 which is not possible which is not possible here; that means, we cannot
solve this system we cannot solve this system for 4 𝜆1 𝑎𝑛𝑑 𝜆2 there is no solution because
our equations are inconsistent.
So, we cannot get any solution of this given a system of equations and therefore, the answer
1
4
3
to this question is that this vector [0]or does not lie in this span of the vectors [2] and [1].
0
7
4
So, here the system is inconsistent and we cannot write down this as linear combination
given there.
708
(Refer Slide Time: 05:52)
So, there is another nice result here we will not go through the formal proof, but we will
see at least some intuition how this is happening. So, if as is a subset of a vector space 𝑉
then this is SPAN(𝑆) yes so, what is SPAN(𝑆)? All the linear combinations of the vectors
that is the SPAN(𝑆), is a subspace of 𝑉 and what is the nice property? This span has 𝑆 so,
we take any two any vectors collection of vectors and then the span of this will give you
the vector space.
So, this is SPAN(𝑆) is the subspace meaning this a vector space of this 𝑉 the subspace of
𝑉 whose elements we have taken as three subsets and this is the smallest another important
information that this is the smallest subspace which contains this set S. So, there cannot
be any smaller subset of this which contain s and is a vector space. So, this is the smallest
vector space containing the set of vectors.
So, just to see the idea of the proof here so, we take let us say this is our set 𝑆 =
{𝑣1 , 𝑣2 , … , 𝑣𝑛 } these are the elements of this 𝑆 and what we will show that the span of this
S is a is a vector space is a vector space of 𝑉. So, for showing the vector space we need to
show those two properties because we are talking about the subspace here we need to
absolutely show those two properties that closure properties; that means, you take any two
elements of this set and their vector addition should also belong to the same set and also
when we multiply this set by a number a scalar number that the new element should also
belong to this set 𝑆.
709
So, here let us suppose this 𝑢, 𝑣 ∈ SPAN(𝑆) and now because u belongs to the SPAN(𝑆)
and v belongs to the SPAN(𝑆) so, we can write down this u and v as a linear combination
of the vectors 𝑣’𝑠. So, here the 𝑢 = ∑𝑛𝑖=1 𝜆𝑖 𝑣𝑖 so, this is a linear combination. Here also
the 𝑣 = ∑𝑛𝑖=1 𝜇𝑖 𝑣𝑖 we have written as a linear combination of the vector 𝑣𝑖 ’s.
Note that these 𝜆′𝑠 and this 𝜇′𝑠 will be different and the other scalars, but this 𝑢 is another
vector so, this is a linear combination, a different linear combination here 𝑣 is another
vector so, we have a different linear combination there. So, these two vectors which we
have taken from the SPAN(𝑆) they must be the linear combination of the 𝑣’𝑠.
And now if we add them so, 𝑢 + 𝑣 if we add them so, what will happen when we add
them? This will be 𝑢 + 𝑣 = ∑𝑛𝑖=1(𝜆𝑖 + 𝜇𝑖 )𝑣𝑖 ∈ SPAN(𝑆). So, again this is a linear
combination of the 𝑣𝑖 is because when 𝜆𝑖 ’s and 𝜇𝑖 ’s are real their sum is also real number.
So, we have again this as a linear combination so, this will also belong to SPAN(𝑆) because
this SPAN(𝑆) contains all linear combination of the 𝑣𝑖 ’s and this is a linear combination so,
definitely this will also belong to the SPAN(𝑆).
And another property the closure property what we check when we multiply by any
constant any scalar like here we have taken the 𝑐 as a scalar. So, if we multiply are this 𝑐
to this vector 𝑢 then what we will get? We will get so, here this 𝑐 not 𝜆𝑖 . So, when we write
down this 𝑢 because 𝑢 is a linear combination well correct. So, this 𝑢 is a linear
combination so, which we have already written here 𝑢 is written as the 𝜆𝑖 𝑣𝑖 and when we
multiply by a constant here 𝑐 by a scalar 𝑐 then what will happen the 𝜆𝑖 𝑣𝑖 .
So, this c 𝜆𝑖 when we have multiplied this scalar 𝑐 here two 𝜆𝑖 that is another real number
and this is another linear combination of 𝑣𝑖 So, once we have this linear combination and
all the linear combinations belongs to this SPAN(𝑆) so, this element will also belong to
SPAN(𝑆). So, what we have seen here the closure properties are satisfied; that means, this
is SPAN(𝑆) is a vector subspace so, SPAN(𝑆) is a vector subspace and what else we need to
show that that any subspace containing the element of 𝑆 is also that set will also contain a
SPAN(𝑆) and the reason is clear because this is SPAN(𝑆) contains all the linear
combinations.
So, if you have a suppose you consider a vector space you we have a vector space for
instance which contains this set 𝑆 here , the definitely because if this is a vector space that
710
say w is a vector space which contains 𝑆. So, definitely because this is a vector space all
the all linear combinations because when we add two elements of this must be there. So,
eventually all the linear combinations must be there in this set w; that means, this will also
have SPAN(𝑆).
And that itself tell us that SPAN(𝑆) is the smallest subspace which contains this 𝑆 because
anything else which contains s will also contain this SPAN(𝑆). So, it cannot be that this is
not the smallest subspace. So, what we have learned here? That this is SPAN(𝑆); the
SPAN(𝑆) you take any vectors, from a vector space and having the span of this these vectors
that will form a vector subspace of that vector 𝑉.
(Refer Slide Time: 12:19)
And there is another one this is called a spanning set. So, what is the spanning set? The
set here 𝑣1 , 𝑣2 , … , 𝑣𝑛 is set to form a spanning set of vector space V.
So, we will call that this is a spanning set of this vector 𝑣 if for any 𝑉, if for any v take any
element from that vector space V then there exist the scalars this 𝛼1 , 𝛼2 , … , 𝛼𝑛 such that
we have 𝑣 is equal to this linear combination of these 𝑣′𝑠 here. So, basically we call that
this is a spanning set of a vector space 𝑉 if any element of this vector space, we can write
down as a linear combination of this given set 𝑉 or in other words which we have written
here the vectors 𝑣1 , 𝑣2 , … , 𝑣𝑛 in 𝑉 are said to a span V if every 𝑣 in 𝑉 is a linear
combination of the vectors 𝑣1 , 𝑣2 , … , 𝑣𝑛 .
711
So, this is spanning set is very important once we have this spanning set basically we can
write down any vector of that vector space in terms of these given or as a linear
combination of these given vectors. So, the example here we are considering the vectors
1
0
0
[0] , [1] & [0] this form a spanning set of ℝ3 meaning what we have to show? That if
0
0
1
we take any element of this ℝ3 any element of this ℝ3 , then we must be able to write down
as a linear combination of these three vectors and that is what we mean this spanning set.
1
0
0
So, let us check whether how this [0] , [1] & [0] form a spanning set for the vector space
0
0
1
𝑣1
3
𝑣
ℝ . So, for any vector so we take a general vector this [ 2 ] which we have denoted by this
𝑣3
𝑣1
𝑣1
3
[𝑣2 ] that can be any vector from ℝ , we are not restricting anything on [𝑣2 ]. They can be
𝑣3
𝑣3
any real number meaning that this vector belongs to this ℝ3 and it’s a general vector yet
can it can be any vector from this ℝ3 and what we will, what we are supposed to do now?
We will show that this vector which is a general vector from this ℝ3 without any
restriction. So, any vector from ℝ3 we can write down as a linear combination of these
𝑣1
𝑣
given vectors and why this is trivial in this case? This is [ 2 ].
𝑣3
And the structure this is called the later on we will come up with another name this called
the standard vectors or rather standard basis which you will refer later on this these
1
0
terminologies. So, here we have [0] , [1] &
0
0
0
[0], we what we need to do? We need to just
1
multiply this first vector by the number 𝑣1 , the second one 𝑣2 and the third one 𝑣3 and as
we see clearly
712
𝑣1
1
0
0
[𝑣2 ] = 𝑣1 [ 0 ] + 𝑣2 [ 1 ] + 𝑣3 [ 0 ]
𝑣3
0
0
1
So, this any vector of this vector space ℝ3 we can write down as a linear combination of
1
0
0
this a these vectors [0] , [1] & [0] which a was trivial at to see in this particular case.
0
0
1
(Refer Slide Time: 16:07)
1
1
1
Another example we will be talking about these vectors here [1] , [1] & [0] and this also
0
0
1
form a spanning set of this ℝ3 , how this forms a spanning set? So, we have seen already
1
1
1
one example where the vectors were taken as [1] , [1] & [0]. This is another set and then
0
0
1
again we are claiming that this form is spending set of this ℝ3 .
So, naturally this spanning set is not unique, we can have we can have a different set of
vectors and that spanned the same vector space. So, here this is another example where we
will see that this set can also span ℝ3 ; that means, any element of this ℝ3 we can write
down as a linear combination of these three vectors. To see this we will take a general
𝑣1
𝑣1
3
vector here [𝑣2 ] from this ℝ and we will try to write down this [𝑣2 ] as a linear
𝑣3
𝑣3
combination of the given vectors meaning that
713
𝑣1
1
1
1
[𝑣2 ] = 𝜆1 [1] + 𝜆2 [1] + 𝜆3 [0]
𝑣3
0
0
1
Now the question is can we write down as a linear combination here because here is not
as trivial to see as the earlier case though this is also not difficult. So, what do we do? We
again we will solve these three equations the system of equations
𝜆1 = 𝑣3
𝜆2 = 𝑣2 − 𝑣3
𝜆3 = 𝑣1 − 𝑣3 − (𝑣2 − 𝑣3 ) = 𝑣1 − 𝑣2
So, without indeed reducing to this row reduce echelon form we can directly also solve
this equation, we do not have to do that because that this equation the third equation tells
that
𝜆1 = 𝑣3 .
So,
from
here
itself
we
get
direct
answer
that
𝜆1 = 𝑣3 . Now you will take the second equation which tells us that is
𝜆2 = 𝑣2 − 𝑣3 . So, the second equation will give 𝜆2 = 𝑣2 − 𝑣3 and from this equation
number 1 which is 𝜆2 = 𝑣2 − 𝑣3 .
(Refer Slide Time: 18:50)
𝑣1
So, we got this linear combination here that this any vector [𝑣2 ]we can write down as a
𝑣3
linear combination of these given vectors in the form that 𝑣3 the first vector, 𝑣2 − 𝑣3 the
714
second vector and 𝑣1 − 𝑣2 this third vector. So, again we have seen in this example that
this was a different set here from than the earlier example, but again this is also a spanning
set of this ℝ3 .
(Refer Slide Time: 19:56)
So, what we have now? We have this example number 3, where we will show now we
1
1
have taken from the previous example yeah so, these vectors [1] , [1] ,
0
1
1
are from the previous example and we have taken one more that [0] &
0
1
1
[0] & [0] these
0
1
1
[0]. So, we have
1
added one more vector here now our set here is bigger and now again this also forms a
spanning set of ℝ3 that is what we will observe now.
1
So, the existing set and we have added one more this vector [0] and this is also forming a
1
spanning set. So, you can have really a different number of elements in this spanning set
exactly, I mean quite different elements also in the in the in the set. So, here we will check
how this forms a spanning set.
715
𝑣1
So, for any general vector [𝑣2 ] what we have to again consider this linear combination that
𝑣3
𝑣1
𝑣
whether we can write down this [ 2 ] as a linear combination of these four vectors.
𝑣3
𝑣1
1
1
1
1
𝑣
[ 2 ] = 𝜆1 [1] + 𝜆2 [1] + 𝜆3 [0] + 𝜆4 [0]
𝑣3
0
0
1
1
1
[𝐴|𝐵] = [1
1
1
1
0
1 1 𝑣1
0 0 | 𝑣2 ]
0 1 𝑣3
𝑣1
1 1
1
1
[𝐴|𝐵] ~ [0 −1 −1 0 | 𝑣3 − 𝑣1 ]
0 0 −1 −1 𝑣2 − 𝑣1
So, we have this reduced form here and now what we will observe out of this reduced form
whether we can get such lambdas or we cannot get such 𝜆′s now.
(Refer Slide Time: 22:45)
So, getting again back to this pivot elements so, the first column has a pivot, second column
has also pivot element and the third column also has a pivot element. So, we have the three
pivot elements and here we do not have this pivot element this is not the pivot element.
So, this 𝜆 corresponding to this 4 so, lambda four is a free variable right so, 𝜆4 is a free
variable and all other lambdas will depend on this choice of this 𝜆4 .
716
So, what do we get now? So, the question is the answer to this question is that we do have
𝑣1
𝑣
such lambdas which will add up to this [ 2 ] and indeed now in this case we have ; we have
𝑣3
𝑣1
𝑣
many choices; we have many choices for these lambdas to give this [ 2 ]. In fact, this it is
𝑣3
not unique now this representation is not unique while in the earlier cases one can closely
observe and doing this echelon form. So, what we will get? We will not get this column
for instance, if we consider just with these three examples.
So, we will get these every column will have a pivot here and that will be equal to the
number of these unknowns the number of 𝜆′s in that case and you will get a unique
representation. So, the earlier example when we did not have this vector there, we have
𝑣1
the unique representation of the of this [𝑣2 ] in terms of the given vectors.
𝑣3
But what happened now in this case we do not have a unique representation, but we can
represented, we can write down any vector from this ℝ3 in terms of these given four
vectors. So, indeed this is a spanning set the only difference from the earlier example to
this example here is that that we have a non unique representation, that we have to choose
now 𝜆4 and then compute 𝜆1 , 𝜆2 , 𝜆3 .
We have basically infinitely many solutions now so, this is a non unique representation in
the first two examples we have a unique representation. So, that was the difference here
and now in this case we have a non unique representation.
717
(Refer Slide Time: 25:10)
Coming back to this example 4, we will see that again we have three vectors here
1
1
1
[2] , [3] , & [5] these are three different vectors. So, but they do not span R 3 in example
3
5
9
1 in example 2 we have three vectors and they span ℝ3 means any vector of ℝ3 we can
write down in terms of those given vectors. So, in the example 1 we have this, these set
1
0
0
here [0] and [1] and then we had [0]. So, these were the three vectors from example 1.
0
0
1
1
1
1
Example 2 I guess it was [1] , [1] & [0] so, this was from example number 2.
0
0
1
So, with these though they have the three different elements from ℝ3 , they span; they span
R 3 means any element of ℝ3 we can write down in terms of these vectors or in terms of
these vectors, but now in this case what we will observe that this is not possible that you
take any three elements from ℝ3 and that will form this spanning set.
718
(Refer Slide Time: 26:30)
𝑣1
How to check this? We will take a general vector here [𝑣2 ] as usual and we will try to
𝑣3
𝑣1
𝑣
write down this [ 2 ] in terms of the 𝜆1 , 𝜆2 , 𝜆3 . So,
𝑣3
1
1
1
𝜆1 [2] + 𝜆2 [3] + 𝜆3 [5]
3
5
9
we have the vector the augmented matrix here now with the coefficient matrix.
1
[𝐴|𝐵] = [2
3
1
[𝐴|𝐵] ~ [0
0
1
1
0
719
1 1 𝑣1
3 5 | 𝑣2 ]
5 9 𝑣3
𝑣1
1
𝑣
−
2𝑣1 ]
3|
2
0 𝑣3 − 2𝑣2 + 𝑣1
(Refer Slide Time: 28:21)
So, now what do we observe in this case? That here we have pivot here also we have a
pivot and this forget about the pivot we have the inconsistency here, while inconsistency
𝑣1
𝑣
because this [ 2 ] they these are they can be any real number because our vector here is
𝑣3
from ℝ3 and say any arbitrary vector from ℝ3 . So, having so, we cannot; we cannot just
observe that this will be 0 this will be a non zero number as well in many situation and
then 0 is equal to something nonzero we are getting.
𝑣1
𝑣
So, this system we cannot solve for general [ 2 ] from ℝ3 ; that means, the system is
𝑣3
inconsistent and hence we cannot write down in this case as a linear combination of these
given vectors. So, what we have observed? That just the number is not important that we
have taken here three elements from ℝ3 in this, in the third example we have taken four
elements from ℝ3 in first two examples we have taken again only three elements.
In the first two cases we had a unique representation for a given vector from ℝ3 in terms
of the given vectors, in the third example where we have taken 4 that was also spanning
set, but there you are getting non unique representations of a given vector from ℝ3 and
now in this example though we have three vectors from ℝ3 , but they are not sufficient to
represent a general element or general vector from ℝ3 . So, this does not form a spanning
set of ℝ3 .
720
So, there is a lot more to discuss on this spanning set and that will follow in the next
lectures.
(Refer Slide Time: 30:21)
Here the conclusion is that this linear span is nothing, but all the all linear combinations of
the given vectors and the spanning set which (𝑣1 , 𝑣2 , … , 𝑣𝑛 ) of 𝑣 we will call that this is a
spanning set, if any vector of this 𝑣, we can represent in the form of these vector these then
we call that this is a spanning set.
(Refer Slide Time: 30:46)
721
So, these are the references use for preparing these lectures.
And thank you very much for your attention.
722
Engineering Mathematics - I
Prof. Jitendra Kumar
Department of Mathematics
Indian Institute of Technology, Kharagpur
Lecture - 42
Vector Spaces – Basis and Dimension
So, welcome back and this is lecture number 42 and we will continue our discussion on
Vector Spaces again.
(Refer Slide Time: 00:24)
In particular today will be talking about the basis of a vector space and also the dimension
of a vector space.
723
(Refer Slide Time: 00:32)
So, just to recall from the previous lecture; so, we have seen that these vectors the set of
1
1
1
1
these 3 vectors are [1] , [1] , [0] & [0], these 4 vectors form a spanning set of ℝ3 . The
0
0
1
1
meaning was that any element of this vector space ℝ3 we can write as a linear combination
of these 4 vectors, this is what we have seen in previous lecture with the help of this
𝑣1
augmented matrix which we can get out of this linear combination equal to this [𝑣2 ].
𝑣3
1
[𝐴|𝐵] = [1
1
1
1
0
1 1 𝑣1
0 0 | 𝑣2 ]
0 1 𝑣3
And from this augmented matrix when we reduce to this rho echelon form we observe the
following. What we have observed that there are these 3 pivots in column 1 we have pivot,
in column 2 also we have pivot, column 3 also has a pivot while the column 4 does not
have a pivot and this is what we call the free variable.
𝑣1
1 1
1
1
[𝐴|𝐵] ~ [0 −1 −1 0 | 𝑣3 − 𝑣1 ]
0 0 −1 −1 𝑣2 − 𝑣1
So, with these system of equations what we observe that lambda 4 is free variable; is free
variable and meaning that we can assign whatever value we want to this lambda 4. So now,
we have the flexibility to give this linear combinations here. So, choosing a different value
724
𝑣1
of lambda we have a different linear combination that will give us exactly this vector [𝑣2 ].
𝑣3
So, this forms a spanning set because any vector any arbitrary vector we can pick from
this ℝ3 space and we can write down as a linear combination of these given 4 vectors.
What is interesting here that we have a non-unique representation because of this free
variable.
So, when we have a free variable we can assign any value we want to this lambda 4 and
then our representation of this given vector will also change. So, this linear combination
will also change. What was also seen in the previous lecture that, if we do not consider for
example, this vector the fourth-one if we do not consider this vector; that means, we will
not have this column here, this column we can delete now. So, in that case this was a
situation and what we observe now for this case that we have the every column has a pivot
and there is no free variable.
Meaning when we do not have a free variable we will get basically the unique
representation, the unique solution of this system. And this is what we will discuss today
that having these 3 vectors we have a spanning set having this 4 vectors, that also form a
spanning set. Then we will come up with the idea now that how many actual minimum
vectors do we need so, that we can span the given vector space.
So, for instance in this case we have observed that taking these 3 vectors we can span the
whole vector space and also the moreover the main point is that we have also the unique
representation of the vector form ℝ3 if we take these 3 vectors. But, if we have taken the
4 vectors still it is forming a spanning set, but we are not getting a unique representation.
So, keeping these facts in mind now we will continue with the definitions of the basis and
the dimensions.
725
(Refer Slide Time: 05:03)
So, the basis is a linearly independent set that span a vector space. So, the simple definition
it is a spanning set because it should span the whole vector space 𝑉, but what is in addition
that the linearly independent set. So, the basis is nothing, but spanning set which has all
linearly independent vectors that we call basis. Like in the earlier example we have
possibility in this spanning set taking all those 4 vectors or taking only those 3 vectors.
And but if we take those 3 vectors we can easily figure out their those vectors are linearly
independent, but having those 4 vectors in the set the set becomes linearly dependent, that
also we can observe with the theory we have already developed in previous lectures. So,
those 3 vectors they are linearly independent and for those vectors we can call that they
will form a basis not the 4 vectors because 4 are also spanning set, but that will not be the
basis of ℝ3 in that example.
So now, the dimension so, the number of elements in a basis is called the dimension of the
vector space. So, here the number how many elements are there, how many elements are
there in that basis that is called the dimension of the vector space. So, these two numbers
are the basis that is the set of the vectors and that is that is span the given vector space and
this number here which is which tells about the dimension of the vector space both are
very important.
Just a note here that every vector in 𝑉 can be written uniquely as a linear combination of
the basis of vectors and this fact also we can explain from the previous example itself, that
726
when we have the base is there; the basis means you are linearly independent vectors. And
when you have linearly independent vectors there will be no free variable when we talk
about that linear combination of the given vectors to represent a vector from this 𝑉.
So, that representation will be also unique whenever we have the basis our spanning set is
having linearly independent vectors than the representation will be also unique. And the
vector space 0 so, this is the special case when we have only one element that is 0 and we
define this dimension as 0.
(Refer Slide Time: 08:06)
So, let us take the example here the vector space ℝ𝑛 which we have already studied. Now,
we will talk about what are the basis of ℝ𝑛 and what is the dimension of this vector space
ℝ𝑛 . So, we consider these vectors now very special vector which we call the standard basis
of ℝ𝑛 . So, consider these vectors 𝑒1 = (1,0,0, … ,0)𝑇 , 𝑒2 = (0,1,0, … ,0)𝑇 and in 𝑒𝑛 =
(0,0,0 … ,0,1)𝑇 .
So, in this way we have defined these elements these elements are from ℝ𝑛 of course. So
now, we will notice here that these vectors are linearly independent; first that we can easily
check out and we have I think checked before as well. So, we have to just see that 𝜆1 𝑒1 ,
𝜆2 𝑒2 and so, 𝜆𝑛 𝑒𝑛 when we set to 0 vector then we should get out of this that 𝜆1 is 0, 𝜆2
is 0 and all are this 0’s which we can easily see because of this structure here itself we will
get from this 𝜆1 = 0 from here we will get 𝜆2 = 0 and so on. So, this trivial to see that this
vector this set of vectors form a linearly independent set.
727
(Refer Slide Time: 09:42)
So, all these vectors are linearly independent. What else we have to observe? So, we have
to we are going to find the basis and the dimension of the vector space. So, for that we
need spanning set basically that can span any that can represent any vector from our vector
space in the form of this given vector. So, whether here it is possible or not we will check
now. So, given these vectors here n vectors are given and now any element if we take from
this ℝ𝑛 so, any this is ℝ𝑛 .
So, this is the element of this ℝ𝑛 . So, if we take a general any vector from this ℝ𝑛 and
which we are calling here (𝑣1 , 𝑣2 , … , 𝑣𝑛 )𝑇 then we can see that we can represent this vector
with the help of these given vectors (𝑒1 , 𝑒2 , … , 𝑒𝑛 )𝑇 . The idea is again simple in this case
and that is for these are called the standard basis. So, this v can be represented as the
𝑣1 𝑒1 + 𝑣2 𝑒2 + ⋯ + 𝑣𝑛 𝑒𝑛 .
And no where else we will get anything at this place because all other vectors have 0 as
first component; for the second 𝑣2 only the 𝑒2 has 1 at the second place all others are 0.
So, naturally we will get just 𝑣2 from this one at the second position and so on. So, we can
represent or we have already represented here any vector 𝑣 from this ℝ𝑛 as a linear
combination of these given vectors (𝑒1 , 𝑒2 , … , 𝑒𝑛 )𝑇 .
So, it satisfies all the properties of the basis we have this is a spanning set because, we can
span any vector of the given vector space in the form of as a linear combination of the
728
given vectors and also these vectors are linearly independent. So, they form the basis so,
we got a basis for this ℝ𝑛 , this set of vectors here this forms a basis forℝ𝑛 .
(Refer Slide Time: 11:57)
And the dimension now it is a number of these elements in this set. So, here we have
(𝑒1 , 𝑒2 , … , 𝑒𝑛 )𝑇 there are 𝑛 elements and we got therefore, this dimension here of this
vector space is 𝑛.
(Refer Slide Time: 12:12)
729
This example where we are talking about the vector space of all 2 × 3 or in general also
we can discuss like for 𝑟 × 𝑠 matrices. So, for simplicity let us take this 2 × 3, but the idea
we can extend for 𝑟 × 𝑠 matrices as well. So, here the vector space of all 2 × 3 matrices
we are talking about and if we consider these vectors here, the 1 at this position and all
other positions are 0; similarly now at the second position is 1 all others are 0.
And so on we continue with this we get these 6 matrices where this 1 is sitting here at the
first position only in this matrix, in the second case 1 is sitting here and all others are 0 and
so on. With this special structure again these are the standard basis for this vector space of
this 2 × 3 matrices we have already seen that this format a vector space in the last lecture.
So, here we are talking about or finding the basis and its dimension.
So, these vectors are linearly independent and span the vector space of 2 × 3 matrices.
Why they are linearly independent? Again the same argument which we had in the
previous lectures. So, if we call them vector 𝑣1 = [
0
0
and this 𝑣3 = [
[
0
0
0
0
0
0
1
0
] here this is 𝑣4 = [
0
1
1 0
0 0
0
0
0 1
0
] 𝑣2 = [
0 0
0
0
0
] and 𝑣5 = [
0
0
0
1
0
]
0
0
] and 𝑣6 =
0
0
] and again the same thing we do that, we take here this linear combination
1
𝜆1 𝑣1 + 𝜆2 𝑣2 + ⋯ + 𝜆𝑛 𝑣𝑛 = 0.
So, this is nothing, but the [0 ,0, 0, 0, 0, 0]𝑇 when we compute this 𝜆1 𝑣1 + 𝜆2 𝑣2 + ⋯ +
𝜆𝑛 𝑣𝑛 we will get matrix here with 𝜆1 this is 𝜆2 = 𝜆3 = ⋯ = 𝜆6 = 0. And now comparing
these entries there we will get 𝜆1 = 0, 𝜆2 = 𝜆3 = ⋯ = 𝜆6 = 0. So, they are linearly
independent vectors so, these are linearly independent vectors. The second we have to
check that they span the vector space of this matrices 2 × 3.
