s
’
a
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s
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K
Engineering
Mathematics - III
for
B.E./B.Tech./B.Arch. students of first semester of all Engineering coleges affiliated
to U.P. Technical University, Lucknow
&
Uttrakhand Technical University, Dehradun
By
Dr. Manoj Kumar
Dr. S. B. Singh
Assistant Professor
Department of Mathematics
R.K. College Shamli (Muzaffarnagar)
Associate Professor
Deptt. of Maths, Stats. & Computer Science
G.B. Pant University of Agri. & Tech.
Pant Nagar (U.A.)
KRISHNA Prakashan Media (P) Ltd.
KRISHNA HOUSE, 11, Shivaji Road, Meerut-250 001 (U.P.), India
JAI SHRI RADHEY SHYAM
Dedicated
to
LORD
KRISHNA
Authors
and
Publishers
(v)
About the Authors
S.B. Singh
Received his Ph.D. in Mathematics from Institute of Advance Studies, C.C.S.
University, Meerut. Presently, he is Associate Professor in the Department of
Mathematics, Statistics and Computer Science, G. B. Pant University of Agriculture
and Technology, Pantnagar. He has around 15 years of teaching experience to
Undergraduate and Post Graduate students at different Engineering Colleges and
University. He is a member of Indian Mathematical Society, Operations Research
Society of India and National Society for Prevention of Blindness in India. He is a
regular reviewer of many books and some International Journals. He is Editor of the
Journal of Reliability and Statistical Studies. He has authored and coauthored eight
books on different courses of Applied/ Engineering Mathematics. He has been
conferred with four national awards. He has published his research works at
national and international journals of repute. His area of research is Reliability
Theory.
Manoj Kumar
received his Ph.D. in Cryptography from Dr. B. R. A. University Agra. Presently, he
is Assistant Professor in the Department of Mathematics, R. K. College ShamliMuzaffarnagar- U.P. He has around 10 years of teaching experience to
Undergraduate and Post Graduate students at different Engineering and degree
Colleges. He is a member of Indian Mathematical Society, Indian Society of
Mathematics and Mathematical Science, Ramanujan Mathematical society and
Cryptography Research Society of India. He is a regular reviewer of eight
International peer reviewed Journals. He is also working as a Technical Editor of
some International Journals: Asian Journal of Mathematics & Statistics, Asian
Journal of Algebra, Trends in Applied Sciences Research, Journal of Applied
Sciences. He has published his research works at national and international . Peer
review Journals. His area of research is cryptography.
Preface
This book Engineering Mathematics - III has been written in conformity with the revised syllabus
for the third/ fourth semester students of the UPTU Lucknow. An ample part of the material in this book
has been rigorously class tested over the past several years. This book is basically the result of our
teaching experiences to the engineering and science students. The principal goal of this book is to
provide a study material to the student with a thorough knowledge of fundamental concepts and
methods of applied mathematics utilized in different engineering disciplines. A different aim of this text
book is to provide the students with the basic knowledge in the subject. We have given the definitions,
methods, theorems and observations, followed by typical problems and the step by step solution. Each
topic is covered in great detail, followed by several meticulously worked-out examples, and a problem
set containing a large number of additional related exercises including objective type problems. This
book aims at an exhaustive coverage of the syllabus and there is definitely an attempt to kindle the
student's creative ability. While preparing for the examination students should not restrict themselves
only to the questions / problems given in the self evaluation. They must be prepared to answer the
questions and problems from the entire text.
The authors feels rewarded for the labor and pains that they have been taken while writing this
book, if it is provides adequate study material covering all topics in details and enables the students to
secure better marks in the examination.
—Authors
Acknowledgment
Authors feel great pleasure in presenting the revised edition of the book Engg. Maths-III. We would
like to thank all those who have used and adopted various editions of this book in past.
We are thankful to Prof. Sunder Lal, Pro V.C. of Dr. B. R. A. University, Agra. for his valuable
suggestions. Authors are particularly obliged to Dr. Jai Kishore, Mr. Vijayveer Singh, Dr. Anil Agarwal, Dr.
Ekata, Dr. Amit Awasthi, Dr. Atul Chaturvadi, Dr. Geetam Sharma, Dr. Vijayveer, Dr. Mukesh Rathor and
Miss Vandana Rathor and many more for their source of inspiration. This list is incomplete and we
apologize to those faculty members whose names have not been included here, as the list is very long.
Without mentioning the names, a particular gratitude is to the all family members who have
supported us at home. SB Singh expresses his appreciation to his wife Mrs. Neelam Singh for her
constant support and encouragement through out the time of revision of the book. Manoj Kumar
especially appreciate the ongoing interest and support of his wife Mrs. Chhaya, father in law Dr. Subey
Singh; mother in law Smt Sureshvati, brother in law Mr. Anurag Verma and his wife Mrs. Meenu.
Finally, authors owe a debt of appreciation to MD Mr. Satyendra Rastogi 'Mitra' for his constant
motivation and SMM Mr. M. R. Sharma of Krishna Prakashan Media Pvt. Ltd. Meerut for his enthusiasm
in the project. We are also thankful to the printing, publishing and editorial staff of Krishna Prakashan
Media (P) Ltd., for their patience with us.
At last, but not the least, we thank to everyone who has lent a helping hand to provide a shape to
our bricks, cement etc. in the preparation of this book.
We welcome valuable remarks and suggestions from students, teachers and academicians so that
this book may further be improved upon. To make this new edition as complete and user friendly as
possible, the reader's remarks will be appreciated always. Please contact us via electronic mail at the
address listed below for remarks, suggestions, questions, intimation of errors, misprints and feedback
regarding this book.
S.B. Singh
drsurajbsingh@yahoo.com
Manoj Kumar
yamu_balyan@yahoo.co.in
(vii)
CONTENTS
Unit-1 Function of Complex Variables…………………….(1-176)
1.
2.
3.
Analytical Functions……………………………………………………….3
Complex Integration……………………………………………………..43
Power Series and Calculus of Residues …………………………………75
Unit-2 Statistical Techniques …………………………...(177-298)
4.
5.
6.
7.
Statistics ………………………………………………………………...179
Curve Fitting ……………………………………………………………217
Correlation and Regression…………………………………………….239
Probability Theory ……………………………………………………..275
Unit-3 Statistical Techniques -II ………………………..(299-558)
8.
9.
10.
11.
12.
Binomial Poisson and Normal Distribution…………………………….301
Sampling theory ………………………………………………………..347
Tests of Significations and its Applications …………………………….447
Time Series and Forecasting …………………………………………...471
Statistical Quality Control ……………………………………………...499
Unit-4 Number Techniques-I……………………………..(559-674)
13.
14.
Solutions of Algebraic and Transcendental Equation………………….561
Final Differences and Interpolation…………………………………….593
Unit-5 Numerical Techniques-II…………………………(675-774)
15.
16.
17.
Solution of system of Linear Equation………………………………….677
Numerical differentiation and Integration………………………………701
Solution of differential Equation……………………………………….733
Appendix A Errors in the Numerical Computations………………………...(777-806)
Appendix B Question Bank……………………………………………………(807-820)
(viii)
Unit-1
1. Functions of Complex Variable Analytic Functions
2. Complex Integration
3. Power Series and Calculus of Residues
Unit-1
Chapter
1
Functions of Complex
Variable– Analytic Functions
1.1 Introduction
Complex analysis is the branch of mathematics investigating functions of complex
numbers. It is enormously useful in many branches of mathematics, including
number theory and applied mathematics. The method of complex analysis are
widely applicable in treating and solving many engineering problems. Many of the
physical problems like heat conduction, fluid dynamics, electrostatics etc. are solved
with the help of complex analysis. The present chapter deals with the functions of a
complex variable and their properties.
1.2 Some Basic Definitions
Before starting the study of complex variable, let us introduce some basic terms
which are important in complex analysis :
(i) Complex plane : The xy-plane in which complex numbers z = x + iy are
represented are complex plane. Here x and y are called real and imaginary axis
respectively.
(ii) Point set : Any collection of points in the complex plane is called a point set
and each point is called an element of the set.
(iii) Circle : z − z 0 = ρ represents a circle with centre at the point z0 and of radius ρ.
Open and closed disk : The set of points which satisfies the equation
z − z 0 < ρ defines an open disk while the set of point which satisfies the
equation z − z 0 ≤ ρ is called a closed disk.
Neighborhood : The neighborhood of a point z0 consists of all points z lying
inside but not on the circle of radius δ with centre at z 0 , i.e. z − z 0 < δ.
A deleted neighorhood of a point z0 in the set of points in z − z 0 < δ except
the point z0 itself.
Annulus of z0 is given by ρ1 < z − z 0 < ρ 2 .
(iv) Interior point : A point z0 is said to be an interior point of a set S if there exist
some neighborhood of z0 which contains only points of S.
(v) Boundary point : If every neighborhood of z0 contains points not belonging
to set S then z0 is called boundary points of S.
4
Engineering Mathematics-III
(vi) Exterior Point - If a point is neither interior nor boundary, then it is called
exterior point of S.
(vii)Open and closed Set - If a point set contains only interior points then it is
called an open set. For example a set defined by z < r is an open set. If,
however, z ≤ r, then the set is called a closed set.
(viii) Bounded Set - A set S is said to be bounded if there exists a finite positive
number k such that z ≤ k for every point z belonging to S. If a set is not
bounded, then it is called unbounded set. For example, z − z 0 < 5 is a
bounded set whereas z − z 0 > 5 is unbounded.
(ix) Connected Set - A set S is said to be connected if any two of its points are
joined by a polygonal line (a path which consist of finitely many straight line
segments), all of whose points belong to S. For example, the set z < 2 is a
connected set.
(x) Domain - An open connected set is called a domain. For example, the annulus
3 < z < 4 is a domain. Thus any neighborhood is a domain.
Boundary of a domain is the collection of all boundary points of S. If boundary
is included to an open domain, then it is called a closed domain.
(xi) Region - A domain together with all, some or none of its boundary points is
called a region.
1.3 Some Basic Results
1.
Polar Form of Complex Number
x = r cos θ, y = r sin θ, z = x + iy = r (cos θ + i sin θ)
2.
Euler 's Theorem
(a) eiθ = cos θ + i sin θ (b) e −iθ = (cos θ − i sin θ) (c) z= reiθ =r (cos θ+ i sin θ)
3.
Exponential Function
(a) ez = ex+iy = ex. iiy = ex(cos y+ i sin y)
(b) eiθ =
cos 2 θ + sin 2 θ = 1
(c) e z = e x + iy = e x . eiy = e x
(d) ez=1= e nπ α if and only if z =2nπ i, n is any integer.
(e) e2nπ i = cos 2nπ + i sin 2nπ = 1,
Thus, we can write ez = e z + 2n
4.
πi
, n is any integer.
Trigonometric Functions
(a) sin x =
eix – e −ix
2i
(b) cos x =
eix + e −ix
2
Functions of Complex Variable– Analytic Functions
5.
5
Hyperbolic Functions
(a) sinh x =
e x – e− x
2
(b) cosh x =
(c) cosh x + sinh x = e x
e x + e− x
2
(d) cosh x − sinh x = e – x
(e) cosh2 x − sinh2 x = 1
6.
Trigonometric and hyperbolic function in terms of z=x+iy
1
1
1
(i) (a) sin z = [ eiz – e – iz ] = [ ei ( x + iy ) – e −i ( x + iy ) ] = [ e − y eix – e y e −ix ]
2
2
2
1
= [ e − y (cos x + i sin x ) − e y (cos x − i sin x )]
2
= cos x cosh y + i sin x sinh y
1
(b) cos z = [ ei ( x + iy ) + e −i ( x + iy ) ] = cos x cosh y– i sin x sinh y
2
(c) sin (− z) = − sin z and cos (− z) = cos z
(d) sin z = sin z
1
1
(ii) (a) sin iz = [ ei (iz ) – e −i (iz ) ] = [ e − z – e z ] = i sinh z
2i
2i
(b) cos iz = cosh z
(c) sinh z = − i sin iz = − i (sin{i(x+iy)}
= i [sin(y − ix)] = i [sin y cosh x − i cos y sinh x]
= sinh x cos y + i cosh x sin y
(d) cosh z = cosh x cos y + i sinh x sin y
7. Logarithmic Functions
(a) If z = reiθ and w = u+iv in z = ew , then we get reiθ = eu+iv
or
rei (θ + 2nπ ) = eu + iv = e w
Thus
e w = eu = r or u = ln r
[since ei θ = ei (θ + 2nπ ) ]
Hence logarithm of real variable r = z and v = θ + 2nπ, n is any integer,
or
w = ln z = ln r + i (θ+2nπ), n is any integer .
(b) Ln z = ln z + iθ, −π< θ ≤ π (principal argument)
(c) Ln z = ln (x2+y2)1/2 + i tan -1 y/x
(d) Ln (1) = 0, ln (1) = 2nπ i.
(c) Ln (−1) = πi, ln (−1) = (2n+1) πi
6
Engineering Mathematics-III
1.4 Functions of a Complex Variable
If a symbol z takes any one of the values of a set of complex numbers, then z is called a
complex variable. If for each value of the complex variable z (=x+iy) in a given region
R, there exist one or more values of w (=u+iv) then w is said to be a function of z and
is written as w=u(x, y)+iv (x, y)= f(z) where u, v are real functions of x and y.
If to each value of z, there exists one and only one value of w, then w is said to be a
single valued function of z otherwise a multi-valued function; e.g., w=(1/z) is a
single valued function whereas w = z is a multi-valued function of z.
Limit - A number l is said to be the limit of a function w = f (z) as z tends to z0 and is
denoted by
lim w = lim f ( z ) = l
z→ z 0
z→ z 0
if for every ε > 0, there exist a number δ > 0 depending upon ε such that
f ( z ) − l < ε whenever 0 < z − z 0 < δ, z ≠ z0.
Here z may approach z0 along any path, straight or curved since the two points
joining z and z0 in a complex plane can be joined by infinite number of curves.
When we say z → z0 we mean x→ x0 and y → yo.
Illustrative Examples
z3 − 8
.
z→ 2 z − 2
Ex. 1 : Find the lim
Sol : Since z ≠ 2, we have
Therefore,
Further,
z 3 − 2 3 ( z − 2)( z 2 + 2z + 4)
=
= z 2 + 2z + 4
z−2
z−2
lim z2+2z+4 = 12
z→ 2
z2+2z+4 = (x + iy)2 +2 (x + iy) +4
= x2 + 2x − y2 + i(2y + 2xy) + 4
= (x2 + 2x − y2 + 4) +i(2y + 2xy)
Since z → 2 implies x + iy → 2 or x → 2, y → 0
lim z 2 + 2z + 4 = lim ( x 2 + 2 x − y 2 + 4) + i( 2 y + 2 xy ) =12
Hence ,
x→ 2
y→ 0
x→ 2
y→ 0
Continuity of f (z). A single-valued function w= f (z) is said to be continuous at
z=z0, if lim f ( z ) = f(z0) . f(z) is said to be continuous in any region R of the z-plane,
z→ z 0
if it is continuous at every point of the region R. When w = f (z) = u (x, y)+iv (x, y) is
continuous at z=z0, then u (x, y) and v (x, y) are also continuous at z = z0, i.e. at x =
x0 and y = y0; and conversely if u (x, y) and v (x, y) are continuous at (x0, y0), then
f(z) will be continuous at z = z 0 .
Functions of Complex Variable– Analytic Functions
7
A function which is not continuous at z0 then either f(z0) does not exist or lim f(z) ≠
z – z0
f(z0).
Note : If f(z) and g(z) are two continuous functions of complex variable then their
sum, difference and product are also continuous. The quotient f(z)/g(z) is also
continuous at points where g(z) does not vanish. Continuous function of continuous
function is also continuous.
Ex. 2 : Discuss the continuity of the function f(z) at z =0, where
 x3 + y 3
x3 − y 3
+
i
,z ≠ 0

f(z) =  x 2 + y 2
x2 + y 2

z=0
 0,
Sol : We have f (z) = u(x , y) + i v (x, y)
Hence at z ≠ 0, i.e. (x, y) ≠ (0, 0)
u(x, y) =
and
Now
and
v (x, y) =
x3 + y 3
x2 + y 2
x3 − y 3
x2 + y 2
lim u ( x, y ) = lim y = 0
x→ 0
y→ 0
y→ 0
lim u ( x, y ) = lim x = 0
y→ 0
x→ 0
x→ 0
Further, when y =mx, we have
lim u( x, y ) = lim
y = mx
x→ 0
Similarly,
x→ 0
x 3 + m3 x 3
x 2 + m2 x 2
= lim
x→ 0
x(1 + m 3 )
(1 + m 2 )
=0
lim v ( x, y ) = 0 along any path in which x and y tend to zero.
x→ 0
y→ 0
Therefore u(x, y) and v(x, y) are continuous at (0,0). Hence f(z) is continuous at
z=0.
Ex. 3 : Show that the function
 Im (z )
,z≠0

f (z ) =  z

 0 ,z=0
is not continuos at z=0.
8
Engineering Mathematics-III
Im ( z )
z
Sol : Since
Now,
=
I m ( x + iy )
2
x +y
Im( z )
= lim
z→ 0 z
x→ 0
lim
y→ 0
Im( z )
= lim
z→ 0 z
y→ 0
lim
x→ 0
2
=
y
x2 + y 2
y
x2 + y 2
y
2
x + y2
= lim 1 = 1
y→ 0
= lim 0 = 0
x→ 0
Here we see that different paths give different limit, hence the function is not
continuous at z=0. This can be shown alternatively as follows :
Consider the path y=mx, then,
Im( z )
lim
= lim
z→ 0 z
x→ 0
y = mx
y
x2 + y 2
= lim
x→ 0
mx
x 2 + m2 x 2
m
=
1 + m2
Since this limit depends on m. Hence the limit does not exist at z=0.
Derivative of f (z). Similar to the case of limits and continuity,
the concepts of differentiation of a function of complex variable is
defined in the same way as the real variable.
B=
z +∆ z
If w=f(z) is a single-valued function of the variable z=x+iy, then
the derivative of w=f(z) at z=z0 is defined by
f (z ) − f (z 0 )
dw
= f ′ ( z 0 ) = lim
z→ z 0
dz
z − z0
= lim
∆z→ 0
A=z
Fig. 1.1
f ( z 0 + ∆z ) − f ( z 0 )
, where ∆z = z – z 0
∆z
provided the limit exists and has the same value for all the different paths in which
∆z tends to zero.
Let A (z) be a fixed point and B ( z + ∆z ) be a neighboring point of A as shown in Fig
1.1. The point B may approach A along any straight or curved path in the given
region i.e. ∆ z may tend to zero in any manner and (dw/dz) may not exist. Hence, it
becomes a fundamental problem to determine the necessary and sufficient
condition for (dw/dz) to exist. This problem is settled by the theorem 1.1.1.
Note :
(i)
If a functions f(z) is not continuous at z = z0 then it cannot be differentiable at z = z0.
(ii) If a function f (z) is continuous at a point z = z0 then it may or may not be
differentiable at z = z0.
(iii) If a function f(z) is differentiable at z = z0 then it must be continuous at z=z0 .
Proof : We have
Functions of Complex Variable– Analytic Functions
lim [f(z) − f(z0)] = lim
z→ z 0
z→ z 0
= lim
z→ z 0
9
[ f ( z ) − f ( z 0 )][ z − z 0 ]
z − z0
f (z ) − f (z 0 )
lim ( z − z 0 ) = f ′ ( z 0 ). 0 = 0
z→ z 0
z − z0
Hence, lim f ( z ) = lim f ( z 0 ) = f ( z 0 ), which shows that the function is continuous at z0.
z→ z 0
z→ z 0
(iv) The rule of differentiation of function of a real variable x holds for a function of
complex variable z as well.
If (z) and g(z) are differentiable function at z, then the function f(z) ± g(z),
f(z) g(z) and f(z)/g(z) are also differentiable at z.
(v) As real calculus, the following rules for differentiation of complex variable also satisfy.
d
d
d
(a)
[c1 f (z ) + c 2 g (z )] = c1 dz (f(z))+c2 dz (g (z ))
dz
d
(b)
{ f (z ). g (z )} = f ′ (z ) . g (z ) + f (z ) . g ′ (z )
dz
f ′ ( z )g ( z ) − f ( z ) . g ′ ( z )
d
(c)
{ f (z ) / g (z )} =
dz
[ g ( z )]2
d n
( z ) = nz n −1
dz
d
(e)
[ f ( g ( z ))] = f '[ g ( z )] . g ' ( z ) (chain rule).
dz
(d)
Ex. 4 Show that the function f (z) = z is continuous at z=0 but not derivable.
Sol : Since
Now,
z = x+iy, hence z = x − iy
lim f(z) = lim x − iy = 0 = f (0)
z→ 0
x→ 0
y→ 0
Hence the function is continuous at z =0
Now
f ( z + ∆z ) = z + ∆z = x + ∆x − i( y + ∆y )
Therefore
f ( z + ∆z ) − f ( z ) = x + ∆x – i( y + ∆y ) – ( x – iy ) = ∆x − i∆y
f ( z + ∆z ) − f ( z )
∆x − i∆y
f ′(z)= lim
= lim
∆z→ 0
∆z→ 0 ∆x + i∆y
∆z
Hence
Now, at z=0
f ′(0) = lim
∆z→ 0
f ( ∆z ) − f ( 0)
∆x − i∆y
–i∆y
= lim
= lim
∆x→ 0 ∆x + i∆y
∆y → 0 i∆y
∆z
∆y → 0
= lim− 1 = −1
∆y → 0
and
f ′(0) = lim
∆z→ 0
= lim
∆x→ 0
f ( ∆z ) − f ( 0)
∆x − i∆y
∆x
= lim
= lim
∆y → 0 ∆x + i∆y
∆x→ 0 ∆x
∆z
∆x→ 0
1=1
10
Engineering Mathematics-III
Since the limits are not equal, hence the functions is not differentiable at z=0. In fact
this function is not differentiable anywhere.
Theorem 1.1.1
The necessary and sufficient conditions for the derivative of the function
w = u (x, y) + iv (x,y) = f(z) to exist for all values of z in a region R, are
(i)
∂u ∂u ∂v ∂v
,
, ,
are continuous functions of x and y in R
∂x ∂y ∂x ∂y
(ii)
∂u ∂v ∂u
∂v
i.e. , ux = vy , uy = − vx
=
,
=–
∂x ∂y ∂y
∂x
These relations are known as Cauchy-Reimann equations or briefly as C – R
equations.
Proof : Condition is necessary. Let ∆u and ∆v be small increments of u and v
corresponding to the increments ∆x and ∆y of x and y, so that ∆ z = ∆ x + i∆y.
Now if f (z) possesses a unique derivative at P (z), we have
f ( z + ∆z) − f ( z )
f ′ ( z ) = lim
∆z → 0
∆z
(u + ∆u ) + i( v + ∆v ) − (u + iv )
= lim
∆z → 0
∆z
∆u
∆v
= lim 
+ i 

∆z → 0 ∆z
∆z 
...(1.1)
Now ∆z → 0 in any manner; we first assume ∆z to be wholly real implying that
∆y = 0 and ∆z = ∆x. Then, we have
∆ v  ∂u
∂v
 ∆u
f ' ( z ) = lim 
+i
+i
 =
Λ x→ 0  ∆ x
∆ x  ∂x
∂x
... (1.2)
Further, when ∆z is wholly imaginary, then ∆ x =0 and ∆z = i∆y . Then
∆ v  1 ∂u ∂v
∂v
∂u
 ∆u
f ' ( z ) = lim 
+i
+
=
−i
 =
∆y → 0  i∆y
i∆y  i ∂y ∂y ∂y
∂y
But the existence of f ' (z) requires that (1.2) = (1.3) ⇒
... (1.3)
∂u
∂v ∂v
∂u
+i
=
−i
∂x
∂x ∂y
∂y
Equating real and imaginary parts from both sides, we get
∂u ∂v ∂u
∂v
or vx=uy, uy =− vx
=
,
=−
∂x ∂y ∂y
∂x
... (1.4)
These are the necessary conditions for the existence of the derivative of f(z) i.e. C —R
equations should be satisfied.
Functions of Complex Variable– Analytic Functions
11
Condition is sufficient : Let f(z) be a single-valued function possessing partial
∂u ∂u ∂v ∂v
derivatives
at each point of the region and let the C — R equations
,
, ,
∂x ∂y ∂x ∂y
(4) are satisfied. Then using Taylor's theorem for functions of two variables, we get
f ( z + ∆z ) = u( x + ∆x, y + ∆y ) + iv( x + ∆x, y + ∆y )
∂u 
 ∂u
= u( x, y ) +  ∆x +
∆y  +...
∂y
 ∂y



∂v 
 ∂v
+ i v( x, y ) +  ∆x +
∆y  +....
∂y
 ∂x



∂u
∂v
∂v 
 ∂u
= f ( z ) + 
+ i  ∆x + 
+ i  ∆y
 ∂x
∂x 
∂y 
 ∂y
[neglecting terms beyond first powers of ∆x and ∆y]
⇒
∂u
∂v
∂v 
 ∂u
f ( z + ∆z) − f(z) = 
+ i  ∆x + 
+ i  ∆y
 ∂x
∂x 
∂
y
∂
y

∂u
∂v
∂v
∂u
=  + i  ∆x + –
+ i  ∆y
 ∂x


∂x 
∂x 
 ∂x
(∴ u y = − v x , v y = u x )
∂u
∂v
∂v ∂u 
=  + i  ∆x + i
+
i∆y
 ∂x
 ∂x ∂x 
∂x 
∂u
∂v
∂u
∂v
=  + i ( ∆x + i∆y ) =  + i  ∆z
 ∂x
 ∂x
∂x 
∂x 
f ' ( z ) = lim
⇒
∆z → 0
f ( z + ∆ z − f ( z ) ∂u
∂v
∂v
∂u
=
+i
or
–i
∆z
∂x
∂x
∂y
∂y
Thus, (1.2) or (1.3) proves the sufficiency of conditions.
Note : (i) f '(z) can be obtained directly by using (1.2) or (1.3).
(ii) The derivative f ' ( z ) may not exist even when f(z) satisfies C – R equations.
1
Ex. 5 : Using Cauchy-Riemann equation show that the function f(z) = , z ≠ 0, is
z
differentiable at all points except at the point z=0.
Sol : Since
f(z) =
—
x − iy
1
z
=
=
,z ≠0
—
2
z
z z
z
=
x − iy
2
x + y2
Now comparing this with f(z) = u(x,y) +iv(x,y), we have
12
Engineering Mathematics-III
u(x,y) =
x
2
x +y
2
, v(x, y) = −
y
2
x + y2
Now, we obtain
y 2 − x2
∂u
∂v
=
=
2
2
2
∂x ( x + y )
∂y
and
2 xy
∂u
∂v
=−
=− .
2
2 2
∂y
∂x
(x + y )
Hence Cauchy-Riemann equations are satisfied.
Since, the partial derivative u x , u y , v x and v y are continuous at all points except 0
and Cauchy-Riemann equations are also satisfied at all all points except 0. Hence
the function is differentiable at all points except 0.
Ex. 6 : Find the derivative of f ( z ) = e z
Sol : We have f(z)= u + iv = e z = e x + iy = e x . eiy .
= e x (cos y + i sin y)
Which gives
Now
u = e x cos y, v = e x sin y
∂u
∂u
= e x cos y ,
= − e x sin y
∂x
∂y
∂v
∂v
= e x sin y,
= e x cos y.
∂x
∂y
Which shows
∂u
∂v
∂u
∂v
=
and
=−
,
∂x
∂y
∂y
∂y
i.e., Cauchy-Riemann equations are satisfied.
Since u, v, ux , uy, vx and vy are continuous and C-R equations are satisfied hence f '(z)
exist.
∂u
∂v
f '(z) =
+i
= e x cos y + ie x sin y =e x . eiy = e x + iy = e z .
∂x
∂x
We can find f '(z) alternatively by using f '(z) =
∂v
∂u
= e x cos y − i (−e x sin y)
−i
∂y
∂y
= e x (cos y+ i sin y) = e x . eiy = e z .
Ex. 7 : Prove that the function f(z) = u+iv,
where
f(z) =
x 3 (1 + i ) − y 3 (1 − i )
x2 + y 2
( z ≠ o) and f ( 0) = 0
is continuous and that Cauchy-Riemann equations are satisfied at the origin, yet
f ' ( z ) does not exist there.
Functions of Complex Variable– Analytic Functions
Sol : Here
u=
x3
− y3
x2 + y 2
,v=
x3
13
+ y3
x2 + y 2
where z ≠ 0.
We see that both u and v are rational and finite for all values of z ≠ 0, so u and v are
continuous at all those points for which z ≠ 0. Hence f(z) is continuous when z ≠ 0.
At the origin u = 0, v = 0.
[since f(0) = 0]
Hence u and v are both continuous at the origin, consequently f (z) is continuous at
the origin. Now, at the origin
u( x, 0) − u( 0, 0)
∂u
x
= lim
= lim   = 1
x→ o  x 
∂x x→ o
x
u( 0, y ) − u( 0, 0)
∂u
−y
= lim
= lim   = −1
y→ o  y 
∂y y → o
y
v( x, 0) − v( 0, 0)
∂v
x
= lim
= lim   = 1
x→ o  x 
∂x x→ o
x
v( 0, y ) − v( 0, 0)
∂v
 y
= lim
= lim   = 1
y→ 0  y 
∂y y → o
y
Hence
∂u ∂v
∂u
∂v
=
and
=−
∂x ∂y
∂y
∂x
Therefore Cauchy-Riemann equations are satisfied at z=o.
f ( z ) − f ( 0)
Again
f '(0) = lim
z→ 0
z
x3
= lim 
z → 0

− y 3 + i( x 3 + y 3 ) 1 
.
 (∵ f ( 0) = 0)
x + iy 
x2 + y 2
Let z → 0 along y = x, then we have
f '(0) = lim
z→ 0
x 3 − x 3 + i( x 3 + x 3 )
2
2
x +x
2i
1
= lim
= (1 + i )
z→ 0 2(1 + i )
2
.
1
x + ix
Further, let z → 0 along y = 0, then we have f ' ( 0) = lim
x→ 0
x 3 (1 + i )
x3
= 1+i
Hence f '(0) is not unique. Thus, f '(z) does not exist at the origin.
1.5 Analytic Function
A function f(z) which is single-valued and possesses a derivative with respect to z at
all points of a region R is said to be analytic (or regular or holomorphic)
function of z in that region. A function f(z) is said to be analytic in a domain D if it is
analytic at every point in D. A function which is analytic at every point of the finite
14
Engineering Mathematics-III
complex plane is called an entire function. For example polynomial of any degree is an
entire function.
A point at which an analytic function ceases to have a derivative is called a singular
point of the function.
Theorem 1.1.2
If u and v are real single-valued functions of x and y such that( ∂u / ∂x ),( ∂u / ∂y ),
( ∂v / ∂x ), ( ∂v / ∂y ), are continuous throughout a region R, then the Cauchy-Riemann
equations
∂u ∂v
∂u
∂v
=
and
=−
∂x ∂y
∂y
∂x
are both necessary and sufficient conditions for the function f(z) = u + iv to be
analytic in R.
For Proof Refer Theorem 1.1.1
Note :
(i) The real and imaginary parts of the analytic function f(z) are called conjugate
functions.
(ii) If the function f(z) does not satisfy the C – R equations at a point, it is not
differentiable and hence not analytic at that point.
(iii) Analyticity implies diffrentiability but the converse is not always true. For
example f (z) =z2 is differentiable at z = 0 but nowhere else. Therefore, this
function is differentiable at z=0 but not analytic anywhere.
(iv) A function f (z) is analytic at z=∞, if the function f (1/z) is analytic at z=0.
(v) If two functions f(z) and g(z) are analytic in a domain D, then their sum,
difference, product are also analytic in domain D.
(vi) If two functions f(z) and g(z) are analytic in domain D, then f (z)/g(z) is also
analytic provided g (z) does not vanish at any point in D.
Ex. 8 : Show that the function ex (cos y+i sin y) is analytic and find its derivative.
Sol : Let e x (cos y + i sin y) ≡ f (z) = u+iv.
So, here u = e x cos y, v = e x sin y, then
∂u
∂v
= e x cos y ,
= e x sin y ,
∂x
∂x
∂u
∂v
= − e x sin y ,
= e x cos y
∂y
∂y
∂u ∂v ∂u
∂v
Here we see that
, that is, Cauchy-Riemann conditions are
=
,
=−
∂x ∂y ∂y
∂x
satisfied and u,v along with their partial derivatives are continuous.
Hence the function f (z) ≡ e x (cos y+i sin y) is analytic.
∂u
∂v
We have
f ' (z) =
+i
= e x cos y + ie x sin y
∂x
∂x
= ex (cos y + i sin y) = e x eiy = e( x + iy ) = e z .
Functions of Complex Variable– Analytic Functions
15
1.6. Polar Form of Cauchy-Riemann Conditions
We have x = r cos θ, y = r sin θ, so that r2= x2+y2 and θ = tan-1 (y/x).
⇒
sin θ
∂r x
∂r
∂θ
1
∂θ cos θ
 y 
≡ = cos θ,
= sin θ,
=
and
=
− 2  = −
2
2
∂x r
∂y
∂x 1 + ( y / x )  x 
r
∂y
r
Now,
Similarly,
and
∂u ∂u ∂r ∂u ∂θ ∂u
=
.
+
.
=
cos θ −
∂x
∂r ∂x ∂θ ∂x ∂r
∂u ∂u ∂r ∂u ∂θ ∂u
=
.
+
.
=
sin θ +
∂y
∂r ∂y ∂θ ∂y ∂r
∂u sin θ
∂θ r
∂u cos θ
.
∂θ
r
∂v ∂v
∂v sin θ
=
cos θ −
.
∂x ∂r
∂θ r
∂v ∂v
∂v cos θ
=
sin θ +
.
∂y
∂r
∂θ
r
Cauchy-Riemann equations in cartesian form are
...(1.1)
... (1.2)
...(1.3)
... (1.4)
∂u ∂v ∂u
∂v
=
,
=−
∂x ∂y ∂y
∂x
Substituting (1.1), (1.2), (1.3) and (1.4) in C – R equations, we have
∂u
∂u sin θ ∂v
∂v cos θ
cos θ −
=
sin θ +
.
∂r
∂θ r
∂r
∂θ
r
∂u
∂u cos θ
∂v
∂v sin θ
and
sin θ +
=−
cos θ +
∂r
∂θ r
∂r
∂θ r
... (1.5)
... (1.6)
Multiplying (1.5) by cos θ and (1.6) by sin θ, and adding, we get
∂u 1 ∂v
=
∂r r ∂θ
Again, multiplying (1.5) by sin θ and (1.6) by cos θ and subtracting, we have
∂u
∂v
= −r
∂θ
∂r
∂u 1 ∂v
... (1.7)
⇒
=
∂r r ∂θ
and
∂u
∂v
= −r
∂θ
∂r
...(1.8)
Relations (1.7) and (1.8) are Cauchy-Riemann equations in polar form.
1.7 Derivative of w in Polar Form
∂w ∂u
∂v
=
+i
∂x
∂x
∂x
dw ∂w ∂w ∂r ∂w ∂θ
=
=
.
+
.
dz
∂x
∂r ∂x ∂θ ∂x
We have w=u+iv, therefore
But
=
∂w
∂u
∂v sin θ
cos θ − 
+ i 
 ∂θ
∂r
∂θ r
(∵ w = u + iv )
16
Engineering Mathematics-III
=
∂w
∂v
∂u sin θ
cos θ −  − r
+ i. r 
 ∂r
∂r
∂r  r
[From C – R equations in polar form]
=
Hence
∂w
∂u
∂v
∂w
∂w
cos θ − i 
+ i  sin θ =
cos θ − i
sin θ
 ∂r
∂r
∂r 
∂r
∂r
dw
∂w
= (cos θ − i sin θ)
dz
∂r
This is a derivative of w.
Again,
dw ∂w ∂r ∂w ∂θ
=
.
+
.
dz
∂r ∂x ∂θ ∂x
∂u
∂v
∂w sin θ
= 
+ i  cos θ −
.
 ∂r

∂r
∂θ
r
(∵ w = u + iv )
1 ∂v
1 ∂u 
∂w sin θ
= 
−i
.
 cos θ −
 r ∂θ

r ∂θ
∂θ
r
[From C – R equations in polar form]
=–
i  ∂u
∂v
∂w sin θ
+ i  cos θ −
.
, since w = u + iv

r  ∂θ
∂r 
∂θ
r
i
∂w
= – (cos θ – i sin θ)
r
∂θ
This also is a derivative of w.
Hence,
dw
∂w
i
∂w
= (cos θ − i sin θ)
= − (cos θ − i sin θ)
.
dz
∂r
r
∂θ
Ex. 9 : If n is real, show that r n (cos n θ +i sin n θ) is analytic except possibly at r =
0, and that its derivative is n r n –1 [cos (n −1) θ +i sin (n −1) θ].
Sol : Let w= r n (cos n θ+i sin n θ) = f (z) = u+iv
So here u=r n cos nθ, v= r n sin nθ, then
∂u
= nr n −1 cos nθ,
∂r
∂u
= − nr n sin nθ,
∂θ
Thus,
∂v
= nr n –1 sin nθ,
∂r
∂v
= nr n cos nθ
∂θ
∂u 1 ∂v
1 ∂u
∂v
=
, and
=−
∂r r ∂θ
r ∂θ
∂r
Hence, Cauchy-Riemann equations are satisfied.
Therefore, The function w=r n (cos nθ+i sin n θ) is analytic, if
values of z.
dw
exists for all finite
dz
Functions of Complex Variable– Analytic Functions
17
dw
∂w
= (cos θ − i sin θ)
dz
∂r
We have ,
= (cos θ − i sin θ). nr n –1 (cos n θ + i sin n θ)
= nr n –1 {cos(n−1) θ + i sin (n−1) θ}
Thus
dw
exists for all finite values of r including zero, except when r=0 and n ≤ 1.
dz
Ex. 10 : Test the analyticity of the function w=sin z and hence derive that
d
(sin z ) = cos z.
dz
Sol : f(z) = w =sin z = sin (x + iy) = sin x cos iy +cos x sin iy = sin x cosh y + i cos x
sinh y = u + iv
u = sin x cosh y and v = cos x sinh y
∂u
∂u
=cos x cosh y,
= sin x sinh y
∂x
∂y
⇒
⇒
and
∂v
∂v
= cos x cosh y
= − sin x sinh y,
∂x
∂y
⇒
∂u ∂v
∂u
∂v
and
=
=−
∂x ∂y
∂y
∂x
so C–R equations are satisfied. Hence sin z is an analytical function.
d
d
Now
(sin z) =
[sin x cosh y+i cos x sinh y]
dz
dz
∂
=
(sin x cosh y+i cos x sinh y)
∂x
= cos x cosh y− i sin x sinh y
= cos x cos iy − sin x sin iy = cos (x+iy) = cos z.
Ex. 11: For what value of z the function w defined by the following equations ceases
to be analytic; where z = e − v (cos u + i sin u ) , w = u +iv.
Sol : We have
z= e (cos u + i sin u) =e – v . eiu
or
z = ei (u + iv ) = eiw
or
Thus
iw = log z
dw
1
=
dz
iz
Hence, w ceases to be analytic when iz = 0, i.e. z=0.
Ex.12 : Find the values of C1 and C2 such that the function
f(z) = x2+C1 y2 −2xy +i (C2 x2 − y2+2xy) is analytic. Also find f '(z).
[UPTU 2002]
18
Engineering Mathematics-III
Sol : Let w =f (z) = u+ iv = x2+C1y2 − 2 xy +i(C2 x2 − y2+2 xy)
Equating real and imaginary parts, we have u =x2+C1y2 − 2xy and v = C2x2 − y2+2xy
∂u
∂v
∂u
= 2x − 2y,
=2C2 x+2y,
= 2C1 y − 2x
⇒
∂x
∂x
∂y
and
∂v
= −2y+2x
∂y
Now C–R equations are :
 ∂u = ∂v
 ∂x ∂y

 ∂u
∂v
=−

∂
y
∂x

⇒ 2 x − 2 y = −2 y + 2 x
... (1.1)
⇒ 2 C1 y − 2 x = −2 C 2 x − 2 y
... (1.2)
Then from (2), equating the coefficients of x and y, we get
2C1=−2 ⇒ C1=−1 and −2 =−2C2 ⇒ C 2 =1
⇒
∂u
∂v
= 2x−2y,
= 2x+2y.
∂x
∂x
Then
dw
∂u
∂v
= f ' (z ) =
+i
dz
∂x
∂x
= ( 2 x − 2 y ) + i( 2 x + 2 y ) = 2[( x + ix ) + ( − y + iy )]
= 2[ x(1 + i ) + i( y + iy )] = 2[( x + iy ) + i( x + yi )].
= 2( x + iy )(1 + i ) = 2(1 + i )z
Ex.13 : Find the point where the Cauchy-Riemann equations are satisfied for the
function f (z) = xy2 + ix2y. Where does f ' (z) exist? Where is f (z) analytic?
Sol : We have w = f (z) = xy2 + i x2y = u + iv
∂u
∂v
∂u
∂v
⇒
= y 2,
= 2 xy ,
= 2 xy ,
= x2
∂x
∂x
∂y
∂y
Now w=f (z) is an analytic function, if it satisfies C−R equations, i.e.when
∂u ∂v
... (1.1)
=
i. e. y 2 = x 2
∂x ∂y
and
∂u
∂v
=−
i. e. 2 xy = −2 x y
∂y
∂x
or
4x y = 0
... (1.2)
Solving these, we get x = y = 0.
Thus, at origin C–R equations are satisfied, f '(z) exists at origin only and nowhere
else. Hence f (z) is analytic at origin only.
Ex.14: Show that the function z z is not analytic anywhere.
Sol : w=z z = f ( z ) = u + iv, where z=x+iy and z =
x2 + y 2
Functions of Complex Variable– Analytic Functions
19
2
u + iv = ( x + iy ) x
Now
w=z z
⇒
u = x x 2 + y 2 and v = y x 2 + y 2
⇒
∴
∂u
=
∂x
x2 + y 2 +
x.2 x
and
∂v
=
∂y
x2 + y 2 +
∴
∂u ∂v
except when x=y
≠
∂x ∂y
Also
x.2 y
∂u
=
=
∂y 2 x 2 + y 2
2 x2 + y 2
y .2 y
2 x2 + y 2
xy
x2 + y 2
y .2 x
∂v
=
=
∂x 2 x 2 + y 2
⇒
=
=
+ y2
x2 + y 2 + x2
x2 + y 2
x2 + y 2 + y 2
x2 + y 2
,
xy
x2 + y 2
∂u
∂v
≠−
∂y
∂x
Ex .15: Prove that the function f (z) = z
2
is continuous everywhere but nowhere
differentiable except at the origin, hence non–analytic.
Sol : Now
f ' (z) = lim
∆ z→ 0
z + ∆z
2
− z
2
( z + ∆z )( −z + ∆ −z ) − z −z
∆ z→ 0
∆z
= lim
∆z


∆ −z 
–
= lim z + ∆ −z + z .
∆ z → 0
∆ z




∆ −z 
–
= lim z + z
∆ z → 0
∆ z


(∵ ∆z → 0 as ∆z → 0)
...(1.1)
Let
∆ z = ρ (cos θ − i sin θ),
⇒
∆ −z = ρ (cos θ − i sin θ)
⇒
∆ z cos θ − i sin θ
=
= (cos θ − i sin θ) × (cos θ + i sin θ) −1
∆ z cos θ + i sin θ
= (cos θ − i sin θ) 2 = cos 2 θ − i sin 2 θ
[By D'Moivre's theorem]
20
Engineering Mathematics-III
∆ −z
does not tend to unique limit as it depends upon arg ∆ z, where z is a
∆z → 0 ∆ z
Now lim
non-zero point. Hence from (1), it follows that f '(z) is not unique, i.e., f (z) is
non-differentiable for any non-zero values of z, hence non-analytic at these points.
When z = 0, the value of f '(z) i.e. f '(0) is zero as
f ' (0) = lim
∆z → 0
∆z
2
∆z
= lim ∆ z− = 0
−
∆ z=0
and is unique implying thereby that function is differentiable at z=0.
1.8 Orthogonal System
Two curves are said to be Orthogonal to each other, when they intersect at right
angle at each of their points of intersection.
To show that, if w = f (z) = u+iv be analytic function of z = x + iy, then the family of
curves u(x, y)= c1 and v(x, y) = c 2 form an orthogonal system.
For the curve u(x, y) = c1, we have on differentiating,
∂u
∂u
du = 0 ⇒
dx +
dy = 0
∂x
∂y
i.e.
dy
∂u / ∂x
=−
= m1 ( say )
dx
∂u / ∂y
and for the curve v (x, y) = c 2 , we have on differentiating
∂v
∂v
dx +
dy = 0
∂x
∂y
i.e.
Now
∂y
∂v / ∂x
=−
= m 2 ( say ).
∂x
∂v / ∂y
 ∂v / ∂x   ∂v / ∂x  ∂v / ∂y  ∂v / ∂x 
m1m2 =  −
× −
 × −
 =
 = −1
 ∂v / ∂y   ∂v / ∂y  ∂v / ∂x  ∂v / ∂y 
(since ux= vy . uy=−vx )
Hence the two curves u=c1 and v=c2 are orthogonal, and since c1 and c2 are
parameters therefore we say that the two family of curves u = c1 and v = c2 form an
orthogonal system.
1.9 Harmonic Functions
Any function of x,y which possesses continuous partial derivatives of the first and
second orders and satisfies Laplace's equation is called a harmonic functions.
Theorem 1.1.3 : If f (z) = u+iv is an analytic function, then u and v are both
harmonic functions.
Proof : Let f (z) = u+iv be an analytic function, then we have
Functions of Complex Variable– Analytic Functions
21
∂u ∂v
∂u
∂v
[Cauchy-Riemann equations]
=
and
=−
∂x ∂y
∂y
∂x
... (1.1)
Also, because u and v are the real and imaginary parts of an analytic function,
therefore, derivatives of u and v of all orders exist and are continuous functions of x
and y, so we have
∂ 2v
∂ 2v
=
∂x ∂y ∂y ∂x
... (1.2)
Again differentiating (1.1) with respect to x and y respectively, we have
∂ 2u
∂x 2
=
∂ 2v
∂ 2u
∂ 2v
and
=−
∂x ∂y
∂y ∂x
∂y 2
Now, adding these, we have
∂ 2u
∂x 2
∂ 2v
Similarly,
∂x 2
+
+
∂ 2u
∂y 2
∂ 2v
∂y 2
= 0 by virtue of (1.2)
=0
Hence both u and v satisfy Laplace's equation.
Therefore, both u and v are harmonic functions. Such functions u and v are also
called conjugate harmonic functions or simple conjugate functions.
Remark : If harmonic functions u and v satisfy Cauchy-Riemann equations in
domain D, then u+iv need not be analytic function in domain D.
1.10 Determination of the Conjugate Function and Analytic Function
Method 1 : If f(z) = u+iv is an analytic function where both u (x, y) and v (x, y) are
conjugate functions. Being given one of these say u (x, y), to determine the other
v (x, y) we follow the following procedure :
Since v is a function of x, y therefore, we have
∂v
∂v
∂u
∂u
dv =
dx +
dy = −
dx +
dy
∂x
∂y
∂y
∂x
(by C–R equations) ... (1.1)
But the right hand side of (1.1) is of the form Mdx+Ndy, where M=−
So
∂M
∂
=
∂y
∂y
∂u
∂u
and N= .
∂y
∂x
∂ 2u
∂N
∂  ∂u  ∂ 2u
 ∂u 
=
  =
 −  = − 2 and
∂x
∂x  ∂x  ∂x 2
 ∂y 
∂y
Again u is a harmonic function, therefore, it satisfies Laplace's equation,
i.e.,
∂ 2u
∂x 2
+
∂ 2u
∂y 2
=0
or
∂ 2u
∂x 2
=−
∂ 2u
∂y 2
22
Thus
Engineering Mathematics-III
∂M ∂N
=
∂y
∂x
This implies that equation (1.1) satisfies the condition of exact differential equation
and hence it can be integrated so as to obtain v.
Method 2 : Milne—Thomson's Method
Milne's-Thomson method is an elegant method to obtain function f (z) directly when
one of the conjugate function's is given. In this method if one of the conjugate
functions say u(x, y) is given then we don't have to find v(x, y) as done in method(1).
We know
x=
1
 1
−
−
 z + z and y =   ( z − z )
2
 2i 



−
−
−
−
z + z z − z
z + z z − z
f ( z ) = u( x, y ) + i v( x, y ) = u 
,
,
 + iv 

2i 
2
2i 
 2




∴
... (1.1)
This relation can be regarded as a formal identity in two independent variables z and
−
z. Putting −z = z, we get
f (z) = u (z,0) + iv (z, 0).
Again
w = f (z) = u + iv
∂w ∂w ∂u
∂v ∂u
∂u
f ' (z ) =
=
=
+i
=
−i
∂z
∂x
∂x
∂x ∂x
∂y
⇒
(by C–R equations)
Hence, if we write
∂u
∂u
= φ1 ( x, y ) and
= φ 2 ( x, y ), we have
∂x
∂y
∂w
= f ' ( z ) = φ1 ( x, y ) − iφ 2 ( x, y ) = φ1 ( z,0) − iφ 2 ( z,0)
∂z
(∵ −z = z ⇒ x = z, y = 0)
Integrating with respect to z, we have
f ( z ) = ∫[ φ1(z,0) − iφ 2(z,0)]dz + C , where C is an arbitrary constant
... (1.2)
Thus function f (z) is constructed when u (x, y) is given.
Similarly if v (x, y) is given, it can be shown that
f ( z ) = ∫[ ψ1 ( z, 0) + iψ 2 ( z, 0)]dz + C ,
where
ψ1 ( x, y ) =
∂v
∂v
and ψ 2 ( x, y ) =
∂y
∂x
Procedure :
Step (1) (a) If u(x , y) is given, take f ' (z) = ux − iuy.
... (1.3)
Functions of Complex Variable– Analytic Functions
23
(b) If v (x, y) is given, take f ' (z) = vy+ivx.
Step (2) Replace x by z and y by 0 in f ' (z)
Step (3) Integrate f ' (z) with respect to z.
Ex. 16 : Show that the function u= cos x cosh y is harmonic and find its harmonic
conjugate.
Sol : We have u=cos x cosh y
∂u
∂u
∴
= − sin x cosh y ,
= cos x sinh y
∂x
∂y
Now,
∂ 2u
∂x 2
+
∂ 2u
∂y 2
= − cos x cosh y + cos x cosh y = 0.
⇒ u is a harmonic function.
Let v be its conjugate harmonic function, then we have
∂v
∂v
∂u
∂u
dv =
dx +
dy = −
dx +
dy
∂x
∂y
∂y
∂x
= − cos x sinh y dx − sin x cosh y dy
= −[cos x sinh y dx + sin x cosh y dy].
Integrating, v = − sin x sinh y+c, where c is a real constant.
Ex.17 : Prove that u = y 3 – 3 x 2 y is a harmonic function. Determine its harmonic
conjugate and find the corresponding analytic function f (z) in terms of z.
Sol : We have,
So,
u = y 3 – 3x 2 y
... (1.1)
∂u
∂u
= − 6 xy ,
= 3 y 2 − 3 x 2 , so
∂x
∂y
∂ 2u
∂y 2
∂ 2u
∂x 2
= − 6y
... (1.3)
= 6y
Adding (1.2) and (1.3), we have
... (1.2)
∂ 2u
∂x 2
+
∂ 2u
∂y 2
= − 6y + 6y = 0
Hence, u satisfies Laplace's equation; thus u is a harmonic function.
Further, let v be harmonic conjugate to u, then we have
∂v
∂v
∂u
∂u
dv =
dx +
dy = −
dx +
dy
∂x
∂y
∂y
∂x
[since u x = v y and uy=− vx]
= − ( 3 y 2 − 3 x 2 )dx − 6 xy dy = − ( 3 y 2 dx + 6 xy dy ) + 3 x 2 dx.
Integrating,
v = −3 xy 2 + x 3 + c.
24
Engineering Mathematics-III
This is the harmonic conjugate to u.
Now,
f ( z ) = u + iv = y 3 − 3 x 2 y + i( −3 xy 2 + x 3 + c )
= i( x + iy ) 3 + ic = iz 3 + ic.
Ex. 18 : Find the regular function w=u+iv, where u=e – x {(x 2 – y 2 ) cos y +2xy sin y}.
u = e − x {( x 2 − y 2 ) cos y + 2 xy sin y }
Sol : Given
... (1.1)
∂u
= − e − x {( x 2 − y 2 ) cos y + 2 xy sin y } + e – x
∂x
⇒
{( 2 x cos y + 2 y sin y }
= φ1 ( x, y ), say
and
∂u
= e − x { −( x 2 − y 2 ) sin y − 2 y cos y + 2 x sin y + 2 xy cos y } = φ 2 ( x, y ), say
∂y
By Milne's method, we have
f ' ( z ) = φ1 ( z, 0) − iφ 2 ( z, 0) = e − z ( − z 2 + 2z )
Integrating,
f ( z ) = ∫ e − z ( − z 2 + 2z )dz + c = − ∫ e − z z 2 dz + 2∫ ze − z + c
= e − z z 2 − 2∫ ze z dz + 2∫ ze − z dz + c = z 2 e − z + c
∴
f ( z ) = z 2 e − z + c,
where z = x + iy .
Ex.19 : Show that the function u = sin x cosh y + 2 cos x sinh y + x 2 − y 2 + 4 xy
satisfies Laplace's equation and determine the corresponding analytic function
f (z) = u+iv.
Sol : u = sin x cosh y + 2 cos x sinh y + x 2 − y 2 + 4 xy
⇒
and
∴
and
∂u
= cos x cosh y − 2 sin x sinh y + 2 x + 4 y = φ 1( x, y ), say
∂x
∂u
= sin x sinh y + 2 cos x cosh y − 2 y + 4 x = φ 2 ( x, y ), say
∂y
∂ 2u
∂x 2
∂ 2u
∂y 2
Adding (1.1) and (1.2),
= − sin x cosh y − 2 cos x sinh y + 2
... (1.1)
= sin x cosh y +2 cos x sinh y − 2
... (1.2)
∂ 2u
∂x 2
+
∂ 2u
∂y 2
= 0.
∴ u satisfies Laplace's equation. In other words u is a harmonic function.
By Milne's method, we have
Functions of Complex Variable– Analytic Functions
25
f ' ( z ) = φ1( z, 0) − iφ 2 ( z, 0) = (cos z + 2z ) − i( 2 cos z + 4z )
Integrating the above, we get
f ( z ) = ∫ (1 − 2 i ) (cos z + 2z ) dz + c
f ( z ) = (1− 2 i ) (sin z + z 2 ) + c.
or
Ex. 20 : If (z) = u+iv is an analytic function of z = x+iy and
u − v = e x (cos y − sin y), find f (z) in terms of z.
Sol : We have
u+iv = f (z)
...(1.1)
∴
iu − v = if ( z )
...(1.2)
Adding (1.1) and (1.2)
...(1.3)
(u − v ) + i (u + v ) = (1 + i ) f ( z ) = F( z ), say
Let us put u − v = U and u + v = V,
Since f ( z ) = u + iv is an analytic function, hence
F (z) = U + iV is an analytic function.
U = u − v = e x (cos y − sin y)
∂U
= e x (cos y − sin y) = φ1 (x, y), say
∂x
∂U
= e x ( − sin y − cos y ) = φ 2 ( x, y ), say.
∂y
Now,
∴
and
By Milne's method, we have
F ' ( z ) = φ1 ( z, 0) − iφ 2 ( z, 0) = ( e z + ie z ) = (1 + i )e z .
Integrating it, we get F( z ) = ∫ (1 + i )e z dz + c = (1 + i )e z + c
i.e.,
(1 + i ) f ( z ) = (1 + i )e z + c
f ( z ) = e z + c1 , where c1 =
or
[From (1.3)]
c
1+i
Ex. 21: If f (z) = u + iv is an analytic function of z, and
u−v=
cos x + sin x − e − y
y
2 cos x − e − e
−y
π
; find f ( z ) subject to the condition f   = 0.
 2
Sol : We have u + iv = f ( z )
∴ iu – v = if ( z ).
On adding, we have
(u − v ) + i (u + v ) = (1 + i ) f ( z ) = F ( z ) , say
Let u − v = U and u+v = V, then U+iV = F (z) is an analytic function.
Now,
U=
sin x + sinh y 
cos x + sin x − e − y
1
= 1 +
2 cos x − 2 cosh y
2
cos x − cosh y 
26
Engineering Mathematics-III
∴
∂U 1  1 + sin x sinh y − cos x cosh y 
= 
 = φ1 ( x, y ).
∂x
2
(cos x − cosh y ) 2

and
∂U 1  −1 + sin x sinh y + cos x cosh y 
= 
 = φ 2 ( x, y )
∂y
2
(cos x − cosh y ) 2

By Milne's method, we have
F ' ( z ) = φ1 ( z, 0) − iφ 2 ( z, 0)
1
1
i
1
1
= .
+ .
= (1 + i ) cosec 2 ( z / 2).
2 1 − cos z 2 1 − cos z 4
Now, integrating it, we get
i.e.
or
1
F ( z ) = − (1 + i ) cos ( z / 2) + c
2
1
(1 + i ) f ( z ) = − (1 + i ) cot( z / 2) + c
[∵ F ( z ) = (1 + i ) f ( z )]
2
1
f ( z ) = − cot( z / 2) + c1 , where c1 = c /(1 + i )
2
But given that when
z = π / 2 , f ( π / 2) = 0 ∴ c1 =
f (z ) =
∴
1
2
1
{1 − cot( z / 2)}.
2
Ex. 22 : If φ and ψ are functions of x and y satisfying Laplace's equation, show that
(s+it) is analytic, where s =
∂φ ∂ψ
−
∂y ∂x
and
t =
∂φ ∂ψ
+
.
∂x ∂y
Sol : Since both φ and ψ satisfy Laplace's equation
∴
Given
⇒
Substracting
⇒
∂ 2φ
∂x 2
+
∂2 φ
∂y 2
s=
= 0 and
∂ 2ψ
∂x 2
+
∂ 2ψ
∂y 2
=0
∂φ ∂ψ
∂φ ∂ψ
−
and t =
+
∂y ∂x
∂x ∂y
∂2 φ
∂2 ψ
∂2 φ
∂2 ψ
∂s
∂t
=
−
and
=
+
∂x ∂x ∂y ∂x 2
∂y ∂y ∂x ∂y 2
 ∂ 2ψ ∂ 2ψ
∂s ∂t
–
= − 
+
 = 0 (∵ ψ is harmonic)
∂x ∂y
 ∂x 2 ∂y 2 
∂s
∂t
=
∂x ∂y
Functions of Complex Variable– Analytic Functions
∂ 2φ
∂ 2ψ
∂ 2φ
∂ 2ψ
∂s
∂t
and
=
−
=
+
∂y ∂y 2 ∂y ∂x
∂x ∂x 2 ∂x ∂y
Also
Adding
27
∂ 2φ ∂ 2φ
∂s ∂t
+
=
+
= 0 (since φ is harmonic)
∂y ∂x ∂x 2 ∂y 2
∴
∂s
∂t
.
=−
∂y
∂x
Hence
∂s
∂t
∂s
∂t
and
=
=−
∂x ∂y
∂y
∂x
i.e., Cauchy-Riemann equations are satisfied. Hence (s+it) is analytic function.
Ex. 23 : Find analytic function f (z) = u (r, θ) +iv (r, θ) such that
v( r, θ) = r 2 cos 2θ − r cos θ + 2
Sol : Given
So,
and
v = r 2 cos 2θ − r cos θ + 2
∂v
= −2r 2 sin 2θ + r sin θ
∂θ
∂v
= 2r cos 2θ − cos θ
∂r
Using C–R equations in polar coordinates, we have
∂u ∂v
r
=
= −2r 2 sin 2θ + r sin θ
∂r ∂θ
∂u
or
= −2r sin 2θ + sin θ,
∂r
1 ∂u ∂v
and
−
=
= 2r cos 2θ − cos θ
r ∂θ ∂r
∂u
or
= −2r 2 cos 2θ + r cos θ
∂θ
Then we have
du =
... (1.1)
... (1.2)
... (1.3)
... (1.4)
... (1.5)
∂u
∂u
dr +
dθ
∂r
∂θ
= ( −2r sin 2θ + sin θ) dr + ( −2r 2 cos 2θ + r cos θ) dθ
= − [( 2r dr ) sin 2θ + r 2 ( 2 cos 2θ dθ)] + [ sin θ. dr + r( cos θdθ)]
= − d( r 2 sin 2θ) + d ( r sin θ)
Integrating, we have
u = − r 2 sin 2θ + r sin θ + c
Thus
f(z) = u + iv = (– r 2 sin 2θ + r sin θ + c ) + i( r 2 cos 2θ − r cos θ + 2)
= ir 2 (cos 2θ + i sin 2θ) − ir(cos θ + i sin θ) + 2 i + c
28
Engineering Mathematics-III
= ir 2 e 2 iθ − ir eiθ + 2 i + c
[By D'Moivre's theorem]
= i ( r 2 e 2 iθ − reiθ ) + 2 i + c.
 ∂2
∂2 
∂2
Ex. 24: Show that 
.
+
 ≡ 4
∂z ∂z
 ∂x 2 ∂y 2 
Sol : We have x + iy = z, x − iy = z
1
∴
x = ( z + z ), y = − (i / 2) ( z − z )
2
∂x 1 ∂x 1 ∂y
i ∂y
i
Now
= ,
= ,
=− ,
=
∂z 2 ∂z 2 ∂z
2 ∂z
2
...(1.1)
Again
∂
∂ ∂x
∂ ∂y 1  ∂
∂
≡
.
+
.
=  −i 
∂z ∂x ∂z ∂y ∂z 2  ∂x
∂y 
and
∂
∂ ∂x
∂ ∂y 1  ∂
∂
≡
.
+
.
=  +i 
∂z ∂x ∂z ∂y ∂z
2  ∂x
∂y 
∂2
1∂
∂ 1∂
∂
≡  −i  .  +i 
∂z ∂z 2  ∂x
∂y  2  ∂x
∂y 
Hence
4
⇒
∂2
∂2
∂2
≡
+
.ds
∂z ∂z ∂x 2 ∂y 2
Ex. 25 : If f (z)=u + iv is an analytic function of z = x + iy, and ψ is any function of
x and y with differential coefficients of the first two orders, then
 ∂Ψ 


 ∂x 
2
∂ 2Ψ
and
∂x
Sol : We have
and
2
 ∂Ψ 
+

 ∂y 
+
∂ 2Ψ
∂y
2
2
 ∂Ψ  2  ∂ψ  2 
=  
 +  
 ∂v  
  ∂u 

 ∂ 2Ψ ∂ 2Ψ 
=
+

 ∂u 2
∂v 2 
f ' (z )
f ' (z )
2
2
∂Ψ ∂Ψ ∂u ∂Ψ ∂v
=
.
+
.
∂x
∂u ∂x ∂v ∂x
∂Ψ ∂Ψ ∂u ∂Ψ ∂v
∂Ψ ∂v ∂Ψ ∂u
=
.
+
.
=−
.
+
.
∂y
∂u ∂y
∂v ∂y
∂u ∂x ∂v ∂x
...(1.1)
...(1.2)
(as u x = v y , u y = − v x )
Squaring and adding (1.1) and (1.2), we have
 ∂Ψ 


 ∂x 
2
 ∂Ψ 
+

 ∂y 
2
 ∂Ψ  2  ∂Ψ  2 
= 
 +
 
 ∂v  
 ∂u 

 ∂Ψ  2  ∂Ψ  2 
= 
 +
 
 ∂v  
 ∂u 

 ∂u  2  ∂v  2 
  +   
 ∂x  
 ∂x 

f ' (z )
2
∵ f ' ( z ) = ∂u + i ∂v 

∂x
∂x 
Functions of Complex Variable– Analytic Functions
2nd Part. Result (1) can be written as
29
∂
∂u ∂
∂v ∂
and also
≡
+
∂x ∂x ∂u ∂x ∂v
∂
∂v ∂
∂u ∂
=−
+
∂y
∂x ∂u ∂x ∂v
⇒
∂
∂
∂u ∂
∂v ∂ 
∂v ∂
∂u ∂ 
+i
= 
+
.  + i  − .
+
. 
 ∂x ∂u ∂x ∂v 
∂x
∂y  ∂x ∂u ∂x ∂v 
∂u
∂v ∂
∂u
∂v ∂
= 
− i 
+ i 
− i 
 ∂x
 ∂x
∂x  ∂u
∂x  ∂v
Similarly
∂
∂
∂u
∂v
∂
∂
+i
≡ 
− i  
– i 
∂x
∂y  ∂x
∂x   ∂u
∂v 
...(1.3)
∂
∂
∂u
∂v
∂
∂
−i
≡ 
+ i  
– i 
∂x
∂y  ∂x
∂x   ∂u
∂v 
...(1.4)
Multiplying (1.3) and (1.4) columnwise, we have
2
2
2
2
∂ ∂
∂   ∂u 
∂
 ∂v    ∂ + ∂ 
+
i
−
i
≡
+
   2

 
  

 ∂x    ∂u
∂y   ∂x
∂y   ∂x 
 ∂x
∂v 2 

i.e
 ∂2
∂2 

+
 ≡ f ' ( z )
 ∂x 2 ∂y 2 
or
 ∂2
∂2 

+
 Ψ = f ' ( z )
 ∂x 2 ∂y 2 
or
∂ 2Ψ
∂x 2
+
∂ 2Ψ
∂y 2
2
 ∂2
∂2 
+


 ∂u 2 ∂v 2 
2
 ∂2
∂2 
+

 Ψ
 ∂u 2 ∂v 2 
 ∂ 2Ψ ∂ 2Ψ 
2
=
+
 f ' (z ) .
 ∂u 2
∂v 2 
Ex. 26 : If f (z) = u+iv is a regular function of z in any domain prove that
 ∂2
∂2 

+

 ∂x 2 ∂y 2 
f (z )
p
= p2
f (z )
p− 2
f ' (z )
 ∂2
∂2 
∂2
Sol : We know that 
+
 ≡ 4
∂z ∂z
 ∂x 2 ∂y 2 
 ∂2
∂2 

+

 ∂x 2 ∂y 2 
f (z )
p
=4
2
.
(by Ex.24)
∂2
∂z ∂z
f (z )
p
=4
∂2
∂2
{ f ( z ) f ( z )} p / 2 = 4
{ f ( z )} p / 2 × { f ( z )} p / 2
∂z ∂z
∂z ∂z
=4
∂
[{ f ( z )} p / 2 .( p / 2){ f ' ( z )} ( p / 2) −1 f ' ( z )]
∂z
= 4 [( p / 2){ f ( z )} ( p / 2) −1 f ' ( z ) × ( p / 2) { f ( z )} ( p / 2) −1 f ' ( z )]
30
Engineering Mathematics-III
= p 2 { f ( z ) f ( z )} ( p / 2) −1 { f ' ( z ) f ' ( z )}
= p 2 { f (z )
2 ( p / 2) −1
}
f ' (z )
2
= p 2 f (z )
p −2
f ' (z )
2
.
1.11 Application to Flow Problems
The real and imaginary parts of the analytic function are the solutions of the
Laplace's equation in two variables. The conjugate functions provide solutions to a
number of field and flow problems. Let us consider the irrotational motion of an
incompressible fluid in two dimensions. Assuming the flow to be in planes parallel to
the xy-plane, the velocity v of a fluid particle can be expressed by
... (1.1)
v = vxi + vy j
Since the motion is irrotational so there exists a scalar function φ( x, y ) such that
∂φ
∂φ
... (1.2)
v = ∇φ ( x, y ) =
i+
j
∂x
∂y
[The function φ (x, y) is called the velocity potential and curves φ (x, y) = c are
known as equipotential lines.]
From (1.1) and (1.2), we then get
∂φ
vx =
∂x
∂φ
and
vy =
∂y
But the fluid is incompressible so div v = 0 ⇒
... (1.3)
∂v x ∂v y
+
=0
∂x
∂y
... (1.4)
Substituting the values of v x and v y from (1.3) in (1.4), we get
∂ 2φ
∂x 2
+
∂ 2φ
∂y 2
=0
i.e. the velocity potential φ is harmonic hence there exist a conjugate harmonic
function ψ( x, y ) such that
w( z ) = φ ( x, y ) + i ψ( x, y ) is analytic.
... (1.5)
Now the slope at any point of the curve ψ( x, y ) = a is given by
dy
( ∂ψ / ∂x ) ( ∂φ/ ∂y )
[Using C–R equations]
=−
=
dx
( ∂ψ / ∂y ) ( ∂φ/ ∂x )
This implies that the velocity of the fluid particle is along the tangent to the curve
ψ( x, y ) = a, i.e., the particle moves along this curve. Such curves are known as
stream lines and ψ( x, y ) is called the stream function.
The equipotential curves φ( x, y ) = c and the stream curves Ψ( x, y ) = a cut each
other orthogonally.
Functions of Complex Variable– Analytic Functions
Now (5)
31
∂φ ∂φ
∂φ
dw ∂ψ
=
+i
=
−i
= v x − iv y
dz
∂x
∂x ∂x
∂y
⇒
[by C–R equations]
whence, the magnitude of the fluid velocity = ( v 2x + v 2y ) = dw/dz .
Hence we can say that the flow pattern is fully represented by the function w (z).
This is known as complex potential. Also, the complex potential w (z) can be taken
to represent other type of 2-dimensional steady flow.
In electrostatic and gravitational fields the curves φ ( x, y ) = c and Ψ ( x, y ) = a are
called equipotential lines and lines of force, whereas in heat flow problems,
the curve φ( x, y ) = c and Ψ ( x, y) = a are known as orthonormal and heat flow
lines respectively. For a given φ( x, y ), we can always find Ψ( x, y ) and vice-versa.
Ex. 27 : If w = φ + iψ represents the complex potential for an electric field and
Ψ = x2 − y 2 +
x
2
x + y2
, determine the function φ.
Sol : Evidently, Ψ satisfies the Laplace's equation. But φ and Ψ must satisfy the
Cauchy-Riemann equations
i.e.
∂φ ∂Ψ
=
∂x
∂y
Now

∂φ
2 xy
∂  2
x
2
=
x − y + 2
 = −2 y − 2
2
∂x ∂y 
x + y 
( x + y 2 )2
and
∂φ
∂Ψ
=−
∂y
∂x
Integrating (1.1) w.r.t. x, we have φ = −2 xy +
y
2
x + y2
+ f ( y ) where f ( y ) is an
arbitrary function of y.
Making use of ( ∂φ/ ∂y ) = − ( ∂Ψ / ∂x ), we get
− 2x +
⇒
Hence
x2 − y 2
( x 2 + y 2 )2
+ f ' ( y ) = − 2x +
...(1.1)
x2 − y 2
( x 2 + y 2 )2
f ' ( y ) = 0, i. e., f ( y ) = c, an arbitrary constant
y
φ = −2 xy +
+ c.
2
x + y2
32
Engineering Mathematics-III
Problem Set
b
Complex functions
1.
Classify the following region :
(i)
z −1 = 5
(ii)
Re ( z ) ≤ 0
(iii)
Im (z ) < 0
(iv)
− 3 < Re ( z ) < 7
z−5 >4
(vi)
3≤ z −1−i ≤ 4
(v)
2.
Find the domain of the following functions :
(i)
(iii)
1
x
+
i
x 2− y
(ii)
z +1
z −i
(iv)
5z 5 + z 4 + 3z 2 + iz + 7
z
2
( z + 1) 2
3. Find the real and complex parts u,v of f (z)= u+iv, where f (z) is :
z
(i)
(iii)
z3
(ii)
Re( z 2 + i )
, z≠0
z
(iv)
b
Limit
4.
Find the limit of the following functions :
(i)
(iii)
1+ z
z
(ii)
z→ 0
Lim [ z 2 − z 2 ]
(iv)
Lim
z→1
z→1−i
, z ≠1
1/ 2
z2 −1
z −1
Lim
2
Lim
z→ ∞
z
z
iz 3 + iz − 1
( 2z + 3 i ) ( z − i ) 2
b
Continuity
5.
Determine the value of z for which function is not continuous :
z
z +7
(i)
(ii)
2
2
z −1
z − 4z + 5
6.
Examine the continuity of the following functions :
Functions of Complex Variable– Analytic Functions
33
2
 Re( z )
, z ≠ 0 at

2
(i) f (z) =  z

 0 ,
z=0
z=0
 z 2 + iz + 2, z ≠ i
(ii) f ( z ) = 
 0 , z = 0
 x 3 y ( y − ix )
f ( z ) − f ( 0)
, z ≠ 0, prove that
→0
 6
2
z
7. (i) If f ( z ) =  x + y

z=0
 0,
along any radius vector but not as z → 0 in any manner
(ii) Show that the function f (z) = xy + iy is everywhere continuous but not analytic.
b
Differentiability
8. Show that functions f(z) = Im( z ) is continuous, examine the differentiability
of the function.
9. Find the derivative of f(z) = zn, where n is a positive integer.
[Hint : To find with basic principle, use Binomial theorem.]
10. Show that the function f(z) = sin x cosh y+i cos x sinh y is continuous as well
as analytic.
11. Show that continuity is the necessary but not the sufficient condition for the
existence of derivative of function f(z).
12. If f ( z ) is analytic at z0, prove that it must be continuous at z0. Give an example
to show that the converse is not necessarily true.
Hint : Start with f(z0 + h) – f(z0) =
f (z 0 + h ) − f (z 0 )
.h
h
Example : f ( z ) = −z is continuous but not analytic anywhere.
b
Analyticity
13. Using Cauchy-Riemann equations show that f ( z ) = z 3 is analytic in the entire zplane.
14. Determine which of the following functions are analytic.
(i) 2xy+i ( x 2 – y 2 )
(ii) ( x – iy ) / ( x 2 + y 2 )
15. Show that the function f(z)=xy+iy is continuous everywhere but is not
analytic.
34
Engineering Mathematics-III
 x 2 y 5 ( x + iy )
,z ≠ 0

16. Consider the function f ( z ) =  x 4 + y 10

z=0
 0,
Examine the nature of f(z).
17. Show that an analytic function f(z) whose derivative is identically zero is
constant.
Hint : f '(z)=ux+ivx = vy− i u y =0 ⇒ u x = u y = 0 and v x = v y = 0. Thus u
and v are constant, hence f (z) = u+iv is a constant .
18. Find the values of c1, c2, c3 and c4 so that the function
f ( z ) = x 2 + c1 y 2 + c 2 xy + i( c 3 x 2 + y 2 + c 4 xy ) is analytic.
19. If f ( z ) =
1+ z
(i) Find f '(z), (ii) Determine where f (z) is non-analytic.
1– z
20. For what values of z, do the function w defined by the following equation
ceases to be analytic.
(i) z = sin u cosh v − i sin u sinh v, w = u +iv
(ii) z = log ρ + iϕ,
b
where w = ρ( cos ϕ + i sin ϕ)
Harmonic Functions and Conjugate Functions
21. Prove that u = e x cos y is a harmonic function. Determine its harmonic
conjugate v(x, y) and analytic function f (z)= u + iv.
22.
If u = x3 − 3xy2, show that there exists a function v (x, y) such that w=u+iv is
analytic in a finite region.
23. If u = ( x − 1) 3 − 3 xy 3 + 3 y 2 , determine v so that w=u+iv is a regular function
of x + iy.
24. Find an analytic function w=u+iv given that
(i) v =
x
2
x + y2
+ cosh x cos y
(ii) u = e x cos y
(iii) u = x sin x cosh y − y cos x sinh y
25. Let f ( z ) = u( r, θ) + iv( r, θ) be an analytic function and u = − r 3 sin 3θ then
construct the corresponding analytic function f(z) in terms of z.
26. Prove that the function u = x 3 − 3 xy 2 + 3 x 2 − 3 y 2 + 1 satisfies Laplace's
equation and determine corresponding analytic function u+iv.
Functions of Complex Variable– Analytic Functions
27. If u =
35
sin 2 x
, find the corresponding analytic function f (z) =u+iv.
cosh 2 y + cos 2 x
28. If u − v = ( x − y ) ( x 2 + 4 xy + y 2 ) and f ( z ) = u + iv is an analytic function of
z = x + iy , find f (z).
29. If u + v =
2 sin 2 x
e
2y
− e −2 y − 2 cos 2 x
, and f ( z ) = u + iv is an analytic function of
z=x+iy. Find f (z) in terms of z.
30. Show that u+iv =(x − iy)/(x − iy + a), where a ≠ 0, is not an analytic function
of z = x + iy , whereas u – iv is a such function.
31. If f(z) =u+iv is an analytic function of z = x+iy and u − v =
find f(z) subject to the condition f ( π / 2) =
e y − cos x + sin x
,
cosh y − cos x
3−i
.
3
 ∂2
∂2 
2
2
32. If f(z) is an analytic function of z prove that 
+
 Re z = 2 f ' ( z ) .
2
2
∂y 
 ∂x
33. If f(z) = u+iv is a regular function of z in any domain prove that
 ∂2
∂2 