730
(Refer Slide Time: 14:53)
So, naturally they do because if we consider any vector any matrix from this 2 × 3 matrices
or the elements we call vectors only. So, anything we take here for example, 2 by 3 so, this
is a general matrix from this space 2 × 3 and with the help of this given vectors we can
easily expand in terms of like a should be multiplied to the first one now and this b is
should be multiplied to the second one and so on. If we continue this so, that is the
representation now of this general vector from this vector spaces and that shows that we
can represent any vector from this vector space in terms of these given vectors which are
6 in number.
(Refer Slide Time: 15:53)
731
So, what is the conclusion that these vectors form a basis for the this vector space and the
dimension is nothing, but the product here 2 × 3 or we have this 6 matrices. So, the
dimension is 6 which we can generalize for 𝑟 × 𝑠 matrices also the idea is same.
(Refer Slide Time: 16:11)
Another example where we are talking about the vector space this 𝑃𝑛 (𝑡) of all polynomials
of degree less than equal to 𝑛; again this example we have seen that this form a vector
space and now we are talking about this set here {1, 𝑡, 𝑡 2 , … , 𝑡 𝑛 }. So, these 𝑛 + 1
polynomials rights these are n plus 1 polynomials, we have taken and we claim that this is
a basis actually for 𝑃𝑛 (𝑡).
And why this form a basis? The reason is again clear we have to check that these are
linearly independent which we can check as we have done for other examples 𝜆1 1 + 𝜆2 𝑡 +
⋯ + 𝜆𝑛 𝑡 𝑛 . This will be set to 0 and this is only possible when all these lambdas are 0. So,
naturally these are linearly independent vectors of 𝑃𝑛 (𝑡). The second we have to check that
any polynomial from this 𝑃𝑛 (𝑡) we take can be represent that polynomial with the help of
these vectors.
So, again here if we take any polynomial that say this 𝑎0 1 + 𝑎1 𝑡 + ⋯ + 𝑎𝑛−1 𝑡 𝑛 that is a
general polynomial we have in this space 𝑃𝑛 (𝑡) they are the polynomials of the degree less
than or equal to n. So, for instance we have taken this general polynomial and with the
help of these given polynomials we can easily use this. So, 𝑎0 with be multiplied by this 1
and this 𝑎1 with this 𝑡 here 𝑎2 with 𝑡 2 and so on. So, that is a linear combination of these
732
given vectors. So, we can represent any element of this set or this space here 𝑃𝑛 (𝑡) as a
linear combination of these given vectors.
So therefore, this set forms a basis for 𝑃𝑛 (𝑡) again which we will also come later on to
more clarification, the basis are not unique. So, this is a set which form a basis we are
writing a basis. There can be other set which can also form the basis like in the example
of the first example which we started with we had those 3 vectors which form the basis
1
0
and they were not the standard basis. So, we can also choose for example, [0] and [1] for
0
1
0
3
ℝ and this [0].
1
1
So, that will also form the basis for ℝ3 and the example we have seen those [1] that was
1
1
0
one vector another one was [1] and the third one was [0]. So, that was not the standard
0
1
basis, but they were also the basis. So, basis are not unique we can have a different sets,
the main properties are the elements must be linearly independent that is one property and
the second one should be that they should span the whole vector space.
(Refer Slide Time: 19:36)
733
So, here we have seen that these are the basis for this 𝑃𝑛 (𝑡) and the dimension here the
number of elements in this basis which is 𝑛 + 1 at present. So, we have the dimension
𝑛 + 1.
(Refer Slide Time: 19:45)
Another example where we will revisit the our problem here 𝐴𝑥 = 0, the system of
homogeneous linear equations and we know that the solutions set of a homogeneous linear
system also forms a vector space. So, here the 𝐴 was this one this example we have already
discussed before with this 𝐴 and we can reduce it to the this echelon form loaded used
𝑥1
𝑥2
echelon form and which tells us that these 𝑥3 by choosing because, we have to we have
𝑥4
[𝑥5 ]
free variables here.
So, this pivot here there is a there is a pivot, but this column does not have a pivot here
also it does not have a pivot. So, we have to choose this 𝑥2 and this 𝑥5 as free variables by
𝑥1
𝑥2
choosing these two as 𝛼1 and 𝛼2 we can compute all other 𝑥3 .
𝑥4
[𝑥5 ]
734
𝑥1
−2
𝑥2
1
And when we combine we got this equation here that 𝑥3 are nothing, but 𝛼1 0 +
𝑥4
0
[0 ]
[𝑥5 ]
−9.5
0
𝛼2 −4 . So, what we observed that with the help of these two vectors we can generate
−0.5
[ 1 ]
the whole set whole solution set of this 𝐴𝑥 = 0. Because, we need to just keep on changing
these alphas the numbers here and every for any alpha here whatever we choose this is this
solution of 𝐴𝑥 = 0.
So, all possible solution of this 𝐴𝑥 = 0 we can generate using these two vectors and these
two vectors looking at them we can easily identify that these are the linearly independent
vectors which we can formally also prove we know with the help of this linear combination
set to 0 and that will imply that those coefficients 𝜆1 = 𝜆2 = 0.
So, these two vectors are linearly independent and their linear combination we can
generate any vector of the solution set. So; that means, these two vectors form a basis
because, that is a property of the basis. We can generate any element of this solution set
with the help of these two or as a linear combination of these two vectors and these two
vectors are linearly independent. So, they span the solution set and these two vectors are
linearly independent and therefore, they form a basis. So, these vectors can the vectors that
generate the solutions are these and this and these vectors are linearly independent.
735
(Refer Slide Time: 22:33)
Therefore, these vectors form a basis of the null space here remember so, we call this
solution set there was a special name, which we call null space. This is a vector space
which is called null space. So, these vectors form a basis for the null space of 𝐴 and the
dimension of this null space of 𝐴 in this particular case its 2 which is again has a name is
called nullity. So, nullity is nothing, but its a dimension of the null space of 𝐴.
(Refer Slide Time: 23:20)
There are some useful results based on the observations what we have regarding these
basis and dimension. So, let 𝑉 be a vector space of this finite dimension 𝑛. So, the
736
dimension of the vector space is 𝑛, then any 𝑛 + 1 or more vectors in 𝑉 are linearly
dependent. So, that is a another nice results we are not going to prove all these results, but
one can with the help of some simple examples one can at least get some idea that how
these results are true.
So, here any n plus 1 in the in our example also which we started this lecture today we had
those 4 vectors, but they were linearly dependent and that was a spanning set ah, but it was
a linearly dependent set. Therefore, it was not a basis so, their vectors were 4 while we
have seen that taking those 3 vector first 3 vectors that form a linearly independent set and
also a spanning set. So, that was the basis so, the vectors in the basis were 3 there for ℝ3 .
And so, if there are 4 vectors which is span ℝ3 they cannot be they cannot be basis because,
4 vectors from ℝ3 they have to be linearly dependent this is the result here.
So, any n plus 1 or more vectors we take in 𝑉 they must be linearly they must be linearly
dependent. So, for a vector space whose dimension is n we cannot get more than n linearly
independent vectors. They will be linearly dependent if we take any 𝑛 + 1 or more. Any
linearly independent set when 𝑛 is a basis of 𝑉 with 𝑛 element this is a basis. Another
beautiful result that we do not have to worry about picking up the elements for the basis
we can take any 𝑛 elements, but that set should be linearly independent.
So, any set which has n linearly independent vectors that will be a basis because for the
for the basis these will automatically will be the spanning set. We pick any 𝑛 vectors which
are linearly independent that will be the basis and any spanning set {𝑣1 , 𝑣2 , … , 𝑣𝑛 } with 𝑛
elements. So, again the same thing spanning set with n elements so, that will be also the
basis. So, if a vector space has finite basis then all of its basis have the same number of
elements, that is another beautiful result that if we have the 𝑛 number of elements in the
basis.
So, whatever you have different different set of basis, but they will have the same number
of elements because that is the n that is a dimension. So, whatever basis we take the
dimension is n of the vector space then there has to be n elements in the set. And the
number of elements in any basis of a vector space is called the dimension which we have
already discussed.
737
(Refer Slide Time: 26:45)
So,
for
instance
we
take
this
example
of
4
vectors
{(1, 1, 1, 1)𝑇 , (0, 1, 1, 1)𝑇 , (0, 0, 1, 1)𝑇 , (0, 0, 0, 1)𝑇 } and they forms the basis of ℝ4
the set this forms a basis of ℝ4 , while note that the give vectors are linearly independent
that we can easily check they are linearly independent whenever we have such a special
structure. So, all 1 and then these three 1’s two 1’s and 1’s so, they form a basis which we
have seen for instance in the case of this ℝ3 . So, this forms a basis because these are
linearly independent and we know that the dimension of ℝ4 is 4.
So, we do not have to check we have to check that they are linearly independent and they
are 4 in number because, the dimension of ℝ4 also we know. So, if we have 4 linearly any
4 linearly independent vectors from ℝ4 not only this 4 vectors we can pick any 4 linearly
independent vectors that will form a basis. So, here are the given vectors they are linearly
independent and then they it has to be a basis because, the dimension is 4 and we have
these 4 linearly independent vectors. Consider this 4 vectors in this ℝ3 ; so, if we consider
these 4 vectors are from ℝ3 and we know the dimension of ℝ3 is 3.
So, when the dimension is 3 these vectors must be linearly dependent, they cannot be
linearly independent. Because, if they are linearly independent then they can form a basis
also, if they are linearly independent and they span the ℝ3 then they can be also basis, but
basis will have only 3 elements not more than 3 elements. So, here we have 4 elements.
738
So, definitely they are linearly dependent which we can conclude from the dimension of
ℝ3 which is 3. So, this cannot have the basis cannot have 4 elements in its set ok.
(Refer Slide Time: 28:58)
So, the conclusion is that we have the dimension which is the maximum number of linearly
independent vectors in a vector space 𝑉 and the basis is a set of linearly independent
vectors that span the vector space.
(Refer Slide Time: 29:16)
So, these are the reference here used for preparing the lectures.
739
And thank you very much.
740
Engineering Mathematics - I
Prof. Jitendra Kumar
Department of Mathematics
Indian Institute of Technology, Kharagpur
Lecture - 43
Rank of a Matrix
So, welcome back and this is lecture number 43. And today we will discuss Rank of a
Matrix from this topic of the matrix which are linear algebra.
(Refer Slide Time: 00:22)
And the definition we start with which we have mentioned in one of our previous lecture
that is the rank of a matrix is the number of nonzero rows or the number of pivots in its
reduced row echelon forms.
741
(Refer Slide Time: 00:28)
So, this that is that is what this row reduced echelon form is very important, it tells us lot
about the matrix and that is one important concept which is used at very applications about
this rank. So, rank of the matrix is the number of the pivots which we see in our reduced a
row echelon forms. So, for instance we take this 𝐴, which was already used in previous
lectures we have also seen its row reduced echelon form which is given here again.
1
2 −2
2
4 −4
𝐴=[
−1 −2 3
3
6 −7
1
0
𝐴~ [
0
0
2
0
0
0
−2
1
0
0
−1
0
3
1
−1
2
2
0
1
3
]
4
1
1
5
]
1
0
So, this is the row reduced echelon form of this matrix. And what is now we need to count
the pivots here. So, the 1 is the pivot and also this 1 is the pivot and this 2 is the pivot. So,
we have 1, 2, 3, there are 3 pivots here in this row reduced echelon form and therefore, the
rank of this matrix is 3. So, there are 3 pivots and the rank as per the definition that rank
of the matrix is nothing but it is a number of pivots in reduced row echelon form which is
3 in this example.
742
Just a consequence of this definition we can easily make it out that the rank of an 𝑚 × 𝑛
matrix cannot be greater than 𝑛 or 𝑚. So, m is the number of rows and number of columns,
and the rank cannot be greater than m or n it has to be the less than the minimum of 𝑚 and
𝑛. The reason is clear because our each row or each column cannot have more than one
pivot, the column can have at most one pivot or the row also it can have at most one pivot,
one row cannot have a two pivots or more, column cannot have a two pivots or more. So,
we are talking about the number of pivots. So, it cannot be the rank, cannot be greater than
m the minimum it has to be less than or equal to the minimum of this 𝑚 and 𝑛 this number.
(Refer Slide Time: 03:15)
1
2 −2
2
4 −4
𝐴=[
−1 −2 3
3
6 −7
−1
0
3
1
1
3
]
4
1
𝐴𝑥 = 0
𝑥1
−2
−9.5
𝑥2
1
0
𝑥3 = 𝛼1 0 + 𝛼2 −4
𝑥4
0
−0.5
[𝑥5 ]
[0 ]
[ 1 ]
Write down the solution by taking these free variables that is the 𝑥2 here and also this 𝑥5 .
So, taking this free variables we can write down this 𝐴𝑥 = 0, the solution of this 𝐴𝑥 = 0 in
this form which we have already observed because now what we are going to do we are
743
actually we are looking for the other definitions of the rank because that is not the only
definition which we have given in terms of the number of pivots. So, for that we just need
some preparations here.
So, if we notice now the case 1, if we take this 𝛼1 = 1 & 𝛼2 = 0. So, that will be one
solution of this system by choosing this alphas basically we are looking for different
different solutions. So, for this particular case we are taking choosing this 𝛼1 as 𝛼2 as 0.
−2
1
So, our solution of this 𝐴𝑥 = 0 will become 0 . So, this is the solution of our 𝐴𝑥 = 0 of
0
[0 ]
the system.
And then this 𝐴𝑥 = 0, this equation we can write down in this form. This we have already
𝑥1
𝑥2
talked before that this product of this matrix 𝐴 with this 𝑥 here, the
𝑥3
. So, this product
𝑥4
[𝑥 ]
5
A and with this 𝑥 we can also write down as this 𝑥1 and then the first column plus this 𝑥2 ,
the second column 𝑥3 , the second third column 𝑥4 , this forth column 𝑥5 and this is equal
to 0.
1
2
−2
0
−1
1
2
4
0
−4
0
3
𝑥1 [ ] + 𝑥2 [ ] + 𝑥3 [ ] + 𝑥4 [ ] + 𝑥5 [ ] = [ ]
−1
−2
3
0
3
4
3
6
0
−7
1
1
So, that is another way of looking at this matrix vector product we can write down in this
vector forms. So, having this now and what we have observed that this is one solution of
this system of equations with minus 2 4 x 1 and this here 1 4. So, this is −2 and this 𝑥2 is
1 and this we can take 0 all these 0s.
744
(Refer Slide Time: 05:28)
So, what we observed now with this that minus this two times the 𝐶1 denotes this column
1 of our matrix. So, this ⟹ −2𝐶1 + 𝐶2 = 0. So, what we observed at least from this relation
that this column 1 and column 2 are dependent. We can write down this column 2 here in
terms of column 1, so column 2 is nothing but the two times the column 1. So, that is a
relation which we can clearly see from directly from the matrix itself because, but
sometimes it is not easy to see when we have more vectors into picture here there were
two vectors only, so we can see easily. So, that is the one relation we have seen that 𝐶2 the
column 2 is nothing but the two times 𝐶1 . So, these two columns are dependent columns.
One more observation quickly we will have now we will take another case where we will
choose this 𝛼1 = 0 & 𝛼2 = 1.
745
(Refer Slide Time: 06:37)
−9.5
0
That means, our solution now of this 𝐴𝑥 = 0 is given by this −4 with this vector and
−0.5
[ 1 ]
that this was given already. So, again this first column third and forth they are the columns
which have the pivots in their reduced in its reduced form. So, here 𝐴𝑥 = 0. Again, with
𝑥1
−9.5
𝑥2
0
this observation we will fix the solution now. So, the solution says that this 𝑥3 = −4 .
𝑥4
−0.5
[𝑥5 ] [ 1 ]
1
2
𝐴=[
−1
3
2
4
−2
6
−2
−4
3
−7
−1
0
3
1
1
3
]
4
1
𝐴𝑥 = 0
1
2
−2
0
−1
1
2
4
0
−4
0
3
⟹ 𝑥1 [ ] + 𝑥2 [ ] + 𝑥3 [ ] + 𝑥4 [ ] + 𝑥5 [ ] = [ ]
−1
−2
3
0
3
4
3
6
0
−7
1
1
−9.5 𝐶1 − 4 𝐶3 − 0.5 𝐶4 + 𝐶5 = 𝟎
⟹ 𝐶5 = 9.5 𝐶1 + 4 𝐶3 + 0.5 𝐶4
746
Another observation which we can find out from the from that reduced echelon form that
these vectors with red here the column number 1, column number 3 and column number
4. So, these vectors here one can easily prove directly from the reduced form that these are
linearly independent, these are linearly independent that form the reduced form one can
talk about this. And we have seen that these columns here corresponding to those free
variables, they are linearly dependent. And this is the general result also not just for this
particular example that the columns which we have the pivots in its reduced form they are
linearly independent always and the other columns which have these free variables or
where they do not have the pivots, they actually are linearly dependent.
So, now we can rewrite our definition which says for this rank that this is nothing but a
number of pivots, but now we can write down as the number of independent columns. The
rank is nothing but the number of independent columns in because this column is
dependent and this column also dependent, there are 3 independent columns and they
basically correspond to this pivot column. So, now our definition for the rank is that rank
is nothing but the number of independent columns in a matrix.
And again, the same observation which we have done here with the columns we can do
with the rows as well. So, those rows where we got these pivots there are there are the
same number of rows which are independent in this matrix. So, we can have also this
definition that the rank is nothing but the number of independent rows and they are the
same the column because the number of pivots are basically fixed in its reduced form.
747
(Refer Slide Time: 10:40)
So, what we have, we can now give another definition the rank of a matrix is the number
of linearly independent rows or number of linearly independent columns in a matrix that
is the rank. One more definition we are going to give in a minutes, before that we just
define these two spaces the column space is nothing but the span of the column vectors of
A.
So, for a given matrix 𝐴 we take its column as vectors and then span them, so that is what
we call the column space because any span of any vectors that is a vector space which we
have already seen in previous lectures. So, here also the span of these vectors of or the
columns, column vectors of this a will also form a vector space which we call the column
space. So, column space is a vector space that is nothing but the span of the columns of 𝐴.
Similarly, we have the rows space row space is nothing but the span of the row vectors of
𝐴 similar to the column space if we take the rows there and span them, so this space which
we call the rows space. And the another definition which we have now for the rank, the
rank of the matrix is the dimension of the row space because when we are talking about
the rows for instance, so what will be the dimension of these rows space. The dimension
will be the number of linearly independent rows in that span in that set here. So, the number
of linearly independent rows which is actually the definition of again with the rank; so
here, the dimension of the row space is nothing but the number of linearly independent
rows in the matrix or the dimension of the column space. So, the dimension of the column
748
space is nothing but the its a number of linearly independent columns in the in the matrix
𝐴.
So, we have already three definitions so far. The rank is nothing but the number of pivots
in a reduced row reduced echelon form. The definition number 2 the rank of a matrix is
nothing but the number of linearly independent rows or number of linearly independent
columns of the matrix and we in the terms of the dimension we can also say that the rank
is nothing but the dimension of the row space or the dimension of the column space of 𝐴.
(Refer Slide Time: 13:24)
There is a rank nullity theorem which we will just observe with the help of the example.
So, again we will continue with the same example. So, we have this 𝐴 here and its reduced
form we have these pivot elements. We had also the solution 𝐴𝑥 = 0 which was written
𝑥1
−2
−9.5
𝑥2
1
0
in this form 𝑥3 = 𝛼1 0 + 𝛼2 −4 which generate the solutions.
𝑥4
0
−0.5
[𝑥5 ]
[0 ]
[ 1 ]
So, here the rank of 𝐴 is 3 because it s a number of linearly independent rows or number
of linearly independent columns or the number of pivots we have this rank 3. What is the
nullity of 𝐴? Nullity remember the nullity was the dimension of the null space and the null
space here is the solutions set of this 𝐴𝑥 = 0. So, the solution set of 𝐴𝑥 = 0 we have the
dimension 2, because these are the two vectors which can generate any solution of this set
749
and these two vectors are linearly independent. So, here the nullity of 𝐴 is 2 because of
this is a precisely the number of this free variables in row reduced echelon form.
In general, or first let us talk about this example once again. So, here we have rank 3 and
this nullity is 2 when we add these two, so 3 + 2 will we will give we will get the number
of these variables here number of unknowns in 𝐴𝑥 = 0. The reason is clear because the
rank defines exactly the number of pivots here and this nullity is nothing but the number
of this known pivots. So, we are covering each column for instance here. Here we have
pivot so that will go count to the rank here. So, 1 here 1 plus 1 and 1 this plus 1. So, we
have the 3 rank because these 3 columns have pivots and these two columns do not have
pivot. So, they will be free variables and that is nothing but the nullity of 𝐴. So, this will
add to the end here which are the number of columns or the number of unknowns in 𝐴𝑥 is
equal to 𝑏.
So, this two 3 + 2 will add to that 5 in this case we have 5 columns. In general, how do
we read this? The nullity of 𝐴 is nothing but the dimension of the null space which was 2
in the this previous case, meaning the number of free variables or in our notation we also
take this number of free variables as n minus r, because we have n columns and the r are
the rows where we have the pivots. So, this n minus r or this r is the rank basically of the
matrix because these are the rows where the pivot is sitting. So, the rank of A is r that is
the number of the of the pivot elements. So, here the rank is 𝑟 and this nullity is 𝑛 − 𝑟. So,
this rank nullity theorem is nothing but it says that the rank of a matrix plus nullity of the
matrix. So, r plus this 𝑛 − 𝑟 must add to 𝑛, the number of the columns in the matrix. So,
this is called the rank nullity theorem, Rank (𝐴) + Nullity (𝐴) = 𝑛.
750
(Refer Slide Time: 17:01)
3
0
2
Now, here we will find the rank of the matrix 𝐴 = [−6 42 24
21 −21 0
2
54 ]. So, rank of
−15
this matrix is here 3 0 2 2 and then. So, what do we need to do? For finding the rank we
have to convert to the row reduced echelon form. So, in this case we will try to set this to
0 here with two times the row 1 we will add and that will give us 0 here. This 42 will
remain as it is and here the 24 and minus this 2, so this will be 28 and then we have 58
there. So, we are adding actually it is a two times of two times of this row one we are
adding to this one So, this was 24 and then to this is 28, here also 54 plus 2 plus 4 it is a
58 and similarly here we will be multiplying here by 7 and then subtracting, so we will get
this row.
3
0
2
2
𝐴~ [0 42
28
58 ]
0 −21 −14 −29
3 0
~ [0 42
0 0
2
2
28 58]
0
0
⇒ Rank(𝐴) = 2.
So, this is the row reduced echelon form, and in this row reduced echelon form we have
to count now the number of the pivots. So, we have one pivot here, we have another pivot
as 42, so, the total pivots here we have 2 and that is what the rank here is of this matrix is
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2. So, this is a simplest way of getting the rank. We just take the matrix and try to reduce
it to the row reduced form and then count the number of pivots and that is precisely the
rank of the matrix.
Let us take the another example where we will find the rank of 𝐴 is given by this
1
𝐴 = [2
1
2
6
4
0 −1
−3 −3].
4
5
1
~ [0
0
2
2
2
0
−3
4
−1
−1]
6
1
~ [0
0
2
2
0
0
−3
7
−1
−1]
7
therefore, the Rank(𝐴) = 3.
(Refer Slide Time: 20:10)
There is another way of determining this rank in terms of the determinant. So, that we will
shortly now give. So, first the definition here of the submatrix. Suppose, this 𝐴 is any
matrix of order 𝑚 × 𝑛, then a matrix obtained by leaving some rows and some columns
from 𝐴 is called a submatrix. So, matrix is given we remove let us say first row, the third
column then whatever left this matrix is a submatrix or we can remove more rows more
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columns. So, whatever this smaller matrix by removing some rows and some columns,
that is called the submatrix.
So, here the rank is defined can be also defined from this determinant as the 𝑛, 𝑚 ×
𝑛 matrix 𝐴 has rank 𝑟, this will have rank 𝑟 when if and only if a has 𝑟 × 𝑟 submatrix, so
the square, a square are submatrix with nonzero determinant. So, we have to now get out
of these given matrix 𝑟 × 𝑟 submatrix which has the nonzero determinant whereas, the
determinant of every square determinant or every square submatrix of 𝑟 + 1 or more rows
is 0. So, we have to start with the whole matrix if for example, it is 2 by 3 matrix then 2
by 2 submatrix is we have to look and see if they all are 0 then we have to go to the one
level and so on. So, we have to see now this submatrix with nonzero determinant.
In a particular case when 𝐴 is a square n cross n matrix it has the rank 𝑛, if the det(𝐴) ≠ 0
say if we have a square matrix and its determinant is not equal to 0 then it has a full rank,
so the rank is n because determinant itself is not 0. If the determinant is 0 then we have to
reduce to this different submatrices of order 𝑛 − 1 we have to check all the submatrices of
order n and their determinant if they all are 0 then we have to reduce to 𝑛 − 2, so on. But
for instance, this 𝑛 − 1 submatrices even one has nonzero determinant then the rank will
be that 𝑛 − 1, and rank of a 0 matrix is 0 because we do not have for example, here this
any submatrix with nonzero determinant any submatrix we take, even up to level 1 also
because all the entries are 0. So, the all the submatrices you take whatever order the
determinant is going to be 0. So, this is what we call the rank of 0 matrix is 0.
753
(Refer Slide Time: 23:26)
3 1
Here we will take this simple example 𝐴 = [6 2
3 1
2
4]. So, in this example we can observe
2
that this determinant of this 𝐴 is 0 which is clearly visible because here [3 1 2] and this
row is also [3 1 2]. So, the determinant of this will be 0. So, the determinant of A is 0; that
means, now the rank cannot be 3 because we are not getting this 3 × 3 matrix, it should
be nonzero the determinant should be nonzero then the rank will be 3
So, now the rank will be less than 3. So, we have to check now the submatrices the
determinant of submatrices of order 2 × 2. So, if we take any this 2 × 2 submatrix for
example, we take this one by leaving this second row third column and the third row. So,
we have this determinant. And now the determinant here will be the 6 − 6, so again this
0. We take any other 1 2, 2 4, one more submatrix here also this is 0 when we take the
determinant.
Any determinant we take this 6 4, 6 and this 4 and 3 and 2. So, here also 12 − 12, 0. So,
any determinant we take of order 2 × 2 out of this matrix the answer is 0.
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(Refer Slide Time: 24:56)
So, all 2 × 2 submatrices have 0 determinant and that is the reason here we now have to
look for this determinant of order 1, but that is a nonzero matrix anyway. So, whatever
value we have, so the rank has to be 1 in the that case because it is not the 0 matrix.
1
So, here the rank is 1 and in this example 2 𝐴 = [3
4
2
4
5
3
5], we take another example
6
where we check that again this determinant of 𝐴 is 0. So, when determinant 𝐴 is 0 the rank
has to be less than or equal to the order of the matrix here it is a square matrix. So, the
Rank(𝐴) < 3 and in this case we also observe for instance this determinant itself we take
1
,|
3
2
| ≠ 0, the value if 4 − 6 = 2. So, this is not equal to 0. So, we have a submatrix
4
whose determinant the submatrix of order 2 whose determinant is not 0 and there we can
decide now the rank has to be 2.