+
 u
 ∂x 2 ∂y 2 
b
p
= p ( p − 1) u
p− 2
2
f ' (z ) .
Applications
34. If the potential function is log ( x 2 + y 2 ), find the flux function and the
complex potential function.
35. In a two-dimensional fluid flows, the stream function ψ is given, find the
velocity potential φ.
(i) ψ = − y ( x 2 + y 2 ) 2 ,
(ii) ψ = tan −1 ( y / x ).
36. An electrostatic field in the xy-plane is given by the potential function
φ = 3 x 2 y − y 3 , find the stream function.
37. Two concentric circular cylinders of radii a, b (a<b) are kept at potentials φ1
and φ 2 respectively. Use the complex function w=A log z+c to prove that the
capacitance per unit length of the capacitor formed by them is 2πk /log( b / a),
where k is the dielectric constant of the medium.
[Hint. Capacitance = charge/Potential difference.]
38. Prove that ψ = log[( x − 1) 2 + ( y − 2) 2 ] is harmonic in every region which
does not include the point (1, 2). Find a function φ + iψ such that φ + iψ is an
36
Engineering Mathematics-III
analytic function of the complex variable z=x+iy. Express φ +iΨ as a function
of z.
Objective Type Questions
Multiple Choice Questions
Tick the correct answer :
1. An annulus C1 < z − a < c 2 is :
(a) connected
(b) disconnected
(c) semi-connected
(d) none of these.
2. The set S is closed if :
(a) it does not contain its boundary points
(b) it has no boundary points
(c) it contains its boundary points
(d) none of these.
3. The equation z − 1 = z + i respresents :
(a) line through origin whose slope is one
(b) a line through origin whose slope is – 1
(c) an ellipse whose foci are at z =1, – i
(d) a circle through origin.
4. Cauchy-Riemann equations for w = u + iv = f ( z ) are :
(a) u x = v x , u y = v y
(b) u x = v y , u y = v x
(c) u x = v y , u y = − v x
(d) none of these.
5. An analytic function with constant modulus is :
(a) variable
(b) may be variable or constant
(c) constant
(d) none of these.
6. If a function is analytic at all points of a bounded domain except finitely many
points, then these points are called :
(a) singular points
(b) simple points
(c) continuous points
(d) none of these.
7. At z = 0, the function f ( z ) = z is :
(a) not analytic
(b) not differentiable
(c) not continuous
(d) none of these.
8. Which of the following is not correct for analytic functions f (z) and g (z) in a
region R ?
(a) f ( z ) + g ( z ) is analytic in R
(b) f ( z ) − g ( z ) is analytic in R
Functions of Complex Variable– Analytic Functions
(c) f ( z ) g ( z ) is analytic in R
37
(d) f ( z ) / g ( z ) is analytic in R.
9. Which of the following functions f (z) satisfies Cauchy-Riemann equations ?
2
(a) f ( z ) = z = x − iy at z = 1 + i
(b) f ( z ) = z
(c) f ( z ) =
(d) none of these.
xy at z = 0
at z ( z ≠ 0)
10. Which of the following function is not analytic ?
(a) f ( z ) = z 6
(b) f ( z ) = 1/ z 4 ( z ≠ 0)
(c) f ( z ) = 1/( z − 1) 3 , z ≠ 1
(d) f ( z ) = 1/( z − 1) 3 .
11. The function f (z) =tan z is :
(a) continuous everywhere
(b) analytic in finite complex plane
(c) analytic everywhere except the points where cos z = 0
(d) none of these.
12. Which of the function f (z) is analytic in complex plane where f (z) =
(a) Re z
(b) Im z
(c) cot z
(d) e z .
13. A single-valued function f (z) defined in a domain D is said to be analytic at a
point z 0 of D if it is differentiable :
(a) at z 0
(d) at the origin
(c) in some neighbourhood of z 0
(d) in some deleted neighbourhood of z 0 .
14. Cauchy-Riemann equations in polar form are :
∂u 1 ∂v
1 ∂u
∂v
(a)
= .
and .
=−
∂r r ∂θ
r ∂θ
∂r
∂u
∂v
∂v
∂v
(b)
=r
and r
=−
∂r
∂θ
∂θ
∂r
∂u
1 ∂v
1 ∂u ∂v
(c)
=− .
and .
=
∂r
r ∂θ
r ∂θ ∂r
∂u
∂v
∂u ∂v
(d)
= −r
and r
= .
∂r
∂θ
∂θ ∂r
15. The derivative of a function w = f ( z ) in polar form is given by :
dw ∂w iθ
dw
∂w iθ
(a)
(b)
=
.e
=−
e
dz
∂r
dz
dθ
dw ∂w − iθ
dw ∂w − iθ
(c)
(d)
=
e
=
e .
dz
dr
dz
∂θ
16. Which of the following functions f (z) is not analytic where f (z) =
(a) sin z
(b) cos z
38
Engineering Mathematics-III
(c) any polynomial of degree more than one
(d) none of these.
17. The function f ( z ) = w = u + iv is independent of z ( = x − iy ) if :
∂w
∂w
(a)
(b)
=0
=∞
∂z
∂z
∂w
(c)
(d) none of these.
=1
∂z
18. The function f ( z ) = z = x − iy is
(a) analytic everywhere
(b) analytic at origin only
(c) not analytic for any z
(d) none of these.
19. Which of the following functions f (z) is not analytic ? Here f (z) =
1
(a) z 3
(b)
1− z
(c) z + z
20. The value of the derivative f ( z ) =
(d) z
2
.
iz + z
at z is:
3z − 6 i
(a) 0
(b) 6z
(c) z
(d) none of these.
21. A function of x and y possessing continuous partial derivatives of the first and
second orders is called a harmonic function if it satisfies :
(a) Euler equation
(b) Laplace equation
(c) Homogeneous equation
(d) Lagrange equation.
22. The analytic function whose real part is e x cos y is :
(a) e z + ci
(b) e 2z
(c) xe z
(d) none of these.
23. The function f ( z ) = xy + iy is :
(a) everywhere continuous
(b) analytic
(c) everywhere continuous but not analytic
(d) none of these.
Fill in the Blanks
1. A single-valued continuous function f (z) is analytic in a domain D if u x , v x , u x , v y
are continuous and satisfy .................. equations.
2. The function f (z) = z is not ................ at z = 0.
3. The function f (z) = arg z is not .................... at z = 0.
Functions of Complex Variable– Analytic Functions
39
4. If harmonic functions u and v satisfy Cauchy-Riemann equations, then
f ( z ) = u + iv is an ................... function.
5. A function f ( z ) = u ( x, y ) + iv ( x, y ) is analytic at a point z 0 in the complex
plane if the partial derivatives u x , u y , v x , v y exist, .................. and satisfy
Cauchy-Reimann relations at z 0 .
6. The function f ( z ) = xy
1/ 2
is not .............. at z = 0.
7. If u = x 2 − y 2 , which is harmonic, then
∂ 2u
= .................. .
∂z ∂z
8. The function f ( z ) = z sec z is not analytic at the points z = .............. .
9. The function f ( z ) = 1/ z ( z − 3) is not ................ at z = 0, 3.
10. Let f ( z ) = u + iv be an analytic function in a domain G. Then f (z) is ........... in G
if Im { f ( z )} is constant.
11. If f ( z ) = u + iv be an analytic function of z = x + iy , then the families of curves
u =constant, v = constant are ............. to each other
12. If f ( z ) = u + iv is an analytic function, then u and v are both .............. functions.
State True/False
1. Continuity is the necessary but not the sufficient condition for the existence of a
finite derivative.
2. If a function f ( z ) is analytic at any point z = x + iy , then C-R equations are
satisfied at that point.
3. Cauchy-Riemann equations are sufficient for a function to be analytic.
4. An analytic function with constant modulus is constant.
5. The function w = z
2
is continuous everywhere but nowhere differentiable
except at the origin.
6. If f ( z ) = u + iv is an analytic function, then being given one of u and v, the other
can be determined.
ANSWERS
Problem Set
1. (i) Circle with radius 5 and centre (1, 0).
(ii) Left half plane x ≤ 0 (including the y-axis), it is closed connected and
unbounded, not a domain.
(iii) Upper half plane y ≥ 0 (excluding the x-axis), it is open, connected and
unbounded, defines a domain.
(iv) Open, connected and unbounded, defines a domain.
(v) Open, connected unbounded region.
40
Engineering Mathematics-III
(vi) Annulus, not open, connected bounded, not a domain.
2. (i) Entire complex, plane except x = 0, y = 2.
(ii) Entire complex, plane.
(iii) Entire z plane except z = i.
(iv) Whole of the complex plane except z =± i.
3. (i) u = x 3
−
3 xy 2 , v = 3 x 2 y − y 3 .
x
(ii) u =
2
1+ x + y
(iii) u =
(iv) u =
x2 − y 2
x2 + y 2
2
, v=
y
2
1+ x + y2
.
, v = 0.
r cos (θ/ 2), v =
r sin (θ/ 2), where x = r cos θ, y = r sin θ.
4. (i) 2
(ii) limit does not exist.
(iii) − 4i.
(iv) i/2, put z =
1
so that when z → ∞, t → 0.
t
5. (i) Continuous everywhere except where the denominator z 2 − 1 = 0 or z = ±1.
6. (i) Function is not continuous at z = 0.
(ii) Continuous.
8. Nowhere differentiable.
9. nz n −1 ,
f ( z ) is
continuous everywhere.
14. (i) and (ii) non-analytic.
16. C–R equations satisfied, f ' (0) does not exist, f (z) is not analytic at the origin
although C–R equations satisfied there.
18. c1 = −1, c2=2, c 3 = −1, c4=2.
19.
f(z) =
2
(1 − z ) 2
, f ( z ) is analytic for all values of z, except z=1.
20. (i) when z = ± 1
(ii) ρ = ∞.
21. v = e x sin y + c, f ( z ) = e z + d, where d = ic.
22. v = 3 x 2 y − y 3 + c, f ( z ) = z 3 + ic.
23. v = 3 x 2 y − 6 xy + 3 y − y 3
24. (i) u =
y
x2 + y 2
+
c
− sinh x sin y + c, w =
i z−
z
2
+ i cosh z + c.
Functions of Complex Variable– Analytic Functions
41
z
(ii) f ( z ) = e − iz.
(iii) z sin z.
25. f ( z ) = i( ri iθ ) 3 + ic = iz 3 + ic.
26. f ( z ) = z 3 + 3z 2 + c.
27. f ( z ) = tan z + c.
1+i
28. f ( z ) = −iz 3 +
c.
2
1
29. f ( z ) = (1 + i ) cos z + c1 .
2
1
3−i
31. f ( z ) = cot( z / 2) + (1 − i ), f ( π / 2) =
.
2
2
34. 2 tan −1 ( y / x ), 2 log + c
35. (i) x /( x 2 + y 2 )
(ii)
1
log ( x 2 + y 2 ).
2
36. Ψ = 3 xy 2 − x 3 + c.
38. −2 tan −1 [( y − 2) / ( x − 1)], 2 i log ( z − 1 − 2 i ).
Multiple Choice Questions
1.
(a)
2. (c)
3.
(b)
4. (c)
5. (c)
6.
(a)
7. (b)
8.
(d)
9. (c)
10. (d)
11.
(c)
12. (d)
13.
(c)
14. (a)
15. (c)
16.
(d)
17. (a)
18.
(c)
19. (a)
20. (d)
21.
(b)
22. (a)
23.
(c)
Fill in the Blanks
1.
Cauchy-Reimann
2. continuous
3.
analytic
4. analytic
5.
continuous
6. differentiable
7.
0
8. 2 nπ ±
9.
differentiable
10. constant
orthogonal
12. harmonic
11.
π
, n∈ N
2
42
Engineering Mathematics-III
True/False
1.
True
2. True
3. False
4. True
5. True
6. True
❑❑❑
Unit-1
Chapter
2
Complex Integration
2.1 Complex Integration
Here we shall consider the integral of a function of a complex variable along a plane
curve. In the case of a real variable, the integration is considered from two points of
view, namely the indefinite integration as an operation inverse to that of
differentiation and the definite integration as the limit of a sum. The concept of
indefinite integral as the process of inverse of differentiation in case of a function of
a real variable also extends to a function of a complex variable if the complex
function f (z) is analytic.
Thus, if f (z) is an analytic function of complex variable z, and if
∫ f (z )dz = F(z ),
then differential of F (z) is equal to f (z) i.e. F ' (z) = f ( z ).
The concept of definite integral of a function of a real variable does not extend to the
domain of complex variables. For example, in the case of real variables for the
b
definite integral ∫ f ( x ) dx the path of integration is always along the x-axis from
a
x = a to x=b. But in the case of a complex function f (z) the path of the definite
b
integral ∫ f ( z ) dz can be along any curve from z=a to z=b; hence its value depends
a
upon the path (curve) of integration. This variation in values can be made to
disappear if the different curve (path) from a to b are regular curves.
2.2 Some Important Definitions
1. Partition : Let [a, b] be a closed interval where a, b are real numbers. Then the set of
points P = {t 0 , t1 ,.... t n } where a = t 0 < t1 < t 2 < t 3 .... < t n = b is called a partition
of the interval [a, b]. The greatest of the numbers t1
− t 0 , t 2 − t1 ,...., t n − t n −1 is called
the norm of the partition P and is denoted by P .
2. Continuous arc : If a point z on an arc is such that z = φ (t ) + iψ(t )
then we may write x = φ(t )
y = Ψ (t )
...(A)
... (2.1)
... (2.2)
If φ (t) and ψ(t ) are continuous functions of the real variable t defined in the range
α ≤ t ≤ β, then the arc is called a continuous arc.
44
Engineering Mathematics-III
3. Multiple point : If the equation (A) or say the equation (2.1) and (2.2) are
satisfied by more than one value of t in the given range, then the point z [or say the
point (x, y)] is called a multiple point of the arc.
4. Jordan arc : A continuous arc without multiple points is called Jordan arc.
Thus for a point z on a Jordan arc, z as expressed in equation (A) is one valued and
φ(t ), Ψ (t ) are continuous; in addition if φ ' (t) and Ψ '(t) are also continuous in the
range α ≤ t ≤ β, then the arc is called a regular arc of a Jordan curve.
Contour : By contour we mean a Jordan curve consisting of continuous chain of a
finite number of regular arcs.
If A be the starting point of the first arc and B the end point of the last arc, then
integral along such a curve is written as ∫ f ( z )dz.
AB
If the starting point A of the arc coincides with the end point B of the last arc then the
contour AB is said to be closed. The integral along such closed contour is written as
∫
f ( z )dz, and is read as integral f(z) taken over the closed contour C. Although
∫
( f ( z ) dz does not indicate the direction along the curve, but it is conventional to
C
C
take the direction positive which is anticlockwise, unless indicated otherwise.
2.3 Connected Region
Simple closed path. A simple closed path is a path that does not intersect or
touch itself .
Connected Region. A region is said to be connected region if any two points of the
region D can be connected by a curve which lies entirely within the region.
Simply-Connected. A connected region
is said to be a simple-connected region if all
the interior points of a closed curve C
drawn in the region D are the points of the
region D. See Fig 2.2(a).
Doubly Connected
Triply Connected
Fig. 2.1
Example : Interior of a circle, ellipse are
simply-connected.
Multi-Connected Region. If all the points of the area bounded by two or more
closed curves drawn in the region D, are the points of the region D, then the region D
is said to be multi-connected region. For example, let there be a number of closed
curves C, C1, C2, C3....all drawn in a certain region D. In other words, we can say if all
the points of the area lying between the closed curves C, C1, C2,......., the area which
is interior to C and exterior to the other curves C1,C2,C3,C4 are the points of the
region D, then the region D is said multi-connected region. For example : An annulus.
doubly-connected, triply-connected. See Fig. 2.1.
Complex Integration
Cross cut (or cut). The lines drawn in a
multiply connected region without intersecting
any one of the curves, which make a
multi-connected region a simply-connected one
are called cuts or cross cuts.
45
C
C2
C1
Simply connected region (a)
C3
C4
Multi connected region (b)
A B
C
C1
Thus, let there be a multi-connected region lying
C V
P C
U
Q
C
between the curves C, C1, C2, C3, C4, where C1, C2,
S
R
C3, C4, lie inside C. In this region draw lines
Fig. 2.2 (c)
joining C to all the curves C1, C2, C3, C4, then the
same region which lie between several curves C,
C1, C2, C3, C4, can also be said to lie between a simple curves whose boundary
consists of the boundaries of C, C1, C2, C3, C4 and the lines AB, PQ, RS,UV....etc. See
Fig 2.0(c)
4
2
3
Such lines AB, PQ, RS,UV....etc. are called cuts or cross-cuts..
By such a manipulation the same region which is multi-connected is made a
simply-connected region. This device makes different curves C, C1 , C2, C3, C4, as
parts of the continuous curves consisting of these curves and the cross-cuts.
2.4 Rectifiable Curve
Let z = z (t ) = x (t ) + iy (t ) be any given curve and let t
take up any value between a and b, i. e. a ≤ t ≤ b. Let
P ={t0, t1, t2,....,tn,} be a partition of [a, b]. If P0 , P1,
P2,.....,Pn be the points on the curve corresponding to the
points t 0 , t1 , t 2 ... t n , then the length of the polygonal
line P0 , P1 , P2 ,.... Pn is the sum of the lengths of the lines
P0 , P1 , P2 , P3 , ... Pn −1 , Pn .
zn–1
zr
er
Pr – 1
Pn zn = b
Pr zr – 1
z3
P3
z0 =a
z
z1 2 P
P0
2
P1
Fig. 2.3
Let z 0 , z1 , z 2 ,.... z n , be the points on the curve
corresponding to the values t 0 , t1 , t 2 ,.... t n , i . e. z(t r ) = z r , then the length of the
n
polygonal line =
∑ z r − z r −1 .
r =1
The value of this sum depends upon the mode of sub-division and is called the length
of an inscribed polygon.
If the curve is such that this sum (the length of the inscribed polygon) have a finite
upper bound l, for all modes of the subdivision, the curve is said too be rectifiable
and l is called the length of the curve.
46
Engineering Mathematics-III
2.5 Complex Line Integrals (Riemann's definition of integration)
joining a and b and let f (z) be a function of a complex
variable z defined and continuous on C.
Consider the partition P = { a = t 0 , t1 , t 2 ,... t n , = b} of
the interval [a, b]. Let z 0 , z1 , z1 , z 2 ,......, z n , be the
points on the curve corresponding to the values
t 0 , t1 , t 2 ,..... t n , i. e. z (t r ) = z r . On each arc joining
z r−1 to z r choose a point e r where r = 1, 2....., n, i.e.
.. . .
.
.
.
.
en
Let z = z (t ) = x (t ) + iy (t ), a ≤ t ≤ b be a given curve C
zn – 1 zn = b
zr
er
zr – 1
..
z0 = a e1 e2
z1
z2
Fig. 2.4
z r −1 ≤ e r ≤ z r form the following sum S p for the
partition P,
S p = f ( e1 )( z 1 − z 0 ) + f ( e 2 )(z 2
− z1 )+ .......+ f ( e r )(z r − z r −1 )+...
+ f (e n ) (z n − z n − 1 )
or
Sp =
where ∆ z r =
n
n
r =1
r =1
∑ f (er ) .(z r − z r –1 ) = ∑ f (er ) . ∆z r
... (2.1)
z r – z r −1 ⋅
As n → ∞, i. e., the largest of the chord lengths ∆ z r approaches to zero and if for
every partition P and for every choice of points e r the sum S p tends to a unique limit,
then the function f (z) is said to be integrable from a to b along C and this limit is
denoted by
b
lim S p = ∫ f ( z ) dz or
n→ α
a
.... (2.2)
∫ f (z ) dz
C
and is called the complex line integral or briefly the line integral of f (z) along
the curve C or the definite integral of f (z) from a to b along the curve C.
n
Thus
∑ f (e r ) (z r − z r −1 ),
∫ f (z )dz = nlim
→∞
... (2.3)
where z r −1 ≤ e r ≤ z r
r =1
C
2.6 Real Line Integral
Let P (x, y) and Q (x, y) be real functions of x and y and continuous at all points of a
curve C. Then the real line integral of Pdx+Q dy along the curve C can be defined in a
similar manner as in article 2 ⋅ 5 and is denoted by
∫ {P( x, y )dx + Q( x, y )dy }
C
or
∫ (P dx + Qdy )
... (2.1)
C
If the curve C is smooth and its parametric equations are given by
x = φ(t ), y = Ψ (t ), t1 ≤ t ≤ t 2 , then the real line integral (2.1) is given by
Complex Integration
47
t2
∫
[ P{ φ (t ), ψ(t )} φ' (t ) + Q{ φ (t ), ψ (t ) }Ψ ' (t )] dt
... (2.2)
t1
2.7 Connection Between the Real and Complex Line Integrals
If f ( z ) = u( x, y ) + iv( x, y ) = u + iv, where z = x + iy , then the complex line integral
∫ f (z ) dz can be expressed in terms of real line integrals as
C
∫ f (z )dz = ∫ (u + iv )(dx + i dy )
C
[∵ dz = dx + idy ]
C
= ∫ (udx − vdy ) + i ∫ ( vdx + udy )
C
C
2.8 Some Elementary Properties of Complex Integrals
In article 2.7, we have seen that a complex integral is the combination of two real
integrals. Some elementary properties of real integrals hold good in case of the
complex integrals also. Some essential properties are as follows :
(1)
∫ [ f (z ) + g (z )] dz = ∫ f (z ) dz + ∫ g (z ) dz
C
(2)
C
C
∫ k. f (z ) dz = k∫ f (z ) dz, where k is a constant.
C
C
(3) If C1 , C 2 are two parts of the curve C, i.e., if C1 , C 2 represent the curves from a
to b as a to m and m to b respectively, then we can consider C = C1 + C 2 and
∫ f (z ) dz = ∫
C
(4)
f ( z ) dz =
C1 + C2
∫ f (z ) dz + ∫ f (z ) dz.
C1
C2
∫ f (z ) dz = − ∫ f (z ) dz
C
–C
Where – C indicates the direction opposite to that indicated by C.
2.9 Evaluation of Complex line integral
1. By indefinite integration : If f (z) is analytic function in a simply connected
domain D and f (z) =F ' ( z ) =
dF
,
dz
z2
then
∫
f ( z )dz = F( z 2 ) − F( z1 ) .
z1
This property is due to independence of path.
2. By Use of Paths : If the function f(z) where z=z(t) is defined in c; a ≤ t ≤ b then
b
∫
C
f ( z ) dz =
∫
a
f [ z(t )]
dz
dt .
dt
48
Engineering Mathematics-III
Here the line integral is converted into ordinary integral by applying the property of
the curve.
Illustrative Examples
Evaluation of integrals by definition (ab-initio)
It is difficult to find the integrals of all the complex functions by the direct
application of the definition of the integral of a complex function. However, we can
find the integrals of some simple functions by direct definition.
Ex.1 : Evaluate
dz.
∫
C
Sol. Refer Figure 2.4. by the definition of complex integral, we have
n
∑ f (e r ) . (z r − z r −1 )
n→ ∞
∫
r =1
C
Here,
... (2.1)
f ( z ) = lim
f ( z ) = 1 ∴ f ( e r ) = 1. Therefore, from (1), we have
n
∫
dz = lim
n→ ∞
C
∑ 1.(z r − z r −1 )
r =1
= lim [( z1
n→ ∞
−
z 0 ) + ( z 2 − z1 ) +....+ ( z n
−
z n −1 ]
= lim ( z n − z 0 ] = b − a [Since z 0 = a and z n = b.]
n→ ∞
Ex. 2 : Evaluate
∫
dz .
C
Sol : By the definition of complex integral, we have
n
∑ f (e r ) (z r − z r − 1)
∫ f (z ) dz = nlim
→∞
... (2.1)
r =1
C
Here, f ( z ) = 1, therefore f ( e r ) = 1.
Substituting
f(z) = 1, f (er) = 1 and replacing dz by dz , we have
n
∫
∑ 1. z1 − z r −1
n→ ∞
dz = lim
C
r =1
= lim [ z1 − z 0 + z 2 − z1 +....+ z n − z n −1 ]
n→ ∞
= l (length of the curve C between z0 = a to zn = b)
Ex. 3 : Evaluate
∫ z dz.
C
Sol. By the definition of complex integral, we have
n
∑ f (e r ).(z r − z r −1 )
∫ f (z ) dz = nlim
→∞
C
r =1
... (2.1)
Complex Integration
49
where z r −1 ≤ e r ≤ z r . Here f (z) = z , therefore f ( e r ) = e r
Substituting in (2.1), we have
n
∫
∑ e r .(z r − z r −1 )
z dz = lim
n→ ∞
C
... (2.2)
r =1
Taking e r = z r −1 in (2.2) we have
n
∫
z dz = lim
n→ ∞
C
∑ z r −1 (z r − z r −1 )
...(2.3)
r =1
Again taking er = zr in (2.2), we have
n
∑ z r (z r − z r −1 )
n→ ∞
∫
...(2.4)
z dz = lim
r =1
C
Adding (2.3) and (2.4), we have
n
∑ {(z r )
n→ ∞
2∫ z dz = lim
C
∴
∫
2
− ( z r −1 ) 2 } = b 2 − a 2
r =1
1 2
( b − a 2 ).
2
z dz =
C
Note : If the curve C is closed then the end points a and b coincide i.e. b=a.
Hence
∫ z dz = 0 (for closed curve C).
C
Complex line integral as the sum of real line integrals.
Ex. 4 : Evaluate ∫ z− dz from z = 0 to z = 4 +2i along the curve C given by
C
(a) z=t2 + it and
(b) the line from z= 0 to z = 2i and then, the line from z=2i to z=4+2i.
Sol. We have z = x + iy, so z = x − iy and dz = dx + i dy
Thus
∫
C
z− dz = ∫ ( x − iy )( dx + idy )
... (2.1)
C
(a) If C is the line given by z = t2 + it, then x = t2, y = t, hence dx = 2t dt and dy = dt.
The points z=0 and z=4+2i on C correspond to t=0 and t=2 respectively.
∴ From (2.1), we have
2
∫
C
z− dz =
∫
2
(t 2 − it )( 2t + i )dt =
t= 0
∫
t= 0
2
1
t2
1
=  t 4 +  − i t 3

2
 3 
2
0
2
= 10 −
0
2
( 2t 3 + t ) dt + i ∫ ( −t 2 ) dt
0
8
i.
3
50
Engineering Mathematics-III
(b) The line from z = 0 to z = 2i is the line OA along
imaginary axis and the line from z = 2i to z = 4 + 2i is
the line AP parallel to the axis.
y
A2
We can take C = C1+ C2, where C1 is the line OA and C2
P(Z = 4 + 2i)
C1
the line AP.
∴
C2
4
From (1), we have
−
∫ z dz
x
O
∫ ( x − iy )(dx + i dy )
=
C
Fig.2.5
C1 + C1
=
∫ ( x − iy )(dx + idy ) + ∫ ( x − iy )(dx + idy )
C1
C2
But on C1, x = 0 so dx = 0 and y varies from 0 to 2 and on C2, y=2 so dy = 0 and x
varies from 0 to 4.
2
∴
4
−
∫ z dz = ∫ (− iy )(idy ) + ∫ ( x − 2i ) dx
C
0
0
22
y
x2

=  +
− 2i x
 2  0  2

4
= 2 + ( 8 − 8i ) = 10 − 8i.
0
1+ i
Ex. 5 : Find the value of the integral
∫ (x − y + i x
2
) dz.
0
(a) Along the straight line from z = 0 to z =1+i.
(b) Along the real axis from z = 0 to z =1 and then along a line parallel to the
imaginary axis from z = 1 to z =1+i.
Sol. Let A be the point of affix 1+i and N be the point of affix 1.
(a) Let OA be line from z = 0 to z = 1 + i. On OA, y = x so z = x+iy = x+ix and
dz = (1+i) dx
1
x3 
Hence, ∫ ( x − y + i x ) dz = ∫ i x (1 + i ) dx = ( −1 + i )  
 3
OA
0
2
1
2
=
0
(b) The real axis from z = 0 to z = 1 is the line ON along real
axis and the line from z=1 to z = 1+i, a line NA parallel to
imaginary axis.
1
( −1 + i).
3
y
A(z =1 + i)
1
So here the contour of integration C consists of the lines ON
and NA. i.e. say C= C1+C2.
On C1 (ON), y = 0 so z = x + iy = x and dz = dx, thus x
varies from 0 to 1 and on C2 (NA), x = 1, so here z = 1 + iy,
dz = i dy and y varies from 0 to 1.
C2
O
C1
Fig.2.6
N(z=1)
x
Complex Integration
51
1+ i
Hence
∫ (x − y + i x
2
) dz =
0
∫ (x − y + i x
2
) dz
C1 + C1
∫ (x − y + i x
=
2
) dz +
C1
1
∫ (x − y + i x
2
) dz
C2
1
= ∫ ( x + i x 2 )dx + ∫ (1 − y + i )dy
0
0
1
3 1

x2
y2
x
=
+i

 + i. (1 + i ) y −
3
2 

 2
0
0
1 i
1 
1 5


=  +  + i  + i = − + i.
 2 3
2 
2 6
Ex. 6 : Integrate z2 along the straight line OM and also
along the path OLM consisting of two straight line
segments OL and OM, where O is the origin, L is the point z
= 3 and M the point z = 3 + i. Hence, show that the
integral of z2 along the closed path OLMO is zero.
y
(z =3 + i)
M
1
O
Sol. We have, z = x + iy.
∫z
C
2
1
3
L(z =3)
x
Fig. 2.7
dz = ∫ ( x + i y ) 2 (dx + i dy )
C
= ∫ ( x 2 − y 2 + 2i xy )( dx + i dy )
... (2.1)
C
Now on the line OM, x = 3y so dx = 3dy and as z moves from O to M, y varies from 0
to 1. Therefore, from (2.1), we have
1
∫
OM
z 2 dz = ∫ ( 9 y 2 − y 2 + 2i.3 y 2 )( 3 + i ) dy
0
1
=
1
2
2
∫ (8 + 6i )(3 + i )y dy = ∫ (18 + 26i )y dy
0
0
3 1
y
= (18 + 26 i )  
 3 
Again
∫
2
z dz =
OLM
=
∫ (x
OL + LM
2
∫ (x
2
−y
2
0
=
1
26
(18 + 26 i ) = 6 +
i ... (2.2)
3
3
+ 2 i xy ).( dx + idy )
− y 2 + 2ixy ) ( dx + idy )
OL
+
∫ (x
2
− y 2 + 2ixy ) ( dx + idy )
...(2.3)
LM
But on OL, y = 0 so dy = 0 and x varies from 0 to 3 on LM, x = 3 so dx = 0 and y
varies from 0 to 1.
Therefore from (2.3), we have
52
Engineering Mathematics-III
z 2 dz =
∫
OLM
3
∫
x 2 dx +
0
1
∫
( 9 − y 2 + 6i ). idy
0
1
3


y3
+ i 9 y −
+ 3iy 2 
3

 0
0
1
26
= 9 + i 9 − + 3i = 6 +
i


3
3
x3 
= 
 3
...(2.4)
Also the integral of z2 along the closed path OLMO is given by
z 2 dz =
∫
OLMO
z 2 dz =
∫
∫
OLM + MO
=
∫
2
z dz −
OLM
OLM
∫
z 2 dz +
∫
z 2 dz
MO
2
z dz
OM
26  
26 
=  6 +
i −  6 +
i = 0.

3  
3 
[substituting from (2.2) and (2.4)]
Ex.7: Using the definition of the integral of f (z) on a given path, evaluate
5+ 3i
∫
z 3 dz.
− 2+ i
Sol. f (z)=z3 is analytic for all finite values of z, so its integration, along a curve
joining two fixed points will be the same, whatever be the path. Here, we have to
integrate z3 between two point (−2, 1) and (5, 3). Let the path of integration joining
these points be along the curve made up of :
1. A line parallel to the real axis from the point (−2, 1) to the point (5, 1). On this
line z = x + i, dz = dx and x varies from −2 to 5.
2. Followed by a line parallel to the axis of imagineries from the point (5, 1) to the
point (5, 3). On this line z = 5 + iy, dz = i dy and y varies from 1 to 3.
5+ 3i
Thus
∫
z 3 dz =
− 2+ i
5
∫
–2
3
( x + i ) 3 dx + ∫ ( 5 + i y ) 3 idy along the chosen path
1
5
3
1
1
=  ( x + i )4  +  (5 + i y )4 
 4

 4

–2
1
1
1
4
4
= [( 5 + i ) − ( −2 + i ) ] + [( 5 + 3i ) 4 − ( 5 + i ) 4 ]
4
4
1
= [ 984i − 637].
4
Ex. 8 : Evaluate
x = a (θ + sin θ),
∫ C (z
2
+ 3z + 2) dz where C is the arc of the cycloid
y = a (1 − cos θ) between the points (0, 0) and (π a, 2a).
Complex Integration
53
Sol. The function f(z) = z2 + 3z + 2 is a polynomial and therefore, analytic in
z - plane; hence its integral between two points (0, 0) and (π a, 2a) is independent of
the path joining these points.
Choose the path of integration a curve C consisting of :
1. The part of real axis from the point (0, 0) to the point (π a, 0). On this line z = x,
y = 0 so dz = dx, and x varies from 0 to π a.
2. Followed by a line parallel to the axis of imagineries from the point (πa, 0) to the
point (π a, 2 a). On this line x = π a so z = π a + i y, dz = i dy, and y varies from 0 to 2a.
Hence
2
∫ (z + 3z + 2) dz =
C
Ex. 9 : Prove that ∫
C
πa
∫
( x 2 + 3 x + 2)dx +
0
2a
∫
[( πa+i y ) 2 + 3( πa+i y )+2] i dy
0
πa
2a
1
3
1
3
=  x 3 + x 2 + 2 x +  ( πa+ iy ) 3 + ( πa + iy ) 2 + 2iy 
 3



2
2
0
3
0
1
3


= ( πa) 3 + ( πa) 2 + 2πa
 3

2
1
3
1
3
+  ( πa + i 2a) 3 + ( πa + i 2a) 2 + 4ia − ( πa) 3 − ( πa) 2 
 3

2
3
2
1
3
= 2πa + ( πa + i 2a) 3 + ( πa + i 2a)2 + 4ia.
3
2
1
dz = 2πi, where C is given by the equation z − a = R.
z−a
Sol. Since C is a circle z − a = R, therefore z − a = Reiθ , where θ varies from 0 to 2π .
So that, dz = Reiθ . idθ.
Hence
∫
C
1
dz =
z−a
2π
∫
0
1
Reiθ
Reiθ . i dθ = i
2π
∫ dθ = 2πi.
0
1
Ex. 10 : Prove that the value of the integral along a semicircular arc z =1 from
z
− 1 to 1 is −πi or πi according as the arc lies above or below the real axis.
Sol: The equation of the circle z = 1 is z = eiθ , so that dz = ieiθ dθ where
0 ≤ θ ≤ 2π.
As z varies along the semi-circular arc z = 1 above
y
the real axis from −1 to 1, i.e. along arc CBA, θ varies
B
from π to 0.
∴
∫
CBA
x'
1
dz =
z
0
∫
π
1
e
iθ
ieiθ dθ = iθ] 0π = – π i
Again as z varies along the semi-circular arc z = 1
below the real axis from −1 to 1, i.e. along the arc CDA,
θ varies from π to 2π.
P
0 A(1,0)
(-1,0)C
O
D
y'
Fig. 2.8
x
54
Engineering Mathematics-III
∴
∫
CDA
2π
1
dz =
z
∫
θ= π
2+ i
Ex. 11 : Evaluate
e −iθ . ieiθ dθ = i [θ]2ππ = πi.
∫ (z )
2
dz along the real axis from z = 0 to z = 1 and then along a
0
line parallel to y-axis from z = 2 to z = 2 + i.
2+ i
Sol.
∫
2+ i
( z ) 2 dz =
∫ ( x − iy )
0
0
2+ i
∫ (x
=
2
2
(U.P.Tech. 2002)
y
( dx + i dy )
− y 2 − 2ixy ) ( dx + i dy )
B 2+i
0
=
∫
( x 2 )dx +
OA
∫ (− y
2
1
)i dy
AB
[Along OA, y = 0, along AB, x=0 ]
x3 
= 
 3
2
0
O
2
x
A
Fig. 2.9
3 1
y
8
1 1
− i   = − i = ( 8 − i ).
3
3 3
 3  0
2.10 An Upper Bound for a Complex Integral
Theorem. If a function f (z) is continuous on a contour C of length l and if M be the
upper bound of f (z) on C, then
∫ f (z )dz
≤ Ml.
C
Proof. By definition of complex line integral we have
n
∑ f (ξ r ).(z r − z r −1 )
n→ ∞
...(2.1)
f ( z )dz = lim
∫
r =1
C
where ξ r is any point on the arc between zr-1 to zr.
n
n
∑
Now
f ( ξ r ) ( z r − z r −1 ≤
r =1
∵
∑
f ( ξ r ) .|z r − z r −1|
r =1
M is the upper bound of f (z) on C. Therefore f ( ξ r ) ≤ M .
n
∴
n
∑
r =1
f ( ξ r )( z r − z r −1 ) ≤ M . ∑ z r − z r −1
r =1
Taking limit on both sides as n → ∞ and using (1), we have
n
∫
C
n→ ∞
where l is length of the contour C
Hence
∫
C
n
f ( z )dz ≤ lim M . ∑ z n − z r −1 = M . lim
f ( z ) dz ≤ Ml.
r =1
n→ ∞
∑ z r − z r −1
r =1
= M . l,
Complex Integration
55
Problem Set (2.1)
✜
Evaluate the following integrals.
1.
∫ ( x − iy
3
) dz, where c is the straight line path from z=2 to z = 2 + 2i.
C
3+ i
2.
z 2 dz, along
∫
0
3.
(i) the line y = x/3
(ii) real axis to z and then vertically to 3 + i
(iii) the parabola x = 3y2.
∫ Re (z ) dz, where c is : (i) Shortest path from 1+i to 3+2i (ii) Along the
C
straight line from (1, 1) to (3, 1) and then from (3, 1) to (3, 2) (iii) Are the
integrals (i) and (ii) equal? If not give reason.
4. ∫ z n dz, n = 0, ± 1, ± 2,..., where c : z = r traversed in the anticlockwise direction.
C
5.
∫ (z / z ) dz, where C is defined by ythe fig.28.
C
2
C1
1
C2
C
–2
D
A
0
–1
B
1
2
x
Fig. 2.10
6.
∫ cos z dz, where C is the semicircle
z = π, x ≥ 0 from −π to π.
C
7.
∫ ze
2z
dz, where C is defined by z = 0 to z = 1.
C
8.
∫z
2
sin 4z dz, where C is defined by z = 0 to z = 2π.
C
9.
∫
sin 2 z dz, where C is z = − π i to z = πi.
C
2
10. Evaluate ∫ z dz around the square with vertices at (0,0), (1,0) (1,1) (0,1).
C
11. Evaluate ∫ z 2 dz around the circle z − 1 = 1.
C
12. Evaluate ∫ ( z 2 + 2z ) dz along :
C
(i) the circle z = 2 from (2, 0) to (0, 2) in a counterclowckwise direction.
(ii) the straight lines joining (2, 0) to (2, 2) and then from (2, 2) to (0, 2).
56
Engineering Mathematics-III
2.11 Cauchy's Integral Theorem or Cauchy's Theorem
If f (z) is an analytic function of z and f ' (z) is continuous at each point within and
on a closed contour C, then ∫ f ( z ) dz = 0.
C
Proof. Let D be the region which consists of all points within and on the contour C.
∂Q ∂p
If P (x, y), Q (x, y),
are all continuous functions of x and y in the region D,
,
∂x ∂y
then by Green's theorem
 ∂Q
∫ (Pdx + Qdy ) = ∫ ∫  ∂x
C
∂P 
 dx dy .
∂y 
−
D
Since f (z) = u + iv is continuous on the simple curve C and f ' (z) exists and is
continuous in D, therefore, u, v, u x , u y , v x , v y are all continuous in D. The
conditions of Green's theorem are thus satisfied. Hence
∫
C
f ( z ) dz = ∫ (u dx − v dy ) + i ∫ ( v dx + u dy )
C
C
 ∂v ∂u 
 ∂u ∂v 
= − ∫ ∫  +  dx dy + i ∫ ∫ 
−  dx dy
 ∂x ∂y 
 ∂x ∂y 
D
D
[By Green's theorem]
= −∫
D
 ∂v
∂v 
 ∂v
∂v 
∫  ∂x − ∂x  dx dy + i ∫ ∫  ∂y − ∂y  dx dy = 0
D
[By Cauchy-Riemann equations]
Hence,
∫ f (z ) dz = 0.
C
Remark. Goursat showed that for the truth of the theorem the assumption of the
continuity of f ' (z) is unnecessary and that Cauchy's theorem holds if and only if f (z)
is analytic within and on C.
2.12 Extension of Cauchy's Theorem to Multi-Connected Region
Theorem. If C is closed curve and C1, C2, C3,.....Cn, are the other closed curves
which lie inside C, and if function f (z) is analytic in the region between these curves,
and continuous on C, then
∫ f (z )dz = ∫ f (z )dz + ∫
C
C1
C2
f ( z )dz +
∫
f ( z ) dz +...
C3
where integral along each curve is taken in the anti-clockwise direction.
Proof. Refer Fig. 2.2 (c) Connect the curves C1, C2, C3, C 4 ,....each to the curve C by
means of the narrow cuts AB, PQ, RS, UV respectively. In this way curve C is connected
to C1 by a line AB going from a point A on the curve C to a point B on C1 and by another
line BA going from the point B on C1, to the point A on C, thus creating a narrow
Complex Integration
57
channel cut of parallel sides AB and BA. Similarly, we introduce other cuts PQ, RS,UV,
etc. As a result of the introduction of all these cut curves to the already existing curves,
the sum of the lengths of these curves is a single closed curve, say Γ, in the interior of
which f (z) is analytic and on whose boundary f (z) is at least continuous. Hence by
Cauchy's theorem ∫ f ( z ) dz = 0.
... (2.1)
Γ
where the integration along Γ is taken in such a way that while taking a complete
round over Γ, the area enclosed by Γ is always to the left of the mover.
When we go round Γ in the above manner the rounding over C is anti-clockwise, the
rounding over each of the curves, C1, C2, C3, C 4 ,.... is clockwise and rounding over
the each cut is repeated twice in opposite directions. Thus when going in this
manner along Γ, we have integral over C as ∫ f ( z )dz, the integral over C1, C2, C3,.....
C
as the integral over the cut AB once as
∫
f ( z ) dz and again as
AB
∫
f ( z )dz = −
BA
∫ f (z )dz
AB
with similar two integrals for each of the remaining cuts. The two integrals for
each cut cancel being opposite in sense. The same is applicable to all other cuts.
Hence (1) is equivalent to
∫ f (z ) dz − ∫
C
i.e
f ( z ) dz −
C1
∫ f (z ) dz = ∫ f (z ) dz + ∫
C
∫
f ( z ) dz −
C2
C1
f ( z ) dz +
C2
∫
f ( z ) dz... = 0
C3
∫
f ( z ) dz +...
C3
Where all integrals are taken in anticlokwise direction. Thus, the theorem is proved.
Particular Case. If there is only one closed curve C1 within C, then
∫ f (z )dz = ∫ f (z ) dz where integration over each curve is in anticlockwise
C
C1
direction.
2.13 Independence of Path
.
Z2
Let f (z) be an analytic function in a simply connected domain
D. Let C1 and C2 be any two paths in D joining any two points
z1 and z2 in D and having no further points in common. Then
∫
f ( z ) dz =
C1
∫
C2
where both the paths C1 and C2 have the same directions.
or
C2
f ( z ) dz,
C1
.
Z1
Fig. 2.9
If f (z) be an analytic function in a simply connected domain D,
then the integral of f (z) from z1 to z2 is independent of the path joining them. The
integral depend on the points z1 and z2.
58
Engineering Mathematics-III
Proof. We can observe that the two curves C1 and C2 form a simple closed curve, say
C in D. By Cauchy's theorem
∫
f ( z )dz =
∫
f ( z )dz = −
C
or
∫
C1
C1
or
f ( z ) dz +
∫ f (z )dz = 0
C2
∫
f ( z ) dz
C2
∫ f (z )dz = ∫
C1
f ( z ) dz
C2
when C1 and C2 are both traversed in the same direction.
Remark : From independence of path, it seems as if the path C2 is obtained by
continuous deformation of path C1.
Ex.12 : Find the value of the integral ∫
C
3z 2 + 7 z + 1
dz, where
z +1
1
C is the circle |z|=
2
Y
C1
O
X=
P
1
2
X
Sol. Poles of the integrand are obtained by putting the
denominator equal to zero i.e., z + 1 = 0 or z = −1.
1
1
with centre at z = 0 and radius
2
2
does not contain any singularity of the given function since
z = −1 lies outside this circle, so we have
Fig. 2.10
Now the given circle |z|=
3z 2 + 7 z + 1
dz = 0.
z +1
∫
C
Ex.13 : Find the value of ∫
C
z+4
2
z + 2z + 5
(by Cauchy's Integral theorem)
dz, if c is the circle |z + 1|= 1.
Sol. Poles of integrand are given by putting the denominator equal to zero, i.e.,
or
z 2 + 2z + 5 = 0
−2 ± 4 − 20 −2 ± 4i
z=
=
= −1 ± 2 i.
2
2
Given circle |z + 1|= 1 with centre at z = −1 and radius unity does not contain any
z+4
singularity of the function
.
2
z + 2z − 5
z+4
So
(by cauchy's Theorem)
∫ z 2 + 2z + 5 dz = 0.
C
Complex Integration
59
Ex.14 : Show that
∫ (z − z 0 )
n
C
z 0 inside c 
 0, if n ≠ −1,


dz =  2πi, if n = −1, z 0 inside c 


 0, if n = −1, z 0 outside c
where orientation of C is anticlockwise and z0 is a complex number.
(i) C is the circle with centre at z0 and of radius R.
(ii) C is any arbitrary simple closed curve.
Sol : (i) We have circle C : z − z 0 = R or z – z0 = Reiθ . So, dz = i Reiθ dθ.
n
Thus,
I = ∫ ( z − z 0 ) dz =
C
=iR
n +1
) (i Reiθ dθ)
y
C
e'
2π
e
∫
iθ n
∫ Re
i ( n + i )θ
R
dθ
0
When
C
Z0
n ≠ −1, we have
I = i R n +1
ei ( n +1)θ
i( n + 1)
X
2π
0
=
Fig. 2.11
R n +1
[cos (n+1) 2π−cos 0]
n +1
=0
When
n = − 1, I =
∫
C
1
dz = i
z − z0
2π
∫
dθ = 2πi
y
0
i
is analytic
z − z0
within c, hence by Cauchy's integral Theorem
1
I= ∫
dz = 0 .
z − z0
When n = − 1 and z0 lies outside c, then
.
z0
R
X
C
(ii) When C is simple closed curve enclosing a circle
C* : z−z 0 =Re iθ.
Fig. 2.12
When z0 lies inside C, then (z−z0)n for any n, is analytic on and between C and
C*. By Cauchy's theorem for multiply connected domain, we have
∫ (z − z 0 )
n
C
dz =
∫
C*
n ≠ –1
 0,
( z − z 0 ) n dz = 
 2πi, n = −1
When z0 lies outside C, then ( z – z 0 ) n is analytic within C, hence by Cauchy's
integral theorem
∫ (z − z 0 )
n
dz = 0.
C
Note :
∫
C
dz
= 2πi, where C is the unit circle z = 1 . This is obtained when n = − 1
z
and z0 = 0.
60
Engineering Mathematics-III
Problem Set (2.2)
✜
Verify Cauchy's integral theorem for the following problems.
1.
∫ (3z
2
+ iz − 4)dz, C is the square with vertices at 1 ± i, − 1 ± i.
C
2.
∫ (z
5
− iz 2 − 5z + 2 i ) dz, C is the circle z = 1.
C
3.
∫ cos z dz, C is the rectangle with vertices −2 ± i,
2 ± i.
C
4.
z3 + 4
∫ cosh z dz, C is the unit circle.
C
5.
e 5z
∫
6.
dz, C is the unit circle.
z 3 + 27
C
sinh z
∫ (z + 1)(z − 2)(z + 3) dz, C is
z − i = 1.
C
7.
cosh 2 3z
∫
( z + 4 i )( z 2 + 9)
C
8.
dz, C is z − 1 =1.
5

∫  sin z + z 3  , C is |z|= 1 .
C
9.
dz
∫ z − 2 around (i) the circle z − 2 = 4
C
(ii) z − 1 = 5 (iii) square with vertices at 2 ± 2 i, − 2 ± i.
Evaluate the following using the extension of Cauchy's integral theorem for multiply
connected domain.
2+ i
10.
∫z
4
dz
1
3i
11.
e −5z dz.
∫
–i /3
i
12.
( z 2 + 1) 3 dz.
∫
0
1+ i
13.
∫
0
dz
( z + 2) 3
1– 2i
14.
∫ z cos(z
0
2
.
) dz.
Complex Integration
61
2.14 Cauchy's Integral Formula
If f (z) is analytic within and on a closed curve C and a is any point within C, then
f ( a) =
1
2πi
∫
C
f (z )
dz.
z−a
Let z=a be a point within a closed curve C. Describe a circle γ, with centre at the
point z=a of radius ρ such that it lies entirely within C as
C
.
shown in the Fig. 2.14.
γ
Z=a
f (z )
. Obviously this function is
z−a
analytic in the region between γ and C. Hence by Cauchy's
Consider the function φ (z)=
Fig. 2.14
theorem for multi-connected region, we have
∫
φ ( z ) dz =
∫
f (z )
dz =
z−a
∫
f ( z ) dz
f ( z ) − f ( a)
f ( a)
=∫
dz + ∫
dz
z−a
z−a
z−a
C
or
C
or
C
∫ φ (z ) dz
γ
f (z )
dz
z−a
∫
γ
...(2.1)
γ
γ
On the circle γ, z − a = ρeiθ , so that dz = ρieiθ dθ and θ varies from 0 to 2π, hence
∫
γ
f ( a)
dz = f ( a)
z−a
2π
ρieiθ dθ
∫
ρeiθ
θ= 0
2π
= f ( a) ∫ idθ = 2πif ( a)
0
Substituting it in (1), we have ∫
C
or
∫
C
Now∫
γ
f ( z ) dz
− 2πi f ( a) =
z−a
f ( z ) − f ( a)
dz =
z−a
2π
∫
0
f ( z )dz
f ( z ) − f ( a)
=∫
dz + 2πi. f ( a)
z−a
z−a
γ
f ( z ) − f ( a)
dz
z−a
∫
γ
f ( a + ρ e θ i ) – f ( a)
ρe
θi
...(2.2)
iρeθi dθ
[ z − a = ρeθi ⇒ dz = iρeθi dθ]
2π
=
∫ [ f ( a + ρe
θi
) − f ( a)] i dθ = 0
[ as ρ → 0]
...(2.3)
0
In the limiting form, as the circle γ shrinks to the point a i.e., ρ → 0, then from (2.2)
and (2.3).
62
Engineering Mathematics-III
∫
C
f (z )
dz = 2πi f ( a)
z−a
Hence
f ( a) =
1
2πi
∫
C
f (z )
dz.
z−a
which is the required Cauchy's Integral formula.
2.15 Extension of Cauchy's Integral Formula to Multi-connected
Regions
If f (z) is analytic in the region bounded by two closed curves C and C ' and a is a
f ( z )dz
f ( z ) dz
1
1
point in the region, then f ( a) =
−
, where C is the outer
2πi ∫ z − a
2πi ∫ z − a
C
C'
curve.
Proof. Draw a small circle γ with its center at point a. Now
C
f (z )
consider the function
which is analytic in the region
z−a
γ
C'
a
bounded by the three contours C, C ' and γ (because z−a is not
zero for any value of z in this annulus) and it is also analytic
Fig. 2.15
on these curves.
Hence by Cauchy's theorem of multi-connected region
∫
C
f ( z ) dz
f ( z )dz
f ( z ) dz
=∫
+∫
z−a
z−a
z−a
C'
γ
where integration round curve is taken in such a way that the annular region always
lies to the left.
f ( z ) dz
Thus,
∫ z−a =
C
or
f ( a) =
∫
C'
f ( z )dz
+ 2πif ( a)
z−a
1
2πi
∫
C
f ( z )dz
1
−
z−a
2πi
[by Cauchy's integral formula]
∫
C'
f ( z )dz
.
z−a
In general, if there be more curves C'', C''',...etc., then we have similarly
f ( a) =
Ex. 15 : Evaluate ∫
C
(i)
z =3
1
2πi
∫
C
f ( z ) dz
1
−
z−a
2πi
∫
C'
f ( z ) dz
1
−
z−a
2πi
e −2z
dz, where C is the circle:
z+2
(ii) z =
1
2
Sol : f ( z ) = e −2z is an analytic function.
∫
C ''
f ( z )dz
...
z−a
Complex Integration
63
(i) Denominator z + 2 = 0 or z = −2 is the point of singularity. This point lies
inside the circle z = 3 .
Therefore by Cauchy's integral formula
f (z )
1
f ( −2) =
dz
∫
2πi z − ( −2)
C
e −2( −2) =
or
1
2πi
e − 2z
dz
z+2
∫
C
[∵ f ( z ) = e −2( −2) ]
0
–3 –2
3
− 2z
or
∫
C
e
dz = 2πie 4
z+2
Fig. 2.16
(ii) The point z = –2 lies outside the circle z =
Therefore, the function
1
.
2
e − 2z
is analytic within and on C.
z+2
Hence by Cauchy's integral theorem , we have
∫
C
e − 2z
dz = 0
z+2
Ex. 16 : Evaluate ∫
C
dz
, where C is the circle z + 3i = 1.
z( z + πi )
Sol : Here, z=0 and z=− πi are the singular points of integrand but z=0 lies outside
the circle z + 3 i = 1, hence take f(z)=
1
which is analytic within and on C.
z
Since z = −πi lies inside the circle z + 3i = 1 hence by Cauchy's integral formula
f (z )
∫ (z − (– πi )) dz = 2πi f (− πi )
C
or
1
1
∫ z (z + πi )dz = 2πi . (− πi ) = −2 .
C
Ex. 17 : Evaluate ∫
C
sin πz 2 + cos πz 2
dz, where C is the circle z = 3.
( z − 1)( z − 2)
[UPTU, 2002]
64
Engineering Mathematics-III
Sol : Singularities are (z − 1) (z − 2) = 0, or z = 1, z = 2.
Y
Both the points z = 1, 2 lie in the circle z = 3.
X
O
Take f (z) = sin πz 2 + cos πz 2 and let the given integral be
–1
denoted by I. Then
–2
I =
∫
C
f (z )
dz
( z − 1)( z − 2)
–3i
–i
z + 3i =1
A
B 
= ∫ 
+
 f ( z )dz.
 z − 1 z − 2
C
Now,
1
A
B
=
+
( z − 1)( z − 2) z − 1 z − 2
Fig. 2.17
[By partial fraction]
1
A=
= −1 at z = 1
z−2
1
B=
= 1 at z = 2
z −1
Y
z =3
1 2
By Cuachy's integral formula
sin πz 2 + cos πz 2
I =∫
dz
( z − 1)( z − 2)
=2π [ Af (1) + Bf ( 2)]
0
3
X
Fig. 2.18
= 2πi [( −1){sin π + cos π} + (1){sin π 2 2 + cos π 2 2 }]
Ex. 18 : Evaluate ∫
C
= 2πi [( −1)( −1) + 1] = 4πi
dz
where C is the circle z = 6.
2
z + 16
Sol : Given integrand has two singular points at z = ± 4i,
both of which lie within the circle z = 6 . Readers note
that this problem can be solved by the help of partial
fractions. It can also be solved alternatively by the
following way. Let the given integral is denoted by I. Both
the singular points z = ± 4i lie within C : z =6. Let us
construct two circles C1 and C2 with the points z = ± 4 i as
Y
4i
c1
z =6
6
0
c2
4i
X
centres in such a way so that they do not intersect each
1
other and lie withing C. Now f ( z ) =
is analytic in
Fig. 2.19
2
z + 16
multiply connected domain in between C, C1, and C2.
Using the extension of the Cauchy's integral theorem for multiply connected
domain, we have
1
I =∫
dz
2
z
+
16
C
Complex Integration
65
=
∫
C1
[1/( z − 4 i]
dz +
z + 4i
∫
C2
By Cauchy's integral formula taking f (z) =
[1/( z + 4i]
dz
z – 4i
1
1
for C1 and f (z) =
for C2
z − 4i
z + 4i
respectively.
I = 2π i f ( −4 i ) + 2π i f ( 4 i )
 1 
 1 
= 2π i 
+ 2π i 


 z − 4 i  z = – 4i
 z + 4 i  z = 4i
= 2π i
Ex.19 : Evaluate ∫
C
1
i
π π
+ 2π i
= − + = 0.
−4 i − 4 i
4i + 4i
4 4
e 5z
dz, where C is the ellipse z − 2 + z + 2 = 6.
(z − 2i )
Sol : Given z − 2 + z + 2 = 6
or
[( x − 2) 2 + ( y ) 2 ]1/ 2 + [( x + 2) 2 + ( y ) 2 ]1/ 2 = 6
or
[( x − 2) 2 + ( y ) 2 ]1/ 2 = 6 − [( x + 2) + ( y ) 2 ]1/ 2
Squaring, we have
x 2 + y 2 + 4 − 4 x = 36 + ( x 2 + y 2 + 4 + 4 x ) − 12[( x + 2) 2 + ( y ) 2 ]1/ 2
or
12( x 2 + y 2 + 4 + 4 x )1/ 2 = 36 + 8 x
or
3( x 2 + y 2 + 4 + 4 x )1/ 2 = 9 + 2 x
Y
Again squaring, we have
2.2i
9( x 2 + y 2 + 4 + 4 x ) = ( 9 + 2 x ) 2 = 81 + 4 x 2 + 36 x
or
or
3
5 x 2 + 9 y 2 = 45
x2 y 2
+
=1
9
5
Comparing this with
2
1
3
2.2i
Fig. 2.20
x2
a2
+
y2
b2
= 1, we have a = 3, b =
5 = 2.2 appr.
Since the integrand has singular point at z=2i, which lies inside the ellipse, hence by
Cauchy's integral formula, taking f (z) = e5z, we have
∫
C
e 5z
dz = 2π i f ( 2 i ) = 2πi e 5.2i = 2πie10i .
z − 2i
66
Engineering Mathematics-III
Problem Set (2.3)
✜
Evaluate the following:
1.
2.
1
2πi
∫
C
4.
∫
C
5.
∫
C
6.
C
ez
dz, where C is the circle z = 3.
z−2
sin 6 z
, where C is the circle z = 2
z − ( π / 6)
∫
C
3.
∫
∫
C
ez
dz, where C is the circle |z − 0 . 5|= 2
( z − 1)( z − 4)
z3 + z +1
2
dz, where C is the ellipse 4x2+9y2=1.
z − 7z + 2
2z + 7
z 2 + 4z
sin πz
z2 −1
dz, where C: (i) z = 0 ⋅ 5, (ii) z = 4.
dz, where C :
(i) rectangle with vertices 2 ± i, −2 ± i
7.
(ii) rectangle with vertices − i, 2 − i, 2 + i, i.
z+5
∫ z 2 + 9 dz, where C: (i) z − 3i = 2
(ii) z + 3i = 1 .
8.
9.
∫
∫
C
e πiz
dz, where C is the circle z − 1 − 2 i = 2 .
z 2 − 4z + 5
e z + cos z
dz, where C is the boundary of triangle with vertices at
( z − 5)( z + πi )
−1, 1 and −3.5 i.
10.
∫
C
11.
e 2z + tan −1 z
( z − 1)( z − 3)( z + 4) 2
dz, where C is the circle z = 2 .
Using Cauchy's integral formula, show that ∫
C
1
2
z +4
dz =
π
, where C is the
2
circle z − 1 =2.
12. Show that ∫
C
ez
2
z 2 (z − 1 − i )
and z = 1 clockwise.
dz = πe 2i , where C consists of z = 2 anticlockwise
Complex Integration
67
2.16 Derivative of an Analytic Function
If a function f (z) is analytic in a region D, then its derivative at any point z = a of D is
f ( z ) dz
1
also analytic in D, and is given by f ' ( a) =
∫
2πi
( z − a) 2
C
where C is any closed curve in D surrounding the point z = a.
f ( z ) dz
1
Proof. By Cauchy's integral formula f (a) =
.
2πi ∫ z − a
C
D
Differentiating this partially w.r.t. ''a'', we get
f ' ( a) =
Similarly
1
2πi
∂  1 
∫ ∂a  z − a .
f ( z ) dz
C
=
f ( z ) dz
1
∫
2πi ( z − a) 2
C
f " ( a) =
2 ! f ( z ) dz
2πi ∫ ( z − a) 3
C
and in general f ( n ) ( a) =
C
a
.
P
Fig. 2.21
f ( z ) dz
f ( z ) dz
n!
2π ! ( n )
or ∫
dz =
f ( a)
n +1
2πi ∫ ( z − a) n +1
ni
( z − a)
C
Hence, we conclude if a given function f (z) is analytic at any point on a simple
closed curve C then the values of the function and all its Dirivatives can be obtained
at any point C. We can note that the derivatives of all orders are analytic themselves.
1
Ex. 20 : Eraluate ∫
dz, where C is the circle z − i = 3.
2
2
(
z
+
9
)
C
Sol : Given integrand has singular points at z = ± 3i. Out of these two points, the
point zi lies within the circle z − i = 3.
Let
I =
1
∫
2
( z + 3i ) ( z − 3i ) 2
C
=
[1/( z + 3i ) 2 ]
∫
( z − 3i ) 2
C
dz
dz.
Taking f (z) = 1/(z + 3i)2 and applying Cauchy's integral formula for derivative (n = 1),
we have
I =
∫
C
f (z )
( z − 3i ) 2
dz = 2πi [ f ' ( z )] z = 3i
 −2 
= 2πi 
3
( z + zi )  z = 3i
68
Engineering Mathematics-III
 −2 
π
= 2πi 
.
=
3
54
( 6 i ) 
Ex. 21 : Evaluate ∫
C
ez
z 2 ( z − 1) 3
dz, where C is the circle z = 2 ⋅ 5.
Y
Sol : Given integrand has singular points at z = 0 and z = 1.
Both of these points lie within the circle z = 2.5. Let us
C2
construct two circles C1 and C2 with centers 0 and 1
–1
respectively in such a way so that they do not intersect each
0
1 2
3
X
C1
other and lie within C. Applying the extension of the Cuachy
integral theorem for multiply connected domain, we have
I =
ez
∫
C
=
∫
z 2 ( z − 1) 3
dz
e z /( z − 1) 3
C1
z
Fig. 2.22
z2
dz +
∫
C2
e z /z 2
( z − 1) 3
dz = I1 + I 2 , say.
3
Taking f (z) = e /( z − 1) for the evaluation of I1 and f (z) = e z / z 2 for I2 and
applying Cauchy's integral formula for derivatives for n =1 and n=2 respectively,
we have
 d  e z 
2πi  d 2
I = 2πi  
+


3
2 !  dz 2
 dz  ( z − 1)   z = 0
 e z 
 
 z 2   z =1
 d  ( z − 2)e z  
( z − 1)3 e z − 3( z − 1)2e z 
 
= 2πi 
+ πi  

3
( z − 1)6
 
 dz  z

 z=0
z =1
 z 3 ( z − 1)e z − 3z 2 ( z − 2)e z 
= 2πi [ −4] + πi 

z6

 z =1
= – 8πi + 3πei = ( −8 + 3e) πi.
2
Ex. 22 : If f ( η) =
5z + 3z + 1
dz, where C is the circle x 2 + y 2 = 4, find the values
z−η
of f(4), f ' (1− i) and f '' (1− i).
Sol : The given circle C is x2+ y2 = 4, which can be written as z
Point z = 4 lies outside the circle z = 2, hence for η = 4,
f ( 4) =
5z 2 + 3z + 1
which is analytic everywhere in z = 2.
z−4
Therefore, by Cauchy's integral theorem, we have
= 2.
Complex Integration
69
f ( 4) =
∫
C
5z 2 + 3z + 1
dz = 0.
z−4
Now let g (z) = 5z 2 + 3z + 1, which is analytic everywhere.
Therefore, if η is a point within C then by Cauchy's integral formula, we have
g ( η) =
1
2πi
g (z )
∫ z − η dz
C
2
or
2πi g ( η) =
∫
5z + 3z + 1
dz = f ( η)
z−η
f ( η) = 2πi ( 5 η2 + 3 η + 1)
Thus
f ' ( η) = 2πi (10 η + 3)
Therefore
[By differentiation]
f ' (1 − i ) = 2πi{10(1 − i ) + 3} = 2πi(13 − 10 i )
Similarly,
f '' (η) =2πi(10) = 20πi
Therefore,
f ' ' (1 − i ) = 20πi
2.17 Morera's Theorem (Converse of Cauchy's theorem)
If f (z) is continuous function in a simply connected domain D and if ∫ f ( z )dz = 0
C
where C is simply closed curve in D, then f (z) is analytic in D.
Problem Set (2.4)
✜
Integrate the following functions in the domain C.
eiz
1.
∫ z3
2.
∫
C
3.
∫
C
4.
∫
C
5.
∫
C
6.
∫
C
dz, where C is the circle z = 2 .
z3 + 3
( z − 1)z 2
sin z
( z − πi ) 2
dz, where C : z = 0.5.
dz, where C : z = 4.
ez
3
dz, where C : z − 1 = .
4
( z − 1) ( z + 4)
2
2
e − zi sin z
(z − π )2
3e z
(z + i )3
dz, where C is the circle z = 3. 5.
dz, where C is z + i = 2.
70
7.
Engineering Mathematics-III
∫
sin 2z
( z + 3) ( z + 1) 2
dz, where C is the rectangle with vertices at 3 + i, − 2 + i,
− 2 − i, 3 −i .
8.
∫
C
dz
2
( z + 4) 2
9. Show that ∫
, where C : z − i = 2 .
C
dz
2
z ( z + 1) 3
dz = (11e −1 − 4) πi where C : z = 2 .
Objective Type Questions
Multiple Choice Problems
Tick the correct answer :
1.
If L represents a square bounded by x = ± a, y = ± a, then the value of ∫
L
2.
(a) 2πi
(b) πi
(c) 0
(d) 4πi.
∫
dz
is :
z
z dz from z = 0 to z = 4 + 2 i along the curve L defined by z = t 2 + it , is :
L
8
i
3
8
(c) 8 + i
3
(a) 10 −
3.
(b) 10 +
8
i
3
(d) none of these.
∫ dz, where L is any rectifiable arc joining the points z = a and z = b is equal to :
L
(a) z
(b) z − a − b
(c) a − b − z
(d) b − a.
b
4.
The path of the definite integral ∫ f ( z ) dz is :
a
(a) the line segment joining the points z = a and z = b
(b) any curve joining the points z = a and z = b
(c) any circle such that the points z = a and z = b lie on it
(d) any rectangle whose two vertices are the points z = a and z = b.
5.
If f (z) is analytic in a simply connected domain D and C is any closed
continuous rectifiable curve in D, then ∫ f ( z ) dz is equal to :
C
(a) 0
(b) 1
(c) C
(d) D.
Complex Integration
6.
71
If f (z) is analytic in a simply connected domain D enclosed by a rectifiable
Jordan curve C and f (z) is continuous on C, then for any points z 0 in D, we
have f (z 0 ) =
(a)
f (z )
1
dz
2π ∫ z − z 0
(b)
C
f (z )
1
(c)
dz
∫
2π z − z 0
8.
Value of ∫
z =3
f (z )
dz.
z − z0
ez
dz :
( z − 2)
(a) 0
(b) 2 π ie 3
(c) 2 π ie 2
(d) 2 π i.
If L is circle z + 3 i = 1, then the value of ∫
L
9.
∫ z − z 0 dz
(d) 2πi ∫
C
7.
f (z )
1
2πi
dz
is :
z ( z + πi )
(a) 0
(b) π i
(c) 2π i
(d) none of these.
The value of the integral
n!
2πi
∫ z =2
(a) 1
(c) 2πi
e 2 dz
z n +1
is :
(b) πi
(d) 0
10. If L is a closed curve and z = a is outside L,then ∫
L
(a) 0
(b) π i
(c) 2 π i
11. Value of
dz
is :
z−a
(d) ∞.
1
2π i
e
z
∫ z = 4 (z + 2)2
dz is :
(a) e 3
(b) e 2
(c) 0
(d) none of these.
12. Value of
2!
2πi
∫ z =3
z 2 + 3z + 4
( z − 1) 3
(a) 2
dz is :
(b) 0
(c) π i
13. When c : z − 5 = 1 then ∫
c
(d) none of these
1
dz = ...........
z ( z − 2)
(a) 2π
(b) 0
(c) –2 π i
(d) none of these
72
Engineering Mathematics-III
14. Which of the following is true :
(a)
∫
f ( z ) dz = ∫
f ( z ) dz
∫
f ( z ) dz < ∫
f (z )
L
(c)
L
(b)
L
∫
f ( z ) dz ≤
∫
f (z )
dz
∫
f ( z ) dz > ∫
f (z )
dz .
L
(d)
dz
L
L
L
L
15. If f (z) is analytic in a domain D, then
(a) f ( n ) ( z ) exist in D
(b) f ( n ) ( z ) does not exist in D
(c) f ( n ) ( z ) = 0 for all n in D
(d) none of these.
16. If L is circle z − a = r, then ∫
L
dz
is :
z−a
(a) 2 πi
(b) π i
(c) 0
(d) − 2 π i.
17. If L is a circle z = 1, then ∫ z dz is :
L
(a) π i
(b) 2 π i
(c) 0
(d) none of these.
Fill in the Blanks
1.
Let f (z) be single-valued analytic function in a simply connected domain D if
b
a, b ∈ D, then ∫ f ( z ) dz is ................... on path.
a
2.
3.
4.
Complex line integrals can be represented in terms of two .................. line
integrals.
Simply connectedness is ........................ for Cauchy's integral theorem.
If C is a semi-circnlar arc z = 1 from – 1 to 1 above the real axis then ∫ z dz =
C
........................ .
5.
If C is a circle z = r > 0, then ∫
C
dz
z2
is ........................ 0 .
State True/False
1.
Let f (z) be analytic in a simply connected domain D. Then the integral
∫ f (z ) dz depends upon path of integration.
C
b
2.
∫e
a
z
dz = e( b − a )i
Complex Integration
73
ANSWERS
Problem Set 2.1
2
1.
∫
( 2 − iy 3 ) idy = 4 + 4 i
0
2. (i) 6 +
26
i
3
(ii) 6 +
26
26
i (iii) 6 +
i
3
3
3. The shortest path is given by the line y =
x +1
, (i) 4 + 2 i (ii) 4 + 3i
2
(iii) Integral values along the path (i) is not equal to the integral value along
the path (ii) since the integrand f (z)=Re (z) = x is not analytic.
4. Represent
z = r in parametric form as z (t) = r (cos t + i sin t), 0 ≤ t ≤ 2π.
Then z' (t ) = r ( − sin t + i cos t ) and f [ z(t )] = [ z(t )]n = r n (cos nt +sin nt).
2π
Thus
I =
∫
f ( z )dz =
C
∫ f [z(t )] z
′
(t ) dt .
0
 2πi, n = −1
I =
 0 , n ≠ −1.
5. Along AB : y = 0, z = x, f (z) =
z
= 1, 1≤ x ≤ 2, dz = dx along C1 : z =2eiθ ,
z
f (z) = e 2iθ , 0 ≤ θ ≤ π, dz = 2 ieiθ dθ. Find similarly for CD and C2. Ans 4/3.
6. 2i sinh π.
7.
1 2
( e + 1).
4
8. −π 2
10. −1 + i
11. 4πi
Problem Set 2.2
Problems 1- 8 answer is 0 since all the integrand is analytic within C.
9. (i) 2πi
(ii) 2πi
(iii) 2πi
74
Engineering Mathematics-III
Problem Set 2.3
1. e2
2. πi /32
3. −2/ 3 πie
4. 0
5. (i)
7 πi
, (ii) 4πi
2
5
7. (i) π (i + ) (ii) π
3
9.
2πi( πi − 5)
π 2 + 25
6. (i) 0 (ii) 0
i − 5


 3
8. e − π + 2i
 π 

πi  2+  4  
e

10. −
25 



( 2 cos πi − i sin πi )
Problem Set 2.4
1. −πi
2. −6πi
3. 2πi. cos πi
4. 6eπ / 25
5. 2πi. e −2 πi
6. πi( 2 − i )e −i
7.
πi( 4 cos 2 + sin 2)
2
8.
π
16
Multiple Choice Problems :
1.
(a)
2. (a)
3.
(d)
4. (b)
5. (a)
6.
(b)
7. (c)
8.
(d)
9. (a)
10. (a)
11.
(b)
12. (a)
13.
(b)
14. (b)
15. (a)
16.
(a)
17. (b)
Fill in the Blanks :
1.
Independent
2. real
3.
essential
4. − π i
5.
0
True/False :
1.
False
2. False
❑❑❑
Unit-1
Chapter
Power Series and
Calculus of Residues
3
3.1 Power Series
Let w1 , w 2 ,...., w m ,...... be a sequence of numbers, complex or real. Then the
infinite series or series is given by
∞
∑ w m = w1 + w 2 + w 3 +.....
...(3.1)
m =1
The w m S are called the terms of the series. The sum of first n terms is given by
...(3.2)
S n = w1 + w 2 +.....+ w n
The expression (3.2) is called the nth partial sum of the series (3.1). Here, the terms
omitted from S n is called remainder term, denoted by R n .
A sequence {S n } is said to be convergent, if
Lt S n = finite quantity.
n→ ∞
A sequence [Sn] is said to be divergent if Lt S n is not finite.
n→ ∞
Further let ( a1 + ib2 ) + ( a2 + ib2 )+....+( an + ibn )+.....
...(3.3)
be an infinite series of complex terms, where a's and b's being real numbers. If the
series Σan and Σbn converges to the sums say A and B respectively, then the series
(3.3) is said to be convergent and it converges to A + iB.
3.1.1 Representation of a function by power series
A power series in the powers of ( z − z 0 ) is a series of the form
∞
∑ an (z − a)n = a0 + a1 (z – a) + a2 (z – a)2 +.......
...(3.1)
∑ an z n = a0 + a1 z + a2 z 2 +.......
...(3.2)
n=0
∞
or
n=0
where an s are complex or real constants and z is a complex variable and a is called
the centre of the series in (3.1).
76
The form (3.1) can be represented to form (3.2) by substituting z = ξ + a so that
∑ a n ( z – a) n = ∑ a n ξ n .
3.1.2 Some tests for convergence of series
Here we give some tests for examining convergence of series for reference purpose.
(i) If Σ u n is convergent, then Lim u n = 0.
n→ ∞
(ii) Comparison Test : Σ u n is absolutely convergent if u n  ≤ v n  and Σ v n is
convergent.
(iii) Cauchy Root test : Let Lim  u n 1/ n = l.
n→ ∞
The series Σ u n is convergent (absolutely) or divergent according as
l<1
or
l>1
This test fails if l = 1.
1
(iv) The series Σ
is convergent if p > 1 and divergent if p ≤ 1 .
np
u
(v) The series Σu n is convergent or divergent according as Lim n +1 < 1 or > 1.
n→ ∞ u n
3.1.3 Region of convergence of power series
Region of convergence of a power series is the set of all points z for which the series
converges.
Consider the power series Σan z n = Σu n ( z ), say Σu n is convergent, if
Lim  u n 1/ n < 1 .
n→ ∞
[ By root test]
Lim  an z n 1/ n < 1 or an 1/ nz< 1
n→ ∞
Taking
Lim  an 1/ n =
n→ ∞
1
, we have
R
z < 1 or z < R.
 
R
Therefore, corresponding to every power series Σan z n , there exist a non-negative
real number R such thatz< R if the series is convergent andz> R if the series is
divergent.
Now if we draw a circle of radius R with centre at origin (or at a), then
(1) The power series Σan z n [ or Σ an ( z – a) n in case of centre a ] is convergent for
every z within the circle.
Power Series and Calculus of Residues
77
Such a circle is called circle of convergence and its radius R is called radius of
convergence of the power series.
(2) The power series Σan z n [ or Σ an ( z – a) n ] is divergent for every z outside the
circle.
There arise following three cases for R:
(i) R = 0
In this case series is convergent at the point z = 0(or z = a).
(ii) R is finite
In this case series is convergent at every point within the circle (or circular disk
 z – a < R) of radius R and diverges everywhere outside the circle (or circular disk
 z – a > R)
(iii) R is infinite
The series converges for all z.
Note: From Cauchy's theorem, we have
a
1
= Lim  an 1/ n = Lim  n +1 .
n→ ∞
R n→ ∞
an
Ex. 1 : Find the radius of convergence of the following power series:
∞
(i)
∑
n=0
∞
(ii)
 z
 
 n
n
∑ (–1)n (z – 3i )n / n
n=0
Sol : (i) Here an =
1
nn
Radius of convergence is given by
1
 1 
= Lim  an 1/ n = Lim 

n→ ∞  n n 
R n→ ∞
Therefore
1/ n
R=∞
(ii) Comparing the given series with Σan ( z – a) n , we have
an =
(–1) n
(–1) n +1
, a n +1 =
n
n +1
Radius of convergence is given by
a
1
n
= Lim  n +1  = Lim –

n→ ∞
R n→ ∞ a n
n +1
= Lim
n→ ∞
1
=0
n
78
= Lim
n→ ∞
Therefore
n
1
= Lim
= 1.
1
n + 1 n→ ∞ 
1 + 

n
R=1
Problem Set (3.1)
✜
Find the radius of convergence of the following power series :
∞
1.
∑
n=0
∞
2.
n!
n
n
zn
zn
∑
n=0 n !
∞
3.
∑ (3 + 4 i )n z n
n=0
4.
∑
1
n
p
zn
5. Find the radius of convergence of the series
z 1. 3 2 1. 3. 5 3
+
z +
z +.....
2 2. 5
2. 5. 8
∞
6. Find the domain of convergence of the series
n
iz – 1
∑  2 + i  .
n =1
Hint. Find Lim 
n→ ∞
u n +1

un
3.2 Taylor's Theorem
If a function f (z) is analytic at all points within and on a circle C, with its center at
the point a and radius R, then at every point z within C, f (z) can uniquely be
represented by a power series known as Taylor's series which is given by
∞
f (z ) =
∑
an ( z – a) n , where an =
n=0
f n ( a)
n!
Proof. Let the function f (z) be analytic at all points within and
C
on the circle C with centre at a and radius = R. Let z be any
point within C. Draw a circle C1 with centre at a and radius r,
smaller than that of C such that z is an interior point of C1 . Let t
denote any point on the circle t – a= ρ where r < R.
1
1
1
=
=
t – z t – a + a – z t – a – ( z – a)
.
.
a
r P
z
t
Fig 3.1
C1
R
Power Series and Calculus of Residues
=
79
1
1
1 
z – a
=
1 –

z – a t – a 
t–a
t – a
1 –


t – a
–1
2
n
1
1 
z – a  z – a
 z – a +.....
=
1
+
+
+
.....
+






 t – a
t – z t – a 
t – a  t – a

Now
[By Binomial expansion]
=
1
z–a
( z – a) 2
( z – a) n
+
+
+....+
+ ...(3.1)
t – a (t – a) 2 (t – a) 3
(t – a) n +1
z – a
As z – a< t – a therefore
< 1.
t – a 
Hence the series converges uniformly, i.e., the series is integrable.
Multiplying (3.1) by f (t), we have
f (t )
f (t )
f (t )
f (t )
f (t )
=
+ ( z – a)
+ ( z – a) 2
+. ....+( z – a) n
+.....
2
3
t–z t–a
(t – a)
(t – a)
(t – a) n +1
Then integrating w.r.t. ''t'', round the contour C1 , we get
f (t )
f (t )
f (t )
∫ t – z dt = ∫ t – a dt + (z – a) ∫ (t – z )2 dt
C1
+( z – a) 2
C1
C1
f (t )
f (t )
n
∫ (t – z )3 dt +....+(z – a) ∫ (t – a)n +1 dt +. ....
C1
... (3.2)
C1
But by Cauchy's integral formula
f (t )
f (t )
f (t )
∫ t – z dt = 2πi f (z ), ∫ t – a dt = 2πi f (a)and ∫ (t – a)2 dt = 2π if ' (a), .....
C1
C1
C1
Substituting these values in (3.2), we get
f ( z ) = f ( a) + f ' ( a)( z – a) +
∞
=
f ' ' ( a)
f n ( a)( z – a) n
( z – a) 2 +.....+
+.....
2!
n!
∑ an (z – a)n , where an =
n=0
f n ( a)
n!
80
Particular case. When a = 0, then the Taylor's series reduces to
∞
∞ n

z
f ( z ) = ∑ an . z n = a0 + ∑
f n ( 0)
∵ an

n=0
n =1 n !
=
f n ( a)

n! 
which is known as Maclaurin's Series.
Remark 1. Since the function f (z) is analytic at all points inside the circle C, hence
the convergence of Taylor's series to f (z) is assumed and no test for the convergence
of the series is required.
Remark 2. For the truth of Taylor's theorem, it is not necessary that f (z) be
analytic on the boundary of the circle C, what is required is that f (z) should be
analytic in the region inside the boundary.
Remark 3. Function f (z) must be non-analytic at the point z = a to obtain Taylor series.
Remark 4. If the function f ( z ) is analytic throughout a domain D except at the points
z1 , z 2 ,...... z n then the radius of convergence of Taylor's series about the point z = a
is the distance between a and the nearest point at which f ( z ) is not analytic.
Remark 5. It is advisable to obtain Taylor series expansion by rearrangemants,
manipulation and using the standard series (like binomial expansion, etc.) rather
then using the Taylor series formula.
Remark 6. Complex analytic function can always be represented by power series.
3.3 Some Useful Results
We can easily find the expansion of the following complex function by Maclaurin's
series expansion.
(i)
ez = 1 + z +
(ii)
sin z = z –
z2 z3
+
+.....+ =
2! 3!
z 3 z 5 z7
+
–
+..... =
3! 5! 7 !
z2 z4
(iii) cos z = 1 –
+
–... =
2! 4!
∞
∑ (–1)n
n=0
∞
∑ (–1)n
n=0
∞
z3 z5
(iv) sinh z = z +
+
+..... =
3! 5!
(v)
zn
forz< ∞
n=0 n !
∞
∑
z 2n
forz< ∞
( 2n )!
z 2n + 1
∑ (2n + 1)! forz< ∞
n=0
∞
z2 z4
cosh z = 1 +
+
+....... =
2! 4!
z 2n
∑ (2n )! forz< ∞
n=0
(vi) (1 – z ) –1 = 1 + z + z 2 + z 3 +...... =
z2 z3
(vii) In (1 + z ) = z –
+
–...... =
2
3
(viii) (1 – z ) – m
z 2n + 1
forz< ∞
( 2n + 1)!
∞
∑ z n forz< 1
n=0
∞
∑ (–1)n –1
n =1
zn
forz< 1
n
m( m + 1) 2
= 1 + mz +
z +......... =
2!
∞
( m + n – 1)! n
z forz< 1
( m – 1)! n !
n=0
∑
Power Series and Calculus of Residues
(ix) (1 + z ) m = 1 + mz +
81
m( m – 1) 2
z +...... =
2!
∞
( m + n – 1)! n
z forz< 1.
n = 0 ( m – 1)! n !
∑
n +1

n n 
2
n n
n +1 z
(–
1
)
z
=
1
–
z
+
z
–....
+
(–
1
)
z
+
(–
1
)
....

∑
z +1


n=0
2
n
n +1


1
1 1  1
z z
z
z
1
(xi)
=– 
– 1 + +
+.....+
+
.

n
n
z–a
a 1 – z / a a 
a a2
z
+ a
a
a
1
1
z z2
(–1) n z n (–1) n +1 z n +1
1 
(xii)
= 1 – +
+.....+
+
.

2
n
n
z +a a
a a
z + a
a
a
(x)
1
=
1+ z
∞
Ex. 2 : Expand cos z about the point z = ( π / 2).
Sol : Here f ( z ) = cos z, which is analytic function for all values of z.
∴ By Taylor's theorem the expansion of f ( z ) about the point z = π/2 is given by
∞
∞
f n ( π / 2)
f ( z ) = ∑ an ( z – π / 2) n = ∑
( z – π / 2) n
n!
n=0
n=0
f n ( π / 2)
( z – π / 2) n
n!
n =1
∞
= f ( π / 2) + ∑
Since
∴
...(3.1)
nπ 
f( π / 2) = cos π/2 = 0 and f n ( z ) = cos  z +


2
π nπ 
 nπ 
f n ( π / 2) = cos  +
 = − sin   which is zero if n is even.
2
 2
2
Hence substituting in (3.1), we have
∞
( z – π / 2) n
f ( z ) = cos z = 0 – ∑
sin( nπ / 2), when n is odd integer
n!
n =1
= –( z – π / 2) +
Aliter. Let
Then
u = z – π / 2 or z = u + π / 2


u3 u5
cos z = cos(u + π / 2) = – sin u = – u –
+
.....
3! 5!


= –( z – π / 2) +
Ex. 3 : Expand
Sol : Here,
( z – π / 2) 3 ( z – π / 2) 5
–
+.......
3!
5!
(z – π / 2)3 (z – π / 2)5
–
+.....
3!
5!
1
in a Taylor's series round the origin.
( z – 1)( z – 2)
1
f (z ) =
( z – 1)( z – 2)
The function f ( z ) is not analytic at z = 1 and z = 2 .
...(3.1)
82
∴ f ( z ) is analytic within a circle C of radius r < 1 and centre at the origin, hence f ( z )
can be expanded in a Taylor's series round the origin in C.
∞
f (z ) =
∞
∑ an (z – 0)n = f (0) + ∑ z n .
n=0
n =1
∞
= f ( 0) +
where f( 0) =
Now, we have
f (z ) =
f n ( 0)
n!
n
z . n
f ( 0)
n =1 n !
∑
...(3.2)
1
2
1
1
1
1
z
=–
+
= (1 – z )–1 – 1 – 
( z – 1)( z – 2)
z–1 z–2
2
2
–1
∴
n
– n –1 

1
1
z
f n ( z ) = (–1) n n ! (–1) n (1– z ) – n –1 – .  –  × 1 – 


2  2
2


or
f n (z )
1
= (1 – z ) –( n +1) –  

n!
2
n +1
1 –


z

2
–( n +1)
f n ( 0)
1
= 1–
n +1
n!
2
1
Also from (3.1), f( 0) =
2
∴
Substituting in (3.2), we have
f (z ) =
1 ∞ 
1  n ∞ 
1  n
+ ∑ 1 –
 z = ∑ 1 – n +1  z .
n
+
1



2 n =1 
2
2
n=0
It can alternatively be obtained by using binomial theorem after partial fraction
without applying Taylor series which is a better option.
Ex. 4 : Find the four terms of the Taylor series expansion of the complex variable
z +1
function f ( z ) =
about z = 2 .
( z – 3)( z – 4)
Sol : Let f ( z ) =
z +1
.
( z – 3)( z – 4)
Y
Consider the centre of the circle at
z = 2, then the distances of the
singularities z = 3 and z = 4 from the
z–2 =1
centre are 1 and 2 respectively. Thus if
.
a circle is drawn with centre z = 2 and
1
O
1
.
3
2
2
radius 1, then within the circle
z – 2= 1; the
function
f ( z ) is
analytic, hence it can be expanded in a
(a)
(b)
Fig. 3.2
4
X
Power Series and Calculus of Residues
83
Taylor's series. This is, therefore, the circle of convergence.
Now
f (z ) =
–4
z +1
5
=
+
( z – 3)( z – 4) z – 3 z – 4
[By Partial fraction method]
–1
–4
5
5  z – 2
=
+
= 4[1–( z – 2)]–1 – 1–
( z – 2) – 1 ( z – 2) – 2
2 
2 
= 4[1 + ( z – 2) + ( z – 2) 2 + ( z – 2) 3 +....]

5
z – 2 ( z – 2) 2 ( z – 2) 3
[By binomial exp.]
– 1 +
+
+
+.....
2
2
4
8

5 
5
5
5


=  4 –  +  4 –  ( z – 2)+  4 –  ( z – 2) 2 +  4 –  ( z – 2) 3 +....



2 
4
8
16
3 11
27
59
2
3
= +
( z – 2) +
( z – 2) +
( z – 2) +.... .
2 4
8
16
Aliter. We have
f (z ) =
z +1
2+1
3
, f ( 2) =
= .
( z – 3)( z – 4)
( 2 – 3)( 2 – 4) 2
To make the differentiation easier let us convert the given fraction into partial
fractions i.e.,
–4
5
+
z–3 z–4
4
5
4
5
11
f ' (z ) =
–
so f ' ( 2) =
–
=
2
2
2
2
4
( z – 3)
( z – 4)
( 2 – 3)
( 2 – 4)
–8
–8
10
10
27
f ' ' (z ) =
+
so f ' ' ( 2) =
+
=
3
3
3
3
4
( z – 3)
( z – 4)
( 2– 3)
( 2– 3)
24
30
24
30
177
f ' ' ' (z ) =
–
so f ' ' ' ( 2) =
–
=
4
4
4
4
8
( z – 3)
( z – 4)
( 2 – 3)
( 2 – 4)
f (z ) =
⇒
Taylor series is
( z – a) 2
( z – a) 3
f ' ' ( a) +
f ' ' ' ( a) +.....
2!
3!
z +1
3
11 ( z – 2) 2  27  ( z – 2) 3 177
⇒
= + ( z – 2)
+
+......
 +
 4
( z – 3)( z – 4) 2
4
2
3!
8
3
11
27
59
= + ( z – 2)
+ ( z – 2) 2 .
+ ( z – 2) 3
+.......
2
4
8
16
Note. Readers are advised to follow binomials expansion method rather then
applying Taylor's (or Laurent's series to be discussed in next section) directly to
obtain any of these expansions.
f ( z ) = f ( a) + ( z – a) f ' ( a) +
Ex. 5 : Find the Taylor series expansion of a function of the complex variable
1
about the point z = 4
f (z ) =
( z – 1)( z – 3)
84
Sol : Let f ( z ) =
1
( z − 1)( z − 3)
Let the centre of the circle be at z = 4, then the
distances of the singularities z = 1 and z = 3 from the
centre are 3 and 1. Hence, if a circle is drawn with
center at z = 4 and radius 1; then within the circle
 z – 4= 1, the function f ( z ) is analytic. Hence, it can
be expanded in Taylor's series within the circle
 z – 4= 1. This, is therefore, the circle of convergence.
Now
1
1
1
1 
f (z ) =
= 
–

( z – 1)( z – 3) 2  z – 3 z – 1
=
O
1
2
.
3
X
Fig. 3.3
1
1
1

–

2  z – 4 + 1 ( z – 4) + 3
–1
1
1
z – 4
[1 + ( z – 4)]–1 – 1 +

2
6
3 
1
= [1 – ( z – 4) + ( z – 4) 2 – ( z – 4) 3 +.....]
2

1
z – 4 ( z – 4) 2 ( z – 4) 3
– 1 –
+
–
+....
6
3
9
27

=
1 1
1 1
1 1
=  –  +  – +  ( z – 4) +  –  ( z – 4) 2
 2 6  2 18
 2 54
1
1 
3
+  – +
 ( z – 4) +.....
 2 162
1 4
13
40
f ( z ) = – ( z – 4) +
( z – 4) 2 –
( z – 4) 3 +.....
3 9
27
81
Ex. 6 : Find the first three terms of the Taylor series expansion of f ( z ) =
1
2
z +4
about z = – i. Find the region of convergence.
1
Sol : f ( z ) =
2
z +4
[UPTU 2005]
Equating denominator to zero z 2 + 4 = 0
⇒
z 2 = – 4 or z = ±2 i.
y
If the centre of the circle is z = –i, then the singularities are
z = 2 i and z = –2 i. Hence if a circle of radius 1 is drawn
with centre at z = –i, then within the circlez + i= 1, the
function f ( z ) is analytic and hence the function can be
expanded by Taylor's series within the circle z + i= 1.
This is, therefore, the circle of convergence.
Now
.
.
.
.
.
2i
i
x'
O
–i
–2 i
y'
Fig. 3.4
x
Power Series and Calculus of Residues
1
1  1
1 
=
–


z + 4 ( z + 2 i )( z – 2 i ) 4 i  z – 2 i z + 2 i 




1 
1
1
1
1
1
1
1

=
–
= –
–



z + i i 
z + i 
4 i ( z + i ) – 3i ( z + 1) + i  4i  3i 
1
−
1
+







3i 
i  
f (z ) =
1
85
2
=



1 1
1
1
= 
+
i
4 3
1 – i( z + i )
 1 + (z + i )



3
–1
=
1 1 
i

 1 + ( z + 1
4  3 
3


+ {1 – i( z + i )} –1 

=
2
3
1 1 
i
 
i
2 i
3
 1 – ( z + i ) +   ( z + i ) –   ( z + i ) +.....  +  1 + i( z + i )




4 3 
3
3
3
 


+(i ) 2 ( z + i ) 2 + (i ) 3 ( z + i ) 3 +.....  

=
1  1 i
1
i

– (z + i ) –
(z + i )2 +
i( z + i ) 3 +.....  {1 + i( z + i )


4  3 9
27
81


– ( z + i ) 2 – i( z + i ) 3 +.....} 

1  4 8i
28
80

=
+ (z + i ) –
(z + i )2 −
i( z + i ) 3 +....