Though it is little bit difficult than the earlier process of getting the rank where we reduced
to the row reduced echelon form and then we can easily identify what is the rank. Here
you have to look for these different orders of determinants and check whether something
is nonzero. For example, this was a easy we have easily seen that this submatrix has
determinant nonzero. So, the rank is 2.
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1
If we take the identity matrix 𝐴 = [0
0
0 0
1 0], we know that the |𝐴| = 1 ≠ 0 and
0 1
therefore, the rank of this matrix is 3.
(Refer Slide Time: 26:50)
Coming to the conclusion what we have seen the various definitions of the rank, one was
the number of pivots in the in the row reduced echelon form, the other one was the number
of linearly independent rows. There was a number of linearly independent columns, and
we can also talk about the dimension of the column space that is nothing but the rank
dimension of the row space that also is the rank and we have also discussed the rank in
terms of the determinants.
756
(Refer Slide Time: 27:27)
So, these are the reference used here.
And, thank you for your attention.
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Engineering Mathematics - I
Prof. Jitendra Kumar
Department of Mathematics
Indian Institute of Technology, Kharagpur
Lecture - 44
Linear Transformations
Welcome back and this is lecture number 44 and we will be talking about Linear
Transformations.
(Refer Slide Time: 00:22)
And in particular we will also talk about the rank and nullity theorem for these linear
transformations and also the kernel and image of linear transformations.
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(Refer Slide Time: 00:31)
So, let me start with what is linear mapping or linear transformation. So, here we talk about
these two vector spaces. So, 𝑋 and 𝑌 be two vector spaces and the mapping 𝐹: 𝑋 → 𝑌 is
called linear transformation or linear mapping if it satisfies the following two conditions.
So, what are these conditions? For any two vectors u and v from this vector space 𝑋, if we
apply this transformation 𝐹(𝑢 + 𝑣) = 𝐹(𝑢) + 𝐹(𝑣).
So, that is one conditions for out of these two conditions this is one which is required to
say that this is a linear transformation. The second one for any scalar 𝑘 here from the set
of real numbers and a vector any vector 𝑢 ∈ 𝑋 this 𝐹 should also satisfies that 𝐹 of the
multiplication of this 𝑘 scalar multiplication of this 𝑘 to 𝑢. So, 𝐹(𝑘𝑢) = 𝑘 𝐹(𝑢). So, we
have these two conditions required for the linearity. One is this 𝐹(𝑢 + 𝑣) = 𝐹(𝑢) + 𝐹(𝑣)
and the second condition we have that this map should also satisfies that 𝐹(𝑘𝑢) = 𝑘 𝐹(𝑢).
So, here 2 remarks, one the 2 conditions given above. So, these 2 conditions which we
have just discussed this 𝐹(𝑢 + 𝑣) = 𝐹(𝑢) + 𝐹(𝑣)and 𝐹(𝑘𝑢) = 𝑘 𝐹(𝑢). And these two
conditions can be combined into one and as follows that 𝐹(𝑘1 𝑢 + 𝑘2 𝑣) = 𝑘1 𝐹(𝑢) +
𝑘2 𝐹(𝑣). So, here we are combining basically the two. So, one was this with the addition
of this u plus v which is already here that 𝐹(𝑘1 𝑢 + 𝑘2 𝑣) must be equal to 𝑘1 𝐹(𝑢) +
𝑘2 𝐹(𝑣). So, in the first step for example, we will think it as 𝑘1 𝐹(𝑢) and then plus this
𝑘2 𝐹(𝑣), this is exactly the condition number 1 and the condition number 2 gives now that
this should be equal to 𝑘1 𝐹(𝑢) + 𝑘2 𝐹(𝑣),.
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So, both the conditions are satisfied. Once this condition here that 𝐹(𝑘1 𝑢 + 𝑘2 𝑣) and we
apply if it is equal to 𝑘1 𝐹(𝑢) + 𝑘2 𝐹(𝑣), then we can call that the given transformation is
a linear transformation.
(Refer Slide Time: 03:41)
So, again note that that for 𝑘 = 0, so, in the second condition for instance if we put 𝑘 = 0
that will give us that 𝐹. So, 0 into u that is a property of the vector space this should be 0
vector then. So, 𝐹(0) = 0. So, here again we have this 0 into this 𝐹(𝑢). So, when we
multiply this 0 to 𝐹(𝑢), 𝐹(𝑢) is an element in this vector space 𝑋 and again this 0 into this
element 𝐹(𝑢) must give 0 element. So, here we are getting again this 𝐹(0) = 0 must be
equal to 0. So, this 𝐹(𝑢) is a element of 𝑌. So, again the same thing should hold when we
have this 0 and something from this 𝑌. So, this should give again the 0 vector in this 𝑌.
So, here 𝐹(0) = 0 so; that means, that every linear map takes the 0 vector from this domain
𝑋 to the 0 vector in this range 𝑌.
So, that is another important property which we can quickly look for the vector spaces;
that means, this 𝐹(0) = 0. So, 0 should map to the 0 vector in the vector space 𝑌.
760
(Refer Slide Time: 05:05)
So, we go through some of the examples where we do see this linear maps. So, first here
let F is a linear map here 𝐹: ℝ3 → ℝ3 with 𝐹(𝑥, 𝑦, 𝑧) = (𝑥, 𝑦, 0). So, here every element of
this ℝ3 which is denoted by this (𝑥, 𝑦, 𝑧) and it maps to this (𝑥, 𝑦, 0). So, it is a projection
of of that point (𝑥, 𝑦, 𝑧) to this 𝑥𝑦 plain. So, whether this map is a linear map or not we can
verify this. So, we take two elements from this ℝ3 . So, 2 members of this ℝ3 ; one is 𝑢 =
(𝑎, 𝑏, 𝑐) and other one we have taken this 𝑣 = (𝑎′ , 𝑏 ′ , 𝑐 ′ ). And now these two properties of
the first property is this that when we apply 𝐹(𝑢 + 𝑣) they should give us 𝐹(𝑢) + 𝐹(𝑣)
and that we will check here.
𝐹(𝑢 + 𝑣) = 𝐹(𝑎 + 𝑎′ , 𝑏 + 𝑏 ′ , 𝑐 + 𝑐 ′ )
= (𝑎 + 𝑎′ , 𝑏 + 𝑏 ′ , 0)
= (𝑎, 𝑏, 0) + (𝑎′ , 𝑏 ′ , 0)
= 𝐹(𝑢) + 𝐹(𝑣)
So, the first condition of this linearity is satisfied and now we will check for the second
one which is also trivial in this case. So, for any scalar number this 𝑘, we consider this
𝐹(𝑘𝑢) = 𝐹(𝑘𝑎, 𝑘𝑏, 𝑘𝑐)
= (𝑘𝑎, 𝑘𝑏, 0)
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= 𝑘(𝑎, 𝑏, 0)
= 𝑘𝐹(𝑢)
(Refer Slide Time: 09:00)
So, here both the conditions 𝐹(𝑢 + 𝑣) = 𝐹(𝑢) + 𝐹(𝑣) and also 𝐹(𝑘𝑢) = 𝑘 𝐹(𝑢). So, both
the properties of the linear map a satisfied are satisfied and therefore, this given map here
which maps the point (𝑥, 𝑦, 𝑧) to (𝑥, 𝑦, 0); it is a projection map and that is linear map
which we have just proved here ok.
(Refer Slide Time: 09:48)
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So, another example we will take that this function 𝐹: ℝ2 → ℝ2 is defined the map is
defined as 𝐹(𝑥, 𝑦) = (𝑥 + 1, 𝑦 + 2).. Now the question is whether this is linear map or it is
not a linear map? And this is clear from here because one of the properties of the linear
map is that it always maps 0 to 0 and we have here in ℝ2 our 0 element is nothing but
(0,0), that is the element in ℝ2 .
So, this 𝐹(0,0) element to the (0,0) element if it is a linear map that is a necessary
condition of this linear map. But what we observed here then when we apply this 𝐹(0,0),
what we are getting? We are getting (1,2).
(Refer Slide Time: 10:45)
So, in this way this cannot be a linear map because it is not mapping (0,0) element to (0,0)
element, but it is mapping (0,0) element to (1,2) element which cannot be a linear map.
So, another very important example we will see that these matrices are also like linear
maps because of the reason we take any 𝑚 × 𝑛 matrix and 𝐴 is the transformation from
this ℝ𝑛 to X m by this rule here that if we multiply 𝐴 to this 𝑥; 𝑥 is an element from this
ℝ𝑛 then we will get element a in ℝ𝑚 .
So, here the A is like map from this ℝ𝑛 to ℝ𝑚 because it is taking by this definition here
𝐴𝑥. So, our function is like 𝐴𝑥. So, this 𝑦 = 𝐴𝑥 is mapping from this 𝑥 ∈ ℝ𝑛 , 𝑦 ∈ ℝ𝑚 . So,
if this A is 𝑚 × 𝑛 matrix then it maps element form ℝ𝑛 to one element in ℝ𝑚 and this one
can see easily.
763
(Refer Slide Time: 12:06)
Because if we consider here that is 𝐴 one matrix here of order this 𝑚 × 𝑛. So, this is like
𝑎11 𝑎12 𝑎𝑛𝑑 𝑎1𝑛 𝑎21 𝑎22 𝑎𝑛𝑑 𝑠𝑜 𝑜𝑛 𝑎2𝑛 and then this 𝑚-th row. So, 𝑎𝑚1 𝑎𝑚2 and we have
𝑎𝑚𝑛 at the last element. So, this 𝐴 is a matrix of order this 𝑚 × 𝑛 and if we consider this
𝐴𝑥; that means, this 𝐴 will be multiplied now by this 𝑥 here. So, 𝑥1 𝑥2 𝑎𝑛𝑑 𝑥𝑛 is from ℝ𝑛
that means it has n components; so, 𝑥1 , 𝑥2 , … , 𝑥𝑛 . And now if we discuss this
multiplication, so, what will happen? This row will be multiplied to this column. So, we
will get here then this will be multiplied to this and so on.
So, we will get these m rows. So, this is the 𝑚th row this is the first row. So, as a product
here we will get a vector from this ℝ𝑚 because this will have m component and so, this
vector will belong to ℝ𝑚 . So, when we have a matrix of m cross n order and if we multiply
vector from ℝ𝑛 , so, this 𝐴𝑥 will be a vector in this ℝ𝑚 space.
So, all matrices of order this 𝑚 × 𝑛 we can think as linear map. Why linear map? Because
of the nice properties of such of matrices. So, first what we have seen that this 𝐴𝑥 is nothing
but it is a mapping now the element this 𝑥 which was from ℝ𝑛 and it is giving us an element
in this 𝑦 in this ℝ𝑚 .
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(Refer Slide Time: 14:05)
So, it maps from ℝ𝑛 to ℝ𝑚 that is clear and these linear property, those are two properties
when we apply this 𝐴(𝑢 + 𝑣) = 𝐴𝑢 + 𝐴𝑣 . So, 𝑢 and 𝑣 those are the elements from this ℝ𝑛 .
So, when we apply 𝐴(𝑢 + 𝑣) what we will get? This is the property of the matrix itself. So,
𝐴𝑢 + 𝐴𝑣 and that is what we were looking for the linear map. This is one of the properties
which must satisfy for the linear map and the second one that the (𝜆𝑢) = 𝜆 𝐴𝑢 . So, 𝐴(𝜆𝑢)
should be equal to the 𝜆 𝐴𝑢 and that is another property of the matrix which we can easily
verify if we want.
So, both the properties of the linear mapping satisfied for matrix for a matrix of order 𝑚 ×
𝑛. So, every 𝑚 × 𝑛 matrix maps and maps and these entries of these matrices are real of
course then it maps an element from ℝ𝑛 to an element in ℝ𝑚 . So, this is another important
point which we will be exploring further in this lecture that this matrices are nothing but
they are linear map.
765
(Refer Slide Time: 15:21)
So, the example here for instance we see this 𝐹: ℝ2 → ℝ3 and given by this the relation
2𝑠 + 3𝑡
that 𝐹(𝑠, 𝑡) = [−𝑠 + 5𝑡]. So, this 𝐹 is a linear map and that we have to check it and one
4𝑠 − 3𝑡
way of checking would be that we take the two elements form ℝ2 and then see whether
this 𝐹(𝑢 + 𝑣) gives us 𝐹(𝑢) + 𝐹(𝑣)and the second condition that 𝐹(𝜆𝑢) = 𝜆𝐹(𝑢). So, we
can check those 2 conditions separately on this map. The other possibility what we will in
fact look here that this 𝐹(𝑠, 𝑡)
And now this if we if we recall the properties of the matrix which we can put this in the
form of matrix vector multiplication because 𝑠 is multiplied to this column here, 𝑡 is
multiplied to this column here and that is exactly the property of the matrix vector
multiplication. So, if we put these two columns as the first and the second column of a
matrix and then this s t again column vector there, so that multiplication which exactly
give this 𝑠 times this vector plus t times this vector. Meaning that we can write down this
as this matrix whose columns are these given vectors are 2 minus 1 4 and 3 5 minus 3 and
this s t we can put again as a column vector there.
So, this is exactly when we look into this product here,
3
2
𝐹(𝑠, 𝑡) = 𝑠 [−1] + 𝑡 [ 5 ]
−3
4
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3 𝑠
2
= [−1 5 ] [ ]
𝑡
4 −3
then what we have seen that these matrices are linear map.
So, we do not have to check anything else because we can write down this as this in terms
of the matrix and therefore, this has to be a linear maps. So, without that fundamental
derivation of this to show that this is a linear map we have taken this way that this we can
represent as a matrix map and therefore, this 𝐹 is a linear map because we have written in
terms of the matrix and matrixes are linear map.
(Refer Slide Time: 18:56)
So, another point here to be discussed that is called the kernel and the image of the linear
mapping. And this let 𝐹: 𝑋 → 𝑌 again be a linear map which maps the elements from X
to the elements in Y. And the kernel F is defined as that is the definition of the Ker =
{𝑥 ∈ 𝑋 : 𝐹(𝑥) = 0} . So, this is the kernel here. So, we have this vector space X we have this
vector space Y and what is the kernel?
We will collect all these points here in 𝑋 and if they map to this 0 element in 𝑌. So, then
this set will be called the kernel of this mapping 𝐹. So, this is the definition of the kernel
of 𝐹; all the elements in 𝑋 which map to the 0 element in 𝑌. Naturally, the 0 element must
be there because the 0 must map to the 0 for the linear map. So, definitely the 0 will be
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there in this kernel, but there can be many other elements than the 0 elements in this kernel
here.
(Refer Slide Time: 20:22)
There is another one that is called the image,
Im = { 𝑦 ∈ 𝑌 : there exists 𝑥 ∈
𝑋 for which 𝐹(𝑥) = 𝑦} . So, we are exactly this is the image which we usually discuss
which we considered here. So, we will collect all only those elements of 𝑦 which are the
map of some elements from this 𝑋 ok.
So, with this 2 definitions the Ker 𝐹 and the Im 𝐹, we let us just look at this simple example
with 𝐹(𝑥, 𝑦, 𝑧). This again its a projection map. So, this any element here in ℝ3 it maps to
this point in 𝑥𝑦 plain that is (𝑥, 𝑦, 0). So, what is the Im 𝐹?
So, image F what we are looking for the all the elements in y whose pre image is there in
𝑋. That means, all the points is (𝑎, 𝑏, 𝑐) in ℝ3 whose the third component is 0 because they
are the candidates of the Y and corresponding to when the c is 0 the corresponding element
is also there in 𝑋 and that is nothing but this 𝑎 and 𝑏. So, here the image of this mapping
is nothing but all the points (𝑎, 𝑏, 𝑐) whose third component is 0; that means, all the points
(𝑎, 𝑏, 0); for a and b this belongs to again the real number.
So, this is the image here and when we have set here all the points where this third
component is 0 this is nothing but the 𝑥𝑦 plane. So, the image as we discussed already that
this mapping is nothing but it maps the point in this ℝ3 to the to the 𝑥𝑦 plane and that that
768
is exactly the projection map we are talking about. And therefore, this image is nothing
but this 𝑥𝑦 plane because this maps the point to the 𝑥𝑦 plane only. So, the image is there
in the 𝑥𝑦 plane.
The kernel all are those points from 𝑋 whose map is 0, they map to the 0. So, here as per
the definition of this linear map, any point here (𝑥, 𝑦, 𝑧) it maps to (𝑥, 𝑦, 0). So, the third
will be set to 0. So, we want all those elements whose who those elements in X whose map
is as a 0 element. So, if we take a point here with (𝑎, 𝑏, 0), so, (𝑎, 𝑏, 0) then and when we
apply the; so, the point is here that if we apply this map where or to any element where we
have taken this first 0 and the second component 0. So, this will be nothing but this will
be the (0,0,0) element. So, that is a 0 element in this (𝑥, 𝑦, 0).
So, we are looking for those points in this ℝ3 because their element their image is nothing
but the 0 element in 𝑦. So, this will form the kernel of this mapping.
(Refer Slide Time: 23:53)
And this is exactly the 𝑧 axis because on the 𝑧 axis the 𝑥 is 0 and 𝑦 is 0. So, the whole 𝑧
axis will be mapped to this 0 element in ℝ3 and that is natural here because the 𝑧 axis
when we take the projection of the 𝑧 axis is nothing but the origin that is means a 0 element
in ℝ3 . So, here the image is the 𝑥𝑦 plane and the kernel of 𝐹 is nothing but the 𝑧 axis
because the whole 𝑧 axis is mapping to this origin and the image means here that all the
points in 𝑥𝑦 plane that is the image of this mapping.
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(Refer Slide Time: 24:34)
So, there is a nice theorem here that 𝐹: 𝑋 → 𝑌 be a linear mapping. Then the kernel of 𝐹,
so, this kernel of 𝐹 and is a subspace of 𝑋 and also image of 𝐹 is a subspace of 𝑋.
So, we are not going to formally prove this here, but this is very simple as per the definition
of the kernel and this image. We can think that they will form a subspace because the
kernel was nothing but all the elements here which maps to 0. So, we have to see basically
those closer properties of the subspace. So, if we take any two elements and they map to
the 0, so, their sum will also map to the 0. So, the sum is addition will be also belong to
the same kernel here and similarly for the image also we can consider exactly the exactly
the same consideration.
So, here the point is that these kernel and the image they form subspace and there is another
nice theorem which will be used later. That suppose this 𝑥1 , 𝑥2 , … , 𝑥𝑚 is span a vector
space 𝑋 and suppose this 𝐹 is a linear map, then their image of these points here what these
vectors from 𝑥 which is span the vector space 𝑋 then this 𝐹(𝑥1 ), 𝐹(𝑥2 ), … , 𝐹(𝑥𝑚 ) will also
span will span the image 𝐹.
So, that is a very nice point here if we know the vectors from this space x and we know
that these vectors span the vector space x and we know the mapping here 𝐹 then this
𝐹(𝑥1 ), 𝐹(𝑥2 ), … , 𝐹(𝑥𝑚 ) they will just span the image 𝐹 and the idea of the proof is very
simple. So, if we take a point in the image 𝐹 now and there exists a 𝑋 because this is a
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point in the image, so, there must exists a X in this vector space X such that this F x is
equal to y.
And then we take this x because any element of this vector space 𝑥 can be represented as
a linear combinations of 𝑥1 , 𝑥2 , … , 𝑥𝑚 because these vectors span the vector space X. So,
by taking this we have represented this x in terms of the 𝑥𝑖 ’s; that means, the linear
combination of these 𝑥𝑖 ’s. And now we apply this linear map 𝐹 on 𝑥 which will give us
the element 𝑦 naturally. So, this F on x and due to the linearity of the map we can take this
𝐹 here inside this 𝑥𝑖 ’s because these are the constant that is the definition of the linear
map.
𝑚
So, we have the 𝑥 = ∑𝑚
𝑖=1 𝛼𝑚 𝑥𝑚 ⟹ 𝑦 = 𝐹(𝑥) = ∑𝑖=1 𝛼𝑚 𝐹(𝑥𝑚 ). So, this is because of the
linearity and now what we see that this 𝑦 element we have written as a linear combination
of this x F x m and this is exactly that any element y in the vector space 𝑦 we can write as
a linear combination of these vectors 𝐹(𝑥𝑖 )’s. Therefore, these vectors 𝐹(𝑥𝑖 ) will span the
image 𝐹.
(Refer Slide Time: 28:11)
So, here the kernel and image of the matrix mapping; so, because this matrix are also linear
maps. So, if you consider just quickly this simple example of this 3 rows and 4 columns
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𝑎1
𝐴 = [𝑏1
𝑐1
𝑎2
𝑏2
𝑐2
𝑎3
𝑏3
𝑐3
𝑎4
𝑏4 ] and we take for instance the usual basis here 𝑒1 , 𝑒2 , 𝑒3 , 𝑒4 from
𝑐4
ℝ4 these are the standard basis of ℝ4 which we have already discussed before.
Then as per the previous theorem, we know that 𝐴𝑒1 , 𝐴𝑒2 , 𝐴𝑒3 , 𝐴𝑒4 will span the image of
𝐴.
𝑎4
𝑎3
𝑎1
𝑎2
𝐴𝑒1 = [𝑏1 ], 𝐴𝑒2 = [𝑏2 ], 𝐴𝑒3 = [𝑏3 ] , 𝐴𝑒4 = [𝑏4 ]
𝑐1
𝑐2
𝑐3
𝑐4
these are the vectors in ℝ3 , these are the vectors in image and we know from the previous
theorem that these vectors here will these vectors will span the image.
So, we know already the spanning vector of this image ℝ3 which is in ℝ3 . So, image of
this 𝐴. So, what are these? These are the precisely the columns here. These are the columns
of this given matrix. So, the image is nothing but image is nothing but the column the
column space the column spaces exactly the span of these columns that is the column space
we have already discussed and what we have observed that the image is nothing but the
span of these columns.
So, thus the Im(𝐴) is precisely the column space of 𝐴. So, when we talk about the matrices
here the column space is nothing but they will span the column space is nothing but the
Im(𝐴). Similarly, the kernel because the kernel will be all the points here whose who map
is to 0s; that means, we are looking for all the points 𝑥 here in ℝ4 and whose map to the 0
in ℝ3 . So; that means, what we are looking exactly? We are looking for the null space of
this null space of this matrix A because that was the definition of the null space that all the
I mean the solution of this system of (Refer Time: 30:58) equation gives us the null space.
So, here the null space in the form of the kernels is same as the null space of a that is the
kernel of the matrix mapping.
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(Refer Slide Time: 31:11)
Last one here the rank and the nullity of a linear map. So, if we remember the image A is
the column space of 𝐴 and then the dimension of the column space when we have discussed
the rank of the matrix. So, that was the dimension of the column space that was the rank
there.
So, here in terms of the linear map, the rank of a linear map will be the dimension of the
image of 𝐹. So, image of 𝐹 will be the rank. So, this is a more general definition of the
rank which the in case of the matrix that is the special case and we have separately
discussed the rank concept of the matrix. So, the rank of the matrix was the dimension of
the column space, but here we have the more general terminology that is called the image
of the mapping 𝐹.
So, this kernel 𝐹 is nothing but the dimension of the image 𝐹. Similarly, the kernel 𝐴 was
null space of 𝐴 and here again this nullity which we call the dimension of the null space,
so that is the nullity which we have discussed already before. So, the nullity is nothing but
the dimension of the kernel. So, these two again we have the two more definitions of about
the kernel which is more general than the definition of the rank we have discussed for
matrices.
So, here the rank of 𝐹 is the dimension of the image of 𝐹 and nullity is nothing but the
dimension of the kernel of 𝐹. And there is a theorem that let 𝑋 be a vector space of finite
dimension then and let this 𝐹 be a linear map then this rank plus nullity is equal to
773
dimension of 𝑋. So, this rank nullity theorem we have also discussed for matrices where
rank plus nullity was the number of variables n which is here in this case that is because
this 𝐴 maps from ℝ𝑛 to ℝ𝑚 ; so that exactly the dimension of this domain here the
dimension of X. So, the rank(𝐹) + nullity(𝐹) = dim(𝑋) that is a more general result than
what we have for the matrices.
(Refer Slide Time: 33:20)
Coming to the conclusion here; so, we have discussed the linear map and to prove the
linear map we just need to apply this 𝐹(𝑘1 𝑢 + 𝑘2 𝑣) = 𝑘1 𝐹(𝑢) + 𝑘2 𝐹(𝑣) then we call that
the given map is the linear map. And what we have also seen that these matrices are also
linear map and if matrix is of order 𝑚 × 𝑛 then they map the elements of ℝ𝑛 to ℝ𝑚 . And
this image of 𝐴 is nothing but the column space of 𝐴 and this image the kernel of this
matrix this linear mapping 𝐴 is nothing but the null the null space of 𝐴.
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(Refer Slide Time: 34:04)
So, these are the references used here and thank you for your attention.
775
Engineering Mathematics - I
Prof. Jitendra Kumar
Department of Mathematics
Indian Institute of Technology, Kharagpur
Lecture - 45
Linear Transformations (Contd.)
So, welcome back. And this is lecture number 45 and we will be talking about or continue
our discussion on Linear Transformations.
(Refer Slide Time: 00:25)
And today, we will see some worked problems where we will find out the Kernel and the
image and their dimensions.
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(Refer Slide Time: 00:31)
So, here just to recall from the previous lecture what we have seen there that 𝐹: 𝑋 → 𝑌 is
a linear map than this Ker 𝐹 = {𝑥 ∈ 𝑋 : 𝐹(𝑥) = 0} and the Im 𝐹 = { 𝑦 ∈ 𝑌 : there exists 𝑥 ∈
𝑋 for which 𝐹(𝑥) = 𝑦} .
So, these were the two definitions for the Ker 𝐹 and also for the Im 𝐹 . And we have also
seen that this rank(𝐹) is nothing but the dimension of the Im 𝐹 and the nullity we define
as the dimension of the Ker 𝐹.
(Refer Slide Time: 01:21)
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So, with these definitions we start now here with the example, where we have taken this
𝐹: ℝ4 → ℝ3 is a linear mapping which is defined by this relation 𝐹 maps this element
which is from ℝ4 . So, 𝐹(𝑥, 𝑦, 𝑧, 𝑡) = (𝑥 − 𝑦 + 𝑧 + 𝑡, 2𝑥 − 2𝑦 + 3𝑧 + 4𝑡, 3𝑥 − 3𝑦 +
4𝑧 + 5𝑡). So, this is the element from ℝ3 and this is the element from ℝ4 and that is what
we said here this F maps an element from ℝ4 to an element in ℝ3 .
And what we want to now find? We want to find basis a basis and the dimension of first
the Im(𝐹) and then second for the Ker(𝐹). So, we need to find what is the image and what
is the Ker(𝐹). So, here first let us talk about the Im(𝐹). What we know from the previous
lecture that once we have the elements in vector space 𝑥 which span this vector space 𝑥.
So, for instance here we have this ℝ4 . So, we know that these a standard basis because
they are simple to work with, so let us take whether standard basis here that this 𝑒1 , 𝑒2 , 𝑒3 ,
𝑒4 these are the standard basis from ℝ4 and we know that the theorem that result from the
previous lecture that these e s span this 𝑥 ∈ ℝ4 than this 𝐹(𝑒1 ), 𝐹(𝑒2 ), 𝐹(𝑒3 ), 𝐹(𝑒4 ) these
are the vectors in ℝ3 and they will span actually the image of this mapping 𝐹.
So, what we know now that this 𝐹(𝑒1 ), 𝐹(𝑒2 ), 𝐹(𝑒3 ), 𝐹(𝑒4 ) they will span the image 𝐹.
So, that is the result we have now that these are the vectors which span the image 𝐹. And
now let us compute what are these 𝐹(𝑒1 ), 𝐹(𝑒2 ), 𝐹(𝑒3 ), 𝐹(𝑒4 ).
(Refer Slide Time: 03:43)
So, here 𝐹(𝑒1 )𝑣so 𝑒1 just remember that 𝑒1 is nothing but in this case these are the standard
vectors from ℝ4 . So, these are that is the (1, 0, 0, 0), from ℝ4 the 𝑒2 will be (0, 1, 0, 0), 𝑒3
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will be (0, 0, 1, 0) now and 0, and 𝑒4 will be (0, 0, 0, 1). So, we have these standard basis
from ℝ4 and now 𝐹 if we apply on this 𝑒1 here. So, what will happen? So,
𝐹(𝑒1 ) = (1,2,3)
𝐹(𝑒2 )=(-1,-2,-3),
𝐹(𝑒3 ) = (1,3,4)
𝐹(𝑒4 ) = (1,4,5)
from this ℝ3 and we know now with this with the theorem we have studied already that
these vectors will span image 𝐹 and what we want to find out we want to find out what is
the image 𝐹, what is what are the basis of image F and what are the what is the dimension
of image 𝐹. So, we already know that these 4 vectors will span image ℝ3 , but they are they
are not the basis because this is this is the spanning set, and spanning set may have some
linearly dependent vectors. So, what we have to now extract out of these vectors what we
have to collect only the linearly independent vectors. So, we have to see now how many
linearly independent vectors we have in this spanning set. And the number of those linearly
independent vectors will be the dimension of the image F and those linearly independent
vectors among all these 4 will form basis of the image 𝐹.
(Refer Slide Time: 06:11)
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So, here now let us do this. So, these may not be the basis now we have to just find out
that how many of these vectors are linearly independent.
(Refer Slide Time: 06:19)
So, to do so,
1
[2
3
−1
−2
−3
1
3
4
1
4]
5
Having this vector here whose columns are the given vectors we can now reduce it to this
row reduced echelon form and from that row reduced echelon form we can find out that
which vectors are linearly independent and which vectors are linearly dependent and we
have seen one such example before in previous lectures.
1
~ [0
0
−1
0
0
1
1
0
1
2]
0
So, now to find out the linearly independent vectors out of this what we can do we can
place them in the in this matrix here as the columns, and then we can reduce it to the row
reduced echelon form which should not be difficult here in this case because as a first step
we will set these two elements to 0. So, the first will first row will be as it is, the second
one here will become 0, the two times of that this is also 0, two times this will give 1, here
two times will get 2, here now the 3 times of the row 1 we are subtracting here so this will
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be 0, this will be again 0 and the 3 times this will be 1 and 3 times this will be 2. So, this
is the one step of this row reduction process and now we will said this 0 with the help of
this row 2 and we will get this matrix which is given already there.
(Refer Slide Time: 08:07)
So, what we will have that this is the row reduced echelon form and in this case now, we
will find out these pivots here. So, we have this one as a pivot and this is also pivot here,
only there are two pivots. So, this column number 1 and the column number 3 they have
the pivots. So, a straight forward we can pick the column number 1 here and the column
number 3 and they will be linearly independent. These column here does not have a pivot
this does not have a pivot and one can show that those two columns depend on this column
number 1 and column number 3. So, this was a very systematic approach to found out the
linearly independent vectors here
So, now, with this we have we can take this column number 1 which has the pivot here,
so the corresponding this column number one and we can take this column number 3 and
they will form the basis.
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(Refer Slide Time: 09:09)
So, the dimension of this image is going to be true and the basis these two vectors. So,
(1, 2, 3) & (1, 3, 4) these will form the basis of the image 𝐹 and the dimension of the
image 𝐹 is 2.
(Refer Slide Time: 09:20)
Now, getting to the Kernel of F; so, what was the Kernel of F, 𝐹(𝑥, 𝑦, 𝑧, 𝑡) = 0? Now, we
are looking for the elements in x whose map is 0 in y, so that means, we are looking for
these elements (𝑥, 𝑦, 𝑧, 𝑡) I mean these vectors or such vectors whose image is 0. And now
to look into this now what is 𝐹(𝑥, 𝑦, 𝑧, 𝑡) from the definition we have that
782
𝑥−𝑦+𝑧+𝑡 =0
2𝑥 − 2𝑦 + 3𝑧 + 4𝑡 = 0
3𝑥 − 3𝑦 + 4𝑧 + 5𝑡 = 0
So, from here we get actually these 3 equations, these 3 equations.
(Refer Slide Time: 10:34)
And we want to now solve these equations to get all possible values of these (𝑥, 𝑦, 𝑧, 𝑡) and
this is nothing but the null space of the coefficient matrix of the coefficient matrix of this
of this system of equations and that is nothing but the Kernel of F. So, the null space here
of this coefficient matrix will be the Kernel of F, because what is the Kernel of F all these
vectors here from ℝ4 whose image is 0 they are not 3. So, what we need to solve? We need
to solve this system of linear equations and we will find out how many solutions are there
or how many linearly independent solutions are there or in other words we call the
generators of the solution of this system of homogeneous equations and from there we will
find out what is the what is the dimension of the Kernel, what is the what are the basis of
the Kernel.
So, we need to solve finally, this system of equations. So, let us put in this matrix form.
1 −1
𝐴 = [2 −2
3 −3
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1 1
3 4]
4 5
So, the matrix
1
𝐴~ [0
0
−1
0
0
1
1
0
1
2]
0
And again we need to just convert into the reduced echelon form which we have done just
before and now we can easily find it out because this is the pivot here and this is the pivot.
So, corresponding to this x this is corresponding to y this is corresponding to z and this is
corresponding to y and this is corresponding to 𝑡. So, here these are the columns where we
do not have pivots. So, these are the free variables which we call, so there are two free
variables and that will define exactly the nullity or the dimension of the null space that will
be 2 here. So, the dimension of this Kernel is clear that is going to be 2 and we have to
find the basis for that we have to write down solution now.
(Refer Slide Time: 12:49)
So, writing the solution out of this. So, first we will assume these free variables. So, that
was 𝑦 and this 𝑡 these are the free variables. So, we will take some values for this free
variables and then, so let us take for this 𝑡 = 𝜇1 & 𝑦 = 𝜇2 some number arbitrary number
because we can choose these 𝑦 𝑎𝑛𝑑 𝑡 whatever we like these are the free variables. So, we
have chosen this 𝜇1 and 𝜇2 and then we can compute the others. So, for example, the z
form this equation z plus 2 t is equal to 0. So, 𝑧 = −2𝜇1 , so that will be the z. And from
the equation number 1, we can get this 𝑥 = 𝜇1 + 𝜇2 .
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So, we have this t, z and x all and also 𝑦 = 𝜇2 . So, we can write down in a matrix in this
vector form x, y, z, t will be this 𝜇1 the constant here and this we can collect now for x, x
is one the coefficient here for y it is 0, for z it is minus 2 and from 𝑡 it is 1. So, similarly
for with 𝜇2 also we can also collect the coefficients here. So, in𝑥 = 𝜇1 + 𝜇2 and then here
𝑦 = 𝜇2 which is 1 and this 𝑧 = −2𝜇1 . So, here 0 and also this 𝑡 does not have 𝜇2 . So, with
this coefficient is 0.
So, we have written this 𝑥, 𝑦, 𝑧 the solution in terms of this 𝜇1 and 𝜇2 we can play now 𝜇1
and 𝜇2 can give any values to 𝜇1 and 𝜇2 and we can keep on generating the solution here
of this system of equation and the solution that is nothing but the Kernel of 𝐹. So, here
these are the two generators which generates the solution or the they span the solution,
also these two vectors are linearly independent. So, we have this linearly independent
vector which can span the whole Kernel of 𝐹. So, naturally these two vectors will form the
basis and the dimension of this basis here will be two. So, these two vectors form the basis
of the of the Kernel 𝐹 and the dimension or the nullity which we call is already there the
dimension of the Kernel or the nullity of this 𝐹 which was clear also from the number of
these free variables which are two here. So, we can just get the dimension or the nullity
and also the basis simply.
𝑥
1
1
𝑦
1
0
⟹ [ ] = 𝜇1 [ ] + 𝜇2 [ ]
𝑧
0
−2
𝑡
0
1
1
1
1
0
⟹ [ ] & [ ] form a basis of ker(𝐹)
0
−2
0
1
785
(Refer Slide Time: 15:37)
Now, the another example where we have 𝑢 = (1,1,3), 𝑣 = (3,2, −2) and it is also given
that the 𝐹(𝑢) = (4,1,1,1), 𝐹(𝑣) = (−5,1, −3,3).. And we assume that this 𝐹 is a linear
map from this ℝ3 → ℝ4 because it maps this element ℝ3 , this u and v are from ℝ3 and the
output is in ℝ4 . So, this maps from ℝ3 to ℝ4 . And what is the question now that if 𝑤 =
(5, 4, 4) & 𝑦 = (2, 1, 7), then can we find this 𝐹(𝑤) & 𝐹(𝑦), yeah. We know only that
𝐹(𝑢) is given and 𝐹(𝑣) is given, but we want to find now what will be the 𝐹(𝑤) the 𝑤 =
(5, 4, 4) & 𝑦 = (2, 1, 7) and we want to find this 𝐹(𝑤) & 𝐹(𝑦) and 𝐹 is a linear map.
So, what do we know about the linear map this additive property; that 𝐹(𝑢 + 𝑣) = 𝐹(𝑢) +
𝐹(𝑣).
So, what we can think here? That if this w we can write down in terms of this u and v if.
So, if we can write down this (5, 4, 4) in terms of the vectors 𝑢 and 𝑣 then we can find
out what is the 𝐹 of 𝑤. If we can if we can write this w as a linear combination of this 𝑢
and 𝑣 then we can also find out the 𝐹 and similarly with 𝑦 also if we can write down this
y as a linear combination of 𝑢 and 𝑣, then we can apply this transformation 𝐹 on 𝑦 and we
can write down in terms of 𝑢 and 𝑣 and that will be the image of this y.
1
[1
3
1 3 5
3 5
~
|
]
[
0 1 | 1]
2 4
0 0 0
−2 4
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So, let us try now how we can write we can express this 𝑤 as a linear combination of 𝑢
and 𝑣; that means, the w can we write as 𝜇1 and 𝜇2 for some scalars 𝜇1 and 𝜇2 . So, to do
so, we have to now again form this augmented matrix to solve this system of equations
here for 𝜇1 and 𝜇2 . So, here 𝜇1 = 1, 𝜇2 = 2 ⇒ 𝑤 = 2𝑢 + 𝑣 , so that will be placed here in
the column when we write down these system of equations. S
So, out of those that is system of equations here from this linear combinations. So, from
those, these system of equations we can write down this augmented matrix and this
augmented matrix is given here.
(Refer Slide Time: 18:40)
So, now out of this augmented matrix, we have to get this row reduced forms. So, this 1 3
5 and this is simple here. So, we can subtract it and then further. So, we will get this
reduced form, and from this reduced form we will now observe that this is the pivot here
this is also the pivot.
1
~ [0
0
3 5
1 | 1]
0 0
𝐹(𝑤) = 2𝐹(𝑢) + 𝐹(𝑣)
= 2 ⋅ (4,1,1,1) + (−5,1, −3,3)
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= (3, 3, −1, 5)
(Refer Slide Time: 20:08)
Now, if we try to do for the y again the same steps so now the 𝑦 = (2, 1, 7) and we have
this 𝑢 𝑎𝑛𝑑 𝑣. So, again we will try to express as a linear combination of 𝑢 𝑎𝑛𝑑 𝑣 to this 𝑦
1
and again we will get this augmented matrix that is [1
3
3 2
2 | 1]. And again, we will
−2 7
convert into the row reduced form, so we got this row reduced form now.
1
[1
3
3 2
2 | 1]
−2 7
Now, the problem here is the inconsistency the last column the right most column here has
a pivot element here this 12 which should not happen. In this case and now because of this
we have this inconsistency in the system and which leads to the nonexistence of the
solution. So, here we cannot express this 𝑦 as a linear combination or this 𝑢 and 𝑣 and
therefore, the system is inconsistence, it is inconsistent and this as a reason that we cannot
we cannot compute this 𝐹 of 𝐹(𝑦) because the information given is not sufficient here.
So, we cannot compute this 𝐹(𝑦) from the given information.
788
(Refer Slide Time: 21:28)
Well, so now, there is another nice result that if T is a linear map from this ℝ𝑛 → ℝ𝑚 then
there is an always m cross n matrix a such that this T x you can represent in terms of this
matrix A. So, whatever we do with the 𝑇(𝑥) = 𝐴𝑥 because working with the matrices are
much easier and we know many operations and many other properties of the matrices. So,
in speed of working with linear map which is not given in the in the form of the matrix it
is better to have a matrix form and then we can work with the matrix we can do all
operations whatever we want on this map with this matrix here 𝐴.
So, that is a proof is very simple here we can define the standard basis in ℝ𝑛 which given
as this 𝑒1 = (1,0, … ,0), 𝑒2 = (0,1,0, . . ,0), … , 𝑒𝑛 = (0,0, … ,1). These are the standard basis
from this vector space ℝ𝑛 . And this 𝑇 is a for any x now in ℝ𝑛 , so any vector we take from
this 𝑥 fromℝ𝑛 what we can because these are the basis here these are the standard basis
we can represent this x in the form of this basis; that means, this 𝑥 = 𝑥1 𝑒1 + 𝑥2 𝑒2 + ⋯ +
𝑥𝑛 𝑒𝑛 . So, x 1, x 2, x 3, x n are the components of this 𝑥 vector. So, we can represent with
the help of these standard basis this x as 𝑥 = 𝑥1 𝑒1 + 𝑥2 𝑒2 + ⋯ + 𝑥𝑛 𝑒𝑛 or in short we can
write down in the form of summation that ∑𝑛𝑖=1 𝑥𝑖 𝑒𝑖 .
So, this x which is a general element in ℝ𝑛 we have represented as a linear combination
of that those standard basis and now if we apply this 𝑇, the linearity of 𝑇 will implies that
𝑇(𝑥) = 𝑇(∑𝑛𝑖=1 𝑥𝑖 𝑒𝑖 ) and basically we can take we can apply on this 𝑇 i’s. So, we have
this ∑𝑛𝑖=1 𝑥𝑖 𝑇(𝑒𝑖 ) here because of the linearity this is the linearity property that
789
𝑇(𝑥1 𝑒1 + 𝑥2 𝑒2 + ⋯ + 𝑥𝑛 𝑒𝑛 ) will be 𝑥1 𝑇(𝑒1 ) + 𝑥2 𝑇(𝑒2 ) + ⋯ + 𝑥𝑛 𝑇(𝑒𝑛 ). So, that will be
the will be the relation here and how we define now the 𝐴 which will be exactly
corresponding to this map 𝑇 or will function as this map 𝑇 there. So, A now we can define
whose columns are these vectors here 𝑇(𝑒1 ), 𝑇(𝑒2 ), … , 𝑇(𝑒𝑛 ). So, these columns if we put
into this matrix A then this Ax will be nothing but exactly this product which is the 𝑇(𝑥).
So, if you remember from this definition of this matrix product which we have seen several
times. So, that is nothing but this is column 1 or will be multiplied to this 𝑥1 , the column
2 will be multiplied to 𝑥2 and so on. The column n will be multiplied to 𝑥𝑛 that is another
way of looking at the matrix vector product and here exactly we have used that idea that
this 𝑥1 multiplied by this vector 𝑇(𝑒1 ). So, 𝑇(𝑒1 )if we place as a first column in our matrix
and similarly this 𝑥2 and then this 𝑇(𝑒2 ) placed in the matrices second column. So, this
product here 𝐴𝑥 will be exactly this one which is nothing but the 𝑇(𝑥). So, the 𝑇(𝑥) =
𝐴𝑥.
So, by this simple idea what we have seen that any linear map which maps the elements
of ℝ𝑛 → ℝ𝑚 we can have matrix here corresponding to this 𝑇(𝑒) map we can have a
matrix 𝐴 of order again this m cross n and whose columns we can easily compute with the
help of this 𝑇(𝑒𝑖 )’s and this will give exactly the map which is given as this from ℝ𝑛 →
ℝ𝑚 .
(Refer Slide Time: 26:06)
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So, now let us just quickly go through some examples here. So, this was the one example
which we have already discussed before. So, this 𝑇: ℝ2 → ℝ3 given as 𝑇(𝑥, 𝑦) = (2𝑥 +
3𝑦, −𝑥 + 5𝑦, 4𝑥 − 3𝑦). So, this is an element from ℝ2 and then we are getting this element
ℝ3 . So, now, how to get this corresponding matrix 𝐴? The idea was that we compute this
𝑇(𝑒1 ). So, 𝑇(𝑒1 )means the 𝑒1 here is nothing but this 1 and 0.
So, the 𝑇(𝑒1 ), so 1 and 0, so y will be set to 0. So, we will get 𝑇(𝑒1 ) = (2, −1,4). Similarly,
we can get the 𝑇(𝑒2 ) = (3,5, −3). So, that will be the second element. And then in 𝐴 =
2
[−1
4
3
5 ] has the second column of this matrix 𝐴. And we note that now that this T x the
−3
given linear map here which was defined in this way we can also defined as 𝐴 and this
element of this ℝ2 because this a into this x y will exactly give us this linear map because
2
3
if we multiply 𝐴 with this x y. So, here we have [−1 5 ] and with this x y what do we
4 −3
get here 2𝑥 + 3𝑦 the second component −𝑥 + 5𝑦 then we have 4𝑥 − 3𝑦.
(Refer Slide Time: 27:35)
So, we are getting exactly the same mapping which was defined there with the help of this
𝑇(𝑥, 𝑦) = (2𝑥 + 3𝑦, −𝑥 + 5𝑦, 4𝑥 − 3𝑦). So, the given linear map we have defined with the
help of this matrix 𝐴.
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(Refer Slide Time: 28:15)
One more example we can look quickly here this 𝐹: ℝ3 → ℝ3 which was which is given
here by this element from ℝ3 and then here also we have this element from ℝ3 ,
𝐹(𝑥1 , 𝑥2 , 𝑥3 ) = (9𝑥1 + 5(𝑥2 + 7𝑥3 ), 5𝑥1 − 12 𝑥2 + 27𝑥3 , 55 𝑥1 − 42 𝑥3 ). This is the first
component then we have we have a this first component and then we have the second
component, we have the third component. So, this maps ℝ3 to this ℝ3 and here the same
idea which we can use again, which we have used before that this 𝑇(𝑒1 ) = (9, 5, 55).
𝑇(𝑒2 ) = (5, − 12, 0), 𝑇(𝑒3 ) = (35, 27, − 42).
9
So, we have this 𝐴 here, the matrix 𝐴 = [ 5
55
5
−12
0
35
27 ] and that will define this linear
−42
map the given linear map as this matrix vector product. So, again this working with the
matrices are much easier. So, this is one way where we can, where we can represent a
linear map with the help of this matrices.
792
(Refer Slide Time: 29:36)
So, here what we have done today, with this from the linear map we have done some
examples where we have computed the Kernel F also the dimension of the Kernel F as
well as the image F and its dimension.
Ker 𝐹 = {𝑥 ∈ 𝑋 : 𝐹(𝑥) = 0}
Im 𝐹 = { 𝑦 ∈ 𝑌 : there exists 𝑥 ∈ 𝑋 for which 𝐹(𝑥) = 𝑦}
Further, we have also looked at this using matrices how to define the linear maps.
Whenever there is a linear map 𝑇 : ℝ𝑛 → ℝ𝑚
elements, we can always define
corresponding matrix A which will work as this 𝑇(𝑥) = 𝐴𝑥 . So, that is very useful as I said
before this working with matrices are much easier.
793
(Refer Slide Time: 30:15)
So, here these are the references here we have used for preparing the lectures.
And, thank you very much.
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Engineering Mathematics - I
Prof. Jitendra Kumar
Department of Mathematics
Indian Institute of Technology, Kharagpur
Lecture - 46
Eigenvalues & Eigenvectors
So, welcome back this is lecture number 46 and today we will continue with Eigenvalues
and Eigenvectors another very important topic in linear algebra.
(Refer Slide Time: 00:26)
So, we will go through the introduction of these eigenvalues and eigenvectors and also
their geometrical interpretation and, then some simple examples to evaluate eigenvalues
and eigenvectors.
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(Refer Slide Time: 00:39)
So, here what are the eigenvalues and eigenvectors of a matrix; here let us consider this
3 −2
]. And, then we consider these two vectors one is this 𝑢 =
1 0
simple matrix here 𝐴 = [
[
−1
] and another one is 𝑣 = [2]. And, with this if we compute this product here 𝐴 and u
1
1
so, here we have this 𝐴 and then we have this 𝑢.
3
1
𝐴𝑢 = [
=[
−2 −1
][ ]
0
1
−5
]
−1
but if we do this product with this vector v then what will happen?
𝐴𝑣 = [
3 −2 2
][ ]
1 0 1
4
=[ ]
2
2
= 2[ ]
1
So, that is exactly the point which will take us to this introduction to eigenvalues and
eigenvectors.
796
(Refer Slide Time: 02:37)
So, if we look at this geometrically what is happening. So, here we have this x axis and y
axis. So, this vector v which is given here as is exactly this one and then the Av after this
product it is just the 2 times. So, that is the vector this width length double of this length
of this v. So, we have this vector Av; what is interesting that this Av and v they have the
same direction and their magnitudes are different. So, here now it is just the double of this
earlier vector v after the multiplication, but in this case of this u vector this was the vector
u and this A times u has become this one.
So, we do not have such relation that after multiplication the vector direction remains the
same, we have a completely different vector now Au. So, our interest is not exactly this u
with this A, our interest is for such vectors whose multiplication with that matrix does not
change its direction its a parallel to the original vector v. And, that is what we are looking
for and these are called actually the eigenvectors and this number which has come here
that will be called as eigenvalues. So, that is the topic of today’s lecture and we will go
little more into the detail now about these eigenvalues and eigenvectors.
797
(Refer Slide Time: 04:12)
So, the definition the mathematical formal definition here: let A be any square matrix, the
entries can be real or the entries of the matrix A can be complex. A scalar 𝜆 is called
eigenvalue of A if there exists nonzero vector yeah, that is also important here this nonzero;
vector if there exists a nonzero vector such that such that this 𝐴𝑥 = 𝜆𝑥. So, exactly it is a
parallel to what we have just seen in previous examples. So, if we have such a vector x
whose multiplication with this given matrix a 𝐴𝑥 is nothing, but the lambda time x and
this lambda is some scalar some real number or a complex number.
So, here this 𝐴𝑥 = 𝜆𝑥 that is a very important equation which we will be talking about
today. So, that is the definition now of this eigenvector this 𝑥 is called the eigenvector and
this lambda is called the eigenvalue. So, this is the eigenvector and this has to be always
nonzero otherwise, the 0 will be satisfied always. So, we are looking for the nonzero vector
here which is called the eigenvector and this is called the eigenvalue ok.
798
(Refer Slide Time: 05:45)
So, now the vector x is the eigenvector associated with this eigenvalue lambda and the two
important points now, the geometrically which we have already seen with the help of
earlier example that an eigenvector of a matrix A is a nonzero vector, 𝑥 in this ℝ𝑛 such
that the vectors here x and this 𝐴𝑥 are parallel. This is what we have seen in previous
example that this vector v and the vector Av they were just the parallel, the length was
different. So, here also the geometrical meaning of this eigenvalues eigenvector in general
is that of this vector this x and this 𝐴𝑥 both are parallel and their magnitude will change
and therefore, we have this eigenvalue lambda.
Algebraically, an eigenvector x is a non-trivial solution because we are looking for the
eigenvector x which is nonzero. So, meaning this non-trivial solution because x is equal to
0 will always satisfy this equation. So, we are not interested in the 0 solution, our interested
in nonzero solution; meaning the non-trivial solution of this equation Ax is equal to lambda
x. Or, this eigenvector x here is a nonzero vector in the null space of this A minus lambda
I, because this equation which we have a is equal to l 𝐴𝑥 = 𝜆𝑥. So, this 𝐴𝑥 = 𝜆𝑥 what we
can do we can bring this 𝜆𝑥 term to the left hand side. So, we have 𝐴 − 𝜆𝐼. Then we have
to also introduce this identity matrix and then x is equal to 0.
So, basically what we are looking for, we are looking for the non-trivial solution. This x
of this equation 𝐴 − 𝜆𝐼 or rather the system of linear equations (𝐴 − 𝜆𝐼)𝑥 = 0 and this is
799
exactly the definition of the null space of this 𝐴 − 𝜆𝐼. So, here the matrix is 𝐴 − 𝜆𝐼 and if
we look for the null space of this 𝐴 − 𝜆𝐼.
So, this x vector which we are looking for is in the null space of this matrix 𝐴 − 𝜆𝐼. And,
since it is a nonzero I mean the null space also has a now has a 0 vector, but we are looking
for the nonzero vector in the null space of this. So, x is a nonzero vector in the null space
of this 𝐴 − 𝜆𝐼. Now, the natural question is how to compute this eigenvector and how to
compute the eigenvalues associated with this with the matrix of order n.
(Refer Slide Time: 08:49)
So, how to find the eigenvalues and eigenvectors we will discuss now; so, consider this
equation (𝐴 − 𝜆𝐼)𝑥 = 0 , that is 𝐴𝑥 = 𝜆𝑥 equation written in this form. And, there are two
unknowns here, the unknowns are the 𝜆 we need to compute lambda and also we need to
compute x. There is a one equation here (𝐴 − 𝜆𝐼)𝑥 = 0 or it is a system of linear equation
and we have these two unknowns the 𝜆 and x. So, how to compute these two unknowns
so, that those unknown satisfy this equation (𝐴 − 𝜆𝐼)𝑥 = 0 ? The lambda is a scalar and
this 𝑥 is a vector whose components will be exactly equal to this 𝑛, if the matrix is n cross
n matrix. So, what is other information we have that we are looking for this non-trivial
solution x.
So, this equation (𝐴 − 𝜆𝐼)𝑥 has a non-trivial solution which we have already studied in
previous lecture. If and only if this satisfy the equation, which equation that the
determinant of this (𝐴 − 𝜆𝐼); if this determinant is 0 then we will have a non-trivial
800
solution. If the determinant is not equal to 0 then there will be a unique solution and that
will be the trivial solution meaning this 𝑥 will be 0.