4  3 9
27
81
1 2i
7
20
= + (z + i ) −
(z + i )2 −
i( z + i ) 3 +......
3 9
27
81
Then region of convergencez + i< 1.
Ex.7 : Find the expansion of e z sin z about z = 0.
Sol : We have
f ( z ) = e z sin z
= (1 + z +
z2 z3
z3 z5
+
+.... )( z –
+
–.... )
2i
3i
3i
5i
[by using expansion of e z and sin z]
= z + z2+
3
5
z
z
–
+.....
3
30
86
Ex. 8 : If the function f ( z ) is analytic and one valued in  z – a< R , prove that for
0< r < R
f ' ( a) =
1
πr
2π
∫
P(θ) e – iθ dθ
0
where P(θ) is the real part of f ( a + reiθ ).
Sol : Since f ( z ) is analytic in z – a< R, therefore it must be analytic in
z – a = r ( r < R ).
Hence f ( z ) can be expanded as Taylor's series about z = a in the circlez – a= r,
i.e.,
∞
f (z ) =
∞
∑ am (z – a)m = ∑ am (reiθ )m
m=0
∞
or
∴
f (z ) =
———
f (z ) =
[∵
z – a = reiθ ]
m=0
∑ am r m eimθ
m=0
∞ ———
am r m
m=0
∑
e – imθ
...(3.1)
By Cauchy's formula for derivative of an analytic function, we have
f ' ( a) =
1
2πi
f ( z )dz
∫
...(3.2)
2
C ( z – a)
∞ ———
∑ am r m e – imθ
———
Therefore
f (z )
1
1
dz =
2πi ∫ ( z – a) 2
2πi
C
=
=
1
2πi
2π ∞
∫ ∑
∫
0 m=0
2π
1 ∞ — m –1
∑ am r
2π m = 0
∫
( z – a) 2
C
am r m e – imθ
( reiθ ) 2
m=0
dz
rieiθ dθ
e – ( m +1) iθ dθ
0
= 0 for all values of m.
or
0=
1
2πi
———
f (z )
∫
2
C ( z – a)
...(3.3)
dz
Adding (3.2) and (3.3), we have
f ' ( a) =
1
2πi
———
∫
C
f (z ) + f (z )
( z – a)
2
dz =
1
2πi
∫
C
2 . real part of f ( z )
( z – a) 2
dz
Power Series and Calculus of Residues
2π
1
πi
=
real part of f ( a + reiθ )
∫
iθ 2
( re )
0
2π
1
πr
=
∫
87
rieiθ dθ
[since z – a = reiθ on C ]
P(θ). e – iθ dθ, where P(θ) = real part of f ( a + reiθ ).
0
Ex. 9 : The function f ( z ) is analytic whenz< R and has the Taylor's expansion
∞
2π
1
∑ an z n . Show that if r < R, 2π ∫
0
iθ 2
 f ( re ) dθ =
0
∞
∑
0
2 2n
an  r . Hence prove that,
∞
if f ( z )≤ M whenz< R . ∑ an 2 r 2n < M 2 .
0
Sol : Since f ( z ) is analytic within the circle z = r ( r < R ), hence f ( z ) can be
expanded in a Taylor's series within the circlez= r.
f (z ) =
so that
———
∞
∞
n=0
∞
n=0
∑ an z n = ∑ an r n einθ where z = reiθ
∑ am r m e – imθ
f (z ) =
m=0
since
∴
 ∞

———
2
n inθ
 f ( z ) = f ( z ) f ( z ) =  ∑ an r e 
n=0

2π
2π
∫
∫
2
 f ( z ) dθ =
0
0
 ∞

 ∑ am r m e – imθ 


m =0

 ∞
  ∞


n inθ
m – imθ  
dθ
  ∑ an r e  ×  ∑ am r e

  n = 0

 m =0

...(3.1)
The two series on the right are absolutely convergent, therefore, their product is
uniformly convergent for 0 ≤ θ < 2π, so term by term integration is justified.
2π
It is easy to see that
∫
eikθ dθ = 0 for k ≠ 0.
0
Hence, all the integrals on the right hand side of the above vanish except for which
m = n.
Therefore from (3.1), we have
2π
∫
2
 f ( z ) dθ =
n=0
0
or
1
2π
2π
∫
2π
∞
∑ a n a n r 2n ∫
iθ 2
 f ( re ) dθ =
0
This proves the first results.
0
∞
∑an2 r 2n
n=0
∞
dθ =
∑  an2 r 2n 2π
n=0
88
∞
Now
2 2n
∑an
r
n=0
1
=
2π
=
∞
or
∑ |an|2 r 2n
2π
∫
0
2π
M2
2π
2π
1
 f ( z ) dθ ≤
2π
2
M 2 dθ [ since| f ( z )|≤ M ]
0
M2
2π = M 2
2π
dθ =
∫
∫
0
≤ M2
n=0
2
 an 
1
Ex.10 : If C is a closed contour around origin, prove that   =
n
!
2π
i
 
∞
Hence deduce
∑
n=0
2
 an 
1
  =
2π
 n !
2π
∫e
2a cosθ
a n e az
∫ n !z n +1 dz.
c
dθ.
0
Sol : Let f ( z ) = e az , ∴ f n ( z ) = a n e az
so that a n = f n ( 0)
Putting z = 0 in the above relation
f ( z )dz
1
f n (z ) =
n! ∫
n +1
2πi
C z
an
1
=
n ! 2πi
or
Multiplying by
z n +1
C
an
on both sides, we have
n!
 an 
 
 n !
∞
Hence
e az dz
∫
∑
n=0
2
=
2
 an 
  =
 n !
=
=
1 1
a n e az
.
dz
∫
n +1
2πi n !
C z
∞
1 1
∑ 2πi . n !
n=0
∫
C
a n e az
z n +1
a n e az
∞
dz
∞
1
2πi
 
az
∫ ∑ n !z n +1 dz = 2πi ∫ e ∑  z 
1
1
2πi
∫e
C0
C

C
az
e a/ z
dz
1
=
z
2πi
C
n
0
e a( z +1/ z )
∫
a
1  dz

n ! z
dz
z
[Taking the circle C asz= 1, ∴ z = eiθ on]
=
iθ
1
1
2a cosθ ie dθ
e
=
∫
i
θ
2πi
2π
e
C
2π
∫
0
e 2a cosθ d θ.
Power Series and Calculus of Residues
89
Problem Set (3.2)
1.
Expand the following functions in Taylor's series and determine regions of
convergence of the corresponding Taylor's series.
(a) f ( z ) = log(1 + z ), about z = 0
(b) f ( z ) = sin z, about z = ( π / 4)
1
(c)
, about z = i
(3 – z )2
z–1
2. Obtain the expression
in a Taylor's series in powers of z − 1.
z2
3. Find the Taylor series expansion of the following function about the point
z = a. Find also the region of convergence.
π
cos z, a = –
e–z , a = 0
(i)
(ii)
2
1
, a = 0,z< 1
(iii) Lnz, a = 1
(iv)
1+ z4
z+2
2
, a = 0,z< 1
(v)
(vi)
e z – z , a = 0,z< ∞
1– z2
(vii)
cos 2z
2
(viii)
,a = 0
1 – 4z
(ix)
Ln( 2 + iz ), a = i
4. Find the Taylor's series expansion of
5.
4 – 3z
(1 – z ) 2
, a = 0,z< 1
1
2
z + (1 + 2 i ) z + 2 i
about z = 0. Also find
the radius of convergence.
Obtain the first three terms in the Taylor series expansion for the following
function at a = 0.
(i) e z cos z
(ii) e z /(1 + e z )
(iii) e z cosh z
3.4 Laurent's Theorem
If a function f ( z ) is analytic in the annulus (ring-shaped region) between two
concentric circles C1 and C 2 with centre at the point z = a and radii R1 and
R 2 ( R1 > R 2 ) , respectively, then at any point z of the annulus,
[UPTU 2002]
∞
f (z ) =
∑
n=0
where
1
an =
2πi
∞
a n ( z − a) n + ∑ b n ( z − a) − n
n =1
∫
C1
f (t )dt
(t − a)
n +1
and bn =
1
2πi
∫
C2
f (t )dt
(t − a) − n +1
.
Proof. Let z be any point in the region bounded by two concentric circles C1 and
C 2 with centre at the point a and radii R1 and R 2 ( R1 > R 2 ) respectively. If z – a= r
90
where R 2 < r < R1 , then by Cauchy's integral formula for multi-connected region, we
have
f (t )dt
f (t )dt
1
1
...(3.1)
f (z ) =
–
∫
∫
2πi
t–z
2πi
t–z
C1
C2
where integrals on C1 and C 2 both are taken in counter-clockwise direction.
In the first intergral t denotes a point on C1 and in the second integral it denotes a
point on C 2 .
For the first integral,
f (t )
can be expanded by Taylor's series.
t–z
z – a
Since 
≤ 1 becouse t lies on C1 .
t – a
f (t )dt
1
1
Thus
=
∫
2πi
t–z
2πi
C1
+
where,
an
( z – a) 2
2πi
∫
C1
∫
C1
f (t )
( z – a)
dt +
t–a
2πi
f (t )
(t – a) 3
= a0 + a1 ( z – a) + a2 ( z – a) 2 +....
f (t )
1
=
dt
∫
2πi (t − a) n +1
In the second integral, t lies on C 2 .
⇒
C1 (t
– a) 2
dt
dt +.....
C1
Therefore
f (t )
∫
...(3.2)
.
R1
z r
.
a B
R2
t – a < 1
t – a<z – a or
z – a
1
1
1
=
=
t – z (t – a) + a – z (t – a) – ( z – a)
C2
Fig. 3.5
–1
1
1
1 
t – a
=–
1 –

t – a
z–a
z – a
z – a
1 –

z – a

2
n +1

1 
t – a  t – a
 t – a
=–
+
+.....
 +.....+ 

1 +
 z – a
z – a 
z – a  z – a

=–
f (t )
, we get
2πi
1 f (t )
1 f (t )
1 (t – a)
1 (t – a) 2
–
=
+
f (t ) +
f (t )+.....
2πi t – z 2πi z – a 2πi ( z – a) 2
2πi ( z – a) 3
f (t )
1  1
1
1
= 
f (t ) +

2 2πi
 z – a 2πi
( z – a)
(t – a) –1
f (t )
1
1
+
+.....
3 2πi
( z – a)
(t – a) –2
Multiplying both sides by –
A
C1
Power Series and Calculus of Residues
⇒
–
1
2πi
∫
C2
91
f (t )
1  1
dt = 

 z – a 2πi
t–z
1
1
+
2 2πi
( z – a)
f (t )dt
∫
C2
f (t )dt
∫
C2 (t – a)
–1
+
1
( z – a)
3
.
1
2πi
b1
b2
b3
+
+
+.....
z – a ( z – a) 2 ( z – a) 3
f (t )dt
1
.
bn =
∫
2πi
(t − a) − n +1
=
where
∫
f (t )dt
C2 (t – a)
–2
+....
...(3.3)
C2
Substituting these values from (3.2) and (3.3) in (3.1), we get
f ( z ) = { a0 + a1 ( z − a) + a2 ( z − a) 2 +.....}
+ { b1 ( z − a) −1 + b2 ( z − a) −2 +.....}
or
f (z ) =
where an =
1
2πi
∫
C1 (t
∞
∞
n=0
n=0
∑ a n ( z − a) n + ∑ b n ( z − a) − n
f (t ) dt
– a)
n +1
, bn =
1
2πi
f (t ) dt
∫
C2 (t
– a) – n + 1
...(3.4)
and t in the first integral
denotes a point on C1 whereas in the second integral it denotes a point on C2.
Note:
1. Laurent's series of a given analytic function f (z) in its annulus of convergence is
unique. There may be different Laurent's series of f ( z ) in different annulii with
the same centre. The number of Laurent's series of a function depends upon the
number of points at which function is not analytic.
2. If f ( z ) is analytic throughout the region inside C1 then by Cauchy's theorem
bn = 0 for all n – 1 ≥ 0. Hence, in this case Laurent's series is same as Taylor's
series.
3. (i) The region of convergence of Laurent's series is R 2 <z – a< R1 .
(ii) If R 2 → 0 in the Laurent's series, then f ( z ) is analytic inz – a< R1 except at
the point z = a. Hence, in this case region of convergence of Laurent series is
0 <z – a< R1 .
(iii) If R1 → ∞, then the region of convergence of Laurent's series z – a> R 2 .
4.
It is advisable to obtain Laurent's series expansion by rearrangement,
manipulation and using the standard series expansion rather than using the
Laurent's series formula.
Ex. 11 : Expand f ( z ) =
Sol : f ( z ) =
1
for 1 <z< 2
( z – 1)( z – 2)
1
1
1
=
–
= ( z – 2) –1 – ( z – 1) –1
( z – 1)( z – 2) z – 2 z – 1
z
= (–2) –1 1 – 

2
–1
–
1
( z ) –1 1 – 

z
–1
92
wherez /2< 1 and1/z< 1
–1
–1
1
z
1
1
1 –  – 1 – 
2
2
z
z
2
3
 1
1
z z
z
1
1
1

= – 1 + +
+
+... – 1 + +
+
+...
2
3
2
2
4
8
z z

z
 z
2
3
1 z z
z
1
1
1
1
=– – –
–
..... – –
–
–
.....
2
3
2 4
8
16
z z
z
z4
f (z ) = –
Ex. 12 : Represent function f ( z ) =
z
by a series of positive and
( z – 1)( z – 3)
negative powers of ( z – 1) which converges to f ( z ) when 0 <z – 1< 2 .
Sol : Here f ( z ) =
z
( z – 1)( z – 3)
Putting z – 1 = u so that z = u + 1, since 0 <z – 1< 2 ⇒ 0 < u < 2
u +1
f (z ) =
u(u – 2)
1
3
=–
+
2 u 2(u − 2)
=–
1
3
– 1 –
2u 4 
u

2
–1
[ ∵ Binomial expansion is valid as
u
< 1]
2
n
=–
1
3 ∞ u
– ∑  
2 u 4 n = 0 2
=–
1
3 ∞ z – 1
– ∑ 

2( z – 1) 4 n = 0 2 
n
[ Replacing u by (z–1)]
Ex. 13 : Find Laurent series about z = 1 for the function f ( z ) =
e 2z
( z – 1) 3
when
0 <z – 1< 1 . 5 . Also find the region of convergence of the series.
Sol : Here f ( z ) =
e 2z
( z – 1) 3
Putting u = z – 1 so that z = u + 1,
Then
f (z ) =
e 2(u +1)
u3
=
=
e2
u3
e2
u3
e2u =
+
2e 2
u2

( 2 u ) 2 ( 2 u ) 3 ( 2u) 4
e2 
1
+
2
u
+
+
+
+....

2!
3!
4!
u3 

+
2e 2 4e 2 2e 2
+
+
u +....
u
3
3
Power Series and Calculus of Residues
=
e2
( z – 1) 3
93
+
2e 2
( z – 1) 2
2e 2
4e 2 2e 2
+
+
( z – 1) +....
( z – 1)
3
3
+
The function f ( z ) is analytic everywhere except at z = 1.
∴ The series converges for all values of z ≠ 1.
Ex. 14 : Obtain expansions for
( z – 2)( z + 2)
which are valid, for the regions
( z + 1)( z + 4)
(i)z< 1
(ii) 1 <z< 4 (iii)z > 4
(iv) 1 <|z + 1|< 3
( z – 2)( z + 2)
1
4
Sol : Here, f ( z ) =
= 1−
–
.
( z + 1)( z + 4)
z +1 z + 4
...(3.1)
Clearly the function f ( z ) is not analytic at z = –1 and z = – 4.
(i) Whenz< 1
From (3.1), we have
z
f ( z ) = 1 – (1 + z ) −1 – 1 + 

4
= 1–
∞
∞
n=0
n=0
−1
z
∑ (−1)n z n − ∑ (−1)n  4
[Since Bianomial expansion of (1 + z ) –1 and 1 +

z
< 1]
z< 1 and
4
 ∞

= 1 – 1+ ∑ ( −1) n z n  –
 n =1

z

4
n
−1
are valid when z< 1, i. e.,
∞

n –n n
1 + ∑ ( −1) 4 z 
 n =1

∞
= 1+
∑ (−1)n +1 (1 + 4– n )z n
n =1
(ii) When 1 <z< 4
Now we shall arrange the partial fractions in (1) so that the binomial expansions
involved may be valid in the given region.
f (z ) = 1 −
1
1 +
z
1

z
–1
– 1 +

z

4
–1
n
= 1−
∞
1. ∞
z n
 1
( −1) n   – ∑ ( −1) n  
∑
 4
 z
z n=0
n=0
[Binomial expansions are valid ∵
1
z
< 1 and
< 1]
z
4
∞
∞

( −1) n n 
= 1 + ∑ ( −1) n +1 z –( n +1) – 1 + ∑
z 
n
 n =1 4

n=0
94
∞
∞
∑ (−1)n z – n +
=
∑ (−1)n +1
n=0
⋅
n =1
zn
4n
(iii) Whenz > 4
Arranging the partial fractions in (1), so that the binomial expansions involved may
be valid in the given region.
f (z ) = 1 −
1
1 +
z
1

z
−1
−
4
1 +
z
n
= 1−
1 ∞
1
( −1) n  
∑
 z
z n=0
= 1+
∑ (−1)n +1  z 
∞
 1
−
−1
4
 1

∴ z < 1 and z < 1


4 ∞
4
( −1) n  
∑
 z
z n=0
n +1

n=0
4

z
4
+  
 z
n
n +1 


n
∞
 1 n
4 
= 1 + ∑ ( −1) n   +   
 z 
 z 
n =1

(iv) When 1 <z + 1 < 3
Writing z + 1 = u, we have 1 < u < 3
Then
f (z ) = 1 −
1
4
1
4
−
= 1− −
z +1 z + 4
u u+3
−1
1 4
u
− 1 + 

u 3
3

1 4
u u2 u3
= 1 − − 1 − +
−
+......
u 3
3 32 33

= 1−

4
1
4  z + 1 ( z + 1) 2 ( z + 1) 3
−
+ 
−
+
+......
3 z +1 3 3
9
27

2
3


1
1
4 z + 1 ( z + 1)
( z + 1)
=− −
+ 
−
+
....
3 z +1 3 z
9
27

= 1−
Ex. 15 : For the function f ( z ) =
2z 3 + 1
z2 + z
find,
(i) A Taylor's series valid in the neighbourhood of the point z = i.
(ii) A Laurent's series valid within the annulus of which centre is origin.
Sol : Here f ( z ) =
2z 3 + 1
2
= 2( z − 1) +
1
1
+ . Try yourself
z +1 z
z +z
Ex. 16 : Find the Taylor's or Laurent's series which represent the function
1
(1 + z 2 )( z + 2)
Power Series and Calculus of Residues
(i) whenz < 1
Sol : We have
95
(ii) when 1 <z < 2
f (z ) =
1
=
2
(1 + z )( z + 2)
(iii)z > 2
1 1
z−2
−

5  z + 2 1 + z 2 
Now, we find below expansions of f ( z ) under specified values of z:
(i) Whenz < 1.
We have
f (z ) =
1 1
z−2
−

5  z + 2 1 + z 2 
Since Binomial expansion of the form (1 + z ) −1 is valid ifz < 1, therefore, partial
fractions of f ( z ) are to be so arranged that the Binomial expansions involved may be
valid forz < 1.
f (z ) =
1 1
⋅ 1 +
5 2
z

2
−1
+
2− z
(1 + z 2 ) −1
5
Obviously Binomial expansions are valid forz < 1, hence
f (z ) =
1 ∞
z
( −1) n  
∑
 2
10 n = 0
n
2− z ∞
∑ (−1)n (z 2 )n .
5 n=0
+
This represents series in positive powers of z, in other words, it is an expansions of
f ( z ) in Taylor's series within the circlez = 1.
(ii) When 1 <z < 2
We have
f (z ) =
1 1
z−2
−

,
5 z + 2 1 + z2
The partial fractions of f ( z ) are to be so arranged that the Binomial expansions may
be valid for 1 <z < 2 .
Thus
1 1
1
f ( z ) = . 1 + z

5 2
2 
−1
z–2 1 
1
–
.
1 + 2 
2
5 z 
z 
−1
Binomial expansions are obviously valid for 1 <z < 2 , hence
f (z ) =
1 ∞
z
∑ (−1)n  2
10 n = 0
n
+
2− z
5z 2
∞
n
 1
∑ (−1)n  z 2  .
n=0
These are the expansions in positive and negative powers of z, that is, it is a Laurent's
expansion of f ( z ) within the annulus 1 <z < 2 .
(iii) Whenz  > 2.
f (z ) =
1
1
1 z−2
⋅
−
5 z + 2 51+ z2
96
=
1 1
2
⋅ 1 + 
5 z
z
−1
1
1
( z − 2)
5
z2
–
1

1 + 2 

z 
−1
arranged suitable to make the Binomial expansions valid forz > 2
f (z ) =
11 ∞
2
( −1) n  
∑

5 z n=0
z
n
−
1 1
2 ∞
n 1
 − 2  ∑ ( −1)
5  z z  n=0
z 2n
This is also Laurent's expansion within the annulus 2 <z < R, where R is large.
Ex. 17 : Find different development of
1
in powers of z; according to
( z − 1)( z − 3)
the position of the point z in the plane. Expand the function in a Taylor's series about
z = 0 and indicate the circle of convergence.
Sol : Here f ( z ) =
1
1 1
1 
= 
−


( z − 1)( z − 3) 2 ( z − 3) ( z − 1)
Thus, function is regular everywhere except at z = 1, z = 3. We give below different
expansions of f ( z ) according to the position of z in the plane.
(i) When 0 <z < 1.
We have
f (z ) =
1 1
1  –1 
−

 =
1 −
2  z − 3 z − 1
6 
z

3
−1
+
1
(1 − z ) −1
2
arranged suitably in order to make Binomial expansions valid for 0 <z < 1
f (z ) = −
1 ∞  z
∑  
6 n = 0  3
n
+
1 ∞
∑ (z )n =
2 n=0
∞
∑
n=0
1
1  n
1 − n +1  z

2
3
which is a Laurent's series within the annulus 0 <z< 1
(ii) When 1 <z < 3
Then
f (z ) =
1 1
1 
1
z
−
= − 1 − 
2  z − 3 z − 1
6
3
–1
−
1 
1
1 − 
2z 
z
–1
[arranged suitably]
∞
=−
1
∑
6 n=0
 z
 
 3
n
−
1
2z
∞
∑
n=0
 1
 
 z
n
which is a Laurent's series within the annulus 1 <z < 3.
(iii) Whenz > 3
Then
f (z ) =
=
1 1
1 
1 
3
−
=
1 − 
2  z − 3 z − 1 2z 
z
1
2z
∞
3
∑  z 
n=0
n
–
1
2z
∞
∑
n=0
 1
 
 z
n
–1
−
1 
1
1 − 
2z 
z
–1
Power Series and Calculus of Residues
=
97
1 ∞ n
n +1
∑ (3 − 1)(1 / z )
2 n=0
which is a Laurent's series, within the annulus 3 <z < R.
Last part. If centre of a circle is at z = 2, then distance of z = 1 and z = 3 (the points
where f ( z ) is not regular) from this centre is 1. Hence, if a circle is drawn with centre
z = 2 and radius 1, then within this circle, i.e., within the circle z – 2= 1, the
function f ( z ) is analytic, hence it can be expanded in a Taylor's series within the
circlez – 2 = 1, which is therefore, the circle of convergence.
Now,
f (z ) =
1
1
1
=
=
2
( z − 1)( z − 3) ( z − 4z + 3) ( z − 2) 2 – 1
expansion is now valid forz – 2< 1
∞
f (z ) =
∑ (z – 2)2n
0
which is obviously a Taylor's series in positive powers of ( z − 2) within a circle
z – 2= 1.
1
Ex. 18 : Express, f ( z ) =
in a Laurent's series in the region
2
z( z + 1) ( z + 2) 3
 5 ≤ z ≤  7  .
    
 4
 4
Sol : f ( z ) =
1
2
z( z + 1) ( z + 2)
3
=

1  −3
1
3
3
1
+
+
+
–


2
2
3
z  z + 1 ( z + 1)
( z + 2) ( z + 2)
( z + 2) 
It is clear that the f ( z ) is analytic everywhere except at z = 0, z = −1, z = −2 .
The region
5
7
≤z≤ is part of the annulus between two concentric circlesz = 1
4
4
andz = 2 within which f ( z ) is analytic; hence we can express it in Laurent's series
at any z within the annulus. As seen above, we have

1  3z + 2
1
f ( z ) = −
+ ( 3z 2 + 15z − 17 )

2
3
z  ( z + 1)
( z + 2) 
=
1  3z + 2 
1 +
−
z 
z2 
1

z
−2
1
+ ( 3z 2 + 15z − 17 ) 1 +
8
z

2
–3 


Both Binomial expansions are valid since 1 <z < 2
n
 3
∞
2 ∞
1
1
17
( −1) n ( n + 1)( n + 2)  z  


n

= −  +
(
−
1
)
(
n
+
1
)
+
3
z
−
+
15




 ∑
∑
 2 
z
2
  z z 2  n=0
zn 8
0
=
n


∞
n
2 ∞
1
1
17
(
−
1
)
(
n
+
1
)(
n
+
2
)
3
 3z −
n +1
= +
( n + 1)
+
+ 15 ∑
zn

 ∑ ( −1)
n

 z z 2  n=0
z
z n 16 
2
n=0
98
This being a series in positive and negative powers of z is Laurent's expansion in the
5
7
annulus 1 <z< 2; or we may say as desired in the region <z< .
4
4
Note : The region given in the question is a part of the annulus 1 <z< 2 , hence
above expansion is true for the region
5
7
≤z≤ .
4
4
1
Ex. 19 : Expand f ( z ) = cosh  z +  .

z
Sol : f ( z ) is analytic everywhere except z = 0.
Thus f ( z ) can be expanded by Laurent's theorem.
∞
⇒
where
Now
f (z ) =
an
∞
∑ a n z n + ∑ ( bn / z n )
n=0
n =1
1
=
2πi
f ( z )dz
an =
=
∫ (z − 0)n +1
and bn =
C
1
2πi
∫ f ( z )z
n −1
dz
C
1
cosh  z +  dz

z
∫
C
2π
1
2πi
1
2πi
∫
z n +1
cosh( 2 cos θ) dz
[since z + z –1 = 2 cos θ]
z n +1
0
Put z = eiθ ⇒ dz = ieiθ dθ
=
=
=
Similarly
Let us write
2π
cosh(2 cos θ) eiθ dθ
1
2π
∫
1
2π
∫ cosh(2 cosθ)
1
2π
1
bn =
2π
ei( n+1 )θ
0
=
1
2π
2π
0
cos nθ dθ –
i
2π
2π
– niθ
∫ cosh(2 cos θ) e
dθ
0
∫ cosh (2θ) sin nθ dθ
C
2π
∫ cosh (2 cos θ ) cos nθ dθ + 0
0
2π
∫ cosh(2 cos θ) cos(nθ) dθ
0
∞
1
1
for z in the given function ⇒ f ( z ) = a0 + ∑ a n  z +  .

z
z
n =1
1 

Ex. 20 : Expand f ( z ) = sin  c ( z +   .

z 

Sol : f ( z ) is not analytic at z = 0 , hence f ( z ) can be expanded by Laurent's theroem.
Power Series and Calculus of Residues
99
∞
⇒
f (z ) =
∞
n =1 z
n=0
where
Now
an =
an =
=
=
bn
∑ an z n + ∑
1
2πi
1
2πi
1
2π
1
=
2π
f ( z )dz
∫
( z − a)
C
1
2πi
n
1
sin c  z +  dz

z
∫
z
C
2π
1
2π
=
e( n+1 )iθ
0
2π
∫ f (z )z
n −1
dz
C
[ put z = eiθ ⇒ dz = ieiθ dθ]
n +1
sin[ c(2 cos θ) ieiθ dθ
∫
1
2πi
and bn =
n +1
i
2π
− niθ
∫ sin(2 cos θ )e
dθ
0
2π
∫ sin(2 cos θ )cos nθdθ − 2π ∫ sin(2 cos θ )sin nθ dθ
0
2π
0
∫ sin(2 cos θ )cos nθdθ + 0
0
2π
If
f ( 2a − x ) = − f ( x ) then
∫ f ( x )dx = 0
0
Similarly
bn =
2π
1
2π
sin( 2 cos θ) cos nθ dθ
∫
0
1
.
z
bn
The function remains unaltered by putting z for
∞
So
∑
f ( z ) = ao +
∞
an z n +
n =1
∑
z
n =1
Ex. 21 : Show that if c > 0 then e z + ( c
3
/ 2z 2 )
1

= a 0 + Σa n  z n +


zn 
( ∵ a n = bn )
n
∞
=
∑ an z n
n = −∞
where
an =
2π
e −(1/ 2)c
2πc
3
∫
n
2
e c(cosθ + cos
θ)
× cos{ c sin θ (1− cosθ) − nθ} dθ.
0
2
Sol : Here f ( z ) = e z + ( c / 2z ) is analytic except at z = 0, therefore, it can be
expanded in the annulus r < z < R in a Laurent's series.
∞
3
2
f (z )
1
Thus f ( z ) = e z + ( c / 2z ) = ∑ an z n where an =
dz
∫
2π z n +1
n = −∞
C
and C is a circlez = c with centre at the origin and radius c, so that z = ceiθ on this circle.
Now
an
1
=
2πi
=
1
2πi
f ( z )dz
∫
C
2π
∫
0
z n +1
1
dz =
2πι
∫
e z + (c
C
c
exp  ceiθ + e –2iθ 


2
c
n +1 i ( n +1)θ
e
3
/ 2z 2 )
z n +1
ceiθ ⋅ idθ
dz
[ as z = ceiθ ]
100
=
=
2π
1
2πc
n
n
iθ
+
0
2π
1
2πc

∫ exp  ce
c –2iθ
e
– inθ dθ

2
c
exp  c cos θ + cos 2θ


2
∫
0
c


exp i  c sin θ – sin 2θ – nθ  d θ

2
 
=
2π
1
2πc
n
∫
0
c
exp  c cos θ + cos 2 θ


2
c
c


× cos  c sin θ – sin 2θ – nθ + i sin  c sin θ – sin 2 θ – nθ  dθ




2
2


=
2π
1
2πc
n

c

c


∫ exp  c cos θ + 2 cos 2θ × cos  c sin θ – 2 sin 2θ – nθ d θ
0
The second integral vanishes on account of the property of definite integrals. Since
F( 2π − θ) = − F(θ).
an =
∴
2π
1
2πc
n
∫
0
c
exp  c cos θ + ( 2 cos 2 θ – 1
2


× cos{ c sin θ(1 − cos θ) − nθ} dθ
=
e
− c/2
2πc
n
2π
∫
2
e c(cosθ + cos
θ)
× cos{ c sin θ( 1 − cos θ) − nθ} dθ.
0
Problem Set (3.3)
1.
Expand each of the following function in a Laurent's series.
( z − 1)
(i)
, forz – 1> 1
z2
sin 4z
(ii)
, which converges for 0 <z< R
z2
(iii)
(iv)
z2 −1
z 2 + 5z + 6
z
about z = 0 in the region 2 <z< 3
for ( a)  z< 1, ( b) 1 < z< 2 ,
( z 2 − 1)( z 2 + 4)
1
(v)
which converges for 0 <z< 1
5
z +2
(vi)
z 2 − 6z – 1
in the region 3 <z + 2< 5
( z − 1)( z − 3)( z + 2)
( c )  z> 2
Power Series and Calculus of Residues
2.
101
Expand each of the following function in Laurent's series about the given point.
1
(i)
, a=1
2
z + ( 3 i − 1) z − 3i )
z2 +1
(ii)
z( z 2 + 3z − 10)
, a=1
(iii) z 2 e1/ z , a = 0
3.
Find all Taylor's series and Laurent's series with centre z = a.
1
(i)
, a = i, 0 <  z − i< 2 and  z − i> 2
2
z +1
1
(ii)
, a = 1;  z − 1< 1 and  z − 1> 1
z3
4
(iii)
, a = 0, z< 1 and  z> 1
1− z4
(iv)
(v)
4.
7 z 2 − 9z − 18
z 3 − 9z
ez
( z − 1) 4
, a = 0, 0 < z< 3 and  z> 3
, a = 1,  z − 1> 0
Find the Laurent's series of the function f ( z ) = 1 / [ 2z 2 − iz] about a = 0 which
converges at z = i.
Objective Type Questions
Multiple Choice Questions
Tick the correct answer :
∞
1. The radius of convergence of the series
∑ n n z n is :
n =1
(a) 0
(c) 1
2.
3.
(b) ∞
(d) none of these.
For the series Σ (1/ n p )z p , the radius of convergence is :
(a) 2
(b) 1
(c) 3
(d) ∞.
The radius of convergence of the series Σ 2 n z n ! is :
(a) 1
(c) ∞
(b) 2
(d) none of these.
∞
4. The radius of convergence for the series
∑ z n ! is :
n=0
(a) 1
(b) 0
(c) ∞
(d) none of these.
102
∞
5. The series f ( z ) =
∑ an (z − z 0 )n where an
=
n = −∞
(a) Laurents series
(c) Maclaurin's
f ( z )dz
1
represents :
∫
2πi ( z − z 0 ) n +1
C
(b) Taylor's series
(d) none of these.
Fill in the Blanks
1.
When 0 < z < 4 , the expansion of
1
2.
is ................... .
4z − z 2
Every analytic function can be represented by .................. .
3.
The series
z 2n
∞
∑ (−1)n (2n )! for
z < ∞ represents ............... .
n=0
ANSWERS
Problem Set 3.1
1. R = e
2. R = ∞
1
5
4. R = 1
3. R =
5. R = 3 / 2
6. R = 5, centre of convergence z = –i
Problem Set 3.2
1.
z2 z3 z4
+
–
....,z = 1 , except for z = ±1.
2
2
2
1 
1
1
(b)
1 + ( z – π / 4) – ( z – π / 4) 2 − ( z − π / 4) 3 ....., z< ∞.


2!
3!
2
(a) z –
(c)
1
(3 – i )2
∞
2.
∞
∑
n=0
n
z – i
( n + 1)
 , z – i<3 – i.
 3 – i 
∑ (–1)n +1 n( z – 1)
n
n =1
3.
z2 z3
–
+..... Region of convergence R = ∞.
2! 3!
3
5
π 1
π
1
π
(ii) z + –  z +  +  z +  –....., R = ∞.
2 3! 
2
5! 
2
1
1
(iii) ( z – 1) – ( z – 1) 2 + ( z – 1) 3 +...., R = 1.
2
3
(i) 1 – z +
(iv) 1 – z 4 + z 8 – z12 +.....,
(v) 2 + z + 2z 2 + z 3 + 2z 4 +.....,
Power Series and Calculus of Residues
(vi) 1 – z +
103
3 2 7 3
z – z +.....,
2
6
(viii) 4 + 5z + 6z 2 + 7 z 3 +.....
∞
(ix)
∑
(–1) n –1 i n ( z – i )
n
n =1
4.
5.
n
, R = 1, Function is not analytic at x = 0, y ≥ 2 .
  1  n +1  
1  ∞
 ∑ (–1) n   
– 1 z n .


(1 – 2 i n = 0
2
i

 

Radius of convergence R = 1, which is the distance between the point z = 0 and
the narest point z = –1, where function is not analytic.
z2
2
z z3
(ii) 1 + –
4 48
(i) 1 + z +
(iii) 1 + z + z 2 +
5 3
z
6
Problem Set 3.3
1.
(i) ( z − 1) −1 − 2( z − 1) −2 + 3( z − 1) −3 − 4( z − 1) −4 +.......
∞
∑
(ii)
n=0
(iii) 1 +
( −1) n 4 2n +1 z 2n − 3
, z> 0
( 2n + 1)!
 8

3
2 22 23
z z2
−
+..... − 1 − +
−....
1 − +
2
3
2
z
z z
3 3
 3

z
z 5z 2 21 5
−
+
z −.....
4 16
64

1 1
1
z z3 z5
(b) 
–
– +
−
+.....
3  z 5 z 3 4 16 64

(iv) (a)
(c)
1
z
∞
∑
(v)
3
−
5
z5
+.....
( −1) n z 4n −1
n=0
(vi)
2.
(i)

2
3
32
1
( z + 2) ( z + 2) 2
+
+
+....+ 1 +
+
+......
2
3
2
z + 2 ( z + 2)
5
5

( z + 2)
5
∞
(1 − 3 i )  1
( −1) n ( z − 1) n 
−

 , 0 <z − 1< 10;
∑
10  z − 1 n = 0 (1 + 3 i ) n +1 
1
( z − 1) 2
∞
∑
( −1) n (1 + 3 i ) n
n=0
( z − 1) n
,z − 1> 10
104
(ii) −
∞
1
+ ∑
10z n = 0
− 5 2 − n + 26 ( −1) n 5 − n  z n , 0 < z < 2;

 28

175
n
−
1
5 ∞
2n
26 ∞
z
+
+
( −1) n   , 2 <z< 5
∑
∑
n
+
1

10z 14 n = 0 z
175 n = 0
5
−
∞
1
26
5
1
+ ∑  ( −5) n +
( 2) n 
, z> 5
10z n = 0  35
14
 z n +1
(iii) z 2 + z +
3.
∞
i
(i) − ∑  
 2
(ii)
1 1 1 1 1
+
+
+......
2 3! z 4! z 2
n +1
( z − i ) n −1 , 0 < z − i< 2 and
n=0
∞
∞
∑
∑
n=0
( n −3 )( z − i ) n and
n=0
∞
(iii) 4 ∑ z 4n and –
(iv)
∑
n=0
4.
∞
∑
n=0
( z − 1) n + 2
(z − i ) > 2
1
( z − 1) n + 3
4
z
4n + 4
∞
2
( −1) n − 4 n
2
+ ∑
z and +
n +1
z n=0
z
3
∞
(v)
( n −3 )
n=0
n=0
( −2 i ) n
∞
∑
∞
∑
3 n ( 4 + ( −1) n )
n=0
z n +1
e
( z − 1) n − 4
n!
Function is not analytic at z = 0 and i/2. For convergence at z = i obtain
∞
(i ) n
Laurent series in the regions z> 1/ 2, f ( z ) = ∑
n +1 n + 2
z
n=0 2
Multiple Choice Questions
1. (a)
2. (b)
3. (a)
4. (a)
5. (a)
Fill in the Blanks
∞
1.
∑
n=0
z n −1
4 n +1
2. power series
3. cos z
Power Series and Calculus of Residues
105
3.5 Zeros, Singularities and Meromorphic Functions
The Zeros of an Analytic Functions
A zero of an analytic function f ( z ) is a value of z such f ( z ) = 0.
If z = a is a zero of the analytic function f ( z ) and f ( a) = f ' ( a) = f ' ' ( a) =.... =
f
( m −1)
( a) = 0 but f ( m ) ( a) ≠ 0, then z = a is called a zero of order m. When m = 1,
i.e., if f ( a) = 0 and f ' ( a) ≠ 0 then zero of f ( z ) is said to be simple. In other words
we can say the order of the zero of an analytic function f ( z ) is the order of first
non-vanishing derivative of f ( z ).
If the function is analytic at a point a, then there exists a neighborhood of the point
z = a, throughout which the function is represented by a Taylor's series.
∞
f n ( a)
f ( z ) = ∑ an ( z − a) n where an =
n!
n=0
when z = a is a zero of f ( z ) then f ( a) = 0, therefore a0 = f ( a) = 0. If, in addition to
this, f ' ( a) = 0 = f ' ' ( a) =.... = f ( m −1) ( a), but f m ( a) ≠ 0, then f ( z ) is said to have a
zero of order m at z = a.
It may also be seen that since am =
f m ( a)
, hence at a zero of order m, we have
m!
a0 = 0 = a1 = a2 ..... = am −1 , but am ≠ 0.
Thus, in the neighbourhood of the zero z = a of order m, the function f ( z ) is
represented as the series
f ( z ) = am ( z − a) m + am +1 ( z − a) m +1 +.....
= ( z − a) m [ am + am +1 ( z − a)+...]
= ( z − a) m
∞
∑
a m + n ( z − a) n
n=0
f ( z ) = ( z − a) m φ ( z )
or
∞
where φ( z ) =
∑
am + n ( z − a) n = am + am +1 ( z − a)+..... is analytic and non-zero
n=0
in the neighbourhood of z = a .
As a matter of fact φ( z ) is equal to am at z = a. This is an alternative way of defining a
zero of order m. For m = 1, the zero is called a simple zero.
3.6 Zeros are Isolated
Theorem. Let f ( z ) be analytic in a domain D. Then unless f ( z ) is identically zero,
there exists a neighbourhood of each point in D throughout which the function has
106
no zero, except possibly at the point itself. In other words, the zeros of an analytic
function are isolated.
Proof. Let z = a be a zero of order m of the function f ( z ) which is analytic at a and
in its neighbourhood, then
f ( z ) = ( z − a) m
∞
∑
a m + n ( z − a) n = ( z − a) m φ ( z )
... (3.1)
n=0
∞
where
φ( z ) =
∑ am + n (z − a)n = am + am +1 (z − a)+... and
φ ( a) = a m ≠ 0
n=0
Clearly φ (z) is analytic at z = a from which it follows that φ (z) is continuous at z = a
and since φ ( a) ≠ 0, there exists a neighbourhood of a, for no point z of which φ (z) is
zero.
Also ( z − a) m ≠ 0 for values of z ≠ 0. So there is no other point in the neighbourhood
of z = a at which f ( z ) is zero.
Hence, the zero z = a is isolated. The same is true for every zero of f ( z ). Hence, all
the zeros of f ( z ) are isolated.
3.7 Singularities of an Analytic Function
Singularities (or Singular Points)
A point at which a function f ( z ) ceases to be regular (or analytic) is called the
singularity [or singular point of the function f ( z )].
Isolated Singularity. The point z = a is called an isolated singularity or isolated
singular point of f ( z ) if we can find δ > 0 such that the circle  z − a< δ encloses no
singular point other than a. If no such δ can be found, we call z = a, a non-isolated
singularity.
For example,
(i) The function f ( z ) =
( z + 3)
is analytic everywhere except at z = 1 and
( z − 1)( z − 2)
z = 2. Thus, z = 1, z = 2 are the only singularities of this function. There are no
other singularities of f ( z ) in the neighbourhood of z = 1 and z = 2, therefore
z = 1, z = 2 are the isolated singularities of this function .
(ii) Consider another function
1
. This function is not analytic at the points
tan( π / z )
π
π
where tan   = 0, i.e., at the points = nπ, i.e, at the points
 z
z
Power Series and Calculus of Residues
z=
107
1
1
1
( n = 1, 2, 3,.... ). Thus z = 1, z = , z = ,....., z = 0 are the singularities of
n
2
3
the function all of which lie on real axis. All the singularities are isolated except
z = 0, because in the neighbourhood of z = 0 there are infinite number of other
singularities z = (1/ n ) when n is large. Therefore z = 0 is the non-isolated
singularities of the function 1 / tan( π / z ).
Note: A function whose only singularities in the entire plane are poles, is called a
meromorphic function.
3.8 Kinds of Singularity and Residue
Consider a function f ( z ) which is analytic within a domain D, except at the point
z = a, which is an isolated singularity. Thus, f ( z ) is analytic
in the annulus
ρ < z − a< R, where ρ can be taken as small as we please and R as large as we please
such that the annulus lying wholly within the domain D. Hence, if z is any point of
this annulus, then by Laurent's series, we have
∞
f (z ) =
∑
a n ( z − a) n +
n=0
∞
The second term
∑
∞
∑
b n ( z − a) − n
... (3.1)
n =1
bn ( z − a) − n on the right hand side is called the principal
n =1
part of f ( z ) at the isolated singularity z = a. In relation to the principal part there
are three possibilities which give rise to three kinds of singularities namely
removable singularity, essential singularity and pole.
(i) Removable Singularity
When all the coefficients bn of (3.1) are zero i.e, if the principal part consists of no
terms, then z = a is said to be a removable singularity.
∞
Here f ( z ) =
∑
an ( z − a) n which is analytic for  z − a< R, except at the point
n=0
∞
z = a. Let φ (z) be the sum function of the power series
∑
an ( z − a) n . Now φ (z)
n=0
differs from f ( z ) only for z = a where there is a singularity. To avoid this singularity
z = a we can suitably define f ( z ) at z = a, so that, we have
 f ( z ) for 0 <z − a< R
φ (z ) = 
z =a
 a0 for
z=
1
( n = 1, 2, 3,.... ).
n
108
This type of singularity, which can be made to disappear by defining the function
suitably, is called Removable Singularity. This is rather artificial sort of
singularity and is of little importance in the theory of functions.
sin( z − a)
has removable singularity at z = a because
z−a


sin( z − a)
1
( z − a) 3 ( z − a) 5 ( z − a)7
f (z ) =
=
(
z
−
a
)
−
+
−
+.....

z−a
( z − a) 
3!
5!
7!

2
4
6
( z − a)
( z − a)
( z − a)
= 1−
+
−
+.....
3!
5!
7!
has no term containing negative powers of ( z − a).
For example, function f ( z ) =
However, this singularity can be removed and the function be made analytic by
sin( z − a)
defining
= 1 at z = a.
z−a
(ii) Essential Singularity (or isolated essential singularity)
If there are infinite number of terms in principal part, that is, if the series
b1 ( z − a) −1 + b2 ( z − a) −2 +.... does not terminate, then z = a is called an isolated
essential singularity of the function f ( z ).
1 
For example, the function sin 
 has essential singularity at z = a because
 z − a
1 
1
1
1
sin 
−
+
+....
 =
3
 z − a ( z − a) 3 !( z − a)
5 !( z − a) 5
has infinite number of the terms in negative powers of ( z − a).
(iii) Pole
If there are finite number of terms in the principal part, that is, if the series
b1 ( z − a) −1 + b2 ( z − a) −2 +.....+ bm ( z − a) − m
consists of finite number of terms such that bm ≠ 0, bm +1 = 0 = bm + 2 +.....
then z = a is said to be a pole of order m of the function f ( z ) .
When m = 1, the pole is said to be a simple pole.
For example, the function
sin( z − a)
( z − a)
4
=
1
sin( z − a)
( z − a) 4
has a pole at z = a because
( z − a) − 1 ( z − a) 3 + 1 ( z − a) 5 − 1 ( z − a)7 −.....

3!
5!
7!
( z − a) 
1
1
1
1
=
−
+ ( z − a) − ( z − a) 3 +.....
3
3 !( z − a) 5 !
7!
( z − a)
4
Power Series and Calculus of Residues
109
has finite number of terms (first two terms only) in negative powers of ( z − a). Since,
1
here b3 = coefficient of
, maximum negative power of ( z − a) = 3 ≠ 0 and
( z − a) 3
b4 = 0 = b5 =.....
Therefore, the point z = a is a pole of order 3 of the function f ( z ) =
sin( z − a)
( z − a) 4
.
If z = a is a pole of order m of the function f ( z ), then
∞

b
b2
bm
f ( z ) =  ∑ a n ( z − a) n  + 1 +
+.....+
, bm ≠ 0
2
 0
 z − a ( z − a)
( z − a) m
=
or
where
 ∞


 a ( z − a) n + m  { bm + bm −1 ( z − a)+...+ b1 ( z − a) m −1 }
m  ∑ n

( z − a)  0

1
f (z ) =
φ (z )
( z − a) m
∞

φ( z ) = ∑ an ( z − a) n + m + { bm + bm −1 ( z − a)+.....+ b1 ( z − a) m −1 
0

1
Obviously φ ( z ) → bm as z → a. Hence φ( z ) is analytic in the neighbourhood of the
pole z = a including z = a itself and φ ( a) = bm ≠ 0.
(iv) Residue
If z = a is a pole of function f ( z ) then b1 the coefficient of
1
in the Laurent's
( z − a)
expansion of f ( z ) around z = a is called the residue at z = a.
Theorem. If f ( z ) has a pole at z = a, then f ( z )→ ∞ as z → a.
Suppose z = a is the pole of order m, then by Laurent's theorem
∞
b
b2
bm
f ( z ) = ∑ a n ( z − a) n + 1 +
, +.....+
, bm ≠ 0
2
z − a ( z − a)
( z − a) m
n=0
∞
=
∑ a n ( z − a) n
+
n=0
1
( z − a) m
[ bm + b1 ( z − a) m −1 + b2 ( z − a) m − 2 +....+ bm −1 ( z − a)]
The expression within the square brackets on right hand side tends to bm while
1
tends to ∞ as z → a so that the whole right hand expression tends to
( z − a) m
infinity as z → a.
Hence
 f ( z )→ ∞ as z → a.
110
3.9 Poles are Isolated
Theorem. The poles of a function are isolated.
Proof. Let the point z = a be a pole of order m of the function f ( z ), then in the
neighbourhood of z = a, by Laurents' theorem, we have
∞
b
b2
bm
f ( z ) = ∑ a n ( z − a) n + 1 +
+...+
2
z
−
a
( z − a)
( z − a) m
n=0
 ∞

1
=
an ( z − a) n + m +{ bm + bm −1 ( z − a)+...+ b1 ( z − a) m −1 }

∑
( z − a) m n = 0

1
=
φ (z )
( z − a) m
∞
where
φ (z ) =
∑ a n ( z − a) n + m
+{ bm + bm −1 ( z − a)+.....+ b1 ( z − a) m −1 }.
n=0
Clearly φ (z) does not tend to infinity for any finite value of z near z = a i.e, there is
no pole of f ( z ) in the neighbourhood of z = a. Thus, the pole z = a is isolated. Hence,
the poles of a function are isolated.
3.10 Limit-Points of Zeros and Poles
Limit Points of Zeros
Theorem. Let f ( z ) be analytic in a domain D and let a set of zeros of f ( z ) have a
limit point in D. Then either f ( z ) vanishes identically in D, i.e, f ( z ) vanishes for all
z ∈ D or the limit point is an isolated essential singularity of f ( z ).
Proof. Let z1 , z 2 , z 3 ,.... be an infinite set of zeros of f ( z ) which have a limit point.
Suppose α be this limit point. Since α is the limit point of zeros of f ( z ), f ( z ) vanishes
at an infinite number of points in every small neighborhood of α i.e, f ( z ) has zeros as
near as we please to α. But α is a point in D, therefore f ( z ) is continuous at z = α and
hence we must have f (α ) = 0 i. e., z = α is a zero of f ( z ) having an infinite number
of zeros in its small neighborhood. Since we have proved that zeros of f ( z ) are
isolated, hence α cannot be a zero of f ( z ) unless the function vanishes identically in D.
If the function f ( z ) does not vanish identically in D, then z = α is not a zero of f ( z ).
Since this limit point α is surrounded by an infinite number of zeros, it shows that α
is a singularity of f ( z ).
Here z = α cannot be a pole since f ( z ) does not tend to infinity as z → α in any
manner. Thus, we conclude that α is an essential singularity of f ( z ). Since the
function f ( z ) is analytic in the neighborhood of α, this singularity is isolated.
Power Series and Calculus of Residues
111
Hence, the limit point of zeros is an isolated essential singularity if it does not vanish
identically in D.
Theorem. Identity theorem. If f ( z ) and g ( z ) are analytic function in a domain
D and if f ( z ) = g ( z ) on a subset of D which has a limit point in D, then f ( z ) = g ( z ) in
whole of D.
Proof. Let f ( z ) and g ( z ) be analytic in domain D. If F( z ) = f ( z ) − g ( z ), then F( z ) is
analytic function in D. Since f ( z ) = g ( z ) on a subset of D, hence F( z ) vanishes at all
points of D. If α is the limit point of this subset then F( z ) vanishes at infinite number
of points in the small neighborhood of α . But the limit point α is in D, therefore, F( z )
is continuous at α and hence F(α ) = 0. Since the zeros of f ( z ) are isolated, hence α
cannot be a zero of F( z ) unless F( z ) vanishes identically in D. Hence F( z ) vanishes
identically in D i.e, f ( z ) = g ( z ) in whole of D.
Limit Point of Poles
'Theorem. The limit point of the poles of a function f ( z ) is a non-isolated essential
singularity.
Proof. Let z1 , z 2 , z 3 ,.... be an infinite set of poles of f ( z ) having a limit point.
Suppose α be this limit point, then f ( z ) becomes unbounded at an infinite number of
points in the small neighborhood of z = α, and hence f ( z ) cannot be analytic at
z = α, i.e, z = α is a singularity of f ( z ).
This limit point z = α cannot be a pole, since it has an infinite number of poles in
small neighborhood around itself while the poles are isolated. Also this limit point
z = α cannot be a zero of f ( z ) because the function f ( z ) is not analytic in its
neighborhood. Thus, we conclude that α is an essential singularity of f ( z ). Also this
singularity is not isolated as there are poles in its neighborhood. Therefore, z = α is
a non-isolated essential singularity of f ( z ).
Hence, the limit point of poles is a non-isolated essential singularity.
3.11 The Point at Infinity
In the theory of complex numbers, infinity is regarded as a point. To examine the
nature of the function f ( z ) at infinity, we substitute z = (1 / ζ ) and examine the
nature of function f(1/ζ ) at ζ = 0.
To study the behavior of function f ( z ) at infintiy, go through the following theorem.
Theorem. The function f ( z ) : (i) is analytic, (ii) has a zero, (iii) has a pole, (iv) has
a removable singularity or (v) has an essential singularity at z = ∞ , according as the
function f(1/ζ ) has the corresponding behaviour (property) at ζ = 0.
112
 1
Proof. (i) Let f   be analytic at ζ = 0
 ζ
Since
1
 1
= z, ∴ f ( z ) = f   is also analytic at z = 0
 ζ
ζ
 1
(ii) Let f   has a zero of order m at ζ = 0
 ζ
 1
∴ By Taylor's expansion of f   at ζ = 0
 ζ
 1
We have f   =
 ζ
Replacing
∞
∑ an ζ n
n=m
1
by z , f ( z ) =
ζ
∞
 an 
 n
 
n=m z
∑
which neither contains constant nor positive powers of z. Hence f ( z ) has a zero of
order m at z = ∞.
 1
(iii) Let f   has a pole of order m at ζ = 0
 ζ
 1
∴ By Laurent's expansion of f   at ζ = 0, we have
 ζ
 1
f  =
 ζ
Replacing
∞
∑
∞
an ζ n +
n=0
∑
bn
n =1
ζn
, bm ≠ 0
1
by z, the expansion of f ( z ) at z = ∞ is of the form
ζ
∞
f (z ) =
∑
n=0
1
an  
 z
n
m
+
Since the principal part of f ( z ) at z = ∞ is
∑
( bn . z n )
∑
bn z n , where bm ≠ 0, hence f ( z ) has a
n =1
m
n =1
pole of order m at z = ∞.
 1
(iv) Let f   has a removable singularity at ζ = 0
 ζ
 1
∴ The expansion of f   in the neighborhood of 0 will be of the form, given by
 ζ
 1
f  =
 ζ
∞
∑
an ζ n
n=0
1
Replacing ζ by , the expansion of f ( z ) at z = ∞ is given by
z
Power Series and Calculus of Residues
∞
f (z ) =
∑
n=0
an
z
n
113
= a0 +
a1 a2
+
+.....
z
z2
Since lim f ( z ) = a0 (finite) ∴ f ( z ) has a removable singularity at z = ∞.
z→ ∞
 1
(v) Let f   has an essential singularity at ζ = 0.
 ζ
 1
∴ By Laurent's expansion of f   at ζ = 0, we have
 ζ
 1
f  =
 ζ
Replacing
∞
∞
n=0
1
by z, we have f ( z ) =
ζ
∞
Since the principal part
 1
∑ an ζ n + ∑ bn  ζ n 
∑
n =1
∞
∑
n=0
an
z
n
∞
+
∑
bn z n
n =1
bn z n of the expansion of f ( z ) at z = ∞ have infinite
n =1
number of terms,
∴ f ( z ) has an essential singularity at z = ∞.
3.12 Some Important Theorems of Singularities
(i)
Let f ( z ) be a function such that for some positive integer m, a value φ (a) exists
and φ( a) ≠ 0 such that the function φ( z ) = ( z − a) m f ( z ) is analytic at a, then f
has pole of order m at a.
(ii) Let f ( z ) have a pole of order m at z = a, then the function φ (z) defined by
φ ( z ) = ( z – a) m f ( z ) has a removable singularity at a and that φ (a) ≠ 0.
(iii) Riemann's Theorem : Let a be an isolated singularity of f ( z ) and if f ( z ) is
bounded on some neighborhood of a, then a is a removable singularity.
1
,
(iv) If an analytic function f ( z ) has a pole of order m at z = a, then φ ( z ) =
f (z )
has a zero of order m at z = a; and conversely.
1
, where
(v) Let a function f ( z ) be analytic in a domain D and let φ ( z ) =
f (z )
f ( z ) ≠ 0. Then f ( z ) has a zero of order m at z = a if and only if φ has a pole of
order m at z = a.
(vi) Liouvill'e Theorem : A function which has no singularity in the finite part of
the plane or at infinity is constant.
Or
A function having no singularity anywhere in the extended complex plane is a
constant.
(vii) Polynomial of degree n has a pole of order n at infinity.
114
be a polynomial of degree n, i. e, f ( z ) = a0 + a1 z + a2 z 2
Proof. Let f ( z )
+.....+ an z n , an ≠ 0.
Replacing z by 1 / ζ, the corresponding polynomial
a
a
a
 1
F(ζ) = f   = a0 + 1 + 2 +.....+ n , an ≠ 0.
 ζ
ζ ζ2
ζn
which is Laurent's expansion of F(ζ) about ζ = 0, having finite number n of terms in
its principal part.
 1
∴ F(ζ) = f   has a pole of order n at ζ = 0. Hence f ( z ) has a pole of order n at z = ∞.
 ζ
3.13 Method for Detecting Singularities
A function f ( z ) at any point z = a, has a removable singularity or a pole or an
essential singularity has already been made in the foregoing articles of this chapter.
However, we restate those tests below:
I. Removable Singularity
If a single valued function is not defined at z = a, but lim f ( z ) exists finitely, then
z = a is a removable singularity.
z→ a
z 2 − a2
has removable singularity at z = a, because
z−a
z 2 − a2
lim
= 2a.
z→ a z − a
For example, the function
II. Poles
(a) If lim f ( z ) = ∞, then z = a is a pole of f ( z ).
z→ a
(b) If there are finite number of terms in the principal part of f ( z ).
In general, If there are n terms in the principal part i.e, if coefficient of
1
( z − a) n
is not zero then pole is of order n.
III. Essential Singularity
(a) If lim f ( z ) does not exist, then z = a is an essential singularity. For example,
z→ a
lim e1/( z − a ) does not exist, so f ( z ) has an essential singularity at z = a.
z→ a
(b) Limit point of zeros is an isolated essential singularity.
1 
For example, consider the function sin 
.
 z − a
1
Its zeros are given by sin
= 0 i. e.,
z−a
1
1
= nπ or z = a +
( n = ± 1, ± 2 ,..... )
z−a
nπ
Power Series and Calculus of Residues
115
Thus, we see that there is a sequence of zeros of f ( z ) which has z = a its limit
1 
point. Hence z = a is an essential singularity of the function sin 
.
 z − a
(c) Limit point of the poles is a non-isolated essential singularity.
1
For example, consider the function f ( z ) =
.
1 

sin 

 z − a
Putting the denominator equal to zero, the poles of f ( z ) are given by
1
1
sin
= 0 i. e., z = a +
( n = ±1, ± 2,..... )
z−a
nπ
1
1
1
Thus, we have a sequence of poles a + , a +
, a+
..... etc.
π
2π
3π
The limit of these poles is z = a. Hence z = a is a non-isolated essential
singularity of the given function.
(d) If there are infinite number of terms in the principal part of f ( z ).
1
1
1
1
For example, sin
=
− .
,+......
z − a z − a 3 ! ( z − a) 3
1
, there are infinite
z−a
number of terms in the negative powers of z − a or in other words are infinite
1
number of terms in principal part of the expansion of sin
. Hence z = a is
z−a
1
an isolated essential singularity of the function sin
.
z−a
Here we see that in the expansion of the function sin
(e) Singularities at infinity : Replacing z by 1 / ζ in f (z) we obtain the function
f (1 / ζ) = F(ζ ). Then nature of singularity at z = ∞ is defined to be the same as
that of F(ζ ) at ζ = 0.
Illustrative Examples
Ex. 22 : Find the order of all zeros of the given functions:
z2 − 2
(i) z 2 + 16
(ii)
z4
ez−a
(iii) (1 + e z )( z 2 + 9) 4
(iv)
( z − a) 2
Sol : (i) The zeros of f ( z ) = z 2 + 16 are given by z 2 + 16 = 0 or z = ± 4 i
Hence z = ± 4 i are simple zeros.
(ii) Zeros are obtained by z 2 − 2 = 0 or z = ± 2 are the simple zeros. z = ∞ is a
zero of second order.
116
To obtain this put z =
1
then
ζ
(1 / ζ 2 ) − 2
4
(1 / ζ )
=
1 − 2ζ 2
ζ
2
ζ 4 = ζ 2 (1 − 2ζ 2 )
Therefore ξ = 0 is a zero of second order, so z = ∞ is a zero of second order.
(iii) Here (1 + e z )( z 2 + 9) 4 = 0 gives the pole.
z = ± 3 i is a pole of order 4. z = 2nπ, n = 0, ± 1, ± 2,..... are zeros of first order.


ez−a
1
( z − a) 2
(iv)
=
1 + ( z − a) +
+.....

2
2
2!
( z − a)
( z − a) 

1
1
z−a
=
+
+
+.....
2!
( z − a) 2 z − a
We can observe that there are only two terms in negative powers of z − a. Hence, the
function has pole of order 2.
2
Ex. 23 : Show that e −1/ z has no singularities.
2
Sol : Here f ( z ) = e −1/ z .
2
The zeros of f ( z ) are given by e −1/ z = 0 or z 2 = 0 or z = 0 (repeated twice)
∴ z = 0 is a zero of order two.
Since there is no limit point of zeros, hence there is no singularity.
Also pole of f ( z ) are given by
2
1
= e1/ z = 0
f (z )
which is impossible for any value of z, real or complex. Therefore, there are no poles.
Hence, the given function has no singularities.
Obviously there are infinite number of terms in the principal part. Hence z = 0 is an
isolated essential singularity.
Ex. 24 : Find the kind of singularities of the following functions:
1
π
cot πz
(i)
(ii)
at z =
at z = 0 and z = ∞.
sin z − cos z
4
( z − a) 2
Sol : (i) Here f ( z ) =
1
.
sin z − cos z
Poles of f ( z ) are given by putting the denominator equal to zero.
i.e., by sin z − cos z = 0 or tan z = 1
π
or
z = nπ + ( n = 0, 1, 2 ..... )
4
Obviously z =
π
is a simple pole.
4
Power Series and Calculus of Residues
(ii) f ( z ) =
117
cot πz
( z − a) 2
Poles of f ( z ) given by {(sin ( πz )}( z − a) 2 = 0 ⇒ sin πz = 0 or ( z − a) 2 = 0
Now sin πz = 0 gives πz = nπ or z = n ( n = 0, ± 1, ± 2, ..... )
Obviously z = ∞ is the limit point of these poles, hence z = ∞ is a non-isolated
essential singularity.
( z − a) 2 = 0 gives z = a repeated twice. Hence z = a is pole of order 2
Ex. 25 : Find the kind of singularity of the following functions:
(i) ( z 2 + 4) / e z at z = ∞
(ii) cos z − sin z at z = ∞
1
(iii) sin
at z = 1
1− z
Sol : (i) Here
f ( z ) = ( z 2 + 4) / e z
Replacing z by
1
,
ζ
1
 1 
f   =  4 +  e ( − 1/ ζ )
 ζ 
ζ2 

1
1
1
1
1
 1 
f   =  4 +  1 − +
−
+
.....
2
2
3
4
 ζ 
ζ 2 !ζ

ζ 
3 !ζ
4 !ζ
or
 4
1 
= 4 + − + (1 + 2)
+  −1 −
ζ2 
 ζ
2 1  1
+ +

3 ζ 3  2

1 1
 4 +.....
6 ζ

 4 3

5
2
= 4 + − +
−
+
+.....
2
3
4
3ζ
3ζ
 ζ ζ

Since there are infinite number of terms containing negative powers of ζ
 1
∴ ζ = 0 is an isolated essential singularity of f   .
 ζ
Hence z = ∞ is an isolated essential singularity of f ( z ).
(ii) f ( z ) = cos z − sin z
Zeros of f ( z ) are given by cos z − sin z = 0 i.e, by tan z = 1,
∴ z = nπ +
π
, ( n = 0, ± 1, ± 2 ,..... )
4
Obviously z = ∞ is the limiting point of these zeros.
Hence z = ∞ is an isolated essential singularity of f ( z ) .
(iii) f ( z ) = sin
1
1− z
Zeros of f ( z ) are given by sin
1
1
= 0 i.e,
= nπ
1− z
1− z
118
or
z = 1−
1
,( n = 0, ± 1, ± 2,..... )
nπ
Obviously z = 1 is the limit point of these zeros.
Hence z = 1 is an isolated essential singularity of f ( z ).
Ex. 26 : What kind of singularity have the following functions :
1
1 − ez
(i) tan at z = 0
(ii)
at z = ∞
z
1 + ez
1
(iii) z cosec z at z = ∞
(iv) cosec   at z = 0.
 z
Sol : (i) f ( z ) = tan
1 sin(1/ z )
=
.
z cos(1/ z )
Poles of f ( z ) are given by cos (1/ z ) = 0 or 1/z = 2nπ + π / 2
1
or
z=
{ n = 0, ± 1, ± 2,.....}
 2n + 1 π



2
Obviously z = 0 is the limit point of these poles.
Hence z = 0 is non-isolated essential singularity.
(ii) f ( z ) =
1 − ez
1 + ez
.
Poles of f ( z ) are given by putting the denominator equal to zero.
i. e,
by 1 + e z = 0
or e z = −1 = eiπ = e 2nπi
∴
z = ( 2n + 1)πi { n = 0, ± 1, ± 2,....}
+ πi
Obviously z = ∞ is the limit point of the sequence of these poles.
Hence z = ∞ is a non-isolated essential singularity.
(iii) f ( z ) = z cosec z = z / sin z.
Poles of f ( z ) are given by sin z = 0 or z = nπ ( n = 0, ± 1, ± 2,.... )
Obviously z = ∞ is the limit point of the sequence of these poles.
Hence z = ∞ is a non-isolated essential singularity.
(iv) f ( z ) = cosec
1
=
z
1
sin
1
z
Poles of f ( z ) are given by sin
(1/ z ) = 0 or
1
= nπ
z
i. e,
z = 1 / nπ ( n = 0, ± 1, ± 2 ,..... )
Obviously z = 0 is the limit point of the sequence of these poles.
Hence z = 0 is a non-isolated essential singularity.
Power Series and Calculus of Residues
119
Ex. 27 : Show that the function e z has an isolated essential singularity at z = ∞ .
Sol : Let f ( z ) = e z .
To know the behavior of f ( z ) at z = ∞, let us know the behavior of f(1/ζ) at ζ = 0.
We have f (1/ζ) = e1/ ζ
 1

1
 1
Now lim f   = lim e1/ ζ = lim 1 + +
+..... = limit does not exist.
2
 ζ  ζ→ 0
ζ→ 0
ζ→ 0
 ζ 2!ζ

∴ ζ = 0 is an essential singularity of e1/ζ .
Hence z = ∞ is an isolated essential singularity of e z .
Ex. 28 : Show that the function e1/ z actually takes every value except zero, an
infinity of times in the neighbour of z = 0 .
Sol : Let f ( z ) = e1/ z .
Here it is required to prove that this function has an isolated essential singularity at z = 0 .
The function f ( z ) = e1/ z is analytic except at z = 0, that is f ( z ) is analytic for
r <z< R.
Thus, f ( z ) is analytic in the annulus r <z< R., where R is finite and r is small,
hence it can be expanded in a Laurent's series. Thus,
∞
∞
1 1
1 1
f ( z ) = e1/ z = ∑
.
= 1+ ∑
.
n
n
n=0 n ! z
n =1 n ! z
∴ f ( z ) has an isolated essential singularity at z = 0.
Ex. 29 : Find out the zeros and discuss the nature of singularities of
f (z ) =
z−2
z
2
sin
1
.
( z − 1)
( z − 2) sin
Sol : Here
where
f (z ) =
z
2
1
( z − 1)
=
P( z )
Q( z )
1 
2
P( z ) = ( z − 2) sin 
 and Q( z ) = z .
 z − 1
1 
Now the zeros of f ( z ) are given by P( z ) = 0 or ( z − 2) sin 
=0
 z − 1
⇒
or
1 
z = 2 or sin 
 = 0 = sin nπ
 z − 1
1
1
= nπ giving z = 1 +
( n = 0, ± 1, ± 2 ,..... )
( z − 1)
nπ
120
Hence the zeros of f ( z ) are z = 2, z = 1 +
The limiting point of zeros z = 1 +
1
( n = 0, ± 1, ± 2,..... )
nπ
1
,( n = 0, ± 1, ± 2, ..... ) is z = 1
nπ
∴ z = 1 is an isolated essential singularity.
Also the poles of f ( z ) are given by Q( z ) = 0 or z 2 = 0 or z = 0 (repeated twice)
∴ z = 0 is a pole of order two of f ( z ).
Problem Set (3.4)
Determine zeros and order of zeros of the following functions.
1. e z
2. ( z 4 + 16) 3
1 − cot z
3.
z
4. sin z
5. Show that cosec z has a simple pole at z = 0.
6. Determine and classify the singularities of the following functions.
ez
1 1
(i)
(ii) e tan(1/ z )
(iii) cot −
2
z z
1+ z
7. Find the kinds of the singularity of the following functions.
1
1
(i) sin
(ii)
at
z=0
at z = 0
z
cos(1 / z )
1
(iii) cot z at z = ∞
(iv)
at z = 2πi
1 − ez
(v) cos z − sin z at z = ∞
Objective Type Questions
Multiple Choice Questions
Tick the correct answer
1. The function f ( z ) =
1
zn
(a) analytic everywhere
is :
(b) singularities at z=0
(c) singularities at n =0
(d) none of these.
2. The zero of first order is known as :
(a) complex pole
(b) simple pole
(c) singular point
(d) none of these.
3. If f (z) is entire, then
(a) f ( z ) is analytic for all
(b) f ( z ) has finite singular points
Power Series and Calculus of Residues
121
(c) f ( z ) diverges for all z
(d) none of these.
4. A point z = z 0 is singular point of analytic function f ( z ), if at z = z 0
(a) f ( z ) is analytic
(b) f ( z ) is not analytic
(c) f ( z ) =0
(d) none of these .
∞
5. Let f ( z ) =
∑ an (z − z 0 )n be analytic in a domain. If a0 = a1 = a2 = … = am −1 = 0 and
n=0
am ≠ 0, then f ( z ) is said to have a :
(a) pole of order m − 1 at z = z 0
(b) zero of order m − 1 at z = z 0
(c) pole of order m at z = z 0
(d) zero of order m at z = z 0 .
1
6. Number of zeros of the function f ( z ) = sin   is :
 z
(a) 3
(b) 4
(c) infinite
(d) none of these.
1
7. Number of poles of the function f ( z ) = tan is :
z
(a) 2
(b) 4
(c) infinite
(d) none of these.
z + 30
8. The number of isolated singular points of f ( z ) =
is :
2 2
z ( z + 12)
(a) 1
(b) 2
(c) 3
(d) 4.
9. For the function f ( z ) = e z , z = ∞ is :
(a) isolated essential singularity
(b) pole
(c) ordinary point
(d) none of these.
z − sin z
10. For the function f ( z ) =
, z = 0 is :
z3
(a) essential singularity
(b) pole
(c) removal singularity
(d) none of these.
11. z = 0 for the function f ( z ) = log z is :
(a) isolated singularity
(b) pole
(c) non-isolated singularity
(d) none of these.
Fill in the Blanks
1. A zero of order one is said to be a .......... zero.
2. If there are infinite number of terms in the principal part of f ( z ) at z = z 0 then
z 0 is called an ..................... of f ( z ).
3. If f ( z ) is analytic at each point of some neighbourhood of a singularity z 0 , then
z 0 is called an ..................... singularity.
2z
4. f ( z ) =
has an .................... at z = − 3.
2 2
z ( z + 3)
122
1
has an ....................... singularity at z = 0.
z
6. If the principal part of f ( z ) at z = z 0 consists of a finite number of terms, say
m, z 0 is said to be a ............... of order m of f ( z ) .
5. f ( z ) = sin
7. f ( z ) =
z
has a pole of order ................... at z =1.
z –1
8. A function which has no singularity in the finite part of the plane or at infinity is
.......... .
z+3
9. f ( z ) =
has an ..................... at z = 0.
2 2
z ( z + 2)
1
10. The function f ( z ) = ( z − 4) sin
has ..................... singularity at z = 4.
( z − 4)
11. The function f ( z ) = e1/ z has ..................... singularity at z = 0.
2
12. The function f ( z ) = e z /( z − 1) 3 has a pole of order ................ at z = 1.
13. The function f ( z ) =
sin z
has ................ singularity at z = 0.
z
State True/False
1. Poles are not isolated.
2. z = 0 is an isolated essential singularity of the function e1/ z .
3. If a function has no singularity in the finite part of the plane and has a pole of
order n at infinity is a polynomial of degree n.
4. The order of a zero of a polynomial is not equal to the order of its first
non-vanishing derivative.
ANSWERS
Problem Set 3.4
1.
no zero
2.
z = ± 2 i, ± 2 i are poles of order
3.
z=
4.
6.
simple zeros at z = 0, ± π, ± 2π, ......
(i) z = ± i simple pole, z = ∞ essential singularity
(ii) z = 2[( 2n + 1)π], n = 0, ± 1, ± 2,.........., essential singularity
1
(iii) z =
, n = ± 1, ± 2,.... are poles of first order, z = 0 is a limit point for
nπ
poles, z = 0 is a non-isolated singular point, z = ∞ is a pole of order one.
π
+ nπ, n = 0, ± 1, ± 2 simple zeros
4
Power Series and Calculus of Residues
7.
123
(i) isolated essential singularity
(ii) non-isolated essential singularity
(iii) non-isolated essential singularity
(iv) simple pole
(v) isolated essential
Multiple Choice Questions
1. (b)
2. (b)
3. (a)
4. (b)
5. (d)
6. (c)
7. (c)
8. (c)
9. (a)
10. (c)
11. (c)
Fill in the Blanks
1. simple
2. isolated essential singularity
3. isolated
4. isolated singularity
5. isolated essential
6. pole
7. one
8. constant
9.
isolated singularity
11. essential
10. essential
12. 3
13. removable
True or False
1. False
2. True
3. True
4. True
124
3.14 Residues
The coefficient of ( z − a) –1 in the expansion of an analytic
a2
C2
C1
a1
D
function f ( z ) around an isolated singularity is called the
residue of f ( z ) at that point. Thus, from Laurent's series, the
a3
residue f ( z ) at z = a is b1 which is given by
1
Residue f ( a) =
2πi
∫ f (z )dz = 2πi
⇒
C3
....(3.1)
∫ f (z )dz
C
Cn
an
Fig. 3.6
Res f ( a),
C
where C is any closed contour enclosing a.
3.15 Residue Theorem
If f ( z ) analytic in a closed curve C except at a finite number of singular points within
C, then
∫
f ( z )dz = 2πi ( sum of the residues at the singular points within C).
C
Proof. Let us surround each of the singularities a1 , a2 ,...., an by the small circles
such that it encloses no other singular point. Then these circles C1 , C 2, .........., C n
together with C, form a multiple connected region where f ( z ) is analytic.
Then applying Cauchy's theorem, we get
a1 C1
∫ f (z )dz = ∫
C
f ( z )dz +
C1
a2 C2
an Cn
∫ f (z )dz +.....+ ∫ f (z )dz
C2
Cn
= 2πi[ Res f ( a1 ) + Res f ( a2 )+..... Res f ( an )]
3.16 Calculation of Residues
(1) If f (z ) has a simple pole ( i.e., pole of order 1) at z = a, then
Res f ( a) = Lim( z − a) f ( z )
z→ a
Laurent's series in this case is given by
f ( z ) = a0 + a1 ( z − a) + a2 ( z − a) 2 +.....+ b1 ( z − a) −1
Then multiplying throughout by z − a , we get
( z − a) f ( z ) = a0 ( z − a) + a1 ( z − a) 2 +.....+ b1
Now taking limits as z → a, we get
lim[( z − a) f ( z )] = b1 = Res f ( a)
z→ a
Aliter to calculate Res f ( a)
Let f ( z ) = φ ( z )/ ψ( z ), where ψ( z ) = ( z − a)F( z ), F( a) ≠ 0
... (3.1)
Power Series and Calculus of Residues
Then [ lim ( z − a) φ ( z )/ ψ ( z )] = lim
z→ a
⇒
125
z→ a
( z − a)[ φ ( a) + ( z − a) φ' ( a)+.....]
ψ( a) + ( z − a) ψ' ( a)+.....
[By Taylor's Theorem]
φ ( a) + ( z − a) φ' ( a) +.....
, since ψ( a) = 0
= lim
z→ a ψ' ( a) + ( z − a) ψ' ' ( a) +....
φ ( a)
Res f ( a) =
ψ' ( a)
(2) If f (z ) has a pole of order n at z = a, then
 d n −1

1
n
Res f ( a) =
 n −1 [( z − a) f ( z )]
( n − 1)!  dz

z=a
We have f ( z ) = a0 + a1 ( z − a) + a2 ( z − a) 2 +.....+ b1 ( z − a) −1 +.....+ bn ( z − a) − n .
Then multiplying throughout by ( z − a) n , we have
( z − a) n f ( z ) = a0 ( z − a) n + a1 ( z − a) n +1 + a2 ( z − a) n + 2 +.....
+ b1 ( z − a) n −1 + b2 ( z − a) n − 2 +.....+ bn
Now differentiating both sides w.r.t. z, n–1 times and putting z = a, we get
 d n −1

n
= ( n − 1)! b1
 n −1 [( z − a) f ( z )]
 dz
 z=a
or
Res f ( a) =
1
d n –1
[( z – a) n f ( z )]z = a
( n – 1)! dz n –1
Note : Residue at a pole z = a of any order (simple or of order n) is given by
1
Res f ( a) = coefficient of , where t = z − a.
t
Let f ( z ) have a pole of order m, then by Laurent's theorem
f ( z ) = a0 + a1 ( z − a) + a2 ( z − a) 2 .....+
b1
b2
bm
+
+..... +
2
z − a ( z − a)
( z − a) m
Putting z − a = t , i. e, z = a + t , we get
f ( a + t ) = a0 + a1t + a2t 2 +
b1
b
b
+.....+ 2 +.....+ m
t
t2
tm
1
Now Res f ( a) = b1 or Res f ( a) = coefficient of .
t
To obtain residue at z = a put z = a + t in f ( z ) and expand it in powers of t. Then
coefficient of (1/t) gives the residue of f ( z ) at z = a.
126
Ex. 30 : Find the residue of
Sol : Let f ( z ) =
z2
z2
(z 2 + a 2 )
=
(z 2 + a 2 )
at its poles.
z2
( z – ai )( z + ai )
The poles of f ( z ) are given by ( z − ai )( z + ai ) = 0
∴
z = ai and z = − ai are simple poles of f ( z ).
Residue at the simple pole ( z = ai )
z2
( ai ) 2 1
=
= ai.
z→ ai ( z + ai )
2ai
2
= lim ( z − ai ) f ( z ) = lim
z→ ai
Similarly residue at simple pole z = − ai is −
Aliter. Let
f (z ) =
z2
2
z +a
2
=
1
ai.
2
φ (z )
where φ ( z ) = z 2 and ψ ( z ) = z 2 + a 2 .
ψ (z )
∴ Residue at the simple pole ( z = ai )
=
φ ( ai )
( ai ) 2
( ai ) 2 1
=
=
= ai.
ψ' ( ai ) [ 2z]z = ai
2( ai ) 2
Similarly , we get residue at (z = − ai ) = −
Ex. 31 : Find the residue of
Sol : Let f ( z ) =
1
ai.
2
z 2 − 2z
( z + 1) 2 ( z 2 + 4)
z 2 − 2z
( z + 1) 2 ( z 2 + 4)
=
at all its poles in the finite plane.
z 2 − 2z
( z + 1) 2 ( z + 2i )( z − 2 i )
The poles of f ( z ) are given by ( z + 1) 2 ( z + 2 i )( z − 2 i ) = 0 i.e, z = 2 i, − 2 i, − 1, − 1.
∴ f ( z ) have simple poles at z = 2 i and z = −2 i and a double pole at z = −1.
Residue at simple pole z = 2 i
= lim ( z − 2 i ) f ( z ) = lim
z→ 2 i
=
z 2 − 2z
z→ 2 i
( 2 i ) 2 − 2.2 i
2
( 2 i + 1) ( 2 i + 2 i )
=
( z + 1) 2 ( z + 2 i )
7 +i
25
Residue at simple pole z = −2 i
= lim ( z + 2 i ) f ( z ) = lim
z→ –2 i
=
z → −2 i
( −2 i ) 2 + 2.2 i
2
( −2 i + 1) ( −2 i − 2 i )
=
z 2 − 2z
( z + 1) 2 ( z – 2 i )
7 −i
25
Power Series and Calculus of Residues
127
To find residue at double pole at z = −1.
we can write
f (z ) =
φ (z )
( z + 1) 2
, where φ ( z ) =
z 2 − 2z
( z 2 + 4)
∴ Residue at double pole at z = −1
1 d ( z + 1) 2 ( z 2 − 2z )
Res f ( −1) =


1 ! dz  ( z + 1) 2 ( z 2 + 4) 
z = −1
 d  z 2 − 2z  
 2z 2 + 8z − 8
14
= 
=
=−
.


2
2
2
25
 dz  z + 4   z = −1  ( z + 4)  z = −1
Aliter. Putting z = −1 + t in the given function f ( z ), we have
f ( z ) = f ( −1+ t ) =
( −1 + t ) 2 − 2( −1 + t )
( −1 + t + 1) 2 {( −1 + t ) 2 + 4}
2t – t 2 
1
4 3
= 1 − +  1 −

5
t t2 
5 
=
t 2 − 4t + 3
t 2 (t 2 − 2 t + 5)
−1
2


2t – t 2  2t – t 2 
1
4 3
= 1 − +  1 +
+
 +.....
5
t t2 
5

 5 


1
4 3
2
 4 − 1 t 2 +.....
...(3.1)

1 − + 2  1 + t + 

5
t t 
5
25 5

1
1
6
14
∴ Residue at ( z = −1) = coefficient of in the expansion (1) =  −4 +  = −
.
t
5
5
25
=
Ex. 32 : Find the residue at the pole of the function
Sol : f ( z ) =
z4
(c 2 + z 2 )4
=
z4
(c 2 + z 2 )4
.
z4
( z − ic ) 4 ( z + ic ) 4
Clearly z = ic and z = − ic are the two poles of f ( z ) each of order 4.
To find residue at the pole z = ic (of order 4), put z = ic + t in f ( z ) where t is very small
∴
f ( z ) = f (ic + t ) =
=
1
4 4
16c t
(ic + t ) 4
t 4 (t + 2 ic ) 4
=
(ic + t ) 4 . 1
t4
( 2 ic ) 4
t 

1 +


2 ic 
−4
(t 4 + 4 t 3 . ic + 6 t 2. i 2 c 2 + 4 t . i 3 c 3 + i 4 c 4 )
it
× 1− 
 2c 
=

4 ic 6c 2 4 ic 3 c 4 
−
−
+ 
1 +
t
16 c 4 
t2
t3
t4
1
−4
128


4 ⋅ it 10 ⋅ i 2t 2 20 ⋅ i 3t 3
× 1 +
−
+
+......
2c


4c2
8c 3
=

4 ic 6c 2 4 ic 3 c 4 
−
−
+ 
1 +
t
16c 4 
t2
t3
t4
1


2 it 5t 2 5 it 3
× 1 +
−
−
+.....
c


2c 2
2c 3
1
in the expansion (3.1)
t
1 
5ic 
i
=
 4 ic − 12 ic + 10 ic −
 =−
4 

2
16c
32 c 3
...(3.1)
∴ Residue at the pole ( z = ic ) = coefficient of
...(3.2)
Replacing c by – c in (3.2), the residue at the pole z = − ic which is also a pole of order
i
4, is
.
32c 3
Ex. 33 : Find the residues of f ( z ) = e z cosec 2 z at all its poles in the finite plane.
Sol : We have f ( z ) =
ez
sin 2 z
.
The poles of f ( z ) are given by sin 2 z = 0 or sin z = 0 (twice).
Hence z = mπ, where m = 0, ± 1, ± 2,..... are the poles of f ( z ) each of order 2, if m is
finite. The limit point of the poles is z = ∞ which is, therefore, a non-isolated
essential singularity.
Putting z = mπ + t in f ( z ), we have
e mπ + t
e mπ . e t
f ( z ) = f ( mπ + t ) =
=
sin 2 ( mπ + t )
sin 2 t
=


t2
e mπ 1 + t +
+....
2!


 t3 t5

+
.....
t −
3! 5!


2
= e mπ

1 
t2
.....
1 + t +
2
2
!

t 
 t 2 t 4

−
..... 
1 − 
 
  3 ! 5 !
= e mπ

1 
t2
+......
1 + t +
2
2!

t 
–2
2


t 2 t 4

t 2 t 4

× 1 + 2 
−
.... + 3 
−
.... +.....


 3! 5!

 3! 5!



Residue at ( z = mπ ) = coefficient of
...(3.1)
1
in the expansion (3.1) = e mπ .1 = e mπ .
t
Power Series and Calculus of Residues
129
Ex. 34 : Evaluate the following integral using residue theorem
1+ z
∫ z(2 − z ) dz, where
C
C is the circlez= 1.
Sol : The poles of the integrand are given by : z( 2 − z ) = 0 ⇒ z = 0, 2 .
Hence, the integrand is analytic at all points insidez= 1 except z = 0.
1+ z
Then by residue theorem
dz = 2πi [ Res f ( 0)]
z( 2 − z )
∫
....(3.1)
C
Now
Res f ( 0) = lim z.
Hence
∫
z→ 0
C
1+ z
1+ z 1
= lim
=
z( 2 − z ) z → 0 2 − z 2
1+ z
 1
dz = 2πi Res f ( 0) = 2πi   = πi
 2
z( 2 − z )
Ex. 35 : Evaluate
∫
z−3
dz where C is the circle
2
C z + 2z + 5
(i)z = 1
(ii)z + 1 − i = 2
Sol : The poles of
f (z ) =
z=
⇒
z−3
2
z + 2z + 5
(iii)z + 1 + i = 2
are given by z 2 + 2z + 5 = 0
−2 ± ( 4 − 20)
= −1 ± 2 i
2
(i) Both the poles z = −1 + 2 i and z = −1 − 2 i lie outside the circlez = 1. Therefore,
f ( z ) is analytic everywhere within C.
Hence, by Cauchy's theorem, we have
z−3
∫
2
z + 2z + 5
C
dz = 0.
(ii) Since one pole z = −1 + 2 i lies inside the circle C : z + 1 − i = 2 . Therefore, f ( z )
is analytic within C except at this pole.
Now
Res f ( −1 + 2i ) =
=
lim
[{ z − ( −1 + 2 i )} f ( z )]
z → − 1+ 2 i
lim
z → − 1+ 2 i
( z + 1 − 2 i )( z − 3)
2
z + 2z + 5
− 4 + 2i
1
=
=i+ .
4i
2
Whence by residue theorem , we have
f ( z ) dz = 2πi Res f ( −1 + 2 i )
∫
C
= 2πi(i + 1/ 2) = π(i − 2)
=
lim
z → − 1+ 2 i
z−3
z + 1 + 2i
130
Engineering Mathematics-III
(iii) The pole z = −1 − 2 i lies inside the circle C :z + 1 + i = 2 . Therefore, f ( z ) is
analytic within C except at this pole.
Now
Res f ( −1 − 2 i ) =
=
( z + 1 + 2 i )( z − 3)
lim
z → − 1− 2 i
lim
z → − 1− 2 i
z 2 + 2z + 5
− 4 − 2i 1
z−3
=
= −i
z + 1 − 2i
− 4i
2
whence by residue theorem, we have
1

∫ f (z )dz = 2πi Res f (−1 − 2i ) = 2πi  2 − i = π(2 + i ).
C
Ex. 36 : Evaluate
∫
C
Sol : Let f ( z ) =
sin πz 2 + cos πz 2
( z − 1) 2 ( z − 2)
sin πz 2 + cos πz 2
( z − 1) 2 ( z − 2)
dz, where C is the circlez = 3.
. Then function is analytic within the circlez = 3
excepting the poles z = 1 and z = 2. Here z = 1 is a pole of order 2 while z = 2 is a
simple pole.
Now
Res f (1) =
 d  sin πz 2 + cos πz 2  
1d

{( z − 1) 2 f ( z )}
=  



1 !  dz
z−2
 z =1  dz 
  z =1
( z − 2)( 2πz cos πz 2 − 2πz sin πz 2) − (sin πz 2 + cos πz 2)
=

( z − 2)2

 z =1
= ( −1)( −2π) − (−1) = 2π + 1
and
Res f ( 2) = lim [( z − 2) f ( z )] = lim
z→ 2
sin πz 2 + cos πz 2
( z − 1) 2
z→ 2
= 1.
Then by residue theorem , we get
∫ f (z )dz = 2πi [Res f (1) + Res f (2)] = 2πi(2π + 1 + 1) = 4π (π + 1) i
C
Ex. 37 : Evaluate the complex integral
coth z
∫ (z − i ) dz, C :z= 2.
C
Sol : We have
∫
C
coth z
dz =
(z − i )
∫
C
ez + e−z
( e z − e − z )( z − i )
dz
The poles of the integrand are given by ( e z − e − z )( z − i ) = 0
⇒ e z − e − z = 0 and z – i = 0 i. e, e 2z = 1 and z = i ⇒ z = 0 and z = i,
Both the poles are inside C :z= 2.
Power Series and Calculus of Residues
Residue at (z = i ) = lim ( z − i )
z→i
131
ez + e−z
( e z − e − z )( z − i )
=
e i + e −i
e i − e −i
= coth i
 φ ( 0) 
Residue at ( z = 0) = 

 ψ' (0) 
1+1
−1
= 0−i =
=i
1+1
i

ez + e−z
ez + e−z
 φ (z )
φ (z )
∴
= z −i ,
= z −i
z
−
z
,
ψ' ( z ) e z + e − z
 ψ (z ) e − e

Now sum of the residues = coth i + i.
Hence, by residue theorem
∫ f (z ) = 2πi × Sum of the residues
C
⇒
∫
C
coth z
dz = 2πi [coth i + i].
z −i
Ex. 38 : Evaluate the complex integral
∫
C
Sol :
f (z ) =
dz
, where C isz= 2.
cosh( z )
1
2
2e z
=
=
cosh( z ) e z + e − z
e 2z + 1
Poles of f ( z ) are given by e 2z + 1 = 0 or ( e z + i )( e z − i ) = 0 ⇒ e z = i, − i
π
−πi
i and
2
2
π
−πi
The poles which lie within the contourz= 2 are z = i and
2
2
πi 

∴ Residue of f ( z ) at z = 

2
⇒
e z = e πi / 2 , e z = e − πi / 2 ⇒ z =




2e z
=

d
2
z
 ( e + 1)
 dz
z= π i
φ ( a) 

∵ Residue = ψ' ( a)


2
z
 2e 
= 
= {e − z }
2z 
z=
 2e  z = π i
2
π
i
2
= e − πi / 2 = − i
πi 

and Residue of f ( z ) at z = – 

2




2e z
=
= { e − z } − π = e πi / 2 = i

d
z=
i
 ( e 2z + 1)
2
−
π
 dz
 z= i
2





132
Engineering Mathematics-III
∫
⇒
f ( z )dz =
C
dz
∫ cosh z = 2πi [sum of residues]
C
Ex. 39 : Evaluate
∫
C
= 2πi(i − i ) = 0
dz
where C is the unit circle about origin.
z sin z
1
1
1
=
=
3
5
2
z sin z




z
z
z
z4
z z −
+
−..... z 2 1 −
+
−.. ...
3! 5!
3! 5!




Sol :
1 
=
1 –
z 2 
z2

z4


 6 − 120 ..... 

 
−1
2


1  z2
z4  z2
z4 





=
1+ 
−
+
−
.....
 


z 2   6 120  6 120



1 
z2
z4
z4
1
1 z2
z2
1+
−
+
+..... =
+ −
+
.....