But, we are looking for a non-trivial solution of this equation (𝐴 − 𝜆𝐼)𝑥 = 0 and in that
case we have this condition that this determinant of this (𝐴 − 𝜆𝐼) matrix this must be 0.
So, we got another condition which is leading us to at least get now something out of this
condition determinant of (𝐴 − 𝜆𝐼) is equal to 0. So, when we expand this determinant
because, this 𝜆 is the unknown now A is a given matrix and I is the identity matrix.
So, here this lambda is unknown. So, when we expand this det(𝐴 − 𝜆𝐼) = 0 determinant is
nothing, but this polynomial equation there 𝑐0 𝜆𝑛 + 𝑐1 𝜆𝑛−1 + ⋯ + 𝑐𝑛 = 0, these are the
coefficients and they will be naturally the given when this 𝐴 is given. So, we have this
polynomial equation and then if we solve this equation we will get at most these n roots of
this equation; that means, the 𝑛 values of the 𝜆′𝑠. They may be distinct and they may not
be distinct, they may be real, they may not be real so whatever. So, the solution of this
equation which is called the characteristic equation. So, this equation is called
characteristic equation of the matrix 𝐴 and after solving this equation we can get the
possible lambdas.
Once we have the 𝜆 here then we have this equation (𝐴 − 𝜆𝐼)𝑥 = 0 and for each lambda
we can find the solution of this (𝐴 − 𝜆𝐼)𝑥 = 0. And, note that our lambda which we will
get with this condition that the determinant is 0. So, naturally we will get non-trivial
solution of this (𝐴 − 𝜆𝐼)𝑥 = 0 because, the trivial solution will come when this
determinant is not equal to 0. But, we have we will get our lambda such that this
determinant (𝐴 − 𝜆𝐼) will become 0. And therefore, automatically for these lambdas we
will get the non-trivial solution of these (𝐴 − 𝜆𝐼)𝑥 = 0.
So, here the roots of this characteristic equation are exactly called the eigenvalues because
this lambda is the eigenvalue. So, once we solve this characteristic equation we will get all
the eigenvalues and the eigenvectors of 𝐴 can be determined by this solving the
homogenous system of this linear equation; that means, this (𝐴 − 𝜆𝐼)𝑥 = 0. So, we need
to solve again the system of linear equations. So, from the beginning of this lectures in
linear algebra I am emphasizing again and again on this system of linear equations because
at each and every step finally, we are solving the system of linear equations. So, it is very
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important now and here for each value of this lambda we need to solve the system of
equations.
So, if we have 3 distinct lambdas for example so, for each lambda. So, for 3 times we have
to solve the system of linear equations and they will be different equations because we
have the different lambda. So, this matrix is going to be different and we will get different
eigenvectors. So, we will be talking more on this now and this null space of this 𝐴 − 𝜆𝐼.
So, the null space means these 𝑥 𝐼 and which includes the 0 also because the null space
will also include the 0. So, the null space here called the eigenspace of 𝐴 corresponding to
eigenvalue 𝜆.
So, what is in the null space? In the null space carries all the eigenvectors plus the 0 vector
because the 0 is not the eigenvector. So, here in the null space which is also called the
eigenspace, it contains all eigenvectors including 0 vector because 0 is naturally the
solution of this (𝐴 − 𝜆𝐼)𝑥 = 0 or 0 will be there in the null space. But, 0 is not the
eigenvector because the eigenvector we define as the nonzero, nonzero x the non-trivial
solution of this equation. So, another terminology here which we may use later that is
eigenspace; so, eigenspace is nothing, but the set of all these eigenvectors including the 0
vector ok.
(Refer Slide Time: 14:58)
802
So, let us go through the problem here. The problem number 1 it is a find the eigenvalues
and the eigenvectors of this matrix 𝐴 = [2
4
1
]; it is a very simple example we start with
−1
the evaluation of these eigenvalues and eigenvectors. So, here first we have to write down
the characteristic equation and we need to solve the characteristic equation always to find
the eigenvalues. And, then for each characteristic value or each eigenvalue we have to
solve that system of equation that now we will get in that way the eigenvectors. So, here
the characteristic equation is the determinant of this 𝐴 − 𝜆𝐼.
So, 𝐴 − 𝜆𝐼; that means so, the eigenvalues are nothing, but the this solution of this
characteristic equation. So, we have this 𝐴 − 𝜆𝐼. So, 𝐴 = [2
4
1
] and then we have this
−1
lambda I that will be the determinant at the end. So, lambda I means the lambda 0 0 and
the lambda that is the product of lambda and this identity matrix and this we want to solve
know the determinant here. So, what is the matrix? The matrix here is 2 minus lambda and
then we have here 1 and here we have 4 and then minus 1 and the minus lambda, that is
the determinant here which we can directly always we can read write down for this given
matrix A.
So, the 𝐴 = [2
4
1
]. So, how to write the characteristic equation? Just the determinant of
−1
this matrix subtracting lambda from the diagonal; so, ⟹ (𝜆 − 3)(𝜆 + 2) = 0 ⟹ 𝜆1 =
3, 𝜆2 = −2. So, that is our characteristic equation here.
(Refer Slide Time: 18:00)
803
And its root that will be the eigenvalue; so, eigen values are the 1 eigenvalue here which
we are calling 𝜆1 = 3 because, this is the solution of this equation. And, the another
eigenvalue will be 𝜆2 = −2 and the eigenvector now corresponding to each here. So, first
let us take this 𝜆1 = 3. So, while taking this we have to now form the system of equation
𝐴 − 𝜆𝐼 and times this 𝑥 here so, that will be the system of equations. So, we have to
subtract this 3 from the diagonal entries and that will be our matrix here.
(Refer Slide Time: 18:51)
Meaning so, if we do so, so here (𝐴 − 3𝐼)𝑥 = 0. So, 𝑥 = [1, 1]𝑇 .
𝜆2 = −2: (𝐴 + 2𝐼)𝑥 = 0, 𝑥 = [1, − 4]𝑇
804
(Refer Slide Time: 23:58)
So, another example here find the eigenvalues and these eigenvectors of A which is given
here again a very simple matrix we have taken just for demonstration 𝐴 = [1 1]. So, if
0
1
we write down the characteristic equation. So, that will be |𝐴 − 𝜆𝐼| = 0 so; that means, we
will subtract here from the diagonal entries lambda.
(Refer Slide Time: 24:38)
So, 𝜆1 = 1, 𝜆2 = 1. So, it is a repeated root the case of the repeated root we have only this
one eigenvalue which is repeated 2 times and now corresponding to this eigenvalue we
need to compute the eigenvector. So, the eigenvector corresponding to corresponding to
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these values here 𝜆1 = 1, 𝜆2 = 1 we can again form the system of linear equation that is
𝐴 − 𝜆𝐼.
So, lambda is 1 here so, (𝐴 − 𝐼)𝑥 = 0. So, this system of equation we have to solve now
and what is the system now (𝐴 − 𝐼)𝑥 = 0. So, our matrix will be now the 0 and 1 and 0 and
then here also this 0 that is our system of equation 𝑥 = [1, 0]𝑇 . So, this is the row reduced
echelon form already and we have here this pivot element; that means, this 𝑥2 is not a free
variable, but here this 𝑥1 because the first column does not have a pivot element.
So, this is what we call the free variable and this free variable we can assign any value we
like. So, this 𝑥1 we are assigning some alpha for instance and this 𝑥2 equation that is
already given from the first equation that 𝑥2 is 0. So, we do not have even dependency on
this alpha, the 𝑥2 is always 0 whatever alpha we take that is the freedom we have now. So,
our solution of this system of equation is 𝑥1 and 𝑥2 this 𝛼 and this [1, 0]𝑇 . So, any alpha
we can we can choose of course, here and that is the solution. So, this [1, 0]𝑇 is the is the
generator here for the solution and now that is what we have this [1, 0]𝑇 is one of the
eigenvectors corresponding to 1 and any multiple of this will be also the eigenvector.
So, here we have seen there were two different eigenvalues, but we get only one
eigenvector. So, that is also possible which can be seen here with the help of this simple
example; more on this we will be talking about later that what are the possibilities
corresponding to 1 this eigenvalues how many eigenvectors are possible and so on.
(Refer Slide Time: 27:41)
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So, there is a Cayley-Hamilton theorem which says that every square matrix satisfy its
own characteristic equation, that is another result which directly coming from the
characteristic equation that every square matrix satisfies its own characteristic equation.
So, we know what is the characteristic equation that is the determinant of this 𝐴 − 𝜆𝐼;
meaning this such polynomial equation is the characteristic equation. And, this result says
which we will not go through the proof that this u this matrix itself will satisfy this
characteristic equation; that means, if instead of the 𝜆 if we replace this by 𝐴.
det(𝐴 − 𝜆𝐼) = 0 ⟹ 𝑐0 𝜆𝑛 + 𝑐1 𝜆𝑛−1 + ⋯ + 𝑐𝑛 = 0
So, here 𝐴𝑛−1 and then this and so on up to 𝑐𝑛 this will be in that case 𝑐𝑛 𝐼, 𝐼 will be the
identity matrix of the same size. So, that this Cayley-Hamilton theorem says that every
square matrix also satisfies its characteristic equation and that means, that 𝐴 will also
satisfy this equation. So, it is just the 𝜆 is replaced by this 𝐴 here and with the c n to make
it consistent because, now we are working with the matrices. So, there should be a matrix
here of the same order.
𝑐0 𝐴𝑛 + 𝑐1 𝐴𝑛−1 + ⋯ + 𝑐𝑛 𝐼 = 0
(Refer Slide Time: 29:00)
So, that is the result of the Cayley-Hamilton theorem which we can verify for instance for
11 −6𝑖
] and we will verify this
4𝑖
1
this very simple example which have taken 𝐴 = [
characteristic polynomial by this or de Cayley’s-Hamilton theorem. So, the characteristic
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polynomial characteristic equation we have to write corresponding to this 𝐴 which will be
just by subtracting this 𝜆 from the diagonal entries and now we can write down in terms
of this 𝐼 mean this determinant here. So, that we will get this equation
11 − 𝜆
|
4𝑖
−6𝑖
|=0
1−𝜆
2
. ⟹ 𝜆 − 12𝜆 − 13 = 0
And, now what we will see the Cayley-Hamilton theorem says that this 𝐴2 − 12𝐴 − 13 =
0 this should be just 0. So, this should satisfy the characteristic equation and if we compute
this 𝐴2 that is coming to be here. So, 𝐴2 − 12𝐴 − 13.
So, when we when we determine this one so, here this minus so, these two will be added
and that this will be subtracted; we are getting actually 0 matrix. And, that is what the
characteristic equation and this Cayley-Hamilton theorem says that every square matrix
satisfy its characteristic equation. So, we have just replaced here 𝜆 by this 𝐴 and then we
have seen that this right side instead of the 0 we got the 0 matrix.
𝐴2 − 12𝐴 − 13𝐼 = [
145
48 𝑖
132
−72𝑖
]−[
48 𝑖
25
0
0
=[
(Refer Slide Time: 31:03)
808
0
]
0
−72𝑖
−13
]+[
12
0
0
]
−13
So, that is what the Cayley-Hamilton theorem is; another use of this Cayley-Hamilton
theorem we can quickly look from this example we can use this Cayley-Hamilton theorem
to prove this A inverse. So, again very simple example we have started with this 𝐴 is equal
2 4
]. So, if we write down its characteristic equation. So, again this 𝜆 is subtracted
3 5
to [
from the diagonal and we can simplify this determinant. So, we will get this
2−𝜆
|
3
4
| = 0 ⟹ 𝜆2 − 7𝜆 − 2 = 0
5−𝜆
And, by Cayley-Hamilton theorem we know that this
𝐴2 − 7𝐴 − 2𝐼 = 0
⟹ 𝐴(𝐴 − 7𝐼) = 2𝐼
1
⟹ 𝐴−1 = (𝐴 − 7𝐼)
2
1 −5
= [
2 3
4
]
−2
So, that is the value of this A inverse using the Cayley-Hamilton theorem.
(Refer Slide Time: 33:00)
So, coming to the conclusion what we have done here, we have studied, we have
introduced the eigenvalues and eigenvector and basically this equation was very important
809
this 𝐴𝑥 = 𝜆𝑥. This lambda is the eigenvalue and the corresponding eigenvector will be
given by x. And, we have also studied this Cayley-Hamilton theorem which says that every
square matrix satisfy its own characteristic equation. So, here the right hand side this is a
0 matrix so, the same order having all the entries 0.
(Refer Slide Time: 33:34)
And, these are the references used for this for preparing these lectures.
Thank you very much for your attention.
810
Engineering Mathematics - I
Prof. Jitendra Kumar
Department of Mathematics
Indian Institute of Technology, Kharagpur
Lecture - 47
Eigenvalues & Eigenvectors (Contd.)
Welcome back and this is a lecture number 47, we will continue our discussion on
Eigenvalues and Eigenvectors.
(Refer Slide Time: 00:25)
In particular today we will focus more on evaluation of the eigenvalues and some good
worked out problems will be discussed.
811
(Refer Slide Time: 00:33)
So, let us start with this problem here find eigenvalues and eigenvectors of this matrix,
again a very simple matrix we have taken 2 square root minus 2 and then square root 2 and
again here square root 2 and this one there. So, the characteristic equation we have to write
down as a first step to find the eigenvalues; that means, the determinant of this 𝐴 − 𝜆𝐼 = 0
that is the characteristic equation we have.
⇒ |2 − 𝜆
√2
√2 | = 0
1−𝜆
⇒ 𝜆(𝜆 − 3) = 0
So, here the 𝜆 will be just subtracted from the diagonal entries otherwise this is just the
matrix 𝐴. So, here we have this characteristic equation which tells that there are two
eigenvalues 𝜆1 = 0, 𝜆2 = 3.
812
(Refer Slide Time: 02:23)
So, the eigenvalues here are 𝜆1 = 0, 𝜆2 = 3. So, now, we will compute the eigenvector
corresponding to 𝜆1 , the first eigenvalue that is 0 here. So, we have to solve the system of
linear equation (𝐴 − 𝜆𝐼)𝑥 = 0. So, the 𝜆1 = 0, so this is simply this a so we are solving
basically this 𝐴𝑥 = 0 or we are trying to get the null space of this 𝐴𝑥 = 0 and the generator
of the null space or the basis of the any basis of the null space will be the eigenvector.
So, here getting to this
⇒ [−1
√2
√2 ] [𝑥1 ] = [0]
0
−2 𝑥2
𝑥1
⇒ [𝑥 ] = 𝛼 [√2]
2
1
So, this is the rho reduce a echelon form of the given matrix. So, we have basically this
first equation which says that 2 𝑥1 − √2𝑥2 = 0 and we have one free variable which we
can take whichever we want, so let us take this 𝑥2 is a free variable because as per our
notations here this is the first the p both here. So, this 𝑥1 is not free and then we can take
as 𝑥2 free, but any case in any case we can take any one of them and the other one will
depend on the given chosen number. So, here the 𝑥2 we can choose and then we can get
the 𝑥1 .
813
So, out of these many possibilities we can just pick one vector that we have done here for
−
instance. The x 2 was taken as the free variable alpha and then 𝑥1 is coming [
can also take the vector here the [
−
1
√2].
So, we
1
1
√2],
so this will become 1.
1
1
1
−
So, this vector also we can take [ √2] or we can take [ √2 ]. So, we can multiply by any
1
−1
number here that will be the characteristic a vector here or the eigenvector of the given
equation. So, here the one characteristic vector or the eigenvector we got corresponding to
𝝀𝟏 = 𝟎.
(Refer Slide Time: 05:19)
Similarly, we can also look for the eigenvector corresponding to 𝜆2 = 3. So, again we
have to solve the system of equation (𝐴 − 𝜆𝐼)𝑥 = 0 this time and then, so again this 𝜆 will
be subtracted from this matrix 𝐴.
⇒ [−1
√2
√2 ] [𝑥1 ] = [0]
0
−2 𝑥2
𝑥1
⇒ [𝑥 ] = 𝛼 [√2]
2
1
−1
] & [√2]
√2
1
Eigenvectors: [
814
(Refer Slide Time: 06:51)
So, in this case the eigenvectors corresponding to the 0 was this − 1 and √2 and here we
have this is √2 and 1. Again we can note that these eigen vectors are linearly independent
and as I said in the previous lecture that we will also proof formally this result that
corresponding to distinct eigenvalues we have the eigenvectors, the set is linearly
dependent the set of eigenvectors is linearly independent or the eigenvectors
corresponding to distinct eigenvalues are always linearly independent. So we can check
here also that these two vectors are linearly independent.
(Refer Slide Time: 07:35)
815
3 −2 0
Another problem this is 𝐴 = [−2 3 0] So, here again we want to find the eigenvalues
0
0 5
and eigenvectors of this given matrix which is 3 × 3 now. So, we have to again follow
these steps that first we have to write down the characteristic equation from there we can
get the eigenvalues. So, the characteristic equation will be determinant of this 𝐴 − 𝜆𝐼, so
we have to subtract this lambda from all the diagonal entries which is 3 × 3 and 5 there.
So, that is the characteristic equation of this matrix and then we have to evaluate this
determinant which is not so difficult. So, we have this
3−𝜆
⇒ | −2
0
−2
3−𝜆
0
0
0 |=0
5−𝜆
⇒ (𝜆 − 1)(𝜆 − 5)2 = 0
Eigenvalues:
𝜆1 = 1, 𝜆2 = 𝜆3 = 5
(Refer Slide Time: 10:23)
So, eigenvalues are now 1 and 5, these are the two eigenvalues and we need to compute
the eigenvectors corresponding to each.
816
(Refer Slide Time: 10:35)
So, when we take this 𝜆1 = 1 what do we get this system of equation (𝐴 − 𝜆𝐼)𝑥 = 0 and
3
our a was this here 𝐴 = [−2
0
−2
3
0
0
0]. So, here a minus this 𝜆𝐼 and 𝜆1 = 1 now, so 1 will
5
be subtracted from the diagonal and that will be our matrix here with this
2
⇒ [−2
0
−2
2
0
0 𝑥1
0
0] [𝑥2 ] = [0]
4 𝑥3
0
𝑥1
1
⇒ [𝑥2 ] = 𝛼 [1]
𝑥3
0
817
(Refer Slide Time: 12:45)
𝑥1
1
We get this as the solution here of this equation that [𝑥2 ] = 𝛼 [1]. So, in this case the
𝑥3
0
1
solution of this system is 𝛼 [1] and that is exactly the generator of the null space of this
0
matrix.
So, we can pick any vector, any nonzero vector from this null space. So, we can take four
1
instances [1] or any multiple of it that will be the eigenvector corresponding to this
0
eigenvalue 𝝀𝟏 = 𝟏.
818
(Refer Slide Time: 13:29)
Moving to the other one, so we have this 𝜆2 = 𝜆3 = 5 and then if we set up the system of
equations for this given A which will be in this case because the 5 will be subtracted from
the diagonal entry,
−2
⇒ [−2
0
−2
−2
0
0 𝑥1
0
0] [𝑥2 ] = [0]
0 𝑥3
0
𝑥2 = 𝛼1 & 𝑥3 = 𝛼2 , 𝑥1 = −𝛼1
𝑥1
−1
0
𝑥
[ 2 ] = 𝛼1 [ 1 ] + 𝛼2 [0]
𝑥3
0
1
−1
0
So, here we have these 2 generators [ 1 ]and 4 alpha to the [0]. So, what do you observe
0
1
in this case? That we are getting this is either 1 is satisfying the given differential equation
−1
0
the given system of the question, so here [ 1 ] satisfies the given equation, also [0] is
0
1
satisfying the given equation. And these two are also linearly independent which we can
see from the structure itself, any linear combination like 𝛼1 + 𝛼2 = 0 2 is equal to if we
set to 0 the 𝛼1 , 𝛼2 has to be 0.
819
So, that says easy check for the linear independency in for this case. So, what we are
getting corresponding to this 𝜆 = 5 which was the repeated eigenvalue and we are also
getting here the two linearly independent vectors which satisfy this (𝐴 − 𝜆𝐼)𝑥 = 0. So,
basically corresponding to 𝜆 = 5 we are getting 2 eigenvectors or rather to say 2 linearly
independent eigenvectors, we are getting in this case.
(Refer Slide Time: 17:15)
Now, if we look into another problem here the eigenspace of this matrix that is
2
0
𝐴=[
0
0
1
2
0
0
0
1
2
0
0
0
]
1
2
Characteristic equation:
det(𝐴 − 𝜆𝐼) = 0
2−𝜆
0
⇒|
0
0
1
2−𝜆
0
0
820
0
1
2−𝜆
0
0
0
|=0
1
2−𝜆
(Refer Slide Time: 18:29)
⇒ (2 − 𝜆)4 = 0
𝑬𝒊𝒈𝒆𝒏𝒗𝒂𝒍𝒖𝒆𝒔: 𝜆 = 2, 2, 2, 2
(Refer Slide Time: 19:29)
So, we got this as our system of equation and here we do not have to do anything to get
the reduced form because we can clearly see this structure here this is already the reduced
form the reduced echelon form.
821
Here we have only one free variable at though the 2 was repeated several times the 2 was
repeated 4 times in this particular example, but what we are getting here we are getting
only one linearly independent eigenvector. Whereas, in the previous example the
eigenvalue was repeated two times and we were also getting two linearly independent
solution of this equation, but now though the 2 was repeated four times, but we are getting
only 1 eigenvector corresponding to this 4 times repeated eigenvalue 2.
So, what we want to discuss now with this example that that anything is possible just based
on this eigenvalue whether it’s repeated several times we cannot claim anything about I
mean directly looking at the eigenvalue, we cannot claim that how many linearly
independent eigenvectors we will get as this is the case here the eigenvalue was repeated
4 times, but we are getting only one linearly independent eigenvector and that is this
1
0
[ ]in this case.
0
0
Eigenvectors (𝜆 = 2):
2
0
𝐴=[
0
0
1
2
0
0
0
1
2
0
0
0
]
1
2
(𝐴 − 𝜆𝐼)𝑥 = 0
0
0
⇒[
0
0
1
0
0
0
0
1
0
0
822
0 𝑥1
0 𝑥2
][ ] = 0
1 𝑥3
0 𝑥4
(Refer Slide Time: 22:53)
𝑥1
1
𝑥2
0
⇒ [𝑥 ] = 𝛼 [ ]
0
3
𝑥4
0
Thus a basis of eigenspace:
{(1,0,0,0)𝑇 }
(Refer Slide Time: 23:35)
823
So, the another example where we will look for the eigenvalues and the eigenvectors of
this
matrix
𝐴=[
1
−2
1
].
3
So,
here
if
we
compute
this
det(𝐴 − 𝜆𝐼) = 0, so what will happen?
1−𝜆
⟹|
−2
1
|=0
3−𝜆
⟹ 𝜆2 − 4 𝜆 + 5 = 0
⟹𝜆 =2±𝑖
but we are getting here 2 complex values for the eigenvalues in this case.
(Refer Slide Time: 27:11)
So, that again shows that the matrix entries may be real we can have a real matrix, but still
we may get the complex roots of this characteristic equation meaning the eigenvalues may
be complex. So, a real matrix may have complex eigenvalues that is the remark and then
eigenvectors corresponding to now this 𝜆1 = 2 + 𝑖 the first eigenvalue. So, we have 1 eigen
value 2 + 𝑖 another one is 2 − 𝑖 and we will see later on that these eigenvalues will always
appear in conjugate form as the root always appears there and not only that we will have
something more interesting for the eigenvectors corresponding to these conjugate
eigenvalues.
So, if we take this 2 + 𝑖 and again the same system of equation this
824
(𝐴 − 𝜆𝐼)𝑥 = 0
𝑥1
0
1
] [𝑥 ] = [ ]
0
3−𝜆 2
1−𝜆
−2
⟹[
(Refer Slide Time: 28:19)
Eigenvector corresponding to 𝜆1 = 2 + 𝑖:
1−𝜆
−2
𝐴=[
−1 − 𝑖
−2
=[
1
]
3−𝜆
1
] , 𝑅2 ← 𝑅2 − 𝑅1 × (1 − 𝑖)
1−𝑖
−1 − 𝑖
~[
0
1
]
0
(1 + 𝑖)𝑥1 = 𝑥2 ⟹ 𝑥2 = (1 + 𝑖) & 𝑥1 = 1
A eigenvector corresponding to 𝜆1 :
[
1
]
(1 + 𝑖)
Eigenvector corresponding to 𝜆1 = 2 − 𝑖:
1−𝜆
−2
𝐴=[
−1 + 𝑖
1
−1 + 𝑖
1
]=[
]~[
−2
1+𝑖
0
3−𝜆
825
1
] , 𝑅2 ← 𝑅2 − 𝑅1 × (1 + 𝑖)
0
(1 − 𝑖)𝑥1 = 𝑥2 ⟹ 𝑥2 = (1 − 𝑖) & 𝑥1 = 1
A eigenvector corresponding to 𝜆1 :
1
[
]
(1 − 𝑖)
(Refer Slide Time: 30:07)
So, the eigenvector or one of the eigenvectors here we are getting exactly the conjugate of
what we have got for the eigenvector corresponding to its conjugate there. So, that is a
nice result here in general we have this then 𝐴 is a real matrix and has complex eigenvalues.
Then the conjugate ; then the conjugate this 𝜆̅ , so 𝜆 is an eigenvalue, then the conjugate
will be also the eigenvalue and, but that is natural which is coming from this characteristic
equation. So, the conjugate will be always there as the root.
So, here and what is interesting? The interesting is that this we have from this 𝐴𝑥 = 𝜆 𝑥
if we take the conjugate both the side what we will get because 𝐴 is a real matrix, so 𝐴𝑥̅ =
𝜆̅ 𝑥̅ . So, again we have this equation, the eigenvalues eigenvector equation is satisfied for
this 𝑥̅ , the conjugate of x and the right hand side is just the 𝜆̅ 𝑥̅ .
So, what this tells now that this 𝑥̅ here the conjugate of 𝑥 will be the eigenvector
corresponding to this conjugate of 𝜆̅. So, indeed we do not have to compute this since we
know this result from this calculation that once we have eigenvector corresponding to one
complex value of this 𝜆 and its conjugate will be will be the eigenvector of the other
826
conjugate eigenvalue. So, here that is the interesting result we have about the conjugate
roots or about the conjugate eigenvalues complex conjugate here.
(Refer Slide Time: 33:41)
So, in this case the eigenvectors corresponding to distinct eigenvalues are linearly
independent this is what we have observed at least for the calculations we had in numerical
calculations and also what we have observed here that a real matrix may have a complex
eigenvalues. So, that is interesting here and both the eigenvalues and eigenvectors are
correct complex conjugate pair.
So, the once we have the complex value as the eigenvalue its conjugate will be there and
not only that the eigenvectors will be also the conjugate pairs. So, that is interesting result
we have seen and that is true in general not for the example we have just shown.
827
(Refer Slide Time: 34:27)
So, these are the references used. Thank you for your attention.
828
Engineering Mathematics - I
Prof. Jitendra Kumar
Department of Mathematics
Indian Institute of Technology, Kharagpur
Lecture - 48
Eigenvalues & Eigenvectors (Contd.)
So, welcome back and this is lecture number 48 and today we will talk about the properties
of Eigenvalues and Eigenvectors.
(Refer Slide Time: 00:26)
Properties of Eigenvalues and Eigenvectors:
Let 𝜆 be an eigenvalue of 𝐴 and 𝑥 be its corresponding eigenvector. Then,
 𝛼𝐴 has eigenvalue 𝛼𝜆 and corresponding eigenvector is 𝑥.
𝐴𝑥 = 𝜆𝑥 ⇒ (𝛼𝐴)𝑥 = (𝛼𝜆)𝑥
 𝐴𝑚 has eigenvalues 𝜆𝑚 and corresponding eigenvector is 𝑥 for any positive
integer 𝑚.
𝐴𝑥 = 𝜆𝑥 ⇒ 𝐴(𝐴𝑥) = 𝐴(𝜆𝑥)
⇒ 𝐴2 𝑥 = 𝜆(𝐴𝑥) = 𝜆(𝜆𝑥) = 𝜆2 𝑥
829
⇒ 𝜆2 is eigenvalue of 𝐴2
(Refer Slide Time: 04:17)
Now the next property so here we have seen this lambda square is the eigenvalue of A.
Now, the two eigenvectors of a square matrix A corresponding to two distinct eigenvalues
of A are linearly independent. So, what we will prove here that two eigenvectors
corresponding to two distinct eigenvalues are always linearly independent. This
observation we already have seen for numerical examples, but we can prove here for more
general matrix for any matrix we can prove this result theoretically.
So, what do we consider now so
Let 𝑥1 , 𝑥2 be the eigenvectors of 𝐴 corresponding to two distinct eigenvalues
𝜆1 , 𝜆2 respectively.
Then 𝐴𝑥1 = 𝜆1 𝑥1 & 𝐴𝑥2 = 𝜆2 𝑥2 .
Consider 𝑐1 𝑥1 + 𝑐2 𝑥2 = 0,
𝑐1 , 𝑐2 ∈ ℝ.
Then, 𝑐1 𝐴𝑥1 + 𝑐2 𝐴𝑥2 = 0
⇒ 𝑐1 𝜆1 𝑥1 + 𝑐2 𝜆2 𝑥2 = 0
830
(Refer Slide Time: 07:54)
With this so we have two equations now this
𝑐1 𝑥1 + 𝑐2 𝑥2 = 0
𝜆1 𝑐1 𝑥1 + 𝜆2 𝑐2 𝑥2 = 0 ⟹ 𝜆1 𝑐1 𝑥1 + 𝜆1 𝑐2 𝑥2 = 0
(𝜆2 − 𝜆1 ) 𝑐2 𝑥2 = 0 ⇒ 𝑐2 = 0, since (𝜆1 − 𝜆2 ) ≠ 0, 𝑥2 ≠ 0
𝑐1 𝑥1 + 𝑐2 𝑥2 = 0, ⟹ 𝑐1 = 0, since 𝑥1 ≠ 0
Hence, 𝑥1 and 𝑥2 are linearly independent.
So, this was the case when we have considered two distinct eigenvalues, but we can also
generalize this case for more eigenvalues for instance here, we have eigenvalues
𝑥1 , 𝑥2 , … , 𝑥𝑟 are corresponding to our distinct eigenvalues here. So, these are the
eigenvectors corresponding to these eigenvalues 𝜆1 , 𝜆2 , … , 𝜆𝑟 respectively and in that
case also we can use the similar trick to prove that these eigenvectors are linearly
independent. So, we have this very nice result that corresponding to distinct eigenvalue
the eigenvectors are linearly independent.
831
(Refer Slide Time: 10:53)
So, another result we have they said if 𝑥 is an eigenvector of 𝐴 corresponding to the
eigenvalue 𝜆, then this 𝑘𝑥 is also the eigenvector corresponding to the same eigenvalue 𝜆.
So, this we have also seen before that for the given eigenvector you can multiply by any
constant and that will also remain the eigenvector and this is what we will see here, and
more formally theoretically that this is true for any matrix.
So, here 𝐴𝑥 = 𝜆𝑥 that is a relation, so it tells us that 𝜆 is one of the eigenvalue and the
corresponding eigenvector is 𝑥 and then this 𝑘 time, so we multiplied the equation by k
here both the sides and then we have ⇒ 𝑘(𝐴𝑥) = 𝑘(𝜆𝑥) and then we can combine this like
𝐴(𝑘𝑥) = 𝜆(𝑘𝑥).
So, we have this relation that 𝐴 some vector here is equal to 𝜆 the same vector kx which
tells us that this kx is the eigenvector again, if the x was the eigenvector the k times x is
also the eigenvector for any this k and nonzero scalar. Here if x is the eigenvector of the
matrix A, then x cannot correspond to more than one eigenvalue of A. So, another
important result that this eigenvector is like a unique.
So, if you have an eigenvector corresponding to let us say the lambda, then this x cannot
correspond to any other eigenvalue, so it is a unique in that sense. So, here if we assume
that for a given matrix here we. have
𝐴𝑥 = 𝜆1 𝑥
&
832
𝐴𝑥 = 𝜆2 𝑥
⇒ 𝜆1 𝑥 = 𝜆2 𝑥
⇒ (𝜆1 − 𝜆2 )𝑥 = 0
⇒ 𝜆1 = 𝜆2 , since 𝑥 ≠ 0
So, these two eigenvalues have to be the same eigenvalues you cannot have 2 distinct
eigenvalues which can correspond to the same eigenvector.
(Refer Slide Time: 14:07)
(𝐴 − 𝑘𝐼) has eigenvalue (𝜆 − 𝑘) and corresponding eigenvector is 𝑥
𝐴𝑥 = 𝜆𝑥
⇒ 𝐴𝑥 − 𝑘𝐼𝑥 = 𝜆𝑥 − 𝑘𝑥
⇒ (𝐴 − 𝑘𝐼)𝑥 = (𝜆 − 𝑘)𝑥
So, this is another result that if we have this new matrix which is just the 𝐴 − 𝑘𝐼, then we
know about the eigenvalues from the eigenvalues of A. And this 𝐴−1 again important here
that if it exists of course, then only we are talking about this result. So, 𝐴−1 exists for a
matrix, then this 𝐴−1 will have eigenvalue here or eigenvalues one over lambda.
1
 𝐴−1 (if it exists) has eigenvalue and corresponding eigenvector is 𝑥
𝜆
833
𝐴𝑥 = 𝜆𝑥
⇒ 𝐴−1 𝐴𝑥 = 𝐴−1 𝜆𝑥
⇒ 𝐴−1 𝑥 = (1/𝜆)𝑥
So, 𝐴 and 𝐴𝑇 have the same eigenvalues, 𝐴 and 𝐴𝑇 have same eigenvalues and which we
can again easily see because the determinant of 𝐴 − 𝜆𝐼 that is the characteristic equation
which actually gives the eigenvalues
det (𝐴 − 𝜆𝐼) = det (𝐴 − 𝜆𝐼)𝑇
= det(𝐴𝑇 − 𝜆𝐼)
So, here the result is that , 𝐴 and 𝐴𝑇 have the same will have the same eigenvalue. So, , 𝐴
and 𝐴𝑇 have same eigenvalue, so that is another important result which easily we can find
out with the help of this determinant property.
(Refer Slide Time: 20:02)
Next theorem, so here the characteristic roots I mean the eigenvalues so sometimes we
also call the characteristic roots. So, the eigenvalues of Hermitian matrix are real. So, what
is the Hermitian matrix? So, we know that 𝐴 is Hermitian when 𝐴∗ = 𝐴 meaning the
conjugate transpose, 𝐴∗ means here that we are taking the transport and also we are taking
the complex conjugate of the matrix 𝐴.
834
So, this complex conjugate of this transpose is equal to 𝐴, then we call that 𝐴 is Hermitian
matrix. So, what is this result that for Hermitian matrices the roots are real because what
we have also seen that though the matrix having all real entries, but we can get the
characteristic roots as complex number we have seen in previous lectures one of the
examples where we had a very simple 2 by 2 matrix with real entries. And its characteristic
polynomial or I mean the characteristic roots the eigenvalues were non real so the complex.
So, here we have at least the results for the Hermitian matrix that all the characteristic roots
are real in this case. So, if 𝜆 be a characteristic root of A and lambda the corresponding
eigenvector and then we will show that this 𝜆 has to be real, how? So, we have this 𝐴𝑥 =
𝜆𝑥 that is the property of the relation of the eigenvalues eigenvector again.
𝐴𝑥 = 𝜆𝑥
⇒ 𝑥 ∗ 𝐴𝑥 = 𝑥 ∗ 𝜆𝑥 = 𝜆𝑥 ∗ 𝑥
Taking conjugate transpose on both sides
(𝑥 ∗ 𝐴𝑥)∗ = (𝜆𝑥 ∗ 𝑥)∗
⇒
𝑥 ∗ 𝐴∗ 𝑥 = 𝜆̅𝑥 ∗ 𝑥
⟹ 𝜆𝑥 ∗ 𝑥 = 𝜆̅𝑥 ∗ 𝑥
⇒ (𝜆 − 𝜆̅) 𝑥 ∗ 𝑥 = 0
⇒ 𝜆 = 𝜆̅,
since 𝑥 ∗ 𝑥 ≠ 0.
⇒ 𝜆 is real.
So, here the 𝜆 must be equal to 𝜆̅ because this quantity cannot be 0, so this has to be 0. So,
here what we have seen that the 𝜆 = 𝜆̅ and that is what we want to see here that the 𝜆′ 𝑠 are
835
real. So, if 𝜆 is the eigenvalue of Hermitian matrix, then the 𝜆 = 𝜆̅ meaning it is a real
number, it cannot be a complex number ok.
(Refer Slide Time: 24:39)
So, another similarly we can prove these following results which have the similar lines of
the proof which we have just done, that the eigenvalues of this real symmetric matrix are
also real that is what we can also do and the eigenvalues of real a skew symmetric matrix.
So, here we have; that means, this a transfer is minus of the A. So, here this real a skew
symmetric matrix are purely imaginary.
So, this is also interesting here that eigenvalues of such matrices are either purely
imaginary or 0 that is the two possibilities, which again if you follow the earlier proof we
can also do this one and the eigenvalues of the a skew Hermitian matrix. So, for Hermitian
matrices we have seen, but now there is a skew Hermitian matrix; that means, this A star
is equal to minus A.
So, for those cases the eigenvalues are purely imaginary or 0 again. So, these are the
consequence of the earlier proof which we can easily see here.
836
(Refer Slide Time: 25:50)
Now, another important result that the eigenvalues of unitary matrix are of unit modulus,
so this also we can prove in general that if you have this unitary matrix; that means, this
𝐴∗ 𝐴 = 𝐼 , so such matrices are called the unitary matrix.
So, if we have unitary matrix then we will prove now the eigenvalues the modulus of the
eigenvalues is 1; that means, if you consider here 𝐴𝑥 = 𝜆𝑥 and then taking the complex
conjugate here again (𝐴𝑥)∗ = (𝜆𝑥)∗ what we will get so this again we will use this property.
So,
𝑥 ∗ 𝐴∗ = 𝜆̅𝑥 ∗
(𝑥 ∗ 𝐴∗ )(𝐴𝑥) = (𝜆̅𝑥 ∗ )(𝜆𝑥)
⇒ 𝑥 ∗ (𝐴∗ 𝐴)𝑥 = 𝜆𝜆̅𝑥 ∗ 𝑥
⇒ 𝑥 ∗ 𝑥(1 − 𝜆̅𝜆) = 0
⇒ 𝜆̅𝜆 = |𝜆|2 = 1, as 𝑥 ∗ 𝑥 ≠ 0
So, meaning this we got that this absolute value of 𝜆 has to be 1. So this unitary matrix the
eigenvalues of the unitary matrix are of unit modulus. Same results we can also use for the
orthogonal matrices because they are also having the same property a transpose A is equal
837
to I. So, for orthogonal matrices also now we can prove thus the similar all absolutely all
same steps here and we can again prove the there eigenvalues are also of unit modulus.
(Refer Slide Time: 29:00)
The location of the eigenvalues now what we have just seen in previous slide. So, if you
have a skew Hermitian matrix their eigenvalues are imaginary, purely imaginary here the
unitary matrix they lie on this modulus 1 and for the Hermitian matrix or the symmetric
matrix the values are sitting on the real axis, so meaning they are the real numbers.
So, here for a skew Hermitian and exclusive metric the same thing unitary and orthogonal
we have the same result, that they are of a unit modulus for Hermitian and symmetric we
have also the same result for both that they are the real entries.
838
(Refer Slide Time: 29:48)
Getting to the conclusion, so we have seen several properties of this eigenvalues and
eigenvectors of a matrix, we have considered different; different types of matrices where
we can tell about whether the eigenvalues will be real imaginary if your imaginary you are
0.
So, here in all these properties the simple idea was to use this 𝐴𝑥 = 𝜆𝑥 and we played with
this equation only to prove all these properties. And now they can be used now without
doing all these numerical calculations we can compute directly also with the help of these
properties.
839
(Refer Slide Time: 30:26)
So, these are the references we have used to prepare these lectures.
Thank you for your attention.
840
Engineering Mathematics - I
Prof. Jitendra Kumar
Department of Mathematics
Indian Institute of Technology, Kharagpur
Lecture - 49
Eigenvalues & Eigenvectors (Contd.)
(Refer Slide Time: 00:22)
Welcome back, so this is lecture number 49 and we will be talking about today the
geometric and algebraic multiplicity and the similarity of matrices.
(Refer Slide Time: 00:29)
841
So, what is the algebraic multiplicity? So, algebraic multiplicity of lambda as a root of the
its a multiplicity of lambda as a root of the characteristic equation. And the geometric
multiplicity is nothing but, the dimension of the eigenspace of lambda; that means, the
number of linearly independent eigenvectors corresponding to an eigenvalue lambda.
So, these are the two numbers which we will now use for telling about the multiplicity of
this lambda, because we have seen in several examples the characteristic, roots all the
eigenvalues were repeated. So, that we can now quantify with the help of this algebraic
multiplicity; so, algebraic multiplicity if for instance one root is repeated 3 times. So, then
it is algebraic multiplicity of that particular root is 3 and the geometric multiplicity will be
the dimension of the eigenspace or the number of linearly independent vectors we have
corresponding to that particular eigenvalue lambda.
So, with these two classification we will move further, but before that there is a note here,
that this geometric multiplicity is always less than or equal to the algebraic multiplicity,
that is a important result which one can formally prove. But, it is it requires little more
knowledge of a diagonalization etcetera.
So, we will not prove this result now, but we will keep in mind that this geometric
multiplicity is always less than or equal to the algebraic multiplicity. Meaning that for
example, one a particular root; one particular eigenvalue is repeated 3 times than the
corresponding geometric multiplicity meaning they are number of linearly independent
eigenvectors cannot be more than 3, they have to be less than or equal to 3.
842
(Refer Slide Time: 02:31)
So, first we will see with the help of many examples that what are the situations arises
here. So, in this case we find the eigenvalues and the eigenvectors of this matrix 𝐴 =
6
[−2
2
−2
3
−1
2
−1]. So, for this eigen, for this matrix we will compute the eigenvalue and
3
eigenvectors we have already computed for several matrices in the last lecture. So, we are
on now the familiar with the computation of the eigenvalues.
So, here first we need to write down the characteristic equation for this given matrix which
is the det(𝐴 − 𝜆𝐼) = 0 and for this matrix we can compute this determinant here. So, the
determinant would be like the
⇒ (2 − 𝜆)(𝜆 − 2)(𝜆 − 8) = 0
So, I skip this portion here because in this lecture that is not important, we have already
seen for several examples in the last lecture.
843
(Refer Slide Time: 04:18)
So, what we have? We have this characteristic equation of this matrix here as this ⇒
(2 − 𝜆)(𝜆 − 2)(𝜆 − 8) = 0. So, what we observe now that there are two distinct eigenvalues
and one is repeated 2 times so; that means, the 𝜆 = 2, 2, 8. So, this eigenvalue 2 is repeated
two times and this 8 is repeated one times, and exactly that is what we have discussed
about this algebraic multiplicity.
So, the algebraic multiplicity of this 𝜆 = 2this eigenvalue 2 is because it is repeated 2
times here. So, that I the algebraic multiplicity 𝜆 = 2 is 2 and the algebraic multiplicity of
8 because this is repeated only once. So, here the algebraic multiplicity of this 𝜆 = 8 is 1.
So, this is how the algebraic multiplicity and the geometric algebraic multiplicity is defined
to define the geometric multiplicity corresponding; to 𝜆 = 2 or 𝜆 = 8 , we need to get the
Eigen space of these vectors of these eigenvalues.
844
(Refer Slide Time: 05:32)
So, the eigenvector corresponding to this 𝜆 = 8. So, remember this 𝜆 = 8was repeated
once. Though, we have already the result that the eigen the geometric multiplicity cannot
be more than 1 now in this case, because the algebraic multiplicity of this 𝜆 = 8 is 1. So,
without calculation of the eigenvectors as well we can claim that the geometric multiplicity
will be 1, because definitely there will be one linearly independent eigenvector
corresponding to this 𝜆 = 8.
So, there cannot be two linearly independent eigenvectors, that was that result says where
we have that these algebraic multiplicity is always bigger than the geometric multiplicity.
So, here we know though beforehand that this there will be there cannot be two linearly
independent vectors, it has to be only one because the algebraic multiplicity of this 𝜆 = 8
is 1.
And it had it just must to have at least 1 eigenvectors because that is what the foundation
says so, we have already this (𝐴 − 𝜆𝐼)𝑥 and the determinant is equal to 0 we have nontrivial solution always. For this equation here (𝐴 − 𝜆𝐼)𝑥 = 0. So, there will be definitely
one linearly independent eigenvector, but there cannot be two this is what we will see in
this case as well.
So, here this is (𝐴 − 𝜆𝐼)𝑥 ; so, this 𝜆 = 8 here was subtracted from the diagonal entries of
𝑥1
0
𝑥
A and then we have [ 2 ] and the right hand side this [0] vector.
𝑥3
0
845
−2 −2 2 𝑥1
0
⇒ [−2 −5 −1] [𝑥2 ] = [0]
2 −1 −5 𝑥3
0
−2 −2 2 𝑥1
0
𝑥
⇒ [ 0 −3 −3] [ 2 ] = [0]
0
0
0 𝑥3
0
So, this is the situation, this is the row reduced echelon form for the system of equations
for this matrix.
(Refer Slide Time: 08:21)
And now, we observe here that this is the pivot element and here also we have the pivot
element. So, the first two columns have pivot element that third column does not have
pivot element. So, that corresponds to this 𝑥3 component of this vector and which we can
take as the free variable. So, there will be only one free variable which was clear from
there also because the algebraic multiplicity was one and corresponding to that we cannot
get two free variables. So, the number of free variables tells about the number of linearly
independent eigenvectors.
So, here we cannot have two linearly independent eigenvector. So, we know in advance
that there will be only one free variable in this case, there cannot be two free variables. So,
this 𝑥3 is the free variable which we can choose again as 𝛼; having that 𝛼 we can compute
the 𝑥1 and 𝑥2 in terms of 𝛼.
846
(Refer Slide Time: 09:20)
𝑥1
2
⇒ [𝑥2 ] = 𝛼 [−1] ,
𝑥3
1
𝛼 ≠ 0, 𝛼 ∈ ℝ.
So, that is the only one eigenvector which is linearly independent, any other eigenvector
which we get out of taking this value alpha where they are the dependent eigenvectors on
2
this [−1]. So, in this case we got only one linearly independent eigenvector and therefore,
1
we say that the geometric multiplicity. The geometric multiplicity that is the number of
linearly independent eigenvectors corresponding to a given eigenvalue here the 𝜆 = 8 is
1. So, the geometric multiplicity of this 𝜆 = 8 is 1. So, here that is the number of linearly
independent eigen vectors.
847
(Refer Slide Time: 10:48)
When we come to this eigenvalues 𝜆 is an; eigenvalue 𝜆 = 2 and remember it was repeated
2 times meaning the algebraic multiplicity of this 𝜆 = 2 was 2. So, in this case we have
the possibility that the corresponding eigenvectors the corresponding linearly independent
eigenvectors there may be 2, I mean at most 2, but there may be 1 as well, we do not know
now in advance we have to compute them.
Because, looking at this eigenvalue we cannot just tell how many eigenvectors will be
linearly independent corresponding to a given eigenvalue, but what we can tell now
because the multiplicity of this 2 was 2 or the algebraic multiplicity was 2 and we know
that the geometric multiplicity will be less than or equal to 2. So, we know now that the
number of linearly independent eigenvectors could be 1 or it could be 2 also now in this
case.
So, like let us compute this. So, this (𝐴 − 𝜆𝐼)𝑥 = 0 So, we get this equation the system of
linear equation and then by reducing to this echelon form.
0
4 −2 2 𝑥1
𝑥
⇒ [−2 1 −1] [ 2 ] = [0]
0
2 −1 1 𝑥3
−2
⇒[ 0
0
1
0
0
−1 𝑥1
0
0 ] [𝑥2 ] = [0]
0 𝑥3
0
848
We have written got this row reduced echelon form of this system of linear equation. And
then and now we can identify that how many linearly independent eigenvectors we are
going to have in this particular case.
(Refer Slide Time: 13:10)
So, here this is the pivot element which is minus 2 in this case and the column number two
does not have a pivot here also we do not have pivot. So, there is only one pivot that is in
the column number 1.
So, here 𝑥2 and the 𝑥3 ; 𝑥2 and 𝑥3 will be will be free variables so, there will be free
variables now free variables. So, we can assign any value to them; that means, we are
going to have now two linearly independent eigenvectors because a number of free
variables decide exactly how many linearly independent eigenvectors we will get.
849
(Refer Slide Time: 13:55)
So, in this case we will get two linearly independent eigenvectors, and that is what we
write.
1
1
𝑥1
−
⇒ [𝑥2 ] = 𝛼1 [2] + 𝛼2 [ 2]
1
0
𝑥3
0
1
So, we got this two linearly independent eigenvector. So, this one and this one, one can
check where they are linearly independent.
So, corresponding to those 𝜆 =2 because it was repeated this 2 times the algebraic
multiplicity was 2, this algebraic multiplicity of this was 2 and we also got now the
geometric multiplicity. So, the geometric multiplicity is also 2 in this case. So, we have
the algebraic multiplicity 2 and as well as the geometric multiplicity 2; geometric cannot
be more than the algebraic one again, but in this case we got at least the equality.
850
(Refer Slide Time: 15:01)
So, the geometric multiplicity of this 𝜆 =2 is 2, because we have two linearly independent
eigenvectors corresponding to this lambda is equal to 2 ok.
(Refer Slide Time: 15:13)
So this example 2, where we determine the eigenvalues and eigenvectors of this 𝐴 =
2
[4
0
0 0
2 0] and in this case one can compute easily the eigenvalues will be the diagonal
0 3
entries because it is a lower triangular matrix and for the triangular matrices, we have all
the eigenvalues sitting on the diagonals here.
851
So, we have this 𝜆 = 2, 2, 3 there these are the eigenvalues and the eigenvalues of a
triangular matrix are always it is a diagonal element; so, we have these eigenvalues 𝜆 =
2, 2, 3. The eigenspace now we will compute for 𝜆 = 2. Again I mean here, the if we want
to know the algebraic multiplicity so it is 2 for 2 and the algebraic multiplicity of 3 is 1.
So, here the eigenspace you want to compute now to get the geometric multiplicity.
So, here the eigenspace; so, 𝐴 − 𝜆𝐼 so here 2 will be subtracted from the diagonal entry.
So, we will get 0 there, 0 there, 1 there. So, this is the now the matrix (𝐴 − 𝜆𝐼)𝑥 = 0. So,
what do we see here, we can actually just take this we can interchange the row and then
we have this echelon form; row reduced echelon form the 0 we can bring to the bottom if
we like.
−1
[4
0
0 0 𝑥1
0
−1 0] [𝑥2 ] = [0]
0 0 𝑥3
0
So, we can easily convert to this echelon form here and then we will see there will be 2;
there will be 2 pivot elements here. So, this will be the pivot element and this will be also
the pivot element when we convert into this echelon form. And this middle one so, here
the first column will have a pivot and the third column has a pivot and this 𝑥2 is going to
be the free variable. So, this 𝑥2 is going to be the free variable; that means, only one free
variable and we will get only one linearly independent eigenvector.
𝑥1
0
𝑥
⇒ [ 2 ] = 𝛼 [0].
𝑥3
1
but now we got the geometric multiplicity as 1.
So, not surprising as I said before that for given eigenvalues we cannot predict in advance
at how many eigenvalue vectors will be linearly independent so, we have to compute them.
What we know from that result that algebraic multiplicity is less than or equal to the, or
the geometric multiplicity is less than equal to the algebraic multiplicity that the algebraic
multiplicity of this 2 was 2. So, we know that there will be at most 2 linearly independent
eigenvectors. There cannot be three linearly eigen linearly independent eigenvectors for
example, in this case, but we do not know whether there will be 2 or there will be 1. So,
what we have observed? In the previous example though the algebraic multiplicity was 2
852
and the geometric multiplicity was also 2. In this case we have the algebraic multiplicity
2, but geometric multiplicity is just 1 in this case.
(Refer Slide Time: 18:57)
So, here the geometric multiplicity is 1, because we have 1 linearly independent
eigenvector and the algebraic multiplicity of 2 is 2 because this 2 was repeated 2 times.
(Refer Slide Time: 19:11)
Coming to the eigenspace of this 𝜆 = 3. So, we know already that there would be only
one. So, in this case we know that there will be only one free variable definitely because
853
the algebraic multiplicity is 1. So, we cannot have more than 1 linearly independent
eigenvector.
So, here if we compute this (𝐴 − 𝜆𝐼)𝑥 = 0. So, we have this and then when we solve this
system. So, we will observe that there are 2 pivots in this case, when we just we can just
make this to 0 and then this will become also a pivot because this will not be 0 in that case.
So, you will have 2 pivot elements and this 𝑥3 will be the free variable in this case.
So, therefore, this alpha is corresponding to 𝑥3 and this 𝑥2 will be 0 and 𝑥1 will be also 0
0
from this structure of the matrix. So, we will get the solution 𝛼 [0] and as expected or there
1
is only one linearly independent eigenvector corresponding to this 𝜆 = 3. So, the
geometric multiplicity of this 𝜆 = 3 is 1 and the algebraic multiplicity of this 𝜆 = 3 was
also 1 in this case.