2
6 120 36
6 120 36
z2 
 z
1
1
7 2
=
+ +
z .....
2
6
360
z
=
∴ z = 0 is a pole of order 2 for the function
1
and the residue at the pole is zero,
z sin z
(because coefficient or (1/z) is 0) and the pole at z = 0 lies within C .
1
Hence
dz = 2πi . 0 = 0.
z sin z
∫
C
Problem Set (3.5)
Find the residues of the following functions at each pole.
1.
3.
5.
7.
3e z
2
2
2. z /( z − 1) ( z + 2)
z4
z
4.
4
z −1
z 2 − 2z
( z + 1) 2 ( z 2 + 1)
2z − 3
3
z + 3z
2
,
which lie inside the circlez= 2
6.
8.
cos z
z5
zeiz
z 2 + a2
1 − e 2z
z4
Power Series and Calculus of Residues
9.
3z + 6
2
,
( z + 1)( z + 16)
133
10.
z4
(c 2 + z 2 )4
which lie inside the circlez= 2
11.
13.
z3
( z − 1) 4 ( z − 2)( z − 3)
cot πz
12.
ze z
( z − a) 3
14. cot z
( z − a) 2
Evaluate the following integrals.
15.
∫
C
4 − 3z
dz,
z( z − 1)( z − 2)
16.
C
3
where C is the circlez=
2
17.
∫
C
ez
z
e −1
dz, where C:z= 1
z2
∫ (z − 1)2 (z + 2) dz,
where C:z= 2 ⋅ 5
18.
z
∫ (z − 1)(z − 2)2 dz,
C
where C:z − 2= 0.5
 1
19.
∫ C sin z  dz, where C:z= 1
21.
∫C e
− 1/ z
20.
sin (1/ z ) dz where C:z= 1 22.
z dz
∫ C sin z (1 − cos z ), where C:z= 5
2dz
2
z + 4i z − 1
, where C:z= 1.
3.17 Evaluation of Real Definite Integrals by Contour Integration
The evaluation of real definite integrals, can easily be obtained by using residues
theorem. For this we take a suitable closed contour C and find the residues of the
function f ( z ) at all its poles which lie within C. Then using residues theorem, we
have
∫
f ( z )dz = 2πi( ΣR + ) , where ∑ R + is the sum of residues at all its poles with
C
in R.
The process of integration along the contour is called contour integration.
3.18 Integration Round the Unit Circle
2π
i.e, Evaluation of the integrals of the type
∫
φ(cos θ, sin θ) dθ,
0
where, φ(cos θ sin θ) is a rational function of cos θ and sin θ.
To evaluate the above integral we take a unit circle C with its centre at the origin
z = 0. Now for any point z on the circle, we can write z = eiθ so that
dz
dz = ieiθ dθ = iz dθ i.e., dθ =
, 0 ≤ θ ≤ 2π
iz
134
Engineering Mathematics-III
Thus
cosθ =
eiθ + e −iθ
1
1
= z + 
2
2
z
sin θ =
eiθ − e −iθ
1 
1
=
z − 
2i
2i 
z
Substituting these values in the given integral, we have
2π
∫ φ (cos θ, sin θ) dθ = ∫
0
C
=
∫
1 1 
1  1
1 
φ   z +  ,  z −   dz
z  2i 
z   iz
2 
f ( z )dz, (say), where f ( z ) is a rational function of z.
C
= 2πΣR + ,
[By residue theorem]
+
where ΣR is the sum of residues of f ( z ) at all its poles inside the unit circle C, whose
centre is at the origin z = 0.
Illustrative Examples
2π
Ex. 40 : Use residue calculus to evaluate the following integral
∫
0
2π
Sol : Let I =
∫
0
putting
we have
1
dθ =
5 − 4 sin θ
2π
∫
0
5− 4
eiθ = z ⇒ dθ =
I =
∫
C
=
∫
C
2π
1
e
iθ
−e
2i
−iθ
dθ =
∫
0
dθ
5 + 2i e
Only pole z =
z=
iθ
− 2 i e – iθ
dz
,
iz
1
dz
, where C is the unit circle z = 1.
2 i iz
5 + 2 iz −
z
dz
5iz − 2z 2 + 2
Poles of integrand are given by −2z 2 + 5iz + 2 = 0
⇒
1
dθ.
5 − 4 sin θ
−5i ± −25 + 16 −5 i ± 3i
i
=
= 2 i,
−4
−4
2
i
lies inside C.
2
i

∴ Residue at the simple pole at  z = 

2
Power Series and Calculus of Residues
135
i 
1


= lim  z −  × 

i

2
( 2z − i )( − z + 2 i )
z→
2
=
lim
i
z→
2
1
=
2(− z + 2i )
1
1
=
 i
 3i
2  − + 2 i
 2

Hence by residue theorem I = 2πi × sum of residues within the contour C.
1
2π
= 2πi ×
=
.
3i
3
This example can alternatively be solved by the method given in the following
examples.
π
Ex 41 : Evaluate
∫
0
ad θ
2
a + sin 2 θ
, a > 0.
π
Sol : Let
I =
2
2
a + sin θ
0
2π
∫
0
=
∫
C
∫
0
ad φ
2a + 1 − cos φ
0
∫
=
2
2π
=
π
ad θ
∫
=
[UPTU 2002]
2ad θ
2
2a + 1 − cos 2θ
[putting 2θ = φ, ⇒ 2dθ = dφ]
2ad φ
2
2( 2a + 1) − ( eiφ + e −iφ )
2a
. dz
1

 iz
2( 2a 2 + 1) −  z + 

z
[writing eiφ = z then d φ =
=
∫
C
where
f (z ) =
2ai
2
z − 2 ( 2a 2 + 1)z + 1
2ai
2
dz
, where C is a unit circle ]
iz
2
z − 2( 2a + 1)z + 1
dz =
∫
f ( z )dz
C
= 2πiΣR +
...(3.1)
(By residue theorem)
+
where ΣR = sum of residues at poles within C.
The poles of f ( z ) are given by z 2 − 2( 2a 2 + 1)z + 1 = 0.
∴
z = (1 + 2a 2 ) + 2a (1 + a 2 ) = α (say)
and
z = (1 + 2a 2 ) − 2a (1 + a 2 ) = β (say)
Both are simple poles. Here αβ = 1 or αβ= 1 or αβ= 1 and sinceα> 1 ∴β< 1
∴ The pole z = β of f ( z ) lies within C, which is a simple pole.
136
Engineering Mathematics-III
∴ Residue at simple pole
( z = β ) = lim ( z − β ) f ( z ) = lim ( z − β )
z→ β
z→ β
i
=−
2
2ai
2ai
=
( z − α )( z − β ) β − α
= ΣR +
2 (1 + a )
π
∴ from (3.1), we have
I =
∫
0
2π
Ex. 42 : Prove that
∫
0


i
= 2πi  −
 =
2
a 2 + sin 2 θ
 2 (1 + a ) 
ad θ
2π
I =
∫
0
where f ( z ) =
(1 + a 2 )
.
sin 2 θ dθ
2π
=
{ a − ( a 2 − b 2 )} if a > b > 0.
a + b cos θ b 2
Sol : If we write z = eiθ , then cos θ =
and so
π
1
1
1 
1
dz
 z +  , sin θ =
 z −  , dθ =
2
z
2i 
z
iz
sin 2 θ d θ
1
=
a + b cos θ 2b
i( z 2 − 1) 2 dz
=
2a

2 2
Cz  z +
z + 1


b
∫
∫ f (z ) dz
C
i( z 2 − 1) 2
and C denotes the unit circle z= 1 = 2πi ΣR +
2
a


2bz 2  z 2 +
z + 1


b
...(3.1)
+
where ΣR = sum of residues at the poles within C.
2a


The poles of f ( z ) are given by z 2  z 2 +
z + 1 = 0, or z 2 ( z − α )( z − β ) = 0


b
where
i.e,
α=
− a + (a 2 − b 2 )
− a − (a 2 − b 2 )
,β=
b
b
z = 0 is a double pole and z = α, z = β are the simple poles of f ( z ).
Obviously β> 1, so β lies outside C, since α β = 1,
 α < 1, so α lies inside C.
∴
Hence inside the contour C there is a simple pole z = α and a pole z = 0 of order two.
Residue at the simple pole z = α is
lim ( z − α ) f ( z ) = lim ( z − α ).
z→ α
z→ α
i
( z 2 − 1) 2
2b z 2 ( z − α )( z − β )
1

α − 
i (α 2 − 1) 2
i 
α
=
=
2b α 2 (α − β ) 2b α − β
2
=
i (α − β ) 2
2b α − β
1


 because αβ = 1, so = β


α
Power Series and Calculus of Residues
=
137
i
i 2 (a 2 − b 2 ) i (a 2 − b 2 )
(α − β ) =
=
2b
2b
b
b2
and residue at the double pole z = 0 is the coefficient of
1
i
in
z
2b
( z 2 − 1) 2
,
2a


z 2 z 2 +
z + 1


b
where z is small.
i
Now,
2b
( z 2 − 1) 2
i 
1
=
1 − 2 
2a

 2b 
z 
z 2 z 2 +
z + 1


b
=
i 
2
1
1 − 2 + 4 

2b
z
z 
2
  2a 1
1 
× 1 +  . +

2 
  b z z 
−1
2
  2a 1

1   2a 1
1
1−  . + 2 +  . + 2 . . . . .




b z z
b z z


1
i  2a
ia
=
−  = − 2
z 2b  b 
b
∴ Residue at pole z = 0 which is the coefficient of
I = 2πi (sum of residucs)
ia 
π
 i
∴ from (3.1), we have I = 2πi 
(a 2 − b 2 ) −
=2
{ a − ( a 2 − b 2 )}
2
2
2
b
b 
b
Ex. 43 : Use the reside theorem to show that
2π
∫
0
dθ
( a + b cos θ)
2
=
2πa
2
( a − b 2 ) 3/ 2
2π
Sol : Let
I =
∫
0
writing eiθ = z, dθ =
dz
zi
I=
∫
2π
dθ
( a + b cos θ)
2
∫
C
f (z ) =
∫
=
1
1 
1 
 a + b z +  

2
z 

− 4 iz dz
2
2
( bz + 2az + b)
2
dθ
 a + 1 b ( eiθ + e −iθ )


2


0
C
=
where
, where a > 0, b > 0, a > b.
2
dz
where C is the unit circle z= 1
iz
=
∫
f ( z )dz = 2πiΣR + ...(3.1)
C
− 4iz
2a
b 2 (z 2 +
z + 1) 2
b
and ΣR + = sum of residues at the poles within C.
2a


The pole of f ( z ) are given by  z 2 +
z + 1


b
where
α=
2
= 0 or ( z − α ) 2 ( z − β ) 2 = 0
− a + (a 2 − b 2 )
− a − (a 2 − b 2 )
,β=
b
b
138
Engineering Mathematics-III
Therefore, z = α, z = β are the poles of f ( z ) each of order 2.
Clearly β > 1 since a > b and α β= 1 or α β = 1 ∴α | < 1.
i.e, the double pole z = α is the only pole within C.
− 4iz
φ (z )
4 iz
We can write f ( z ) =
=
, where φ ( z ) = –
2
2
2
2
2
b ( z − α ) ( z − β)
(z − α )
b (z − β )2
 d − 4iz 
d
Residue at the double pole z = α is =  φ ( z )
=
 dz

2
z =α
 dz ( z − β)  z =α
4 i ( z − β ) 2 − 2z( z − β) 
4i α + β
=
.

2 
4
b 
( z − β)
b 2 (α – β) 3
 z =α
=−
ai
=–
2
= ΣR + (putting α and β)
2 3/ 2
(a − b )
Hence from (3.1), we have I = 2πi ×
− ai
2
2 3/ 2
(a − b )
=
2πa
2
( a − b 2 ) 3/ 2
Ex. 44 : By the method of contour integration, prove that
π
∫
0
cos 2 θ d θ
1 − 2a cos θ + a 2
=
πa 2
1 − a2
π
Sol : Let
I =
∫
0
cos 2 θ d θ
1 − 2a cos θ + a

∵


=
=
( −1 < a < 1)
2a
∫
=
0
1
real part of
2
where
f (z ) =
∫
0
2π
1
real part of
2
∫
0
∫
C
1
real part of
2
∫
1
real part of
2
∫
2
1
2
a
f ( x )dx = 2
[writing eiθ = z, dθ =
=
2
=
C
C
–iz
(1 − az )( z − a)
2π
∫
0
cos 2 θ d θ
1 − a( eiθ − e −iθ ) + a 2

f ( x )dx if f ( 2a − x ) = f ( x )


e 2iθ dθ
(1 − aeiθ )(1 − ae( −iθ ) )
z2

(1 − az ) 1 −

dz
.
a iz

z
dz
, where C is the unit circlez= 1 ]
iz
− iz 2
dz
(1 − az )( z − a)
f ( z )dz
...(3.1)
Power Series and Calculus of Residues
By residue theorem
∫
139
f ( z )dz = 2πiΣR +
...(3.2)
C
+
where ΣR = sum of residues at pole of f ( z ) within C.
Poles of f ( z ) are given by (1 − az )( z − a) = 0. Thus z =
1
and z = a are the simple
a
poles. Only z = a lies within the unit circle C as a < 1.
The residue of f ( z ) at ( z = a) = Lim( z − a) f ( z ) = Lim( z − a)
z→ a
∴ from (3.2)
z→ a
− iz 2
− ia 2
=
(1 − az )( z − a) 1 − a 2
 –ia 2 
2πa 2
 =
f ( z )dz = 2πi 

1 − a2  1 − a2
∫
C
∴ from (3.1) I =
1
real part of
2
 2πa 2 
πa 2

 =
.


1 − a2  1 − a2
Ex. 45 : Use the method of contour integration to prove that
2π
∫
0
(1 + 2 cos θ) n cos nθ
2π
dθ =
(3 − 5 )n .
3 + 2 cos θ
5
2π
Sol : Let I =
∫
0
(1 + 2 cos θ) n cos nθ
dθ
3 + 2 cos θ
2π
= real part of
∫
0
(1 + eiθ + e −iθ ) n einθ
3 + eiθ + e −iθ
dθ
[since eiθ = cos θ + i sin θ]
n
= real part of
∫
C
1 n

1 + z +  z

dz
z
1
iz
3+ z +
z
where C is the unit circlez= 1,
Writing
eiθ = z, dθ =
dz
iz
= real part of
1
i
= real part of
∫
C
∫
C
where
f (z ) =
(1 + z + z 2 ) n
i(1 + 3z + z 2 )
(1 + z + z 2 ) n
1 + 3z + z 2
f ( z )dz
dz
...(3.1)
140
Engineering Mathematics-III
By residue theorem
f ( z )dz = 2πiΣR +
∫
... (3.2)
C
+
where ΣR = sum of residue of f ( z ) at its poles within C.
Poles of f ( z ) are given by ( z 2 + 3z + 1) = 0 i. e., by z =
or by
z=
−3+ 5
= α (say)
2
and
z=
−3 − 5
= β (say)
2
−3 ± 5
2
Both the poles are simple poles. Clearly β> α and since αβ = 1
i.e.,
. β= 1 ∴ α < 1
 αβ= α
Hence, the simple pole z = α is the only pole which lies within C.
Residue at the simple pole ( z = α ) = Lim ( z − α ) f ( z )
z→ α
= Lim ( z − α )
z→ α
(1 + z + z 2 ) n
(1 + α + α 2 ) n
=
i( z − α )( z − β)
i(α − β )
2

1 + −3 + 5 +  −3 + 5  


2
2  



=
i 5
∴ from (3.2),
f ( z )dz = 2πi .
∫
(3 − 5 )
I = real part of
2π
Ex. 46 : Evaluate
∫e
− cosθ
=
i 5
C
Hence from (3.1),
n
2π
5
2π
5
n
=
(3 − 5 )n
i 5
= ΣR +
(3 − 5 )n
(3 − 5 )n =
2π
5
(3 − 5 )n .
cos( nθ + sin θ) dθ, where n is a positive integer.
0
2π
Sol : Consider the integral I =
∫
e − cosθ [cos( nθ + sin θ) − i sin( nθ + sin θ)] dθ
0
2π
∫
=
e − cosθ e −i ( nθ + sin θ) dθ
0
2π
∫
=
e −( cosθ + i sin θ) e −in θ dθ
0
2π
=
∫
0
iθ
e− e e– inθdθ =

∫  e
C
–z
1  dz 
dz
iθ
 writing e = z, dθ = 

iz 
z n  iz 
Power Series and Calculus of Residues
141
where C denotes the unit circlez= 1.
Then,
I =
1
i
e−z
∫
z n +1
C
= 2π i ΣR
dz =
e−z
∫ f (z )dz, where f (z ) = iz n +1
C
+
[By residue theorem]
Obviously the only pole of f ( z ) within the contour C is z = 0 of order n + 1.
At z = 0, the residue of f ( z ) =
1  d n

n !  dz n
I = 2πi ×
∴
π
i.e,
∫
 n +1 e − z  
( −1) n
z

=
= ΣR +


i(n )!

iz n +1   z = 0
( −1) n
2π
=
( −1) n
i( n )!
n!
e − cosθ [cos( nθ + sin θ) − i sin( nθ + sin θ)] dθ =
0
Equating real parts, we have
2π
∫
e − cosθ [cos( nθ + sin θ)]dθ =
0
2π
( −1) n .
n!
Problem Set (3.6)
2π
1. Show that
∫
0
2π
2. Show that
∫
0
2π
3. Show that
∫
0
dθ
2π
=
2 + cos θ
3
dθ
=
1 + a cos θ
dθ
=
a + b sin θ
2π
4. Show that (i)
∫
0
π
(ii)
∫
0
2π
5. Show that
∫
0
2π
6. Evaluate
∫
0
2π
1– a
2
2π
(a − b 2 )
dθ
=
a + b cos θ
2
cos θ
dθ
3 + sin θ
,a > b
2π
2
a − b2
dθ
π
=
3 + 2 cos θ 2 5
dθ
π
=
5 + 3 cos θ 2
, a2 < 1
( a > b > 0)
2π
( −1) n
n!
142
Engineering Mathematics-III
2π
7. Evaluate
sin 2 θ
dθ
5 − 4 cos θ
∫
0
2π
8. Show that
cos 2 3θ
3π
dθ =
5 − 4 cos 2θ
8
∫
0
π
9. Evaluate
1 + 2 cos θ
dθ
5 + 4 cos θ
∫
0
2π
10. Evaluate
sin 2 θ − 2 cos θ
dθ
2 + cos θ
∫
0
π
11. Evaluate
a dθ
∫
2
a + cos 2 θ
0
2π
12. Show that
∫
0
π
13. Prove that
dθ
( 5 − 3 sin θ)
dθ
∫
0 ( a + cos θ)
2π
cosθ
14. Show that
e
∫
0
2π
15. Evaluate
∫
2
=
2
∫
0
2
( a − 1) 3/ 2
. cos(sin θ − nθ) dθ =
dθ
2π
5π
32
πa
=
1 − 2 p sin θ + p
0
16. Show that
where a > 0
2
1 − 2 p cos 2θ + p 2
dθ = π
2π
17. By contour integration evaluate
∫
0
18. Show that if a > 1,
∫
0
n
, where P 2 < 1
cos 2 3θ
2π
2π
dθ
3 – 2 cos θ + sin θ
dθ
2
1 − p + p2
,0< p < 1
1− p
1 + a – 2a cos θ
=
2π
2
a –1
3.19 Theorem
Let AB be the arc α ≤ θ ≤ β of the circle z − a = r. If
where
k
is
constant,
then
Lim( z − a) f ( z ) = k,
.
.
z
B
z→ α
Lim
r→ 0
r
∫
f ( z )dz = i(β − α )k , where the integration along AB
AB
is taken in anti-clockwise direction.
3.20 Theorem
Let AB be the arc α ≤ θ ≤ β of the circle z= R.
a
Fig. 3.7
.
A
Power Series and Calculus of Residues
143
If Lim z f ( z ) = k where k is constant; then Lim
z→ ∞
∫
f ( z )dz = i(β − α) k the
R → ∞ AB
integration along AB is taken in anti-clockwise direction.
B
z
Proof. Since Lim z f ( z ) = k ∴ lim { zf ( z ) − k} = 0 as k is
z→ ∞
R
z→ ∞
A
constant.
or z f ( z ) − k< ε for large values of z or z f ( z ) − k = η( z )
η( z ) + k
Thus f ( z ) =
z
Hence
η( z ) + k
dz =
z
∫ f (z )dz = ∫
AB
AB
η( z )
dz + k
z
∫
=
AB
∫
AB
β
∫α
η( z )
dz + k
z
z=0
Fig. 3.8
∫
AB
dz
z
iReiθ dθ
Reiθ
[since on AB, z = Reiθ , dz = iReiθ dθ]
∫
=
AB
or
∫ f (z )dz − i(β − α) k = ∫
AB
∫
AB
f ( z )dz − i(β − α) k =
AB
η( z )dz
+ ki(β − α)
z
η( z )dz
z
η( z )dz
z
∫
AB
η( z ) dz dz < ε
 
z
AB  
∫
<
β
=ε
∫
α
dz
z
AB  
∫
[sinceη( z ) < ε]
Rd θ
]
R
[since on AB, z = Reiθ , z= R  dz = R dθ]
= ε (β − α) → 0 as ε → 0 and ε → 0 as z → ∞.
Thus, when R → ∞, ε → 0, and so we have


Lim 
R→ ∞ 

or
Lim
R→ ∞
∫


f ( z )dz − i(β − α )k = 0

AB
∫
AB
f ( z )dz = i(β − α )k.
144
Engineering Mathematics-III
Jordan's Inequality
If
0< θ<
π
2 sin θ
, then <
< 1.
2
π
θ
Proof. Consider the relation y = cosθ. As θ increases cos θ decreases
and therefore the ordinate y decreases.The mean ordinate between
θ = 0 to θ =
1
θ
θ
cos θ d θ =
∫
0
sin θ
. When θ = 0, the ordinate is clearly
θ
1
π, the mean ordinate is equal to
2
cos θ, i. e., equal to 1; and when θ =
y
P
y
1
π
2 , i. e, equal to π .
o
Fig. 3.9
1
2
π
2
1
2
2 sin θ
Hence when 0 < θ < π the mean ordinate lies between 1 and , i. e, <
< 1.
2
π
π
θ
sin
3.21 Jordan's Lemma
If f ( z ) → 0 uniformly as z → ∞, and f ( z ) is mesomorphic in the upper half plane
then,
lim
R→ ∞
∫
eimz f ( z )dz = 0,
( m > 0)
CR
where C R denotes the semi-circle z = R, I ( z ) > 0, i. e, y > 0
Proof. Here R is large and can be made larger so as to include within it all the
singularities and none on its boundary.
Since
lim f ( z ) = 0
R→ ∞
 f ( z ) < ε for z on the large circlez = R
∴
Now

∫
CR
eimz f ( z )dz≤
∫
imz
e  f ( z )dz
CR
π
≤ε
∫
e
im ( R cosθ + iR sin θ)
iθ
Rie dθ
0
π
≤ε
∫
e − mR sin θ Rdθ = 2εR
0
Since
∫
0
iθ
 f ( z )< ε and z = Re on C R
π /2
≤ 2εR
π /2
∫
0
e −2mR (θ / π ) dθ
e − mR sin θ dθ
Power Series and Calculus of Residues
=
145
επ
(1 − e − mR ) → 0 when R → ∞ as ε → 0.
m
[By Jordan's inequality]
lim
R→ ∞
∫
eimz f ( z )dz = 0.
CR
3.22 Contour Integration of Functions Having no Poles on the
Real Axis
Theorem. Let f ( z ) be a function , analytic in the upper half of the z-plane except at
a finite number of poles in it and have no pole on the real axis. If z f ( z ) tends to zero
∞
as z tends to infinity, then
∫
f ( x )dx = 2πiΣR + , where ΣR + is the sum of residues
–∞
of f ( z ) at its poles in the upper half plane.
∞
3.23 Evaluation of Integrals of the Form
∫
–∞
f (x )
dx
F (x )
xf ( x )
→ 0 as x → ∞ where F( x ) has no zero on the real axis and degree of
F( x )
polynomial f ( x ) is less than polynomial F( x ).
f (z )
Consider the integral
dz, taken round the closed contour C consisting of the
F( z )
Such that
∫
C
real axis from – R to R and semi-circle C R of radius R in the upper half plane. Take
the radius R large enough so that all the poles of F( z ) lie in the upper half plane
within C R .
By residue theorem, we have
f (z )
dz = 2πiΣR + , where ΣR + is the sum of residues.
F( z )
∫
C
R
or
∫
–R
f ( x)
dx +
F( x )
∫
CR
f (z )
dz = 2πΣR +
F( z )
...(3.1)
[ since for a real axis z = x]
146
Engineering Mathematics-III
Taking z = Reiθ so that dz = Reiθ idθ in the second
integrand of left hand side, then as R is constant on C R
y
and as z moves along C R , θ varies from 0 to π.
Then
π
f (z )
dz =
F( z )
∫
CR
π
When R is large,
∫
0
∫
0
iθ
f ( Re )
F( Reiθ )
F( Re )
π
f ( Reiθ )
∫
F( Reiθ )
0
CR
Reiθ idθ
Reiθ idθ is of order
iθ
Therefore
f ( Reiθ )
–R
Rf ( R )
F( R )
O
C1
x
R
Fig. 3.10
Reiθ idθ → 0 when R → ∞
Hence from (3.1), we get
∞
∫
–∞
f ( x)
dx = 2πi ΣR +
F( x )
Illustrative Examples
∞
Ex. 47 : Show by contour integration that
∫
0
dx
1+ x
2
=
π
.
2
Sol : Consider the integral
1
∫ f (z )dz where f (z ) = 1 + z 2 .
C
Poles of f ( z ) are given by 1 + z 2 = 0 i. e, z = ± i,
y
which are simple poles, since there is no pole on the
real axis, therefore we may take the contour C
consisting of upper half C R of a large circle z = R and
CR
real axis from −R to R.
i
Hence by residue theorem, we have
–R
C
+R
x
Fig. 3.11
R
∫ f (z )dz = ∫
O
f ( x )dx +
–R
∫
+
f ( z )dz = 2πiΣR ,
CR
where ΣR + = sum of residues of f ( z ) at its poles within C
R
or
∫
–R
dx
1+ x
2
+
∫
CR
dz
1+ z
2
= 2πiΣR +
The simple pole z = i lies within the contour C .
... (3.1)
Power Series and Calculus of Residues
147
Residue f ( z ) at ( z = i ) = lim ( z − i ) f ( z ) = lim ( z − i )
z→i
z→i
z→i
1
∫
2
(1 + z )
CR
1+ z2
1
1
=
(z + i ) 2i
= lim
and
1
dz ≤
∫
CR
iθ
dz ≤
2
1 + z 
iθ
∫
CR
dz
2
z −1
[since z = Re , dz = iRe dθ= R dθ and |z + 1| ≥|z|2 –1 = R 2 – 1]
π
∫
=
R dθ
2
R −1
0
=
2
πR
2
R −1
→ 0 as R → ∞.
Hence when R → ∞, from (3.1), we have
∞
1
∫
–∞ 1 + x
∞
1
∫
∴
1+ x
0
∞
Ex. 48 : Evaluate
∫
0
2
dx
4
x + a4
Sol : Consider the integral
2
dx = 2π i
dx =
1
2i
2
∫
0
dx
1 + x2
=π
1
π
2
( a > 0).
f ( z )dz, where f ( z ) =
∫
∞
or
C
1
4
z + a4
.
The poles of f ( z ) are given by
z 4 + a 4 = 0 i. e, z 4 = − a 4 = a 4 e πi = a 4 e( 2n ) πi + πi or z = ae( 2n +1) πi / 4 ,
n = 0, 1, 2, 3. Since there is no pole on the real axis, therefore we may take the close
contour C consisting of the upper half C R of a large circlez= R and the real axis,
from −R to R.
By Cauchy's residues theorem, we have
R
f ( z )dz =
∫
C
f ( x )dx +
∫
–R
∫
f ( z )dz = 2πiΣR +
CR
+
where ΣR = sum of residues of f ( z ) at its poles within C.
R
or
∫
–R
1
4
x +a
4
dx +
∫
CR
1
4
z + a4
dz = 2πiΣR +
... (3.1)
The poles z = ae πi /4 and z = ae 3 πi / 4 ( for n=0 and 1) are the only two poles which
lie within the contour C.
Let α denotes any one of these poles then α 4 + a 4 = 0 or α 4 = − a 4 .
148
Engineering Mathematics-III




1
Residue of f ( z ) at z = α is = 

d 4
4
 ( z + a )
 dz
 z =α
=
1
=
4α 3
α
=
4α 4
y
CR
α
i
− 4 a4
+R
O
–R
x
Fig. 3.12
φ( a)

Res f ( a) = ψ' ( a) , where ψ( z ) = ( z – a)F( z ), F( z ) ≠

−1 πi / 4
∴ Residue at ( z = α = ae πi / 4 ) =
e
4 a3
and residue at ae 3πi /4
=−
∴ Sum of residues
=−
=−
Now
∫
CR
1
4
z +a
4
dz ≤
e 3πi / 4
4 a3
=
– e πi . e −πi / 4
4 a3
=

0

e −πi / 4
4 a3
1 eiπ / 4 − eiπ / 4
2
2 a3
1
2a
3
. i sin
π
i
=−
= ΣR +
3
4
2 2a
dz ≤
4
4
CR z + a 
∫
π
dz
=
4
4
CR z −a 
∫
∫
0
Rdθ
4
R − a4
[ ∴ z = Reiθ ]
=
πR
4
R − a4
which → 0 as R → ∞.
Hence, when R → ∞, relation (3.1) becomes
∞
∫
–∞
∞
or
∫
0
dx
4
x +a
4
dx
4
x +a
4

i  π 2
= 2πi.  –
 =
 2 2a 3  2a 3
=
π 2
4a 3
.
∞
Ex. 49 : Apply the calculus of residues to prove that
∫
0
Sol : Consider the integral
∫
x2
2
2 2
(x + a )
dx =
π
4a
f ( z )dz, where
C
f (z ) =
z2
(z 2 + a 2 )2
.
Poles of f ( z ) are given by ( z 2 + a 2 ) 2 = 0. Thus z = ai and z = − ai are the two poles
each of order 2. Since there is no pole on the real axis, therefore we may take closed
Power Series and Calculus of Residues
149
contour C consisting of the upper half C R of a large circlez= R and the real axis
from − R to R.
Hence, by residue theorem, we have
f ( z )dz = 2πiΣR + ,
∫
C
+
where ΣR = sum of residues of f ( z ) at its poles within C.
R
i.e.,
∫ f ( x ) dx + ∫ f (z ) dz = 2πiΣR
–R
...(3.1)
CR
R
or
+
x2
∫
2
2 2
(x + a )
–R
dx +
z2
∫
2
2 2
(z + a )
CR
dz = 2πiΣR +
The only pole z = ai lies within the contour.
d
Risidue at ( z − ai ) = lim
[( z − ai ) 2 f ( z )]
z → ai dz
d
dz
= lim
z → ai
 z2

i
=−

2
4a
( z + ai ) 
Now let R > a , then
2
z
f (z ) =
z2+ a
Hence,
∫
f ( z ) dz ≤
CR
2 2
R2
(R 2 − a 2 )2
≤
R2
2
2 2
(R − a )
, since z 2 + a 2 ≥ R 2 − a 2
π R → 0 as R → ∞
When R → ∞,( 3.1) becomes
∞
∫
−∞
∞
or
∞
x2
2
2 2
(x + a )
dx = 2
x2

i 
∫ ( x 2 + a 2 )2 dx = 2 π i  − 4 a
0
x2
π
∫ ( x 2 + a 2 )2 dx = 4 a
0
∞
Ex. 50 : Prove that
∫
dx
2
2
2
2 2
– ∞ ( x + b )( x + c )
Sol : Consider the integral
∫
C
=
π( b + 2c )
2bc 3 ( b + c ) 2
f ( z )dz, where f ( z ) =
where b > 0, c > 0.
1
2
2
( z + b )( z 2 + c 2 ) 2
Poles of f ( z ) are given by ( z 2 + b 2 )( z 2 + c 2 ) 2 = 0 i. e, z = ± i b are the two simple
poles and z = ± ic are two double poles. Since there is no poles on the real axis
therefore we may take the contour C consisting of the upper half C R of a large circle
 z= R and the real axis, from −R to R.
150
Engineering Mathematics-III
Hence, by Cauchy's residue theorem, we have
∫
f ( z )dz = 2πi ΣR +
C
+
where ΣR = sum of residues of f ( z ) at its poles within C.
R
i.e.,
f ( x )dx +
∫
–R
CR
R
or
∫
f ( z )dz = 2πi ΣR +
∫
dx
2
–R
2
2
2 2
( x + b )( x + c )
+
dz
∫
2
2
2
2 2
(z + b )(z + c )
CR
= 2πi ΣR +
...(3.1)
The poles which lie within the contour C are : z = ib (simple pole ) and z = ic (double
pole).
Residue at simple pole ( z = ib) is = lim ( z − ib) f ( z )
z→ ib
= lim ( z − ib)
z→ ib
=
Residue at ( z = ic ) =
1
( z − ib)( z + ib)( z 2 + c 2 ) 2
1
2
−i
=
2 2
2 i b( c − b )
.
2
2b( c − b 2 ) 2

1 d 
1
2
( z − ic ) . 2

2
2
2
1 ! dz 
( z + b )( z − ic ) ( z + ic ) 
= −2z ( z 2 + b 2 ) −2 ( z + ic ) −2 − 2 ( z + ic ) –3 ( z 2 + b 2 ) −1
=
=
=
−2 ic
2
2 2
2
( − c + b ) ( −4 c )
−2 ic
2
2 2
2
( − c + b ) ( −4 c )
i
2c ( b 2 − c 2 ) 2
−
−
−
2
3
− 8 ic ( − c 2 + b 2 )
2
3
− 8 ic ( − c 2 + b 2 )
i
4c 3 ( b 2 − c 2 )
(3 c 2 − b 2 )i
=
4 c 3 (b 2 − c 2 )2
The sum of residues at the poles within C
=
=
Now
−i
2b( c 2 − b 2 ) 2
4c 3 ( b 2 − c 2 ) 2
−i( c − b ) 2( 2c + b )
3
2
2 2
4bc ( b − c )
lim z f ( z ) = lim z.
z→ ∞
( 3c 2 − b 2 ) i
+
z→ ∞
= lim
z→ ∞
=
=
− i{i 2c 3 − b( 3c 2 − b 2 )}
4bc 3 ( b 2 − c 2 ) 2
−i( b + 2c )
3
4bc ( b + c )
2
1
2
2
( z + b )( z 2 + c 2 ) 2
1

b2  
c2 
 1 +

z 5 1 +
 


z2 
z2
2
=0
= ΣR +
Power Series and Calculus of Residues
∴ By (3.22)
151
dz
∫
= lim
R→ ∞
2
( z + b )( z 2 + c 2 ) 2
CR
2
=0
Hence when R → ∞ (3.1) becomes
∞
∫
– ∞( x
dx
2
2
2
2 2
+ b )( x + c )
= 2πi
− i( b + 2c )
3
4bc ( b + c )
2
=
π( b + 2c )
2bc 3 ( b + c ) 2
3.24 Evaluation of the Integrals of the Form
∞
∞
f ( x ) sin mx dx,
∫
f ( x ) cos mx dx, m > 0
∫
–∞
–∞
where f ( x ) is a rational function of x, such that
y
(i) the degree of the denominator of f ( x ) exceeds that
of the numerator,
(ii) the denominator of f ( x ) does not have a real zero.
Consider the integral
∫
CR
eimz f ( z )dz, taken round the
C
–R
O
R
x
Fig. 3.13
contour C. The given conditions ensure the
convergence of the improper integral eimz f ( z )dz.
∫
C
By residue theorem, we have
∫
eimz f ( z )dz =
C
R
∫
eimx f ( x )dx +
∫
–R
eimz f ( z )dz = 2πi ΣR +
CR
where ΣR + is the sum of residues of f ( z ) at its poles within C.
Taking limit as R → ∞, we have
R
lim
R→ ∞
∫
eimx f ( x )dx + lim
R→ ∞
–R
∫
eimz f ( z )dz = 2πi ΣR +
CR
Now consider the contribution of integral
∫
eimz f ( z )dz.
CR
∫
eimz f ( z )dz ≤
CR
But
e
∫ e
imz
  f ( z )  dz
CR
imz
im ( x + iy )
imx − my
− my
≤ 1.
=e
=e e
= e
Since m > 0 and y ≥ 0 and | f ( z )eimz|=| f ( z )||eimz|≤| f ( z )|
Hence, by Jordan's lemana
...(3.1)
152
Engineering Mathematics-III
eimz f ( z )dz = 0
∫
lim
R→ ∞
CR
R
and
R→ ∞
∞
eimx f ( x )dx =
∫
lim
∫
–R
–∞
∞
∴ From (3.1) , we have
eimx f ( x )dx.
∫
eimx f ( x )dx = 2πiΣR +
–∞
∞
or
∫
∞
f ( x ) cos mx dx + i
f ( x ) sin mx dx = 2πiΣR + .
∫
–∞
–∞
Equating real and imaginary parts, we shall get the values of the integrals
∞
∞
f ( x ) cos mx dx and
∫
–∞
∫
f ( x ) sin mx dx.
–∞
Ex. 51 : Use the method of contour integration to prove that
∞
(i)
∫
0
∞
(ii)
∫
cos mx
2
a +x
2
sin mx
–∞ a
2
+ x2
∞
Deduce that
∫
0
dx =
π − ma
e
and
2a
dx = 0
x sin mx
2
x +a
dx =
2
Sol : Consider the integral
∫
π − ma
e
( m ≥ 0).
2
f ( z )dz, where f ( z ) =
C
eimz
a2 + z 2
.
The pole of f ( z ) are given by z 2 + a 2 = 0 i. e, z = ai and z = − ai are the simple
poles of f ( z ). Since there is no pole on the real axis, therefore we may take the closed
contour C consisting of the upper half C R of a large circle z= R and the real axis,
from −R to R as given in fig. 3.13
∴ By residues theorem, we have
R
∫
f ( z ) dz =
C
∫
f ( x )dx +
–R
∫
f ( z )dz = 2πiΣR +
CR
where ΣR + = sum of residues of f ( z ) at its poles within C.
R
or
∫
–R
eimx
a2 + x 2
dx +
CR
The pole z = ai lies within the contour C.
∴ Residue at the simple pole ( z = ai ) is
∫
eimz
a2 + z 2
dz = 2πiΣR +
...(3.1)
Power Series and Calculus of Residues
153
eimz
e − ma
=
= ΣR +
( z − ai )( z + ai )
2ai
lim ( z − ai ) f ( z ) = lim ( z − ai )
z→ ai
Now,
∫e
z→ ai
miz
f ( z ) dz ≤
CR
Since
c miz
∫
f ( z ) dz ≤
CR
2
e miz = e − my ≤ 1 and z 2 + a 2 ≥ z
Therefore,
∫
f ( z ) dz =
CR
dz
∫
2
z +a
CR
≤
2
∫
−∞
∞
cos mx
∫
−∞
a2 + x 2
− a2 = R 2 − a2
dz
∫
2
R −a
CR
∞
Hence when R → ∞, (3.1) becomes
or
f ( z ) dz
∫
CR
eimx
a2 + x 2
∞
dx + i
∫
−∞
2
πR
=
2
R − a2
dx = 2πi.
sin mx
a2 + x 2
→ 0 as R → ∞
e − ma
πe − ma
=
2ai
a
dx =
πe − ma
a
Equating real and imaginary parts, we have
∞
cos mx
∫
2
a +x
−∞
∞
cos mx
∫
∴
a2 + x 2
0
∞
and
2
∫
dx =
sin mx
−∞
π − ma
e
or 2
a
dx =
∞
∫
0
cos mx
2
a +x
2
dx =
πe − ma
2a
π − ma
e
a
...(3.2)
= 0.
a2 + x 2
Deduction. Differentiating (3.2). w.r.t. m, we get
∞
∫
a2 + x 2
0
∞
or
∫
0
∞
Ex. 52 : Evaluate
∫
0
− x sin mx
x sin mx
2
a +x
2
dx =
dx =
x 3 sin x
( x 2 + a 2 )( x 2 + b 2 )
Sol : Consider the integral
∫
−πae − ma
2a
π − ma
e
2
dx,( a > 0, b > 0).
f ( z )dz, where f ( z ) =
C
z 3 eiz
.
( z 2 + a 2 )( z 2 + b 2 )
Poles of f ( z ) are given by ( z 2 + a 2 )( z 2 + b 2 ) = 0. i. e., z = ± ai, z = ± bi are the
four simple poles. Since there is no pole on the real axis, therefore we may take the
closed contour C consisting of the upper half C R of the large circlez= R, and the
real axis from −R to R as shown in fig. 3.13
154
Engineering Mathematics-III
∴ By residues theorem, we have
f ( z )dz = 2πiΣR +
∫
C
where ΣR
+
= sum of residue of f ( z ) at its poles within C.
R
or
f ( x )dx +
∫
−R
CR
x 3 eix
R
or
f ( z )dz = 2πiΣR +
∫
∫ − R ( x 2 + a 2 )( x 2 + b 2 ) dx
+
z 3 eiz
∫ CR (z 2 + a 2)(z 2 + b 2 ) dz = 2πiΣR
+
...(3.1)
The poles which lie within the contour C are z = ia and z = ib, both simple poles.
Residue of f ( z ) at z = ia, is
lim ( z − ia) f ( z ) = lim ( z − ia)
z→ ia
z→ ia
Similarly residue at z = ib is =
z 3 eiz
( z − ia)( z + ia)( z 2 + b 2 )
=−
a2e −a
2( b 2 − a 2 )
−b 2 e − b
2( a 2 − b 2 )
1
So that sum of these two residues =
2
2
2( a − b )
[ a 2 e − a − b 2 e − b ] = ΣR +
For second integral of right hand side of (1), we have
∫
CR
z 3 dz
( z 2 + a 2 )( z 2 + b 2 )
3 iz
z e  dz
≤
2
2
2
2
z
+
a
z
+
b



CR
dz ≤
π
R3
=
≤
∫
∫e
( R 2 − a 2 )( R 2 − b 2 )
2
CR
− R sinθ
(z 2|− a 2 )(  z2 − b 2 )
R dθ
0
π /2
R4
2
∫
|z|2z 3 eiz dz
2
2
2
( R − a )( R − b )
∫
e −2Rθ / π dθ
0
[by Jordan's inequality]
πR
=
2
2
3
2
2
( R − a )( R − b )
[1 − e − R ] which → 0 as R → ∞
Hence when R → ∞, relation (3.1) becomes
∞
∫
−∞ (x
x 3 eix
2
2
2
2
+ a )( x + b )
Equating imaginary parts, we have
dx =
πi
2
2
(a − b )
(a 2 e − a − b 2 e − b )
Power Series and Calculus of Residues
155
∞
x 3 sin x
∫
2
−∞ (x
2
2
+ a )( x + b )
∞
or
2
x 3 sin x
∫
2
2
2
2
( x + a )( x + b )
0
dx =
dx =
π
2
2
(a − b )
π
2
2
2( a − b )
(a 2 e − a − b 2 e − b )
(a 2 e − a − b 2 e − b )
Ex. 53 : Use the method of contour integration to prove that
∞
cos x 2 + sin x 2 − 1
∫
x2
−∞
dx = 0.
2
Sol : Consider the integral
∫
f ( z )dz, where f ( z ) =
eiz − 1
C


i 2z 4
2
+..... − 1
1 + iz +
2!


2
or
eiz − 1
f (z ) =
z2
=
z2
z2
= i +i2
z2
+.....
2!
Obviously f ( z ) has no poles on the real axis, therefore we may take the closed
contour C consisting of the upper half C R of a large circlez= R and real axis from
−R to R as shown in fig. 3.13
Hence, by residue theorem, we have
f ( z ) dz = 2πi ΣR +
∫
C
R
i.e,
∫
f ( x )dx +
–R
∫
f ( z )dz = 2πi × 0, since there is no pole within C
CR
2
R
or
eix − 1
∫
x2
–R
∫
CR
eiz − 1
z2
∫
eiz − 1
z2
CR
...(3.1)
dz = 0
2
2
Now
2
dx +
dz ≤
∫
CR
=
≤
1
R
2
1.
2
R
{eiz +1}dz
2
z 
π
∫
(e − R
2 sin 2θ
+ 1)Rdθ
(∵ z = Reiθ )
0
π /2
∫
( e −4R
2θ / π
+ 1)dθ
[ by Jordan's inequality ]
0
2 
π
π
= +
(1 − e −2R ) which → 0 as R → ∞.
3
 R 2R

∞ i x2
Hence when R → ∞,( 3.1) becomes
∫
–∞
e
x
2
−1
dx = 0
156
Engineering Mathematics-III
∞
or
∫
cos x 2 + i sin x 2 − 1
x2
–∞
dx = 0
Equating real and imaginary parts, we have
∞
∫
cos x 2 – 1
x2
0
∞
Adding these, we have
∞
dx = 0
x2
0
sin x 2
x2
0
cos x 2 + sin x 2 − 1
∫
∫
and
dx = 0
dx = 0
Problem Set (3.7)
Use the method of contour integration to evaluate the following
integrals:
∞
1.
∫
0
∞
3.
∫
0
∞
5.
∫
0
∞
7.
∫
0
∞
9.
∫
0
∞
dx
2
x +a
2.
2
0
∞
x 2 dx
2
( x + 1)
4.
2
∫
0
∞
x 2 dx
( x 2 + 1) 3
dx
1+ x
∫
6
6.
∫
8.
∫
dx
2
( x + a 2 )2
dx
(1 + x 2 ) 3
x 2 dx
2
−∞ ( x
∞
2
4
( x + 1)
2
10.
dx
+ 1)( x 2 + 4)
x dx
x6 + 1
0
∞
x6
,a > 0
cos mx
dx
(a + x 2 )2
π
=
(1 + ma)e − ma ,( a > 0, m > 0)
3
4a
∫
2
0
∞
hence evaluate
∞
11.
13.
∫
−∞ ( x
∞
3
∫
0
∞
15.
17.
∫
0
∞
∫
0
2
2
+ a )( x + b )
x sin mx
4
2
x +a
4
dx, m > 0
cos 3 x dx
( x 2 + 1)( x 2 + 4)
cos 2πx
x4 + x2 + 1
dx
0
∞
cos x dx
2
a> b> 0
12.
∫
∫
0
( x 2 + 1) 2
2
x − 2x + 5
x sin πx
2
−∞
∞
16.
cos x dx
sin x dx
∫
−∞
∞
14.
∫
dx
x + 2x + 5
cos 2 x
( x 2 + 9) 2 ( x 2 + 16)
dx
18. By integrating e z / ( z − ai ),( a > 0)
round a suitable contour, prove that
∞
a cos x + x sin x
dx = 2πe − a .
2
2
x +a
−∞
∫
Power Series and Calculus of Residues
157
19. Prove that
∞
∫
0
cos 2 x
2 2
(1 + x )
dx =
[Hint. Take f ( z ) =
π
(1 + 3e −2 )
2
1 + e 2iz
(1 + z 2 ) 2
, so that f ( x ) =
1 + cos 2 x
(1 + x 2 ) 2
=
2 + cos 2 x
(1 + x 2 ) 2
. Integrate
over the contour C].
3.25 Contour Integration of Function Having Poles on the Real Axis
When the poles of f ( z ) lie on the real axis then these poles which lie on the real axis
can be avoided by drawing small semicircles C r , C r' etc. about these poles as centre
and small radii r, r' etc. in the upper half of the plane.
The method is said to be ''indenting the semi-circular contour. Thus, the function
f ( z ) is analytic along this modified contour and hence the integral f ( z )dz can be
∫
C
evaluated by Residue theorem. The following examples will illustrate the method.
∞
Ex. 54 : By contour integral prove that
∫
0
Sol : Consider
∫
f ( z )dz, where f ( z ) =
C
e
sin mx
π
dx =
x
2
imz
z
The singularity of f ( z ) is at z = 0.
Let C be the closed contour consisting of the upper half
C R of a large semi-circlez= R and the real axis from
CR
Cr
−R to R indenting the path at z = 0 by a semi-circle C r
centred at O and of a small radius r. Thus, the closed
contour C consists of the upper half C R of large circle
–R
–r O r
Fig. 3.14
z= R, real axis from −R to −r, ( semi-circular are C r of
circlez= r, where r is very small) and again the real axis from r to R.
Hence, by residue theorem, we have
∫
f ( z )dz = 2πiΣR +
C
where ΣR + = sum of residues of f ( z ) at its poles within C.
Since there is no pole of f ( z ) within C, therefore ΣR + = 0.
–r
i.e.,
∫
–R
R
f ( x )dx +
∫
Cr
f ( z )dz +
∫
f ( x )dx +
r
By letting r → 0 and R → ∞ , we have
∫
CR
f ( z )dz = 0
R
X
158
Engineering Mathematics-III
0
∞
f ( x )dx + lim
∫
∫
r→ 0
–∞
f ( z )dz +
∫
Cr
f ( x )dx + lim
R→ ∞
0
∫
f ( z )dz = 0
CR
∞
or
f ( x )dx = – lim
∫
π
Now,
f ( z ) dz =
∫
CR
f ( z )dz − lim
∫
r→ 0
–∞
R→ ∞
Cr
eimr(cosθ + i sinh θ )
∫
Re
0
iθ
...(3.1)
f ( z )dz.
∫
CR
R iei θ d θ
eimR (cosθ + i sin θ ) = e − mR sin θ + imR cosθ = e − mR sin θ
Since
π
Therefore
f ( z ) dz ≤
∫
∫
CR
e − mR sin θ dθ = 2
0
∫e
− mR sin θ dθ
0
π /2
=2
π /2
∫e
−2mR θ / π
dθ [Since 0 ≤ π / 2 , sin θ/θ = 2/ π]
0
z = reiθ in c r , hence
Similarly
=
∫
π
(1 − e − mR ) → 0 as R → ∞
mR
Cr
i.e.,
0
0
∫
∫
f ( z ) dz = i eimr (cosθ + i sin θ ) dθ → – i dθ as r → 0
π
π
∫ f (z ) dz = − i zπ as r→ 0
Cr
Hence from (3.1), we have
∞
∫
f ( x )dx = − ( − πi )
–∞
∞
or
∫
–∞
∞
or
∫
–∞
eimx
dx = πi
x
cos mx
dx + i
x
∞
∫
–∞
sin mx
dx = πi
x
Equating the imaginary parts, we have
∞
∫
–∞
sin mx
dx = π
x
∞
or
∫
0
sin mx
π
dx =
x
2
∞
Ex. 55 : Show that, if a ≥ b ≥ 0, then
∫
0
cos ax − cos bx
x
2
dx =
π
( b − a).
2
Power Series and Calculus of Residues
Sol : Consider the integral
159
f ( z )dz, where f ( z ) =
∫
eiaz − eibz
z2
C
, whose pole is at
z = 0. Let C be the closed contour consisting of the upper half C R of the large circle
z= R and the bounding diameter indented at z = 0, and let r be the radius of
indentation. The modified contour C is the upper half C R of the large circlez= R,
the upper half C r of the small circlez= r, and the two lines along the real axis from
−R to −r and from r to R as shown is fig. 3.14
There are no poles of f ( z ) within the modified contour C.
Hence by residue theorem, we have
∫
f ( z )dz = 2πiΣR + = 0
C
r
or
R
f ( x )dx +
∫
–R
∫
f ( z ) dz +
Cr
∫ f ( x ) dx + ∫ f (z ) dz = 0
r
CR
When r → 0 and R → ∞ it gives
0
∞
f ( x ) dx + lim
∫
∫
f ( z ) dz +
∫
f ( z )dz + lim
r→ 0
–∞
∫ f ( x ) dx + ∫ f (z ) dz = 0
Cr
0
CR
∞
or
f ( x )dx + lim
∫
r→ 0
–∞
Now
Cr
lim z. f ( z ) = lim
z→ 0
∫
R→ ∞
e
iaz
z→ 0
−e
z
f ( z )dz = 0
CR
ibz
 

1 
i 2a 2z 2
i 2b 2z 2
+. . . . . − 1 + ibz +
+. . . . . 
1 + iaz +
z→ 0 z 
2!
2!
 


= lim


(a 2 − b 2 )
= lim i( a − b) −
z +..... = i( a − b) = k1
z→ 0 
2

∴ we have
lim
r→ 0
Also lim
z→ ∞
1
z2
lim
R→ ∞
f ( z )dz = ik1 (β − α ) = i . i( a − b). ( 0 − π ) = π( a − b) [see theorem .3.20]
∫
Cr
= 0 therefore By Jordans' Lemma, we have
∫
CR
f ( z )dz = lim
R→ ∞
∫
CR
∞
Hence from (1), we have
∫
–∞
eiaz
z2
dz − lim
R→ ∞
eiax − eibx
x2
∫
CR
eibz
z2
dz = 0 − 0 = 0
dx + π( a − b) + 0 = 0
160
Engineering Mathematics-III
∞
or
∫
(cos ax − cos bx ) + i(sin ax − sin bx )
x2
–∞
∞
Equating real parts, we get
∫
(cos ax − cos bx )
x2
–∞
∞
or
∫
(cos ax − cos bx )
0
x
2
dx = π( b − a)
dx =
π
( b − a).
2
∞
Ex. 56 : Use the method of contour integration to prove that
∫
0
Sol : Consider the integral
∫
dx = π( b − a)
f ( z )dz, where f ( z ) =
C
z4
( z 6 − 1)
x 4 dx
6
( x − 1)
=
π 3
.
6
.
The poles of f ( z ) are given by z 6 − 1 = 0
or
z 6 = 1 = e 2nπi ∴ z = e nπi / 3 ( n = 0, 1, 2, 3, 4, 5)
The poles z = 1 and z = e πi = −1 (for n = 0, 3) lie on real axis. Therefore to avoid
the poles on real axis, consider the contour C consisting of the upper half C R of a
large circlez= R and the real axis from −R to R indented at z = −1 and z = 1 and let
r and r' be the radii of indentations respectively.
Clearly the poles of f ( z ) within C are at z = e πi /3 = α and z = e 2 πi / 3 = β (for
n = 1, 2) both simple poles.




z4
α4
1
Residue at simple pole ( z = α ) = 
=
=
.

5
d
6α
6α
 ( z 6 − 1
 dz
 z =α
Similarly residue at simple pole (z = β) =
1
.
6β
Sum of residues of f ( z ) at these poles within C
1
1
1  1 1 1 −iπ / 3
+
=
+
= [e
+ e −2 π / 3 ]
6 α 6 β 6 α β  6
1
1
= [ e −iπ / 3 + e −iπ . eiπ / 3 ] = ( e −iπ / 3 − eiπ / 3 )
6
6
=
=
1
π − i 3
 −2 i sin  =

6
3
6
Hence by residue theorem, we have
–(1+ r ')
∫
C
f ( z )dz =
∫
–R
1– r
f ( x )dx +
∫
Cr'
f ( z )dz +
∫ f ( x )dx
– (1– r ')
Power Series and Calculus of Residues
161
R
f ( z )dz +
∫
+
f ( x )dx +
∫
Cr
f ( z )dz = 2πiΣR +
∫
1+ r
CR
Where ΣR + = sum of residues of f ( z ) at its poles within C =
–i 3
.
6
∴ When r' → 0, r → 0 and R → ∞, we have
–1
1
f ( x )dx + lim
∫
f ( z )dz +
∫
r '→ 0
–∞
Cr′
f ( x )dx
∫
–1
∞
f ( z )dz +
∫
+ lim
r→ 0
f ( x )dx + lim
∫
Cr
∫
R→ ∞
1
CR

3
f ( z )dz = 2πi  − i

6

∞
or
∫
f ( x )dx + lim
–∞
f ( z )dz + lim
∫
r '→ 0
r→ 0
Cr'
f ( z )dz + lim
∫
R→ ∞
Cr
∫
f ( z )dz =
CR
π 3
3
...(3.1)
Now lim { z − ( −1)} f ( z ) = lim ( z + 1)
z → −1
z → −1
4
5z + 4z
= lim
z → −1
and
6z
6
z −1
5
= lim
= lim
z
= lim
6z
z→1
0

 form 

0
z −1
2
=−
1
= k1
6
5z 4 − 4z 3
z 6 − 1 z→1
5z − 4 1
= lim
= = k2
z→1 6z 2
6
z→1
4
6
5z + 4
z → −1
4
z +z
z → −1
3
5
lim ( z − 1) f ( z ) = lim ( z − 1)
z
4
6z 5
∴ By theorem 3.19 we have
Lim
r '→ 0
and
Also
Lim
r→ 0
∫
1
πi
f ( z )dz = i (β − α )k1 = i( 0 − π) ( − ) =
.
6
6
∫
1
πi
f ( z )dz = i (β − α )k2 = i( 0 − π) ( ) = − .
6
6
Cr '
Cr
Lim zf ( z ) = Lim
z→ ∞
z5
z→ ∞
∴ By Jordan's temma Lim
R→ ∞
6
z −1
∫
= Lim
z→ ∞
1

1
z 1 −


z6
=0
f ( z )dz = 0
CR
Cr'
Hence from (3.1), we have
∞
∫
–∞
f ( x )dx +
πi πi π 3
−
=
6
6
3
–R
–1
Cr
O
Fig. 3.15
1
R
162
Engineering Mathematics-III
∞
or
x4
∫
–∞
∞
or
6
x −1
x 4 dx
∫
=
6
x −1
0
dx =
π 3
3
π 3
.
6
3.26 Contour Integration of Many Value Functions
The integrals of the many valued functions such as x a , where a is not integer, log z
etc, can also be evaluated by the method of contour integration. In such cases the
closed contour C is so chosen that the interior of this contour C do not contain any
branch (singular) points. This is illustrated by the following examples.
Ex. 57 : Use the method of contour integration to prove that
∞
x a −1
∫
1+ x
0
Sol : Consider the integral
2
dx =
π
 π a
cosec   , 0 < a < 2
 2
2
f ( z )dz where f ( z ) =
∫
C
z a −1
1+ z2
, 0 < a < 2.
Clearly z = 0 is a singular point (branch point) of the given function and the poles of
f ( z ) are given by z = i and z = − i and are the simple poles.
To avoid the singularity at z = 0, consider the closed contour C consisting of the
upper half C R of the large unit circlez= R and the real axis from −R to R indented
at z = 0. Let r be the radius of indentation.
Clearly the simple pole z = i lies within C.
∴ Residue at pole ( z = i ) = lim( z − i ) f ( z ) = lim( z − i )
z→ i
=
z→ i
z a −1
( z + i )( z − i )
(i ) a −1 i −1 . i a
1
1
1
=
= − i a = − ( eiπ / 2 ) a = − eiπa / 2
2i
2i
2
2
2
Hence by residue theorem, we have
–r
∫
f ( z )dz =
C
∫
R
f ( x )dx +
–R
∫
f ( z )dz +
Cr
∫
f ( x )dx +
r
where ΣR + = sum of residues of f ( z ) at its poles within C = −
∫ f ( z )dz = 2πiΣR
CR
1 iπa / 2
e
2
∴ When r → 0, R → ∞, we have
0
∫
∞
f ( x )dx + lim
–∞
r→ 0
∫
f ( z )dz +
Cr
∫ f ( x )dx + RLim
∫
→∞
0
f ( z )dz
CR
 1

= 2πi  − eiπa / 2  = − πieiπa / 2
 2

...(3.1)
+
Power Series and Calculus of Residues
lim zf ( z ) = lim z.
Now
z→ 0
z→ 0
163
z a −1
1+ z2
za
= lim
z→ 01 +
z2
(∵ a > 0 )
= 0= k
∴ we have
f ( z )dz = i(β − α ). k = 0
∫
lim
r→ 0
Cr
and
za
lim z f ( z ) = lim
z→∞ 1 +
z→∞
z
2
1
=0
z→∞ ( 2− a )  1

z
 2 + 1
z

( ∵ a < 2 ).
= Lim
∴ we have
lim
R→∞
∫ f (z )dz = 0
CR
0
Also
∫
0
f ( x )dx = −
–∞
∫ f (− x )dx [Putting − x for x ]
∞
Cr
∞
∫ f (e
=
iπ
–r O r
–R
x )dx
0
∞
Hence from (1), we have
∫
Fig. 3.16
∞
f ( eiπ x )dx +
∫
0
∞
or
∫
0
∞
or
∫
( eiaπ . e −iπ + 1)
0
∫
1 + ( xeiπ ) 2
or
(− e
iaπ
∞
+ 1)
∫
0
∞
or
x a −1
x a −1
1 + x2
0
∞
( xe )
( ei ( a −1) π + 1)
∞
f ( x )dx = − πieiπa / 2
iπ a −1
0
or
x a −1
1+ x
2
dx +
∫
0
x a −1
1 + x2
x a −1
1 + x2
dx = − πieiπa / 2
dx = − πieiπa / 2
dx = − πieiπa / 2
dx = − πieiπa / 2
− πieiπa / 2
∫ 1 + x 2 dx = − eiπa/ 2 (eiπa/ 2 − e −iπa/ 2 )
0
=
CR
i
πi
π
 π a
= cosec  
 2
2
 π a
2 i sin  
 2
R
X
164
Engineering Mathematics-III
Ex. 58 : By integrating
∞
(i)
∫
0
(log x ) 2
1+ x
2
(log z ) 2
1+ z2
round a suitable contour, prove that
∞
π3
dx =
8
(ii)
(log x )
∫
1 + x2
0
Sol : Consider the integral
∫
dx = 0.
f ( z )dz, where f ( z ) =
(log z ) 2
1+ z2
C
. Clearly z = 0 is a
singular point of f ( z ) and its poles are given by
z 2 + 1 = 0 i. e, z = i, z = − i, are the simple poles of
f ( z ).
To avoid the singular point at z = 0, consider the
closed contour C consisting of the upper half C R of a
.
–R
large circle z= R and the real axis from −R to R
–r
Cr
O
R
r
x
Fig. 3.17
indented at z = 0. Let r be the radius of indentation.
The only pole within the contour C is a simple pole at
z = i,

 (log z ) 2  (log i ) 2
(log z ) 2 
The residue at ( z = i ) = lim ( z − i )
= lim 

 =
z→ i
2i
1 + z 2  z→ i  z + i 

=
 π
i 
 2
=
2i
(log eiπ / 2 ) 2
2i
Hence, by residue theorem, we have
2
=−
π2
.
8i
f ( z )dz = 2πi × ΣR +
∫
C
–r
or
∫
R
f ( x )dx +
–R
∫
f ( z )dz +
Cr
 −π 2 

8 i 
∫ f ( x ) dx + ∫ f (z ) dz = 2πi × 
r
CR
∴ When r → 0 and R → ∞, we have
0
∫
∞
f ( x )dx + lim
r→ 0
–∞
Now
∫
f ( z )dz +
Cr
lim z f ( z ) = lim
z→ ∞
∫
z→ ∞
f ( x )dx + lim
R→ ∞
0
z(log z ) 2
1+ z2
2
= lim
z→ ∞
f ( z )dz = −
∫
CR
 log z 
z3

 z 
2
1+ z2
 log z 


 z 
log z
= lim
= 0 Since lim
=0
1
z→ ∞ 1
z→ ∞ z
+
z3 z
π3
...(3.1)
4
Power Series and Calculus of Residues
∴
∫ f (z )dz = 0.
lim
R→ 0
CR
Also
z(log z ) 2
lim z f ( z ) = lim
z→ 0
∴
165
lim
r→ 0
∫
z→ 0
1+ z
ζ ( log ζ ) 2
= lim
2
ζ→ ∞
1+ζ
2
1
= 0 [putting z = ]
ζ
f ( z )dz = 0.
Cr
0
Hence, relation (1) becomes
∞
(log x ) 2
∫
dx +
1 + x2
–∞
∫
(log x ) 2
1 + x2
0
 π3

dx =  −

 4
[ putting − x for x in first integral]
∞
or
∫
0
∞
or
∫
0
∞
or
∫
0
∞
or
∫
{log( − x )}
1 + x2
∞
2
dx +
1 + x2
0
∞
{log( xeiπ )} 2
1+ x
(log x )
∫
dx +
2
(log x + iπ ) 2
1 + x2
∫
dx +
dx = −
(log x ) 2
2
0
1+ x
∞
(log x ) 2
∫
1 + x2
0
2(log x ) 2 − π 2 + i 2π log x
1 + x2
0
2
π3
4
dx = −
π3
4
dx = −
π3
4
dx = −
π3
4
...(3.2)
Equating real parts, we have
∞
2
∫
0
∞
i.e,
2
∫
0
∞
or
2
∫
0
∞
∴
∫
0
(log x ) 2
1 + x2
(log x ) 2
1 + x2
(log x ) 2
(1 + x 2 )
(log x ) 2
1 + x2
∞
dx − π 2
dx
∫
1 + x2
0
dx − π 2
dx = −
dx =
dx = −
π3
4
π
π3
=−
2
4
π3 π3
π3
+
=
4
2
4
π3
.
8
(ii) Also equating imaginary parts in (3.2), we have
∞
∫
0
2π log x
1+ x
2
∞
dx = 0 or
∫
0
log x
1 + x2
[since eiπ = −1]
dx = 0.
166
Engineering Mathematics-III
Problem Set (3.8)
∞
1.
Prove that
∫
cos 2ax − cos 2bx
x3
0
∞
2.
Prove that
∫
sin 2 x
x
0
∞

Hint. Write


∫
Prove that
∫
x2
x
4.
∞
dx =
∫
1 − cos 2 x
2
=
dx. Take f ( z ) =
2x 2
0
1 − cos x
0
π
2
dx =
sin 2 x
0
∞
3.
2
dx = − π( a − b), a > b > 0.
1 − e 2iz 
.
2z 2 

π
.
2
Show that, if a and m are positive , then
∞
sin 2 mx
∫
2
2
2 2
x (a + x )
0
dx =
π
8a
5
{ e − ma ( 2ma + 3) + 4ma − 3}


e 2miz
.
Hint. Take f ( z ) = 1 − 2 2
2 2
z (a + z ) 

5.
Show that
∞
(i )
∫
0
xa
1+ x
dx =
2
π
 π a
sec   , − 1 < a < 1
 2
2
∞
6.
Prove that
xa
∫
2
x − x +1
0
∞
7.
Prove that
∫
log(1 + x 2 )
x1+ a
0
8.
dx =
2π
3
sin
∞
(ii )
∫
0
x log x
1 + x2
dx = 0.
2 aπ
cosec aπ, − 1 < a < 1.
3
dx = 0, 0 < a < 1.
By integrating a suitable function of z, round a circular contour cut along the
∞
log x
π
positive real axis or otherwise, prove that
dx = −
.
2
2
3
3
(
x
+
x
+
1
)
0
∫
9.
By integrating
z
round a large semi-circle in the upper half plane
1+ z + z2
indented at the origin, prove that
∞
∫
0
x p −1
1− x + x
2
dx =
 2πp + π 
sin 
 cosec πp.

6 
3
2π
Power Series and Calculus of Residues
∞
10. Prove that
1
∫ (1 + x 2 )2 dx = − 4 π
0
∞
11. Prove that
log x
167
∫
0
∞
x a −1
x a −1
π
dx = π cot aπ and
dx =
− 0 < a < 1.
1− x
1+ x
sin aπ
∫
0
3.27 Integration along Contours other then Circle or Semi Circle
So far we have used the contours, circles or semi-circles, with or without indentation
∞
for evaluating the integrals
∫
f ( x )dx.
–∞
Now we shall illustrate the use of quadrant, sector of a circle and rectangular
∞
contours, with or without indentation, for evaluating the integrals
f ( x )dx.
∫
–∞
2
Ex. 59 : By integrating e − z round the rectangle whose vertices are 0, R, R + ia, ia,
show that
∞
(i)
∫
e–a
2
2
e − x cos 2a x dx =
0
Sol : Consider the integral
∫
2
∞
(ii)
π
∫
2
e − x sin 2ax dx = e − a
0
2
a
∫e
y2
dy
0
f ( z )dz, where f ( z ) = e
−z2
taken round closed contour
C
C which is the perimeter of the given rectangle OABD. Since f ( z ) is analytic within
and on the contour C (i.e, there is no singularity within the contour) hence by
Cauchy's theorem, we have
∫
f ( z )dz = 0 or
C
or
∫e
∫e
−z2
dz = 0
C
−z2
dz +
OA
∫e
−z2
dz +
AB
∫e
−z2
dz +
BD
∫e
−z2
dz = 0
DO
...(3.1)
y
Since on OA, z = x, dz = dx, x varies from O to R.
On AB, z = R + iy , dz = idy , y varies from 0 to a.
ia iR
D
On BD, z = x + ia, dz = dx, x varies form R to O.
R+ia
B
On DO, z = iy , dz = i dy , y varies from a to 0.
R
Hence the equation (3.1), becomes
O
A
x
Fig. 3.18
R
∫
0
2
e – x dx +
a
∫
0
2
e –( R + iy ) i dy +
0
∫
R
2
e –( x + ia ) dx +
0
∫
a
2
e –(iy ) i dy = 0 ...(3.2)
168
Engineering Mathematics-III
a
Now
∫
a
2
2
e −( R + iy ) i dy ≤ e −( R + iy ) i dy
∫
0
0
a
∫
=
e−R
2+ y 2
a
dy ≤
e−R
∫
0
2 + a2
dy ,
[since y ≤ a on AB]
0
= e( − R
2 + a2 )
a which → 0 as R → ∞
Hence, when R → ∞ equation (2) becomes
∞
e
∫
∞
– x2
–
0
∫
e
–( x + ia )2
a
dx – i
0
∫
2
e –(iy ) dy = 0
0
[Interchanging the limits of integral]
∞
∫
2
e −( x + ia ) dx =
0
∞
∫
a
2
e − x dx − i.
0
∫
2
e y dy
0
∞
[since
∫
2
e − x dx =
0
∞
i.e,
∫
e–(x
2 + a 2 − 2aix )
dx =
0
∞
and
∫
(e − x
2 + a2
a
π
−i
2
∫
π
]
2
2
e y dy
0
)(cos 2ax − i sin 2ax ) dx =
0
π
−i
2
a
∫
2
e y dy
0
Equating real and imaginary parts, we have
∞
∫
e
e−a
cos 2ax dx =
2
2
2
0
∞
and
∫
2
− x2
e − x sin 2ax dx = e − a
0
π
a
∫e
y2
dy .
0
Ex. 60 : Apply the calculus of residues to prove that
∞
∫
−∞ e
Sol : Consider
∫
C
e ax
x
+1
dx = π cosec a π,( 0 < a < 1)
f ( z ) dz where f ( z ) =
e az
1 + ez
taking the contour C as the
perimeter of a rectangle whose sides are the lines, real axis, x = ± R and y = 2π.
The pole of f ( z ) are given by 1 + e z = 0 i. e., e z = −1 = ei ( 2n +1) π i. e., z = i( 2n + 1)π
where n = 0, ± 1, ± 2 . Out of these poles only the simple pole z = πi lies within the
contour.
Power Series and Calculus of Residues
169


e az
At z = πi the residue = 
= − e aπ i
z 
( d/ dz )(1 + e ) z = πi
Hence, by residue theorem
f ( z )dz = 2πiΣR +
∫
C
R
or
∫
2π
f ( x )dx +
f ( R + iy ) idy
∫
–R
0
0
–R
+
f ( x + 2πi ) dx +
∫
R
2π
2π
But
2π
f ( R + iy ) i dy ≤
∫
f ( − R + iy ) i dy = 2πi( − e aπi ) ...(3.1)
∫
∫
0
 f ( R + iy ) i dy
y
–R+2 i
D
0
2π
e a( R + iy )
∫
=
1 + e R + iy
0
R+2 i
C
.
 i dy
A
B
O
–R
2π
∫
≤
0
a( R + iy )
e
 dy ≤
R + iy
e
−

 1
e aR
=
eR − 1
e
(1− a )R
0
Similarly
∫
1 − e−R
2π
e
aR
0
Fig. 3.19
eR − 1
dy
2π
2π
=
∫
e aR
→ 0 as R → ∞ since 0 < a < 1.
− e − aR
e − aR
f ( − R + iy ) i dy ≤
2π
which → 0 as R → ∞ since 0 < a <1.
− e −(1− a ) R
Hence when R → ∞, relation (3.1) reduces to
–∞
∞
∫
f ( x )dx +
−∞
∞
i.e.,
or
∫
e ax
–∞ e
∞
e ax
∫
∫
f ( x + 2πi )dx = −2π ie aπi
∞
x
x
R
2π
2π
=
i
+1
x
–∞ e + 1
∞
dx −
∫
–∞ e
∞
dx −
e a( x + 2 πi )
∫
–∞
x + 2 πi
+1
e 2aπi e ax
ex +1
dx = −2πi e a π i
dx = −2πie a πi
170
Engineering Mathematics-III
or
(1 − e 2aπi )
or
eiπa ( e − πai − e πai )
or
e
iπa
∞
∫
x
–∞ e + 1
∞
e ax
∫
( −2 i sin πa)
or
e ax
–∞ e
x
∞
e ax
∫
+1
–∞ e
x
∞
e az
∫
–∞1 +
+1
ez
dx = −2πie a πi
dx = −2πie a πi
dx = −2πie aπi
dz = π cosec ( πa)
Problem Set (3.9)
∞
1.
By contour integration prove that
∫
sin x
0
2.
By integrating
∫
0
cos x
x
dx =
 π
 .
 2
e az
(a real and 0 < a < 1) round the rectangle with corners at
1 − ez
R, R + iπ, − R + iπ, − R indented at the origin prove that
∞
(i)
∫
e ax
–∞1 +
∞
(ii)
3.
x
∞
dx =
∫
ex
e ax
x
–∞1 − e
dx = π cosec aπ and
dx = π cot aπ
By integrating f ( z ) =
∞
∫
0
sinh az
, round a suitable rectangle, prove that
sinh πz
sinh ax
1
1 
dx = tan  a 0 < a < π
2 
sinh πx
2
[Hint: Take the rectangle with vertices at R, R + i,− R + i and −R indented at z = 1]
4.
A contour C is formed by indenting at 0 and i the rectangle with vertices at
e az
round this contour, that
R, R + i, − R + i − R. Prove by integrating
sinh z
∞
sinh αx
1
α
dx = tan . ( 0 < α < π ).
sinh πx
2
2
∫
0
The Calculus of Residues [Contour Integration]
5.
eiz
z
By integrating
∞
Deduce the
171
2
∞
round a suitable contour, prove that
∫
0
sin x 2
1
dx = π .
x
4
sin x π
= .
x
2
∫
0
2
Hint : Consider the integral
∫
f ( z )dz, where f ( z ) =
C
eiz
.
z
Clearly f ( z ) is analytic for all finite values of z except at
iR
z = 0. Therefore we take the closed contour C,
consisting of the positive quadrant C R of a large circle
z= R and the positive real and imaginary axis
CR
Cr
.O
r
indented at z = 0. Let r be the radius of indentation.
Since there are no singularities within the contour,
hence
by
residue
theorem,
we
have
f ( z )dz = 2πiΣR + = 0
x
R
Fig. 3.20
∫
C
∞
6. Prove that
∫
sin x 2 dx =
0
∞
∫
cos x 2 dx =
0
Hint : Consider the integral
π
2 2
.
∫ f (z )dz, where f (z ) = e
−z2
.
C
Clearly f ( z ) is analytic for all finite values of z.
Take the contour C consisting of the are C R of the large
Fig. 3.21
circle z= R, of 0 ≤ θ ≤ π / 4 bounded by two radii OA
and OB as shown in fig 3.21, f (z) is analytic at all points
within and on the contour C.
Therefore, by residue theorem, we have
∫ f (z )dz = ∫ f (z )dz + ∫ f (z )dz + ∫
C
7.
OA
CR
f ( z )dz = 2πiΣR + = 0
BO
eiz
along the boundary
z+a
x = 0, x = R, y = 0, y = R, prove that
Integrating
∞
(i)
∫
0
cos x
dx =
x+a
∞
∫
0
xe − ax
1 + x2
of
the
∞
dx, a > 0
(ii)
∫
0
square
sin x
dx =
x+a
defined
∞
∫
0
e − ax
1 + x2
dx
by
172
Engineering Mathematics-III
∞
8.
Apply the calculus of residue to prove that
∫
0
cosh ax
1
dx = sec( a/ 2) where
cosh πx
2
− π < a < π.
Objective Type Questions
Multiple Choice Questions
Tick the correct answer :
1.
The residue of f ( z ) =
z
at z = 1 is :
( z − 1)( z − 2)
(a) 0
(b) 1
(c) –1
(d) ∞.
3
2.
3.
Residue of
z
at z = ∞ is :
( z − 1)( z − 2)( z − 3)
(a) – 6
(b) +6
(c) 4
(d) 5.
If z = 0 is a simple pole of f ( z ), then the residue at this pole is given by :
(a) lim ( z − a) f ( z )
(b) lim ( z − a) f ( z )
(c) lim zf ( z )
(d) f ( a).
z→ a
z→ 0
z→ 0
4.
Residue of
i ( 2a)
5.
( z + a 2 ) n +1
( 2n )!
(a)
(c)
1
2
2n + 1
( n !)
2
( 2n )!
( 2a) 2n ( n !) 2
If f ( z ) =
z6
(z 4 + a 4 )2
at z = ai is :
(b)
( 2n )!
i ( 2a) 2n +1 . n !
(d) none of these.
, it's poles are z = α1 , α 2 , α 3 , α 4
where α1 = aeiπ / 4 , α 2 = ae 3 πi / 4 , a3 = ae 5 π i / 4 , α 4 = ae7 πi / 4 . Order of each
pole is 2. If α is either of the poles, then the residue of f ( z ) at z = α is :
3α
3
(a)
(b)
16α
16
(c)
1
4α
(d) none of these.
Power Series and Calculus of Residues
6.
The residue of the function
(a)
3
16i
(c) −
(c)
3
16i
( z + 1) 3
at z = i is :
3i
16
(d) none of these.
1
π
at z =
is :
sin z − cos z
4
1
(b) 2
2
1
2
8. If f ( z ) =
1
2
(b)
7. Residue of
(a)
173
(d) none of these.
e miz
(z 2 + a 2 )
, then Res f ( z ) at z = ai is :
(a)
e − ma
2 ia
(b)
(c)
e − ma
2a
(d) none of these.
9. Residue of
1
3
z − z5
(a) 1
e ma
2 ia
at z = 0 is :
(b) 0
3
(d) .
2
(c) 2
 1 
10. Residue of cos 
 at z = 2 is :
 z − 2
(a) 1
1
(c)
2
(b) 0
(d) 2.
Fill in the Blanks
1.
If f ( z ) is analytic, except at a finite number of poles within a closed contour C
and continuous on the boundary of C, then f ( z ) dz = ...................... .
∫
C
2.
If f ( z ) =
3.
If f ( z ) =
4.
sin 2z
( z + 1) 3
log z
, then Residue f ( z ) at z = −1 is ............... .
then Residue f ( z ) at z = i is ......................... .
(1 + z 2 ) 2
A function which is analytic everywhere in the finite complex plane except at a
finite number of poles is called a .............. .
174
Engineering Mathematics-III
State True/False
1.
2.
If f ( z ) has a zero of n th order at z = z 0 then 1/ f ( z ) has a pole of order n at
z = z 0.
Residue of z 2 /( z 2 + 1) 2 at z = i is i /4.
ANSWERS
Problem Set 3.5
1.
1
(at z = 0)
2
2.
4
5
at z = −2 ,
at z = 1
9
9
3.
1 1 1 1
, ,− ,− (at z = 1,−1, i, − i )
4 4 4 4
4.
1
(at z = 0)
24
5.
1 1
1
1
1
, (1 + 2 i ), (1 − 2 i ) (at z − 1, i − i ) 6. e – a , e a (at z = 0, ia,− ia)
2 4
4
2
2
8. −
7. 1 at z = 0
−
3
at z = −1
9.
17
10.
4
at z = 0
3
i
32c
3
at z = ie,
i
32c 3
at z = − ic
11.
13.
101
7
at z = 1, − 8 at z = 2,
at z = 3
16
16
1

aa
12. e  + 1 at z = a
2
14. 1 at z = nπ
π( m − a) 2
15. 2πi
16. 2πi
17. 2πi
18. −4πi
19. 2πi
20. 0(z = ± π poles of order 2 in C)
21. 2πi
22. 2π/ 3 { z = ( −2 + 3 )i simple pole}
Problem Set 3.6
6.
11.
0
10. 2π
3
9. 0
π
1+ a
15.
2
2π
1– p
17. π
2
Power Series and Calculus of Residues
175
Problem Set 3.7
π
1.
π
2a
2.
4 a3
3.
π
4
4.
3π
16
5.
π
16
6.
π
3
8.
π
6
π
2e
πi
e6
3πi
,e 6
5 πi
6
7.
z=
9.
3 2π
16
10.
11.
e −b e −a 
−


a 
(a 2 − b 2 )  b
12.
13.
π − ma
e
2
15.
π  e −2 e −6 
−


2 3
6 
,e
lie inside C.
π
17.
−
π
2 3
−
e
 ma
× cos  
 2
2
π sin 1
2e 2
14.
− πe −2 π
16.
π  31 e −6 e −8 
+


196  27
2 
π
3
Multiple Choice Questions
1. (c)
2. (a)
3. (c)
4. (a)
5. (a)
6. (a)
7. (a)
8. (a)
9. (a)
10. (b)
Fill in the Blanks
1. sum of the residues
3.
1 π 
 + i
42 
2. 2 sin 2
4.
Meromorphic function
176
Engineering Mathematics-III
True or False
1. True
2. False
❑❑❑
Unit-2
4. Statistics
5. Curve Fitting
6. Correlation and Regression
7. Probability Theory
Unit-2
Chapter
4
Statistics
4.1 Statistics
It is that branch of science which deals with the collection of data, organising,
summarising, presenting and analysing data and making valid conclusion and
drawing reasonable decisions on the basis of such study.
4.2 Frequency-Distribution and Graphical Representation
The statistical data is summarised by classifying it into classes or categories with
their frequencies.
Consider the marks obtained by 50 students which is as follows :
Marks
Number of students
0-10
10-20
20-30
30-40
40-50
6
11
18
10
5
Graphical representation is commonly used to represent frequency distribution by
means of a diagram. The common diagrams are as follows :
(a) Histogram
(b) Frequency Polygon
(c) Frequency curve
(d) Cumulative frequency curve or ogive
(e) Bar chart
(f) Pie chart.
We now discuss few of these, e.g., (a), (b) and (c).
y
Histogram
It consists of a collection of rectangles with their
heights proportional to class frequencies of equal
class intervals whereas for unequal class interval,
the rectangular areas are proportional to the
frequencies (see Fig. 4.1).
Frequency Polygon
The graph obtained by connecting mid-points on the
tops of the rectangles in histogram is called
frequency polygon and is a line.
.
20
10
5
0
Frequency
Polygon
.
15
.
10
.
20
30
Fig. 4.1
.
40
50
x
180
Engineering Mathematics-III
Ogive or Cumulative Frequency Curve
Consider the upper limit of the class as x-coordinate and the cumulative frequency
as y-coordinate. Such points are then joined by a free hand smooth curve, known as
an ogive (see Fig. 4.2).
4.3 Measure of Central Tendency (or Average)
4.3.1 Arithmetic Mean
50
45
40
35
30
25
20
15
10
5
0
.
.
. .
Ogi
ve
An average is a value representing a set of data and is also
called as the measure of the central tendency. As a
matter of fact, there exist five types of common averages :
(i) Arithmetic mean or mean, (ii) Median, (iii) Mode
(iv) Geometric mean, (v) Harmonic mean
.
Consider x1 , x 2 , x 3 ..., x n as n numbers then their
arithmetic mean is defined by
10 20 30 40 50
x + x 2 + x 3 +…+ x n
Σx
Fig. 4.2
A. M. = 1
= i
n
n
In case x1 is repeated f1 times, x 2 is repeated f 2 times, x 3 is repeated f 3 times, ...,
x n is repeated f n times then their arithmetic mean is defined by
A. M. =
n
x1 f1 + x 2 f 2 +…+ x n f n n
=∑ xi fi / ∑ fi
f1 + f 2 +…+ f n
i =1
i =1
where fi is the frequency of occurrence of xi
Short cut method : Let the assumed mean to be a, if the deviation of the variable
xi from a be di , then we have xi − a = di
⇒
⇒
Σfi di
Σf ( x − a) Σfi xi − aΣfi
Σf x
= i i
=
= i i − a = A. M. − a
Σfi
Σfi
Σfi
Σfi
A.M. = a +
Σfi di
Σfi
Step deviation method : Let the width of the class interval be i and a be the
assumed mean, then
x −a
Di = i
⇒ xi − a = iDi ⇒ di = iDi
i
Σf d
Then we have A.M. = a + i i
Σfi
or
A.M. = a +
iΣfi Di
Σf D
= a+i i i
Σfi
Σfi
when the width of the class interval i is constant.
Statistics
181
Illustrative Examples
Ex. 1: Find the mean of 20, 22, 25, 28, 30, 55.
20 + 22 + 25 + 28 + 30 + 55 180
Sol : A.M. =
=
= 30
6
6
Ex. 2: Find the mean of the following:
Variable
8
10
15
20
8
Frequency
5
8
8
4
5
Sol :
Σfi xi = 8 × 5 + 10 × 8 + 15 × 8 + 20 × 4 + 8 × 5
= 40 + 80 + 120 + 80 + 40 = 360
Σfi = 5 + 8 + 8 + 4 + 5 = 30
Σf x
360
A.M. = i i =
= 12
Σfi
30
and
Ex. 3: Find the arithmetic mean for the following distribution :
Class
Frequency
0-10
10-20
20-30
30-40
40-50
12
8
15
10
5
xi − a = di
fi di
Sol. Let the assumed mean be 25 i.e., a = 25
Class Interval Mid-value xi Frequency fi
0-10
5
12
– 20
– 240
10-20
15
08
– 10
– 80
20-30
25
15
0
0
30-40
35
10
10
100
40-50
45
05
20
100
Total
50 = Σfi
−120 = Σfi di
Σf d
−120
A.M. = a + i i = 25 +
= 25 − 2 .4 = 22 . 6
Σfi
50
Then
By step deviation method :
xi − a
Σf D
We have
= Di ⇒ A.M. = a + i i i , a = 25
i
Σfi
Class Interval Mid-value xi
Frequency fi D = xi − a
i
i
fi Di
0-10
05
12
–2
– 24
10-20
15
08
–1
– 08
20-30
25
15
0
0
30-40
35
10
1
10
40-50
45
05
2
10
Total
50
−12 = Σfi Di
182
Engineering Mathematics-III
A.M. = a + i
Σfi Di
( −12)
= 25 + 10
= 22 . 6
Σfi
50
4.3.2 Median
Connor emphasized that the median of a series is that value of the variable which
separates the group into two equal parts, one part having all values greater and the
other part having all values lesser than the median. When the items are arranged in
ascending or descending order of magnitude then the median thus becomes the
measure of the central item. In case the total number of the item is odd (say equal to
( n + 1)
n), then the value of
th item gives the median, while the total number of
2
items is even (say equal to n), then there exist two middle items and the mean of the
n th
1

values of
and  n + 1 th item is defined as the median.