(Refer Slide Time: 20:35)
Another example we will find the dimension of this eigenspace of this lambda is equal to
1
0 0
this very special matrix here 𝐴 = [1 1 1]. So, in this case again we need to write the
0
0 1
characteristic equation. So; that means, (𝐴 − 𝜆𝐼)𝑥 = 0;
So, what we will observe in this case that the characteristic equation is (𝜆 − 1)3 = 0. So,
we have these 3 roots; so, 1, 1, 1. That means, this algebraic multiplicity of this lambda is
854
equal to 1 is 3 now. So, we have an example where all these we have the same eigenvalues,
but repeated three times. When we compute the eigenspace here meaning we have to
compute the eigenvector so with this equation (𝐴 − 𝜆𝐼)𝑥 = 0. So, what will happen in this
case? That when we take this (𝐴 − 𝜆𝐼)𝑥 = 0 from the diagonal entries.
0
⇒ [1
0
0
0
0
0 𝑥1
0
𝑥
1] [ 2 ] = [0]
0 𝑥3
0
So, this diagonals will be also 0 and we have this very a simple example and in this case,
there will be only 1 pivot. So, this first column will have pivot and the second and the third
one will be the free variables. So, this is going to be the pivot element and then nothing
else. So, we have the free variable we have the free variable. So, there are two free
variables, meaning 2 linearly independent eigenvectors.
(Refer Slide Time: 22:12)
So, here for ⇒ 𝑥2 = 𝛼1 , 𝑥3 = 𝛼2 , 𝑥1 = −𝛼2 .
So, we have
𝑥1
0
−1
𝑥
⇒ [ 2 ] = 𝛼1 [1] + 𝛼2 [ 0 ]
𝑥3
0
1
so, eigenvalue 1 here, which was repeated 3 times. So, the algebraic multiplicity of this
lambda is equal to 1 was 3 and the geometric multiplicity of this lambda is equal to 1 is 2.
855
So, they are 2 linearly independent vector. So, the dimension of this eigen space is 2 or the
geometric multiplicity of this lambda is equal to 1 is 2, in this case. So, again though it
was repeated 3 times, but we got only this dimension as 2 not 3 and not 1, but the possible
values here could be 3, it could be 2 as this is the case here, but it can be 1 as well.
(Refer Slide Time: 23:49)
In this example again we will take this identity matrix. So, very simple to evaluate so we
have, we want to find the dimension again of the eigenspace of this 𝐴 is equal to this
identity matrix. And if we write down its characteristic equation, we will get this
𝟏
(𝜆 − 1)3 = 0; so, again we have this [1] which the algebraic multiplicity of this eigenvalue
1
is just 3 now.
So, corresponding to this 1 so; algebraic multiplicity is 3 and if we compute the geometric
multiplicity now that is interesting. So, the eigenspace will be computed by this
(𝐴 − 𝜆𝐼)𝑥 = 0 and therefore, when we subtract from the diagonal entries this eigenvalue 1.
𝑥1
𝑥
So, what we will get this 0 matrix here and [ 2 ] is equal to again the 0 matrix.
𝑥3
0
⟹ [0
0
0
0
0
856
0 𝑥1
0
0] [𝑥2 ] = [0]
0 𝑥3
0
So, what do we see now in this case? That there is no there is no pivot here; there is no
𝑥1
𝑥
pivot here and all the variables [ 2 ]they are the free variables. So, we can choose, we can
𝑥3
𝑥1
𝑥
assign any value to [ 2 ]they are free here and that is a very special case which we have
𝑥3
just seen now, that we got the 0 matrix here as A minus lambda I and then we have the
𝑥1
possibility of choosing this [𝑥2 ]freely. So, whatever we like and we have taken ⇒ 𝑥1 =
𝑥3
𝛼1 , 𝑥2 = 𝛼2 , 𝑥3 = 𝛼3 because all three are free variables.
𝑥1
1
0
0
0
0
1
And then this ⇒ [𝑥2 ] = 𝛼1 [0] + 𝛼2 [1] + 𝛼3 [0]. So, we have three linearly independent
𝑥3
eigenvector in this case corresponding to this 𝜆 = 1. So, the algebraic multiplicity of 𝜆 =
1 was 3 and also the geometric multiplicity which is the dimension of the eigenspace that
is also 3 in this case. So, we have seen in this example that, if it is repeated 3 times it is
also possible that we can get the full dimension, here the dimension of the eigenspace that
is 3.
As many times as the 𝜆 was repeated, but what that result says that it cannot be more than
3. And naturally, that is the case here because the dimension that is the full dimension
because the elements belongs to this 𝑅3 and we cannot have the dimension more than 3 in
that sense also we can conclude here.
857
(Refer Slide Time: 26:38)
There is a concept here the similarity of matrices which will introduce here and we will
continue for the discussion in the next lecture. So, an n cross n matrix B is called similar
to an n cross n matrix A, if we have this 𝐵 = 𝑃−1 𝐴𝑃. If we have this relation between the
between the matrix A and B, then we call this P is similar to the matrix A or A is similar
to the matrix B. And what to we have to we this P what is the P before some non singular
matrix P, if there exists a matrix here this 𝑃−1 I mean, this non singular matrix P therefore,
that P inverse make sense.
So, if we have this relation between the two matrices here B and A that 𝑃−1 𝐴𝑃 gives the
B the other matrix, then we call that these two are similar. Why do we use the similar
words some of the properties we will check today itself, that they share away many
common properties this B and A in terms of the eigenvalues, eigenvectors and there are
other considerations as well which we will continue in the next lecture.
858
(Refer Slide Time: 27:46)
So, today we will see that if 𝐵 is similar to 𝐴, then the 𝐵 has the same eigenvalues as A
and if x is an eigenvector of A. Then this 𝑦 = 𝑃−1 𝑥 is the eigenvector of B corresponding
to the same eigenvalue. So, meaning if we know the eigenvalues and eigenvector of one,
we can get the eigenvalues, eigenvectors of the other. In fact, they same they have the
same eigenvalues and the eigenvector also will be just the 𝑦 = 𝑃−1 𝑥𝑃 we have introduced
already in the similar 𝑡 definition. So, what we take that let us say this lambda is the
eigenvalue of this matrix A and the similar to 𝐴, we have the 𝐵 matrix. So, first of a relation
we have for this 𝐴 that x is the eigenvector and 𝜆 is the eigenvalue.
So, we have this relation 𝜆𝑥 = 𝐴𝑥. And what we do now? We multiply by this 𝑃−1 here.
So, the right hand side we have 𝑃−1 , P is that matrix which we are talking about the
similarity there. So, we have 𝑃−1 Ax and here also 𝑃−1 . So, the lambda is constant so, we
have 𝑃−1 x there. And then what we do, we have here the lambda 𝑃−1 x the same, the 𝑃−1
A again we have introduced this identity matrix.
𝜆𝑥 = 𝐴𝑥
⇒ 𝜆𝑃−1 𝑥 = 𝑃 −1 𝐴𝑥
⇒ 𝜆𝑃 −1 𝑥 = 𝑃−1 𝐴(𝑃𝑃 −1 )𝑥 = 𝑃−1 𝐴𝑃(𝑃−1 𝑥)
⇒ 𝜆(𝑃−1 𝑥) = 𝐵(𝑃−1 𝑥)
859
So, 𝜆 is an eigenvalue of 𝐵 and 𝑃−1 𝑥 is an eigenvector corresponding to the eigenvalue 𝜆.
(Refer Slide Time: 30:29)
Another result which actually we have seen already in this first result A and B are the
square similar matrices, then they have the same characteristic polynomial. So, eventually
we have seen already that they have the same eigenvalue. So, if we have the same
eigenvalues meaning they have the same characteristic polynomial, but this is just another
way of looking at it.
So, we take
𝐵 = 𝑃−1 𝐴𝑃
det(𝐵 − 𝜆𝐼) = 𝑑𝑒𝑡(𝑃−1 𝐴𝑃 − 𝑃−1 (𝜆𝐼)𝑃)
= det(𝑃−1 (𝐴 − 𝜆𝐼)𝑃)
= det(𝑃−1 ) det(𝐴 − 𝜆𝐼) det(𝑃)
= det(𝐴 − 𝜆𝐼)
(Refer Slide Time: 31:46)
860
So, what we have seen that the determinant of this det(𝐵 − 𝜆𝐼) is equal to det(𝐴 − 𝜆𝐼) so,
they have the same characteristic polynomial. In other words, we can say again that this A
and B will have the same eigenvalues and we have seen again in the previous slide here
the relation for the eigenvectors as well ok.
(Refer Slide Time: 32:56)
Coming to the conclusion so, in this lecture we have talked about the algebraic multiplicity
and that was nothing but the number of occurrence of an eigenvalue. And we have also
seen the geometric multiplicity that was the number of linearly independent eigenvectors
associated with that eigenvalue. And always this is the case that geometric multiplicity is
less than equal to the algebraic multiplicity and we have also talked about the similar
861
matrices; that means, B and A are called the similar to each other they are the similar
matrices, if we have this relation that 𝐵 = 𝑃−1 𝐴𝑃 for some invertible matrix P.
(Refer Slide Time: 33:40)
So, these are the references used to prepare these lectures and thank you for your attention.
862
Engineering Mathematics - I
Prof. Jitendra Kumar
Department of Mathematics
Indian Institute of Technology, Kharagpur
Lecture - 50
Eigenvalues & Eigenvectors: Diagonalization
So, welcome back and today again we will continue our discussion on these eigenvalues
and eigenvectors. This is lecture number 50 and we will mainly focus on iagonalization
and also it is applications for solving system of linear equations, also for getting the power
of the matrices etcetera. So, what is the diagonalization of the matrix? We will discuss
here now.
(Refer Slide Time: 00:26)
(Refer Slide Time: 00:38)
863
So, A square matrix A is said to be diagonalizable if there exists an invertible matrix P. If
there exists an invertible matrix P such that this 𝑃−1 𝐴𝑃 is a diagonal matrix or in other
words, because this concept we have already introduced the similarity of the matrices.
So, in other words if A is similar to a diagonal matrix, because if the point is here A is
called diagonalizable. If we can write down this A as this 𝑃−1 the 𝑃−1 𝐴𝑃 is a diagonal
matrix then we call that this A is diagonalizable. So, the meaning this A is similar to the
diagonal matrix here; 𝑃−1 𝐴𝑃 and this P is invertible matrix which we have to find, so that
this 𝑃−1 𝐴𝑃 becomes a diagonal matrix.
So, let A be an 𝑛 × 𝑛 matrix and that is a nice result that A is always diagonalizable; that
means, we can find such A P such that this 𝑃−1 𝐴𝑃 is a diagonal matrix. So, the result is
here A is diagonalizable, if A has n linearly independent eigenvector. So, that is a nice
result here, nothing to do with this eigenvalues. Actually we have to look for the
eigenvector. So, if we get n linearly independent eigenvectors then A can be diagonalized.
And if we cannot get these n linearly independent eigenvectors then the A is not
diagonalizable. So, that is the main result of this lecture.
Again we will not go through the formal proof here, but we will see with the help of many
examples, how this is working. And another subsequence of this result we can we can have
here if n is an 𝑛 × 𝑛 matrix here A and it has 𝑛 distinct eigenvalues, then also A is
diagonalizable and then reason is clear, because if we have n distinct eigenvalues, then we
will also get 𝑛 linearly independent eigenvectors; that is a result we have already seen in
864
previous lecture that corresponding to distinct eigenvalues we have a linearly independent
eigenvectors. So, eventually this second result here is again the same as this previous one;
that is diagonalizable, if and only if it has n linearly independent vectors.
So, just a note here that this matrix P which diagonalizes A is called the model matrix of
A and whose columns, I mean the point is how to find this A matrix P. So, here this model
matrix P has the columns that are nothing, but the eigenvectors corresponding to these
different eigenvalues. So, if you have n linearly independent eigenvectors, we will just
place them in this matrix P as the columns, and this is our matrix this P. And when we
check this 𝑃−1 𝐴𝑃 that will be nothing, but the diagonal matrix and entries of these diagonal
matrix will be just the eigenvalues, corresponding to these eigenvectors we have placed in
the sequence as columns of this matrix P.
(Refer Slide Time: 04:11)
So now here let us consider this example with 𝐴 = [5 4]. So, here we will not spend much
1
2
of our time for computing eigenvalues and eigenvectors, because that we have already
seen in previous lectures. So, we can compute the eigenvalues for this example and then it
comes to be 1 & 6, because the idea is simple that this 5 minus lambda and then we have
[
−1
4
] & [ ].
1
1
So, this determinant we have to solve which we can make this product here, the 10 and
then we will have here minus 2 and minus 5. So, minus 7 lambda and plus this lambda
865
square and minus 4 is equal to 0. So, this is lambda square minus this 7 lambda. And then
we have here 6 is equal to 0 and that can factorize to this minus 6 minus 1. So, lambda
minus 6 and lambda minus 1 is equal to 0. So, here we get these eigenvalues as 1 and 6.
So, having these eigenvalues now corresponding to each, we have to find the eigenvector.
And here we have indeed these distinct eigenvalues. So, we will get two linearly
independent eigenvectors and then we can diagonalize this matrix. So, here the
−1
]. So, that is corresponding to
1
4
this one, and corresponding to 6 we will get here [ ]. And now once we have the
1
eigenvectors corresponding to 1 will be coming as this [
eigenvectors we can formulate this model matrix which we call P. So, the P will be we are
placing these matrices are these vectors the eigenvectors as columns of this P.
−1
4
], that is the first column, and then this [ ] that is the second column. So, our
1
1
−1 4
model matrix is ready now and we can verify that how this 𝑃 = [
] look like. So,
1 1
1 −1 4
here we have to get this P −1 = [
] also, that is the inverse. So, for 2 by 2 matrix it
5 1
1
So, here [
is simple, so we have to divide here by this determinant and then we need to change the
sign and determined will be again with minus sign.
1 0
𝑃−1 𝐴𝑃 = [
]
0 6
So, finally, we will get this as the as the 𝑃−1 of this matrix 𝑃 which we can also verify by
multiplying these two and we are getting the identity matrix. So, here the 𝑃−1 and if we
1
0
compute this 𝑃−1 𝐴𝑃, so that is coming to be [
0
] in the diagonal and it is a diagonal
6
matrix and this is exactly the point here. So, we have kept in our model matrix this, these
vector minus 1 as the first column, and that was corresponding to the eigenvalue 1 and that
is the reason here this first eigenvalue is coming and in the second case we have kept this
second column which was corresponding to the 6 here and therefore, the second element
in the diagonal is 6 here. So, this order if we change for instance the order here.
866
−1
1
So, if we change it to 𝑃 = [
4
]. If we change, if we take this P, if we take this model
1
matrix then here 𝑃−1 𝐴𝑃 = [1 0], because here this was corresponding to this 6 and then
0 6
this is corresponding to this number 1 here. So, accordingly that order will change.
So, the order we place here for the eigenvectors corresponding order will be followed in
the diagonal entries as the eigenvalues, so that is important. Second point here which also
needs to be mention that this is not the unique eigenvector for instance; so, we can multiply
by any number to this [−1, 1]𝑇 that will be also the eigenvector. Again here also this [4, 1]𝑇
we can multiply by any scalar that will also be the eigenvector, because eigenvectors are
not unique and by doing so, also there is no problem we can keep any vector here not only
−1 and 1, we can also place for example, −2 and 2 here in the first column.
6
𝐴 = [−2
2
−2
3
−1
2
−1]
3
And in the second column we can place for instance we multiplied by 2 here, so 2, 2 & 8.
So, we can place 8 and 2 in the second column. So, does not matter that will be taken care
by this P inverse and still this product will give us the same diagonal matrix 1 0 0 6. So,
here it is material that whether we multiply here to these eigenvectors by some scalars; it
does not matter with this 𝑃−1 𝐴𝑃 will lead to the same eigen, same diagonal matrix whose
entries will be 1 and 6. The only thing matter again it s the order here we place these
eigenvectors.
So, the order we keep here placing as these columns of these eigenvectors in the same
order these eigenvalues will appear.
867
(Refer Slide Time: 09:42)
6
So, here another example of this 3 by 3 matrix [−2
2
−2 2
3 −1]. So, for this matrix also if
−1 3
you want to check whether it can be diagonalized or not and what will be the diagonal
matrix? What will be the model matrix? So, for that we need to compute the eigenvalues.
So, the eigenvalues for this matrix will be coming 2, 2 & 8.
Eigenvectors:
1 −1
2
[2] , [ 0 ] & [−1]
0
2
1
So, again now for each eigenvalue we need to compute the eigenvector to form this model
𝑃−1 and the corresponding to this eigenvalues this 2, 2 the repeated 1. So, here the
algebraic multiplicity of this 2 is 2 and now we compute the eigenvector corresponding to
this. So, here this 2 minus 2 and minus 2, so that will be the matrix there which we want
to solve as the system of linear equations. So, by doing so what we are actually getting
here we are getting 3 linearly independent eigenvector meaning this geometric multiplicity
of 2 is 2 and also it is; as the algebraic multiplicity is 2, also the geometric multiplicity in
this particular case is coming to be 2.
So, this is corresponding to this 2 this is also corresponding to 2 and here we have this
corresponding to this 8. So, we have 3 linearly independent linearly independent
868
eigenvector and that is the reason now we can actually diagonalize this matrix because we
need 3 matrices. Remember it is easy to remember here the model P here the model matrix
P will be of the same order as A. So, we need this 3 columns to fill the matrix P. So, if we
have 3 linearly independent vectors we can form this P otherwise for example,
corresponding to 2 if it happens that we have only 1 eigenvector then we cannot form this
P in other words the matrix A is not diagonalizable in that case.
So, here the matrix A is diagonalizable because we are getting 3 linearly independent
1
eigenvector and the model matrix 𝑃 = [2
0
−1
0
2
2
−1]. First two columns and the
1
corresponding to 8; we have this 2 minus 1 as a third column. So, this is corresponding to
2. The first column this is also corresponding to 2 and this corresponds to 1 the third
column. So, our order will remain exactly this one and this will become the diagonal entries
of the matrix 𝑃−1 𝐴𝑃. So, if you compute the 𝑃−1 𝐴𝑃 now.
2
𝑃−1 𝐴𝑃 = [0
0
0
2
0
0
0]
8
So, we need to get this 𝑃−1 and then this product we have to make and then we will get
this 2, 2 and exactly the order we have placed here these eigenvectors. So, we are getting
these diagonal entries absolutely the same here 2, 2 & 8. So, that is the diagonal matrix
here which is similar to the matrix A and later on we will observe several good properties
about this matrix because they share many common properties these similar matrices and
some of the applications very important applications one we can once we can diagonalize
the matrix we can we can use them in many applications.
So, that will be the also topic of discussion of this lecture.
869
(Refer Slide Time: 13:31)
So, the last example here we will take another one
2 0
𝐴 = [4 2
0 0
0
0]
3
where we do see this is the lower triangular matrix with entries 2, 2 & 3 in the diagonals
and then we have this 4 in the off diagonal rest everything is 0. So, in this case if we
compute the eigenvalues we know for the triangular matrices. So, this is 2, 2 & 3. So, the
eigenvalues will be 2, 2 & 3 and we have to compute again the eigenvectors corresponding
to the 2 and also corresponding to this 3. What happens in this case that here this algebraic
multiplicity of 2 is 2 and it comes to be that the geometric multiplicity of this 2 is 1 and
that is the point where actually we cannot diagnolize the system because geometric
multiplicity is not equal to the algebraic multiplicity for this eigenvalue 2 then we will get
less number of eigenvectors and we cannot form this model matrix P.
0
2
[1] 𝑎𝑛𝑑 [−1]
0
1
So, here the corresponding to these 2 eigenvalues the repeated eigenvalue we are getting
only 1 eigenvector and the reason is clear because if we formulate this equation A minus
lambda I x is equal to 0. So, what will happen here? We get this 0 0 0 and then we have
here 4 this again 0 0 and 0 0 1. This is the situation of this system of equation for the
870
eigenvector here and the right hand side 0. So, for this system of equation what do observer
what do we observe here that we have the 2 pivots element here 4 and also this 1. These
are the pivot elements and the free variable we have only one that is here x 2.
So, x 2 is free variable free variable; that means, we can choose this x 2 whatever we like.
So, let us take this alpha and then directly from these equations we have observe that the
x 1 is 0 from this second equation and from this third equation we observe that x 3 is equal
to 0. So, the eigenvectors x 1, x, 2, x 3 in this case corresponding to this repeated
eigenvalue is coming to be 0 1 0 and any multiple of this 0 1 0.
So, that is here we have taken just alpha 1. So, this is one of the eigenvectors here and then
corresponding to 3 also we can compute the eigenvector and that is naturally it will come
1 only. So, we have 2 minus 1 and 1. So, with these 2 vectors because we need 3 vectors
to fill the positions of this model matrix P. So, we cannot do in this case because we are
getting 2 linearly independent eigenvectors and therefore, this matrix A is not
diagonalizable. So, the given matrix is not diagonalizable.
So, what we have seen here that every matrix we cannot diagonalize. We can diagonalize
only those matrices when the eigen vectors the set of this eigen vectors is full means if it
is a 𝑛 × 𝑛 matrix then if we get n linearly independent Eigen vectors then we can
diagonalize the matrix, otherwise we cannot diagonalize the matrix.
(Refer Slide Time: 17:09)
871
Now, coming to the applications of the diagonalization, the first application we will
consider that we can easily compute the power of the matrices once we can diagonalize
the matrix. So, why so? What is the connection here or to the to the eigenvalues,
eigenvectors that we will see now. So, this 𝑃−1 𝐴𝑃 what we have seen that A can be
diagonalized then we have this relation 𝑃−1 𝐴𝑃 is equal to D or we can rewrite it that A is
equal to so we multiply by P here first there will be PD and then the right side we will
multiply by 𝑃−1 .
So, we will get out of this relation here A is equal to 𝑃𝐷𝑃−1 . So, having this relation then
if you want to multiply or we want to get this A power 2 or A square in that case. So, we
need to multiply this 𝑃𝐷𝑃−1 with the 𝑃𝐷𝑃−1 and then with this associativity property of
this product we will realize here that this P inverse, P is there which we can put as the
identity matrix and then we will get here 𝑃𝐷𝑃−1 meaning this P and the power of this
diagonal matrix. So, here the A square the power of this matrix is 2 power of this matrix;
it is coming to be that this power is exactly translated to the power of this diagonal matrix.
What is the use here that this diagonal matrix we can easily get this power here because
this power will directly go to this diagonal entry. So, we do not have to actually multiply
these matrices D here for this power, but only the diagonal entries will be squared and that
is a reason here.
𝑃−1 𝐴𝑃 = 𝐷 ⇒ 𝐴 = 𝑃𝐷𝑃−1
So, the 𝐴2 is very simple now the 𝑃𝐷2 𝑃−1 . So, we have to only do these multiplications.
𝐴2 = (𝑃𝐷𝑃−1 )(𝑃𝐷𝑃−1 ) = 𝑃𝐷(𝑃−1 𝑃)𝐷𝑃−1 = 𝑃𝐷2 𝑃−1
Naturally when we have a high power here and not just for the two because in any case
now we have to do this multiplication 𝑃𝐷2 𝑃−1 . So, here naturally the work is more if we
just want to find out this power 2, but in case for example, we want to power to get the
power 1000 then definitely this will be very very useful because D power higher power
would be easier to compute. So, why this A square is coming D square? We can continue
this idea for example, 𝐴3 also the same similar structure will happen.
So, this 𝑃𝐷2 𝑃−1 that is for A square and then again multiplied by A and this P inverse P
will again become the identity matrix and we will have PDQ P inverse. So, what is the
point here that we can see out of these calculations that A power n will be also just D
872
power n here and this PD power and P inverse. So, this will continue this power here on
translating to this matrix T. So, in general also we can prove that this 𝐴𝑛 = 𝑃𝐷𝑛 𝑃−1 . So,
we can use as I said before when we want to compute this very high power of this matrix
a large number here and then this is very very useful because 𝐷𝑛 the computation here is
very simple because if we take 2 diagonal matrix for example, this here and we want to
multiply with the same.
So, what will happen now? So, if you multiply this a square will come and then this product
will be 0 here. Also this will be 0 and b square will come. So, what happens when we do
the product of the diagonal matrix just simply we will this power; this power will go to the
diagonal entries. So, we do not have to do this multiplication as the matrix multiplication
just simply when we have the power n. So, for instance we have this a 0 0 b and we want
to get this power n there.
So, this will be nothing, but the a power n zero and b power n. So, that is the point here.
So, having this relation that 𝐴𝑛 is nothing, but the 𝐷𝑛 here. So, that is very simple to
compute and then finally, we need to make this product with the P and the 𝑃−1 . So, that is
the only computation load here for this matrix multiplication, but for D power n only that
linear relations here. So, these diagonal entries will be powered and nothing else.
(Refer Slide Time: 22:24)
873
1 4
So, let us just go through one example which says this find 𝐴5 for 𝐴 = [ 1 0]. So, but to
2
do so, we have to compute the eigenvalues and also the eigenvectors, because we need that
𝑃 and we also need that diagonal matrix. So, in this case the eigenvalues are coming to be
2
] and
−1
−1 & 2 and then we compute the eigenvectors corresponding to minus 1 it is [
4
corresponding to 2 it is coming as [ ]. So, having this we can now form this P the model
1
matrix P. So, placing these eigenvectors here as the columns.
Take
𝑃=[
2
−1
4
]
1
1 1 −4
⇒ 𝑃−1 = [
]
6 1 2
So, we can actually compute a very high power. Also the computational load will remain
the same. Only thing we have to we have to just do this power here of the diagonal entries.
So, and then later on we have to in any case just multiply by this P and the 𝑃−1 . So, it is
easy now to find having this relation any power of the matrix A.
1 2
6 −1
4 (−1)5
][
1
0
𝐴5 = 𝑃𝐷5 𝑃−1 = [
21
5.5
⇒ 𝐴5 = [
0 1
][
25 1
−4
]
2
44
]
10
So, that is A power 5, but it is usually used when we really want to have we want to
compute a high power of this matrix A then this is computationally very very efficient as
compared to doing the product of matrices A.
So, here was the one of the applications where we use this eigenvalues, eigenvectors or in
particular this idea of the diagonalization of the matrix.
874
(Refer Slide Time: 25:29)
Now, coming to the next application which is the solution of the system of linear
differential equations though the differential equations will be the topic of the next few
lectures, but here just to introduce the idea of this or the application of this diagonalization.
We will be doing very simple example also. So, here we consider the linear differential
equations. So, here it is a system, system means we have a more than one differential
equations and their variables are coupled. So, here we have this; the left hand side for
example, this is the derivative term.
Consider the system of linear differential equations
𝑋̇(𝑡) = 𝐴 𝑋(𝑡)
Let us assume that 𝐴 is diagonalizable.