2
2
1
N −F
For a grouped data, the median is defined as l + 2
. i, where l is the lower
f
limit of the median class, f is the frequency of the class, i is the width of the class
interval and F is the cumulative frequency of the class preceding the median class
and N is the total frequency of the data.
Partition Values
As defined earlier, median, is the value of that item which separates the data into
two equal parts. We could however divide the observations into four equal parts by
three points; into ten equal parts by nine points or into 100 equal parts by 99 points.
The variable values which divide the whole obervations into four equal parts are
called quartiles; which divide the whole observation into ten equal parts are called
deciles while those which divide the total frequency into 100 equal parts are called
percentiles. Thus, there exist only three Quartiles, nine Deciles and ninety nine
percentiles of a series and are denoted respectively by Q, D and P. First, second and
third quartiles are Q1 , Q 2 , Q 3 respectively. Nine deciles are D1 , D2 , D3 ,…, D9 ; while
99 percentiles are P1 , P2 , …, P99 . We now give a mathematical formula to determine
Q, D and P respectively as follows for a continuous series.
N
N
−F
3 −F
Q1 = l + 4
i ; Q 2 = median; Q 3 = l + 4
i
f
f
N
8N
−F
−F
10
D1 = l +
i ; D8 = l + 10
i
f
f
and
N


N
− F
 88
−F


100
100
P1 = l +
i , P88 = l +
i
f
f
Statistics
183
 N

− F
j
 4

Qj = l +
i ; ( j = 1, 2, 3)
f
i.e.,
 N

− F
k
 10

Dk = l +
i ; ( k = 1, 2, ..., 9)
f
N


− F
m
 100

= l+
i; ( m = 1, 2, ..., 99)
f
Pm
Ex. 4: Find the median, lower quartile and upper quartile for the following
observations :
Marks
Number of students
0-10
22
10-20
38
20-30
46
30-40
35
40-50
20
Sol:
Class Interval
Frequency = f
Cumulative frequency = F
0-10
22
22
10-20
38
60
20-30
46
106
30-40
35
141
40-50
20
161
Total
Σf = N = 161
N
= 80.5, which corresponds to cumulative frequency 106. Hence
2
median class is 20-30.
N
−F
80.5 − 60
Now,
median = l + 2
i = 20 +
10
f
46
N = 161 ⇒
= 20 + 4.46 = 24.46
184
Engineering Mathematics-III
(ii) To calculate lower quartiles
N 161
=
= 40.25
4
4
Hence Q1 lies in the class interval (10-20)
N
−F
40.25 − 22
⇒
Q1 = l + 4
i = 10 +
10
f
38
182 . 5
= 10 + 4.8 = 14.8
38
To calculate upper Quartile
3N 3 × 161 488
=
=
= 120.75
4
4
4
⇒ Q 3 lies in the class interval (30-40)
3N
−F
120.75 − 106
Hence
Q3 = l + 4
i = 30 +
10
f
35
= 10 +
= 30 +
147.5
= 30 + 4.21 = 34.21
35
4.3.3 Mode
Mode is that value of the variable which occurs most frequently in a set of
observations. In other words, mode of a frequency distribution is the value of the
variable ( x ) corresponding to maximum frequency.
Mode of a grouped frequency distribution is given by the following formula :
f m − f m −1
Mode = l +
i
2 f m − f m −1 − f m +1
where l = lower limit of the modal class, f m = frequency of the modal class, f m −1 =
frequency of the class preceding the modal class, f m +1 = frequency of the class
succeeding the modal class, i = class interval.
For a symmetrical distribution; mean, median and mode coincide while for a
moderately asymmetrical distribution, mode is given by
Mode = 3 Median - 2 Mean (emperical formula)
Ex. 5: Calculate the mode from the following data :
Wages (in Rs.)
Number of workers
Below 100
8
100-200
12
200-300
25
300-400
15
400-500
10
Above 500
6
Statistics
185
Sol : The maximum frequency is 25, therefore Modal class is 200-300
l = 200, f m = 25 , f m −1 = 12, f m +1 = 15 and i = 100
f m − f m −1
Mode = l +
i
2 f m − f m −1 − f m +1
= 200 +
( 25 − 12) × 100
= 256 . 52
50 − 12 − 15
4.3.4 Geometric Mean
Definition. If x1 , x 2 , x 3 , ..., x n are n values of a variate x, none of them being
zero, their geometric mean G is defined as G = ( x1 . x 2 . x 3… x n )1/ n
⇒
log G = log ( x1 . x 2 . x 3 … x n )1/ n =
⇒
log G =
1
log ( x1 . x 2 . x 3 … x n )
n
1
(log x1 + log x 2 +…+ log x n )
n
While for frequency distribution, the geometric mean of n values x1 , x 2 , ..., x n of a
variate x with frequencies f1 , f 2 , … f n respectively, is given by
G = [ x1f1 . x 2f2 … x nfn ]1/ n
⇒
log G =
1
( f1 log x1 + f 2 log x 2 +…+ f n log x n )
n
⇒
log G =
1
n
n
∑ fi
log xi
i =1
4.3.5 Harmonic Mean
Definition. The Harmonic mean of a number of observations is the reciprocal of
the arithmetic mean of the reciprocals of the given observations.
Let x1 , x 2 , x 3 , … , x n be n observations, then their harmonic mean H is given by
H =
1
1 1
1
1 
+
+…+


n  x1 x 2
xn 
While the Harmonic Mean of a frequency distribution xi fi is given by
 1
1
1
=
fi   where N = Σfi
 xi 
H N
Ex. 6: If x1 , x 2 , ..., x n be non-zero positive numbers, show that
A ≥ G ≥ H (A being A.M., G being G.M. and H being H.M.).
Sol : We have
⇒
Also
x1 + x 2 +…+ x n
≥ ( x1 . x 2 … x n )1 n
n
A.M. > G.M.
1
1
1
+
+…+
1n
x1 x 2
xn
1 1
1 
≥
…

 x1 x 2
n
xn 
186
Engineering Mathematics-III
1
1
≥
⇒ H. M. ≤ G. M. ⇒ G. M. ≥ H. M.
H. M. G. M.
⇒
Hence AM ≥ GM ≥ HM
Ex. 7: Calculate mode from the following data :
Marks
No. of students
Marks
No. of students
Above 0
80
Above 60
28
''
10
77
''
70
16
''
20
72
''
80
10
''
30
65
''
90
8
''
40
55
''
100
0
''
50
43
Sol:
Marks
No. of students
0-10
3
10-20
5
20-30
7
30-40
10
40-50
13
50-60
15
60-70
12
70-80
6
80-90
2
90-100
8
Since maximum frequency is 15, so modal class is 50-60
f m − f m −1
Mode = l +
⇒
×i
2 f m − f m −1 − f m +1
= 50 +
15 − 13
2 × 10
× 10 = 50 +
= 50 + 4 = 54
30 − 13 − 12
5
4.4 Measures of Dispersion
By dispersion, we mean the spread or scatter of the values of the observations around
the measure of central tendency. Also dispersion of the data is the degree to which the
numerical data tend to spread about an average value-according to Spiegel.
For a given dispersion, it is important to know how the variables are scattered away
from the point of central tendency. Consider two groups of students consisting of (i)
Statistics
187
very bright and very dull students (ii) average and marginally intelligent, but having
the same average marks. These two groups differ in variation from the mean. Such
variation is called dispersion or scatter. Some measures of dispersion are as follows :
1. Range, 2. Quartile Deviation (or Semi-interquartile range), 3. Mean Deviation
about (a) mean (b) median , 4. Standard Deviation
1. Range
It is the difference between the value of the smallest item and the value of the largest
item included in a distribution, e.g., in the series 8, 9, 14, 10, 12, 7; range= 14 − 7 = 7.
Coefficient of Dispersion
The relative measure of the range is called the coefficient of dispersion and is obtained
by dividing the range with the sum of the extreme values, i.e., coefficient of dispersion
R − R2
,where R1 is maximum and R 2 is minimum value of the variate.
= 1
R1 + R 2
2. Quartile Deviation
(i) Inter-quartile range. It is defined as the difference between the third and first
1
quartiles i.e., Q 3 − Q1 ; while Q = (Q 3 − Q1 ) is called the quartile deviation or
2
semi-interquartile range.
Q − Q1
Also Quartile Co-efficient of dispersion is defined as 3
Q 3 + Q1
3. Mean Deviation
It is the arithmetic average of the deviations of various items from a measure of
central tendency (mean, median, mode). The mean deviation about any measure of
Σ|d|
central tendency is M.D. =
(for a series of individual values);|d| stands for the
N
absolute deviation of the values from the measure of central tendency and N is the
total number of values. Also for discrete variables of a frequency distribution, we
Σf |d|
have
;
M. D. =
N
while for a continuous series of frequency distribution xi fi , (i = 1, 2, ..., n ).
Mean deviation (M.D.) from a point 'a' is defined by
i
M.D. = ∑ fi | xi − a| where ∑ fi = N
N i
and | xi − a|= absolute value of the deviation ( xi − a).
1
Σ fi | xi − x|
N
Similarly, we can define mean deviation from the median, mode etc.
Also, mean deviation from the mean x is given by M.D. =
188
Engineering Mathematics-III
Ex. 8: Calculate the mean-deviation from the mean for the following data :
Marks
0-10
10-20
20-30
30-40
40-50
50-60
60-70
70-80
Freq.
5
8
7
12
28
20
10
10
Sol: Mean = [ Σf ( x )/ N ] =
Class
Interval
4500
= 45 = x
100
Mid-value
( x)
Freq. (fi )
fi ( x )
| x − x|
x = 45
fi| x − x|
0-10
5
5
25
40
200
10-20
15
8
120
30
240
20-30
25
7
175
20
140
30-40
35
12
420
10
120
40-50
45
28
1260
0
0
50-60
55
20
1100
10
200
60-70
65
10
650
20
200
70-80
75
10
750
30
300
N = 100
4500 = Σfi ( x )
Total
Then mean deviation from the mean =
1400
Σfi | xi − x| 1400
=
= 14
N
100
4. Standard Deviation (Root Mean Square Deviation)
Consider a frequency distribution
xi fi , i = 1, 2, ..., n, Σfi = N ; then we define
1
(i) Variance, as σ 2 =
Σfi ( xi − x ) 2
N
1
(ii) Standard Deviation S.D. = σ =
Σfi ( xi − x ) 2
N
(iii) Root Mean Square Deviation. Root mean square deviation about a point
1
`a' is defined by s =
Σfi ( xi − a) 2
N
S. D.
(iv) Coefficient of Variation. It is defined by C.V. = 100 ×
mean
(v)
Relation between σ and s is given as s 2 = σ 2 + d 2 , where d = x − a
(vi) Different Formulae for Standard Deviation
1
σ2 =
Σfi ( xi − x ) 2
N
Statistics
189
1
Σfi xi2 + x 2 − 2 xi x
N
1
1
1

=
Σfi xi2 + x 2
Σ i fi − 2 x  Σ fi xi 
N

N
N
=
(
)
1
Σfi xi2 + x 2 − 2 x 2
N
1
=
Σfi xi2 − x 2
N
1


∵ N = Σfi ; x =
Σfi xi


N
=
⇒
Variance = σ 2 =
or
S.D.=
1
1

Σfi xi2 −  Σfi xi 
N

N
2
1
1

Σfi xi2 −  Σfi xi 
N

N
2
Short-Cut method
s 2 = σ 2 + d 2 or σ 2 = s 2 − d 2
1
s2 =
Σfi ( xi − a) 2 , d = x − a
N
1
2
s =
Σfi di2 , xi − a = di
N
where
or
2
Σf d 


 Σf d 
d = ( x − a) =  a + i i  − a =  i i 


 N 
N


2
But
2
⇒
Variance = σ 2 =
⇒
S.D. σ =
Σfi di2  Σfi di 
−

 N 
N
2
2
 Σf d 2  Σf d  2 
i i
− i i 

 N  
 N

This shows that standard deviation is independent of change of origin.
Ex. 9: Calculate mean deviation from mean :
Marks
No. of students
Marks
No. of students
0-10
4
40-50
10
10-20
6
50-60
6
20-30
10
60-70
4
30-40
20
190
Engineering Mathematics-III
Sol:
xi − 35
10
fi di
|Di| =| xi − 35|
fi |Di|
5
—3
— 12
30
120
6
15
—2
— 12
20
120
20-30
10
25
—1
— 10
10
100
30-40
20
35
0
0
0
0
40-50
10
45
1
10
10
100
50-60
6
55
2
12
20
120
60-70
4
65
3
12
30
120
Total
60
Class
fi
xi
0-10
4
10-20
di =
0
680
Let assumed mean=35. Then, we have
n
Σ fi di
A.M. = a + i =1
× i = 35 +
n
Σ fi
0
× 10 = 35
60
i =1
M.D. =
Σfi |Di |
n
Σ fi
=
680
= 11 ⋅ 3
60
i =1
Ex. 10: Goals scored by two teams A and B in a football match were as follows :
No. of goals scored in a match
No. of matches
A
B
0
27
17
1
9
9
2
8
6
3
5
5
4
4
3
Find out which team is more consistent.
Sol: First we calculate coeff. of variation of the team A.
No. of goals No. of matches fi
scored xi
di = xi − 2
fi di
fi di2
0
27
–2
– 54
108
1
9
–1
–9
9
2
8
0
0
0
3
5
1
5
5
4
4
2
8
16
Total
N = 53 = Σfi
Σfi di = −50 Σfi di2 = 138
Statistics
191
Let assumed mean=2
Then
Mean = x = a +
⇒
σ=
Σfi di
50
= 2−
= 2 − 0.94 = 1.06
N
53
2
 1
 Σfi di  
2
Σ
f
d
−

 =

i i
 N  
 N

 138  −50 2 
−
  = 1.31

 53  53  
whence coefficient of variation for the team A is
S. D.
σ
1.31 × 100
=
× 100 = × 100 =
= 123.6
mean
x
1.06
Secondly we calculate C.V. for the team B.
No. of goals No. of matches fi
scored xi
di = xi − 2
fi di
fi di2
0
17
–2
– 34
68
1
9
–1
–9
9
2
6
0
0
0
3
5
1
5
5
4
3
2
6
12
Total
N = 40 = Σfi
Σfi di = −32
Σfi di2 = 94
⇒
Mean = x = a +
Σfi di
32
= 2−
= 2 − 0.8 = 1.2
N
40
2
 1
 94  −32 2 
 Σfi di  
2
Σ
f
d
−
=
−
= 1⋅ 3


 


i i
 40  
 N  
40

 N



σ
1.3 × 100
Coefficient of variation for the team B = × 100 =
= 108 ⋅ 3
x
1.2
and
σ=
Since, the coefficient of variation for the team B is less, so we conclude that team B is
more consistent.
Ex. 11: The following table shows the marks obtained by 100 candidates in an
examination. Calculate the mean, median and standard deviation.
Marks obtained
Candidates
1-10
11-20
21-30
31-40
41-50
51-60
3
16
26
31
16
8
192
Engineering Mathematics-III
Sol:
Class interval Mid-point xi Freq.
fi
di =
xi − 25.5
10
f1 di
fi di2
1-10
5.5
3
–2
–6
12
11-20
15.5
16
–1
– 16
16
21-30
25.5
26
0
0
0
31-40
35.5
31
1
31
31
41-50
45.5
16
2
32
64
51-60
55.5
8
3
24
72
65
195
Total
100 = Σfi
Let assumed mean = a = 25.5.
i = 10 ; Σfi di = 65,
N = Σfi = 100; Σfi di2 = 195
Mean, ( x ) = a +
Σfi di
65
× i = 25.5 +
× 10 = 25.5 + 6.5 = 32
Σfi
100
To calculate median
Class interval
Frequency fi
Cumulative frequency Fi
1-10
3
3
11-20
16
19
21-30
26
45
31-40
31
76
41-50
16
92
51-60
8
100
Total
 100
Median = Size of 

 2 
N = 100 = Σfi
th
item = Size of 50th item.
Hence the median class is 31-40.
N
−F
50 − 45
50
Now,
median = l + 2
× i = 31 +
× 10 = 31 +
= 32 .61
f
31
31
and
2
 Σf d 2  Σf d  2 
195  65 
S.D. (σ ) = i ×  i i −  i i   = 10 ×
−

 N  
100  100
 N

Statistics
193
= 10 × 1.95 − ( 0.65) 2
= 10 × 1 ⋅ 95 − 0 ⋅ 4225 = 12 ⋅ 36
Ex. 12: Calculate the mean and standard deviation of the following:
Size of item
6
7
8
9
10
11
12
Frequency
3
6
9
13
8
5
4
Sol:
xi
fi
di = xi − 9
fi di
fi di2
6
3
–3
–9
27
7
6
–2
– 12
24
8
9
–1
–9
9
9
13
0
0
0
10
8
1
8
8
11
5
2
10
20
12
4
3
12
36
Total
48 = Σfi
0 = Σfi di
124 = fi di2
Let assumed mean = a = 9
Σf d
Then,
mean ( x ) = a + i i = 9 + 0 = 9
Σfi
and
 Σf d 2  Σf d  2 
i i
− i i 

 N  
N


S.D. (σ ) =
 124  0  2 
−   =

 48  48 
=
 124

 = 1.607
 48 
4.10 Moments
Let
x
x1
x2
...
xn
f
f1
f2
...
fn
be a discrete frequency distribution. The arithmetic mean of the nth power of the
deviation from any point A is called the rth moment about A and is denoted by µ r ′.
k
∑ fi xir
The rth moment of x about the origin is given by µ ′r =
i =1
N
k
, where ∑ fi = N
i =1
194
Engineering Mathematics-III
Putting r = 1,2,3,..... we get 1 st , 2 nd , 3 rd ,.....moments raspectively
The first moment µ1′ is called the mean of x i.e.,
k
∑ fi x1i
x = µ1′ =
i =1
N
Also, the moments about any arbitrary point a, are defined by
k
∑ fi ( xi
µ ′r =
− a) r
i =1
N
when 'a' is equal to the mean, we get the moments about the mean, called central
moments, denoted by µ r i.e.,
k
∑ fi ( xi
µr =
− x )r
i =1
N
From these definitions, we have
1
µ′0 = µ 0 =
Σf = 1
N
1
1
1
µ ′1 =
Σfi ( xi − a) =
Σfi xi −
Σfi a = x − a
N
N
N
1
µ′2 =
Σfi ( x − a) 2
N
and
k
k
∑ fi
µ1 =
i =1
N
∑ fi
−x
k
∑ fi ( xi
µ 2 = i =1
i =1
N
= x−x=0
− x )2
= σ 2 (variance)
N
Thus, the second moment about the mean is equal to the variance of x.
(i) Moments about mean in terms of moments about any point a :
k
∑ fi ( xi
We have
Writing
where
Then
µr =
− x )r
i =1
N
( xi − x ) = ( xi − a) − ( x − a) = di − d,
xi − a = di and x − a = d
( di − d) r = dir − rC1 dir −1 d +
r
C 2 dir − 2 d 2 − r C 3 dir − 3 d 3 +…+( −1) r d r
Statistics
195
k
∑ fi ( xi
⇒
µr =
k
− x )r
∑ fi (di
i =1
i =1
=
N
− d) r
N
k
∑ fi[ dir − r C1 dir −1 d + r C 2 dir − 2 d 2 − r C 3 dir − 3 d 3 +…+(−1)r d r ]
=
i=1
N
k
k
∑ fidir
=
⇒
− r C1 d
i=1
N
∑ fidir −1
i=1
r
N
r
k
+ r C 2d 2
∑ fidir − 2
i=1
N
2
r
k
…+ (−1)r d r
∑ fi
i=1
N
3
µ r = µ ′ r − C1 d µ ′ r −1 + C 2 d µ ′ r − 2 − C 3 d µ ′ r − 3 +…+( −1) r d r
= µ r ′− rC1 µ ′ r −1 d + rC 2 µ ′ r − 2 d2 − rC 3 µ ′ r − 3 d3 +. . . +( −1)r dr
Putting r = 1, we get µ1 = µ ′1 − d = µ ′1 − µ ′1 = 0
Then putting
r = 2 and r = 3, we get
[Since d = µ ′1 and µ ′ 0 = 1]
µ 2 = µ ′ 2 − 2C1µ ′1 d + d 2 = µ ′ 2 −2 µ ′1 2 + µ ′1 2 = µ ′ 2 −µ ′1 2
and
µ 3 = µ ′ 3 − 3C1 µ ′ 2 d + 3C 2 µ ′1 d 2 − d 3
= µ ′ 3 −3µ ′ 2 µ ′1 +3µ ′1 µ ′12 −(µ ′1 ) 3
= µ ′ 3 −3 µ ′ 2 µ ′1 +3( µ ′1 ) 3 − ( µ ′1 ) 3 = µ ′ 3 −3µ ′ 2 µ ′1 +2( µ ′1 ) 3
Also
µ 4 = µ ′ 4 − 4C1 d µ ′ 3 + 4C 2 d 2 µ ′ 2 − 4 c 3 d 3 µ ′1 + d 4
= µ ′ 4 − 4 µ ′ 3 µ ′1 +6 µ ′ 2 (µ ′1 ) 2 − 4µ ′1 . (µ ′1 ) 3 + (µ ′1 ) 4
= µ ′ 4 −4µ ′ 3 µ ′1 + 6 µ ′ 2 µ ′12 −3 µ ′14
(ii) Moments about any point in terms of moments about mean :
k
∑
We have
Writing
where
But
µ′r =
i =1
∑ fi ( pi
=
N
i =1
N
( pi + d) r = pir + r C1 pir −1 d + r C 2 pir − 2 d 2 +…+ d r
µ′r =
∑ fi [ pir + r C1
d pir −1 + r C 2 d 2 pir − 2 +…+ d r
i =1
∑ fi pir
=
i =1
N
k
∑ fi ( xi
µr =
]
N
k
Again
+ d) r
( xi − a) = ( xi − x ) + ( x − a) = pi + d,
xi − x = pi and x − a = d
k
Therefore,
k
fi ( xi − a) r
i =1
N
k
∑ fi pir −1
+ r C1 d i =1
N
k
+ rC 2 d 2
∑ fi pir − 2
i =1
k
− x )r
∑ fi pir
=
i =1
N
k
and ∑ fi = N
i =1
N
k
+…+ d r
∑ fi
i =1
N
196
Engineering Mathematics-III
µ ′ r = µ r + r C1 d µ r −1 + r C 2 d 2 µ r − 2 + r C 3 d 3 µ r − 3 +…+ d r
⇒
Putting r = 0, 1, 2, … we get various moments about 'a'.
µ ′1 = µ1 + d = d
µ ′2 = µ 2 + 2 µ1 d + d 2
= µ 2 + d2
µ ′ 3 = µ 3 + 3 µ 2 d + 3 µ1 d 2 + d 3
= µ 3 + 3µ 2 d + d 3
µ ′4 = µ 4 + 4 µ 3 d + 6 µ 2 d 2 + 4 µ1 d 3 + d 4
= µ 4 + 4µ 3 d + 6µ 2 d 2 + d 4
4.6 Effects of Change of Origin and Scale on Moments
If variable u is related to x by
x −a
or xi − a = hui
ui = i
h
...(4.1)
∴
x − a = hu
when bar denotes the mean of the respective variable.
Subtracting (2) from (1), we have
∴
Similarly
...(4.2)
[from (1)]
xi − x = h(ui − u )
1
1
µ ′r =
Σfi ( xi − a) r =
Σfi h r uir
N
N
1
= hr
Σfi uir
N
1
1
µr =
Σfi ( xi − x ) r = h r
Σfi (ui − u ) r
N
N
Illustrative Examples
Ex. 13: Calculate the first four moments of the following distribution about the
mean.
x
0
1
2
3
4
5
6
7
8
f
1
8
28
56
70
56
28
8
1
xi
fi
di = xi − 4
fi di
fi di2
fi di3
fi di4
0
1
–4
–4
16
– 64
256
1
8
–3
– 24
72
– 216
648
2
28
–2
– 56
112
– 224
448
Sol:
Statistics
197
3
56
–1
– 56
56
– 56
56
4
70
0
0
0
0
0
5
56
1
56
56
56
56
6
28
2
56
112
224
448
7
8
3
24
72
216
648
8
1
4
4
16
64
256
Total
256 = Σfi
0 = Σdi
0 = Σfi di 512 = Σfi di2 0 = Σfi di3 2816 = Σfi di4
Now moments about the point x = 4 are given by
1
1
512
µ ′1 = ∑ fi di = 0, µ ′ 2 = ∑ fi di2 =
=2
N
N
256
1
µ ′ 3 = ∑ fi di3 = 0
N
1
2816
and
µ ′ 4 = ∑ fi di4 =
= 11
N
256
Moments about the mean are given as under
µ1 = 0, µ 2 = µ ′ 2 − µ ′12 = 2
µ 3 = µ ′ 3 −3 µ ′ 2 µ ′1 + 2 µ ′13 = 0
µ 4 = µ ′ 4 − 4 µ ′ 3 µ ′1 + 6 µ ′ 2 µ ′12 − 3 µ ′14 = 11
Ex. 14: Calculate the first and second moment about zero for the observations 2, 5, 8, 12, 20.
Sol: From the moment's formula about zero, we have
xi
xi2
2
4
5
25
8
64
12
144
20
400
Σxi2
Σxi = 47
µ ′1 =
= 637
Σ xi
47
=
= 9.4
n
5
Σ xi2
637
=
= 127.4
n
5
Ex. 15: Calculate first three moment's about 80 from the following frequency
distribution.
µ′2 =
x
72
78
80
81
82
85
87
f
5
9
13
6
2
1
2
198
Engineering Mathematics-III
Sol:
xi
fi
di = x i − 80
fi di2
fi di
fi di3
72
5
–8
– 40
320
– 2560
78
9
–2
– 18
36
– 72
80
13
0
0
0
0
81
6
1
6
6
6
82
2
2
4
8
16
85
1
5
5
25
125
87
2
7
14
98
686
– 29
493
– 1799
Total
38=N
1
1
Σfi ( xi − 80) =
Σfi di
N
N
−29
µ ′1 =
= −0.7631
38
1
1
493
µ′2 =
Σ fi ( xi − 80) 2 =
Σfi di2 =
= 17
N
N
29
1
1
−1799
µ′3 =
Σ fi ( xi − 80) 3 =
Σ fi di3 =
N
N
29
µ1 ′ =
⇒
⇒
µ ′ 3 = 62 ⋅ 0344
Ex. 16: Calculate the first four moments about mean for the following frequency
distribution.
x
0
1
2
3
4
5
6
7
8
f
1
8
28
56
70
56
28
8
1
Sol: Mean,
x=
Σfi xi
1024
=
=4
Σfi
256
First form the following table.
xi
fi
fix i
xi − 4
f i ( x i − 4) f i ( x i − 4)2
f i ( x i − 4)3
f i ( x i − 4)4
0
1
0
—4
—4
16
— 64
256
1
8
8
—3
— 24
72
— 216
648
2
28
56
—2
— 56
112
— 224
448
Statistics
199
3
56
168
—1
— 56
56
— 56
56
4
70
280
0
0
0
0
0
5
56
280
1
56
56
56
56
6
28
168
2
56
112
224
448
7
8
56
3
24
72
216
648
8
1
8
4
4
16
64
256
Total
256
1024
0
512
0
2816
µ1 =
Σfi ( xi − x ) Σfi ( xi − 4)
0
=
=
= 0 (always zero)
Σfi
Σfi
256
µ2 =
Σfi ( xi − x ) 2
Σf ( x − 4) 2
512
= i i
=
=2
Σfi
Σfi
256
µ3 =
Σfi ( xi − x ) 3
Σf ( x − 4) 3
0
= i i
=
=0
Σfi
Σfi
256
µ4 =
Σfi ( xi − x ) 4
Σf ( x − 4) 4
2816
= i
=
= 11
Σfi
Σfi
256
Ex. 17: Calculate the first four moment's for the following frequency distribution.
Class
Frequency
0-10
10-20
20-30
30-40
2
4
5
3
Sol:
Mid- u = xi − 25
i
10
value
fi ui
fi u i2
fi u i3
fi u 4
–2
–4
8
– 16
32
15
–1
–4
4
–4
4
1
25
0
0
0
0
0
30-40
3
35
1
3
3
3
3
Total
10
–2
–5
15
– 17
39
Class
fi
0-10
2
5
10-20
4
20-30
If we denote the moments of u about 0 by µ i′ then
200
Engineering Mathematics-III
µ1′ =
Σfi ui
−5
=
= − 0⋅ 5
N
10
µ 2′ =
Σfi ui2 15
=
= 1⋅ 5
N
10
µ ′3 =
Σfi ui3
17
=−
= −1 ⋅ 7
N
10
µ ′4 =
Σfi ui4
39
=
= 3⋅ 9
N
10
If we denote the moments of u about means by µ i then applying the formula for
moment about mean in terms of moment about any point, then
µ 2 = µ ′2 − (µ1′ ) 2
= 1 ⋅ 5 − ( −0 ⋅ 5) 2 = 1 ⋅ 5 − 0 ⋅ 25 = 1 ⋅ 25
µ 3 = µ ′3 − 3 µ ′2µ1′ + 2(µ1′ ) 3
= −1.7 − 3(1.5)( −0.5) + 2( −0.5) 3
= −1.7 + 2 . 25 − 0 . 25
= 0 ⋅ 30
µ 4 = µ ′4 − 4µ ′3µ1′ + 6µ ′2 (µ1′ ) 2 − 3(µ1′ ) 4
= 3 . 9 − 4( −1 . 7 )( −0 . 5) + 6(1 . 5)( −0 . 5) 2 − 3( −0 . 5) 4
= 3 ⋅ 9 − 3 ⋅ 4 + 2 ⋅ 25 − 0 ⋅ 1875
= 2 ⋅ 5625
Now the required central moment is given by
µ1 = 0
µ 2 = h 2µ 2 = 10 2 × 1.25 = 1.25 × 10 2
µ 3 = h 3µ 3 = 10 3 × 0.30 = 3.0 × 10 2
µ 4 = h 4µ 4 = 10 4 × 2 . 5625 = 2 . 5625 × 10 4
Ex. 18: The first four moments of a distribution about the value 4 of the variable are
−1 ⋅ 5, 17, − 30 and 108. Find the variance and moments about the mean.
Sol: We are given with
1
µ1′ =
Σfi ( xi − 4) = −1.5
N
1
µ ′2 =
Σfi ( xi − 4) 2 = 17
N
1
µ 3′ =
Σfi ( xi − 4) 3 = −30
N
1
µ 4′ =
Σfi ( xi − 4) 4 = 108
N
For the moment about the mean
µ1 = 0 (always)
Statistics
Variance
201
σ 2 = µ 2 = µ ′2 − (µ1′ ) 2 = 17 − (1.5) 2
= 17 − 2 . 25 = 14 . 75
µ 3 = µ ′3 − 3µ ′2 (µ1′ ) + 2(µ1′ ) 3
= −30 − 3(17 )( −1.5) + 2( −1.5) 3
= −30 + 76.5 − 6.75 = 39.75
and
µ 4 = µ ′4 − 4 µ ′3 (µ1′ ) + 6µ ′2 (µ1′ ) 2 − 3 (µ1′ ) 4
= 108 − 4( −30)( −1.5) + 6(17 )( −1.5) 2 − 3( −1.5) 2
= 108 − 180 + 229.5 − 15.1875
= 142 ⋅ 3125
4.12 Skewness
Measures of central tendency give us an estimate of the representative value of a
series, the measures of dispersion gives an indication of the extent to which the
items cluster around or scatter away from the central value and the skewness is a
measure that refers to the extent of symmetry or asymmetry in a distribution. In
other words, it describes the shape of a distribution.
Literally, skewness mean lack of symmetry. We study skewness to have an idea
about shape of the curve which we can draw with the help of the given data. A
distribution is said to be skewed if :
(i)
(ii)
Mean (x), Median(M) and Mode(Z) fall at different points, i.e., Mean ≠
Median ≠ Mode.
Quartiles are not equi-distant from median, and
(iii) The Curve drawn with the help of the
given data is not symmetrical but
stretched to one side than to the other.
Types of Frequency Distribution
The following are the various types of
frequency distributionsin connection with
Skewness :
(1)
X =M=Z
(Q3 -M)=(M-Q 1)
Normal Frequency Distribution or
Bell Shaped Symmetrical Frequency
Distribution :
The curve of a bell shaped frequency curve is as follows :
Here the mean, median and mode and on one point and third and first quartiles are
at equal distances from median.
(2)
Asymmetrical Frequency Distribution or Skew Frequency Distribution :
A frequency distribution is said to be moderately asymmetrical or skew if it is not
symmetrical, i.e., if class frequencies are not distributed evenly on both sides of the
central maximum.
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Engineering Mathematics-III
Moderately asymmetrical frequency distribution
are of two types :
(a) Positively Skewed Distribution : In the
positively skewed distribution, the value of the
mean is maximum and that of mode least –the
median lies in between the two as is clear from
the adjacent diagram. The curve of such a
frequency distribution has longer tail to the right.
(b) Negatively Skewed : In a negatively
skewed distribution the value of mode is
maximum and that of mean least –the
median lies in between the two. The curve of
such a frequency distribution has a longer
tail to the left.
X >M>Z
(Q3-M)<(M-Q1 )
Positively Skewed Curve
Objectives of Skewness
(1) It helps in finding out the nature and
X>M>Z
(Q 3 -M)<(M-Q1 )
the degree of concentration – whether
Negatively Skewed Curve
it is in higher or the lower values.
(2) The empirical relations of mean,
median and mode are based on a moderately skewed distribution. The measure
of skewness will reveal to what extent such empirical relationship holds good.
(3) It helps in knowing if the distribution is normal. Many statistical measures,
such as the error of mean, are based on the assumption of a normal
distribution.
Test of Skewness
In order to find out whether a particular distribution is skewed, certain testes are
usually applied. They are as follows :
(1) In a skew distribution values of mean, median and mode would not coincide.
The mean and mode would be pulled wide apart and median would usually
lie between them. In a moderately asymmetrical distribution.
Mean – Mode = 3 (Mean – Median)
⇒
3 Median = 3 Mean +Mode – Mean
= 2 Mean +Mode
= 2 Mean + Mode +2 Mode – 2 Mode
(2)
(3)
= Mode + 2 (Mean – Mode)
2
Median = Mode + (Mean – Mode)
⇒
3
2
Median = Mode + (Mean – Mode)
∴
3
In a skewed distribution the two quartiles would not be equidistant from the
median or in other words (Q 3 − M ) − ( M − Q1 ) would not be zero.
A skewed distribution when plotted on a graph paper would not give a
symmetrical bell- shaped curve.
Statistics
203
(4)
The sum of positive deviations from the Median is not equal to the sum to
negative deviations.
(5) Frequencies are not equally distributed at various points which are
equ–distant from the Mode. In an asymmetrical distribution.
Mean–Median =3 (mean–mode) on either side.
Measures of Skewness
Measures of skewness tell us the direction and extent of asymmetry in a series and
permit us to compare two or more series with regard to these. They may either be
absolute or relative.
(1) Absolute Measures of Skewness : Skewness can be measured in absolute
terms by taking the difference between Mean and Mode.
Symbolically,
Absolute skewness =X − Mode
When Skewness is based on quartiles, absolute skewness in given by the formula
absolute skewness = Q 3 + Q1 − 2 Median.
If the value of Mean is greater than Mode skewness will be positive i.e., we shall get a
plus sign in the above formula. Conversely, if the value of mode is greater than mean
we shall get a minus sign meaning thereby that the distribution is negatively
skewed.
Drawback of Skewness
These measures of skewness suffer from some drawbacks. These are :
(i)
These measures are expressed in the units of a distribution (like rupees, tons,
liters, etc.) and as such can not be compared with measures expressed in
another distribution in different units. For example, absolute skewness of
weights of students (expressed in kgs.) cannot be compared with the
skewness in their heights (expressed in centimeters).
(ii) There is considerable variation in different distributions. In one distribution
the difference between Mean and Median may be very large as compared to
similar difference in another distributions, if and yet the two distributions, if
plotted on a graph paper, may give similar curves.
(2) Relative Measures of Skewness : For purpose of comparison it is necessary
to have relative measures of skewness. Relative measures of skewness are obtained
by dividing the absolute measures by any one of the measures of dispersion. Relative
measures of skewness are known as coefficient of skewness.
There are four measures of relative skewness. They are :
(i)
The Karl Pearson's coefficient of Skewness.
(ii) The Bowley's coefficient of Skewness.
(iii) The Kelly's coefficient of Skewness.
(iv) Measure of skewness based on moments.
(i) Karl Pearson's Coefficient of Skewness (First Measure of Skewness) :
Karl Pearson has given a formula for relative measures of skewness. It is known as
Karl Pearson's coefficient of Skewness or Pearsonian coefficient of Skewness. The
204
Engineering Mathematics-III
formula is based on the difference between the mean and mode, divided by the
standard deviation. The formula thus becomes
Mean − Mode
X−M
Karl Pearson's coefficient of Skewness =
=
Standard Deviation
σ
If in a particular frequency distribution, it is difficult to determine precisely the
mode, or the mode is ill-defined, the coefficient of skewness can be determined by
the the following changed formula :
3 (Mean − Mode)
3( X − M)
Karl Pearson's coefficient of Skewness =
or
Standard Deviation
σ
(ii) Bowley's Coefficient of Skewness : An alternative measure of skewness
has been proposed by Professor Bowley. Its measure is based on quartiles. In a
symmetrical distribution first and third quartiles are equidistant from the Median as
can be seen from the following diagram.
Q1
Median
Q3
In an asymmetrical distribution the third quartile is the same distance over the
median as the first quartile is i.e.
Q 3 − Med = Med − Q1 or Q 3 + Q1 − 2 Med = 0
If this distribution is positively skewed the top 25 percent of the values will tend to
be farther from Median than the bottom 25 percent, i.e. Q 3 will be farther from
Median than Q1 is from Median and the reverse for negative skewness. Hence a
possible measure is
(Q 3 − Med) − (Med − Q1 )
Q + Q1 −2 Med.
Skewness B =
or 3
(Q 3 − Med) + (Med − Q1 )
Q 3 − Q1
Skewness B = Bowley's coefficient of skewness.
It must be remembered that the results obtained by these two measures are not to be
compared with one another. Especially, the numerical values are not related to one
another since the Bowley's measure, because of its computational basis, is limited to
values between -1 and +1, while Pearson's measure has no such limits.
(iii) Kelly's Coefficient of Skewness : Bowley measures neglects the two
extreme quartiles. To measure the skewness, it would be better to consider the
entire data or the more extreme items. His measure of skewness is based on two
deciles ; i.e. 10 th and 90 th percentiles (first and Ninth Deciles), symbolically
P + p − 2P50
kelly's coefficient of Skewness = 90 10
P90 − P10
Also kelly's coefficient of Skewness =
D9 + D1 − 2 Median
D9 − D1
This method is rarely used in practice.
Statistics
205
(iv) Measures of Skewness based on Moments : A measure of skewness can
be calculated by using the third moment about the Mean.
Karl Pearson defined the following four coefficients calculated from the moments
about the mean :
β1 =
µ 23
µ 32
µ4
, β2 =
µ 22
, γ 1 = β1 and γ 2 = β 2 − 3.
These β and γ coefficients give some idea about the shape of the curve obtained from
the frequency distribution. For symmetrical distribution (e.g. normal distribution),
it can be proved that all the moments of odd order about the mean vanish and so in
particular µ 3 = 0 ⇒ β1 = 0.
Hence β1 gives a measure of departure from symmetry. β1 is a measure of skewness.
Ex. 1. Calculate Karl Pearson's co-efficient of skewness form the data given below :
Value
10
20
30
40
50
60
70
Frequency
1
5
12
22
17
9
4
di2
f i di
f i di 2
Sol. Calculation of coefficient of Skewness
Value
(x i )
Freque
ncy ( f i )
Deviation from
assumed means (a)
=40
x − 40
di = i
10
10
1
–3
9
–3
9
20
5
–2
4
– 10
20
30
12
–1
1
– 12
12
40
22
0
0
0
0
50
17
+1
1
+17
17
60
9
+2
4
+18
36
70
4
+3
9
+12
36
Σ fi di = 22
Σ fi di 2 = 130
Total
Σfi = 70
 Σf d

22
Mean or X = a +  i i × i = 40 + 
× 10 = 43 .1


70
 Σfi

Mode or Z = 40 by inspection
Standard Deviation or σ =
Σfi di 2  Σfi di 
−

Σfi
 Σfi 
2
× 10
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Engineering Mathematics-III
2
=
130  22
−   × 10 = 13 .26
(70)  70
Karl Pearson's coefficient of Skewness =
Mean − Mode 43 .1 − 40
=
= + 0 .23
S. D.
13 .26
The Distribution is Positively Skewed.
Ex. 2. Calculate coefficient of skewness from the following :
Marks
(above)
No. of
Students
0
10
20
30
40
50
60
70
80
150
140
100
80
80
70
30
14
0
Sol. Calculation of coefficient of Skewness
Marks
M.V.
(X)
Frequency
( fi)
Deviation
from
assumed
means
(a)=45
X − 45
di =
10
0 – 10
5
10
–4
16
– 40
160
10 – 20
15
40
–3
9
– 120
360
20 – 30
25
20
–2
4
– 40
80
30 – 40
35
0
–1
1
0
0
40 – 50
45
10
0
0
0
0
50 – 60
55
40
1
1
40
40
60 – 70
65
16
2
4
32
64
70 – 80
75
14
3
9
42
126
Total
Σ fi = 150
 Σf d

Mean or X = a +  i i × i
 Σfi

86
= 45 − 
× 10
 150

= 45 − 5 .73
= 39 .27
di2
f i di
f i di 2
Σf idi = − 86
Σf idi2 = 830
Statistics
207
Standard Deviation
Σfi di 2  Σfi di 
=
−

Σfi
 Σfi 
2
×i
2
=
830  − 86
−

150  150 
=
=
5 .533 − 0 . 328 × 10
5 .205 × 10 = 2 .281 × 10
× 10
= 22 .81
Mode is ill-defined, therefore we must use median to calculate the coefficient of
skewness :
N 150
Median number, m =
=
= 75 which lies in the class 40–50.
2
2
m−F
Hence,
M = l1 +
×i
f
75 − 70
M = 40 +
× 10 = 40 + 5 = 45
10
3 (Mean − Median)
Coefficient of skewness =
Standard Deviation
3 ( 39 .27 − 45)
17 .19
=
=−
= 0 .75
22 .81
22 .81
Ex. 4. From the data given below, calculate Karl Pearson's, Bowley's and Kelly's
Coefficient of skewness.
Mean = 150, Median = 142, Q1 = 62, Q 3 = 195
D1 = 30, D9 = 230, Standard Deviation = 30
Sol. Karl Pearson's Coefficient of skewness
3 (Mean − Median) 3 (150 − 142)
=
=
= 0 .8
σ
30
Bowley's Coefficient of skweness
Q + Q1 − 2M 195 + 62 − 284
= 3
=
= − 0 .2
Q 3 − Q1
195 − 62
Kelley's Coefficient of skewness
D + D9 − 2M
= 1
D9 − D1
where, D1 and D9 are the first and ninetieth deciles (or 10 th and 90 th percentiles)
30 + 230 − 284
24
=
=−
= −0 .12
230 − 30
200
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Engineering Mathematics-III
4.13 Kurtosis
Kurtosis is yet another
measure which tells us
about the form of a
distribution. It tells us
whether the distribution, if
Lepto-Kurtic
plotted on a graph would
No.2
give us a normal curve, a
curve more flat than the
normal curve or a curve
mor peaked than the
normal curve.
Meso-Kurtic
According to Simpson and
No.1
Kafa, " The degree of Kurtosis
Platy-Kurtic
of a distribution is measured
relative to the peakeness of a
normal curve".
According to Croxton and
Cowden, "A measure of
No.3
Kurtosis indicates the
degree to which a curve of
frequency distribution is peaked or topped".
In the figure curve No. 1 is normal or mesokurtic, curve No. 2 is more peaked than
the normal curve and is leptokurtic and curve No. 3 is more flat than the normal
curve and is platykurtic.
Karl Pearson in 1950 introduced the terms mesokurtic, leptokurtic and platykurtic.
A peaked curve is called 'Leptokurtic' and a flat topped curve is termed platkurtic.
These are evaluated by comparison with an intermediate peaked curve called
'mesokurtic' which is normal curve. These three curve differ widely in regard to
convexity. These three types of curves are shown in the figure.
Measure of Kurtosis
Karl Pearson has defined the following four coefficients.
β1 =
µ 23
µ 32
γ1 = +
, β2 =
µ4
µ 22
or σ 4 =
β1 , γ 2 = β 2 − 3 =
Fourth Moment
Second Moment
µ 4 − 3 µ 22
µ 22
The first two is called Beta distribution and last two are called Gamma coefficient. In
a normal distribution β 2 will be equal to 3. If it is greater than 3, the curve is more
peaked than normal, if less than 3, the curve is flatter at the top than normal.
Statistics
209
Thus :
If β 2 = 3, the curve is normal or Mesokurtic.
If β 2 > 3, the curve is Leptokurtic or more peaked.
If β 2 < 3, the curve is plantykurtic or flat topped.
The measure of kurtosis is also represented by γ 2 . Therefore
If γ 2 or β 2 − 3 = 0, the curve is Mesokurtic.
If γ 2 is positive, the curve is Leptokurtic.
If γ 2 is negative, the curve is platykurtic.
Ex. 19: Calculate the first four moments about the mean of the following
distribution. Also calculate β1 and β 2 . The x values in centimeters are the mid-points
of intervals.
x
2.0
2.5
3.0
3.5
4.0
4.5
5.0
f
3
36
63
90
68
40
10
Sol: Let us take the origin at 3.5 to solve the given problem, then
x − 3.5
di = i
0.5
Applying this transformation, we have following table
xi − 3.5
0.5
fi di
fi di2
fi di3
fi di4
3
–3
–9
27
– 81
243
2.5
36
–2
– 72
144
– 288
576
3.0
63
–1
– 63
63
– 63
63
3.5
90
0
0
0
0
0
4.0
68
1
68
68
68
68
4.5
40
2
80
160
320
640
5.0
10
3
30
90
270
810
Total
310
34
552
226
2400
xi
fi
2.0
di =
µ1′ =
Σfi di
34
=
= 0.1096
N
310
µ 2′ =
Σfi di2
552
=
= 1.7806
N
310
µ ′3 =
Σfi di3
226
=
= 0.7290
N
310
µ ′4 =
Σfi di4
2400
=
= 7.7419
N
310
µ 2 = µ ′2 − (µ1′ ) 2 = 1.7806 − ( 0.1096) 2 = 1.7685
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Engineering Mathematics-III
µ 3 = µ ′3 − 3µ ′2µ1′ + 2(µ1′ ) 3
= 0.7290 − 3(1.7806)( 0.1096) + 2( 0.1096) 3
= 0.1461
µ 4 = µ ′4 − 4 µ ′3 µ1′ + 6 µ ′2 (µ1′ ) 2 − 3(µ1′ ) 4
= 7.7419 − 4 ( 0.1461)( 0.1096) + 6(1.7806)( 0.1096)2 − 3( 0.1096)4
= 7.6775
β1 =
β2 =
µ 23
µ 32
µ4
µ 22
=
=
( 0.1461) 2
(1.7685) 3
7.6775
(1.7685) 2
= 0.0038
= 2 . 4548
Ex. 20: Find the measures of skewness and kurtosis on the basis of moments for the
following distribution and draw your conclusion.
Class
5-15
15-25
25-35
35-45
45-55
2
4
6
3
5
Frequency
Sol: Form the following table.
Class
Midvalue xi
fi
xi − 30
10
a = 30, h = 10
fi ui
fi u i2
fi u i3
fi u i4
5-15
10
2
–2
–4
8
– 16
32
15-25
20
4
–1
–4
4
–4
4
25-35
30
6
0
0
0
0
0
35-45
40
3
1
3
3
3
3
45-55
50
5
2
10
20
40
80
5
35
23
119
Total
ui =
20
First four moments of variable u are given by
Σf u
5
µ1′ = i i =
= 0.25
N
20
µ 2′ =
Σfi ui2
35
=
= 1.75
N
20
µ ′3 =
Σfi ui3
23
=
= 1 .15
N
20
µ ′4 =
Σfi ui4 119
=
= 5 .95
N
20
Now the first four central moments of variable x are given by
Statistics
211
µ1 = 0 (always)
µ 2 = [µ ′2 − (µ1′ ) 2 ]h 2
= [1 .75 − ( 0.25) 2 ](10 2 )
= 1 .6875 × 10 2
µ 3 = [µ ′3 − 3 µ ′2 . µ ′3 + 2(µ1′ ) 3 ]h 3
= [1 . 15 − 3(1.75)( 0.25) + 2( 0.25) 3 ](10 3 )
= −0.13125 × 10 3 = −1 .3125 × 10 2
µ 4 = [µ ′4 − 4µ ′3 . µ1′ + 6 µ ′2 ( µ1′ ) 2 − 3 (µ1′ ) 4 ] h 4
= [ 5 .95 − 4(1 .15)( 0 .25) + 6(1 .75)( 0 .25) 2 − 3( 0 .25) 4 ](10 4 )
= 5 .44 × 10 4
µ 23
β1 =
µ 32
=
( −1.3125 × 10 2 ) 2
(1.6875 × 10 2 ) 3
0.0035 = 0.0591
γ 1 = β1 =
β2 =
µ4
µ 22
=
= 0.0035
5.44 × 10 4
(1.6875 × 10 2 ) 2
= 1.91
Ex. 21: In two frequency distributions the second moments about mean are 25 and
36 respectively, while third moments about mean are 43.2 and 85.75. Compare the
skewness in two frequency distributions.
Sol: Take the case of first frequency distribution.
µ 2 = 25 and µ 3 = 43.2
β1 =
and
µ 23
µ 32
γ 1 = β1 =
µ3
µ2
32
=
43.2
( 25)
32
=
43.2
5
3
=
43.2
= 0.3456
125
Hence, distribution is positively skew.
For the second frequency distribution
µ 2 = 36, µ 3 = 85.75
β1 =
and
µ 23
µ 32
γ1 =
µ3
µ2
32
=
85.75
( 36)
32
=
85.75
63
85.75
= 0.39699
216
Hence, distribution is positively skew.
=
Comparison. Both distribution are positively skew but distribution 2nd is more
skew than distribution 1st.
212
Engineering Mathematics-III
Ex. 22: The fourth moment about mean of a frequency distribution is 243. What
must be the value of its standard deviation in order that the distribution be (i)
Leptokurtic (ii) Mesokurtic (iii) Platykurtic?
µ
µ
243
Sol: Coefficient of kurtosis β 2 = 4 = 4 =
2
4
µ2 σ
σ4
(i) For Leptokurtic, β 2 > 3
243
or
> 3 or σ 4 < 81
σ4
or
σ<3
(ii) For Mesokurtic, β 2 = 3
Thus
σ=3
(iii) For Platykurtic, β 2 < 3
Thus
σ>3
Ex. 23: For a distribution with mean x = 10, variance σ 2 = 16, γ 1 = 1 and β 2 = 4,
find the first four moments about origin.
Sol: Given x = 10, σ 2 = µ 2 = 16
µ3
γ 1 = β1 =
=1
µ 32/ 2
Therefore,
Also given
Therefore,
µ 3 = µ 2 3 2 = (16) 3 2 = ( 4 2 ) 3 2 = 64
µ
β2 = 4 = 4
(µ 22 )
µ 4 = 4 × 16 2 = 1024
Moments about origin are given by
µ ′1 = µ1 + d = 0 + 10 = 10 [where d = x – 0, x = 10]
µ ′2 = µ 2 + ( d) 2 = 16 + 10 2 = 116
µ ′3 = µ 3 + 3 µ 2 d + ( d ) 3
= 64 + 3(16)(10) + (10) 3
µ ′4
= 64 + 480 + 1000 = 1544
= µ 4 + 4 µ 3 d + 6 µ 2 ( d1 ) 2 + d14
= 1024 + ( 4) × ( 64) × (10) + ( 6) × (16)(10 2 ) + 10 4
= 1024 + 2560 + 9600 + 10000 = 23184
Ex. 24: For any frequency distribution, show that β 2 ≥ 1.
Sol: To show that β 2 ≥ 1.
β2 ≥ 1
µ4
≥1
µ 22
If
or
if
1
1
Σ fi ( xi − x ) 4 ≥  Σfi ( xi − x ) 2 
N
N


2
Statistics
213
2
or if
1
1
Σfi y i2 ≥  Σfi y i  , where y i = ( xi − x ) 2
N

N
or if
1
1
Σfi y i2 −  Σfi y i 
N

N
2
≥0
Σf y
1
Σfi ( y i − y ) 2 ≥ 0, where y = i i
N
N
which is true, hence β 2 ≥ 1
or if
Problem Set
1.
Calculate the first and second moments about zero for the observations 3, 8,
11, 12, 20.
Calculate the first three moments about 90 from the following frequency
distribution.
2.
x
82
88
90
91
92
95
97
f
7
11
15
8
4
3
2
3.
Calculate the first three central moments from the following data :
4.
x
5
7
10
12
17
f
1
2
4
2
1
Calculate mean, variance and third central moment from the following data :
x
0
1
2
3
4
5
6
7
8
f
1
9
20
59
72
52
29
7
1
5.
Calculate the first four central moments for the following frequency
distribution.
Class
2-4
4-6
6-8
8 - 10
10 - 12
Frequency
4
2
8
6
1
6.
The first three moments of a distribution about the value 5 of the variable are
2, 20 and 40. Find the mean, the variance and the third moment about the
mean.
The first three moments of a distribution about the value 2 of the variable are
1, 16 and –40. Show that the mean = 3, the variance =15 and µ 3 = − 86.
Compute the quartile coefficient of skewness for the following distribution :
7.
8.
x
3-7
8-12
13-17
18-22
23-27
28-32
33-37
38-42
f
2
108
580
175
80
32
18
5
214
Engineering Mathematics-III
9.
In a certain distribution, the first four moments about a point are –1.5, 17, –30
and 108. Calculate the moments about the mean, β1 and β 2 ; and state whether
the distribution is leptokurtic or platykurtic?
Compute skewness and kurtosis, if the first four moments of a frequency
distribution f ( x ) about the values x = 4 are respectively, 1, 4, 10 and 45.
Calculate the measure of skewness and kurtosis on the basis of moments for
the following distribution:
10.
11.
Marks more than
0
10
20
30
40
No. of students
10
8
6
3
1
12.
13.
The first four central moments of a distribution are 0, 2.5, 0.7 and 18.75.
Calculate the coefficient of skewness and kurtosis.
The following table gives the monthly wages of 72 workers in a factory.
Compute the standard deviation, quartile deviation, coefficients of variation
and skewness.
Monthly wages
(in Rs.)
No. of workers
Monthly wages
(in Rs.)
No. of workers
12.5 – 17.5
2
37.5 – 42.5
4
17.5 – 22.5
22
42.5 – 47.5
6
22.5 – 27.5
19
47.5 – 52.5
1
27.5 – 32.5
14
52.5 – 57.5
1
32.5 – 37.5
3
Objective Type Questions
Multiple Choice Questions
Out of four alternative answers one is correct. Tick mark the correct alternative.
1. The first moment µ ′1 is equal to :
(a) mean
(b) S.D.
(c) variance
(d) none of these.
2. Second moment µ 2 about the mean is
(a) S.D.
(b) variance
(c) variation
3. Third moment µ 3′ about a is
(a) µ 2 + d 2
(c) µ 3 + 3 µ 2 d + d
(d) none of these.
(b) d
3
(d) none of these.
Statistics
215
4. The value of coefficient of skewness in Karl's Pearson's formula lies between :
(a) –3 and 3
(b) –1 and 1
(c) – ∞ and 0
(d) 0 and ∞.
5. A negative coefficient of skewness implies when M: Mean , Me : Median, Mo :
Mode :
(a) M > Mo
(b) M = Mo
(c) M = Me
(d) M < Mo.
6. For a given series the mean, S.D. and mode are respectively 622.08, 3.06 and
625. The coefficient of skewness :
(a) – 0.95
(b) 0.95
(c) – 0.84
(d) 0.84.
7. In a distribution mean = 12.8, mode = 12, S.D.= 12. The Karl Pearson's
coefficient of skewness is :
(a) 0.347
(b) 0.0667
(c) 0.667
(d) none of these.
8. In a distribution mean = 65, median = 70, coefficient of skewness is – 0.6. The
coefficient of variation is :
(a) 45.32%
(b) 38.46%
(c) 37. 96 %
(d) none of these.
Fill in the Blanks
1.
2.
3.
4.
5.
6.
7.
8.
The r th moment of x about a is given by ............... .
The second moment about mean is ................ of x.
Bowley's coefficient of skewness is given by ....................... .
For positive skewness, γ 1 .................. .
Kurtosis β 2 =..................... .
If γ 2 = 0, the frequency curve is ....................... .
The value of coefficient of skewness in Bowley's formula lies between ..............
The S.D. of a distribution is 5. The value of the µ 4 in order that the distribution
be mesokurtic be ................... .
State True/False
1.
For a symmetrical distribution the coefficient of skewness is 1.
2.
For a symmetrical distribution mean, median and mode are equal.
3.
The Karl Pearson's coefficient of skewness is always positives.
4.
In a positively skewed distribution :
5.
Mean > Mode > Median
The limits of quartile coefficient of skewness are – 3 and 3.
6.
The measure of kurtosis is β 2 = 0.
7.
If β 2 < 3 then the frequency curve is platykurtic.
216
Engineering Mathematics-III
ANSWERS
1. µ1′ = 10.8, µ ′2 = 147.6
2. µ1′ = −6.6, µ ′2 = 13.98, µ ′3 = −51.42
3. µ1 = 0, µ 2 = 10, µ 3 = 14.2
4. mean=3.973, µ1′ = −0.027, µ ′2 = 1.980, µ ′3 = −0.145, variance σ 2 = 1.979,
µ 3 = 0.0115
5. µ1 = 0, µ 2 = 5.3, µ 3 = −4.44, µ 4 = 62 . 45
6. Mean =7, variance σ 2 = 16, µ 3 = − 64
8. 0.22; 1.157
9. β1 = 0.493, β 2 = 0.655, platykurtic
10. 0, 2.9
11. γ 1 = 0.074, β 2 = 2 ⋅ 03
12. β1 = 0.63, γ 1 = 0.18 positively skew,
β 2 = 3, γ 2 = 0, the curve is Mesokurtic.
13. 8.85, 5.25, 0.32, 1.09.
Multiple Choice Questions
1. (a)
2. (b)
3. (c)
4. (a)
5. (d)
6. (a)
7. (c)
8. (b)
Fill in the Blanks
1
r
1. µ r′ = n ∑ fi ( xi − a)
i
3.
Q1 + Q 3 − 2 Median
Q 3 − Q1
2. Variance
4. γ 1 > 0
2
5. µ 4 /µ 2
6. Mesokurtic
7.
8. 1875
–1 and 1
True /False
1. False
2. True
3. False
5. True
6. False
7. True
4. False
❑❑❑
Unit-2
Chapter
5
Curve Fitting
5.1 Curve Fitting
Consider, we have a data consisting a set of pairs of points ( x1 , y1 ) , ( x 2 , y 2 )
...( xi , y i )... ( x n , y n ) and we are interested to fit a curve of our choice in this
data. The concept of curve fitting serves our purpose. The fitted curve may be a
straight line, polynomial or any other curve. Since, curve fitting plays an important
role in the study of theoretical and practical aspects of correlation and regression,
therefore this chapter deals with some techniques of curve fitting.
5.1.1 Method of Least Square
Consider, a set of n points ( x1 , y1 )( x 2 , y 2 ) … ( x n , y n ) and we want to fit y=f(x)
in this data. If Yi denotes the approximate value of y i i.e. ( Yi =f ( xi ) , then method
of least square is a technique to minimize the error in the sum of the squares of the
difference of exact and approximate value. Mathematically, we can say that we have
to minimize E, where E is given by.
E = ( Y1 − y1 ) 2 +( Y2 − y 2 ) 2+ ... ( Yn − y n ) 2
= e12+e 22 − + e 22+ … e 22+e n2 , where ei =Yi − y i
n
E =
∑ ei2
...(5.1)
i =1
the value of E always depends upon the nature of curve. The expression for E, may
be linear, transcendental or in the form of a polynomial.
Fitting a straight line
When we want to fit a straight line y=bx+a to the given set of points, then the value
of E will be given as
n
E =
∑ [Yi
− ( bxi +a)]2
...(5.2)
i =1
Now we have to minimize E. Since the parameters are a and b, which minimize or
∂E
∂E
maximize E, therefore
= 0,
= 0, by maxima and minima of functions of two
∂a
∂b
variable. Thus, we have
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Engineering Mathematics-III
n
∂E
= −2∑ (Yi − bxi − a)=0
∂a
i =1
...(5.3)
n
∂E
= −2∑ xi (Yi − bxi − a)=0
∂b
i =1
...(5.4)
Now, approximating Yi ≈ y i and then on simplification, we get
Σ y i = na+b Σx i
and Σ xi y i = ai Σ x i + b Σ x i2
The above equations are known as normal equations for the least square line. On
solving these equations, we get the values of a and b. For the above system of
equations, the values of a and b are given by the following relation.
a=
∑ xi2 ∑ y i − ∑ xi y i ⋅ ∑ xi
...(5.5)
n ( ∑ xi2 ) − ( ∑ xi ) 2
n
n ⋅ ∑ xi y i − ∑ xi ⋅ ∑ y i
and
b=
i =1
...(5.6)
n( ∑ xi2 ) − ( ∑ xi ) 2
Now by the theory of "Maxima and Minima of function of two variables" we can say a
function u = u(x, y) will be minimum if
r > 0 and
r=
∂ 2u
∂x 2
,r=
rt – s 2>0 , where
∂ 2u
∂ 2u
,t =
∂x∂y
∂y 2
Here in this problem E is a function of two variable a and b, therefore
r=
∂ 2E
∂ a2
= 2, s =
∂ 2E
∂ 2E
= 2, t =
= 2 ∑ xi2
2
∂ a ∂b
∂b
It is clear that r = 2 < 0 and rt − s 2 = 4 ⋅ Σx i2 − 4 > 0, Hence the value a and b given
by (5.5) and (5.6) respectively provide a minimum value of E.
Ex. 1 : Find a straight line that can be fitted to the following data
x
1
2
3
4
5
6
y
1200
900
600
200
110
50
Sol : Let the required line is y = bx + a, then a and b are given by (5.5) and (5.6).
The values of various summation are given by the following table :
xi
yi
x i2
xi y i
1
1200
1
1200
2
900
4
1800
3
600
9
1800
4
200
16
800
Curve Fitting
219
5
110
25
550
6
50
36
300
Σx i = 21
Σy i = 3060
Σxi2 = 91
Σ xi y i = 6450
Putting the values of Σ x i , Σy i , Σ x i2 and Σ x i y i in (5.5) and (5.6) we get
a=
and
b=
91 × 3060 − 6450 × 21
6 × 91 − ( 21) 2
6 × 6450 − 21 × 3060
6 × 91 − ( 21) 2
= 1361 ⋅ 97
= −243 ⋅ 42
Now putting the value of a and b in y = bx + a, we get
y = − 243 ⋅ 42 x + 1361 ⋅ 97,
Which is the required straight line.
Fitting a curve to y = ab x
The method to fit y = ab x to the given set of point is given below.
Step 1 : Taking the logarithm of both sides, we get log y=x log b+ log a
Step 2 : Denote log y = Y, log b = B and log a = A, we get Y = Bx+A, which is a
straight line, therefore the given problem is reduced to fitting Y = Bx + A
to the given data.
Step 3 : Find the value of A and B, which are given as,
Σxi2 ΣYi − Σxi Yi Σxi
A=
B=
...(5.7)
n.(Σxi2 ) − (Σxi ) 2
n ⋅ Σxi Yi − Σxi ΣYi
...(5.8)
n.(Σxi2 ) − (Σxi ) 2
where Yi = log y i , a= antilog A, b = antilog B
Step 4 : Putting the value of a and b we get the required fitted curve.
Ex. 2 : Fit the curve y=ab x to the given data
x
2
3
4
5
6
y
144
172 ⋅ 8
207 ⋅ 4
248 ⋅ 8
298 ⋅ 5
Sol : In this problem the values of a and b are given by
a = antilog A and b = antilog B, where A and B are given by (5.7) and
(5.8) respectively . Here Yi = log y i . The required computation of various
summation are given by the following table
220
Engineering Mathematics-III
xi
yi
x i2
Yi = log y i
Σxi Yi
2
144
4
2 ⋅ 1584
4 ⋅ 3168
3
172 ⋅ 8
9
2 ⋅ 2375
6 ⋅ 7125
4
207 ⋅ 8
16
2 ⋅ 3168
9 ⋅ 2672
5
248 ⋅ 8
25
2 ⋅ 3959
11 ⋅ 9795
6
298 ⋅ 5
36
2 ⋅ 4749
14 ⋅ 8494
Σx i2 = 90
ΣYi = 11 ⋅ 5835
Σx i Yi = 47 ⋅ 1254
Σ x i = 20
Thus, we get
A=
90 × 11 ⋅ 5835 − 47 ⋅ 1254 × 20
5 × 90 − ( 20) 2
5 × 47 ⋅ 1254 − 20 × 11 ⋅ 5835
= 2 ⋅ 0014,
and
B=
Now
and
Hence
a = antilog A = antilog 2 ⋅ 0014 = 100 (App)
b = antilog B = antilog 0 ⋅ 12034 = 1 ⋅ 3183 (App)
y = (100) (1 ⋅ 3183) x is the required curve.
5 × 90 − ( 20) 2
= 0 ⋅ 12034
Fitting the curve y = ax b
The method to fit y = ax b to given set of points is given below
Step 1 : Taking the logarithm of both sides, we get log y = b ⋅ log x + log a
Step 2 : Denote log y = Y, log x = X and log a = A, we get Y = b X + A, which is a
straight line, therefore, the given problem is reduced to fitting Y = bX + A
to the given data.
Step 3 : Find the value of A and b, which are given as,
A=
and
b=
ΣX i2 ⋅ ΣYi − ΣX i Yi ΣX i
...(5.9)
n ⋅ (ΣX i2 ) − (ΣX i ) 2
n ⋅ ΣX i Yi − ΣX i ΣYi
...(5.10)
n ⋅ (ΣX i2 ) − (ΣX i ) 2
where Yi = log y i , X i = log xi and a = antilog A.
Ex. 3 : Fit the expontial curve y = ax b to the given data
x
2
3
4
5
y
27 ⋅ 8
62 ⋅ 1
110
161
Sol : In this problem, the value of a = antilog A, where A is given by (5 ⋅ 9) and the
value of b is given by (5⋅ 10). The required terms are given in the following table.
Curve Fitting
221
xi
yi
X i = log xi
Yi = log y i
X i2
X i Yi
2
27 ⋅ 8
0 ⋅ 3010
1 ⋅ 4440
0 ⋅ 0906
0 ⋅ 4346
3
62 ⋅ 1
0 ⋅ 4771
1 ⋅ 7931
0 ⋅ 2276
0 ⋅ 8555
4
110
0 ⋅ 6021
2 ⋅ 0414
0 ⋅ 3625
1 ⋅ 2291
5
161
0 ⋅ 6990
2 ⋅ 2068
0 ⋅ 4886
1 ⋅ 5426
Σy i = 350 ⋅ 9
ΣX i = 2⋅ 0792
Σ Yi =7 ⋅ 4853
ΣX i2 = 1 ⋅ 1693
ΣX i Yi = 4 ⋅ 0618
Thus, we get
A=
1 ⋅ 1693 × 7 ⋅ 4853 − 4 ⋅ 0618 × 2 ⋅ 0792
4 × 1 ⋅ 1693 − ( 2 ⋅ 0792) 2
4 × 4 ⋅ 0618 − 2 ⋅ 0792 × 7 ⋅ 4853
and
b=
Now
a = antilog A = antilog (0 ⋅ 8678) = 7 ⋅ 37
Hence
y = 7 ⋅ 37 x1⋅9 is the required curve.
4 × 1 ⋅ 1693 − ( 2 ⋅ 0792) 2
= 0 ⋅ 8678,
= 1 ⋅ 9311 ≈ 1 ⋅ 9
Fitting the curve y = a ⋅ e bx
The method to fit the curve y=a ⋅ e bx to the given set of points is given below.
Step 1 : Taking logarithm of both side, we get log y=bx log e+ log a
Step 2 : Denote log y = Y, b log e = B, log a = A, thus we get Y = Bx + A, which is a
straight line. Therefore this problem is also reduced to fitting a straight
line y = Bx + A to the given data
Step 3 : In this case the value of A and B are given as,
A=
and
B=
where
Σxi2 ⋅ ΣYi − Σxi Yi Σxi
...(5.11)
n (Σxi2 ) − (Σxi ) 2
n ⋅ Σxi Yi − Σxi ΣYi
...(5.12)
n ⋅ (Σxi2 ) − (Σxi ) 2
Yi = log y i , a = antilog A, b = antilog ( B / log10 e)
Ex 4 : Fit the exponential curve y = ae bx to the given data
x
1
7 ⋅ 25
1 ⋅ 50
1 ⋅ 75
2 ⋅ 00
y
5 ⋅ 10
5 ⋅ 79
6 ⋅ 53
7 ⋅ 45
8 ⋅ 46
Sol : In this problem, the value of a and b are given by
a = anti log A and b = antilog B/loge respectively. Here the value of A and B are
given by the expression (5.11) and (5.12). The summation which are used in this
computation, given in the following table
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Engineering Mathematics-III
xi
yi
Yi = log y i
Σ xi Yi
Σ xi2
1
5 ⋅ 10
0 ⋅ 7075
0 ⋅ 7075
1
1 ⋅ 25
5 ⋅ 79
0 ⋅ 7626
0 ⋅ 9533
1 ⋅ 5625
1 ⋅50
6 ⋅ 53
0 ⋅ 8149
1 ⋅ 2224
2 ⋅ 2500
1 ⋅ 75
7 ⋅ 45
0 ⋅ 8721
1 ⋅ 5269
3 ⋅ 0625
2 ⋅ 00
8 ⋅ 46
0 ⋅ 9273
1 ⋅ 8546
4 ⋅ 000
Σy i = 4 ⋅ 0844
Σ xi y i = 6 ⋅ 2647
11 ⋅ 875
Σx i = 7 ⋅ 5
Thus, we get
A=
11 ⋅ 875 × 4 ⋅ 0844 − 7 ⋅ 5 × 6 ⋅ 2647
5 × (11 ⋅ 875) − (7 ⋅ 5) 2
48 ⋅ 5023 − 46 ⋅ 9852
=
= 0 ⋅ 4185
3 ⋅ 625
5 × 6 ⋅ 2647 − 7 ⋅ 5 × 4 ⋅ 0844 0 ⋅ 6905
B=
=
= 0 ⋅ 1905
3 ⋅ 625
5 × 11 ⋅ 875 − (7 ⋅ 5) 2
Now a = antilog 0 ⋅ 4185 = 2 ⋅ 6212 and b =
antilog 0 ⋅ 1905 1 ⋅ 5506
=
= 3 ⋅ 6 (App)
log10 e
0 ⋅ 4313
Thus, the required curve is y = (2 ⋅ 6212) ⋅ e 3⋅6 x
Fitting of a Parabola y=a 0+a1 x+a 2 x 2
Let we want to fit y=a0+a1 x+a2 x 2 to the given data, ( xi , y i ), where i=1 , 2 ...n. If
Yi is the approximate value when x=xi , then we have to minimize the error E, where
n
E =
∑
(Yi − y i ) 2
i =1
n
=
∑ [Yi
− (a0+a1 x+a2 xi ) 2 ]
...(5.13)
i =1
For the minimum value of E, we have
∂E
= 0,
∂a 0
∂E
∂E
= 0,
= 0.
∂a1
∂ a2
Using these conditions, and taking Yi ≈ y i ,we get
Σ y i =n a0+a1 Σxi +a2 Σxi2
...(5.14)
Σxi y i =a0 ⋅ Σxi +ai Σxi2+a2 Σxi3
Σ xi2 y i =a0 Σxi2 + a1 Σxi3+a2 Σ xi4
...(5.15)
...(5.16)
The above equation (5.14), (5.15) and (5.16) are known as normal equations.
On solving these equations we get the required value of a0 , a1 and a2 and then the
required curve.
Curve Fitting
223
Ex. 5 : Fit the curve y=a0+a1 x+a2 x 2 to the following data
x
1
2
3
4
5
y
5
12
26
60
97
Sol : The normal equations are (5.14), (5.15) and (5.16), so the various summation
values required for computation are given below
xi
yi
xi2
Σ x i3
x i4
xi y i
xi2 ⋅ y i
1
5
1
1
1
5
5
2
12
4
8
16
24
48
3
26
9
27
81
78
234
4
60
16
64
256
240
960
5
97
25
125
625
485
2425
Σxi =15
Σy i = 200
Σxi2 = 55
Σ xi3 = 225
Σxi4 = 979
Σxi y i = 832
Σ xi2 y i = 3672
We know the normal equations are
n a0+a1 ⋅ Σxi +a2 ⋅ Σxi2=Σy i
a0 ⋅ Σxi +a1 ⋅ Σxi2+a2 ⋅ Σxi3=Σxi y i
a0 ⋅ Σxi2+a1 ⋅ Σxi3+a2 Σxi4=Σxi2 ⋅ y i
Now putting the values of
n=5 , Σxi =15 , Σxi2=55 , Σxi3=225
Σxi4=979 , Σy i =200 , Σxi y i =832 , Σxi2 y i =3672 , we get
5a0+15a1+55a2=200
15a0+55a1+225a2=832
55a0+225a2+979a2=3672
On solving for a0 , a1 and a2 , we get
a0=10 ⋅ 40 , a1= − 11 ⋅ .08 and a2=5 ⋅ 71.
Thus the required fitter curve is y=10 ⋅ 4 − 11 ⋅ 08 x+5 ⋅ 71 x 2
Fitting a polynomial of degree n
When we want to fit a polynomial
f(x)=a0+a1 x+a2 x 2+ … an x n
...(5.17)
to the given set of points, the procedure is similar to as discussed earlier. In this case
the normal equations to obtain the values of a 0 , a1 , a 2… a n are given below.
224
Engineering Mathematics-III
na0+a1 Σxi +a2 Σxi2+… +an ⋅ Σxin = Σy i


a0 Σxi +ai ⋅ Σxi2+a2 Σxi3+… +an ⋅ Σxin+1 = Σxi y i


2
3
4
n+ 2
2

a0 ⋅ Σxi + ai ⋅ Σx 2 + a2 Σxi + … + an ⋅ Σxi
= Σxi y i

........... + .......... + .......... + ....... .. + ................. .. = .........
........... + ............ + .......... + ....... .. + ............ ..... .. = ......... 

a0 Σxin+ai Σxin+1+a2 Σxin+2+ ......... . +an Σxin+n=Σxin ⋅ y i 
...(5.18)
On solving, there above simultaneous linear equations. We can get the values of
desire parameters, a0 , a1 ,… an .
5.2 Data Fitting: Cubic Splines
Practically the interpolating polynomial of higher degree passes through the given
set of point, but they are not smooth and oscillate between the end points of the data
range. To solve this problem, instead of single polynomial, some lower order
polynomials are approximated to provide the smoothness to the curve.
Amongst the various available techniques for its solution, spline fitting is the one
recent method. The name spline actually originated from a draft man device to draw
a smooth curve. Since, there smooth curves are cubic, therefore this method is
known as cubic spline : piece-wise cubic interpolating polynomials.
Now consider the problem of interpolating polynomial between the data points
(xi , y i ),i=0 , 1 , 2...n. by means of smooth polynomial curve : cubic spline S(x). To
construct this cubic spline S(x), which is a combination of ( n − 1) cubic polynomials
S j (x ) corresponding to each sub-interval [ x j , x j +1 ], where j = 0, 1, 2...n – 2 The
cubic spline S(x) and cubic polynomial S j (x ) satisfy the following properties
(i) S( xi )= y i ∀ i=0 , 1 , 2...n.
(ii) S( x )=S j ( x ) ∀ j= 0 , 1 , 2..n − 2
(iii) S j ( x j )= S j+1 ( x j ) ∀ j = 0 , 1 , 2 , ...n − 2
(iv) S ′j ( x j )= S' j+1 ( x j ) ∀ j = 0, 1, 2... n − 2
(v) S" j ( x j )= S" j+1 ( x j ) ∀ j = 0 , 1 , 2...n − 2
(vi) S"( x 0 ) = S"( x n )= 0
Now, Since S j (x ) is a cubic polynomial in the interval [ x j , x j+1 ] therefore, S j (x ) can
be written, as
S j ( x )=a j ( x − x j ) 3+b j ( x − x j ) 2 + c j ( x − x j )+d j
...(5.19)
Now using the above conditions, we have the following equations.
when
x= x j , S j ( x j )= S( x j )= y j , therefore (5.19) gives
y j=a j ( x j − x j ) 3+b j ( x j − x j ) 2+c j ( x j − x j )+d j
or
when
therefore
y j=d j
x= x j+1 , S j ( x j+1 )= S ( x j+1 )= y j+1
...(5.20)
y j+1=a j ( x j+1 − x j ) 3+b j ( x j+1 − x j ) 2 +c j ( x j+1 − x j )+d j
...(5.21)
Now let
x j+1 − x j=h j+1 , then (5.21) becomes
y j+1=a j .h 3j+1 + b j h 2j+1+c j ⋅ h j+1 + d j+1
...(5.22)
Curve Fitting
225
To apply, other conditions, find S j′ ( x ) and S j′′( x ), which are given below.
S ′j (x)=3a j (x − x j ) 2 2b j (x − x j )+c j
...(5.23)
S ′′j ( x ) = 6a j (x − x j )+2b j
...(5.24)
Now we denote S ′′(x j )=M j , then at x=x j , (5.24) becomes
Mj
M j=2b j or b j=
2
...(5.25)
Now when x=x j+1 , we have
M j+1=6a j (x j+1 − x j )+2b j
= 6a j ⋅ h j+1+M j
M j+1 − M j
a j=
6h j+1
or
...(5.26)
Now putting the value of a j ⋅ b j and a j back in (5.22), we get
M j+1 − M j 3
Mj
y j+1=
h j+1+
⋅ h 2j+1+C j h j+1+y j
6h j+1
2
on solving for c j , we get
c j=
y j+1 − y j
h j+1
−
h j+1
6
...(5.27)
( 2M j + M j +1 )
Now we shall try to find out the value of c j from a different way, by using the given
condition S ′j (x j )=S ′j+1(x j ) or S ′j −1(x j − 1 )=S ′j ( x j −1 )
or
S ′j −1(x j )=S ′j ( x j ) ⋅ Since, we have (5.23), which gives
S ′j ( x j )=c j .......(5.24), and (5.23) also gives
S j′ −1(x)=3a j
or
−1(x
− x j −1 ) 2+2b j
S ′j −1(x j )=3a j
−1(x j
− xj
S ′j −1 ( x j )= 3a j
−1(h j )
−1 )
2
+2b j
2
−1(x
+2b j
−1
− xj
−1(x j
⋅ h j+c j
−1
−1 )+c j −1
− xj
−1 )+c j −1
...(5.25)
Now the condition S ′j −1 ( x j ) = S ′j ( x j ) gives
3a j
−1
⋅ h 2j +2b j
−1
⋅ h j+c j
−1
= cj
...(5.26)
Putting the value c j from (5.26) into (5.27) we have
y j+1 − y j h j+1
−
( 2M j+M j+1 )
h j+1
6
= 3a j
−1
⋅ h 2j +2b j
−1 h j+c j −1
...(5.28)
226
Engineering Mathematics-III
M j+1 − M j
M j − M j −1

, therefore a j −1
Since a j = 6 h
6h j
j+1


Mj
M j −1
bj =
therefore b j −1=

2
2

y j+1 − yi h j+1
y j − y j −1 h j

cj =
−
( 2M j+M j+1 ),therefore C j −1 =
−
( 2M j −1 − M j )

h
6
hj
6
j+1

Putting the value of a j −1 , b j −1 and C j −1 in (5.28) we get
 M j − M j −1  2
y j+1 − y j h j+1
M j −1
 ⋅ h j +2 ⋅
−
( 2M j+M j+1 )=3 
⋅ hj
h j+1
6
6h j
2


+
yj − yj
−1
hj
−
hj
6
( 2M j
−1
− Mj)
On simplification, we get
hjM j
−1+2
 y j+1 − y j
yj − yj
(h j+h j+1 )M j+h j+1 ⋅ M j+1=6 
−
hj
 h j+1
−1 
 ...(5.29)

where j = 1, 2, 3 ... n − 1
In case of equal interval, we have h j=h j+1=h, then (5.29) becomes
6
...(5.30)
M j −1 + 4M j +M j+1 =
(y j+1 − 2 y j+y j −1 )
h2
where j = 1, 2, 3, 4 ... n − 1
This gives a system of (n − 1) simultaneous lines equations for (n+1) unknown.
S j (x ) for equal interval [ x j −1 , x j ] is given by


1
1
h2
S j ( x )=
⋅ [( x j − x ) 3 M j −1+( x − x j −1 ) 3 ⋅ M j ]+ ( x j − x )  y j −1 −
M j −1 
6h
h
6


+
1
⋅ (x − x j
h
−1 )


h2
⋅Mj
y j −
6


S j (x ) for unequal interval is given by
M j −1
S j ( x )=[( x j − x ) 3 ]
+
6h j
+
1
(x j − x)y j
hj
−1+
...(5.31)
[(x − x j
1
(x − x j
hj
−1 )
−1 )
3
− h 2j ]
Mj
6h j
...(5.32)
yj
For better understanding see the following examples
Ex. 6 : Obtain cubic spline for every sub-interval for the following set of data. It is
given that M 0 = 0, M 3 = 0
x
1
2
3
4
y
1
5
11
8
Also find S(1.6),
Curve Fitting
227
Sol : In this problem, h = 1, n = 3, x 0=1 , x1=2 , x 2=3 , x 3=4 , y 0=1 , y1 = 5,
y 2 = 11 y 3 = 8, By the system of equations
6
M j −1+4M j+M j+1=
( y j+1 − 2 y j+y j −1 ) , where j = 1, 2
h2
when j = 1, we get
M 0+4M1+M 2=6 (y 2 − 2 y1+y 0 )
or
4M1+M 2=6( 11 − 2 × 5+1 )=12…( 1 ) { ∵ M 0 = 0}
Similarly, when j = 2, we get
M1+4M 2+M 3=6 (y 3 − 2 y 2+y1 )
or
M1+4M 2= − 54 ... (2)
{∵ M 3 = 0}
102
228
on Solving (1) and (2), we get M1=
, M 2= −
15
15
The corresponding cubic is given by (5.31). In this case, for the interval [ x 0 , x1 ], i.e.
x 0 = 1, x1 = 2, the cubic is given by
M 
1

S1(x)= [(x1 − x) 3 . M 0+(x − x 0 ).M1 ] + (x1 − x)  y 0 − 0 
6
6 

M 

+ (x − x 0 )  y1 − 1 
6 

102
Putting
x 0=1 , x1=2 , M 0=0 , M1=
, we get
15
1
S1(x)=
[ 17 x 3 − 51 x 2+94 x − 45 ]
15
For the interval [ x1 , x 2 ] i.e. x1=2 , x 2=3 , we get S 2(x ) as
M 
M 
1


S 2(x)= [(x 2 − x) 3 M1+(x − x1 ) 3 M 2 ]+(x 2 − x)  y1 − 1  + (x − x1 )  y 2 − 2 
6
6
6 



1
=
[ −55 x 3 − 381 x 2 − 770 x − 351]
15
Similarly for the interval [ x 2 , x 3 ] i.e. x 2=3 , x 3=4 , we get
M 
1

S 3(x)= .[(x 3 − x) 3 M 2+(x − x 2 ) 3 M 3 ]+(x 3 − x)  y 2 − 2 
6
6 

M 

+ (x − x 2 )  y 3 − 3 
6 

1
=
( 38 x 3 − 456 x 2+1741 x − 1980 )
15
To find the value S (1 ⋅ 6), we shall use S1 ( x ) because 1 ⋅ 6 is lying in the interval
[1, 2] therefore
1
S( 1 ⋅ 6 )=S1( 1 ⋅ 6 )= [ 17 ( 1 ⋅ 6 ) 3 − 51 × ( 1 ⋅ 6 ) 2 + 94 ( 1 ⋅ 6 ) − 45 ]
15
1
=
[ 69 ⋅ 632 − 130 ⋅ 56 + 144 − 45]
15
228
Engineering Mathematics-III
1
× 38 ⋅ 072
15
= 2 ⋅ 53813
 π
Ex. 7 : Find the value of y   by using cubic spline method, for the given set of
 6
=
data. Given that M 0=M 2=0. Find S(1 ⋅ 6)
x
0
π/2
π
y
0
1
0
Sol : In this case x 0 = 0, x1 = π / 2, x 2 = π, y 0 = 0, y1 = 1, y 2 = 0, n = 2, h =
π
.
2
Now by the system of equations
6
M j −1+4M j+M j+1
(y j+1 − 2 y j+y j −1 ) where j = 1, we get
h2
4
48
M 0+4M1+M 2=6 ×
[ 0 − 2+0 ]= −
2
n
π2
12
or M1 = −
{∵ M 0 = M 2 = 0}
π2
In the interval [ x 0 , x1 ] i.e. [ 0, π / 2], the spline is given by


1
1
h2
S1 ( x ) =
⋅ [( x1 − x ) 3 ⋅ M 0 +( x − x 0 ) 3 ⋅ M1 ]+ ⋅ ( x1 − x ) ⋅  y 0 −
⋅ M 0
6h
h
6


+
Putting x 0 = 0, x1 = π / 2, h = π / 2, M 0 = 0, M1 = −
S1 ( x ) =
12
π2
S1 ( x ) =
,
3
 12   2  π
2  π
π2 


.  − x . 0 + ( x − 0) 3 .  −
+  − x  0 −
.0




 π 2  π  2
6 × π  2
24 

+
or


1
h2
[x − x 0 ]  y1 −
⋅ M1 
h
6



2
π 2  12  
.( x − 0) 1 −
. −

π
24  π 2  

2  2x 3 3 
+ x
−
π  π 2
2 
π
 π
Now S   is given by S1 (1 ⋅ 6) = 0 ⋅ 4815 because is lying in the interval [ 0, π / 2]
 6
6
In the interval [ x1 , x 2 ], i.e. [ π / 2, π], the spline is given by
S2 ( x) =


1
1
h2
[( x 2 − x ) 3 ⋅ M1 +( x − x1 ) 3 . M 2 ]+ ( x 2 − x )  y1 −
. M1 
6h
h
6


Curve Fitting
229
+
Now putting x1 = π / 2, x 2 = π, M1 = −
S2 ( x) =
12
π2


1
h2
(x − x1 )  y 2 −
M 2
h
6


, M 2 = 0, h = π / 2, we get
3 

2 
12  
π
2
π 2  12  
3
. −
( π − x ) 1 − 2  +  x −  .0 + ( π − x ) ⋅ 1 −


6π 
2
24  π 2  
π  

 π
2
π 
π2 
+  x −   0 −
. 0
π
2 
24 
3
( x − π )3 + (π − x )
π
π
Ex. 8 : Find S(1 ⋅ 6) by cubic spline method for the given data, if M 0 = 0 = M 4 = 0
S2 ( x) =
4
3
x
0
y
0
π/4
1
2
3
π
4
π/2
1
1
0
2
Sol : In this case the system of equation
6
M j −1+4M j+M j+1 =
[ y j +1 − 2 y j+y j
h2
4M1 + M 2 = − 4
M1 + 4M 2 + 4M 3 = − 5 ⋅ 699
π
−1 ],
where j = 1, 2, 3, we get
{ ∵ M 0 = 0, M 4 = 0}
M 2 + 4M 3 = − 4 ⋅ 029
On solving, we get M1 = − 0 ⋅ 7440, M 2 = −1 ⋅ 053, M 3 = − 0 ⋅ 7440
π
For the interval [ x 0 , x1 ], i.e. 0, , the spline is given by
 4
S1(x)=


1
1
h2
[(x1 − x 0 ) 3 ⋅ M 0+(x − x 0 ) 3 M1 ]+ [x1 − x]  y 0 −
M 0
6h
h
6




1
h2
[ x − x 0 ]  y1 −
M1 
h
6


1
Putting x 0 = 0, x1 = π / 4, M 0 = 0, M1 = − 0 ⋅ 7440, y1 =
, we get
2
4
S1 ( x ) = [ − 0 ⋅ 124 x 3 + 0 ⋅ 7836 x]
π
and S1 (π / 6) = 0 ⋅ 4998. It means for the better approximation the interval should
be small.
+
230
Engineering Mathematics-III
Ex 9 : Fit a cubic spline to the given data and then find S (1 ⋅ 5) and S ′(1).
S(3), S ′(3).
x
1
2
3
y
−8
−1
18
Sol : Here x 0=1 , x1=2 , x 2=3 , y 0= − 8 , y1= − 1 , y 2=18 , n=2 , h = 1 and
M 0 = M 2 = 0. By the system of equation (6.30), we have
M 0+4M1+M 2=6( 18 − 2 x − 1 − 8 )=72 or M1 = 18
For the interval [x 0 , x1 ], i.e. [1, 2] the spline is given by
M 
M 
1


S1(x)= [(x1 − x) 3 M 0+(x − x 0 ) 3 M1 ]+(x1 − x)  y 0 − 0  + (x − x 0 )  y1 − 1 
6
6
6 



Now putting x 0=1 , x1=2 , M 0=0 , M1=18 , y 0= − 8 , y1= − 1 , we get
1
0
18

S1(x)= ⋅ [( 2 − x) 3 ⋅ 0+(x − 1 ) 3 ⋅ 18 ]+( 2 − x) ⋅ − 8 −  + (x − 1 )  −1 − 

6
6
6


S1(x)=3(x − 1 ) 3 − 8( 2 − x) − 4(x − 1 )
Now, since x = 1 ⋅ 5, lying in the interval, therefore S1 (1 ⋅ 5) = −
45
and
8
S1′(x)=3 (x − 1 ) 2+4 , s1′( 1 )=4.
For the interval [ x1 , x 2 ], i.e. [2, 3], the spline is given by
1
M 
M 


S2(x)= ⋅ [(x2 − x)3 ⋅ M1+(x − x1 )3M 2 ]+(x2 − x)  y1 − 1  + (x − x1 )  y 2 − 2 
6
6
6 


Putting x1=2 , x 2=3 , M1=18 , y1= − 1 , y 2=18 , we get
1
18

S 2(x)= [( 3 − x) 3 ⋅ 18+0 ]+( 3 − x)  −1 −  + (x − 2 ) ( 18 − 0)

6
6
S 2(x)=3( 3 − x) 3 − 4( 3 − x)+18(x − 2 )
Now S 2 (3) = 18 and S 2′ (x)= − 9( 3 − x) 2+22 , S 2′ ( 3 )=22
Ex. 10 : Find cubic spine for every sub-interval for the following set of points
x
4
9
16
y
2
3
4
It is also given that M 0 = M 2 = 0 Find S(8) , S(10).
Sol : In this problem
x 0=4 , x1=9 , x 2=16 , y 0=2 , y1=3 , y 2=4 n=2. This is also a problem of unequal
interval because
h1=x1 − x 0=9 − 4=5 and
h 2=x 2 − x1=16 − 9=7
We know that the system of equation for unequal interval is
 y j+1 − y j
y j − y j −1 

h j ⋅ M j −1+2(h j+h j+1 ) ⋅ M j+h j+1 M j+1=6 
−
hj
 h j+1

Curve Fitting
231
where j = 1, 2, ... n–1
Since, in this problem, n = 2, so j = 1, therefore the required system of equation is
 y − y1 y1 − y 0 
h1 ⋅ M 0+2(h1+h 2 ) ⋅ M1+h 2 M 2=6  2
−

h1 
 h2
Putting y 0=2 , y1=3 , y 2=4 , M 0=0=M 2 , h1=5 , h 2=7 , we get
 4 − 3 3 − 2
2 ( 5 + 7 ) ⋅ M1 = 6 
−

 7
5 
or
12
2
 1 1
or M1 = −
24 M1=6  −  = −
= − 0 ⋅ 0143
 7 5
35
35
In this problem, the number of sub-intervals are two, therefore we can find two
splines. Now the corresponding spline for the sub-interval [ x 0 , x1 ] i.e. [4, 9] is given
by
M0
M1
S1(x)=[(x1 − x) 3 − h12 ]
+ [(x − x 0 ) 3 − h12 ]
6h1
6h1
1
1
+
(x1 − x). y 0 +
(x − x 0 ) y1
h1
h1
Putting
x 0= 4 , x1= 9 , M 0= 0 , M1= − 0 ⋅ 0143 , h1= 5 , y 0= 2 , y1= 3
1
 2
S1(x)=[( 9 − x) 3 − 5 2 ] × 0+[(x − 4 ) 3 − 5 2 ]  −  ×
 32 6 × 5
1
1
+ ( 9 − x ) × 2 + ( x − 4) × 3
5
5
2
2
3
3
=−
[( x − 4) − 25]+ ( 9 − x ) + ( x − 4)
1050
5
5
2
2
3
3
Now
S(8) = S1 (8) = −
[( 8 − 4) − 25] + ( 9 − 8) + ( 8 − 4)
1050
5
5
2
2 12
=−
× 39 + +
1050
5 5
= − 0 ⋅ 07428 + 2 ⋅ 8
= 2 ⋅ 72572
Now the spline for the interval [ x1 , x 2 ] i.e. [9, 16] is given by
M1
M
S 2(x)=[x 2 − x) 3 − h 22 ] ⋅
+ [(x − x1 ) 3 − h 22 ] . 2
6h 2
6h 2
1
1
+
(x 2 − x) ⋅ y1 +
(x − x1 )y 2
2
h2
h
2
Putting
x1=9 , x 2=16 , M1= −
, M 2=0 , h 2=7 , y1=3 , y 2=4
35
0
 2 1
S 2(x)=[( 16 − x) 3 − 7 2 ]  − 
+ [(x − 9 ) 3 − 7 2 ]
 35 6 × 7
6 × 42
1
1
+ (16 − x ) × 3 + ( x − 9) × 4
7
7
2
3
4
S2 ( x) = −
[(16 − x ) 3 − 49] + 0 + (16 − x )+ ( x − 9)
1470
7
7
232
Engineering Mathematics-III
2
3
4
× [(16 − 10) 3 − 49] + (16 − 10) + (10 − 9)
1470
7
7
2
3
4
=−
× 167 + × 6 + × 1
1470
7
7
22
= − 0 ⋅ 2272 +
7
Now
S(10) = S 2 (10) = −
= − 0 ⋅ 2272 + 3 ⋅ 1429
S 2 (10) = 2 ⋅ 9157
5.3 Fitting a curve y=c1e m1x + c 2e m2x
By the theory of differential equation, y=C1 e m1 x + C 2 e m 2 x is the solution of
d2 y
dx
2
= a1
dy
+ a2 y (a linear differential equation with constant coefficient).
dx
Now consider that in the given data a0 is the initial value of x then on integrating
d2 y
dx
2
= a1
dy
+ a2 y, from a0 to x, we get
dx
x d2 y 
∫ a  dx 2  dx =
or
a1
x
∫a
0
dy
⋅ dx+a2
dx
x
∫a
ydx
0
dy  dy 
− 
= a1 [y (x) − y (a0 )]+a2
dx  dx  x = a0
x
∫a
...(5.33)
ydx
0
Integrating again form a to x we get
y(x) − y(a0 ) − y ′(a)(x − a0 )=a1
Since
a
a
0
0
x
∫ a ∫ a ... ∫ a
0
f(x)dx =
x
∫a
ydx − a1(x − a0 ) y(a0 )+a2
0
x
x
∫a ∫a
0
x
1
|———
n–1
y(x) − y(a0 ) − y ′(a0 )(x − a0 ) = a1
∫ a (x − t)
n −1
ydx
...(5.34)
0
f(t) dt, therefore (5.34) becomes
0
x
∫a
ydx − a1(x − a) y(a)+a2
0
x
∫ a (x − t) y(t) dt
0
Now to set up a linear relation between a1 and a 2 , we select two points x1 and x 2
such that a0 − x1=x 2 − a0 we get
y(x1 ) − y(a0 ) − y′(a0 ) (x1 − a0 )=a1
x1
∫a
ydx − a1(x1 − a0 ) y(a0 )+a2
0
x1
∫ a (x1 − t) y(t) dt
0
and
y(x2 ) − y(a0 ) − y ′(a0 )(x2 − a0 )=a1
x2
∫a
ydx − a1(x2 − a0 ) y(a) + a2
0
Adding the above two equations, we get
x2
∫ a (x1 − t) yu dt
0
Curve Fitting
233
x1

y (x1 )+y(x 2 ) − 2 y(a0 )=a1 

∫a

+ a2 

ydx +
0
x2
∫a
0
x1
x2
0
0

ydx

∫ a (x1 − t).y(t) dt +∫ a (x 2 − t) y(t) dt
...(5.35)
The above equation is now solved for a1 and a2 . Since the auxiliary equation of
d2 y
dy
= a1
+ a2 y will be m 2 = a1 m+a2 . Now on solving m 2 − a1 m+a2=0 , we get
2
dx
dx
m=m1 , m 2 . After getting m, and m 2 the value of c1 and c 2 can be obtained by
method of least square. For the better understanding see the following example.
Ex. 11 : Fit a function y=c1 e m1 x + c 2 e m 2 x to the given data
x
1⋅ 0
1⋅1
1⋅ 2
1⋅ 3
1⋅ 4
1⋅ 5
1⋅ 6
1⋅7
1⋅ 8
y
1 ⋅ 54
1 ⋅ 67
1 ⋅ 81
1 ⋅ 97
2 ⋅ 15
2 ⋅ 35
2 ⋅ 58
2 ⋅ 83
3 ⋅ 11
Sol : In this problem a0 = 1 ⋅ 2 because if we take x1 = 1 ⋅ 0 , x 2 = 1 ⋅ 4 then
a0 − x1=x 2 − a0= ⋅ 2 . Now using the relation
x2
 x1
y(x1 )+y(x 2 ) − 2 y(a0 )=a1 
y(x) dx +
y(x) dx
a
a
0
 0
∫
∫

+ a2 

x1
x1
0
0
∫ a (x1 − t) y(t) dt+∫ a (x 2 − t) y(t) dt, we get
1

y( 1 ⋅ 0 )+y( 1 ⋅ 4 ) − 2 y( 1 ⋅ 2 )=a1 

1⋅ 4

∫1⋅2 ydx + ∫1⋅2 ydx

+ a2 

1
1⋅ 4
∫1⋅2 (1 ⋅ 0 − t) ⋅ y(t) dt+∫1⋅2 (1 ⋅ 4 − t) y(t) dt
or

1 ⋅ 54 + 2 ⋅ 15 − 2 × 1 ⋅ 81 = a1 −


+ a2 −

1⋅ 2
1⋅ 4

∫1⋅0 ydx+∫1⋅2 ydx
1⋅ 2
1⋅ 4

∫1⋅0 (1 ⋅ 0 − t) y(t) dt+∫1⋅2 (1 ⋅ 4 − t) y(t) dt
1
rule and on simplification, we get
3
1 ⋅ 81 a1 + 2 ⋅ 18 a2 = 2 ⋅ 1
Similarly, again select a0=1 ⋅ 6 , x1=1 ⋅ 4 and x 2 = 1 ⋅ 8, we get
2 ⋅ 88 a1 + 3 ⋅ 1 a2 = 3 ⋅ 0
On solving (5.36) and (5.37), we get a1 = 0 ⋅ 032 and a2 = 0 ⋅ 94
Applying Simpson's
...(5.36)
...(5.37)
Now putting these values in m 2=a1 m+a2 and then solving for m, we get m1 = 0 ⋅ 99
and m 2 = − ⋅ 96 . Again using the technique of least square, we get c1 = 0 ⋅ 499 ≈ 0 ⋅ 5
and c 2 = 0 ⋅ 491 ≈ 0 ⋅ 49 ≈ 0 ⋅ 5 . Putting the value of c1 , c 2 and m1 , m 2 we get
1
y = [ e 0⋅99 x + e −0 ⋅96 x ] ≈ cos h x
2
234
Engineering Mathematics-III
Problem Set 5.1
1. Fit a straight line y = bx + c to the given data
(i)
(ii)
x
1
2
3
4
6
8
y
2⋅ 4
3⋅1
3⋅ 5
4⋅ 2
5⋅ 0
6⋅ 0
x
0
1
2
3
4
y
1
2⋅ 9
4⋅ 8
6⋅7
8⋅ 6
(iii)
(iv)
(v)
(vi)
(vii)
x
0
1
2
3
y
2
5
8
11
x
0
1
2
3
4
y
1
1⋅ 8
3⋅ 3
4⋅ 5
6⋅ 3
x
1
3
5
7
9
y
1⋅ 5
2⋅ 8
4⋅ 0
4⋅7
6⋅ 0
x
0
1
2
3
4
y
1⋅ 8
1⋅ 6
1⋅1
1⋅ 5
2⋅ 3
x
3
5
7
10
y
162
255
360
495
2. Fit a function y = ab x to the given data
(i)
(ii)
x
1
2
3
4
5
6
7
8
y
1⋅ 0
1⋅ 2
1⋅ 8
2⋅ 5
3⋅ 6
4⋅7
6⋅ 6
9⋅ 8
x
2
3
4
5
6
y
144
172 ⋅ 8
207 ⋅ 4
248 ⋅ 8
298 ⋅ 5
3. Fit a function y = ax b to the given data
(i)
(ii)
x
2
4
7
10
20
40
60
80
y
43
25
18
13
8
5
3
2
x
1
2
3
4
5
6
y
2 ⋅ 98
4 ⋅ 26
5 ⋅ 21
6 ⋅ 10
6⋅ 8
7⋅5
Curve Fitting
235
4. Fit a function y=ae bx to the given data
(i)
(ii)
(iii)
(iv)
x
1
2
3
4
5
6
7
8
y
15 ⋅ 3
20 ⋅ 5
27 ⋅ 4
36 ⋅ 6
49 ⋅ 1
65 ⋅ 6
87 ⋅ 8
117 ⋅ 6
x
1⋅ 0
1⋅ 2
1⋅ 4
1⋅ 6
y
40 ⋅ 17
73 ⋅ 196
133 ⋅ 372
243 ⋅ 02
x
0
0⋅ 5
1⋅ 0
1⋅ 5
2⋅ 0
2⋅ 5
y
0⋅1
0 ⋅ 45
2 ⋅ 15
9 ⋅ 15
40 ⋅ 35
180 ⋅ 75
x
1
2
3
4
5
6
y
1⋅ 5
4⋅ 6
13 ⋅ 9
40 ⋅ 1
125 ⋅ 1
299 ⋅ 5
5. Fit the function y=A1 e λ1 x + A 2 e λ 2 x to the given data
(i)
x
1
1 ⋅1
1⋅ 2
1⋅ 3
1⋅ 4
1⋅ 5
1⋅ 6
1⋅7
1⋅ 8
y
1 ⋅ 175
1 ⋅ 336
1 ⋅ 510
1 ⋅ 698
1 ⋅ 904
2 ⋅ 129
2 ⋅ 376
2 ⋅ 646
2 ⋅ 942
6. Fit a second degree polynomial parabola to the following data
(i)
(ii)
(iii)
x
1
2
3
4
5
6
7
8
9
y
2
6
7
8
10
11
11
10
9
x
0
1
2
3
4
y
1
0
3
10
21
x
1⋅ 0
1⋅1
1⋅ 3
1⋅ 5
1⋅ 9
2⋅1
y
1 ⋅ 84
1 ⋅ 96
2 ⋅ 21
2 ⋅ 45
2 ⋅ 94
3 ⋅ 18
7. Find a cubic spline for the following data Given that M 0=M B=0
(i)
(ii)
(iii)
(iv)
x
0
1
2
3
y
1
2
33
244
x
0
1
2
3
y
1
1
6
10
x
0
1
2
3
y
1
2
5
11
x
2
3
4
y
11
49
123
Given that M 0 = M 2 = 0
236
Engineering Mathematics-III
Objective Type Questions
Multiple Choice Questions
Tick the correct answer.
1. In curve fitting, we get
(a) The exact curve
(b) Approximating curve
(c) A scatter diagram
(d) Can not say
2. Which one of the correct result related to curve fitting
(a) By a method of curve fitting we get an approximating curve
(b) By a method of curve fitting we get an exact curve
(c) By a method of curve fitting we get an equation to fit a curve
(d) By the method of curve fitting we get a formula to fit a curve
3. To the given set of points, a smooth curve is drawn passing through the plotted
points. By the method of curve fitting, we get an
(a) The exact curve
(b) Approximating curve
(c) A scatter diagram
(d) Can not say
4. Which one of the following is a method of curve fitting
(a) Graphical method
(b) Method of group average
(c) Method of least square
(d) All of these
5. To the given set of points, the number of fitted curve may be
(a) One
(b) two
(c) three
(d) infinite
6. Any smooth curve passing the plotted point is known as
(a) The exact curve
(b) Approximating curve
(c) A scatter diagram
(d) Can not say
7. Any smooth curve passing the plotted point is known as approximating curve
and its equation is known as
(a) identity
(b) unique equation
(c) fitting equation
(d) empirical equation
8. Principle of least squares provides a
(a) Unique set of values
(b) Different values
(c) Positive values
(d) Negative values
Fill in the Blanks
1 Principle of least squares provides ……………… .
2 To the given set of points, a smooth curve is drawn passing through the plotted
points. By the method of curve fitting, we get an …………………. .
3 Which one of the correct results related to curve fitting ………………….. .
Curve Fitting
237
4
5
6
7
In curve fitting, we get ……………………….. .
Method of curve fitting are …………, ……………….,………… .
To the given set of points, the number of fitted curve may be …………….. .
Any smooth curve passing the plotted point is known as approximating curve
and its equation is known as ………………………… .
8. Any smooth curve passing the plotted point is known as …………………… .
State True/false
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
In curve fitting, we get an exact curve.
By a method of curve fitting we get an approximating curve.
By a method of curve fitting we get an equation to fit a curve.
By the method of curve fitting we get a formula to fit a curve.
To the given set of points, a smooth curve is drawn passing through the plotted
points. By the method of curve fitting, we get an exact curve.
To the given set of points, a smooth curve is drawn passing through the plotted
points. By the method of curve fitting, we get an approximating curve.
To the given set of points, a smooth curve is drawn passing through the plotted
points. By the method of curve fitting, we get a scatter diagram.
Graphical method is a method of curve fitting.
Method of group average is a method of curve fitting.
Method of least square is a method of curve fitting.
To the given set of points, the number of fitted curve may be one.
To the given set of points, the number of fitted curve may be two.
To the given set of points, the number of fitted curve may infinite.
Any smooth curve passing the plotted point is known as an exact curve.
Any smooth curve passing the plotted point is known as an approximating curve.
Any smooth curve passing the plotted point is known as scatter diagram.
Any smooth curve passing the plotted point is known as approximating curve
and its equation is known as fitting equation.
Any smooth curve passing the plotted point is known as approximating curve
and its equation is known as empirical equation.
Principle of least squares provides a unique set of values.
ANSWERS
Problem Set :
1.
(i)
y = 0 ⋅ 503x+ 2 ⋅ 016
(ii)
y = 2x+ 0 ⋅ 8
(iii)
y = 3x + 2
(iv)
y = 1 ⋅ 33x+ 0 ⋅ 72
(v)
y = 0 ⋅ 523x+1 ⋅ 184
(vi)
y = 0 ⋅ 09x+1 ⋅ 48
(vii)
y = 49 ⋅ 9x+18 ⋅ 7
238
Engineering Mathematics-III
2.
(i)
y = (0 ⋅ 6824) (1 ⋅ 3828) x
(ii)
y = (100) (1 ⋅ 2) x
3.
(i)
y = 78x − 8⋅ 8
(ii)
y = 2 ⋅ 978x − ⋅54
(iv)
y = 6 ⋅ 239x 2⋅ 0196
(i)
y = 11 ⋅ 58 e − 0⋅289x
(ii)
y = 2e 3x
(iii)
y = 0 ⋅ 1e 3x
(iv)
y = 0 ⋅ 56e1⋅ 05x
(i)
y = − 27x 2 + 3 ⋅ 35x − 1
(ii)
y = 2x 2 − 3x+1
(iii)
y = 0 ⋅ 24x 2 − 0 ⋅ 2x+1 ⋅ 04
(iv)
y = − 0 ⋅ 011x 2 +1 ⋅ 3x+ 0 ⋅ 6
(i)
S1 (x) = − 4x 3 + 5x+1,
(ii)
S2(x)= 50 x 3 − 162 x 2+167 x − 53 ,
(iii)
S 3 (x) = − 46x 3 + 414x 2 − 985x+715
(iv)
S1 (x) =
4.
6.
7.
(vi)
1 3
1 3
(x + 2x+ 3),
(v) S 2 (x) = (x + 2x+ 3)
3
3
1
S 3 (x) = ( − 2x 3 +18x 2 − 34x+ 27)
3
Multiple Choice Questions
1. (b)
2. (a)
3. (b)
6. (b)
7. (d)
8. (a)
4. (d)
5. (d)
Fill in the Blanks
1. a unique set of values
2. approximating curve
3. by a method of curve fitting we get an approximating curve
4. approximating curve
5. Graphical method, Method of group average, Method of least square
6. infinite
7. empirical equation
8. approximating curve
True or False
1. False
2. True
3. False
4. False
5. False
6. True
7. False
8. True
9. True
10. True
11. False
12. False
13. True
14. False
15. True
16. False
17. False
18. True
19. True
20. True
21. True
22. True
❑❑❑
Unit-2
Chapter
Correlation and
6
Regression
6.1 Correlation
If the variables x and y are such that the change in one is accompanied by a change in
other and vice-versa then the variables x and y are said to be correlated and this
relationship is called Correlation. Correlation may be of following kinds :
(i) Positive Correlation : If the variables concerned show a tendency to move in
the same direction i.e., if the decrease or increase in x is followed by an decrease or
increase in y, then the correlation between x and y is said to be positive, e.g..
x
78
63
70
39
46
98
87
y
63
53
58
36
40
80
69
(ii) Negative Correlation : When the variables move in an opposite direction
i,e., with an increase or decrease in x there is a corresponding decrease or increase in
y then the two variables are said to be negatively correlated e.g.
x
73
63
48
40
35
y
30
35
41
50
60
(iii) Linear Correlation : If the change in x bears a constant ratio with
corresponding change in y, the correlation is said to be linear. In this case, if the
pairs of values of x and y are plotted on the graph paper, we shall get a straight line.
(iv) Non-Linear of Curvilinear Correlation : When the change in x does not
bear a constant ratio with change in y then the correlation between x and y is said to
be non-linear or curvilinear.
(v) Perfect Correlation : When the percentage change in x is followed by same
percentage change in y the correlation between x and y is said to be perfect.
(vi) Simple, Partial and Multiple Correlation : If, we study relationship
between two variables say x and y, then the correlation is said to be simple (or
bivariate). When three or more variables are studied simultaneously then it is the
case of partial correlation or multiple correlation.
6.1.1 Scatter or Dot Diagram
The graphical representation of the corresponding values of two variables (x, y) on a
graph by means of dots is called Scatter or Dot Diagram.
240
Engineering Mathematics-III
Here each pair (x, y) is represented by some dot (.) or cross (×) on the graph paper
and the number of such dots/crosses will be equal to the number of observations
given.
Scatter diagram is the simplest device to study the nature of relationship between
the variables under study. Simply looking at the scatter diagram, one can get some
idea whether the two variables are related or not. The following inferences about
correlation can be drawn by observing the scattering of points in a Scatter diagram.
Y
Y
Y
r=0
X
0 Positive Correlation
0
X
0
Negative Correlation
Y
X
Zero or No Correlation
Y
Coeff. of
Correlation = + 1
0
X
0
X
Coeff. of Correlation = – 1
Methods for measuring Correlation
Fig. 6.1
(a) When the plotted points are haphazardly situated i.e., the points do not show
any particular tendency, then the variables are said to be independent i.e., we
have zero correlation and conclude that no relationship exists between x and y.
(b) If the points show an upward or downward trend, the variables are said to be
correlated. If the tendency of the points rises upwards i.e., from lower left hand
corner to the upper right hand corner, the variables are said to be positively
correlated. If all points lie on a straight line falling from lower left hand corner
to upper right hand corner, the correlation is said to be perfectly positive.
(c) When the tendency of points is to slope downwards from top left hand corner
to lower right hand corner, the correlation is said to be negative. If all the
points lie on straight line sloping downwards from left hand corner to right
hand corner, the correlation is perfectly negative.
Correlation and Regression
241
Method for measuring Correlation
Graphical
Scatter Diagram
Karl Pearsons
Coefficient of
Correlation
Algebraic
Correlation Graph
Spearman's Rank
Difference
Method
Concurrent Deviation's
Method
Other Method
6.2 Correlation Analysis and Determination of Karl Pearson
Coefficient
Covariance : Let the n pairs of observations on two quantitative variables X and Y be
given as under : ( x1 , y1 ), ( x 2 , y 2 ) , … ( x n , y n ) where x1 , x 2 ,… denote observed
values of the variable X and y1 , y 2 ,… those of Y.
The deviations of X-values from their mean x, are given by
x1 − x, x 2 − x, … x n − x.
And the deviation of Y-values from their mean y, are given by
y1 − y , y 2 − y , … y n − y
Then the sum of n products
( x1 − x )( y1 − y ), ( x 2 − x )( y 2 − y )…( x n − x )( y n − y )
divided by n, is defined as the covariance of X and Y i.e., we have
{( x1 − x )( y1 − y ) + ( x 2 − x )( y 2 − y ) +…+( x n − x )( y n − y )}
Cov ( X , Y ) =
n
=
1 n
∑ ( xi − x )( y i − y )
n i =1
Cov (X, Y) can also be expressed as
Cov.( X , Y ) =
1 n
∑ {( xi − x )( y i − y )}
n i =1
n
n
n
1
=  ∑ xi y i + nxy − y ∑ xi − x ∑ y i 
i =1
i =1
i =1
n
n
n

x
.
∑ i ∑ yi
n

i =1
i =1
=  ∑ xi y i + nx y − y . nx −
n
i =1




1

n


242
Engineering Mathematics-III
n
n
∑ xi y i
=
i =1
n
n
∑ xi . ∑ y i
− i =1
i =1
n2
Cov (x, y) is read as covariance of x and y.
Karl Pearson's. Coefficient of Correlation : It is calculated to study the extent or
the degree of correlation that exists between two variables X and Y and is denoted by r
or r ( X , Y ) . Karl Pearson defined the correlation coefficient r ( X , Y ) between two
variables X and Y as
r ( X, Y ) =
Cov ( X , Y )
{Var ( X ) . Var ( Y )}
=
1 n
∑ ( xi − x )( y i − y )
n i =1
n
1 n
2  1
2
 ∑ ( xi − x )   ∑ ( y i − y ) 
 n i =1
  n i =1

n
∑ ( xi
∴
r ( X, Y ) =
− x )( y i − y )
i =1
n
n
2 
2
 ∑ ( xi − x )   ∑ ( y i − y ) 
i =1
 i =1

=
ΣXY
[ ΣX
2
. ΣY 2 ]1 2
…[ A]
where
X = xi − x, Y = y i − y
This is known as Karl Pearson's coefficient of correlation.
Now, denominator of r is always positive, so the value of r is positive or negative
according as the convariance value is positive or negative. Its value lies between −1
and +1.
[A] can also be written as
Σxi y i −
r ( X, Y ) =
( Σxi )( Σy i )
n

( Σxi ) 2  
( Σy i ) 2 
2
2
Σ
x
−
Σ
y
−



i
i
n 
n 

n
n
n
n ∑ xi y i − ∑ xi ∑ y i
=
i =1
i =1
i =1
2
2
 n
 n
n
 
n
 


2
2
 n ∑ xi −  ∑ xi  
n ∑ y i −  ∑ y i  
i =1  
i =1  
 i =1
 i =1


xi − a
yi − b
Putting ui =
, vi =
, where a and b are some arbitrary chosen numbers
h
k
and h is a positive common factor of all deviations xi − a of items in x-series, k being
a positive factor of all deviations y i − b of items in y series; we obtain
r ( X , Y ) = ρ(u, v ) =
nΣui vi − Σui . Σvi
{nΣui2 − ( Σui ) 2 } . {nΣvi2 − ( Σvi ) 2 }
The value of r is independent of the origin and the scale.
Correlation and Regression
243
Here the values of u and v are called the step deviations of the values of x and y
respectively. In practical problems, the choice of common factors h and k would not
create any problem. Even, if we do not care to compute step deviations by dividing
the step deviations xi − a and y i − b by some common factor, the formula would
still work. We can also take h = 1 = k.
Hence, while solving problems, we first calculate deviations of items of x-series and
y-series from some convenient and suitable assumed means a and b respectively.
These deviations of x-series and y-series are then divided by positive common
factors, if at all needed. Positive common factors h and k can be chosen by seeing the
conditions of the problem.
6.3 Coefficient of Correlation for Grouped Data
In case x and y series are both given as frequency distributions, then these can be
represented by a two-way table known as Correlation table. It is a double entry
table with one series along the horizontal and the other along the vertical. The
coefficient of Correlation for such a bivariate frequency distribution is calculated by
the formula
n( Σfi ui vi ) − ( Σfi ui )( Σfi vi )
r=
2
[{nΣfi ui − ( Σfi ui ) 2 × [ nΣfi vi2 − ( Σfi vi ) 2 }]
where
ui = deviation of the central values from the assumed mean of x-series
= [( xi − a)/ h]
vi = deviation of the central values from the assumed mean of y-series
= [( y i − a)/ k]
fi = frequency corresponding to the pair ( xi , y i )
n = Σfi = total number of frequencies.
Note : (i) If r ( X , Y ) = 1, there is a perfect positive correlation between x and y.
(ii)
If r( X , Y ) = −1, there is a perfect negative correlation between x and y.
(iii)
If r( X , Y ) = 0, there is no correlation between the two variables x and y.
(iv)
If r( X , Y ) > 0, the correlation between x and y is positive.
(v)
If r( X , Y ) < 0, the correlation between x and y is negative.
6.4 Spearman's Rank Correlation Coefficient
In 1904, Charles Edward Spearman, a Britsh psychologist obtained the method of
ascertaining the coefficient of correlation by ranks. This measure is useful in dealing
with qualitative characteristics e.g. beauty, morality, character and intelligency etc.
The formula runs as follows :
n
6∑ di2
r = 1−
i =1
n( n 2 − 1)
where r = Rank Coeff. of correlation
di2 = sum of the squares of the difference of two corresponding ranks
244
Engineering Mathematics-III
n = Number of paired observations.
Suppose n students taken Engineering Mathematics-I Paper and the top scorer of the
marks is placed at number 1, the second at number 2, the third at number 3 and so
on. Similarly the same n students are graded according to the marks obtained by
them in Engineering Mathematics II and given the rank as before. Let the ranks
given to Engineering Mathematics-I be xi , where i = 1, 2, … n. Similarly, let the
ranks given to Engineering Mathematics II by y i , where i = 1, 2, 3, .., n and the
above condition hold good for y i also.
 1
1 n
1 n( n + 1) n + 1
Then
x =  ∑ xi  = (1 + 2 + 3+…+ n ) = .
=
n i =1  n
n
2
2
 1
1 n
1 n( n + 1) n + 1
=
 ∑ y i  = (1 + 2 + 3+…+ n ) = .
n i =1  n
n
2
2
y =
σ 2x = σ 2y =
And
2
1  n 2
1
 n + 1
 ∑ y i  − y 2 = (1 2 + 2 2 + 3 2 +…+ n 2 ) − 

 2 
n i =1 
n
2
=
1 n( n + 1)( 2n + 1)  n + 1
.
−

 2 
n
6
=
( n + 1)( 2n + 1) ( n + 1) 2 ( n + 1)
−
=
( 4n + 2 − 3n − 3)
6
4
12
=
( n + 1)( n − 1) n 2 − 1
=
12
12
...(6.1)
If di denotes the difference in ranks of the ith individuals then
[∵ x = y ]
di = xi − y i = ( xi − x ) − ( y i − y )
And
n
n
i =1
i =1
n
∑ di2 = ∑ ( xi
=
∑ [( xi
− x ) 2 + ( y i − y ) 2 − 2( xi − x )( y i − y )]
∑ [( xi
− x ) 2 + ( y i − y ) 2 − 2( xi − x )( y i − y )]
i =1
n
=
i =1
n
=
n
n
∑ ( xi − x )2 + ∑ ( y i − y )2 − 2∑ ( xi − x )( y i − y )]
i =1
n
Thus
− y i )2
i =1
n
i =1
n
1 n

1
1
1
di2 = ∑ ( xi − x ) 2 + ∑ ( y i − y ) 2 –2  ∑ ( xi − x )( y i − y )
∑
n i =1
n i =1
n i =1
 n i =1

...(6.2)
n
But
Var ( x ) =
n
1
1
∑ ( xi − x )2 , Var( y ) = n ∑ ( y i − y )2
n i =1
i =1
Cov ( x, y ) =
1 n
∑ ( xi − x )( y i − y )
n i =1
Correlation and Regression
245
Putting these in (6.2), we have
1 n 2
∑ di = Var ( x ) + Var ( y ) − 2 Cov ( x, y )
n i =1

Cov ( x, y )
= Var ( x ) + Var( y ) − 2σ x σ y . r ( x, y ) ∵ r ( x, y ) =

σ xσ y 

= 2 Var ( r ) − 2 σ 2x . r ( x, y )
[∵ σ x = σ y ]
= 2 σ 2x − 2 σ 2x . r ( x, y )
= 2 σ 2x [1 − r ( x, y )]
 n 2 − 1
= 2
 [1 − r ( x, y )]
 12 
or
6
n
∑ di2 = 1− r( x, y )
n( n 2 −1) i =1
n
6 ∑ di2
or
r( x, y ) = 1 −
i =1
n( n 2 − 1)
This is known as spearman's coefficient of rank correlation.
Illustrative Examples
Ex.1: Find Cov ( X , Y ) between x and y for the following data :
Σxi = 15, Σy i = 36, Σxi y i = 110, n = 5
 n

1n
1 n
Sol: Cov [ X , Y ] =  ∑ xi y i −  ∑ xi   ∑ y i  
n i =1
n i =1  i =1  
1
1
1
2
= 110 − × 15 × 36 = (110 − 108) = = 0.4
 5
5
5
5
Ex.2: Calculate the Karl Pearson's coefficient of correlation from the data given
below:
x
4
6
8
10
12
y
2
3
4
6
10
246
Engineering Mathematics-III
Sol:
yi − y ( xi − x ) × ( xi − x )2 ( yi − y )2
y = 5 ( yi − y )
S.No.
xi
yi
xi − x
x =8
1
4
2
–4
–3
12
16
9
2
6
3
–2
–2
4
4
4
3
8
4
0
–1
0
0
1
4
10
6
2
1
2
4
1
5
12
10
4
5
20
16
25
Total
Σ{( xi − x ) Σ( xi − x ) 2 Σ{( y i − y )2}
=40
× ( yi − y )
=40
=38
Σxi = 40 Σy i = 25
Σxi
Σy i
40
25
=
= 8, y =
=
=5
n
5
n
5
Σ( xi − x )( y i − y )
38
38
r=
=
=
= 0.95
40 40 40
{Σ( xi − x ) 2 } {Σ( y i − y ) 2 }
We have
x=
⇒
which shows that there is high degree positive correlation between the variables x
and y.
Ex. 3: Psychology test of intelligence and of arithmetical ability were applied to 10
children. Here is a record of ungrouped data showing intelligence and arithmetic
ratios. Calculate Karl Pearson's coefficient of correlation.
Child
A
B
C
D
E
F
G
H
I
J
I.R.
105
104
102
101
100
99
98
96
93
92
A.R.
101
103
100
98
95
96
104
92
97
94
Sol: Let x and y denote I.R. and A.R. respectively, then frame the following table.
Case
xi
yi
vi = yi − b u i = xi − a
b = 96
a = 100
k=1
h=1
u i vi
u i2
vi2
A
105
101
5
5
25
25
25
B
104
103
4
7
28
16
49
C
102
100
2
4
8
4
16
D
101
98
1
2
2
1
4
E
100
95
0
–1
0
0
1
Correlation and Regression
247
F
99
96
–1
0
0
1
0
G
98
104
–2
8
– 16
4
64
H
96
92
–4
–4
16
16
16
I
93
97
–7
1
–7
49
1
J
92
94
–8
–2
16
64
4
n = 10
Σui = –10
r=
⇒
=
=
Σvi = 20 Σui vi = 72 Σui2 = 180 Σvi2 = 180
nΣui vi − Σui Σvi
{nΣui2 − ( Σui ) 2 } {nΣvi2 − ( Σvi2 }
10(72) − ( −10)( 20)
10 (180) − ( −10) 2
10(180) − ( 20) 2
720 + 200
1800 − 100 1800 − 400
=
920
1700 1400
= 0.596
which shows that there is moderate degree of positive correlation between the
variables x and y.
Ex.4: Calculate Rank correlation coefficient from the marks obtained by 10
students in Mathematics (x) and Statistics (y).
x:
75
30
60
80
53
35
15
40
38
48
y:
85
45
54
91
58
63
35
43
46
44
Sol:
x
y
1
75
85
2
2
0
0
2
30
45
9
7
2
4
3
60
54
3
5
–2
4
4
80
91
1
1
0
0
5
53
58
4
4
0
0
6
35
63
8
3
5
25
7
15
35
10
10
0
0
8
40
43
6
9
–3
9
9
38
46
7
6
1
1
10
48
44
5
8
–3
n = 10
Total
Rank in Rank in di = xi − yi
x = xi
y = yi
di2
S.No.
9
Σdi2
= 52
248
Engineering Mathematics-III
r = Rank correlation coefficient =1−
6 Σdi2
n( n 2 − 1)
6 × 52
r = 1−
= 1 − 0.315 = 0.685
10 (10 2 − 1)
⇒
Ex.5: The coefficient of rank correlation of marks obtained by 10 students in
English and Economics was found to be 0.5. It was later discovered that the
difference in ranks in the two subjects obtained by one of the students was wrongly
taken as 3 instead of 7. Find the correct coefficient of rank correlation.
6Σdi2
Sol.
r = 1−
Here
r = 0 .5, n = 10 so, 0 . 5 = 1 –
n( n 2 − 1)
6 ∑ di2
10(100 – 1)
6Σdi2
= 1 − 0 .5 = 0 .5
990
Correct value of di2 = 82 . 5 − 3 2 + 7 2 = 122 . 5
Thus correct r = 1 −
6 × 122.5
735
= 1−
= 0.2676
10 × 99
990
Ex.6: Ten competitors in a beauty contest got ranks by three judges in the following
order :
Judge I
1
6
5
10
3
2
4
9
7
8
Judge II
3
5
8
4
7
10
2
1
6
9
Judge III
6
4
9
8
1
2
3
10
5
7
Use rank correlation coefficient to discuss which pair of judges have the nearest
approach to common tastes in beauty.
Sol:
Rank Rank by Rank by
by 1st
2nd
3rd
Judge Judge R 2 Judge
R1
R
D12 =
D23 =
D13 =
R1 − R 2
R2 − R3
R1 − R 3
2
D12
2
D23
2
D13
3
1
3
6
–2
–3
–5
4
9
25
6
5
4
1
1
2
1
1
4
5
8
9
–3
–1
–4
9
1
16
10
4
8
6
–4
2
36
16
4
3
7
1
–4
6
2
16
36
4
2
10
2
–8
8
0
64
64
0
4
2
3
2
–1
1
4
1
1
Correlation and Regression
249
9
1
10
8
–9
–1
64
81
1
7
6
5
1
1
2
1
1
4
8
9
7
–1
2
1
1
4
1
200
214
60
Total
Here n = 10
Rank correlation coefficient between 1st and 2nd judges is given by
r12 = 1 −
= 1−
2
6Σd12
n( n 2 − 1)
6 × 200
2
10(10 − 1)
= 1−
1200
= − 0.21
10 × 99
Rank correlation coefficient between 1st and 3rd judges
6 × 60
i.e.,
r13 = 1 −
10 (10 2 − 1)
= 1−
360
= 0.64
10 × 99
While Rank correlation coefficient between 2nd and 3rd judges is
r23 = 1 –
= 1−
2
6Σd23
n( n 2 − 1)
6 × 214
10(10 2 − 1)
= – 0.29
Hence the correlation between first and second judges and between second and
third judges is negative which means their opinions are mutually opposite to each
other while between first and third judges, the correlation coefficient is positive
implying their opinions are very similar.
Ex.7: Establish the formula σ 2x − y = σ 2x + σ 2y − 2rσ x σ y where r is the correlation
coefficient between x and y.
Sol:We know that σ 2x =
Σ[( x − y ) − ( x – y )]2
Σ( x − x ) 2
therefore σ 2xy =
n
n
where x − y = mean of ( x − y ) series = mean of x – mean of y
= x−y
σ 2x − y =
=
Σ[( x − y ) − ( x − y )]2
Σ[( x − x ) − ( y − y )]2
=
n
n
Σ( x − x ) 2 − ( y − y ) 2 − 2( x − x )( y − y )
n
Σ( x − x ) 2 Σ( y − y ) 2 2( x − x )( y − y )
+
−
n
n
n
2Σ(
x
−
x
)(
y
−
y
)
= σ 2x + σ 2y −
n
=
...(6.1)
250
Engineering Mathematics-III
But
r=
Σ( x − x )( y − y )
Σ( x − x )( y − y )
or
= rσ x σ y
nσ x σ y
n
Substituting this value in (6.1), we have
σ 2x − y = σ 2x + σ 2y − 2rσ x σ y
Ex.8: The correlation table given below shows that the ages of husband and wife of
53 married couples living together on the census night of 1991. Calculate the
coefficient of correlation between the age of the husband and that of the wife.
Age of
husband
Age of wife
Total
15-25
25-35
35- 45
45-55
55- 65
65 -75
15-25
1
1
-
-
-
-
2
25-35
2
12
1
-
-
-
15
35- 45
-
4
10
1
-
-
15
45-55
-
-
3
6
1
-
10
55- 65
-
-
-
2
4
2
8
65-75
-
-
-
-
1
2
3
Total
3
17
14
9
6
4
53
Sol:
Age of husband
xi − 40
10
yi − 40
vi =
10
Age of wife xi series
ui =
15-25 25-35 35-45 45-55 55-65 65-75
yc
Mid
series
pt x i
Age of Mid
group pt y
20
30
40
50
60
70
y i − a x i − a –20
–10
0
10
20
30 Σf i
–2
–1
0
1
2
3
f i vi
f ivi2
f iuivi
-4
8
6
i
ui
vi
15-25 20
-20
-2
4
1
2
1
2
Correlation and Regression
25-35 30
-10
-1
4
2
35-45 40
0
251
12
12
0
10
0
20
30
1
0
6
6
2
15
0
0
0
10
10
10
8
12 8
16
32
32
18 3
9
27
24
1
16
4
3
2
6
17
16
2
4
Σf i 3
15
1
2
65-75 70
-15
0
10
3
55-65 60
15
1
0
4
45-55 50
0
14
9
1
2
6
4
Σ53
Σf iui Σ19
=n
= 16
Thick figures in
small sqs. stand for
f iuivi
f iui -6
-17
0
9
12
12
10
f iui2 12
17
0
9
24
36
98
f iuivi 8
14
0
10
24
30
86
Σ86
Check : from Σf iuivi
both=86 sides
With the help of the above correlation table, we have
n( Σfi ui vi − ( Σfi ui )( Σfi vi )
r=
[{nΣfi ui2 − ( Σfi ui ) 2 } {nΣfi vi2 − ( Σfi vi ) 2 }]
=
=
53 × 86 − 10 × 16
[( 53 × 98 − 100)( 53 × 92 − 256)]
4398
( 5094 × 4620)
=
4398
= 0.91.
4850
Problem Set 6.1
1.
Calculate the covariance between x and y from the following data :
(i) Σxi = 50, Σy i = −30, Σxi y i = −115 and n = 10
(ii) Σxi = 100, Σy i = 140, Σ( xi − 10)( y i − 15) = 60 and n = 10
2.
Compute the covariance between x and y of the following pair of observations :
3.
(3, 5), (6, 7), (9, 9), (12, 11), (15, 13), (18, 15), (21, 17), (24, 19).
Calculate Karl Pearson's coefficient of correlation for the following data :
x
1
3
5
7
9
11
y
10
9
8
7
6
5
252
4.
Engineering Mathematics-III
Calculate the coefficient of correlation for the following ages of husbands and
wives :
Age of husband
23
27
28
28
Age of wife
18
20
20
27 21
5.
29
30
31
33
35
36
29
27
29
28
29
The following table gives the value of iron ore exported and value of steel
imported in India during 1970-71 to 1976-77. Find the value of correlation
coefficient between exports and imports.
Year
1970-71 1971-72 1972-73 1973-74 1974-75 1975-76 1976-77
Export of
iron
ore
('000 Rs.)
42
44
58
55
89
98
66
Import of
steel ('000
Rs.)
56
49
53
58
65
76
58
6.
7.
Draw a scatter diagram for the following data :
x
4
6
8
10
12
17
6
y
20
25
30
40
35
45
50
A sample of five items is taken from the production of a firm. The length and
weight of five items are given below. Compute the value of the coefficient of
correlation.
Length (in inches)
3
4
6
7
10
Weight (in kgs)
9
11
14
15
16
8.
9.
Find the coefficient of correlation from the following data. Also explain, what
does it express.
x
300
350
400
450
500
550
600
65
700
y
800
900
1000
1100
1200
1300
1400
1500
1600
The following table gives the number of students having different heights and
weights.
Correlation and Regression
253
Weights in lbs
Weight in inches 80-90 90-100 100-110 110-120 120-130 Total
50-55
1
3
7
5
2
18
55-60
2
4
10
7
4
27
60-65
1
5
12
10
7
35
65-70
0
3
8
6
3
20
Total
4
15
37
28
16
100
Find the coefficient of correlation.
10. Two independent variates x1 and x 2 have means 5 and 10, variances 4 and 9
respectively. Obtain the correlation coefficient between y1 = 3 x1 + 4 x 2 and
y 2 = 3 x1 − x 2 .
6.5 Regression
If the scatter diagram indicates some relationship between two variables x and y,
then the dots of the scatter diagram are concentrated round a curve, which is known
as the curve of regression. Regression analysis is the method used for estimating
the unknown values of an variable corresponding to the known value of the other
variable.
Lines of Regression : When the curve of regression is a straight line, it is called
line of regression and the regression is said to be linear.
The line used to estimate the value of x for a given value of y is called the regression
line of y on x. Similarly, the line used to estimate the value of x for a given y is called
the regression line of x on y. In regression line of y on x ( x on y), the variable y is
considered as the dependent (independent) variable whereas x is considered as the
independent (dependent) variable. The regression lines are also called estimated
lines. In case of perfect correlation between x and y, the two lines become
coincident. When r, the coefficient of correlation numerically decreases from 1 to 0,
the angle between the regression lines will increase from 0° to 90° and if r = 0, the
regression lines will be perpendicular to each other. Regression lines will be
determined with the help of principle of least squares.
To find the equation of regression line :
y = a + bx regression line of y on x
...(6.1)
254
Engineering Mathematics-III
Fig. 6.2
x = a + by regression line of x on y
...(6.2)
x on y
Fig. 6.3
Summing (6.1) both sides, we have
∴
Σy = na + bΣx
Σy i
bΣxi
= a+
⇒ y = a + bx
n
n
...(6.3)
... (6.4)
where x and y denote the means of x series and y series respectirely. This shows that
( x, y ) lies on the line of regression on ( y = a + bx ).
Now subtracting (6.4) from (6.1), we get
y − y = b( x − x )
Coeff. of correlation between x and y is
Σx y − nx y
r= i i
nσ x . σ y
⇒
Σxi y i = nrσ x σ y + nx y
...(6.5)
... (6.6)
Now from (6.1) and (6.6)
n r σ x σ y + n x y = aΣxi + bΣxi2
⇒
Also
... (6.7)
a
b
Σ xi + Σxi2
n
n
... (6.8)
y = a + bx ⇒ x y = ax + bx 2
...(6.9)
rσ x σ y + x y =
Correlation and Regression
255
From (6.8) and (6.9), we have
r σ x σ y + ax + bx 2 =
a
b
Σ xi + Σ xi2
n
n
a
b
nx + Σ xi2
n
n
 Σx 2

= b i − x 2  = b σ 2x
 n

rσ x σ y + ax + bx 2 =
or
r . σ xσ y
or
b=
r σ xσ y
σ 2x
=
rσ y
σx
Hence, the equation to the line of regression of y on x is
y − y = ( r σ y /σ x )( x − x ) or y − y = b yx ( x − x )
The constant ( rσ y / σ x ) is called the regression coefficient of y on x and is
denoted by b yx .
i.e.,
b yx =
rσ y
...(6.10)
σx
Similarly, the equation to the line of regression of x = a + by on y is
or
x − x = ( rσ x /σ y ) ( y − y )
x − x = b xy ( y − y )
... (6.11)
b xy = ( rσ x / σ y ) is called the regression coeff. of x on y.
Note. 1. It is clear that the two lines of regression pass through the point ( x, y )
1
Σ xy − x y
σx
σ
2.
b xy = r
= n
. x
σy
σx σy
σy
⇒
b xy
Similarly,
b yx
1
1
Σxy − x y
Σxy − x y
n
=
= n
1
σ 2y
Σy 2 − y 2
n
σy
1 Σxy − x y
=r.
=
σx
n 1
Σx 2 − x 2
n
6.5.1 Method of Finding the Value of Regression Coefficient of x on y
when the given Bivariate Distribution are Large :
Let u = x − a and v = y − b where a and b are assumed means, then
b yx = bvu
1
Σ uv − u v
= n
1
Σu 2 − u 2
n
256
and
Engineering Mathematics-III
b xy = buv
1
Σuv − u v
= n
1
Σv 2 − Σv 2
n
6.6 Properties of Regression Coefficients
Prop. 1. The correlation coefficient is the geometric mean between the regression
coefficients.
Proof. Regression coefficients are given by:
σy
σ
b yx = r .
and b xy = r . x
σx
σy
⇒ b yx × b xy = r 2 , where r is the correlation coefficient.
Remark : From the above relation, it is clear that b xy and b yx have the same sign
i.e., either ( b xy > 0 and b yx > 0) or (b xy < 0 and b yx < 0)
b xy =
rσy
rσ x
and b yx =
σy
σx
But σ x > 0 and σ y > 0, it follows that r, b xy and b yx have the same sign.
Prop. 2. If one of the regression coefficients is greater than unity, the other must be
less than unity.
Proof. Let one of the regression coefficients (say) b yx be greater than unity, then
we have to show that b xy < 1.
1
Let
b yx > 1 ⇒
<1
b yx
But
r 2 ≤ 1, so b yx . b xy ≤ 1 ⇒ b xy ≤
1
b yx
⇒ b xy < 1, using (1).
Prop. 3. The arithmetic mean of the regression coefficients is greater than the
correlation coefficient.
1
Proof. We have to prove that ( b yx + b xy ) ≥ r
2
σy σx
σ 
1 σy
⇒
+ r . x  ≥ r or
+
≥2
r .
2 σx
σy 
σx σy
⇒
σ 2y + σ 2x − 2σ x σ y ≥ 0 or (σ y − σ x ) 2 ≥ 0,
which is true.
Prop. 4. The regression coefficients are independent of the change of origin but not
of scale.
y−b
x−a
Proof. Let
u=
, v=
h
k
⇒ x = a + hu, y = b + kv, where a, b, h and k are constants.
Correlation and Regression
257
Now
rxy = ruv, σ 2x = h 2 σ u2 and σ 2y = k 2σ 2v .
and
b xy = rxy
σx
h σu
σ  h
h 
= ruv .
=  ruv . u  = buv .
σy
k σv
k 
σv  k
Similarly, we can prove that b yx = ( k / h )bvu .
Prop 5. If x and y are independent, the regression coefficients are zero.
Proof. x and y are independent to each other, so the correlation coefficient r = 0
σ
i.e.,
b xy = r . x = 0. Similarly b yx = 0
σy
Prop 6. Angle between lines of regression. If α is acute angle between the two
1 − r 2 σ xσ y
regression lines, show that tan α =
. Interpret the cases when
r
σ 2x + σ 2y
r = 0, r = ±1.
Proof. The equation of two lines of regression are given by
σy
σ
y−y =r
( x − x ) and x − x = r x ( y − y )
σx
σy
The slopes of the lines of regression are given by m1 =
rσy
σx
, m2 =
σy
rσx
.
Now, let the angle between the two lines of regression be α, then we have
σy
rσ y
−
rσ x
σx
1 − r2 σ y σ x
tan α =
=
r σ 2x + σ 2y
σ 2y
1+
σ 2x
Fig. 6.4
Particular cases. If r = 0, then tan α = ∞ ⇒ α = 90°.
Hence, if two variables are uncorrelated, the two lines of regression are
perpendicular to each other.
If r = ±1, then tan α = 0 ⇒ α = 0 or π
258
Engineering Mathematics-III
In this case the two lines of regression either coincide or parallel to each other. But
the two lines of regression pass through ( x, y ), they cannot be parallel. Hence, the
two lines of regression coincide.
Illustrative Examples
Ex.9: The following data regarding the heights (y) and the weights (x) of 100
college students are given :
Σx = 15000 ; Σx 2 = 2272500 ;
Σy = 6800 ; Σy 2 = 463025 ; Σxy = 1022250
Find the correlation coefficient between height and weight and state the equation of
regression of height on weight.
[U.P.T.U Tech. 2001]
Σx 15000
x=
=
= 150
n
100
Σy 6800
y =
=
= 68
n
100
Sol: We have
σx =
Σx 2  Σx 
− 
 n
n
2
⇒
σy =
Σy 2  Σy 
− 
 n
n
2
and
=
2272500  15000
−

 100 
100
=
463025  6800
−

 100 
100
2
225 = 15
=
2
=
6.25 = 2 .5
Σxy
1022250
− ( x )( y )
− (150)( 68)
n
100
r=
=
(σ x )(σ y )
15 × 2.5
⇒
=
22 . 5
1. 5
=
= 0. 6
15 × 2 . 5 2 . 5
whence the regression equation of y on x is given by
σy
 2 . 5
y– y =r
( x − x ) ⇒ y − 68 = 0.6 
 ( x − 150)
 15 
σx
y − 68 =
⇒
1
( x − 150) ⇒ 10 y = x + 530
10
Ex.10: Obtain the equations of the lines of regression for the following data :
x
1
2
3
4
5
6
7
8
9
y
9
8
10
12
11
13
14
16
15
Also obtain an estimate of y for x = 6.2
Correlation and Regression
259
Sol: Here h = 1 = k
S.No.
x
y
u = x−a
a=5
v= y−b
b = 12
u2
v2
1
1
9
–4
–3
16
9
12
2
2
8
–3
–4
9
16
12
3
3
10
–2
–2
4
4
4
4
4
12
–1
0
1
0
0
5
5
11
0
–1
0
1
0
6
6
13
1
1
1
1
1
7
7
14
2
2
4
4
4
8
8
16
3
4
9
16
12
9
9
15
4
3
16
9
12
0
0
n=9
Now
∴
Also
rxy = ruv =
r=
b yx =
uv
Σu 2 = 60 Σv 2 = 60 Σuv = 57
1
Σ uv − u v
n
 1 Σu 2 − u 2   1  Σv 2 − ( v ) 2 

  

n
  n 

1
1
( 57 ) − 0
( 57 ) − 0
19
9
= 9
=
= 0 ⋅ 95
1
20
 1 [( 60) − 0]  1 ( 60) − 0
( 60)
9
 9
  9

rσy
σx
1
1
( 57 ) − 0
Σuv − u v
rσ v
19
9
n
=
=
=
=
= 0.95
1
1
σu
20
Σu 2 − u 2
( 60) − 0
n
9
where x = a + u, y = b + v ⇒ x = a + 9Σui and y = b + 9Σvi
The equation to the line of regression of y on x is given by
rσ y
y−y =
( x − x ) ⇒y − 12 = 0.95 ( x − 5)
σx
or
y = 0.95 x + 7.25
When x = 6.2, the estimated value of y = 0.95 × 6.2 + 7.25 = 13.14
1
1
( 57 ) − 0
Σuv − u v
rσ x
σu
19
Also
b xy =
=r
= n
= 9
=
= 0.95
1
1
σy
σv
20
2
2
Σv − ( v )
( 60) − 0
n
9
260
Engineering Mathematics-III
Then the equation to the line of regression of x on y is given by
rσ x
x−x =
(y − y)
σy
or
x − 5 = 0.95 ( y − 12)
or
x = 0.95 y − 6 .4
Ex.11: The following regression equations were obtained from a correlation table :
y = 0.516 x + 33.73, x = 0.512 y + 32 .52
Find the value of (a) the correlation coefficient, (b) the mean of x's and (c) the mean
of y's.
Sol: (a) y = 0.516 x + 33.73 (Regression equation of y on x)
...(6.1)
x = 0 .512 y + 32 .52 (Regression equation of x on y)
...(6.2)
From (1), slope = r
σy
σx
From (6.2) slope =1/r
...(6.3)
= 0.516 = m1 , say
σx
σ
= 1 / 0.512 = m 2 , say or r x = 0.512
σy
σy
...(6.4)
 σ y  σ x 
From (6.3) and (6.4) m1 m 2 =  r
 = ( 0.516)( 0.512)
 
 σ x  σ y 
r 2 = 0.516 × 0.512 or r = 0.514
⇒
i.e., Coefficient of correltion=0.514
(b) Since Equations (1) and (2) pass through ( x, y ).
∴
y = 0.516 x + 33.73 and x = 0.512 y + 32 .52
Solving these, we obtain x = 67.6, y = 68 .61
Ex.12: In the following table, recorded data showing the test scores made by
salesmen on an intelligence test and their weekly sales :
Salesmen
1
2
3
4
5
6
7
8
9
10
Test scores
40
70
50
60
80
50
90
40
60
60
Sales (000)
2.5
6.0
4.5
5.0
4.5
2.0
5.5
3.0
4.5
3.0
Calculate the regression line of sales on test scores and estimate the most probable
weekly sales volume if a salesman makes a score of 70.
Sol: We have x = mean of x (test scores) = 60 + ( 0 / 10) = 60
and y = mean of y (sales) = 4.5 + ( − 4.5) / 10 = 4.05
Then regression line of sales (y) on scores (x) is given by
where
y − y = b yx ( x − x )
σy
1 Σxy − x y
b yx = r
=
σx
n ( Σx 2 / n ) − x 2
Correlation and Regression
261
0 × ( − 4.5)
140
10
=
=
= 0.06
2
2400
2400 − ( 0 / 10)
140 −
∴ The required regression line is y − 4.05 = 0.06( x − 60)
or
y = 0.06 x + 0.45
Then for
x = 70, y = 0.06 × 70 + 0.45 = 4.65
Hence the most probable weekly sales volume for a score of 70 is 4.65
6.7 Standard Error of Estimate
The sum of the squares of the deviation of the points from the regression line of y on x is
S = Σ ( y i − a − bxi ) 2 = Σ( Yi − bX i ) 2 ,
where
X i = xi − x, Yi = y i − y
σy


= Σ  Yi − r
Xi 
σx


2
= ΣYi 2 − 2r (σ y /σ x ) ΣX i Yi + r 2 (σ 2y / σ 2x ) ΣX i2
= n σ 2y − 2r (σ y /σ x ) r. nσ x σ y + r 2 (σ 2y / σ 2x ) . nσ 2x
= nσ 2y (1 − r 2 ) = nS 2y , say where S y = σ y (1 − r 2 ) ... (6.1)
Again S y is the root mean square deviation of the point from the regression line of y
on x, it is called the standard error of estimate of y.
Similarly the standard error of estimate of x is given by
S x = σ x (1 − r 2 )
...(6.2)
But the sum of the squares of deviations cannot be negative, so we can say that
r2 ≤ 1 ⇒ – 1 ≤ r ≤ 1
When r = 1 or −1, the sum of the squares of deviations from either line of regression
is zero. Consequently each deviation is zero and the whole of the points lie on both
the lines of regression. These two lines then coincide and we say that the correlation
between the variables is perfect. The nearer r 2 is to unity the closer are the points to
the lines of regression. Hence the departure of r 2 from unity is a measure of
departure from linearity of the relationship between the variables.
6.8 Error Prediction
The deviation of the predicted value from the observed value is called the standard
error or prediction and is given by
E yx =
Σ( y − y e ) 2
n
Where y is the actual value and y i is the predicted value.
262
Engineering Mathematics-III
6.9 Multiple Regression
Multiple Regression is a statistical technique that allows us to predict someone's
score on one variable on the basis of their scores on several other variables. For
example if we are interested in predicting how much an individual enjoys their job.
Variables such as salary, extent of academic qualifications, age, sex, number of
years in full-time employment and socioeconomic status might all contribute
towards job satisfaction. If we collected data on all of these variables, perhaps by
surveying a few hundred members of the public, we should be able to see how
many and which of these variables gave rise to the most accurate prediction of job
satisfaction. We might find that job satisfaction is most accurately predicted by
type of occupation, salary and years in full-time employment, with the other
variables not helping us to predict job satisfaction. When using multiple regression
in psychology, many researchers use the term independent variables to identify
those variables that they think will influence some other dependent variable. We
prefer to use the term "predictor variables" for those variables that may be useful in
predicting the scores on another variable that we call the "criterion variable". Thus,
in our example above, type of occupation, salary and years in full-time
employment would emerge as significant predictor variables, which allow us to
estimate the criterion variable – how satisfied someone is likely to be with their job.
As we have pointed out before, human behavior is inherently noisy and therefore it
is not possible to produce totally accurate predictions, but multiple regression
allows us to identify a set of predictor variables which together provide a useful
estimate of a participant's likely score on a criterion variable.
6.9.1 Relation between multiple regression, correlation and ANOVA
It is advised to go through this article after studying ANOVA. In a previous articles
we introduced you to correlation and the regression line. If two variables are
correlated, then knowing the score on one variable will allow you to predict the
score on the other variable. The stronger the correlation, the closer the scores will
fall to the regression line and therefore the more accurate the prediction. Multiple
regression is simply an extension of this principle, where we predict one variable on
the basis of several other variables. Having more than one predictor variable is
useful when predicting human behavior, as our actions, thoughts and emotions are
all likely to be influenced by some combination of several factors. Using multiple
regression we can test theories (or models) about precisely which set of variables is
influencing our behavior. We know on Analysis of Variance, human behavior is
rather variable and therefore difficult to predict. What we are doing in both ANOVA
and multiple regression is seeking to account for the variance in the scores we
observe. Thus, in the example above, people might vary greatly in their levels of job
satisfaction. Some of this variance will be accounted for by the variables we have
identified. For example, we might be able to say that salary accounts for a fairly
large percentage of the variance in job satisfaction, and hence it is very useful to
know someone's salary when trying to predict their job satisfaction, and hence it is
very useful to know someone's salary when trying to predict their job satisfaction.
You might now to see that the ideas here are rather similar to those underlying
ANOVA. In ANOVA we are trying to determine how much of the variance is
accounted for by our manipulation of the independent variables (relative to the
Correlation and Regression
263
percentage of the variance we cannot account for). In multiple regression we do not
directly manipulate the independent variables, but instead just measure the
naturally occurring levels of the variables and see if this helps us predict the score on
the dependent variable (or criterion variable). Thus, ANOVA is actually a rather
specific and restricted example of the general approach adopted in multiple
regression. To put another way, in ANOVA we can directly manipulate the factors
and measure the resulting change in the dependent variable. In multiple regression
we simply measure the naturally occurring scores on a number of predictor variables
and try to establish which set of the observed variables gives rise to the best
prediction of the criterion variable.
A current trend in statistics is to emphasize the similarity between multiple
regression and ANOVA, and between correlation and the t-test. All of these
statistical techniques are basically seeking to do the same thing – explain the
variance in the level of one variable on the basis of the level of one or more other
variables. These other variables might be manipulated directly in the case of
controlled experiments, or be observed in the case of surveys or observational
studies, but the underlying principle is the same. Thus, although we have given
separate discussion to each of these procedures they are fundamentally all the same
procedure.
6.9.2 Use of Multiple Regression
1. You can use this statistical technique when exploring linear relationships between
the predictor and criterion variables – that is, when the relationship follows a
straight line. To examine non-linear relationships, special techniques can be used.
2. The criterion variable that you are seeking to predict should be measured on a
continuous scale (such as interval or ratio scale). There is a separate regression
method called logistic regression that can be used for dichotomous dependent
variables (beyond the scope of this book and so not covered here).
3. The predictor variables that you select should be measured on a ratio, interval, or
ordinal scale. A nominal predictor variable is legitimate but only if it is dichotomous,
i.e. there are no more that two categories. For example, sex is acceptable (where
male is coded as 1 and female as 0) but gender identity (masculine, feminine and
androgynous) could not be coded as a single variable. Instead, you would create
three different variables each with two categories (masculine/not masculine;
feminine/not feminine and androgynous/not androgynous). The term dummy
variable is used to describe this type of dichotomous variable.
4. Multiple regression requires a large number of observations. The number of cases
(participants) must substantially exceed the number of predictor variables you are
using in your regression. The absolute minimum is that you have five times as many
participants as predictor variables.
6.10 Regression Plane
As we have already described, that multiple regression reals the study of
dependance of one variable on more than one independent variables. Here, we shall
restricted our self upto the problem of two independent variables. when a variable z
264
Engineering Mathematics-III
depends upon two variables x and y, then the linear relation between x, y and z
given as
...(6.1)
z= ax+by +c
Now consider we have a set of n points with co-codinates (x i , y i , z i ), and try to fit a
plane to this data by the method of least square. It means we have to minimize the sum's
of the squares of the vertical distances form the observations z i to the plane i.e. in fig. 6.5.
Fig. 6.5
n
S=
∑ [zi
− ( axi +by i +c )]2
...(6.2)
i =1
Differentiating partially w.r. to a, b and c we get
n
∂s
= −2∑ xi [ z i − ( axi +by i +c )]
∂a
i =1
n
∂s
= −2∑ y i [ z i − ( axi +by i +c )]
∂b
i =1
n
∂s
= −2∑ [ z i − ( axi +by i +c )]
∂c
i =1
Now for minima
∂s
∂s ∂s
=
=
= 0 and we get the following simultaneous
∂a ∂b ∂c
equation,
Σ xi z i =a ⋅ Σxi2+b Σ xi y i +c ⋅ Σ xi
...(6.3)
Σ y i z i = a ⋅ Σ xi y i +b ⋅ Σ y i2+c ⋅ Σ y i
...(6.4)
Σz i =a Σ xi +b Σ y i +nc
...(6.5)
Correlation and Regression
265
the above equations (6.2), (6.3), (6.4) are known as normal equations and can be
solved by any standard method. After getting the numerical values of a, b and c, we
obtain the required plane. For the better understanding see the following example.
Ex.13: Find a regression plane z= ax+bx+c, by using the following points
x
1
2
3
4
y
0
1
2
3
z
12
18
24
30
Sol: First we shall construct the following table for various summation values,
which are used in normal equations.
xi
yi
zi
xi2
y i2
xi y i
y i zi
z i xi
1
0
12
1
0
1
0
12
2
1
18
4
1
2
18
36
3
2
24
9
4
6
48
72
4
3
30
16
9
12
90
120
Σ x i =10
Σyi =6
Σz i = 84
Σ x 2i = 30
Σ y 2i =14
Σ x i y i = 20
Σ y i z i =156
Σz i x i
= 240
Now putting the values of these summation in normal equations (6.3), (6.4) and
(6.5), we get
30 a + 20 b +10 c = 240
20 a +14 b + 6 c = 156
10 a + 6 b + 4 c = 84
On solving the above equations, we get a = 2 , b = 4 , c = 10 . Hence the required
regression plane is z= 2 x+ 4 y +10
Problem Set 6.2
1. Compute the regression line of y on x for the following data :
x
1
2
3
4
5
y
5
4
3
2
1
266
2.
Engineering Mathematics-III
Compute the regression line of y on x for the following data :
x
1
2
3
4
5
6
y
2
2
2
2
2
2
3. From the following table, form the two regression equations :
Height of father (x) (in inches)
65
66
67
67
68
69
71
73
Height of son (y) (in inches)
67
68
64
68
72
70
69
70
4. Find the regression line of y on x for the data :
x
1
4
2
3
5
y
3
1
2
5
4
5. Find the regression line of y on x if :
x
40
70
50
60
80
50
90
40
60
60
y
2.5
6.0
4.5
5.0
4.5
2.0
5.5
3.0
4.5
3.0
6. The following results were obtained from records of age (x) and systolic blood
pressure (y) of a group of 10 men :
x
y
Mean
53
142
Var.
130
165 and Σ {( x − x ) ( y − y )} = 1220
Find the approximate regression equation and use it to estimate the blood
pressure of a man at the age of 45.
7. In a partially destroyed laboratory record of an analysis of correlation data, the
following results only are legible :
Variance of x = 9 Regression equations : 8 x − 10 y + 66 = 0, 40 x − 18 y − 214 = 0.
What were (a) the mean values of x and y, (b) the standard deviation y, and (c) the
coefficient of correlation between x and y.
8. If two regression coefficients are 0.8 and 0.2, what would be the value of
coefficient of correlation?
9. The following regression equations and variances are obtained from a
correlation table :
20 x − 9 y − 107 = 0, 4 x − 5 y + 33 = 0, variance of x = 9
Find (i) the mean values of x and y (ii) the standard deviation of y.
Correlation and Regression
267
10. Ten people of various heights as under, were requested to read the letters on a
car at 25 yards distance. The number of letters correctly read as given below :
Height (in feet)
5.1
5.3
5.6
5.7
5.8
No. of letters
11
17
19
14
8
5.9 5.10 5.11 6.0
6.1
15
12
20
6
18
Is there any correlation between heights and visual power ?
11. Two random variables have the regression lines with equations 3 x + 2 y = 26
and 6 x + y = 31.. Find the mean values and the correlation coefficients
between x and y.
12. The regression equations of two variables x and y are x = 0.7 y + 5.2, y = 0.3 x + 2 .8.
Find the means of the variables and the coefficient of correlation between them.
13. Two lines of regression are given by x + 2 y = 5 and 2 x + 3 y = 8.
Calculate (i) mean values of x and y; (ii) the coefficient of correlation; (iii) the
ratio of the regression coefficients.
14. What do you mean by regression ?
15. Explain line of regression and write down two regression equations. why
are there two line of regression ?
16. Find the lines of regression of the following points
x
152
114
138
158
144
153
141
117
136
154
y
198
300
414
594
676
549
580
483
481
659
17. Find the equation of line of regression of yield of wheat on water from the
following data
Water in cm
12
18
24
30
36
42
48
yield in tons
5 ⋅ 27
5 ⋅ 68
6 ⋅ 25
7 ⋅ 21
8 ⋅ 02
8 ⋅ 71
8 ⋅ 42
18. Write a short note on
(A) Multiple regression and its application
(B) Regression plane
(C) Difference between interpolation, curve fitting and regression
(D) Regression, correlation and ANOVA
268
Engineering Mathematics-III
Objective Type Questions
Multiple Choice Questions
1. Match list A with list B.
List – A
(a) Increase in one variable followed by decrease of other variable
(b) Simple Correlation
(c) Multiple Correlation
(d) Increase in one variable followed by increase in other variable
List – B
(i) Positive Correlation
(ii) Negative Correlation
(iii) Bi-variate Correlation
(iv) Multi Correlation
The correct match is :
(A)
(B)
(C)
(D)
(a)
i
ii
iii
iv
(b)
ii
iii
iv
ii
(c)
iii
ii
iii
iv
(d)
ii
iii
iv
i
Tick mark the correct alternative .
2. The ranks of two variates are :
x
1
2
3
4
5
y
5
4
3
2
1
The coefficient of correlation is
(a) 0
(b) +1
(c) –1
(d) None of these.
3. The coefficient of correlation between X and Y for the following data :
x
–4
–3
–2
–1
0
1
2
3
4
y
16
9
4
1
0
1
4
9
16
Correlation and Regression
(a) 0
269
(b) 0.5
(c) –1
(d) none of these.
4. A regression line has a slope of 8 and an intercept of 3. If the mean of the
independent variable is 8, then the mean of the dependent variable is :
(a) 19
(b) 11
(c) 67
(d) 20.
5. Equation of two lines of regression are :
4 x + 3 y + 7 = 0 and 3 x + 4 y + 8 = 0, then the correlation coefficient between x
and y is :
(a) 1.25
(b) – 0.25
(c) – 0.75
(d) 0.92.
6. if r is the correlation coefficient, the correct relation is
(a) r ≤ 1
(b) r ≤ 1
(c) r ≥ 1
(d) r ≥ 1.
7. If Σ xi = 30, Σy i = 42, Σ xi y i = 199, Σ xi2 = 184,
Σy i2 = 318 , and n = 6, then the regression coefficient b xy is :
(a) 0.36
(b) 0.28
(c) –0.46
(d) none of these.
8. If the lines of y on x and x on y are respectively y = kx + 4 and x = 4 y + 5, then :
(a) 0 ≤ k ≤ 4
(b) 0 ≤ k ≤ 1/ 4
(c ) k > 1/ 4
(d) −1/ 4 ≤ k ≤ 1/ 4.
9. If a linear equation ax + by + c = 0 exists between x and y and if ab > 0, then
the coefficient of correlation between x and y is :
(a) – 1
(b) 0
(c) 1
(d) none of these.
10. The two lines of regression are 8 x − 10 y = 66 and 40 x − 18 y = 214 and
variance of x is 9. The standard deviation of y series is :
(a) 8
(b) 6
(c) 4
(d) 3.
11. For 12 pair of observations of x and y, the r calculated is 0.674. The standard
error is :
(a) 0.2563
(b) 0.1575
(c) 0.4384
(d) none of these.
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Engineering Mathematics-III
12. The coefficient of correlation
(a) has no limit
(b) can be less than 1
(c) can be greater than 1
(d) lying between 1 and -1
13. Which one of the correct result related to Karl Pearson’s coefficient of correlation
(Pearsonian’s coefficient)
(a) r =
∑ XY
x
(c) r =
∑ XY
xy
(b) r =
∑ xy
Nσ x σ y
(d) r =
∑ XY
y
14. Which one of other correct result related to Karl Pearson’s coefficient of
correlation (Pearsonian’s coefficient)
(a) r =
∑ XY
x
(b) r =
(c) r =
∑ XY
xy
(d) r =
∑ xy
∑ x2 × ∑ y 2
∑ XY
y
15. Probable error is
(a) 0.06745 Standard Error
(b) 0.6745 Standard Error
(c) 0.06475Standard Error
(d) 0.06745 Standard Error
16. Rank correlation coefficient R is equal to
(a) 1 –
(c)1 –
6 ∑ D2
N3– N
6 ∑ D2
N3+ N
(b) 1 +
(d) 1 +
6 ∑ D2
N3– N
6 ∑ D2
N3+ N
17. Coefficient of determinant is equal to
(a) 1 – r 2
(b) 1 + r 2
(c) r 2
(d) 1 ± r 2
18. The limit of the population correlation is
(a) r ± S. E .
(b) r + S. E .
(c) r – S. E .
(d) r ± P. E .
19. The regression lines cuts each other at the point of
(a) average of X and Y
(b) average of X only
(c) average of Y only
(d) can not say
Correlation and Regression
271
20. When coefficient of correlation is zero, it means the regression lines cut each
other at
(a) right angle
(b) zero degree
(c) 180 degree
(d) can not say
21. We know that regression coefficient has two values of same sign and the
regression coefficient and the coefficient of correlation also has the
(a) same sign
(b) opposite sign
(c) positive sign
(d) negative sign
Fill in the Blanks
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
Spearman's rank correlation coefficient is ..................... .
If r = +1, then there is : .................... correlation.
If r is nearer to zero, then there is ................... correlation.
The line of regression of x on y is ........................ .
If the two lines of regression coincide then the correlation between x and y, i.e.
r =................. .
The coefficient of regression of y on x is ........................ .
The correlation coefficient and two regression coefficients have the ..................
sign.
The coefficient of correlation r is independent of ....................... .
Angle between two lines of regression is ..................... .
The …………………. and …………….. are related with each other.
The coefficient of correlation is the under root of …………………………
The coefficient of correlation ……………………
Regression coefficient has two values of …………
Positive correlation means one variable is ………….with other.
Negative correlation means one variable is ………….with other.
When coefficient of correlation is zero, it means the regression lines cut each
other at …………
The product of two regression coefficient is equal to ………..
If both the regression coefficients are negative, the coefficient of correlation
would be ………
The under root of regression coefficient gives the value of ……..
State True/False
1. If r is nearer to zero, then there is no correlation.
2. The Spearman's Rank Correlation Coefficient = 1 −
( 6 Σdi ) 2
n ( n 2 − 1)
.
3. If 0 .75 < r < 1, then there is a high degree of correlation and the value of the
other variable can be estimated.
272
Engineering Mathematics-III
4. The coefficient of regression of y on x is given by b yx =
rσ x
.
σy
5. Correlation coefficient between x and y is geometric mean between two
regression coefficients.
6. Arithmetic mean of regression coefficients is greater than the correlation
coefficient .
7. If one of the regression coefficients is greater than unity numerically, then the
other is less than unity numerically.
8. The coefficient of correlation between x and y for y = −5 x + 4 is –1.
9. If b xy and b yx are the regression coefficient of x on y and y on x respectively
then b xy + b yx ≤ 2r.
10. If ax + by + c = 0 is the line of regression of y on x and a1 x + b1 y + c1 = 0 is the
line of regression of x on y, then a b1 ≤ a1 b .
11. The point that lies on both the lines of regression for a bivariate distribution is
( X , Y ).
12. If two regression lines are coincident, then r = ± 1.
13. The coefficient of correlation lying between 1 and -1.
14. The under root of regression coefficient gives the value of coefficient of
correlation.
15. The product of two regression coefficient is equal to one.
16. The regression coefficient of y on x and x on y are same.
17. When coefficient of correlation is zero, it means the regression lines cut each
other at zero degree.
18. Probable error is 0.06745 Standard Error.
19. The regression lines cuts each other at the point of average of X and Y.
20. If both the regression coefficients are positive, the coefficient of correlation
would be negative.
21. The term regression was first used by Karl Pearson in the 1900.
22. The value of r may be greater than one and less than -1.
ANSWERS
Problem Set 6.1
1. (i) 3.5 (ii) 6
2. 31.5
3. −1
4. 0.8067
5. 0.9042
6. 2
7. 0.9393
8. 1
9. 0.07
10.
r( y1 , y 2 ) = 0
Problem Set 6.2
1. y = 6 − x
Correlation and Regression
273
2. y = 2
3. y = 0.42 x + 3.956, x = 0.525 y + 3 ⋅ 229
4. y = 2 ⋅ 7 + 0.1 x
5. y = 0.55 + 0.0583 x
6. y = 0.94 x + 76.88 Blood pressure=119
7. x = 13, y = 17; y = 0.8 x + 6.69 x = 0 ⋅ 45 y − 5.35; r = 0.6, σ y = 4
8. r = 0.4
9. x = 13, y = 17, σ y = 4
10. 0.7291
11. x = 9.06, y = 5.52, r = 0.46
12. r = 0.7395, x = −0.1034, y = 0.5172
13. x = 4, y = 7, r = −0.5
25. Regression of y on x is y = 3 ⋅ 575x − 34 ⋅ 67
Regression of x on y is x = 0 ⋅ 3y +12 ⋅ 693
26. yield 0 ⋅ 10 × water in cm + 4 ⋅ 08 tons
Multiple Choice Questions
1. (d)
2. (c)
3. (a)
4. (c)
5. (c)
6. (b)
7. (c)
8. (c)
9. (a)
10. (c)
11. (b)
12. (d)
13. (b)
14. (b)
15. (b)
16. (a)
17. (c)
18. (d)
19. (a)
20. (a)
21. (a)
Fill in the Blanks
1. 1 −
3.
6Σdi2
n ( n 2 − 1)
x0
2. perfect positive
4.
x−x =r
σx
(y − y)
σy
5. ± 1
6. rσ y /σ x
7. same
8. both origin and scale
−1  r1 r2 − 1