Then 𝐷 = 𝑃−1 𝐴𝑃, ⇒ 𝐴 = 𝑃𝐷𝑃
−1
∴ 𝑋̇(𝑡) = 𝑃𝐷𝑃−1 𝑋(𝑡)
⇒ 𝑃−1 𝑋̇(𝑡) = 𝐷𝑃−1 𝑋(𝑡)
⇒ [𝑃−1 𝑋(𝑡)]′ = 𝐷[𝑃 −1 𝑋(𝑡)]
Substituting 𝑃−1 𝑋(𝑡) = 𝑌(𝑡) we get
875
𝑌̇(𝑡) = 𝐷 𝑌(𝑡)
(Refer Slide Time: 29:37)
So, the benefit here that our system the reduced system is that here we have the derivatives
from the, here we have this diagonal matrix and then we have the 𝑦 the unknown. Now if
we just look at this multiplication what we are getting? We are getting these n equations
and they are actually decoupled equations now because once we multiply this with the
diagonal entries what we are getting
876
𝑦1̇ (𝑡)
𝑦1 (𝑡)
𝑦 ̇ (𝑡)
𝑦 (𝑡)
⇒ [ 2 ] = Diag(𝜆1 , 𝜆2 , … , 𝜆𝑛 ) [ 2 ]
⋮
⋮
𝑦𝑛̇ (𝑡)
𝑦𝑛 (𝑡)
⇒ 𝑦𝑖̇ (𝑡) = 𝜆𝑖 𝑦𝑖 (𝑡), ∀ 𝑖
⇒ 𝑦𝑖 (𝑡) = 𝐶𝑖 𝑒 𝜆𝑖 𝑡
where 𝐶𝑖 is constant, and 𝑖 = 1,2, … , 𝑛.
𝑃−1 𝑋(𝑡) = 𝑌(𝑡) ⇒ 𝑋(𝑡) = 𝑃 𝑌(𝑡)
|
[𝑣𝑖 ] is the eigenvector corresponding to 𝜆𝑖
|
𝑥1 (𝑡)
|
|
|
𝑥 (𝑡)
[ 2 ] = 𝐶1 [𝑣1 ] 𝑒 𝜆1 𝑡 + 𝐶2 [𝑣2 ] 𝑒 𝜆2 𝑡 + ⋯ + 𝐶𝑛 [𝑣𝑛 ] 𝑒 𝜆𝑛 𝑡
⋮
|
|
|
𝑥𝑛 (𝑡)
So, what we have to do? We have to compute the; this original coefficient matrix A which
was given for the system of equations. We need to just compute the lambdas and the
eigenvalues eigenvectors and then we can find a solution.
877
(Refer Slide Time: 33:41)
So, just to demonstrate this we have taken the simple example here.
𝑑𝑥1
= 3𝑥1 + 2𝑥2
𝑑𝑡
𝑑𝑥2
= 7𝑥1 − 2𝑥2
𝑑𝑡
We can rewrite in the system form here. So,
𝑥1 ′
3
[𝑥 ] = [
7
2
2 𝑥1
][ ]
−2 𝑥2
⟹ 𝑥 ′ = 𝐴𝑥
So, what we have to do? We have to just compute the eigenvalues and eigenvector of this
matrix here.
3
𝐴=[
7
878
2
]
−2
(Refer Slide Time: 34:03)
So, the eigenvalues of this matrix are coming as when we write down this characteristic
polynomial we are getting
det (𝐴 − 𝜆𝐼) = 0 ⟹ 𝜆2 − 𝜆 − 20 = 0
⟹ (𝜆 + 4)(𝜆 − 5) = 0
⟹ 𝜆1 = −4 & 𝜆2 = 5
Eigenvectors:
[
2
1
] & [ ]
−7
1
𝑥1
2
1
[𝑥 ] = 𝐶1 [ ] 𝑒−4𝑡 + 𝐶2 [ ] 𝑒5𝑡
2
−7
(Refer Slide Time: 34:53)
879
1
So, what we have seen here that with the help of this diagonalization; this was very easy
to solve such a system, and the conclusion here is that this diagonalization of the matrix
we have learnt, and in particular we have seen these two applications, the power of the
matrices, we can easily compute with the help of this idea and also the solution of the
system of linear differential equations we can compute with this diagonalization.
(Refer Slide Time: 35:17)
So, these are the references we have used and thank you for your attention.
880
Engineering Mathematics – I
Prof. Jitendra Kumar
Department of Mathematics
Indian Institute of Technology, Kharagpur
Lecture – 51
Differential Equations – Introduction
Welcome back so this is a lecture number 51 and we will now continue with a new topic
now on differential equations.
(Refer Slide Time: 00:26)
So, today’s lecture will be diverted to the introduction and the information of differential
equations.
881
(Refer Slide Time: 00:30)
So, what is the differential equations? So an question which involves the derivates or the
differentials of one or more independent variables with respect to one or more independent
variables is called differential equation. So, here these are the examples of the differential
equations.
So, we have these derivative terms involving in these equations and therefore, we call this
equation as the differential equation. And this is the ordinary differential equation because
this derivatives are ordinary meaning this we have only one a independent variable as x in
this case, 𝑦 is a dependent variable on 𝑥. So, we have the ordinary derivatives here the
second equation this is also the ordinary differential equation. So, we have these
differential terms here 𝑑𝑥 𝑑𝑦 and this is also having these ordinary derivatives only. So,
only one dependent variable that is 𝑦 and one independent variable that is 𝑥.
So, for instance here we have two independent variables the x and t and this we depends
on two variables here x and t. So, here we have the notion of the partial derivates in the
equation. And therefore, this is called the partial differential equation or PDE, because
here we the partial derivatives not the ordinary, but we have the partial derivatives, now
in the equation. So, there are other classifications here which we call the order what is the
order of the differential equation? And order is nothing, but the order of the highest order
derivates involved. So, for example, in this first equation the highest order derivative is 4.
882
So, the order of this equation will be 4 similarly will be talking about the order of these
two equations.
There is another terminology we will using here degree and degree here in these equations
will be the degree of again the highest order derivatives involved, for instance in this case
this is the higher order derivative. And the power here is 1 or the degree of this term is 1.
So, this is the first degree, but the 4th order differential equation.
So, here the order is 4 and the degree is 1 in this equation again the order is 1 because we
have the highest order is the first order like, 𝑑𝑦 𝑑𝑥 term will be present here nothing else
or the first order differentials are present in this case and the degree is also 1. And in this
case here the order is 3 so because third order derivative terms are there and the degree is
2, because the highest order derivative appears in degree 2. So, the degree of this
differential equation is 2 and the order is s 3.
𝑑4𝑦 𝑑2 𝑦
𝑑𝑦 3
+
+ ( ) = 𝑒 𝑥 , 𝑜𝑟𝑑𝑒𝑟 − 4, 𝑑𝑒𝑔𝑟𝑒𝑒 − 1
𝑑𝑥 4 𝑑𝑥 2
𝑑𝑥
𝑦(𝑦 2 + 1)𝑑𝑥 + 𝑥(𝑦 2 − 1)𝑑𝑦, 𝑜𝑟𝑑𝑒𝑟 − 1, 𝑑𝑒𝑔𝑟𝑒𝑒 − 1
2
𝜕2𝑣
𝜕3𝑣
= 𝑘 ( 3 ) , 𝑜𝑟𝑑𝑒𝑟 − 3, 𝑑𝑒𝑔𝑟𝑒𝑒 − 1
𝜕𝑡 2
𝜕𝑥
883
(Refer Slide Time: 03:37)
The further classifications will be talking about like the linear and the non-linear
differential equations. So, here we a differential equation is called linear, if every
dependent variable and every derivative occurs in the first degree only.
So, if we do not see the product or the power, then we have the linear term meaning the no
product of the dependent variable and or the derivatives occur. So, then we have the linear
equation and if the differential equation is not linear then we call the non-linear equation.
So, in the non-linear one the product of the derivative or the dependent variable with the
derivative we will exist. So, every linear equation is a first degree that is clear. So, because
in linear equation there will no product or no power, so it will be first degree.
But every first degree equation may not be linear, because for instance here this is the
second order equation and the degree is one here because the degree is defined as this
higher the degree of the highest order term. But this equation is non-linear because of this
product y and the
𝑑𝑦
𝑑𝑥
. So, this is a non-linear equation and its a first degree, but it is a non-
linear. So, every linear equation is of first degree, but not every first degree equation will
be a linear.
𝑑2𝑦
𝑑𝑦
+𝑦
+𝑦 =0
2
𝑑𝑥
𝑑𝑥
884
(Refer Slide Time: 05:11)
Now coming to the solution of the differential equation, so any relation between the
dependent and independent variable without the derivative terms, because in the solution
we do not want to see the derivatives, the derivatives will be in the equation in the
differential equation.
So, any relation between this dependent and independent variable which satisfies the
differential equation is called a solution or the integral of the differential equation. So, for
instance
we
take
this
function
𝑦=
this
relation
between
𝐴
+𝐵
𝑥
with these two are arbitrary constants, what we can check that this is a solution of this
2
differential equation; 𝑦 ′′ + (𝑥) 𝑦 ′ = 0. Why this is the solution, because this relation
satisfies this differential equation which we can easily verify because if we take the
derivative here of this y we will get minus A over x square and because we have the second
derivative as well in the equation.
So,
we
𝑦′ = −
𝐴
𝑥2
will
we
. So, 𝑦 ′′ =
2𝐴
𝑥3
go
for
the
second
order
derivative
here.
So,
that is the second order derivative and if we substitute this here in
the equation so the first the second here.
885
So, if we have the two
what
2𝐴
𝑥3
do
𝐴
𝑥3
2
𝐴
and then plus this and the first order term here that is− 2. So,
we
𝑥
see
𝑥
here
2𝐴
𝑥3
and
here
also
. So, that will cancel out and the we will get the 0 which is the right hand side of the
equation.
(Refer Slide Time: 07:17)
So,
what
we
have
observed
that
this
𝑦=
𝐴
+𝐵
𝑥
relation
this
given
a
relation
is a solution of this differential equation because this relation set satisfies this differential
equation.
886
(Refer Slide Time: 07:21)
And now we will we are going to the direction of the formation of these differential
equation. So, before that we need to introduce this family of curves.
So, an n parameter family of curves is the set of relations of the form this one, so here the
x
y
there
is
a
relation
in
{(𝑥, 𝑦): 𝑓(𝑥, 𝑦, 𝑐1 , 𝑐2 , … 𝑐𝑛 ) = 0}. So, changing these parameters you will get different
different curves here, that is what it is a family of the curves because different values of
this c’s we have a different curve. And the examples here the first set of this concentric
circles if we take; that means, 𝑥 2 + 𝑦 2 = 𝑐 and c is a parameter. So, one parameter family
and the c takes a non negative real numbers, because here 𝑥 2 + 𝑦 2 so it has to be positive.
So, this a non negative. So, this c has to be non negative. So, it is taking non negative real
values and for different different values of c, these are the equations of the circle of the
same centre. So, here it is a family were we have only one parameter that is c and if we
keep on changing the c we are getting the circles here, with centre same and different
radius. Again the set of the circles if we take here, here now we are also varying with the
centre and as well as the radius.
So, this is the set of the all circles here we can choose the 𝑐1 , 𝑐2 and the 𝑐3 . So, 𝑐1 , 𝑐2 any
real number and the 𝑐3 is a positive, a non negative real number. So, in that case this is a
family of these three parameters 𝑐1 , 𝑐2 , 𝑐3 while the earlier one was the family of one
parameter, because the only the radius was varying in that case here we have different
887
centers and different radius of the of these circles of this family of circle. So, this is the 3
parameter family of circles. And now what we also reserve that the solution of the
differential equation which we have also seen in the previous example, it is the family of
curves nothing, but nothing else, but the family of curves which we will also observe in
next slides.
(Refer Slide Time: 09:59)
So, how to get the how to do this formation of the differential equation for given n
parameter family of curves; so what you have see here if we have this n parameter family
of curves; meaning the solution of a differential equation then from this given family of
curves how to form the differential equation or the associated differential equation
corresponding differential equation whose solution is this given family of curves. So, the
from a given family of curves containing n arbitrary constants we can obtain nth order
differential equation whose solution is the given family.
So, the order of the differential equation will be dependent actually how many parameter
family we have taken. So, here how to get this actually, so differentiate the given equation
n times; and we get an additional equations there was a given n parameter family equation
we differentiate keep on differentiating this family. So, we will get n more equations when
we differentiate this n times and out of this n plus 1 total equations we will eliminate these
constant terms of the parameters here. And then we will get equation which we will contain
only the derivatives terms and the dependent independent variables free from that
888
parameters. So, here by eliminating this from this 𝑛 + 1 equations we will obtain a
differential equation of nth order.
(Refer Slide Time: 11:40)
So, we will see with the help of the examples how this works here the formation, so, we
consider now this example here, so we will obtain the differential equation which is
satisfied by this relation 𝑥𝑦 = 𝑎𝑒 𝑥 + 𝑏𝑒 −𝑥 + 𝑥 2 . So, this is the two parameter family of
curve so we have a and b they are the arbitrary constants here a and b. So, we have this
two parameter family of curves. So, out of this two parameter family of curves, we are
expecting that it will be a corresponding differential equation will be a second order
differential equation, when we try to eliminate these two constants from the equations after
taking the derivative of this relation.
So, here the given family of curves we have the 𝑥𝑦 = 𝑎𝑒 𝑥 + 𝑏𝑒 −𝑥 + 𝑥 2 and if we
differentiate this with respect to x what we get; so we will get x y derivative plus the y
here and the derivative of x 1. So, that is 𝑥𝑦 ′ + 𝑦 = 𝑎𝑒 𝑥 − 𝑏𝑒 −𝑥 + 2𝑥 .
So, we have differentiated with respect to x only once and now we have to differentiate
this once again because, we do see here to arbitrary constants in the equation. So, if we
differentiate once again what we will get out of this 𝑥𝑦 ′′ + 2𝑦 ′ = 𝑎𝑒 𝑥 + 𝑏𝑒 −𝑥 + 2. So, now,
we have already differentiated this two times and now you want to eliminate these
constants. So, what do we see here that 𝑎𝑒 𝑥 + 𝑏𝑒 −𝑥 + 2 that is also there in this given
relation.
889
So, if we substitute for example from here, if you substitute this 𝑎𝑒 𝑥 + 𝑏𝑒 −𝑥 + 2 and put it
here in this equation. Then these this constant terms will be removed and we will get an
equation which contains the derivative terms dependent variables and independent
variables. So, here this using this equation which we call as equation number 1. So, what
do we get now we will get this 𝑥𝑦 ′′ + 2𝑦 ′ = 𝑥𝑦 − 𝑥 2 + 2.
So, now, we got this differential equation here, which corresponds to this family of curves
with two parameters. Or in other words the solution of this differential equation is nothing,
but this given family of two parameter family of curves.
So, that is the how the formation works of this differential equations from a given family
of curves. And mainly in this lectures upcoming lectures you will see actually how to find
this from this given differential equation, how to get this family of curves.
(Refer Slide Time: 15:36)
So, we have the other concept of the general and the particular solution. So, let this is the
nth order differential equation. So, here we have this nth order derivatives and other lower
order derivatives also, this dependent variable x and y since a relation meaning is a
differential equation the nth order differential equation.
So, what do we call as the general solution it is the solution containing n independent
arbitrary constants. So, in the general solution also in the previous slide what we have
observed we started with a two parameter family of curves. And then corresponding
890
differential equation was second order differential equation or other way around whenever
we have a second order differential equation and we find the solution of this equation or
other to say the general solution of this differential equation it will have to arbitrary
constants of two parameters in the solution. So, here that is exactly the definition of the
general solution, the solution containing 𝑛 independent arbitrary constant.
So, any solution of this differential equation of a given differential equation and if it has 𝑛
independent arbitrary constants corresponding to the such nth order, ordinary differential
equation then we call this solution as general solution. Another one what we are talking
about the particular solution, so giving some values to these coefficients that may depend
on some other supplementary, conditions may be given along with the differential
equation. So, based on the other information which we can evaluate these coefficients
because this is the general solution is nothing, but the family of the curves of this n
parameter.
So, but having a particular solutions means a particular curves, so that is possible only
when some other supplementary conditions are supplied with the differential equation.
And then as differential equation together with these supplementary conditions we will get
particular solutions. So, here the particular solution means the solution giving particular
values to one or more of the independent constants. So, usually that is done from the
physical model that some extra conditions, are given based on the physics of the problem
and then we can compute actually some or all of the these constants and the solution here
now does not have any independent arbitrary constants.
So, which call which we call as the particular solution and as we discussed already that the
number of arbitrary constants in a solution of a differential equation depends on the order
of the differential equation. And in that we have seen at least through the examples that a
general solution of nth order we will contain n arbitrary constants ok.
891
(Refer Slide Time: 18:47)
So, here now and let us just go through the example, so here we have for instance this
differential equation
(
𝑑𝑦 2
) − 4𝑦 = 0
𝑑𝑥
it is a general solution.
So, we are not finding the techniques or using any techniques to find the solution that will
be discussed in the next lecture. Here we are just providing the definitions. So, for example,
this 𝑦 = (𝑥 − 𝑐)2 . So, this is the solution this was satisfy this differential equation. So, if
you get here this
𝑑𝑦
𝑑𝑥
for instance that will be two times x minus c. And if we substitute now
in this differential equation what will happen two times this (𝑥 − 𝑐)2 and minus this four
times this 𝑦 = (𝑥 − 𝑐)2 . And here also we have 4(𝑥 − 𝑐)2 they have also 4(𝑥 − 𝑐)2 .
So, this will cancel out and what we observe that, this solution here which a satisfy satisfies
this differential equation. And therefore, this is a solution also this contains one this
arbitrary constant this c for any value of 𝑐 this will satisfy the differential equation which
we have observed. So, c is an arbitrary constant and the order here was 1 the order of this
differential equation was 1 and in the solution we have 1 arbitrary constant. So, therefore,
we are calling it has the general solution.
892
(Refer Slide Time: 20:28)
Particular solution: So if we give any we assign any value to this c we can assign directly
also because for any value of c has a solutions for instance if I call that 𝑦 = 𝑥 2 . So, 𝑦 = 𝑥 2
is also a solution of this differential equation and by setting just this 𝑐 = 0. But now this
is a particular solution say parabola it is a fixed parabola. But here that parabola depends
on this parameter it was a family of the parabolas, but now we have a particular curve out
of this family.
So, we call this as a particular solution or the general solution, this was the general solution.
So, both have the solutions, but one is the general solution the other one is the particular
solution or particular a curve out of this family of one parameter curves.
893
(Refer Slide Time: 21:24)
Another example which we can look here, so this is the example number 2. So, here this
we consider this 𝑦𝑦 ′ − 𝑥(𝑦 ′ )2 = 1. So, we try to find I mean that is the techniques we will
discuss later on. So, this comes to be the general solution of this differential equation 𝑦 =
1
𝑐𝑥 + . So, here we have this one parameter family the c is the only parameter in this in
𝑐
this equation. So, this is a one parameter family of curves and we one can easily verify
that. So, also satisfy this differential equation for any value of this c. So, this is the general
solution because the given differential equation was the first order differential equation.
And we are getting the solution which is also having one arbitrary constant.
So, therefore, we call this as a general solution, but if we give any particular value to this
constant, then we are basically selecting one element of this family or one curve out of this
family. So, that will be called as the particular solution. So, for instance in this case if we
take this 𝑐 = 1. So, here we taken the 𝑐 = 1 we have fix the c is equal to 1 and in that case
this 𝑦 = 𝑥 + 1, that will also satisfy the differential equation and which can again one can
easily see. So, the 𝑦 = 𝑥 + 1 and here y prime will be 1 and minus this x and again y prime
that is 1 whole square is equal to 1.
So, do we see the left hand side this x get cancelled, so 1 is equal to 1 so this is naturally
satisfying, the differential equation because for any c that satisfy the differential equation.
But now this 𝑦 = 𝑥 + 1 that is a one curve the one that is straight line out of this family of
894
lines we got this as a particular solution. So, this is no more a general solution, this is a
particular solution of the given differential equation.
(Refer Slide Time: 23:55)
There is another terminology use for defining the solution. So, called the explicit solution
or we also called as a implicit solution. So, here the explicit solution 𝑦 = 𝑦(𝑥), when the
solution is given in this form that y the dependent variable is given the right hand side
everything contains as the independent variables. So, there is no implicit relation of y and
x is given, but rather we call this as a explicit relation of y in terms of x.
So, we have y is given in terms of x no other implicit relation. So, this is called the explicit
solution. So, if we get out of our differential equation are the solution in this form that y is
equal to function of x. And the in case of the implicit solution we get this relation of the x
and y, relation here which satisfies the given differential equation. And therefore, this is
called the implicit solution.
So, for example, if we take this differential equation the second order differential equation
𝑦 ′′ + 𝑘 2 𝑦 = 0 and we can observe now by substituting. In fact, at this point because we are
now going to discuss the solution techniques, this is a needed a solution because this will
satisfies the given differential equation. And therefore, but what else we observe here
where that is the explicit solution because this y is given in terms of the x right hand side.
Here is a function of x. So, this is then an explicit solution is given y is a stated in terms
of the x.
895
𝑦 = 𝑐1 cos 𝑘𝑥 + +𝑐2 sin 𝑘𝑥
And now for the implicit solution, so this is the explicit solution and there is an another
example where 𝑥 + 3𝑦𝑦 ′ = 0 is given first order differential equation. And in this case what
we observe that this 𝑥 2 + 3𝑦 2 = 𝑐 . This satisfies this differential equation, so if you
satisfies the differential equation, then is solution so which we can easily see. So, if we
differentiate for example, this what relation we are getting 2 x plus 6 y and dy over dx is
equal to 0.
So, from here we can compute, in fact or we just divide here by 2 so we will get here 𝑥 +
3𝑦𝑦 ′ = 0. So, naturally it is satisfying this differential equation so, but what is the different
here. Now from the earlier one here, the function of this x y is given not the as a explicit
relation of y in terms of x is given, but it is an implicit relation here given in terms of x
and y.
(Refer Slide Time: 27:05)
So, we call such solution as implicit solution.
896
(Refer Slide Time: 27:08)
So, with this we come to the conclusion now. So, we have defined the order and also the
degree of the differential equation and also some classification we have done. Mainly
based on this whether the equation is linear or this is a non-linear equation. We have also
seen in this lecture that how to this form differential equation out of the given of parameter
n parameter family of curves.
So, there we have to keep on differentiating this n parameter family of course, n times and
then found the set of 𝑛 + 1 equations we have to eliminate these parameters. And then we
will get a relation of the dependent variables independent variables and their derivatives
which is the differential equation. So, what will be actually coming now in the future
lectures that how to find the solution of given differential equation.
So, how to the find the solution; how to find the family of curves whether a particular
family of curves or the general family of curves, of that will be as a solution of the given
differential equation. In other words what we have also discussed here the general solution
of nth order differential equation which contains n arbitrary constants. So, the general
solution is nothing, but the n parameter family of curves which satisfies the given
differential equation.
897
(Refer Slide Time: 28:42)
And these are the references we have used for preparing the lectures.
Thank you for your attention.
898
Engineering Mathematics – I
Prof. Jitendra Kumar
Department of Mathematics
Indian Institute of Technology, Kharagpur
Lecture - 52
First Order Differential Equations
(Refer Slide Time: 00:22)
So, welcome back and this is lecture number 52. Today will be talking about this First
Order Differential Equations, and mainly we will consider first order and the first degree
differential equations and their solution techniques.
899
(Refer Slide Time: 00:30)
𝑑𝑦
So, today we will focus on these two standard forms; the one we will consider 𝑑𝑥 = 𝐹(𝑥, 𝑦).
So, the function of x y right-hand side and the other form we will be discussing in this
lecture that will be 𝑀(𝑥, 𝑦) 𝑑𝑥 + 𝑁(𝑥, 𝑦) 𝑑𝑦 = 0. So, these are the two types of first order
differential equations we will be covering in this and the next few lectures.
(Refer Slide Time: 01:03)
So, the solution techniques. So, what are the solution methods? The first one is the
separation of variables. So, we will quickly review some of the techniques which are very
900
fundamental of basics and they are required at a every stage for solving the differential
equations. So, the one of them is the separation of variable.
𝑑𝑦
So, here if a differential equation can be written in this form. So, we have 𝑓1 (𝑦) 𝑑𝑥 = 𝑓2 (𝑥).
So, the point here is that everything the function of y if we can collate to one side and the
other side the function of x, in that case we call that this equation is variable separable
because now we can easily integrate this because the left-hand side only the function of y
and the right-hand side everything of function x.
So, if we can put in such a form then we can easily integrate these equations and then we
say that the variable variables are separable in the given differential equation, and the
solution for such differential equation once we separate the variables then it can be written
as we can simply integrate this here or we can first write down this equation as this f 1 y
and dy is equal to this f 2 x and dx and then we can simply integrate this equation with this
constant of integration. So, this is equal here. So,
∫ 𝑓1 (𝑦) 𝑑𝑦 = ∫ 𝑓2 (𝑥) 𝑑𝑥 + 𝑐
(Refer Slide Time: 02:52)
So, see one example of this kind which is given here that
𝑑𝑦
= 𝑒 𝑥−2𝑦 + 𝑥 2 𝑒 −2𝑦
𝑑𝑥
901
So, if we take a close look here at the right-hand side of this function what we observe that
e power x because this we can write down this first here as 𝑒 𝑥 in to 𝑒 −2𝑦 and this term we
have already in the separable form, so 𝑒 −2𝑦 . And what we do now, 𝑒 −2𝑦 we can take as a
common and that can go to the other side of the equation and this one side it will remain
as e𝑥 + 𝑥 2 . So, this equation is variable separable which we can easily do and then we can
integrate it.
𝑒 2𝑦
𝑑𝑦
= e𝑥 + 𝑥 2
𝑑𝑥
(Refer Slide Time: 03:42)
So, here we can rewrite this. So, we multiply it basically 𝑒 2𝑦 to this equation and
automatically the right-hand side becomes now e𝑥 + 𝑥 2 , so free from y. So, we have this
side everything the function of y and the right-hand side we have the function of x.
𝑒 2𝑦
𝑑𝑦
= e𝑥 + 𝑥 2
𝑑𝑥
So, now, we can easily integrate this equation and integrating we will get because
𝑒 2𝑦
𝑥3
= 𝑒 𝑥 + + 𝑐1
2
3
2
𝑒 2𝑦 = 2𝑒 𝑥 + 𝑥 3 + 𝑐
3
902
So, this is the solution which is given in this implicit form. So, this is the implicit solution
of the given differential equation by separating this variable. So, it was easy to integrate
and find the solution. So, this is one of the very basic techniques which we use for solving
differential equations.
(Refer Slide Time: 05:22)
The other one there are equations which can be reduced to the separable of variables. So,
as such in the given in the given equation may not be separated with respect to these
variables, but it can be made by some substitution to separable form and then we can again
repeat the process which we have done for the separable equations.
𝑑𝑦
So, for instance if we consider this 𝑑𝑥 = 𝑓(𝑎𝑥 + 𝑏𝑦 + 𝑐). So, if we have here the function
of this ax plus by plus c or the form is just ax plus by, so as such this given equation may
not be in the separable form, but if we make a substitution here. So, in this case we say it
is a we w