9. tan 
 r1 + r2 
10.
coefficient of correlation,
Regression
274
Engineering Mathematics-III
11. Regression coefficient
12. lying between 1 and -1
13. same sign
14. increasing
15. decreasing
16. right angle
17. one
18. negative
19. coefficient of correlation
True or False
1. True
2. False
3. True
4. False
5. True
6. True
7. True
8. True
9. False
10. True
11. True
12. True
13. True
14. True
15. True
16. True
17. False
18. False
19. True
20. False
21. False
22. False
❑❑❑
Unit-2
Chapter
7
Probability Theory
7.1 Introduction
The words ‘Probability’ and ‘Chance’ are quite familiar to everyone. Many a times, we
come across statements like, “Probably it may rain today”, “Chances of his visit to
the university are very few”, “It is possible that he may pass the examination with
good marks”. In the above statements, the probably, chance, possible etc., convey
the sense of uncertainty about the occurrence of some event. Ordinarily, it appears
that there cannot be any exact measurement for these uncertainties, but in
Mathematical Statistics, we do have methods for calculating the degree of certainty
of events in numerical value, under certain conditions. When, we perform
experiments in science and engineering, repeatedly under identical conditions, we
get almost the same result. There also exist experiments in which the outcome may
be different even if the experiment is performed under identical conditions. In such
experiments, the outcome of each experiment depends on chance.
Also a random experiment is defined as an experiment in which all the possible
outcomes are known in advance and no personal bias is exercised. Throwing of an
unbiased die is a random experiment as any of the six faces of the die may come up.
In this experiment, there exist six possibilities (1 or 2 or 3 or 4 or 5 or 6) This is a
random experiment.
Note : A die is a small cube used in gambling. Plural of die is dice. The outcome of
throwing a die is the number of dots on its upper face.
7.2 Sample Space
The sample space of a random experiment is defined as the set of all possible
outcomes of the experiment. The possible outcomes are called sample points. The
sample space is generally denoted by the letter S.
Random Experiments
Sample space
Throwing of a fair die
S = {1, 2, 3, 4, 5, 6}
Tossing of an unbiased coin
S = {H,T}
Tossing of two unbiased coins
S = {HH, HT, TH, TT}
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Engineering Mathematics-III
7.3 Event
Any subset of the sample space is defined as an event. An event is called an
elementary (or simple) event if it contains only one sample point. In the experiment
of rolling a die, the event A of getting 2 is a simple event. We write A = {2}. Also an
event is called an impossible event if it can never occur. In the above experiment,
event B = {8} of getting 8 is an impossible event. An event which is sure to occur is
called a certain event; e.g. here in rolling a die, the event of getting a number less
than 7 is a certain event. Also in the random experiment of throwing of two dice, the
number of cases favourable to get a sum 6 is 5 viz.(1, 5), (2, 4), (3, 3), (4, 2), (5, 1).
Equally Likely Outcomes
The outcomes of any random experiment are called equally likely, if all of these have
equal preferences. In the experiment of tossing an unbiased coin, the outcomes
‘Head’ and ‘Tail’ are equally likely to occur.
Exhaustive Outcomes
The outcomes of a random experiment are called exhaustive, provided these cover
all the possible outcomes of the experiment. In the experiment of rolling a die, the
outcomes 1, 2, 3, 4, 5, 6 are exhaustive.
Algebra of Events
Consider two events : “The number is even” and “the number is more than 3”
associated with the random experiment of throwing a die. The sets E and F
representing these events are E = {2, 4, 6} and F = {4, 5, 6}. We now define a new
event “E or F ” (or E ∪ F) which occurs when E or F occur, what is the subset which
represents this new event? Obviously the numbers 2, 4, 6 belong to this sub set and
so also the numbers 4, 5, 6; no other outcome i.e., 1 or 3 can belong to this subset.
Thus, the event “E or F ” is represented by the subset E ∪ F= E or F= {2, 4, 5, 6}. We
can also define a new event : “E and F ” which occurs only when E and F both occur.
When the outcome is 2, the event E occurs but F does not. Hence “E and F ” can occur
only when the outcome belongs to E and F both and hence the event “E and F ” i.e., E
∩ F = {4, 6}.
Given an event E, the event which occurs when and only
when E does not occur is called the event “not-E”. If the
event E is represented by the subset E of the sample space
S, the event not-E is represented by the subset consisting
of all these elements of S which do not belong to E. This
event is also called as complementary event of E in S or
negation of E.
S
E
E
C
Fig. 7.1
Now consider the experiment of drawing a card from a pack of 52 playing cards
which are divided into four types; heart, diamond, spade and club. Spades and clubs
cards are black in colour while the other two are red. We now define the following
four events :
1. Card drawn is a spade
2. Card drawn is a heart
3. Card drawn is a diamond
4. Card drawn is a club.
Probability Theory
277
One of these four events must occur as the card drawn is necessarily one of the four
types. Of course at any time, if any one of these events occurs the others cannot
occur. Such a collection of events form a mutually exclusive and exhaustive system of
events. Let E1 , E2 , ....... Ek be the subsets of a sample space S representing a system of
mutually exclusive and exhaustive system of events, then we have Ei ∩ Ej = φ for i ≠ j
and Ε1 ∪ E2 ∪...... ∪ Ek= S. We now give equivalent forms of probability statements
and the corresponding statements in set theory :
Probability theory
Set notation
(i)
Sample space
S
(ii)
Outcome of the random experiment
a
(iii)
event A
A⊂S
(iv)
event A has occurred
a∈A
(v)
event A has not occurred
a∉A
(vi)
event “A or B”
A∪B
(vii)
event “A and B”
A∩Β
(viii)
event not A
Ac
(ix)
event A ⇒ event B
A⊂B
(x)
events A and B are mutually exclusive
A ∩ B= φ
(xi)
events A1, A2, ...... , Am are mutually
exclusive and exhaustive.
Ai ∩ Aj = φ (i ≠ j),
m
∪ Ai
=S
i =1
Also if E and F are represented by the subsets E and F of the sample space S of a
random experiment, then the events : (i) only E occurs (ii) only F occurs (iii) none
of them occurs (iv) at least one of them occurs; are represented by the sets E ∩ Fc,
E c ∩ F, E c ∩ F c , and E ∪ F respectively. Also (E ∪ F)( E ∪ F ) c = E c ∩ F c .
7.4 Probability
Let an event A can happen in m ways, and fail in n ways, where all ways are equally
likely to occur, then the probability of the happening of A is defined as
number of favourable cases
P( A ) =
[Total number of mutually exclusive and equally likely cases]
=
m
= p, say
m+ n
while that of its failing is defined as P (not A) =
n
= q, say
m+n
278
Engineering Mathematics-III
m
n
m+n
+
=
=1
m+n m+n m+n
⇒
p+q=
⇒
P ( A ) + P (not A ) = 1
From above, it may be noted that P (A) = p is such that 0 ≤ p ≤ 1. P (not A) is called
the complementary event. Also 0 ≤ q ≤ 1.
Some More Definitions
(i) Independent Events : When the actual happening of one event does not
influence in any way the probability of the happening of the other, the two events
are called independent events.
(ii) Mutually Exclusive Events : Two events are known as mutually exclusive
when the occurrence of one of them, excludes the occurrence of the other; e.g., while
tossing a coin, we either get a head or tail but not both.
(iii) Compound Event : When two or more events occur in connection with each
other, then simultaneous occurrence is called a compound event; e.g., in throwing a
die, getting a 5 or 6 is called a compound event.
(iv) Favourable Events : The events which entail the required happening are
said to be favourable events.
(v) Conditional Probability : The probability of happening an event A, while
another event B has already happened is called the conditional probability of A such
that B has already happened. It is usually denoted by P (A/B).
Remarks :
(i) Probability of the happening of an event is also known as the probability of
success (say p) and the probability of the non-happening of the event as the
probability of failure, (say q).
(ii) If P(E) = 1, E is called a certain event and if P(E) = 0 , then E is called an
impossible event.
n!
(iii) The number of ways to choose r object from n distinct objects n C r =
r!( n − r )!
(iv) The number of arrangements of r objects in a line out of n distinct objects is
n!
n
Pr =
( n − r )!
Illustrative Examples
Ex. 1 : An urn contains 13 balls numbering 1 to 13. Find the probability that a ball
selected at ranom is a ball with number that is a multiple of 3 or 4.
Sol : Total number of balls in the urn = 13
Now, the favourable cases [i.e., number which is multiple of 3 or 4] are 3, 4, 6, 8, 9, 12
Hence, the total number of favourable cases = 6
6
Thus, the required probability =
13
Ex. 2 : A bag contains 6 red, 5 white and 4 black balls. Two balls are drawn. Find
the probability that none of them is red.
Probability Theory
279
Sol : Total number of balls = 6 + 5 + 4 = 15
Now, two balls can be drawn in 15 C2 ways i.e.,
15 × 14
= 105 ways.
2×1
But total number of non-red balls = 5+4 = 9
9× 8
Hence number of favourable cases = 9 C2 =
= 36.
2×1
36 12
Then the required probability. =
=
105 35
Ex. 3 : A die is thrown three times and the sum of the three numbers thrown is 15.
Find the probability that the number on the first throw is 4.
Sol : The possible numbers on the die in three throws having sum 15 are
(4, 5, 6), (4, 6, 5), (5, 4, 6), (5, 6, 4), (6, 4, 5), (6, 5, 4), (5, 5, 5), (6, 6, 3), (6, 3, 6), (3, 6, 6)
which are 10 in number.
Out of 10 possible outcomes, only first two are favourable outcomes and hence the
2
1
required probability =
=
10 5
Ex. 4 : Find the probability of getting a ' King' or a 'Queen' in a single draw from a
well-shuffled pack of playing cards.
Sol : Let A be the event of getting a King or Queen in the draw, then it is obvious
that the number of the favourable cases for happening of the event A= 4+4=8.
Also total no. of cases = 52
8
2
Whence P(A)=
=
52 13
Ex. 5 : What is the probability of getting an even number in the throw of an
unbiased die?
Sol : In this experiment, there are 6 equally likely possible outcomes, 1, 2, 3, 4, 5, 6.
Hence the sample space is given below :
S= {1, 2, 3, 4, 5, 6}.
3 1
= .
6 2
Ex. 6 : If there are two children in a family, find the probability that there is at least
one girl in the family.
Let A be the required event, then we have A= {2, 4, 6} ⇒ P (A)=
Sol : Let S be the sample space, then we have S= {BB, BG, GB, GG}
where B and G stand for the 'Boy' and 'Girl' respectively. If A is the required event,
3
then we can write A= {BG, GB, GG} ⇒ P (A) = .
4
Ex. 7 : Find the probability that a leap year, selected at random, will contain 53
Sundays.
Sol : There are 366 days in a leap year and we write 366= 7 × 52 +2. This means
that the leap year will contain at least 52 Sundays. The possible combinations for the
remaining two days can be made as follows :
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Engineering Mathematics-III
(i) Sunday and Monday
(iii) Thursday and Friday
(v) Monday and Tuesday
(vii) Friday and Saturday.
(ii) Tuesday and Wednesday
(iv) Saturday and Sunday
(vi) Wednesday and Thursday
If A is the event of getting 53 sundays in the leap year, then only those combinations
will be favourable to the event A which contain 'Sunday'. Here (i) and (iv) are
2
favourable to A ⇒ P(A) = .
7
Ex. 8 : Three coins are tossed; find the probability of getting at least two heads.
Sol : Let S be the sample space , then we can write
S= {HHH, HHT, HTH, THH, THT, TTH, TTT}.
If A is the required event, then A = {HHH, HHT, HTH, THH).
Hence the required probability is given by
No.of cases favourable to A 4 1
P (A) =
= =
Total No.of cases
8 2
7.5 Theorems on Probability
Theorem 1. Addition law of probability. If E and F are two mutually exclusive
events of a random experiment, the probability of occurrence of the events " E or F "
is the sum of the probabilities of the events E and F or P (E or F) = P (E) + P (F).
Proof : Let e, f be the number of elementary events respectively in the sets E and F
representing the mutually exclusive events E and F. Also, let n be the total number of
elementary events in the sample space S of which E and F are two subsets. But E and
F are mutually exclusive events so that E ∩ F = φ and hence the number of
elementary events in E ∪ F = e + f .
e+ f
e f
Then
P( E ∪ F ) = P ( E or F ) =
= + = P (E ) + P (F )
n
n n
Cor : If E 1 , E 2 ,.... E k are mutually exclusive events then
k
P ( E 1 ∪ E 2 ∪....∪ E k ) =
∑
P (Ei )
i =1
Theorem 2. If E and F are the two events associated with a random experiment,
then
P (E or F) = P (E ∪ F ) = P (E) + P(F) – P (E and F).
Proof : Let e , f, g denote the number of elementary
events in E, F and E ∩ F respectively. If n is the total
number of elementary events then E ∩ F contains e + f – g
events.
e+ f −g
e f g
P (E or F)= P ( E ∪ F ) =
∴
= + −
n
n n n
E
E∩F
F
Fig. 7.2
S
= P (E) +P (F) –P (E and F) .
When events E and F are mutually exclusive they cannot occur together and hence
E ∩ F is an impossible event implying P( E ∩ F ) = 0;
Probability Theory
281
i.e,
P(E ∪ F ) = P (E) + P (F).
Theorem 3. For every event E belonging to any random experiment
P (not-E) =1 – P (E).
Proof : The two events E and not-E are mutually exclusive, so that
P(E ∪ not-E)= P(E) + P (not-E).
But one of these events (E and not-E) must occur so that the event “E ∪ not-E ” is the
sure event with probability 1 and hence P(E) + P (not-E) = 1.
Note 1. If the event E implies the event F, then P (E) ≤ P (F).
Note 2. Two events E and F which are such that the information that one of them
has occurred does not change the probability of occurrence of the other are called
independent events and more rigorously E and F defined on sample space S are
independent, if P(E ∩ F) = P (E) × P (F)
Theorem 4. If the events E and F defined on a sample space S of a random
experiment are independent, then P(E )| (F) = P (E) and P (F )|(E) = P (F).
Proof: E and F are given to be independent events,
∴
⇒
P (E and F) = P(E) . P(F)
P ( E ∩ F ) P ( E ). P ( F )
P (E | F ) =
=
= P (E )
P (F )
P (F )
P (F ∩ E ) P (F ) . P (E )
=
= P (F )
P (E )
P (E )
and
P ( F|E ) =
7.6
Use of Permutations and Combinations in Calculation of
Probabilities
Let the random experiment consist of drawing four cards from a pack of 52 playing
cards. Each group of four cards is an elementary event of this experiment ; the
number of elementary events of the sample space of this experiment = 52 C4 .
Consider now the event associated with this experiment which is “the four cards
having the same value” i.e., all four are aces, or kings, or queens and so on. Hence
there are 13 elementary events favourable to the occurrence of this event implying
that probability of this event is given by
13
52
C4
=
13
= 0.000048
(52)!
4!(48)!
Again, we consider the event “all the four cards are of same colour” i.e., either all are
red, or all are black. Here exist 26 red cards from which four can be selected in 26 C4
ways while out of 26 black cards, the four can be selected in
26
C4 ways. Hence, the
probability of the event “all the four cards are of the same colour” is given by
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Engineering Mathematics-III
26
C4 +
52
26
C4
(26) ! × 2
C4
(4) ! . (22)!
=
= 0.11
(52)!
4! . (48)!
Now consider another random experiment of first selecting three numbers out of the
numbers 1, 2, 3, 4 and 5 and then writing down all possible arrangements of these
numbers e.g., if we consider 2, 3, and 4 are selected, we will get six arrangements as
follows : (2, 3, 4), (3, 2, 4), (2, 4, 3), (4, 2, 3), (3, 4, 2) and (4, 3, 2). Three numbers
out of the given five can be selected in 5 C3 ways and each set of three numbers gives
rise to 3!= 6 arrangements. Thus the total number of outcomes is
5! × 3!
5
C3 × 3! =
= 60
(2!) (3!)
Each of the
5
C3 sets of three numbers gives rise to only one arrangement in which
the numbers are in the natural order. For example, if the three selected numbers are
2, 3, 4. These will result in 3! = 6 arrangements out of which only one arrangement
2, 3, 4 will have the three numbers in natural order. Hence, for the event “the
arranged numbers are in natural order” the number of favourable cases is 5 C3 and
hence its probability is
5
5
C3
P3
=
(5! / 3! . 2!) 1
= = 0.33
(5! / 2!)
6
Ex. 9 : From a set of 17 cards numbered 2, 3,..... , 16, 17, one is drawn at random.
7
Show that the chance of appearing such a number which is divisible by 3 or 7 is
.
17
Sol : S= {1, 2, 3,....., 17}, n (S)= 17
Let
E : The number is divisible by 3
Then
E= {3, 6, 9, 12, 15}, n (E) = 5
Let
F : The number is divisible by 7
Then
F = {7, 14}, n (F) = 2
n(E )
n(F )
5
2
and P ( F ) =
P (E ) =
=
=
n ( S ) 17
n ( S ) 17
⇒
Also
P( E ∩ F ) = 0
[∵ E and F are mutually exclusive events]
15 2
7
+
=
17 17 17
Note : In a combination only selection is made whereas in permutation not only
selection is made but also arrangement in a definite order is considered, i.e., in a
combination, the ordering of the selected object is immaterial whereas in
permutation it is essential.
Ex. 10 : A bag contains 4 red, 5 white and 7 black balls. If 2 balls are taken out at
random, what is the probability that both of them will be black or both will be white?
∴
P (E ∪ F ) = P (E ) + P (F ) =
Sol : Let E : Both the balls are white
F : Both the balls are black
Probability Theory
283
The total number of balls in the bag = 4+5+7 = 16
(16)!
16 × 15
But 2 balls can be drawn in 16 C2 ways =
=
= 120
2! (14) !
2×1
Thus, favourable ways of getting 2 white balls =
5
C2 =
Also favourable ways of getting 2 black balls = 7 C2 =
⇒
P (E ) =
5!
5× 4
=
= 10
2!× 3! 2 × 1
7!
7× 6
=
= 21
2!× 5! 2 × 1
10
1
21
7
=
and P ( F ) =
=
120 12
120 40
Here E and F are mutually exclusive events.
∴
P ( E ∪ F ) = P( E ) + P ( F ) =
1
7 10 + 21 31
+
=
=
12 40
120
120
Ex. 11 : A die is thrown and the 6 possible outcomes are assumed to be equally
likely. If E is the event : “The number appearing is a multiple of 3”, and F the event :
“The number appearing is even”. Prove that the event E and F are independent.
Sol : E = “number appearing is a multiple of 3” i.e., E = {3, 6}
F = “number appearing is even” = {2, 4, 6}
∴
E ∩ F = {6}
Also
P (E ) =
2
3 1
; P (F ) = =
6
6 2
Then P ( E ∩ F ) =
1 1 1
= × = P ( E ) × P ( F ). Hence, the events E and F are independent.
6 3 2
Ex. 12 : An urn contains 10 black and 10 white balls. Find the probability of
drawing two balls of the same colour.
Sol : Probability of drawing two black balls =
Probability of drawing two red balls =
10
C2
20
C2
10
C2
20
C2
⇒ Probability of drawing two balls of the same colour
10
=
20
C2
C2
10
+
20
C2
C2
= 2×
(10 × 9) / (2 × 1) 9
=
(20 × 19) / (2 × 1) 19
Ex. 13 : A bag contains four white and two black balls and a second bag contains
three of each colour. A bag is selected at random, and a ball is then drawn at random
from the bag chosen. What is the probability that the ball drawn is white?
284
Engineering Mathematics-III
Sol : There exist two mutually exclusive cases :
Case (i) : when the first bag is chosen.
Case (ii) : when the second bag is chosen.
The chance of choosing the first bag is
1
and if this bag is chosen, then the
2
probability of drawing a white ball is (4/6), thus the probability of drawing a white
ball from first bag is
=
1 4 1
× =
2 6 3
Also probability of drawing a white ball from second bag is =
1 3 1
× =
2 6 4
But the events are mutually exclusive, so the required probability is equal to
1 1
7
= + =
.
3 4 12
Ex. 14 : A bag contains 6 white and 9 black balls. Four balls are drawn at a time.
Find the probability for the first draw to give four white and second draw to give
four black balls in each of the following cases :
(i) The balls are replaced before the second draw.
(ii) The balls are not replaced before the second draw.
Sol : (i) When balls are replaced before the second draw :
The two draws are independent, so we have P ( A ∩ B ) = P ( A ) × P ( B )
Now probability of drawing 4 white balls = P(A) =
and Probability of drawing 4 black balls = P(B)=
6
Then
P ( A ∩ B ) = P ( A ) × P (B ) =
15
6
C4
15
C4
9
C4
15
C4
C4
C4
×
9
C4
15
C4
6× 5× 4 × 3
9× 8× 7× 6
6
=
×
=
15 × 14 × 13 × 12 15 × 14 × 13 × 12 5915
(ii) When balls are replaced before the second draw: Here two events are not
6
independent. Now Probability of drawing 4 white balls =
15
C4
C4
=
1
. The balls
91
drawn have not been replaced, so the bag contains 2 white and 9 black balls.
Probability Theory
285
9
⇒ Probability of drawing 4 black balls P(B|A) =
6
Then P ( A ∩ B ) = P( A ) P ( B| A ) =
15
C4
C4
9
×
11
C4
C4
=
11
C4
=
C4
21
55
1 21
3
3
×
=
=
91 55 13 × 55 715
Ex. 15 : A die is thrown twice and the sum of the numbers appearing is observed to
be 6. What is the conditional probability that the number 4 has appeared at least
once?
Sol : Let event A : The number 4 appears at least once.
Event B : The sum of the numbers appearing is 6. Then, we have
A = {(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (1, 4), (2, 4), (3, 4), (5, 4), (6, 4)}
and B = {(1, 5), (5, 1), (2, 4), (4, 2), (3, 3)}
So that A ∩ B = {(2, 4), (4, 2)}.
But the total number of ways in which a die can be thrown = 6 × 6 = 36
11
5
2
∴
P( A ) =
, P (B ) = , P ( A ∩ B ) =
36
6
36
P ( A ∩ B ) (2 / 36)  2
⇒
P ( A|B ) =
=
=  .
P (B )
(5 / 36)  5
Ex. 16: Two events A and B have probabilities 0.25 and 0.50 respectively. The
probabilities that both A and B occur simultaneously is 0.14. Find the probability
that neither A nor B occurs.
Sol : Here
P(A) = 0.25, P(B)= 0.50 and P ( A ∩ B ) = 0.14
∴
P ( A or B ) = P ( A ) + P ( B ) − P ( A ∩ B ) = 0.25 + 0.50 – 0.14 = 0.61
Then P (neither A nor B) = P ( A '∩ B' ) = P ( A ∪ B )' = 1– P ( A ∪ B ) =1– 0.61= 0.39
⇒ The probability that neither A occurs nor B occurs is 0.39
Ex. 17: Two cards are drawn from a pack of 52 cards, one after another without
replacement. Find the chance that one of these is an ace and the other is queen of
opposite shade.
Sol : There arise the following two possibilities :
(i) First card drawn is an ace and the second drawn is a queen of opposite shade
4
2
(colour), so the probability of this event =
×
52 51
(ii) The first card is a queen and the second is an ace of opposite shade. The
4
2
probability of this event is =
×
52 51
Thus, the required probability =
=
4
2
4
2
2
2
×
+
×
=
+
52 51 52 51 13 × 51 13 × 51
2
2
4
+
=
663 663 663
286
Engineering Mathematics-III
Ex. 18 : A and B take turns in throwing two dice, the first to throw 10 being
awarded the prize. Show that if A has the first throw, their chance of winning are in
the ratio 12 : 11.
Sol : The combinations of getting 10 from the two dice can be taken as (6+4),
(4+6), (5+5), i.e., the number of combinations is =3.
Also total combinations from two dice = 6 × 6= 36
3
1
= p, say
=
36 12
∴ Hence the probability of getting 10 =
1
11
=
= q, say
12 12
As A is to win, he should get 10 in either the first, the third, the fifth, ..... throws.
Their respective probabilities are given as under :
Also, the probability of not getting 10 = 1– p = 1–
p, q 2 p, q 4 p,..... or as
1
,
12
 11
 
 12
2
1  11
,  
12  12
2
4
1
.....
12
4
1  11 1  11 1
+ 
+ 
+..... [ G.P. series]
12  12 12  12 12
1
12
a 
12
=
= . sum of GP =

2
23
1
−
r 
11

1−  
 12
∴ Probability of winning for A is =
Aslo, B can win in either 2nd , 4th, 6th, ... throws.
Hence, the total probability of winning for B is
3
5
11
1
11
1
11
1
=qp + q 3 p + q 5 p +..... =     +     +     +..... [G.P. series]
 12  12  12  12  12  12
=
 11
 
 12
 1
 
 12
11
1 −  
 12
∴ A's probability : to B's probability =
2
=
11
23
12 11
:
= 12:11
23 23
7.7 Baye’s Theorem
Let an event A correspond to a number of exhaustive events B1 B2 ....... Bn , where
P ( Bi ) and P ( A / Bi ) are given, then we have
P ( Bi ) P ( A / Bi )
P ( Bi / A ) =
ΣP ( Bi ) P ( A / Bi )
Proof. By the multiplication law of probability, we have
P ( A Bi ) = P ( A ) P ( Bi / A ) = P ( Bi ) P ( A / Bi )
...(7.1)
Probability Theory
287
P ( Bi / A ) =
⇒
P ( Bi ) P ( A / Bi )
P( A )
...(7.2)
But the event A corresponds to B1 , B2 ,..... Bn by the addition law of probability, we
have
P ( A ) = P ( AB1 ) + P ( AB2 ) +.....+ P ( ABn )
= ΣP ( ABi ) = ΣP ( Bi ) P ( A / Bi )
Then from (7.2), we have P ( Bi / A ) =
[ using (7.1)]
P ( Bi ) P ( A / Bi )
ΣP ( Bi ) P ( A / Bi )
This is known as the theorem of inverse probability.
Note : The probability P(Bi ), i = 1, 2,..., n are called apriori probabilities, while the
probabilities P(ABi ), i= 1, 2...n are called posteriori probabilities.
Ex. 19 : Three machines M1, M2 and M3 produce identical items. Of their respective
output 5%, 4% and 3% of items are faulty. On a certain day, M1 has produced 25% of
the total output, M2 has produced 30% and M3 the remainder. An item selected at
random is found to be faulty. What are the chances that it was produced by the
machine with the highest output?
Sol : Let the event of drawing a faulty item from any of the machines be A, and the
event that an item drawn at random was produced Mi be Bi. Then obtain P(Bi /A).
Prob.
M1
M2
M3
Remarks
P(Bi)
0.25
0.30
0.45
sum=1
P (A/Bi)
0.05
0.04
0.03
P(Bi) P(A/Bi)
0⋅125
0.012
0.0135
sum=0.38
P(Bi/A)
0.0125
0.038
0.012
0.038
0.0135
0.038
by Baye’s theorem
The highest output is from M3 . Hence the required probability = (0.0135/0.038) =
0.355.
Ex. 20 : In a bolt factory, there are four machines A, B, C, D manufacturing 20%,
15%, 25% and 40% of the total output respectively. Of their outputs 5, 4, 3 and 2%
in the same order are defective bolts. A bolt is chosen at random from the factory’s
production and is found defective. What is the probability that the bolt was
manufactured by machine A or machine D?
Ans. 0.3175, 0.254
Ex. 21 : In a bolt factory, machines P, Q and R manufacture 25%, 35% and 40% of
the total. Of their output 5, 4 and 2% are defective bolts. A bolt is drawn at random
288
Engineering Mathematics-III
from the product and is found to be defective. What are the probabilities that it was
manufactured by machines P, Q and R?
Ans. (25/69), (28/69), (16/69).
7.8 Random Variable
Let the real variable X be associated with some outcome of a random experiment.
Then the values, which X takes depend on chance, so it is called a random variable or
a stochastic variable or simply a variate. Let a random experiment E consist of tossing
a pair of dice, then the sum X of the two numbers which turn up have the values 2, 3,
4 ..., 12 depending on chance. Thus, X is the random variable. Its values are real
numbers and depend on chance. If the random variable X takes a finite set of values,
then it is called a discrete variable. On the other hand, if it assumes an infinite
number of values, it is called a continuous variable.
7.9 Discrete Probability Distribution
Let the discrete variable X be outcome of some experiment. If the probability that X
takes the values xi is pi then we can say that P(X = xi) = pi or p(xi) for i = 1, 2, ... where
(i) p(xi) ≥ 0 for all values of i, (ii) Σp( xi ) = 1
Such set of values xi with their probabilities pi form a discrete probability
distribution. Let us further consider the discrete probability distribution for X, the
sum of the numbers which turn on tossing a pair of dice is contained in the table :
X = xi
2
3
4
5
6
7
8
9
10
11
12
p(xi)
1
36
2
36
3
36
4
36
5
36
6
36
5
36
4
36
3
36
2
36
1
36
Now there exist 6 × 6 = 36 equally likely outcomes and hence, each has the
probability (1/36). We have X = 2 for one outcome, i.e., (1, 1) ; X = 3 for two
outcomes (1, 2) and (2, 1) and X = 4 for three outcomes (1, 3), (2, 2) and (3, 1) and
so on. The distribution function F(x) of the discrete variate X is defined by
x
F(x) = P ( X ≤ x ) =
∑
P ( xi ), where x is any integer.
i =1
The distribution function is also sometimes called as cumulative distribution
function.
Ex. 22 : The probability density function of a variate X is
X
0
1
2
3
4
5
6
p(X)
k
3k
5k
7k
9k
11k
13k
(i) Find P(X < 4), P(X ≥ 5), P(3 < X < 6).
(ii) What will be the minimum value of k so that P(X ≤ 2) >3.
6
Sol : (i) If X is a random variable, then we have
∑
i =0
p( xi ) = 1
Probability Theory
289
⇒
k + 3k + 5k + 7k + 9k + 11k + 13k = 1 ⇒ k = (1/49)
Now
P(X< 4) = k + 3k + 5k + 7k = 16k = (16/49)
and
P(x ≥ 5) = 11k + 13k = 24k = (24/49)
Also, we have
P(3<X ≤ 6) = 9k + 11k + 13k = 33k = (33/49)
(ii)
P(X ≤ 2) = k + 3k + 5k = 9k > 0.3 or k >1/30
⇒ minimum value of k = (1/30).
7.10 Continuous Probability Distribution
When a variable X takes every value in an interval, it gives rise to continuous
distribution of X. The distributions defined by the variable like heights, weights are
continuous distributions. The probability distribution of a continuous variable x is
defined by the function f(x) such that the probability of the variable x in the small
1
1
interval x – dx to x+ dx is f(x)dx. Symbolically it can be expressed by
2
2
1
1

P  x − dx ≤ x ≤ x + dx = f ( x ) dx.

2
2 
The function f(x) is called the probability density function and the continuous curve
y=f (x) is called the probability curve. The range of the variable may be finite or
infinite. In case the range is finite, it is convenient to consider it as infinite by
supposing the density function to be zero outside the given range. Let f(x) = φ( x ) be
density function for x in the interval (a, b), then it can be written as
 0, x < a

f ( x ) =  φ ( x ), a ≤ x ≤ b
 0, x > b

∞
The density function f(x) is always positive and
∫
f(x)dx=1
−∞
i.e, the total area under the probability curve and the x-axis is unity. This means that
the total probability of happening of an event is unity.
7.11 Distribution Function or Cumulative Distribution Function
∞
If F(x) = P (X ≤ x) = ∫ f(x)dx, then F(x) is defined as the cumulative distribution
−∞
function or simply the distribution function of the continuous variate X.
It has the following properties :
(i) F′ (x) = f(x) ≥ 0, i.e., F(x) is a non-decreasing function.
(ii) F( −∞ ) = 0
(iii) F( ∞ ) = 1
b
(iv) P(a ≤ x ≤ b) = ∫ f ( x ) dx =
a
b
∫
–∞
a
f ( x ) dx −
∫
–∞
f ( x ) dx = F ( b ) − F ( a ).
290
Engineering Mathematics-III
Ex. 23: (i) Is the function defined below a density function ?
f(x) = e − x ,
x≥0
= 0,
x < 0,
(ii) If so, determine the probability that 1
the variable having this density within
the interval (1, 2) ?
(iii) Also find the cumulative probability
function F (2).
1
F (x)
f (x)
0
1
2
(i)
X
0
1
2
(ii)
3 X
Fig. 7.3
Sol : (i) f(x) ≥ 0, for all x in (1, 2) and
0
∞
f ( x )dx =
∫
∞
0. dx +
∫
–∞
–∞
∫
e − x dx = 1
0
Hence, the function f(x) defines a density function.
(ii) Required probability = P(1 ≤ x ≤ 2)
2
∫
e − x dx = e −1 – e −2 = 0.368 − 0.135 = 0.233 = shaded area in fig. 7.3 (i)
1
(iii) Cumulative probability function
2
F(2)=
∫
0
f ( x ) dx =
–∞
∫
–∞
2
0. dx + ∫ e − x dx = 1– e –2 = 1 − 0.135 = 0.865
0
shown in fig.7.3(ii)
7.12 Expectation
The mean value (µ) of the probability distribution of a variate X commonly known as
its expectation and is denoted by E(X). Let f(x) be the probability density function of
the variate X, then we have
E(X) =∑ xi f ( x ) for discrete distribution.
i
∞
E(X) =
xf ( x )dx for continuous distribution.
∫
–∞
In general, expectation of any function φ (x) is
E [φ (x)] =∑ φ ( xi ) f ( xi ) for a discrete distribution
i
∞
and
E [φ (x)] =
∫
φ ( x ) f ( x ) dx for a continuous distribution.
–∞
Variance of the distribution is given by
Probability Theory
291
σ2 =
∑ ( xi
2
− µ ) f ( xi ) , for discrete distribution
i
σ2 =
or
∞
∫
( x − µ ) 2 f ( x ) dx, for a continuous distribution.
–∞
where σ is the standard deviation of the distribution.
Also rth moment about the mean (denoted by µ r ) is defined by
µ r = Σ( xi − µ ) r f ( xi ) for a discrete distribution
∞
or
µr =
∫
( x − µ ) r f ( x )dx for a continuous distribution.
–∞
Further, Mean deviation from the mean is given by
Σ | xi − µ | f ( xi ) for a discrete distribution
∞
and
∫
| xi − µ | f ( x ) dx for a continuous distribution.
–∞
Ex. 24 : In a lottery, p tickets are drawn at a time out of n tickets numbered from 1
to n. Find the expected values of the sum of the numbers on the tickets drawn.
Sol : Let x1 , x 2 ,....., x n be the variable representing the numbers on the first,
1
second, ...., nth ticket; then the probability of drawing a ticket out of n tickets is in
n
the each case.
1
1
1
1 1
∴
E ( xi ) = 1 . + 2. + 3. +....+ n . = ( n + 1)
n
n
n
n 2
⇒ Expected value of the sum of the numbers on the tickets drawn
= E ( x1 + x 2 +..... + x m ) = E ( x1 ) + E ( x 2 ) +....+ E ( x m )
1
= mE ( xi ) = m( n + 1)
2
7.13 Moment Generating Function
The moment generating function (m.g.f.) or the discrete probability distribution of
the variate X about the value x = a is defined as the expected value of e t ( x − a ) and is
denoted by M a (t). Thus
M a (t ) = Σpi e t ( x i − a )
⇒
M a (t ) = Σpi + tΣpi ( xi − a ) +
+
...(7.3)
r2
Σpi ( xi − a ) 2 +......
2!
rr
Σpi ( xi − a ) r +....
r!
= 1 + tµ'1 +
t2
tr
µ' 2 +....+ µ' r +....
2!
r!
...(7.4)
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Engineering Mathematics-III
where µ' r is the moment of order r about a. Thus, M a (t ) generates moments and
that is why it is called the moment generating function .
Here µ' r = coefficient of (t r / r ) in the expansion of M a (t ).
Atternatively. Differentiating (2) r times with respect to t and then putting t=0,
we get
 dr

...(7.4)
µ r′ = 
M a (t )
r
 dt
 t = 0
This gives us a most convenient formula.
Again (7.3)
⇒ M a (t ) = e − at ΣPi e tx i = e − at . M 0 (t )
Thus the m.g.f. about the point a=e − at (m.g.f. about the origin).
Also, the m.g.f of the sum of two independent variables is the product of their m.g. f. ’s.
Note : If f(x) is the density function of a continuous variate X, then the moment
generating function of continuous probability distribution about x=a is given by
∞
M a (t ) =
∫
e t ( x − a ) f ( x )dx
[Remember]
–∞
Ex. 25 : Find the moment generating function of the exponential distribution
1
f ( x ) = e − x / c , 0 ≤ x ≤ ∞, c > 0
c
Hence find its mean and S.D.
Sol : The moment generating function about the origin is
∞
M 0 (t) =
∫
0
=
∞
1
1
e tx . e –x/c dx= ∫ e(t–1 /c)x dx
c
c
1 [ e( t −1/ c ) x ]∞
0
c [ (t − 1/ c )]
0
∵|t| < 1

c 
= ( 1 − ct) −1 = 1 + ct + c 2t 2 + c 3t 3 + .....
⇒
d
µ'1 =  M 0 (t )
= ( c + 2c 2 t + 3c 3 t 2 +..... ) t = 0 = c
 dt

t= 0
and
 d2

µ' 2 = 
M 0 (t )
= 2c 2
2
 dt
 t = 0
Also
µ 2 = µ 2 '− (µ '1 ) 2 = 2c 2 − c 2 = c 2
Since
µ 2 = σ2 = c2 ⇒ σ = c
Thus, the mean is c and S.D. is also c.
Probability Theory
293
Problem Set 7.1
1. Words from the letters of the word PROBABILITY are by taking all at a time.
Find the probability that both A's are together and both I's are together.
2. Out of 40 consecutive natured numbers, two are chosen at random. What is the
probability that the sum of the numbers is odd ?
3. A four figure number is formed from the digits 0,1,2,3,4,5. What is the
probability that the number is divisible by 5.
4. If A and B are two independent events, the probability that both A and B occur
is 1/8 and probability that neither of them occurs is 3/8. What is the probability
of occurrence of A.
5. Find the standard deviation for the following discrete distribution.
x
8
12
16
20
24
p(x)
1/8
1/6
3/8
1/4
1/12
6. Four coins are tossed. What is the expectation of the number of heads ?
7. Obtain the distribution function of the total number of heads occurring in three
tosses of an unbiased coin.
8. Show that for any discrete distribution β 2 ≥ 1
 ax x ≥ 0
9. Let f(x)=  ce ,
. Determine c, if f (x) is the density of a continuous
0,
x<0

distribution. Find the corresponding distribution function.
10. The frequency distribution of a measurable characteristics varying between 0
and 2 is as under :
f (x) = x 3 ,
0≤ x ≤ 1
3
= (2 − x ) , 1 ≤ x ≤ 2
Calculate the standard deviation and also the mean deviation about the mean.
 0, x < 2
 1
11. A function is defined as follows : f ( x ) =  ( 2 x + 3), 2 ≤ x ≤ 4
 18
 0, x > 4
Show that it is a density function. Find the probability that a variate having this
density will fall in the interval 2 ≤ x ≤ 3.
294
Engineering Mathematics-III
Objective Type Questions
Multiple Choice Questions
1. There coins are tossed simultaneously. The probability of getting at least one
head is :
(a) 3/4
(b) 1/8
(c) 5/8
(d) 7/8.
2. A bag contains 5 brown and 4 white socks. A man pulls out two socks. The
probability that they are of the same colour, is :
(a) 1/6
(b) 5/108
(c) 4/9
(d) 5/18.
3. A person draws a card from a pack of playing cards, replaces it and shuffles the
pack. He continues doing this untill he shows a spade. The chance that he will
fail the first two times, is :
(a) 9/64
(b) 1/64
(c) 1/16
(d) 9/16.
3
. He tries five times, The
4
probability that he will hit the target at least three times, is :
4. The probability that a man can hit the target is
(a) 291/364
(b) 371/464
(c) 471/502
(d) 459/512.
5. A and B are two independent events. The probability that both A and B occur, is
1/6 and the probability that neither of these is 1/3. Then the probability of
occurrence of A is :
(a) 1/2
(b) 1/3
(c) 1/5
(d) 1/6.
6. A pair of dice is thrown and the numbers appearing have the sum greater than
or equal to 10. The probability of getting a sum 10 is :
(a) 5/6
(b) 1/6
(c) 1/2
(d) none of these.
7. In a collection of 6 mathematics books and of 4 physics books, the probability
that 3 particular mathematics will be together, is –
(a) 1/8
(b) 1/10
(c) 1/15
(d) none of these.
Probability Theory
295
8. A box contains 12 tickets, 4 of tickets carry a prize of 5 each, the other right a
prize of Rs. 1 each . If one ticket is drawn the, expected value of the ticket is :
(a) 4/5
(b) 4/3
(c) 7/3
(d) none of these.
9. What is the chance that a leap year should have 53 sundays ?
(a) 1/7
(b) 2/7
(c) 3/7
(d) none of these.
10. If two cards are drawn with replacement, what is the probability that cards will
be of different suits ?
(a) 9/26
(b) 18/26
(c) 3/52
(d) none of these.
11. An urn contains 6 bals of which 3 are white and 3 are black. They are drawn
successively without replacement. What is the probability that colours are
alternate ?
(a) 1/20
(b) 1/10
(c) 1/5
(d) none of these.
12. The odds that a book will be favourably reviewed by three independent critics
are 3 to 2, 4 to 3 and 2 to 3. They probability that of the three review, a majority
will favour is :
(a) 36/175
(b) 47/175
(c) 17/175
(d) none of these.
13. India plays two matches each with West Indies and Australia. In any match the
probabilities of India getting points 0, 1 and 2 are 0.45, 0.05 and 0.50
respectively. Assuming that the outcomes are independent, then the
probability of India getting at least 7 points is –
(a) 0.8750
(b) 0.0875
(c) 0.0625
(d) 0.0250.
14. A random variable X has the following probability distribution :
x
1
2
3
4
P(X =xi )
k
2k
3k
4k
The value of k is –
(a) 0.1
(b) 0.5
(c) 0.3
(d) 0.2.
296
Engineering Mathematics-III
15. The value of k for which function :
 ke −3 x x > 0
F( x) = 
 0 , elsewhere
is a p.d.f. –
(a) 3
(b) 1/2
(c) 2
(d) – 3.
16. For the continuous distribution :
dF = y 0 ( x − x 2 ) dx 0 ≤ x ≤ 1, y 0 being a constant, then A.M. is –
(a) 3/4
(b) 1/2
(c) 4/5
(d) none of these.
17. The probabilities of there mutual exclusive events A, B and C are 2/3, 1/4 and
1/6 respectively. Is the statement :
(a) true
(b) wrong
(c) could be either
(d) do not know.
Fill in the Blanks
1. If the events A and B are mutually exclusive, then P(A+B) will be equal to
..................... .
2. P(A ∪ B) = P(A ∩ B), if and only if the relation P(A) and P(B) is ................ .
3. If A and B are two events and B is a subset of A, then P (A/B) is equal to ........... .
4. A fair dice is tossed repeatedly until six shows up 3 times. The probability that
exactly 5 tosses are needed is ................... .
5. The probability that a radar will detect an object is p. The probability that it will
be detected in n cycles is ......................... .
1 1 1
6. The probabilities that three men hit a target are , , respectively. Each
6 4 3
shoots once the larget then the probability that exactly one of them hits the
target .......................... .
7. If A and B are mutually independent events, then A and B are ................... .
8. If A and B are mutually independent events such that P ( A ) ≠ 0 and P ( B) ≠ 0,
then events A and B have .................... on common sample point.
9. For application of Baye's theorem, it is necessary that the events B1 , B2 , ...........
Bn are .................... and sum of their Probabilities is equal to 1.
Probability Theory
297
State True/False
1. For two events A and B. P(A∩B) is not less than P(A)+P(B) –1.
2. If A and B are two independent events in a sample space, then P(A / B ) equals–
1–P(A / B).
3. If f ( x ) is p.d.f. of continuous random variable X then the function.
f ( x ) = P( X ≤ x ) =
x
∫−∞ f (t ) dt , − ∞ < x < ∞ is called the distribution function.
4. If x represents the number of heads in 4 tosses of a coin. Then the expected
value of x is 2.
5. The probability that the thirteenth day of a randomly chosen month is a Friday
is 1/84.
6. The mathematical expection of sum of points on n dice is 7 n/ 2.
7. A random variable x may have no moments although its m.g.f. exists.
ANSWERS
Problem Set :
1. 2/55
2. 20/39
5. 2
7. F(x) = 0, −∞ < x < 0
= (1/8), 0 ≤ x ≤ 1
3. 9/25
4. 1/2
= (1/2), 1 ≤ x ≤ 2
= (7/8), 2 ≤ x ≤ 3
= 1, 3 ≤ x ≤ ∞
9. c=a, F(x)=1– e − ax
10. σ 2 = (1 / 15), mean deviation about mean = (1/5).
11. 49.
Multiple Choice Questions
1.
(d)
2. (c)
3.
(c)
4. (d)
5. (a)
6.
(c)
7. (c)
8.
(b)
9. (b)
10. (d)
11.
(b)
12. (d)
13.
(b)
14. (a)
15. (a)
16.
(b)
17. (b)
298
Engineering Mathematics-III
Fill in the Blanks
1.
P(A)+P(B)
2.
P(A)+P(B) =2 P(A) P(B/A)
3.
1
4.
25/1296
5.
1 − (1 − p) n 6.
6.
31/72
7.
independent
8.
at least one
9.
mutually exclusive
True/False
1.
False
7.
True
2. False
3. True
4. True
5. True
6. True
❑❑❑
Unit-3
8. Binomial, Poisson and Normal Distributions
9. Sampling Theory
10. Tests of Significance and its Applications
11. Time Series and Forecasting
12. Statistical Quality Control
Unit-3
Chapter
8
Binomial, Poisson and
Normal Distributions
8.1 Binomial Distribution
Consider a random experiment having the sample space S and an event E (subset of S)
associated with it. Let P(E) = p and p ( E c ) = q , (where E c denotes the complement
of E) so that p, q > 0 and p +q =1. When the experiment results in the event E, we say
a “success”, denoted by S, has occurred. If on the other hand, the event E does not
occur then E c occurs and we say that the experiment has resulted in failure. The
probability of the success is p and that of a failure is 1– p. Now, let the experiment be
carried out twice under identical conditions so that we can regard them as
independent experiments. If the two experiments can be thought of as a single
experiment, then following four possibilities of the occurrence of E and E c exist :
(1) E occurs in the first and E occurs in the second.
(2) E occurs in the first and E c occurs in the second.
(3) E c occurs in the first and E occurs in the second.
(4) E c occurs in the first and E c occurs in the second.
In terms of success and failure these outcomes can be written as SS, SF, FS, FF.
But the two experiments are independent, so the probabilities of the above four
outcomes are
P(SS) = P(S) . P(S) = P 2 , P(SF) = P(S) . P(F) = pq
P(FS) = P(F) . P(S) = qp, P(FF) = P(F) . P(F) = q2
∴ Sum of all the probabilities =1
⇒
p 2 + 2pq + q 2 = 1 or ( p + q ) 2 = 1
....(8.1)
We now define a random variable X on S as the number of success in the above four
outcomes. Then we can write
S
P(X = 0 i.e., zero success ) = P(FF) =q 2
P(X = 1 i.e., one success) = P(SF or FS) = 2pq
E
E
P(X = 2 i.e., two success) = P(SS) =p 2
Fig. 8.1
If the experiment is carried out 3 times then for the
combined experiment there are following (2 3 = 8 possible outcomes).
In terms of success and failure, the eight possible outcomes are:
SSS, SSF, SFS, SFF, FSS, FSF, FFS, FFF
C
302
Engineering Mathematics-III
Their corresponding probabilities being
P(SSS) = p 3 , P(SSF) = p 2 q, P(SFS) = pqp = p 2 q, P(SFF) = pqq = pq 2 ,
P(FSS) = qp 2 , P(FSF) = q 2 p , P(FFS) = q 2 p, P(FFF) = q 3
Sum of these probabilities being 1
i.e.,
p 3 + 3p 2 q + 3pq 2 + q 3 = 1
or
( p + q )3 = 1
....(8.2)
Let X now stand for the random variable “number of successes”, then we have
P(X = 0) = P(FFF) = q 3
P(X =1) = P(SFF, FSF, FFS) = 3pq 2
P(X=2) =P(SSF, SFS, FSS)=3p 2 q
P(X=3) = P(SSS)=p 3 .
This gives the probability distribution of X. In the above two examples, the probability
values in the probability distribution of X are equal to the various terms of the
bionomial expansions of (q+p)2 and (q+p)3 respectively for n = 2 and n = 3. The
above results can be easily extended to the case when the experiment is carried out n
times under identical conditions. In this case, the probability values in the probability
distribution of X are equal to the various terms of the binomial expansion of
( q + p ) n = n C0 q n + n C1 q n −1 p + n C2 q n − 2 p 2 + n C3 q n − 3 p 3 +...+ n Cr q n − r p r +...+ p n .
The probability distribution of X is therefore given by
X=0
n
C 0 qn
X=1
nC
qn-1p
1
X=2
n 2 n-2 2
Cq p
......
X=r
nC
r
qn-rpr
......
X=n
pn
Because of this, we can say that the probability distribution of X is Binomial
Distribution or that X is Binomial random variable.
Applications of Binomial distribution. This distribution is mainly applied in
problems concerning :
(i) Number of defectives in a sample from production line.
(ii) Estimation of reliability of systems.
(iii) Number of rounds fired from a gun hitting a target.
(iv) Radar detection
Note : Binomial distribution is concerned with trials of a receptive nature in which
only the occurrence, success or failure, acceptance or rejection, yes or no of a
particular event is of interest.
Binomial, Poisson and Normal Distributions
303
Illustrative Examples
Ex. 1 : A die is thrown 6 times. If “getting an odd number” is a “success”, what is
the probability of :
(i) 5 successes ?
(ii) At least 5 successes?
(iii) At most 5 successes ?
Sol : Here, S = {1, 2, 3, 4, 5, 6}, n(S) = 6
Let 'A' denote “getting an odd number”.
∴
A= {1, 3, 5}, n(A)=3
In one throw
P=
n (A) 3 1
1 1
= =
⇒ q = 1− p = 1− =
n (S ) 6 2
2 2
The die is thrown 6 times, therefore n= 6
(i) P ( X = r ) = n Cr p r q n − r
5
⇒
1
1
1
P ( X = 5) = 6 C5     = 6 ×  
 2  2
 2
6
=
3
32
(ii) For at least 5 successes
P (at least 5 successes)
= P(X = 5 ) + P(X=6) =
3  1
+ 
32  2
6
=
6
1
7
+
=
64 64 64
(iii) P (at most 5 successes)
1
= 1– P (X=6) =1–  
 2
6
=
63
64
Ex. 2 : A pair of dice is thrown 4 times. If getting a doublet is considered a success,
find the probability of 2 successes.
Sol : Here,
11
12

13
S=
14
15

16
21
22
23
24
25
26
31
32
33
34
35
36
41
42
43
44
45
46
51
52
53
54
55
56
61
62

63
; n( S ) = 36
64
65

66
Let ‘doublet’ be a success denoted by ‘A’; then we have A={11, 22, 33, 44, 55, 66}
and n (A) = 6
Now, in one throwp=
n(A)
6
1
=
=
n(S) 36 6
304
⇒
Now
Engineering Mathematics-III
1 5
=
6 6
q = 1− p = 1−
P (X = r) =
n
Cr P r q n – r
2
1
5
P( X = 2) = 4 C 2    
 6  6
2
2
1
25
25
= 6 ×     =
 6  36 216
(∵ n = 4, r = 2)
Ex. 3: The probability that a man aged 60 will live to 70 is 0.65. What is the
probability that out of 10 men, now 60, at least 7 will live to 70?
Sol : The probability that a man aged 60 will live to 70 is
p = 0.65 ⇒ q = 1 − p = 1 − 0.65 = 0.35
Number of men = n = 10
Hence, the probability that at least 7 men will live to 70 = (7 or 8 or 9 or 10)
= P(X=7)+P(X=8)+ P(X=9)+ P(X=10)
= P(7)+ P(8) + P(9) + P(10)
=
10
C7 q 3 p 7 + 10 C8 q 2 p 8 + 10 C9 qp 9 + p10
=
10 × 9 × 8
10 × 9
(0.35) 3 (0.65) 7 +
(0.35) 2 (0.65) 8
1× 2 × 3
1× 2
+10(0.35)(0.65) 9 + (0.65)10
= ( 0.65)7 [120 ( 0.35) 3 + 45 ( 0.35) 2 ( 0.65)
+10 ( 0.35)( 0.65) 9 + ( 0.65)10 ]
= 0.04901×125 [0.04116+ 0.028665+0.011830+0.002197]
= 0.5137.
Ex. 4: If on an average one ship in every ten is wrecked, find the probability that
out of 5 ships expected to arrive, 4 at least will arrive safely.
Sol : Out of 10 ships one ship is wrecked. Hence, we can say that nine ships out of
ten ships are safe.
Thus
p=
9
9
1
, q = 1−
=
10
10 10
Now P (at least 4 ships out of 5 are safe) = P(X= 4) + P(X= 5)
= 5 C4 p 4 q 5 − 4 + 5 C5 p 5 q 0
4
9
1
9
= 5     +  
 10  10  10
5
=
(∵ n = 5)
7 9
 
5  10
4
Ex. 5: If 10% of bolts produced by a machine are defective, calculate the probability
that out of a sample selected at random, of 7 bolts, not more than one bolt will be
defective.
Binomial, Poisson and Normal Distributions
305
Sol : Here probability of defective bolts
= p = 10% = 0.1
⇒ Probability of not defective bolts
= q = 1–P = 0.9
Total number of bolts
=n=7
Then the probability of not more than 1 defective bolt
= P(X=0) + P(X=1)
7
= P (0) + P (1) =
C0 q 7 p 0 + 7 C1 q 6 p1
= q7 + 7 q 6 p = ( 0.9)7 + 7( 0.9) 6 ( 0.1) = ( 0.9) 6 (1.6)
Ex. 6: Ten percent of screws produced in a certain factory turn out to be defective.
Find the probability that in a sample of 10 screws chosen at random, exactly two will
be defective.
Sol :
P=
1
9
where n=10, r =2
⇒q=
10
10
Now
P ( X = r ) = n Cr p r q n − r
∴
1
9
P( 2) = 10 C 2    
 10  10
2
10− 2
2
=
10 × 9  1   9 
   
1 × 2  10  10
8
= 0.1937
Ex. 7: Five cards are drawn successively with replacement from a well-shuffled
deck of 52 cards. What is the probability that
(i) all the five cards are spades ?
(ii) only 3 cards are spades ?
(iii) none is a spade ?
Sol : S = {All the 52 cards } then n(S)=52
Now let “A'’ represent a spade then A={13 spades} and n(A)=13
In one draw, without replacement, we have
p=
n ( A ) 13 1
=
=
n ( S ) 52 4
q = 1− p = 1−
Here
Now
1 3
=
4 4
n= 5, r= 5
P ( X = r ) = n Cr p r q n − r
5
1
3
P ( X = 5) = 5C 5    
 4  4
5
3
0
1
3
C3    
 4  4
Also
P ( X = 3) =
and
1
3
P ( X = 0) = 5C 0    
 4  4
0
1
=  
 4
2
5
5
1
= 10 ×  
 4
3
=  
 4
5
3
1
× 9 ×  
 4
1
= 243 
 4
5
2
1
= 90  
 4
5
306
Engineering Mathematics-III
Ex. 8: The items produced by a firm are supposed to contain 5% defective items.
What is the probability that a sample of 8 items will contain less than 2 defective
items?
5
1
1
1
19
=
⇒ q = 1− = 1−
=
100 20
p
20 20
Sol :
P=
Here
n = 8, r = 0, 1
Now P [a sample of 8 items will contain less than 2 defective items]
= P ( X = 0) + P ( X = 1) = ( q ) 8 + 8 C 1 p1 q 7
19
=  
 20
8
+ 8×
1  19
× 
20  20
7
7
19
19 8
27  19
=    +  =
 


 20  20 20 20  20
7
Ex. 9: The probability that a bulb produced by a factory will fuse after 150 days of
use is 0.05. Find the probability that out of 5 such bulbs
(i) none
(ii) not more than one
(iii) more than one
(iv) at least one, will fuse after 150 days of use.
Sol :
p = 0.05 =
(i) P ( X = r ) =
⇒
n
1
1
19
⇒ q = 1− p = 1−
=
20
20 20
Cr p r q n − r
P( X = 0) =
5
0
1
19
C 0    
 20  20
5
19
=  
 20
5
For P (not more than one fuses), let us find P(X=0) and P(X=1)
19
P(X = 0) =  
 20
5
1
and
1
19
P(X = 1) = 5C1    
 20  20
4
=
5  19
 
20  20
4
Also, P (not more than one bulb fuses after 150 days of use)
5
4
4
5  19 
24 24  19 
 19 
 19 
= P(X=0)+ P(X = 1) =  +
=
  =  ×
 
 20
 20
20  20
20 20  20
Further P (more than one bulb fuses after 150 days of use)
= 1−[P(X = 0)+ P(X = 1)] = 1–
24  19
 
20  20
4
4
Binomial, Poisson and Normal Distributions
307
and P (at least one bulb fuses after 150 days of use)
19
=1− P(X = 0) =1−  
 20
5
Ex. 10: For special security in a certain protected area, it was decided to put three
lighting bulbs on each pole. If each bulb has a probability p of burning out in the first
100 hours of service, calculate the probability that at least one of them is still good
after 100 hours. If p= 0.3, how many bulbs would be needed on each pole to ensure
99% safety that at least one is good after 100 hours?
Sol : Probability of burning out in the first 100 hours of service is
p = 0.3 ⇒ q=1–P =1–0.3 = 0.7
Then the probability that at least one of them is still good after 100 hours
= P(X=1) +P(X=2)+ P(X=3)
=
3
C1 q 2 p1 +
3
C2 q 1 p 2 +
3
C3 q 0 p 3
= [ 3 C0 q 3 p 0 + 3C1 q 2 p1 + 3C2 q1 p 2 + 3C3 q 0 p 3 ] − 3C0 q 3 p 0
= 1 − 3 C0 q 3 p 0 = 1 − q 3 = 1– (0.7) 3 = 0.657
Second part. Probability = 99% = 0.99
Let the number of bulbs required be n.
Now P(at least one bulb is good) = 1 − q n
0.99 = 1 − ( 0.7 ) n ⇒ ( 0.7 ) n = 1 − 0.99 = 0.01
⇒
⇒
log (0.7) n = log (0.01) ⇒ n log (0.7) = log 0.01
or
n=
log 0.01
2 . 000
⇒n=
=
log 0.7
1.8451
—
—2
= 13
0.1549
Ex. 11: A bag contains 10 balls each marked with one of the digits 0 to 9. If four
balls are drawn successively with replacement from the bag, what is the probability
that none is marked with the digit 0 ?
Sol : S= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} ⇒ n(S)=10
Now let A represent the digit ‘0’ then A= {0} and n(A)=1.
In one draw,
p=
n( A )
1
1
9
=
⇒ q = 1− p = 1−
=
n ( S ) 10
10 10
n
Cr p r q n − r
Now
P (X = r) =
⇒
1
9
P( X = 0) = 4 C0    
 10  10
0
4
9
=  
 10
4
308
Engineering Mathematics-III
8.2 Recurrence Formula for the Binomial Distribution
P (r ) =
n
Cr p r q n − r and P ( r + 1) =
P ( r + 1)
=
P (r )
n
Cr +1 p r +1 q n − r –1
We have
⇒
=
P ( r + 1) =
⇒
n
n
Cr +1 p r +1 q n − r –1
Cr p r q n − r
n!
r!( n − r )! p r + 1 q n − r −1 n − r p
×
×
=
.
( r + 1) ! ( n − r − 1)!
n!
r +1 q
pr qn−r
n−r p
. p (r)
r +1 q
This is the recurrence formula for Binomial distribution.
8.3 Mean, Variance and Moment Generating Function of a
Binomial Distribution
For binomial distribution, we have P(r)= n C r p r q n − r .
n
Now Mean
µ=
∑
r P ( r ) = 0. P (0) + 1. P (1) + 2. P (2)....+ n . P ( n )
r= 0
n
= 1. C 1 pq n −1 + 2. n C 2 p 2 q n − 2 +....+ np n
= npq n –1 + 2 .

= np q n–1 + ( n

n ( n – 1) 2 n − 2
p q
+....+ np n
2!
—
1) pq n–2 +
(n
—
1)( n
2!
—
2)

p 2q n− 2 +....+ pq n–1 

= np( q + p ) n −1 = np (1) n −1
[∵ q + p = 1]
= np
Hence, mean of binomial distribution is np=m, say.
Variance. We know that variance σ 2 is given by σ 2 =
n
∑
r 2 p(r) − µ 2
r =0
n
Now
∑
r 2 P (r ) =
r= 0
n
∑ [r + r ( r − 1)] n Cr p r q n − r
r= 0
n
=
∑
r . n Cr p r q n − r +
r= 0
∑
r ( r — 1) n Cr p r q n − r
r= 0
n
= np +
n
∑
r ( r — 1)
n
Cr p r q n − r
...(8.3)
r= 0
 n

∵ ∑ rP ( r ) = np
 r= 0

Binomial, Poisson and Normal Distributions
n
Now
∑
309
n
r (r — 1 ) n C r p r q n − r = 0 + 0 +
r= 0
r(r − 1 ) n C r p r q n − r
∑
r= 2
n

∑ r ( r
=
r= 2
n
1) ×
n ( n — 1) ( n — 2)... ( n — r + 1) r n − r 
p q

r!

n ( n — 1) ( n − 1) ( n — 2)... ( n — r + 1) r n - r
p q
( r — 2)!
∑
=
—
r= 2
[∵ r! = r ( r − 1)( r − 2)!]
n n – 2 Cr
= n ( n − 1) p 2
∑
p r– 2 q n – r
r= 2
= n ( n − 1) p 2 ( q + p ) n − 2 = n ( n − 1) p 2 [∵ p + q = 1]
n
⇒
∑
r 2 P ( r ) = np + n ( n − 1) p 2
r= 0
∴
n
σ2 =
∑
r 2 P ( r ) − µ 2 = np + n ( n − 1) p 2 − n 2 p 2
r= 0
= np + n 2 p 2 − n 2 p − n 2 p 2
= np − np 2 = np (1 − p ) = npq
⇒
S. D. =
Variance =
Also
µ 2 = σ 2 = npq.
[∵ µ = np]
npq
Similarly, we can obtain the following moment generating functions :
µ 3 = npq ( q − p )
µ 4 = npq [1+ 3pq ( n − 2)]
Then moment coefficient of skewness is given by
β1 =
⇒
µ 23
µ 32
=
[ npq( q − p )] 2
r1 = β1 =
( npq ) 3
=
(q − p )2
npq
q− p
npq
Also, coefficient of Kurtosis is given by
β2 =
µ4
µ 22
=
npq [1+ 3pq ( n − 2)]
( npq )
2
= 3+
1 − 6 pq
npq
310
Engineering Mathematics-III
r2 = β 2 − 3 =
⇒
1 − 6pq
npq
Aliter. Moment Generating Function About the Origin.
M 0 (t ) = Ε ( e tr ) = Σ n C r p r q n − r e tr
We have
=Σ
n
C r ( pe t ) r q n − r = ( q + pe t ) n
Differentiating with respect to t and putting t = 0, we get the mean µ'1 = np.
Again M a (t ) = e − at M 0 (t ) = m.g.f. of the binomial distribution about its mean, is
given by
M m (t ) = e − npt ( q + Pe t ) n = ( qe − Pt + Pe qt ) n
(∵ m = np, p + q = 1)
2


t
t
t4
= 1+ pq
+ pq ( q 2 − p 2 ) + pq ( q 3 + p 3 ) +...
2!
3!
4!


1 + µ 1t + µ 2
⇒
3
n
t2
t3
t4
t2
+µ3
+ µ 4 +... = 1 + npq
2!
3!
4!
2!
+ npq ( q − p )
t3
t4
+ npq [1 + 3( n − 2pq] +...
3!
4!
Then equating the coefficients of like powers of t on either side, we have
µ 2 = npq, µ 3 = npq ( q − p ), µ 4 = npq [1 + 3( n − 2) pq].
Also.
β1 =
µ 23
µ 32
=
µ
(q − p )2
(1 − 2p ) 2
1 − 6pq
=
and β 2 = 4 = 3 +
2
npq
npq
npq
µ2
Thus, Mean = np,
Standard deviation= ( npq )
skewness=(1 − 2p ) /
( npq )
Kurtosis= β 2
Ex. 12 : The probability of a man hitting a target is
1
. If he fires 7 times, what is the
4
probability of his hitting the target at least twice?
Sol : p =
1
1 3
⇒ q = 1 − p = 1 − = . Also, n = 7
4
4 4
Then the probability of hitting the target twice
= 1 − {p (0) + p (1)} = 1 −
n
C0 p 0 q 7 −
n
C1 pq 6
Binomial, Poisson and Normal Distributions
311
7
3
= 1 −  
 4
=1−
36
7
4
1 3
− 7  
4  4
6
(3 + 7 ) = 1 −
7290
4547
=
16384 8192
Ex. 13: Eight coins are thrown simultaneously. Find the chance of obtaining at
least six heads.
1
Sol : P, the probability of getting head when a coin is thrown =
2
q=
⇒
Then P (at least 6 heads)
1
and n = 8
2
= P(6 heads)+P(7 heads)+ P(8 heads)
6
1
1
= 8 C 6    
 2  2
2
7
1
1
1
+ 8 C 7     + 8 C 8  
 2  2
 2
8
8
1
=   ( 8 C 6 + 8 C 8 + 8 C 7 )
 2
=
1 8 × 7 8 
37
+ + 1 =

 256
256  2 × 1 1
Ex. 14: State reason to justify whether the following statement is true or false.
“The mean of a binomial distribution is 6 and its standard deviation is 3”.
Sol : Mean of B.D.= np, S.D.= npq
Now np = 6 and npq = 3 i.e., npq = 9
⇒
npq 9
3
= ⇒ q = i.e., q >1
np
6
2
Hence, the given statement is wrong.
Ex. 15: Six dice are thrown 729 times. How many times do you expect at least
three dice to show a five or six ?
2 1
2
Sol : The chance of getting 5 or 6 with one die is p = = ⇒ q = .
6 3
3
Here n = 6 and N = 729.
n
Now Binomial distribution is N ( q + p ) n ⇒ N ∑ P ( r ) = f ( r )
r= 0
Whence the expected number of times at least three dice showing five or six
3
3
2
4
5
6

2
1
2
1
2 1
1 
= 729  6 C 3     + 6 C 4     + 6 C 5     + 6 C 6   
 3  3
 3  3
 3  3
 3 


312
Engineering Mathematics-III
=
729  6!
6!
6!
× 23 +
× 22 +
× 2 + 1 × 1

6  3!3!
4!2!
5!1!
(3)
= 160 + 60 + 12 + 1 = 233
Ex. 16: Four coins are tossed 160 times. The number of times r heads occur (r=0,
1, 2, 3, 4) is given below :
r
0
1
2
3
4
No. of times
8
34
69
43
6
Fit a binomial distribution to this data on the hypothesis that coins are unbiased.
Sol : The coins are unbiased so the probability p, of getting head with a coin is
1
1
⇒q=
2
2
Here
n = 4, N=160 ⇒ f(r) = 160 P(r)
1
f (0) = Nq n = 160  
 2
n−r p
f ( r + 1) =
. . f (r )
r +1 q
Now
f ( r + 1) =
⇒
4
= 160 ×
4−r
. f (r)
r +1
1
= 10
16
[ ∵ p=q and n= 4]
Putting r=0, 1, 2, 3
4−0
f (0) = 4 f (0) = 4 × 10 = 40
0+1
3
3
f (2) = f (1) = × 40 = 3 × 20 = 60
2
2
2
2
f (3) = f (2) = × 60 = 2 × 20 = 40
3
3
f (1) =
and
f (4) = (1 / 4) f (3) = (1/ 4) × 40 = 1 × 10 = 10
Then the expected frequencies are
r
0
1
2
3
4
f (r)
10
40
60
40
10
Ex. 17 : Assuming that the half of population are consumers of rice so that chance
of an individual being a consumer is 1/2 and assuming that 100 investigations each
take 10 individuals to see whether they are consumers, how many investigators
would you expect to report that three people or less were consumers?
Sol : Given
p=1/2, n=10, N=100 ⇒ q=1– p =1– (1/2) = (1/2)
Now,
p( r ) = n Cr p r q n − r = 10 Cr (1/ 2) r (1/ 2)10 − r = 10 Cr (1/ 2)10
Binomial, Poisson and Normal Distributions
where
313
f(r)=100 P(r).
Then the required number = 100 [P(3) + P(2)+P(1) + P(0) ]
= 100 [10 C 3 (1/ 2)10 + 10 C 2 (1/ 2)10 + 10 C 1 (1 / 2)10 + 10 C 0 (1/ 2)10 ]
=
=
100
210
[10 C 3 + 10 C 2 + 10 C 1 + 10 C 0 ]
100  10 × 9 × 8 10 × 9
17600
+
+ 10 + 1 =
= 17 (nearly)


1024  3 × 2 × 1
2×1
124
Ex. 18 : A die is thrown 8 times and it is required to find the probability that 3 will
show (i) Exactly 2 times (ii) At least seven times (iii) At least once.
1
Sol : The probability of throwing 3 in a single trial is given by P = and hence the
6
5
probability of not throwing 3 in a single trial = q =
6
Now, (i) P (getting 3 exactly 2 times) =
8
6
5
1
C 2 q 6 p 2 = 28    
 6  6
2
=
28 × 56
68
(ii) P (getting 3 exactly 2 times )
= P(getting 3, at 7 or 8 times )
=
8
C 7 q1 p 7 +
8
5 1
C 8 q 0 p 8 = 8    
 6  6
7
1
+  
 6
8
=
41
68
(iii) P ( getting 3 at least once) =P(getting 3, at 1 or 2 or 3 or 4 or 5 or 6 or 7 or 8 times)
= 1– P (getting 3, at 0 times)
= 1−
8
8
5
C 0 q 8 p 0 = 1–  
 6
Ex. 19: If the probability of a defective bolt is 0.1, find (a) the mean (b) the
standard deviation for the distribution of bolts in a total of 400.
Sol : n = 400, P = 0.1
Then Mean = np = 400 × 0.1= 40
∴ Standard deviation
= ( npq ) =
400 × 0.1(1 − 0.1) =
400 × 0.1 × 0.9 = 20 × 0.3 = 6
Ex. 20: Find the binomial distribution whose mean is 5 and variance is (10/3).
Sol : Mean = 5, and np = 5 and variance =
⇒
But
10
10
∴ npq =
3
3
npq 10 1
2
=
× ⇒ q = ⇒ p = 1– (2 / 3) = (1/ 3)
np
3 5
3
np = 5
314
So
Engineering Mathematics-III
n = 15
Hence, Binomial distribution is P(r) = n Cr q r p n - r =
15
Cr (1/ 3) r (2 / 3)15 − r .
Ex. 21: Find the probability that in ten tosses of a fair coin, a head appears
(a) at no time
(b) once
(c) twice
(d) three times
(e) four times
(f) five times
(g) six times
(h) seven times
(i) eight times
(j) nine times
(k) ten times .
Show the probability distribution in a diagram.
Sol : Probability of a head in a single toss = p =
1
2
1
and the probability of no head in a single toss = q =
2
Then by Binomial distribution, we have
(a) P (head occurs 0 time )
10
1
= 10 C 0  
 2
 1
 
 2
0
=
1
= 0.001
1024
(b) P (head occurs 1 time )
9
1
1
1
10
= 10 C 1     =
= 0.0102
 2  2
1024
(c) P (head occurs 2 times )
8
1
1
= 10 C 2    
 2  2
2
=
45
= 0.044
1024
=
120
= 0.1172
1024
=
210
= 0.2052
1024
=
252
= 0.2461
1024
(d) P (head occurs 3 times )
7
1
1
= 10 C 3    
 2  2
3
(e) P (head occurs 4 times )
6
1
1
= 10 C 4    
 2  2
4
(f) P (head occurs 5 times )
5
1
1
=10 C 5    
 2  2
5
0.26
0.24
0.22
0.20
0.18
0.16
0.14
0.12
0.10
0.08
0.06
0.04
0.02
0
...
. .
.
.
.
..
.
0 1 2 3 4 5 6 7 8 9 10
Number of heads
Fig. 8.2
Binomial, Poisson and Normal Distributions
315
(g) P (head occurs 6 times )
4
1
1
= 10 C 6    
 2  2
6
=
210
= 0.2052
1024
=
120
= 0.1172
1024
=
45
= 0.044
1024
=
10
= 0.102
1024
(h) P (head occurs 7 times)
3
1
1
= 10 C 7    
 2  2
7
(i) P (head occurs 8 times)
2
1
1
= 10 C 8    
 2  2
8
(j) P (head occurs 9 times)
1
9
1
1
= 10 C 9    
 2  2
(k) P (head occurs 10 times)
0
1
1
= 10 C 10    
 2  2
10
=
1
= 0.001
1024
Ex. 22: In a hurdle race, a player has to cross 10 hurdles. The probability that he
will clear each hurdle is 5/6. What is the probability that he will knock down less
than 2 hurdles?
Sol : Here P =
5
5 1
, q = 1 − = and n=10
6
6 6
The Binomial distribution is P ( r ) =
n
Cr p r q n − r
P (less than 2) = P (0 or 1) = P (0) + P (1) =
1
=  
 6
p (2 or more hurdles)
=1−
10
+ 10 ×
10
C0 p 0 q10 +
5  1
×  
6  6
9
10
1
=
10
C1 p1 q 9
[1 + 50] =
(6)
51
610
51
610
Ex. 23: An electronic component consists of three parts. Each part has probability
0.99 of performing satisfactorily. The component fails if 2 or more parts do not
perform satisfactorily. Assuming that the parts perform independently, determine
the probability that the component does not perform satisfactorily.
Sol : Let three parts be A, B, C.
The probability that the component does not perform
= P (2 or more parts fail) = P (2 or 3)
316
Engineering Mathematics-III
= P(2) + P(3) =
3
C2 p 2 q 1 +
3
C3 p 3
= 3 × (0.01) 2 (0.99) + (0.01) 3 = (0.01) 2 [2.97 + 0.01]
= 0.0001 × 2.98 = 0.000298
Problem Set 8.1
1. The incidence of occupational disease in an industry is such that the work men
have a 25% chance of suffering from it. What is that probability that out of six
workmen 4 or more will contact the disease?
2. An irregular six-faced die is thrown and the expectation that in 10 throws it will
give five even numbers is twice the expectation that it will give four even
numbers. How many times in 10,000 sets of 10 throws would you expect it to
give no even number?
3. A product is supposed to contain 5% defective items. What is the probability
that a sample of 8 items will contain less than 2 defective items?
4. A pack of 100 balloons is known to have 5% defective. If 5 balloons chosen at
random are inflated, what is the probability that none of them will be
defective? What is the probability that exactly 2 defective will be found?
5. Six dice are thrown 729 times. How many times do you expect at least three
dice to show a five or a six?
6. Consider an urn in which 4 balls have been placed by the following scheme: A
fair coin is tossed, if the coin falls head, a white ball is placed in the urn, and if
the coin falls tail, a red ball is placed in the urn. (i) What is the probability that
the urn will contain exactly 3 white balls? (ii) What is the probability that the
urn will contain exactly 3 red balls, given that the first ball placed was red?
7. An insurance salesman sells policies to 5 men, all of identical age in good health.
According to the actual trial table the probability that a man of this particular age
will be alive 30 years hence is 2/3. Find the probability that in 30 years.
(a) All 5 men
(b) At least 3 men
(c) Only 2 men will be alive
(d) At least 1 man will be alive.
8. Out of 800 families with four children each, how many families would be
expected to have :
(a) 2 boys and 2 girls
(b) At least one boy
(c) No girl
(d) At most two girls ?
9. Find the probability that in five tosses of a fair dice a 6 appears (i) twice (ii) at
least two times.
10. If succesive trials are independent and the probability of success on any trial is
p, show that the first success occurs on the tenth trial is p(1−p)n–1 , n=1, 2, 3,.....
11. If on average one ship in every tour is wrecked, find the probability that out of 5
ships expected to arrive, 4 at least will arrive safely.
Binomial, Poisson and Normal Distributions
317
12. The probability of a defective bolt is 0.2. Find the mean and S.D. for the
distribution of defective bolts in a total of 1000.
13. If on an average, 1 ship in every 10 is sunk, find the chance that out of 5 ships
expected at least 4 will arrive safely.
14. If the probability of hitting a target is 10% and 10 shots are fired
independently, what is the probability that the target will be hit at least once?
15. 10% of the bolts produced by a machine are defective. 1000 samples of 5 bolts
each are chosen at random . Find the number of samples containing :
(a) No defective bolt
(b) One defective bolt
(c) At least two defective bolts.
16. If the chance that any one of the 10 telephone lines is busy at any instant is 0.2,
what is the chance that 5 of the lines are busy? What is the probability that all
the lines are busy?
17. If on an average 1 vessel in every 10 is wrecked, find the probability that out of
5 vessels expected to arrive, at least 4 will arrive safely.
18. A sortie of 20 aeroplanes is sent on an operational flight. The chance that an
aeroplane fails to return is 5%. Find the probability that (i) one plane does not
return (ii) at the most 5 planes do not return, and (iii) what is the most
probable number of returns?
19. Out of 800 families with 5 children each, how many would you expect to have
(a) 3 boys (b) 5 girls (c) either 2 or 3 boys? Assume equal probabilities for
boys and girls.
20. A product is 0.5% defective and is packed in cartons of 100. What percentage
contains not more than 3 defective?
8.4 Poisson Distribution
The Poisson distribution is a particular limiting form of the Binomial distribution
when p(or q) is very small and the number of trials is large enough, so that np is
finite, say m. In Binomial distribution , we have
p( r ) = n Cr q n − r p r where np = m
=
=
n
Cr (1 − p ) n − r p r
n
m
Cr 1 − 

n
n−r
 m
 
 n
r
∵ p = m , m is finite




n
r
=
=
n ( n − 1)( n − 2).... ( n − r − 1)  m  m
×   1 − 
 n 
r!
n
n−r
n  n 1   n 2   n r - 1 r  m
 −   −  ....  −
 m 1 − 

n  n n  n n  n
n 
n
n
m
r! 1 − 

n
r
318
Engineering Mathematics-III
=
1
2
r − 1 r  m
11 −  1 −  .... 1 −
 m 1 − 
 n  n 

n 
n
n
r
m
r! 1− 

n
n 
r
r

1m
m
m

When n → ∞ , we have P ( r ) =
. lim 1 −  ∵ lim 1 –  = 1 ...(8.4)
r ! 1 n→ ∞ 
n   n→ ∞ 
n


–m
– n/m 

m
= lim 1 − 
= e−m



n→ ∞ 
n


r
m −m
Thus, we get,
where r = 0, 1, 2,..... ∞
P (r ) =
e
r!
 m m2

mr
Thus the limit of ( q + p ) n is e − m 1 + +
+....+
+....
1!
2!
r!


But
m
lim 1 − 

n→ ∞
n
n
This is called Poisson Distribution.
8.5 Mean of Poisson Distribution
We have, P(r) =
We have
e−m mr
,For this distribution, we construct the following table :
r!
Success r
Frequency f
rf
0
e−m m0
0!
0
1
e − m m1
1!
e−m . m
2
e−m m2
2!
e−m . m2
3
e−m m3
3!
e−m . m3
2!
...
...
...
r
e−m mr
r!
e−m . mr
( r - 1)!


m3
mr
∑ f r = 0 + e − m . m+ e − m . m 2 + e − m .
+....+ e − m
+....
2!
( r − 1)!


Binomial, Poisson and Normal Distributions
319
 m m2

m r −1
= e − m . m 1+ +
+....+
+....
2!
( r − 1)!
 1!

= me − m .[ e m ] = m


m2 m3
∑ f = e − m 1 + m +
+
+.... = e − m . e m = 1
2!
3!


and
Mean =
⇒
∑fr m
=
=m
∑f
1
8.6 Standard Deviation of Poisson Distribution
We have, P ( r ) =
e−m mr
, Let us frame the following table to obtain the standard
r!
deviation of Poisson distribution.
Success r
Frequency f
rf
r2 f
0
e−m m0
0!
0
0
1
e − m m1
1!
e−m . m
e−m . m
2
e−m m2
2!
e−m . m2
2e − m . m 2
3
e−m m3
3!
....
....
....
....
r
e−m mr
r!
e−m mr
( r - 1)!
re − m . m r
( r - 1)!
We have
Now,
∑ f = 1,
e−m .
m3
2!
3e − m
m3
2!
∑ fr = m
∑ fr 2 = 0 + e — m . m + 2e — m . m 2 + 3. e — m .
m3
re − m . m r
+.....+
+.....
2!
( r − 1)!


3m 2 4m 3
r.m r -1
= m . e – m 1+ 2m +
+
+.....+
+.....
2!
3!
( r − 1)!




m2 m3
m r –1
+
+.....+
+.....
= m . e – m 1+ m +
2!
3!
( r − 1)!


320
Engineering Mathematics-III


( r − 1)m r -1
2m 2 3m 3
+ m +
+
+.....+
+..... 
2!
3!
( r − 1)!

 


m2 m3
m r -1
= m . e – m 1+ m +
+
+.....+
+.....
2!
3!
( r − 1)!


 m m2

m r -2
+ m1 + +
+.....+
+.... 
2!
( r − 2 )!
 1!
 
= m. e − m [ e m + me m ] = m + m 2
Hence
σ
2
∑ fr 2  ∑ fr 
=
−

∑f
∑f
2
=
m + m2
− (m )2 = m
1
= Variance of Possion distribution
⇒
σ=
Similarly, we can compute
m
µ 3 = m, µ 4 = 3m 2 + m , β 1 =
1
1
1
1
, β2 = 3 + , γ1 =
,γ 2 =
etc.
m
m
m
m
8.7 Recurrence Formula for the Poisson Distribution
We have
Then
⇒
P(r) =
e−m . mr
e − m . m r +1
⇒ P(r+1) =
r!
( r + 1)!
[U.P.T.U 2001]
e − m . m r+ 1
( r + 1)!
P ( r + 1)
e − m . m r+ 1
r!
m
=
=
×
=
−m
r
−m r
P (r )
(
r
+
1)!
r
+1
e .m
e m
r!
m
P ( r + 1) =
P ( r ), r = 0, 1, 2, 3..... ∞
r +1
This is called the recurrence formula for Poisson distribution.
8.8 Mean Deviation
Show that in a Poission distribution with uniit mean, and the mean deviation about
the mean is (2/e) times the standard deviation.
Sol : We have P ( r ) =
mr −m
e , where mean = m=1
r!
and S.D. = m = 1
⇒
P (r ) =
e −1 1 1
= .
r!
e r!
Binomial, Poisson and Normal Distributions
321
Let us frame the following table :
r
P(r)
|r–1|
P(r) |r–1|
0
1
e
1
1
e
1
1
e
0
0
2
11
e 2!
1
1 1
e 2!
3
11
e 3!
2
1 2
e 3!
4
11
e 4!
3
1 3
e 4!
r
1 1
e r!
r–1
1 r −1
e r!
Now mean deviation about the mean
1
11 12 13
1r −1
= ∑ P(r )| r − 1| = + 0 +
+
+
+.....+
+.....
e
e 2! e 3! e 4!
e r!
1
1 2 3
r −1
= 1 + 0 + + + +.....+
.....

e 
2! 3! 4!
r!
1
1 1
2 1
3 1
4 1
r 1

= 1 +  –  +  −  +  −  +  –  +....+  –  +....
 r! r!
e   1! 1!  2! 2!  3! 3!  4! 4!

1
1 2 3 4
r
1 1 1
1

1 + + + + +.....+ +.....−  + + +.....+ +..... 

e  1! 2! 3! 4!
r!
r!
 1! 2! 3!


1 
1 1 1
1
= 1 + 1 + + + +.....
+.....
e 
1! 2! 3!
( r − 1)!

1 1 1 1
1


–1 + + + + +.....+ +..... + 1
1! 2! 3! 4!
r!



1
= [1 + e − e + 1]
e
2 2
2
2
= = (1) = σ = S. D.
e e
3
e
=
8.9 Applications of Poisson Distribution
This distribution is applied to problems concerning with :
(i) Arrival pattern of ‘defective vehicles in a workshop’, ‘patients in a hospital’ or
‘telephone calls’.
(ii) Demand pattern for certain spare parts.
(iii) Number of fragments from a shell hitting a target.
(iv) Spatial distribution of bomb hits.
322
Engineering Mathematics-III
Illustrative Examples
Ex. 24: Fit a Poisson distribution to the following :
x
0
1
2
3
4
f
192
100
24
3
1
Sol :
We have
x
f
fx
0
192
0
1
100
100
2
24
48
3
3
9
4
1
4
Total
∑ f = 320
∑ fx = 161
mean =
∑ fx 161
=
= 0.5 = m
∑f
320
P (r ) =
e −0.5 . (0.5) r
e−m . mr
=
r!
r!
Then
This is the required poisson distribuition.
Ex. 25: Using Poission’s distribution, find the probability that the aces of spades
will be drawn from a pack of well-shuffled cards at least once in 104 consecutive
trials. use (e −2 = 0.1353).
Sol :
e−m . mr
; r = 0, 1, 2,....
r!
1
m = np = 104 ×
=2
52
P (r ) =
∵ n = 104, p = 1 



52
⇒ Probability of drawing an ace of spades at least once = 1–P (0)
e −2 . (2) 0
= 1 − e −2
0!
=1 – 0.1353= 0.8647
Ex. 26: A manufacturer knows that the condensers he makes contain on an
average 1% of defective. He packs them in boxes of 100. What is the probability that
a box packed at random will contain 3 or more faulty condensers. (use e −1 =
0.3679).
=1 −
1
= 0.01
100
Sol : We have n =100 ,
p=
⇒
m = np = 100 × 0.01 = 1
Binomial, Poisson and Normal Distributions
P (r ) =
Also
323
e − m . (m )r
e −1
=
.
r!
r!
P(r >3) =1− P(0)+P(r) + P(2)}= 1
 e −1 e −1 e −1 
1
 = 1 − e −1 1 + 1 + 
− 
+
+

0
!
1
!
2
!
2


=1− 0 . 3679 × 2.5=0.08025...
Ex. 27: Criticize the statement : The mean of a Poisson’s distribution is 5 while its
standard deviation is 4.
Sol : For a Poisson’s distribution, we have mean = variance
⇒ 5 = (4)2; which is wrong. Thus, the given statement is wrong.
Ex. 28: If the variance of the Poisson distribution is 2 , find the probabilities for r =
1, 2, 3, 4 from the recurrence relation of the Poisson distribution . Also find P(r ≥ 4).
Sol : Variance = m = 2
[U.P.T.U. 2001]
e
−m
r
−2
r
–2 0
e .2
.m
e 2
=
P( 0) =
r!
r!
0!
We have
P(r) =
⇒
P (1) =
e −2 . 21
e −2 . 2 2
= 0.271, P (2) =
= 0.271
1!
2!
P (3) =
e −2 . 2 3
e −2 . 2 4
= 0.180; P (4) =
= 0.090
3!
4!
= 0.135
P ( r ≥ 4) = 1 − [ P (0) + P (1) + P (2) + P (3)]
= 1 − [0.135 + 0.271 + 0.271 + 0.180]
= 1−[0.857] = 0.143
Ex. 29: Assume that the probability of an individual coal miner being killed in a
1
mine accident during a year is
. Use appropriate statistical distribution to
2400
calculate the probability that in a mine employing 200 miners, there will be at least
one fatal accident in a year.
⇒
Sol : We have P =
1
200
1
,n = 200 and m = np =
=
2400
2400 12
Then P(at least one killed )= P(1) + P(2) + P(3) +....+ P(200) =1− P(0)
e−m . m0
= 1 − e −(1/12) = 1 − 0.92 = 0.08
0!
Ex. 30: An insurance company finds that 0.005% of the population dies from a
certain kind of accident each year. What is the probability that the company must
pay off no more than 3 of 10,000 insured risks against such accident in a given year?
0.005
Sol : p = 0.0005% =
= 0.00005, n =10000
100
= 1−
⇒
m = np =10000 × 0.00005 = 0.5
324
Also
and
Engineering Mathematics-III
P (r ) =
e −0.5 . (0.5) r
e−m . mr
=
, r = 0, 1, 2,....
r!
r!
P ( r > 3) = 1 − P ( r ≤ 3) = 1 − [ P (0) + P (1) + P (2) + P (3)]

(0.5) 2 (0.5) 3 

= 1 − e −0.5 1 + 0.5 +
+
2
6 

= 1– 0.6065 (1.646) = 0.0016.
Ex. 31: If the probability that an individual suffers a bad reaction from a certain
injection is 0.001, determine the probability that out of 2000 individuals (a) exactly
3 (b) more than 2 individuals will suffer a bad reaction.
Sol : p = 0.001, n = 2000, m = np = 2000 × 0.001 = 2
Also
P (r ) =
e − m . (m )r
, r = 0, 1, 2, 3,.... .
r!
Now
P (3) =
e −2 . (2) 2
4 −3 4
=
e = × 0.136 = 0.18
3!
3
3
and
P (r>2) = P (3) + P (4) + P (5) = 1 − [ P (0) + P (1) + P (2)]
 e −2 ( 2) 0 e −2 ( 2) e −2 ( 2) 2 
= 1− 
+
+

1!
2! 
 0!
= 1 − (1 + 2 + 2) e −2 = 1 − 5 × 0.136
= 1−0.680 = 0.32
Ex. 32: A car hire firm has two cars which it hires out day by day. The number of
demands for a car on each day is distributed as Poisson distribution with mean 1.5.
Calculate the proportion of days on which neither car is used and the proportion of
days on which some demand is refused. ( e −1.5 = 0.2231).
Sol : Given m = 1.5 , P ( r ) =
e − m . ( m) r
r!
Now Proportion of days on which neither car is used
= P (0) =
e −1.5 . (1.5) 0
= e −1.5 = 0.2231
0!
Hence proportion of the days on which some demand is refused
= P(r>2)=1– [P(0)+P(1)+P(2)]

(1.5) 2 
= 1 − e −1.5 1 + 1.5 +

2 

=1– 0.2231(3.625) = 0.1913
Binomial, Poisson and Normal Distributions
325
Ex. 33: A manufacturer of cotter pins knows that 5 percent of his product is
defective. If he sells pins in boxes of 100 and gurantees that not more than 10 pins
will be defective, what is the approximate probability that a box will fall to meet the
guaranteed quantity.
Sol :
p=
Now
P (r ) =
5
= 0.05, n = 100, m = np = 0.05 × 100 = 5
100
e − m . (m )r
e −5 . 5 r
=
; ( r = 0, 1, 2, 3,.... )
r!
r!
Hence, the probability that the defective pins are more than ten is
10
= 1−
∑
10
P (r ) = 1 −
∑
r= 0
r=0
e −5 . 5 r
r!
Ex. 34: In a certain factory producting cycle tyres, there is a small chance of 1 in
500 tyres to be defective. The tyres are supplied in lots of 10. Using Poisson
distribution, calculate the approximate number of lots containing no defective, one
defective and two defective tyres, respectively, in a consignment of 10,000 lots.
Sol : We have p =
1
1
, n = 10, m = np = 10 ×
= 0.02
500
500
e −0.02 (0.02) 0
= e −0.02 = 0.9802
0!
and number of lots containing no defective =10000 × 0.9802 = 9802 lots
Then probability of no defective = P (0) =
Also probability of one defective = P(1) etc.
Ex. 35: Using Poisson distribution, find the probability that the ace of spades will
drawn from a pack well shuffled cards at least once in 104 consecutive trials.
Sol : We have probability of the ace of spade =
1
1
= P say n =104 , m = np =104 ×
=2
52
52
Ex. 36: Find the probability that at most 5 defective components will be found in a
lot of 200, if experience shows that 2% of such components are defective. Also find
the probability of more than five defective components.
Sol : We have
∴
p = 2% =
2
= 0.02, n = 200
100
mean = np = 200 × 0.02 = 4 = m
Hence P(at most 5 defective components)
= P ( X = 0) + P ( X = 1) + P ( X = 2) + P ( X = 3) + P ( X = 4) + P ( X = 5)
But
P(X= r)
e−m . mr
r!
⇒
P ( X = 0)
e −4 . 4 0
e −4 . 4 1
e −4 . 4 2
; P ( X = 1)
; P ( X = 2)
0!
1!
2!
...(8.1)
326
Engineering Mathematics-III
⇒
P ( X = 3)
e −4 . 4 3
3!
P ( X = 4)
e −4 . 4 4
e −4 . 4 5
; P ( X = 5)
4!
5!
e −4 . 4 0 e −4 . 4 1 e −4 . 4 2 e −4 . 4 3 e −4 . 4 4 e −4 . 4 5
+
+
+
+
0!
1!
2!
3!
4!
5!
2
3
4
5

4
4
4
4 
= e −4 1 + 4 +
+
+
+

2! 3! 4! 5! 

32 32 128
= e −4 1 + 4 + 8 +
+
+
[∵ e −4 = 0.018]
 = 0.784

3
3
15 
P (X ≤ 5) =
⇒
Then P(more than five defective components) = P(X > 5) =1 − 0.784 = 0.216
Ex. 37: Fit a Poisson distribution to the following data and calculate theoretical
frequencies :
Deaths
Frequency
0
1
2
3
4
122
60
5
2
1
n
∑
Sol :
fi xi
We have, Mean i =1
n
∑
=
fi
0 × 122 + 1 × 60 + 2 × 15 + 3 × 2 + 4 × 1
= 0.5 = m say
122 + 60 + 15 + 2 + 1
i =1
Now the Probability of r deaths = 200 ×
e − m . (m )r
r!
(0.5) r
, ( r = 0, 1, 2, 3, 4) ( e −0.5 = 0.61)
r!
Then the various theoretical frequencies are given as under :
(0.5) 0
P(0) = 200 × 0.61 ×
= 200 × 0.6 × 1 = 121
0!
0.5
P(1) = 200 × 0.61 ×
= 200 × 0.61 × 0.5 = 61
1!
(0.5) 2
P(2) = 200 × 0.61 ×
= 100 × 0.61 × 0.25 = 15
2!
(0.5) 3 100 × 0.61 × 0.125
P(3) = 200 × 0.61 ×
=
=2
3!
3
(0.5) 4
25 × 0.61 × 0.0625
P(4) = 200 × 0.61 ×
=
=4
4!
3
⇒
P ( r ) = 200 × 0.16 ×
Binomial, Poisson and Normal Distributions
327
Ex. 38: Data were collected over a period of 10 years showing number of deaths
from horse kicks in each of the 20 army corps. The distribution of deaths was as
follows :
No. of deaths 0
1
2
3
4
Total
Frequency
65
22
3
1
200
109
Fit a Poisson distribution to the data and calculate the theoretical frequencies.
Sol : Mean of P.D. =
Further
∑ fi xi
65 × 1 + 22 × 2 + 3 × 3 + 1 × 4
=
= 0.61 = m say
∑ fi
109 + 65 + 22 + 3 + 1
f ( r ) = N. P( r ) = n
(0.61) r
e−m . mr
= 200. e −0.61 .
[∵ N = ∑ fi = 200]
r!
r!
= 200 × 0.5435.
= 108.7 ×
(0.61) r
r!
[∵ e −0.61 = 0.5435]
(0.06) r
r!
Hence the various theoretical frequencies are given as under :
f( 0 ) = 108.7 ×
( 0.61 ) 0
1
= 108.7 × = 109
0!
1
f( 1 ) = 108.7 ×
( 0.61 )1
= 66.3 = 66
1!
f( 2 ) = 108.7 ×
( 0.61 ) 2
= 20.2 = 20
2!
f( 3 ) = 108.7 ×
( 0.61 ) 3
= 4.1 = 4
3!
(0.61) 4
= 0.7 = 1
4!
Ex. 39: A manufacturer knows that the razor blades he makes contain on the
average 0.5% as defectives. He packs them in packets of 5. What is the probability
that a packet at random will contain 3 or more defective blades.
f (4) = 108.7 ×
Sol : Here
Now
∴
p = 0.5% = 0.005; n = 5; m = np = 5 × 0.005 = 0.025
P (r ) =
e −0.025 (0.025) r
e−m . mr
=
r!
r!
P (3 or more) = P (3) + P (4) + P (5) =
=
e−0.025 (0.025)3 e−0.025 (0.025)4 e−0.025 (0.025)5
+
+
3!
4!
5!
e −0.025 (0.025) 3
[20 + 5(0.025) + (0.025) 2 ]
5!
328
Engineering Mathematics-III
=
0.975 × 0.000015625 × 20.125625
120
= 0.0000025555
Ex. 40: Suppose that a book of 600 pages contains 40 printing mistakes. Assume
that these errors are randomly distributed throughout the book and the number of
errors per page has a Poisson distribution. what is the probability that 10 pages
selected at random will be free of errors?
Sol : Here
p=
P (r ) =
⇒
40
1
1
2
=
, m = np = 10 ×
=
600 15
15 3
e
2
e −2/3  
 3
.m
=
r!
r!
−m
r
r
0
2
e −2/3  
 3
⇒
P (0) =
= e −2/3 = 0.51
0!
Ex. 41: Bortkiewicz has given the following data of men killed by the kick of a horse
in certain Persian army corps in twenty years (1875-1894). Calculate the
theoretical frequencies taking the disribution to be Poisson.
Number of deaths
Frequency
0
109
1
65
2
22
3
3
4
1
Sol :
Number of deaths x
Frequency
fx
0
109
0
1
65
65
2
22
44
3
3
9
4
1
4
Total
∑ f = 200
∑ fx = 122
Mean
∑ fx 122
=
= 0.61
∑f
200
Binomial, Poisson and Normal Distributions
329
Now Poisson distribution is given by P(r) =
⇒
e −0.61 ( 0.61) r
r!
P( 0) =
e −0.61 ( 0.61) 0 −0.61
e
= 0.54335
0!
P(1) =
e −0.61 ( 0.61)1
0.54335 × 0.61 = 0.3314435
1!
P (2) =
e −0.61 (0.61) 2
0.54335 × 0.3721
=
= 0.1010902675
2!
2
P (3) =
e −0.61 (0.61) 3
0.54335 × 0.226981
=
= 0.02049081520
3!
6
P (4) =
e −0.61 (0.61) 4
0.54335 × 0.138458
=
= 0.003124849318
4!
24
Hence, the theoretical frequency is as follows :
Number of deaths
Theoretical Frequency
0
0.54335 × 200 = 108.67
1
0.3314435 × 200 = 66.29
2
0.1010902675 × 200 = 20.22
3
0.02049081520 × 200 = 4.10
4
0.003124849318 × 200=0.63
Ex. 42: An insurance company found that only 0.01% of the population is involved
in a certain type of accident each year. If its 1000 policy holders were randomly
selected from the population, what is the probability that not more than two of its
⋅
clients are involved in such an accident next year ? (given that e −01
= 0.9048)
Sol : Here p = 0.01% =
1
1
1
×
=
, n = 1000,
100 100 10000
m = np = (1000) ×
.
1
1
e – 01
. ( 0.1) r
×
= 0.1 ⇒ P ( r ) =
10000 10
r!
Thus, P(not more than 2)= P(0, 1 and 2) = P(0)+P(1)+P(2)
.
.
.
e −01
( 0.1) 0 e −01
( 0.1)1 e −01
( 0.1) 2
+
+
0!
1!
2!
0.01


= e −0.1 1 + 0.1 +
 = 0.9048 × 1.105 = 0.9998

2 
=
Ex. 43: The random variable x has Poisson distribution with parameter λ . Find the
expected value E(x) of the random variable x.
330
Engineering Mathematics-III
Sol : We have E(x) =∑ r. P ( r ) = ∑
=∑
r. e − λ . λ r
r!
∞
e −λ λr
λ r –1
= λe −λ ∑
= λe −λ . e λ = λ
( r – 1)!
(
r
–
1)!
r =1
Problem Set 8.2
1. In sampling a large number of parts manufactured by machine, the mean
number of defectives in a sample of 20 is 2. Out of 1000 such samples, how
many would be expected to contain at least 3 defective parts?
2. Find the expectation and variance of variable X if the probability P(X=k)=
e −λ λk
, k= 0, 1, 2, .... = 0 otherwise (λ is a positive real number).
k!
3. Ten percent of the tools produced in a certain manufacturing process turn out
to be defective. Find the probability that in a sample of 10 tools chosen at
random, (i) exactly two will be defective and (ii) more than one will be
defective, by using Poisson approximation.
4. In a certain factory producing cycle tyres there is a small chance of 1 in 500
tyres to be defective. The tyres are supplied in lots of 20. Using Poisson
distribution, calculate the approximate number of lots containing no defective,
one defective and two defective tyres respectively in a consignment of 20,000
tyres. Given e −0.4 = 0.9608.
5. Bortkiwiez has given the following data of men killed by the kick of a horse in
certain Persian army corps in twenty years (1875-1884)
No. of deaths
Frequency
0
1
2
3
4
109
65
22
3
1
Calculate the theoretical frequencies taking the distribution to be Poisson.
6. Distribution of typing mistakes committed by a typist is given below. Assuming
a Poisson model, find out the expected frequencies :
Mistakes per page
No. of Pages
0
1
2
3
4
5
142
156
69
27
5
1
7. In sampling a large number of parts manufactured by a machine, the mean
number of defectives in a sample of 20 is 2. Out of 1000 such samples how
many would be expected to contain at least 3 defective parts?
8. Find the probability that at most 5 defective fuses will be found in a box of 200
fuses if experience shows that 2% of such fuses are defective.
Binomial, Poisson and Normal Distributions
331
9. Wireless sets are manufactured with 25 soldered joints each. On the average, 1
joints in 500 defective. How many sets can be expected to be free from
defective joints in a consignment of 1000 sets?
10. In a certain factory turning out razor blades, there is a small chance of 0.002 for
any blade to be defective. The blades are supplied in packets of 10, use Poisson
distribution to calculate the approximate number of packets containing no
defective and two defective blades respectively in a consignment of 10000
packets.
11. A certain screw making machine produces on average of 2 defective screws out
of 100, and packs them in boxes of 500. Find the probability that a box contains
15 defective screws.
12. A car-hire firm has two cars which it hires out day-by-day. The number of
demands for a car on each day is distributed as a Poisson distribution with
mean 1.5. Calculate the proportion of days (i) on which there is no demand,
(ii) on which demand is refused. (e–1.5 = 0.2231).
8.10 Normal Distribution
8.10.1 Continuous Distribution
So far, we have been dealing with those distributions where the variable takes only
the integral values and hence such distributions are called discrete distributions,
whereas on the other hand when the variables (e.g. current, temperature, height,
weight etc.) take all possible values in a given interval. Variables of this type are
known as continuous variables. The probability distribution of a continuous random
variable cannot be presented in the tabular form. The probability distribution of the
continuous random variable x is represented by f(x), called the density function, x is
defined over a continuous sample space, the graph of f(x) will be continuous. For
example, the given figures represent the graphs of Continuous Probability
Distributions.
f (x)
f (x)
f (x)
f (x)
P[x1 <x<x2 ]
0
x
(i)
0
x
(ii)
0
x
(iii)
x1
0
x2
x
(iv)
Fig. 8.3
A probability density function is constructed so that the area under its curve bounded
by the x-axis is 1, when computed over the range of x. If f(x) be represented as in the
figure 8.3(iv), then the probability that x assumes a value between x1 and x 2 is the
shaded area under the density function between the ordinates x = x1 and x = x2
332
Engineering Mathematics-III
x2
i.e.,Probability that x assumes a value between x1 and x2 = P(x1<x<x2) =
∫
f ( x ). dx
x1
∞
Total area =
f ( x )dx = 1,( −∞ ≤ x ≤ ∞ )
∫
–∞
Usually in an experiment, the density function f(x) is unknown and its equation is
assumed.
8.10.2 Normal Distribution
The continuous random variable x is said to have a normal distribution, if its
probability density function is defined by
2
f ( x ) = c . e – (1/2) ( x – a / b ) , – ∞ ≤ x ≤ – ∞
...(8.5)
where c is a constant. It involves two parameters a and b, the constant c is
determined by the condition that total area under curve is unity.
∞
–(1/ 2)( x – a / b )2
∫ ce
dx = 1
–∞
Putting
x−a
= z, we get dx = b d z
b
⇒
c
∞
(–1/2)z 2
∫e
bdz = 1
–∞
∞
or
2
2c ∫ be – (1/2)z dz = 1
( ∵ function is even )
0
2
Again, let us put
or
and
z
= ω , so that z dz = dω
2
2cb
2
∞
But
dω
=
z
dz =
∫
∞
∫
dω
( 2ω )
∞
, hence 2cb
∫
e −ω .
0
ω –(1/ 2) e −ω dω = 1
∞
∫
e − x x n −1 dx = Γ( n )
0
ω −(1/ 2)e −ω dω = Γ(1 / 2) =
( 2ω )
=1
...(8.6)
0
0
∴
dω
π
Binomial, Poisson and Normal Distributions
Hence, from (8.7), 2cb π = 1 or c =
f ( x) =
⇒
1
333
1
b ( 2π )
2
e −(1/ 2)( x − a / b ) ; − ∞ ≤ x ≤ ∞
b ( 2π )
...(8.7)
The mean µ of the normal variable is given by
µ=
∞
1
2
x. e −(1/ 2)[( x − a )/ b] dx
∫
b (2π)
–∞
x−a
= z , dx = b dz , we get
b
Putting
∞
1
µ=
(2π)
∞
∫
(2π)
(2π)
[ − e −( z
2
∞
b
2
e −(1/ 2)z dz +
(2π)
–∞
b
= a+
2
( a + bz )e −(1/ 2)z dz
–∞
a
=
Hence, mean
∫
∫
ze −(1/ 2)z
–∞

= a + 0 ∵

/ 2)] ∞
−∞
2
∞
∫
e −α
2
x2
dx =
–∞
π 

2α 

µ=a
2
The variance σ is given by
σ2 =
∞
1
∫
b 2π
2
( x − µ ) 2 . e −(1/ 2){( x −µ) / b} dx
–∞
x −µ
z=
⇒ dx = b dz
b
Put
σ2 =
∴
σ2 =
∞
1
∫
b ( 2 π)
–∞
∞
1
( 2π )
2
b 2 z 2 . e −(1/ 2)z . b dz
∫
2
b 2 z 2 . e −(1/ 2)z . dz =
2b 2
–∞
Again put
2z dz
z
dW
=W ⇒
= dW ⇒ dz =
=
2
2
z
∴
σ2 =
∞
But
∫
∞
∫
(2π )
2 We −W .
0
dW
(2W)
=
2b 2
(π )
W –1/2 e – w dW = Γ (3/ 2) = (1/ 2)Γ (1/ 2) = (1/ 2) ( π )
0
2b 2
∴
σ2 =
⇒
σ 2 = b2
(π )
.
1
(π ) = b 2
2
dW
(2W)
∞
∫
0
2
z 2 . e −(1/ 2)z dz
0
2
2b 2
∞
∫
( 2π )
W 1/2 e − w . dW
334
Engineering Mathematics-III
The distribution of normal variate x with mean µ and variance σ 2 is given by
dP =
1
σ ( 2π )
2
e −(1/ 2)[( x −µ )/ σ ] dx, − ∞ ≤ x ≤ ∞
When X has the above density function, we write X as N(µ, σ 2 ) and the curve
y = f ( x) =
1
σ ( 2π )
2
e −(1/ 2)[( x −µ )/ σ ] is called normal probability curve.
x −µ
is called standard normal variate, whose mean and
σ
variance are 0 and 1 respectively and we write z as N(0, 1). The distribution of z is
called standard normal distribution and is written as :
2
1
...(8.8)
dP =
e −(1/ 2)z dz, − ∞ ≤ z ≤ ∞
( 2π )
The variate z =
8.10.3 Main Characteristics of Normal Distribution
(i) The mean, median and mode for a normal distribution are identical.
(ii) The curve is smooth, regular, bell-shaped and symmetrical about the line x = µ
2
1
because the expansion of y = f(x) =
. e −(1/ 2)( x −µ / σ ) , does not contain
( 2πσ )
the odd powers of [( x − µ /σ)].
(iii) The ordinate of the curve decreases rapidly as | x| increases. The maximum
1
ordinate at x = µ is given by y max =
and hence the curve is unimodal.
( 2π )σ
(iv) The curve extends from −∞ to + ∞
(v) As σ becomes larger, the ordinate y decreases
i.e., the curve spreads out more but flattened at
the top. On the other hand when σ becomes
smaller, y increases and the curve becomes more
peaked.
(vi) Total area under the curve above x-axis from
−∞ to + ∞ is always unity.
=0.5
=1
=2
Normal Curve
Fig. 8.4
(vii)Area of normal curve between µ − σ and µ + σ
is 68.27 % .
(viii)Area between µ − 2σ and µ + 2σ is 95.45%.
z=0
(ix) Area between µ − 3σ and µ + 3σ is 99.73%.
(x) The area bounded by the curve with x-asis and
z
0
any two ordinates equals to the probability for the
Fig. 8.5
interval marked on x-axis by the two ordinates.
Applications of Normal Distribution
This distribution is widely used in problems concerning with :
(i) Calculation of errors mode by chance in experimental measurements.
(ii) Computation of hit probability of a shot.
Binomial, Poisson and Normal Distributions
335
(iii) Statistical inference in almost every branch of science.
8.10.4 Computation of Area Under the Normal Curve
x −µ
, we get the standard normal curve and the
σ
total area under this curve is unity. The line z = 0 divides the whole area in two
equal parts. Left side area to z = 0 is 0.5 whereas the right side area to z = 0 is 0.5.
The area in between the ordinates z = 0 and z = z1 say can be computed from the
By using the transformation z =
standard table.
Illustrative Examples
Ex. 44: Fit a normal curve to the following data :
Length of
8.60
line (in cm.)
8.59
8.58
8.57
8.56
8.55
8.54
8.53
8.52
Frequency
3
4
9
10
8
4
1
1
2
Sol : Let assumed mean = 8.56
xi − 8 . 56
0.01
fi d ' i
d ' i2
fi d ' i2
4
8
16
32
3
3
9
9
27
8.58
4
2
8
4
16
8.57
9
1
9
1
9
8.56
10
0
0
0
0
8.55
8
–1
–8
1
8
8.54
4
–2
–8
4
16
8.53
1
–3
–3
9
9
8.52
1
–4
–4
16
16
Total
Σfi = 42
xi
fi
8.60
2
8.59
di′ =
∑ fi d ' i = 1
11
× 0.01 = 8.56262
42
2
N
Let the normal curve be y =
e −(1/ 2)( x −µ / σ)
σ 2π
Now, mean, µ = 8.56 +
Here N= 42, µ = 8.563 cms.
∑ fi di2 = 133
336
Also ,
Engineering Mathematics-III
 ∑ f d ' 2    ∑ f d '  2
i i
σ = i ×  i i   − 

∑
f
∑
fi 

i

 
 133  11 2 
= 0.01 × 
 −    = 0.0176
 42   42 
Hence the equation of the normal curve fitted to the given data is
2
2
N
P( x ) =
e −[( x −µ ) / 2σ ] ; − ∞ ≤ x ≤ ∞
σ 2π
where µ and σ are defined above and N= 42
Ex. 45: The incomes of a group of 10000 persons were found to be normally
distributed with mean = Rs. 750 per month and S.D. = Rs. 50 , show that of this group
about 95% had income exceeding Rs. 668 and only 5% had income exceeding Rs. 832.
Sol : µ = 750, σ = 50
(i) P( x > 668) = 0.5 + P( 668 < x < 750)
x −µ
Since, z =
= 0.5 + P{–1 ⋅ 64 < z < 0},
σ
= 0 ⋅ 5 + P{ 0 < z < 1 ⋅ 64} = 0 ⋅ 5 + 0 ⋅ 4495 = 0 ⋅ 9495
Hence, the percentage of persons having income exceeding Rs. 668 = 0.9495 × 100
= 94.95%.
(ii) P(x >832) = 0.5 − P(750 < x < 832)
= 0.5 − P(0 <z <1.64)= 0.5 − 0.4495=0.0505
Hence, the percentage of persons having income exceeding Rs. 832 = 0.0505 × 100 = 5%
Ex. 46: Assume the mean height of solders to be 68.22 inches with a variance of
10.8 inches square. How many soldiers in a regiment of 10000 would you expect to
be over 6 feet tall, given that the area under the standard normal curve between
x = 0 and x = 35 is 0.1363 and between x = 0 and x =1.15 is 0.3746.
Sol : Given µ = 68.22, σ = (10.8) = 3.28
x − µ 72 − 68.22
⇒
z=
=
= 1.15
σ
3.28
Hence, the area between x = 0 and x = 1.15 is 0.3746, implying that the area
beyond z = 1.15 is 0.5 − 0.3746 =1.1254
Thus, the expected number of soldiers above 6 ft.=0.1254 ×10000 =1254
Ex. 47: On a final examination in mathematics, the mean was 72 and the standard
deviation was 15. Determine the standard scores of students receiving grades.
(a) 60 (b) 93 (c) 72.
x − µ 60 − 72
Sol : (a) z =
=
= −0 ⋅ 8
σ
15
93 − 72
(b) z =
= +1 ⋅ 4
15
72 − 72
(c) z =
=0
15
Binomial, Poisson and Normal Distributions
337
Ex. 48: Find the area under the normal curve in each of the following cases :
(a) z = 0 and z = 1.2
(b) z = − 0.68 and z = 0
(c) z = − 0.46 and z = 2.21
(d) z = 0.81 and z = 1.94
Sol : (a) Required area is the shaded area = 0.3849 (see fig. 8.6)
(b) Area = Area between z = 0 and z = 0.68 = 0.2518 (see fig. 8.7)
(c) Required area
z=0 z=1.2
z = 0 . 68 z = 0
Fig. 8.6
Fig. 8.7
= Area between z = 0 and z = 2.21 + Area between z = − 0.46 to z = 0
= (Area between z = 0 and z = 2.21) + (Area between z = 0 and z = 0.46)
= 0.4865 + 0.1772 = 0.6637 (see Fig. 8.8 (a))
(d) Required Area = (Area between z = 0 and z = 1.94) − (Area between z = 0 and z = 0.81)
= 0.4738 − 0.2910 = 0.1828 (see Fig. 8.8(b))
– 0 . 46 0
(a)
2 . 21
0 0 . 81 1 . 94
Fig. 8.8
Ex. 49: Students of a class were given an aptitude test.
Their marks were found to be normally distributed with
mean 60 and standard deviation 5. What percentage of
students scored more than 60 marks?
Sol : x = 60, x = µ = 60, σ = 5
⇒
z=
0
Fig. 8.9
x − µ 60 − 60
=
= 0
σ
5
When , x>60, z>0. Hence area lying to the right of z = 0 is 0.5
Then the percentage of students getting more than 60 marks = 50%.
z
338
Engineering Mathematics-III
Ex. 50: Find the value of z in each of the cases
(a) Area between 0 and z is 0.3770
(b) Area to the left of z is 0.8621
Sol : (a) z = ± 1.16
(b) The area is greater than 0.5
Area between 0 and z = 0.8621− 0.5 = 0.3621
Here z = 1.09
Ex. 51: In a sample of 1000 cases, the mean of a certain test is 14 and standard
deviation is 2.5. Assuming the distribution to be normal, find
[U.P.T.U. 2001]
(i) How many students score between 12 and 15?
(ii) How many score above 18?
(iii) How many score below 8?
(iv) How many score 16?
Sol : n = 1000, µ = 14, σ = 2.5
(i) Here z1 =
x1 – µ 12 – 14
=
= – 0.8
σ
2⋅ 5
and
15 – 14
1
=
= 0⋅ 4
2⋅ 5
2⋅ 5
z 2=
z = 0.8
0 z = 0.4
Fig. 8.11
Thus, the area lying between z = − 0.8 to z = 0.4
=[Area from (z = 0) to (z = 0.8] + [Area (z = 0) to (z = 0.4)]
= 0.2881 + 0.1554 = 0.4435
Hence, the required number of students
= 1000 × 0.4435 = 443.5 = 444
(ii) z1 =
18 – 14
4
=
= 1.6
2.5
2.5
Then area right to z = 1.6 = 0.5 − area between (z = 0) to
(z =1.6) = 0.5 - 0.4452 = 0.0548
z = 1.6
Fig. 8.12 (a)
Hence the required number of students
= 1000 × 0.548 = 54.8 = 55
(iii) z =
8 – 14
6
=
= –2 ⋅ 4
2⋅ 5
2⋅ 5
Now Area left to – 2.4 = 0.5 – area between 0 and 2.4
= 0.5 – 0.4918 = 0.0082
Hence the required number of students
= 100 × 0.0082 = 8.2 = 8
(iv) Between x1 =15.5 and x2 =16.5
z=2. 4
0
Fig. 8.12 (b)
Binomial, Poisson and Normal Distributions
339
15.5 – 14
16.5 – 14
= 0.6 and z2=
=1
2.5
2.5
Then area between 0.6 and 1 = 0.3413 – 0.2257 = 0.1156
Hence, the required number of students = 0.1156 × 1000 = 115.6 = 116
Ex. 52: Five thousand candidates appeared in a certain examination carrying a
maximum of 100 marks. It was found that the marks were normally distributed with
mean 39.5 and with standard deviation 12.5. Determine approximately the number
of candidates who secured a first class for which a minimum of 60 marks is
necessary. You may see the table given below (x denotes the deviation from the
mean). The proportion A of the whole area of the normal curve lying to the left of the
ordinate at the deviation x/σ is :
z1 =
x
=z
σ
1.5
1.6
1.7
1.8
A
0.93319
0.94520
0.95543
0.96407
Sol : Mean = 39.5 = µ and standard deviation = σ = 12.5
60 – 39. 5 20 . 5 41
⇒
z=
=
=
= 1 . 64
12 . 5
12 . 5 25
Area for
x
= z = 1.6 is 0.94520
σ
Also Area for z =
x
= z = 1.7 is 0.95543
σ
0
z = 1. 64
Fig. 8.13
Now difference for 0.1 = 0.01023 then difference for 0.04 = 0.004092
Thus, the area for z =
x
=1.64 is 0.94520 + 0.004092 = 0.949292
σ
Hence, the area right to ordinate 1.64 = 1 – 0.949292 = 0.050708
Then the number of candidates securing marks 60 or more = 5000 × 0.050708 =
253.54 or the candidates securing first division = 254
Ex. 53: The mean inside diameter of a sample of 200 washers produced by a machine is
0.502 cm and the standard deviation is 0.005 cm. The purpose for which these washers
are intended allows a maximum tolerance in the diameter of 0.496 to 0.508 cm,
otherwise the washers are considered defective. Determine the percentage of defective
washers produced by the machine, assuming the diameter are normally distributed.
Sol :
x – µ 0.496 – 0.502
=
= –1.2
σ
0.005
x – µ 0.508 – 0.502
=
=
= +1.2
σ
0.005
z1 =
z2
Hence, the area for non-defective washers
= Area between z = –1.2 and z = + 1.2
= 2 (Area between z = 0 and z = 1.2)
= 2 × (0.3849) = 0.7698 = 76.98%
z = 1.2
0 z = 1.2
Fig. 8.14
340
Engineering Mathematics-III
Then the percentage of defective washers
= 100 − 76.98 = 23.02%
Ex. 54: In a normal distribution, 31% of the items are under 45 and 8% are over 64.
Find the mean and standard deviation of the distribution.
Sol : Let µ be the mean and σ the S.D., then
for x1= 45, z1=
45 – µ
64 – µ
, x2 = 64, z2 =
σ
σ
45 – µ
Now area between 0 and
= 0.50 – 0.31 = 0.19
σ
19%
42%
64 – µ
= 0.5 – 0.08 = 0.42
σ
For area 0.42, z = 1.405, implying
64 – µ
=1.405
σ
1.405
Fig. 8.15
But by the table, for the area 0.19, z = 0.496 implying
45 – µ
= 0.496
σ
Now, area between z = 0 and z =
8%
31%
– 0.496 0
...(8.1)
Then solving (8.1) and (8.2), we get µ = 50, σ =10
...(8.2)
Ex. 55: Assuming that the diameters of 1000 brass plugs taken consecutively form a
machine form a normal distribution with mean 0.7515 cm and standard deviation
0.0020 cm how many of the plugs are likely to be rejected if the approved diameter
is 0.752 ± 0.004 cm?
Sol : Limits of the diameter of non-defective plugs =
0.752 – 0.004 = 0.748 cm and 0.752 + 0.004 = 0.756 cm
x –µ
Now
z=
σ
Let
x1 = 0.748 ⇒ z1 =
0.748 – 0.7515
= –1.75
0.002
and
x2 = 0.756 ⇒ z2 =
0.756 – 0.7515
= 2 . 25
0.002
Then area between z1= −1.75 to z2 = 2.25
–1.75
= (Area from z = 0 to z1= − 1.75)
...(8.1)
2.15
0
Fig. 8.16
+ (Area from z = 0 to z2 = 2.25)
= 0.4599 + 0.4878 = 0.9477
Hence the number of plugs likely to be rejected=1000
(1 − 0.9477)=1000 × 0.0523 = 52.3 i.e., 52 plugs are likely to
be rejected.
–1
0
+1
Fig. 8.17
Binomial, Poisson and Normal Distributions
341
Ex. 56: A random variable x has a standard normal distribution φ. Prove that
P(|x| <k) = 2φ (k)−1.
Sol : We have
∞
φ=
1
∫
( 2π )
–∞
k
φ ( k) =
⇒
e –( x
0
...(8.2)
f ( x ) dx
∫
0
0
f ( x ) dx =
∫
...(8.1)
k
f ( x ) dx
–k
−k
1
∫
( 2π )
−k
e–
( x 2 /2)
dx
x = – t i.e., dx = – dt, we get
0
∫
0
f ( x ) dx =
–k
1
∫
–k
0
1
= –∫
( 2π )
k
k
⇒
/ 2)
0
∫
0
Putting
2
k
0
f ( x ) dx =
∫
I1 =
e –( x
∫
( 2π )
k
–k
Now
1
f ( x ) dx + ∫ f ( x ) dx = 0.5 + ∫ f ( x ) dx
k
P(| x | < k) =
dx
f ( x ) dx , where f ( x ) =
–∞
But
/ 2)
∫
–∞
0
=
2
P( | x | < k) =
∫
0
( 2π )
e– (t
2
e–
/ 2)
( x 2 /2)
k
dt =
∫
0
dx
1
e– (x
( 2π )
k
2
/ 2)
k
dx =
∫
f ( x )dx = I 2
0
k
f ( x )dx + ∫ f ( x )dx = 2∫ f ( x )dx = 2 [φ( k) − 0.5]
0
0
= 2φ (k) – 2 × 0.5 = 2φ(κ) −1
Problem Set 8.3
1. If x is normally distributed with mean 1 and variance 4 (i) find Pr (–3 ≤ x ≤ 3);
(ii) Obtain k if Pr (x ≤ k) = 0.90
2. The scores made by candidate in a certain test are normally distributed with
mean 500 and standard deviation 100. What percentage of candidates receive
scores (i) Less than 400 (ii) between 400 and 600 ?
x−x
(For a standard normal distribution z =
, the area under the curve
σ
between z = 0 and z = 1 is 0.34134).
3. If x is normally distributed with mean 4 and variance 9, find
(i) Pr (2.5 ≤ x ≤ 5.5)
(ii) Obtain k if Pr (x ≤ k) = 0.09. Use Pr (x ≤ 0.5) = 0.691 and Pr (x ≤ 1.3) = 0.90.
342
Engineering Mathematics-III
4. A certain number of articles manufactured in one batch were classified into
three categories according to a particular characteristic being less than 50,
between 50 and 60, or greater than 60. If the characteristic is known to be
normally distributed, determine the mean and S.D. for this batch if 60%, 35%
and 5% were found in these categories.
Given that if P =
1
z
∫
2π
e −( t
2
/ 2)
dt the values of Z corresponding to P = 0.1
0
and P = 0.45 and 0.25 and 1.65 respectively.
5. A large number of measurements is normally distributed with a mean of
65.5 ′′ and S.D. of 6.2 ′′ . Find the percentage of measurements that fall between
54.8 ′′ and 68.8 ′′ .
1

6. If log e x is normally distributed with mean 1 and variance 4, find P  < r < 2
2

given that loge 2= 0.693.
7. Show that the mean deviation from the mean of the normal distribution is
4
about of its standard deviation.
5
8. The average daily sale of 500 branch offices was Rs. 150 thousand and the
standard deviation is Rs. 15 thousand. Assuming the distribution to be normal,
indicate how many branches have sales between
(i) Rs. 120000 and Rs. 145000
(ii) Rs. 140000 and Rs. 165000
9. A random variable x has a standard normal distribution φ. Prove : Pr (| x | > k)
= 2[1– f (k)].
10. The life time (in hours) of a television tube of a certain type has the exponential
distribution f(x) = a exp {– a (x – q)}, x > q with q = 60 and a = 0.013. What
is the probability
(i) that a tube will work for less than 100 hours ?
(ii) that a tube will work for more than 500 hours ?
(iii) that tube will work for between 700 and 1000 hours ?
11. The pdf of X is given by f(X) = λe – λx , x ≥ 0, λ > 0. Calculate Pr (X>E(X)]. If
X~N (75, 25), find Pr [X > 80/X > 77]. If X~N (10, 4) find Pr[| X|≥ 5].
12. The life time of radio tubes manufactured in a factory is known to have an
average value of 10 years. Find the probability that the life time of a tube taken
randomly
(i) exceeds 15 years
(ii) is less than 5 years, assuming that the exponential probability law is followed.
Binomial, Poisson and Normal Distributions
343
Objective Type Questions
Multiple Choice Questions
Tick mark the correct alternative:
1. The mean of a binomial distribution is 5, then its variance has to be :
(a) > 5
(b) = 5
(c) <5
(d) =25.
2. If 3 is the mean and 3/2 is the S.D. of a binomial distribution, then distribution
is :
1

4
12
1 4
(c)  + 
 5 5
12
3
(a)  +
2
1
(b)  +
3
2

3
12
(d) none of these.
3. The mean and variance of a binomial distribution are 5/4 and 15/16
respectively, then p is :
(a) 1/ 2
(b) 15/16
(c) 1/ 4
(d) 3/ 4.
4. In a Poisson's distribution P( x ) for x = 0 is 10%. The mean of distribution is :
(a) 2.32026
(b) 2.308
(c) 1.875
(d) none of these.
5. If x has a Poisson distribution with parameter 3, then P( x ≥ 3) is :
17
(a) 13/e 3
(b) 1 –
2e 3
7
(c) 7 /e 3
(d) 1 –
.
e3
6. A Poisson distribution has a double mode at x = 1 and x = 2 . What is the
probability that x will have one or the other of these two values ?
(a) 2e 2
(b) 2e –2
(c) 4e –2
(d) 2e –2 + 2e 2 .
7. If X is a Poisson variate such that :
P( 2) = 9P( 4) + 90 P( 6), then mean of X is :
(a) ±1
(b) ±2
(c) ±3
(d) none of these.
8. An office switch board receives telephone calls at the rate of 2 calls per minute
on an average. The probability of receiving no call in a minute is :
(a) e 2
(b) 1/ 2
344
Engineering Mathematics-III
(c) e –2
(d) none of these.
9. If X and Y are independent Poisson variates such that P( X = 1) = P( X = 2) and
P( Y = 2)= P( Y = 3). The variance of X – 2Y is :
(a) 9
(b) 12
(c) 11
(d) 14.
10. If X has a Poisson distribution with P( X = 1) = 0.2 and P( X = 2) = 0.4 then
P( X = 0) is :
1
(a) e –2
(b) 1 – e –2
2
3
(c) e –4
(d) e –4 .
4
11. If X is normally distributed and the mean value of X is 12 and S.D. is 4. Then
P( X ≥ 20) is :
(a) 0.0228
(b) 0.4772
(c) 0.5228
(d) none of these.
12. The marks obtained by a number of students for a certain subject are assumed
to be normally distributed with mean value 65 and S.D. 5. The probability that
a randomly selected student gets marks over 70 is :
(a) 0.0635
(b) 0.3413
(c) 0.6587
(d) 0.1587.
13. Records show that the probability of a car breaking down while driving in a
tunnel is 0.0004 . The probability that out of 2000 cars that drive the tunnel at
least one will break is :
(a) e – 4/ 5
(b) 1 – e 4/ 5
(c) 1 – e – 4/ 5
(d) 1 + e –4/ 5 .
14. If X follows a binomial distribution
4P( X = 4) = P( X = 2) then P is :
(a) 1/ 2
(b) 1/ 4
(c) 1/ 6
(d) 1/ 3.
with
parametor
n = 6 and
Fill in the Blanks
1. In binomial distribution standard deviation is ............. .
2. In binomial distribution mean is ........... than variance.
3. In Poisson distribution variance is .............. .
4. Normal distribution is a limiting form of ....................under condition n → ∞,
p → 0.
5. In normal distribution mean and median ....................... and equal to ............. .
6. The mean deviation for normal curve measured from mean value is ............. .
7. The third moment of a Poisson distribution is ................ .
Binomial, Poisson and Normal Distributions
345
State True/False
1. The mean of a Poisson distribution is 10, then β1 and β 2 are 0.1 and 3.1
respectively.
2. For a Poisson distribution µ 2 = 4.2 , then µ 3 is 4.2.
3. In a normal distribution the curve is bell shaped and symmetrical about the line
x = µ.
4. Normal distribution is discrete distribution.
ANSWERS
Problem Set 8.1
1. 0.0376
2. 1
7
5
2
 19
 27 
3.   .  
 20
 20
 19
 1   19
5
4. (i)   , (ii) C2    
 20
 20  20
5. 233
 1
 1
6. (i)   , (ii)  
8
8
7. (a) (32/243) (b) (192/243)
(c) (40/243) (d) (242/243)
8. (i) 300
9. (i) 10 (53 / 65 ) (ii) 1 − 2 ( 5/ 6) 5
(ii) 250
(iii) 50
(iv) 600
11. (i) 1/8
(ii) 1/8
12. 200, 12.6
13.
14. 1 − (0.9)10 = 0.65 nearly
15. (a) 590 (b) 228
16.
10
C5 (0.2) 5 (0.8) 5 , (0.2)10
18. (i)
(ii)
20
C1 (1/ 20) (19/ 20)19
∑
r =0
(iii) 90
20. 99.83
=
20
Cr (1/ 20)r (19 / 20)20− r
45,927/90,000
(c) 81
17. [45927/50000]
19. (a) 250 (b) 25 (c) 500
3
346
Engineering Mathematics-III
Problem Set 8.2
1. 324
2. λ, λ
3. 0.18395, 0.2642
4. 960, 38.4, 0.76864
5. 108.67, 66.29, 20.22, 4.10, 0.63
6. 0.503, 193.6, 97.8, 24.5, 4.1, 0.5
7. 1 – (5101/e100)
8. 0.7845
9. 9512
10. 2 approx.
11. (10)15e–10 /(15) !
12. (i) 0.2231
(ii) 0.1913
Multiple Choice Questions
1. (c)
2. (a)
3. (c)
4. (a)
5. (b)
6. (c)
7. (a)
8. (c)
9. (d)
10. (d)
11. (a)
12. (d)
13. (c)
14. (d)
Fill in the Blanks
1.
npq
5. coincide, µ
2. greater
3. mean
2
6. 3 σ.
7. m
2. True
3. True
4. binomial
distribution
State True/False
1. True
4. False
❑❑❑
Sampling Theory
Unit-3
347
Chapter
9
Sampling Theory
Introduction
In our daily life, 'we come across persons making an assessment of the population
through samples. Thus trader thrusts a conical trowel in a bag of wheat and assesses
the quality of wheat in the bag by having a look upon the samples so drawn. A
house-wife tests a small quantity of rice to see if it has been well cooked and so on.
The importance of the theory of sampling lies in the fact that for a large population,
it is neither practical nor necessary to collect data for each and every member of the
population. Thus in order to have an information about the economic conditions of
the rural population of U.P., it would require a huge establishment to collect data
from each and every individual in the villages and then to tabulate and calculate the
parameters from it. It would be quite sufficient if a sample of village is selected (with
due precautions) and information are collected from it. The information so gathered
may be taken to represent the whole rural population of the state. Most of the
industrial concerns have some sort of quality control over their manufactured goods
before sending them to the market. If each and every item has to be tested, perhaps
in some cases no goods can be sent to the market e.g., in case of the match boxes, a
test would means the burning of all the matches. Thus the purpose of sampling is to
get information about the population from the sample. The population measures
e.g., mean, S.D. etc. are called parameters while the measure obtained from are
samples are called statistics.
9.1 Sampling and Related Terms
In statistics, a statistical population is a set of entities concerning which
statistical inferences are to be drawn, often based on a random sample taken from
the population. For example, if we are interested in generalizations about birds,
then we would describe the set of birds that is of interest. Notice that if we choose a
population like all birds, we will be limited to observing birds that exist now or will
exist in the future. Probably, geography will also constitute a limitation in that our
resources for studying birds are also limited. In 1786 Pierre Simon Laplace
estimated the population of France by using a sample, along with ratio estimator. He
also computed probabilistic estimates of the error. These were not expressed as
modern confidence intervals but as the sample size that would be needed to achieve
a particular upper bound on the sampling error with probability 1000/1001. His
estimates used Bayes’ theorem with a uniform prior probability and it assumed his
348
Engineering Mathematics-III
sample was random. The theory of small-sample statistics developed by William
Sealy Gossett put the subject on a more rigorous basis in the 20th century. However,
the importance of random sampling was not universally appreciated and in USA in
1936 Literary Digest prediction of a Republican win in the presidential election went
badly awry, due to severe bias. More than two million people responded to the study
with their names obtained through magazine subscription lists and telephone
directories. It was not appreciated that these lists were heavily biased towards
Republicans and the resulting sample, though very large, was deeply flawed. The
sampling process comprises following several stages:
♠
Defining the population of concern.
♠
Specifying a sampling frame, a set of items or events possible to measure.
♠
Specifying a sampling method for selecting items or events from the frame.
♠
Determining the sample size.
♠
Implementing the sampling plan.
♠
Sampling and data collecting.
♠
Reviewing the sampling process.
9.1.1 Population
Population is also used to refer to a set of potential measurements or values,
including not only cases actually observed but those that are potentially observable.
Suppose, for example, we are interested in the set of all adult birds now alive in the
state of Utter Pradesh in India want to know the mean weight of these birds. For
each bird in the population of birds there is a weight, and the set of these weights is
called the population of weights.
9.1.2 Subpopulation
A subset of a population is called a subpopulation. If different subpopulations
have different properties, they can often be better understood if they are first
separated into distinct subpopulations. For instance, a particular medicine may have
different effects on different subpopulations, and its effects may be obscured or
dismissed if the subpopulation is not identified and examined in isolation. Similarly,
one can often estimate parameters more accurately if one separates out
subpopulations: distribution of heights among people is better modeled by
considering men and women as separate subpopulations.
9.1.3 Sampling frame
In the most straightforward case, such as the sentencing of a batch of material from
production (acceptance sampling by lots), it is possible to identify and measure
every single item in the population and to include any one of them in our sample.
However, in the more general case this is not possible. There is no way to identify all
rats in the set of all rats. Where voting is not compulsory, there is no way to identify
which people will actually vote at a forthcoming election (in advance of the
election). These imprecise populations are not amenable to sampling in any of the
ways below and to which we could apply statistical theory. As a remedy, we seek a
sampling frame which has the property that we can identify every single element and
Sampling Theory
349
include any one of them in our sample. The most straightforward type of frame is a
list of elements of the population (preferably the entire population) with
appropriate contact information. For example, in an opinion poll, possible sampling
frames include: Electoral register and Telephone directory. Not all frames explicitly
list population elements. For example, a street map can be used as a frame for a
door-to-door survey; although it doesn’t show individual houses, we can select
streets from the map and then visit all houses on those streets. (One advantage of
such a frame is that it would include people who have recently moved and are not
yet on the list frames discussed above.) The sampling frame must be representative
of the population and this is a question outside the scope of statistical theory
demanding the judgment of experts in the particular subject matter being studied.
All the above frames omit some people who will vote at the next election and contain
some people who will not; some frames will contain multiple records for the same
person. People not in the frame have no prospect of being sampled. Statistical theory
tells us about the uncertainties in extrapolating from a sample to the frame. In
extrapolating from frame to population, its role is motivational and suggestive. To
the scientist, however, representative sampling is the only justified procedure for
choosing individual objects for use as the basis of generalization, and is therefore
usually the only acceptable basis for ascertaining truth.
In defining the frame, practical, economic, ethical, and technical issues need to be
addressed. The need to obtain timely results may prevent extending the frame far
into the future. The difficulties can be extreme when the population and frame are
disjoint. This is a particular problem in forecasting where inferences about the
future are made from historical data. In fact, in 1703, when Jacob Bernoulli
proposed to Gottfried Leibniz the possibility of using historical mortality data to
predict the probability of early death of a living man, Gottfried Leibniz recognized
the problem in replying, According to Gottfried Leibniz Nature has established
patterns originating in the return of events but only for the most part. New illnesses
flood the human race, so that no matter how many experiments you have done on
corpses, you have not thereby imposed a limit on the nature of events so that in the
future they could not vary. A frame may also provide additional auxiliary
information about its elements; when this information is related to variables or
groups of interest, it may be used to improve survey design. For instance, an
electoral register might include name and sex; this information can be used to
ensure that a sample taken from that frame covers all demographic categories of
interest. (Sometimes the auxiliary information is less explicit; for instance, a
telephone number may provide some information about location.)Having
established the frame, there are a number of ways for organizing it to improve
efficiency and effectiveness.
9.1.4 Probability and Nonprobability Sampling
A probability sampling scheme is one in which every unit in the population has a
chance (greater than zero) of being selected in the sample, and this probability can
be accurately determined. The combination of these traits makes it possible to
produce unbiased estimates of population totals, by weighting sampled units
350
Engineering Mathematics-III
according to their probability of selection. For example if we want to estimate the
total income of adults living in a given street. We visit each household in that street,
identify all adults living there, and randomly select one adult from each household.
(For example, we can allocate each person a random number, generated from a
uniform distribution between 0 and 1, and select the person with the highest
number in each household). We then interview the selected person and find their
income. People living on their own are certain to be selected, so we simply add their
income to our estimate of the total. But a person living in a household of two adults
has only a one-in-two chance of selection. To reflect this, when we come to such a
household, we would count the selected person’s income twice towards the total. (In
effect, the person who is selected from that household is taken as representing the
person who isn’t selected.) In this example, not everybody has the same probability
of selection; what makes it a probability sample is the fact that each person’s
probability is known. When every element in the population does have the same
probability of selection, this is known as an ‘equal probability of selection’
(EPS) design. Such designs are also referred to as ‘self-weighting’ because all
sampled units are given the same weight. Probability sampling includes: Simple
Random Sampling, Systematic Sampling, Stratified Sampling, Probability
Proportional to Size Sampling, and Cluster or Multistage Sampling. These various
ways of probability sampling have two things in common: Every element has a
known nonzero probability of being sampled and Involves random selection at some
point.
Nonprobability sampling is a sampling method where some elements of the
population have no chance of selection (these are sometimes referred to as out of
coverage), or where the probability of selection can’t be accurately determined. It
involves the selection of elements based on assumptions regarding the population of
interest, which forms the criteria for selection. Hence, because the selection of
elements is nonrandom, nonprobability sampling does not allow the estimation of
sampling errors. These conditions place limits on how much information a sample
can provide about the population. Information about the relationship between
sample and population is limited, making it difficult to extrapolate from the sample
to the population. For example: We visit every household in a given street, and
interview the first person to answer the door. In any household with more than one
occupant, this is a nonprobability sample, because some people are more likely to
answer the door (e.g. an unemployed person who spends most of their time at home
is more likely to answer than an employed housemate who might be at work when
the interviewer calls) and it’s not practical to calculate these probabilities.
Nonprobability Sampling includes: Accidental Sampling, Quota Sampling and
Purposive Sampling. In addition, nonresponse effects may turn a probability design
into a nonprobability design if the characteristics of nonresponse are not well
understood, since nonresponse effectively modifies each element’s probability of
being sampled.
Sampling Theory
351
9.1.5 Replacement of Selected units
Sampling schemes may be without replacement (‘WOR’ - no element can be selected
more than once in the same sample) or with replacement (‘WR’ - an element may
appear multiple times in the one sample). For example, if we catch fish, measure
them, and immediately return them to the water before continuing with the sample,
this is a WR design, because we might end up catching and measuring the same fish
more than once. However, if we do not return the fish to the water (e.g. if we eat the
fish), this becomes a WOR design.
9.1.6 Random Variables
Categorical random variables yield responses such as ‘yes’ or ‘no’. Categorical
variables can yield more than two possible responses. For example: ‘Which day of
the week are you most likely to wash clothes?’ Numerical random variables
yield numerical responses, such as your height in centimeters. There are two types of
numerical variables: discrete and continuous. Discrete random variables
produce numerical responses from a counting process. An example is ‘how many
times do you visit the cash machine in a typical month?’ Continuous random
variables produce responses from a measuring process. Height is an example of a
continuous variable because the response takes on a value from an interval.
Precision of the measurement instruments may lead to tied observations. A tied
observation occurs when the measuring device is not sensitive or sophisticated
enough to detect incremental differences in the experimental or survey data.
Generally continuous random variable requires fewer samples than of discrete
random variable.
9.2 Sampling Methods
Within any of the types of frame identified above, a variety of sampling methods can
be employed, individually or in combination. Factors commonly influencing the
choice between these designs include:
♠
Nature and quality of the frame.
♠
Availability of auxiliary information about units on the frame.
♠
Accuracy requirements and the need to measure accuracy.
♠
Whether detailed analysis of the sample is expected.
♠
Cost/operational concerns.
9.2.1 Simple Random Sampling
In a simple random sample (SRS) of a given size, all such subsets of the frame are
given an equal probability. Each element of the frame thus has an equal probability
of selection: the frame is not subdivided or partitioned. Furthermore, any given pair
of elements has the same chance of selection as any other such pair (and similarly
for triples, and so on). This minimizes bias and simplifies analysis of results. In
particular, the variance between individual results within the sample is a good
indicator of variance in the overall population, which makes it relatively easy to
estimate the accuracy of results. However, SRS can be vulnerable to sampling error
352
Engineering Mathematics-III
because the randomness of the selection may result in a sample that doesn’t reflect
the makeup of the population. For instance, a simple random sample of ten people
from a given country will on average produce five men and five women, but any
given trial is likely to overrepresented one sex and underrepresented the other.
Systematic and stratified techniques, discussed below, attempt to overcome this
problem by using information about the population to choose a more representative
sample. SRS may also be cumbersome and tedious when sampling from an
unusually large target population. In some cases, investigators are interested in
research questions specific to subgroups of the population. For example, researchers
might be interested in examining whether cognitive ability as a predictor of job
performance is equally applicable across racial groups. SRS cannot accommodate
the needs of researchers in this situation because it does not provide subsamples of
the population. Stratified sampling, which is discussed next, addresses this
weakness of SRS. Simple random sampling is always an EPS design, but not all EPS
designs are simple random sampling.
9.2.2 Systematic Sampling
Systematic sampling relies on arranging the target population according to some
ordering scheme and then selecting elements at regular intervals through that
ordered list. Systematic sampling involves a random start and then proceeds with
the selection of every kth element from then onwards. In this case, k= (population
size/sample size). It is important that the starting point is not automatically the first
in the list, but is instead randomly chosen from within the first to the kth element in
the list. A simple example would be to select every 10th name from the telephone
directory (an ‘every 10th’ sample, also referred to as ‘sampling with a skip of 10’). As
long as the starting point is randomized, systematic sampling is a type of probability
sampling. It is easy to implement and the stratification induced can make it efficient,
if the variable by which the list is ordered is correlated with the variable of interest.
‘Every 10th’ sampling is especially useful for efficient sampling from databases. For
example: Suppose we wish to sample people from a long street that starts in a poor
district (house #1) and ends in an expensive district (house #1000). A simple
random selection of addresses from this street could easily end up with too many
from the high end and too few from the low end (or vice versa), leading to an
unrepresentative sample. Selecting every 10th street number along the street
ensures that the sample is spread evenly along the length of the street, representing
all of these districts. (Note that if we always start at house #1 and end at #991, the
sample is slightly biased towards the low end; by randomly selecting the start
between #1 and #10, this bias is eliminated.) However, systematic sampling is
especially vulnerable to periodicities in the list. If periodicity is present and the
period is a multiple or factor of the interval used, the sample is especially likely to be
unrepresentative of the overall population, making the scheme less accurate than
simple random sampling. For example: Consider a street where the odd-numbered
houses are all on the north (expensive) side of the road and the even-numbered
houses are all on the south (cheap) side. Under the sampling scheme given above, it
is impossible’ to get a representative sample; either the houses sampled will all be
Sampling Theory
353
from the odd-numbered, expensive side, or they will all be from the even-numbered,
cheap side. Another drawback of systematic sampling is that even in scenarios
where it is more accurate than SRS; its theoretical properties make it difficult to
quantify that accuracy. (In the two examples of systematic sampling that are given
above, much of the potential sampling error is due to variation between neighboring
houses - but because this method never selects two neighboring houses, the sample
will not give us any information on that variation.)
As described above, systematic sampling is an EPS method, because all elements
have the same probability of selection (in the example given, one in ten). It is not
‘simple random sampling’ because different subsets of the same size have different
selection probabilities - e.g. the set {4, 14, 24... 994} has a one-in-ten probability of
selection, but the set {4, 13, 24, 34...} has zero probability of selection.
9.2.3 Stratified Sampling
Where the population embraces a number of distinct categories, the frame can be
organized by these categories into separate strata. Each stratum is then sampled as
an independent sub-population, out of which individual elements can be randomly
selected. There are several potential benefits to stratified sampling.
1. Dividing the population into distinct, independent strata can enable
researchers to draw inferences about specific subgroups that may be lost in a
more generalized random sample.
2. Utilizing a stratified sampling method can lead to more efficient statistical
estimates (provided that strata are selected based upon relevance to the
criterion in question, instead of availability of the samples). It is important to
note that even if a stratified sampling approach does not lead to increased
statistical efficiency; such a tactic will not result in less efficiency than would
simple random sampling, provided that each stratum is proportional to the
group’s size in the population.
3. It is sometimes the case that data are more readily available for individual,
pre-existing strata within a population than for the overall population; in such
cases, using a stratified sampling approach may be more convenient than
aggregating data across groups (though this may potentially be at odds with
the previously noted importance of utilizing criterion-relevant strata).
4. Since each stratum is treated as an independent population, different sampling
approaches can be applied to different strata, potentially enabling researchers
to use the approach best suited (or most cost-effective) for each identified
subgroup within the population.
5. Focuses on important subpopulations and ignores irrelevant ones.
6.
Allows use of different sampling techniques for different subpopulations.
7.
Improves the accuracy/efficiency of estimation.
8.
Permits greater balancing of statistical power of tests of differences between
strata by sampling equal numbers from strata varying widely in size.
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There are, however, some potential drawbacks to using stratified sampling.
1. Identifying strata and implementing such an approach can increase the cost
and complexity of sample selection, as well as leading to increased complexity
of population estimates.
2. When examining multiple criteria, stratifying variables may be related to some,
but not to others, further complicating the design, and potentially reducing the
utility of the strata.
3. In some cases (such as designs with a large number of strata, or those with a
specified minimum sample size per group), stratified sampling can potentially
require a larger sample than would other methods (although in most cases, the
required sample size would be no larger than would be required for simple
random sampling.
4. Requires selection of relevant stratification variables which can be difficult.
5.
It is not useful when there are no homogeneous subgroups.
6.
It is expensive to implement.
A stratified sampling approach is most effective when three conditions are met
1. Variability within strata are minimized.
2.
Variability between strata are maximized.
3.
The variables upon which the population is stratified are strongly correlated
with the desired dependent variable.
9.2.4 Poststratification
Stratification is sometimes introduced after the sampling phase in a process called
poststratification. This approach is typically implemented due to a lack of prior
knowledge of an appropriate stratifying variable or when the experimenter lacks the
necessary information to create a stratifying variable during the sampling phase.
This is method can provide several benefits in the right situation. Implementation
usually follows a simple random sample. In addition to allowing for stratification on
an ancillary variable, poststratification can be used to implement weighting, which
can improve the precision of a sample’s estimates.
9.2.5 Oversampling
Choice-based sampling is one of the stratified sampling strategies. In choice-based
sampling, the data are stratified on the target and a sample is taken from each strata
so that the rare target class will be more represented in the sample. The model is
then built on this biased sample. The effects of the input variables on the target are
often estimated with more precision with the choice-based sample even when a
smaller overall sample size is taken compared to a random sample. The results
usually must be adjusted to correct for the oversampling.
9.2.6 Probability Proportional to size Sampling
In some cases the sample designer has access to an auxiliary variable or size
measure, believed to be correlated to the variable of interest, for each element in the
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355
population. This data can be used to improve accuracy in sample design. One option
is to use the auxiliary variable as a basis for stratification, as discussed above.
Another option is probability-proportional-to-size sampling, in which the selection
probability for each element is set to be proportional to its size measure, up to a
maximum of 1. In a simple PPS design, these selection probabilities can then be used
as the basis for Poisson sampling. However, this has the drawbacks of variable
sample size, and different portions of the population may still be over- or
under-represented due to chance variation in selections. To address this problem,
PPS may be combined with a systematic approach. For example: we have six schools
with populations of 150, 180, 200, 220, 260, and 490 students respectively (total
1500 students), and we want to use student population as the basis for a PPS sample
of size three. To do this, we could allocate the first school numbers 1 to 150, the
second school 151 to 330 (=150+180), the third school 331 to 530, and so on to the
last school (1011 to 1500). We then generate a random start between 1 and 500
(equal to 1500/3) and count through the school populations by multiples of 500. If
our random start was 137, we would select the schools which have been allocated
numbers 137, 637, and 1137, i.e. the first, fourth, and sixth schools. The PPS
approach can improve accuracy for given samples size by concentrating sample on
large elements that have the greatest impact on population estimates. PPS sampling
is commonly used for surveys of businesses, where element size varies greatly and
auxiliary information is often available - for instance, a survey attempting to
measure the number of guest-nights spent in hotels might use each hotel’s number
of rooms as an auxiliary variable. In some cases, an older measurement of the
variable of interest can be used as an auxiliary variable when attempting to produce
more current estimates.
9.2.7 Cluster Sampling
Sometimes it is cheaper to cluster the sample in some way e.g. by selecting
respondents from certain areas only, or certain time-periods only. (Nearly all
samples are in some sense ‘clustered’ in time - although this is rarely taken into
account in the analysis.) Cluster sampling is an example of two-stage sampling or
multistage sampling: in the first stage a sample of areas is chosen; in the second
stage a sample of respondents within those areas is selected. This can reduce travel
and other administrative costs. It also means that one does not need a sampling
frame listing all elements in the target population. Instead, clusters can be chosen
from a cluster-level frame, with an element-level frame created only for the selected
clusters. Cluster sampling generally increases the variability of sample estimates
above that of simple random sampling, depending on how the clusters differ
between themselves, as compared with the within-cluster variation. Nevertheless,
some of the disadvantages of cluster sampling are the reliance of sample estimate
precision on the actual clusters chosen. If clusters chosen are biased in a certain way,
inferences drawn about population parameters from these sample estimates will be
far off from being accurate.
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9.2.8 Multistage Sampling
Multistage sampling is a complex form of cluster sampling in which two or more
levels of units are imbedded one in the other. The first stage consists of constructing
the clusters that will be used to sample from. In the second stage, a sample of
primary units is randomly selected from each cluster (rather than using all units
contained in all selected clusters). In following stages, in each of those selected
clusters, additional samples of units are selected, and so on. All ultimate units
(individuals, for instance) selected at the last step of this procedure are then
surveyed. This technique, thus, is essentially the process of taking random samples
of preceding random samples. It is not as effective as true random sampling, but it
probably solves more of the problems inherent to random sampling. Moreover, It is
an effective strategy because it banks on multiple randomizations. As such, it is
extremely useful. Multistage sampling is used frequently when a complete list of all
members of the population does not exist and is inappropriate. Moreover, by
avoiding the use of all sample units in all selected clusters, multistage sampling
avoids the large, and perhaps unnecessary, costs associated traditional cluster
sampling.
9.2.9 Matched Random Sampling
A method of assigning participants to groups in which pairs of participants are first
matched on some characteristic and then individually assigned randomly to groups.
The Procedure for Matched random sampling can be briefed with the following
contexts.
1. Two samples in which the members are clearly paired, or are matched
explicitly by the researcher. For example, IQ measurements or pairs of identical
twins.
2. Those samples in which the same attribute or variable is measured twice on
each subject under different circumstances. Commonly called repeated
measures. Examples include the times of a group of athletes for 1500m before
and after a week of special training; the milk yields of cows before and after
being fed a particular diet.
9.2.10 Quota Sampling
In quota sampling, the population is first segmented into mutually exclusive
sub-groups, just as in stratified sampling. Then judgment is used to select the
subjects or units from each segment based on a specified proportion. For example,
an interviewer may be told to sample 200 females and 300 males between the age of
45 and 60. It is this second step which makes the technique one of non-probability
sampling. In quota sampling the selection of the sample is non-random. For example
interviewers might be tempted to interview those who look most helpful. The
problem is that these samples may be biased because not everyone gets a chance of
equal selection. This random element is its greatest weakness and quota versus
probability has been a matter of controversy for many years
Sampling Theory
357
9.2.11 Convenience Sampling
Convenience sampling (sometimes known as grab or opportunity
sampling) is a type of nonprobability sampling which involves the sample being
drawn from that part of the population which is close to hand. That is, a sample
population selected because it is readily available and convenient. The researcher
using such a sample cannot scientifically make generalizations about the total
population from this sample because it would not be representative enough. For
example, if the interviewer was to conduct such a survey at a shopping center early
in the morning on a given day, the people that he/she could interview would be
limited to those given there at that given time, which would not represent the views
of other members of society in such an area, if the survey was to be conducted at
different times of day and several times per week. This type of sampling is most
useful for pilot testing. Several important considerations for researchers using
convenience samples include:
1. Are there controls within the research design or experiment which can serve to
lessen the impact of a non-random, convenience sample whereby ensuring the
results will be more representative of the population?
2. Is there good reason to believe that a particular convenience sample would or
should respond or behave differently than a random sample from the same
population?
3. Is the question being asked by the research one that can adequately be
answered using a convenience sample?
In social science research, snowball sampling is a similar technique, where existing
study subjects are used to recruit more subjects into the sample.
9.2.12 Line-Intercept Sampling
Line-intercept sampling is a method of sampling elements in a region whereby
an element is sampled if a chosen line segment, called a transect, intersects the
element.
9.2.13 Panel Sampling
Panel sampling is the method of first selecting a group of participants through a
random sampling method and then asking that group for the same information
again several times over a period of time. Therefore, each participant is given the
same survey or interview at two or more time points; each period of data collection
is called a wave. This sampling methodology is often chosen for large scale or
nation-wide studies in order to gauge changes in the population with regard to any
number of variables from chronic illness to job stress to weekly food expenditures.
Panel sampling can also be used to inform researchers about within-person health
changes due to age or help explain changes in continuous dependent variables such
as spousal interaction. There have been several proposed methods of analyzing
panel sample data, including MANOVA, growth curves, and structural equation
modeling with lagged effects.
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9.2.14 Event Sampling Methodology
Event Sampling Methodology (ESM) is a new form of sampling method that
allows researchers to study ongoing experiences and events that vary across and
within days in its naturally-occurring environment. Because of the frequent
sampling of events inherent in ESM, it enables researchers to measure the typology
of activity and detect the temporal and dynamic fluctuations of work experiences.
Popularity of ESM as a new form of research design increased over the recent years
because it addresses the shortcomings of cross-sectional research, where once
unable to, researchers can now detect intra-individual variances across time. In
ESM, participants are asked to record their experiences and perceptions in a paper
or electronic diary.
There are three types of ESM:
1. Signal contingent– random beeping notifies participants to record data.
The advantage of this type of ESM is minimization of recall bias.
2. Event contingent – records data when certain events occur.
3.
Interval contingent – records data according to the passing of a certain
period of time.
ESM has several disadvantages. One of the disadvantages of ESM is it can sometimes
be perceived as invasive and intrusive by participants. ESM also leads to possible
self-selection bias. It may be that only certain types of individuals are willing to
participate in this type of study creating a non-random sample. Another concern is
related to participant cooperation. Participants may not be actually fill out their
diaries at the specified times. Furthermore, ESM may substantively change the
phenomenon being studied. Reactivity or priming effects may occur, such that
repeated measurement may cause changes in the participant's experiences. This
method of sampling data is also highly vulnerable to common method variance.
Further, it is important to think about whether or not an appropriate dependent
variable is being used in an ESM design. For example, it might be logical to use ESM
in order to answer research questions which involve dependent variables with a
great deal of variation throughout the day. Thus, variables such as change in mood,
change in stress level, or the immediate impact of particular events may be best
studied using ESM methodology. However, it is not likely that utilizing ESM will
yield meaningful predictions when measuring someone performing a repetitive task
throughout the day or when dependent variables are long-term in nature (coronary
heart problems).
9.3 Sampling and Data Collection
Good data collection involves:
♠
Following the defined sampling process.
♠
Keeping the data in time order.
♠
Noting comments and other contextual events.
♠
Recording non-responses.
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359
Most sampling books and papers written by non-statisticians focus only in the data
collection aspect, which is just a small though important part of the sampling
process.
9.4 Review of Sampling Process
After sampling, a review should be held of the exact process followed in sampling,
rather than that intended, in order to study any effects that any divergences might
have on subsequent analysis. A particular problem is that of non-responses.
9.4.1 Non-Response
In survey sampling, many of the individuals identified as part of the sample may be
unwilling to participate or impossible to contact. In this case, there is a risk of
differences, between (say) the willing and unwilling, leading to selection bias in
conclusions. This is often addressed by follow-up studies which make a repeated
attempt to contact the unresponsive and to characterize their similarities and
differences with the rest of the frame. The effects can also be mitigated by weighting
the data when population benchmarks are available. Nonresponse is particularly a
problem in Internet sampling. One of the main reasons for this problem could be
that people may hold multiple e-mail addresses, which they don’t use anymore or
don’t check regularly.
9.4.2 Survey Weights
In many situations the sample fraction may be varied by stratum and data will have
to be weighted to correctly represent the population. Thus for example, a simple
random sample of individuals in India might include some in remote areas who
would be inordinately expensive to sample. A cheaper method would be to use a
stratified sample with urban and rural strata. The rural sample could be
under-represented in the sample, but weighted up appropriately in the analysis to
compensate. More generally, data should usually be weighted if the sample design
does not give each individual an equal chance of being selected. For instance, when
households have equal selection probabilities but one person is interviewed from
within each household, this gives people from large households a smaller chance of
being interviewed. This can be accounted for using survey weights. Similarly,
households with more than one telephone line have a greater chance of being
selected in a random digit dialing sample, and weights can adjust for this. Weights
can also serve other purposes, such as helping to correct for non-response.
9.4.3 Sampling Error
In statistics, sampling error or estimation error is the error caused by
observing a sample instead of the whole population. An estimate of a quantity of
interest, such as an average or percentage, will generally be subject to
sample-to-sample variation. These variations in the possible sample values of a
statistic can theoretically be expressed as sampling errors, although in practice the
exact sampling error is typically unknown. Sampling error also refers more broadly
to this phenomenon of random sampling variation. The likely size of the sampling
error can generally be controlled by taking a large enough random sample from the
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population, although the cost of doing this may be prohibitive. If the observations
are collected from a random sample, statistical theory provides probabilistic
estimates of the likely size of the sampling error for a particular statistic or
estimator. These are often expressed in terms of its standard error. Sampling error
can be contrasted with non-sampling error. Non-sampling error is a catch-all term
for the deviations from the true value that are not a function of the sample chosen,
including various systematic errors and any random errors that are not due to
sampling. Non-sampling errors are much harder to quantify than sampling error.
Theoretical Examples
Suggest a possible source of bias in the following :
Ex. 1: The mean income per family in a certain town is sought to be estimated by
sampling from motor owners.
Ex. 2:
Readers of newspapers are sampled by printing in it an invitation to them
to send up their observation on some typical event.
Ex. 3:
A sample of household survey which includes several houses in which
none is at home when the investigator first calls.
Ex. 4:
A sample of unemployed obtained by taking every hundredth name in the
register of applicants for appointments arranged in alphabetical order.
Ex. 5:
A survey of incomes of the residents of a locality by interviewing the
owner of tenth house.
Ex. 6:
A barrel of apples is sampled by taking a handful from the top.
Ex. 7:
A mixture of sand and sawdust is sampled by scooping up a quantity from
the bottom.
Ex. 8:
A set of digits is taken by opening a telephone directory at random and
choosing the telephone numbers in the order in which they appear on the
page.
Sol. 1:
Motor car licences are owned by only very rich people and thus the sample
would represent only the wealthy class of the town. The sample is thus
very much biased.
Sol. 2:
The sampling is not unbiased since those readers are more likely to write
on the topic in which they are interested. Some of the readers are
generally lethargic in writing letters and may not be represented in the
sample, moreover, some others may have missed the newspaper of the
date of publication of the invitation and thus are unable to send their
views.
Sol. 3:
The surveyer may have visited at working hours so that the families in
which both husband and wife are employed are absent and they would
not be represented in the sample. Thus the sample is not unbiased.
Sol. 4:
The sample is fairly unbiased since alphabetic arrangement of names is
independent of the status of the individual. It is known as Systematic
Sampling which is fairly random if the attribute considered is
independent of the other.
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361
Sol. 5:
It is possible that the locality consists of ten house blocks so that every
tenth house is a corner house and may belong to comparatively richer
people, in which case it may be biased.
Sol. 6:
Generally in transportation the heaviest apples go down to the bottom
and the lighter and smaller ones remain at the top, so that the apples at
the top are not true representatives of the whole barrel. Moreover, a
shopkeeper is likely to put the best apples at the top to attract the
customer and it is possible that the other apples in the barrel may be of
inferior quality.
Sol. 7:
The density of sand is greater than that of sawdust and hence when a
sample is taken from the bottom, it is likely to consist of a higher
proportion of sand than in the whole mixture.
Sol. 8:
If we take a used directory, the pages frequently used are more likely to be
opened and thus the more popular numbers are apt to be included in the
sample than others.
Short Answers Type Questions
1.
Write a short essay on Sampling in Statistics.
2.
Write brief note on Sampling Theory.
3.
Write a short note on random sampling.
(Meerut , 1983,2000)
(Meerut, 1980, 2005)
9.5 Population and Samples
As already explained that the word population is used to refer to any collection of
objects or results of operations. For example, we may speak of the population of
dairy cows in Meerut district, the population of mileages of automobile tyres, the
population of prices of a commodity in a city. We may also speak of hypothetical
population of heads and tails obtained by tossing a coin an infinite number of times
or the population of all possible values which the bank rate can have in twenty years
time and so on.
The aim of a statistical enquiry is to find out something about some specified
population. It is impossible or impracticable to examine each and every member of
the population since such a process will be too costly in terms of time and money.
Thus the investigator is led to the study of a selected number of individuals from the
population and on the basis of this limited investigation, he makes inferences
regarding the whole population. This selected number of indivituals from a
population is called a sample. The inferences that can be made from a sample about
the whole population can never be of a categorical certainty. They can only be
expressed in terms of probabilities. In order that the theory of probability can be
applied, the sampling should be random. In the case of non-random sample, there is
no way to measure the degree of confidence to be placed in any inference which can
be made from such a sample. Recall that the selection of an individual from a
population is random when each member of the population has the same chance of
being selected. The aims of the theory of sampling are 1.... to find estimates of
certains constants such as mean and standard deviation of the population. 2.... to
determine what degree of confidence can be placed in these estimates when they are
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obtained; in other words, to determine the limits within which the parameters of the
population are expected to lie with a specified degree of confidence.
9.6 The Simple Sampling of Attributes
In sampling of attributes, we are concerned only with the presence or absence of an
attribute. More specially, the sampling of attributes may be thought of as drawing
samples from a population containing A's and not A's. The presence of the attribute A
may be called a success and its absence a failure. For example, in sampling a
population of tosses of a coin for the prepositions of heads and tails we might speak
of a sample of one thousand tosses, seven hundred of which were heads. In other
words, the sample consisted of one thousand events, seven hundered of which were
successes and three hundred failures. By simple sampling we mean random
sampling in which each event has the same chance p of success and in which the
chances of success of different events are independent of whether previous trials
have been made or not. Thus in the throwing of a die, the chance of getting an ace is
not affected by what was obtained on the previous trials and remains the same
1

 e. g .  for subsequent trials, provided, of course, the die does not wear out or is not

6
deliberately manipulated to get some other number by the experimenter. It should
be noted that random sampling is not necessarily simple but simple sampling is
always random as is evident from the above definitions. As a matter of fact, simple
sampling is a particular form of random sampling. For example, if a bag contains 5
5
black balls and 3 white balls, the chance of drawing a black ball at the first trial is
8
and if the ball is not put back in the bag the chance of drawing a black ball at the
4
second trial is which is not the same as before. Hence the sampling is not simple.
7
However, the sampling is random since on the first trial each black ball has got the
5
same chance of being drawn out and at the second trial each black ball has the
8
4
same chance of being selected.
7
9.6.1 Mean and Standard Deviation in Simple Sampling of Attributes
Suppose we draw N samples of size n from a large population. If each individual in a
sample has a chance p for success, (i. e. selection) and a chance q = 1 − p for failure
so that the probabilities of 0, 1, 2,...n successes are given by successive terms of
( q + p) n where p + q = 1.
This means that the probabilities of 0, 1, 2, 3, ...n times in the sample possessing the
attribute under study is q n , n C1 q n − 1 p, n C 2 q n − 2 p 2 ,... p n . We know that the
mean of this distribution is np and standard deviation is √ ( npq). Hence the expected
value of success in a sample of size n is np and the standard error of the number of
the successes in sample of size n is √ ( npq). If instead of the number of successes in
Sampling Theory
363
each sample we take proportion of successes, the mean proportion of successes will
np
be
= p and the standard error of the proportion of successes is
n
 p q
n ⋅ ⋅  =
 n n
 pq
 
 n
Note. If p or q become very small, then pq = p (1 − p) = p approx. Hence
σ = √ ( np) = √ ( M ). It follows that if the proportion of successes be small, the
standard error of the number of successes is the square root of the mean number of
successes.
9.6.2 The Sampling of Attributes
As distinguished from variables where quantitative measurement of a phenomenon
is possible, in case of attributes we can only find out the presence or absence of a
particular characteristic. The sampling of attributes may, therefore, be regarded as
the drawing of samples from a population whose members possess the attribute A or
not A. For exempted in the study of attribute literacy a sample may be taken and
people classified as literates and illiterates. With such data the binomial type of
problem may be formed. The selection of an individual on sampling may be called
event the appearance of an attribute A may be taken as success and its
non-appearance as 'failure'. Thus, if out of 1,000 people selected for the sample, 100
are found literates and 900 illiterates, we would say that the sample consists of 1000
units out of which 100 are successes and 900 failures. The probability of success or
P = 100 / 1000 or 0.1 and the probability of failure or q = 900 / 1,000 = 0.9 so that
p + q = 0.1 + 0.9 = 1.
The various tests of significance for attributes are discussed under the following
heads :
(i)
Test for number of successes.
(ii)
Test for proportion of success. and
(iii)
Test of difference between proportions.
Test for Number of Successes
The sampling distribution of the number of successes follows a binomial probability
distribution. Hence its standard error is given by the formula :
S.E. of no. of successes = npq
where
n = size of sample
p = probability of success in each trial
q = (1 – p), i.e., probability of failure.
9.6.3 Conditions for Simple Sampling

 pq 
The use of the formula √ ( npq)  or    for the standard error of p is justified
 n

only under certain conditions which give rise to simple sampling in practice. There
are three such conditions given below :
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2.
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The probability p for drawing an individual with attribute A on random
sampling must remain constant and must be the same for all samples. This
means that the proportion of individuals with attribute A in the populations
remains constant at the drawing of each sample. Thus the theory of simple
sampling cannot apply to the variations of the death rate in localities having
population of different ages and sexes, or to death rates in successive years
during a period of improved sanitary conditions.
The properties of individuals with attribute A must remain constant at the
drawing of each individual member of the sample. Thus in the case of death
rates mentioned above, the samples must not only be of the same age and sex
composition and being under the same sanitary conditions, but each sample
should contain persons of one age and one sex only. For if in each sample,
there were persons of both sexes and different ages, the condition would be
violated, the probability of death during a given period being different for the
two sexes, or for different age-groups.
It should be noted that the condition for the consistency of p will be satisfied if
(i) The individuals are replaced after each drawing, before making the next
drawing : by such a device the constitution of the population does not
change so that the chance of success remains the same.
(ii) The population is infinite : in this case, the removal of a finite number of
individuals does not affect the proportion of individuals in the population
possessing a specified attribute :
(iii) The population is very large so that p may be taken to be constant without
any appreciable error, provided the sample is not also large. This is very
important case for the application of the theory of simple sampling to
many practical data.
3.
The third condition for simple sampling is that the individual events must be
completely independent of one another like the throws of a coin. For
example, if we were dealing with deaths from an infection or contagious
disease, the theory of simple sampling could not be applied in such a case
even if the sample population consisted of persons of one age and one sex
only. For if one person in a certain sample has contacted the disease in
question, he has increased the possibility of other persons dying from the
same disease.
Having discussed the above concepts let us now discuss the various situations where
we have to apply the various tests of significance. For the sake of convenience and
clarity these situations may be summed up under the following three heads :
(i) The sampling of Attributes
(ii) The sampling of variables (large samples)
(iii) The sampling of variables (small samples)
9.7 Test of Significance for Large Samples
We know that if n is large, the binomial distribution tends to normal, so that in the
case of large samples properties of normal curves can be used. Suppose we wish to
Sampling Theory
365
test the hypothesis that a given large sample of size n is obtained by simple sampling
from a population for which the probability of success is p. For normal distribution,
we know that 99 ⋅ 7% of its members lie within a range ± 3σ i. e. ± 3 √ ( npq) on
either side of the mean np, so that only 0⋅3% of the members lie outside this range.
Again only 5% of the members of a normal population lie outside the range mean
± 2σ [i. e. np ± 2 √ ( npq)]. Hence we have the following test of significance for large
samples :
If the number of successes in a large sample of size n differs from the expected value
np by more than 3 √ ( npq): we call this difference highly significant. Sometimes a
difference of more than 2 √ ( npq) may be called significant whereas a difference of
less than 2 √ ( npq) is called insignificant.
We may put the above test in a mathematical form as follows :
If z is the standard normal variate, we have
x − np
z=
√ ( npq)
where x is the observed number of successes in the sample. Thus
(i)
If |z| > 3, difference between the observed and the expected number of
successes is highly significant.
(ii) If 2 < |z| < 3, the difference may be regarded as significant.
(iii)
If |z| < 2 the difference is not significant.
9.8 Standard Error
The standard deviation of simple sampling is called standard error. The use of the
word error is justified on the ground that we usually regard the expected value as
the true value and divergence from it as errors of estimation due to fluctuations of
sampling. But too much importance must not be attached to the word error. Mostly
the term ‘standard error’ is applied to the standard deviation of simple sampling.
Apart from this, the term has got a wider meaning which we shall discuss when we
deal with the theory of sampling of variable.
9.8.1 Standard Error and Sampling Distribution
If the universe distribution is not normal, then the sampling distribution of sample
means approaches normality as the sample size increases. The word ''error'' is used
in place of ''deviation'' to emphasize that variation among sample means is due to
sampling errors. It is so called because it measures the sampling variability due to
change or random forces. Hence to clarify the term standard error it is necessary to
describe a sampling distribution. If we select a number of independent random
samples of a definite size from a given population and calculate some statistic (like
the mean, standard deviation, etc.) from each sample we shall get a series of value of
these statistics or functions. these values obtained from the different samples can be
put in the form of a frequency distribution. The distribution so formed of all possible
values of a statistic is called the sampling distribution of the probability distribution
of that statistic.Thus if we draw 100 random samples from a given population and
calculate their means, we shall get a series of 100 means which would form a
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Engineering Mathematics-III
frequency distribution. This distribution will be known as the sampling distribution
of the means. An explanation of sampling distribution would be incomplete without
describing the universe distribution, the sample distribution and showing the
relationship of these two with the sampling distribution.
9.8.2 Universe Distribution
Such a distribution emerges when each and every item of the universe is studied,
and we have full knowledge of its mean and standard deviation. The mean of the
universe which is also called the true means is symbolized by µ. And its standard
deviation by σ (lower-case sigma). Greek letters are used for these manures to
emphasize their differences from corresponding measures taken from a sample.
Measures characterizing a universe, such as µ and σ are called parameters.
9.8.3 The Sample Distribution
If instead of all the items of the universe we study only a small part of it, i.e.,
take a sample, we will arrive at a sample distribution. The symbols X and S are
used to designate the mean and standard deviation of the sample distribution.
It may be noted that several sample distributions are possible from a given
universe. The sampling distribution of a statistic reveals some important
features :
1. First, a sampling distribution is generated from a population distribution,
known or assumed.
2. Secondly, the sample population may generate an infinite number of sampling
distributions for the statistic, each for special sample size n.
3. Finally, a population may generate sampling distributions for two or more
different statistics.
Sampling distributions are of great importance in theory and practice of statistics. It
is because of the fact that sampling distribution of a statistic has well defined and by
these properties that we can calculate risks (errors due to change) involved in
making generalisation about population on the basis of samples.
For example, the sampling distribution of means has the following important
properties :
1. The arithmetic means of the sampling distribution is the same as the means of
the universe from which samples were taken. For this reason the mean of the
sampling distribution may be denoted by the same symbol as that the means of
the universe involved, namely µ,
2. The sampling distribution of mean has a standard deviation (a standard error)
equal to the population standard deviation divided by the square root of the
σ
sample size, i. e.,
.
n
3.
The sampling distribution of means is normally distributed. Since the sampling
distribution of the means is normally distributed, we can develop a method in
which we can use the mean of our sample to estimate the population mean. For
Sampling Theory
367
example, if we compute an interval of X ± 1.96 (σ/ n ), the interval will include
95 percent of all the sample means.
9.8.4 Utility of the Concept of Standard Error
The concept of standard error is of great significance in statistical work because of
the following reasons :
(i) It is used as an instrument in testing a given hypothesis. The hypothesis is
generally tested at 5% level of significance. If the difference between observed
and expected means is more than 1.96 standard error (S.E.), we say that the
result of the experiment does not support the hypothesis at 5% level or, in other
words, the difference is regarded as significant i.e., it could not have arisen due
to fluctuations of sampling. On the other hand, if the difference between
observed and expected results is less than 1.96 S.E., it is not regarded as
significant, i.e., it could have arisen due to fluctuation of simple sampling, i.e.,
we say that the result of the experiment does not provide evidence against the
hypothesis. If the difference is more than 2.58 S.E., it is considered to b
significant at 10% level. In practice, quite often a hypothesis is accepted if the
difference is less than 3 S.E., because the probability of a difference is greater
than 3 S.E., arising by chance, is only about 3 in thousand (0.27)% as 99.73 per
cent items are covered between mean ± 3σ on either side of the mean.
However, it must be emphasized that the use of 3σ rules justified only if n is
large. Some people apply a criterion of 2 S.E. in order to determine whether or
not the difference could have arisen due to fluctuations of sampling. However,
instead of 3 S.E. or 2 S.E., it is suggested that we should use either 5% level or
10% level of significance. (In practice, 5% level is more popular.)
(ii) Standard error provide an idea about the unreliability of sample. The greater
the standard error, the greater is the departure of actual frequencies from the
expected ones and hence the greater the unreliability of the sample. The
1
reciprocal of S.E., i.e.,
is a measure of reliability or precision of the sample.
S. E .
The reliability or precision of an observed proportation varies as the square
root of the number of items in the sample. In other words, if we want to double
the precision (which is th same thing as reducing the standard error to one
half), the number of observation should be increased four times.
(iii) With the help of S.E., we can determine the limits within which the parameter
values are expected to lie. This is made possible because for large samples.
Sampling distributions tend to approximate a normal distribution. In a normal
distribution 68.27% of the samples will have their mean values (or any other
constant) within a range of the population mean ± 1 standard deviation or
standard error. Similarly a range of mean ± 2 S.E., will give 95.45 per cent
values and mean ± 3 S.E. will give 99.73 per cent values. Thus a range of ± 3
S.E. should be taken as the determining limit outside which the value of the
parameter probability does not fall. The change of a value lying outside ± 3 S.E.
limits is only 0.27% (i.e., approximately 3 in 1,000)
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Engineering Mathematics-III
9.8.5 Probable error
The term ‘probable error’ is used by some authorities instead of ‘standard error’. The
probable error is 0 ⋅ 67449 times the standard error. The reason for the use of the
term ‘probable error’ lies in the fact that in the normal curve, the quartiles are distant
0 ⋅ 67449 σ from the mean, so that the probability that a deviation is in excess of the
1
probable error is and is equal to probability of a deviation being less than the
2
probable error. If P denotes the probable error, we have
∴
P = ⋅ 67449 σ.
3P
3σ =
= 4 ⋅ 5 P approx.
⋅67449
–3σ –2σ
–σ
P.E Area 25
Area 25 P.E.
Hence the rule that the observed deviation should not be greater than 3 times the
standard error is then roughly equivalent to a rule that it should not exceed 4 ⋅ 5
times the probable error.The probable error is often used as a measure of dispersion,
instead of standard error, since it has the merit of being easily understood : but the
term itself is misleading, for it is not actually an error and its use is being discarded
in favour of standard error. The figure (9.1) is helpful for understanding the nature
of probable error and the relationship between probable error and standard error.
Half of the observations lie between Mean + P. E . and Mean − P. E . and hence the
chance that an observation taken at random will lie between these limits is equal to
its chance of falling outside them.
0
σ
2σ
3σ
Fig. 9.1
9.8.6 Precision
We have seen above that the standard error measures the unreliability of the value
of p. Fluctuations of the observed proportion will increase with an increase in the
standard error. The reciprocal of the standard error is called precision and measures
the reliability of the observed proportion. Since the standard error varies inversely
as the square root of the number of observations in the sample, the precision will
vary as the square root of the number of observations. Thus to double the precision
(or halve the standard error) we should increase the number of observations four
times.
Sampling Theory
369
The following illustrations shall explain this methods.
Numerical Examples
Ex. 1: A dice is thrown 49152 times and of these we get 25145 times, 4, 5 or 6. Is
this consistent with the hypothesis that the dice must be unbiased ?
(ICWA. 1980)
Sol : Let is take the hypothesis that the dice is unbiased. On the basis of this
hypothesis, probability of gettingodd points and even points should be the same,
1
i. e., . Hence the expected number of getting 4, 5 or 6 in 49,152 throws is
2
49152
= 24,576. the deviation of the actual from the expected number is
2
( 25145 – 24576) = 569
S.E. = npq
1
1
n = 49,152, p = , q =
2
2
1 1
S.E. = 49152 × × = 110 ⋅ 85
2 2
Difference
569
=
= 5.13
S. E.
110.85
Since the difference is more than 2.58 S.E. (1% level of significance) the hypothesis
is rejected. Hence it cannot be concluded that the dice must be unbiased.
Ex. 2:. A coin is tossed 400 times and it turns up head 216 times. Is it reasonable to
assume that the coin is unbiased ?
Sol : Let is take the hypothesis that the coin is unbiased. On the basis of this
hypothesis the probability of getting a head is 1/2 and, therefore, the expected
1
number of heads in 400 tosses is 400× =200. The observed number of heads is
2
216. The deviation of actual number of heads from expected is 16 (216–200).
S.E. of no. of heads
= npq
n = 400, p = 1 / 2, q = 1 / 2
1 1
S.E. = 400 × × = 10
2 2
Diff 16
=
= 1.6
S. E. 10
Since the difference between observed and expected number of heads is less than
1.96 S.E. (at 5% level of significance), the hypothesis is accepted. Hence it is not
reasonable to assume that the coin is unbiased.
Ex. 3: In a sample of 500 persons from a village in Rajasthan, 280 are found to be
rice eaters and the rest wheat eaters. Can we assume that both the food articles are
equally popular?
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Engineering Mathematics-III
Sol : Let us take the hypothesis that the food articles are equally popular. The
500
expected frequency for wheat eaters and rice eaters would be
= 250 each.
2
1 1
S.E. = npq = 500 × × = 11.18
2 2
Difference between observed and expected number of rice eaters
= 280 – 250 = 30
Difference
30
=
= 2.68
S. E .
11.18
Since the difference is more than 2.58 S.E. (1% level), it could not have arisen due to
fluctuations of sampling. Hence we cannot assume that both the food articles are
equally popular.
Ex. 4: In a big city 325 men out of 600 men were found to be smokers. Does this
information support the conclusion that the majority of men in this city are smokers ?
Sol : Let us take the hypothesis that there is no significant difference in big and
small city in smoking habits.
1
Expected smokers = 300, i. e., p = q =
2
1 1
S. E . = npq = 600 × × = 12.25
2 2
Diff
25
=
= 2.04
S. E. 12.25
Difference = 325 – 300 = 25
Since the difference is more than 1.96 S.E. (5% level of significance) the hypothesis
is rejected.
Test for Proportion of Successes
Instead of recording the number of successes in each sample, we might record the
1
proportion of successes, th of the number of successes in each sample. As this
n
would amount to dividing all figures of the record by n; the mean proportion of
successes must be p, and the standard deviation of the proportiuon of successes
pq
. Thus we have the following formula :
n
S. E . p =
pq
.
n
Ex.5 : 500 apples are taken at random from a large basket and 50 are found to be
bad. Estimate the proportion of bad apples in the basket and assign limits within
which the percentage most probably lies.
Sol : The proportion of bad apples in the given sample.
50
=
= 0.1
500
Hence p = 0.1 and q = 0.9
Sampling Theory
371
S. E. p =
pq
=
n
0.1 × 0.9
=
500
0.09
= 0.013
500
The limits within which percentage of bad apples lies

pq 
= p ± 3
 × 100
n

= [ 0.1 ± 3( 0.013)] × 100
= [ 0.1 + ( 0.039)] × 100
= [ 0.1 ± ( 0.039)] × 100
= 6.1 to 13.9
Thus, the percentage of bad apples in the consignment almost certainly lies
between 6.1 and 13.9.
Ex. 6:. A wholesaler in apples claims that only 4% of the apples supplied by him are
defective. A random sample of 600 apples contained 36 defective apples. Test the
claim of the wholesaler.
Sol : The wholesaler claims that only 4% of apples are defective. Hence the 95%
confidence limits are given by X ± 1.96 S.E.
pq
0.96 × 0.04
S. E. p =
=
= 0.008
n
600
95% confidence limits = X ± 1.96 S.E.= 0.96 ± 1.96 ( 0.008)
= 0.96 ± 0.0157 = 0.9443 to 0.9757
Out of 600 apples, the good apples may vary between 0.9443 × 600 = 566.58 and
0.9757 × 600 = 585.42, i.e., 567 and 585.
The number of defective is thus expected to lie between 15 and 33. Since the actual
number is 36. the wholesaler's claim that only 4% of its apples are defective cannot
be accepted.
Test for Difference between Proportions
If two samples are drawn from different populations, we may be interested in
finding out whether the difference between the proportion of successes is significant
or not. In such a case we take the hypothesis is that the difference between p1 i. e.,
the proportion of successes in one sample, and p 2 , i.e., the proportion of successes in
another sample, is due to fluctuations of random sampling. The standard error of
the difference between properties is calculated by applying the following formula :
S. E .( p1 – p2 ) =
1
1
pq 
+

 n1 n 2 
the following illustrations shall explain this method.
where p = the pooled estimate of the actual proportion in the population. The value
of p is obtained as follows:
n p + n 2 p2
p= 1 1
n1 + n 2
372
or
Engineering Mathematics-III
p=
x1 + x 2
n1 + n 2
where x1 and x 2 stand for the number of occurrences in the two samples of sizes n1
and n 2 respectively.
p – p2
If 1
is less than 1.96 S.E. (5% level of significance), the difference is due to
S. E .
random sampling variation, i.e., as not significant.
The following illustrations shall explain the method :
Ex. 7: In a random sample of 1,000 persons from town A, 400 are found to be
consumers of wheat. In a sample of 800 from town B 400 are found to be consumers
of wheat. Do these data reveal a significant difference between town A and town B,
so far as the proportion of wheat consumers is concerned ?
Sol : Let us set up hypothesis that the two towns do not differ so far as proportion of
what consumers is concerned, i.e., H 0 : p1 = p 2 against H 2 : p1 , p 2 .
S. E.( p1 p2 ) =
1
1
pq 
+

 n1 n 2 
400
400
= 0.4; n1 = 800, p 2 =
= 0.5
1000
800
(1000 × 0.4) + ( 800 × 0.5) 400 + 400
p=
=
1000 + 800
1800
x1 + x 2
400 + 400
4
p=
=
=
n1 + n 2 1000 + 800 9
n1 = 1000, p1 =
or simply
∴
q = 1–
S. E .( p1 – p2 ) =
4 5
=
9 9
4 5 1
1 
× 
+
 =

9 9 1000 800
20
9
×
= 0.024
81 4000
p1 – p 2 = 0.4 – 0.5 = 0.1
Difference
0.1
=
= 4.17
S. E.
0.024
Since the difference is more than 2.58 S.E. (1% level of significance) it could not
have arisen due to fluctuations of sampling. hence the data reveal a significant
difference between town A and town B so far as the proportion of wheat consumers
is concemed.
Ex. 8 : In a random sample of 500 persons from Maharashtra, 200 are found to be
consumers of vegetable oil. In another sample of 400 persons from Gujarat, 200 are
found to be consumers of vegetable oil. Discuss whether the data reveal a significant
difference between Maharashtra and Gujarat so far proportion of vegetable oil
consumers is concerned.
Sol : Let us take the hypothesis that there is no difference between Maharashtra and
Gujarat so far as the proportion of vegetable oil consumers is concerned, i. e.,
p1 – p 2 (null hypothesis).
Computing the Standard Error of the difference of proportions
Sampling Theory
373
1
1
pq
+

 n1 n 2 
x + x2
p= 1
n1 + n 2
S. E.( p1 – p2 ) =
q = (1 – p)
200
200
n1 = 500
p1 =
= 0.4, n 2 = 400, p 2 =
= 0.5
500
400
p1 – p 2 = 0.4 – 0.5 = 0.1. However, the conclusion remains the same irrespective of
the fact whether the difference is positive or negative.
200 + 200 400 4
4 5
p=
=
= and q = 1 – =
500 + 400 900 9
9 9
S. E. ( p1 – p2 ) =
=
4 5 1
1 
× 
+


9 9 500 400
20
9
×
=
81 2000
0.0011 = 0.003
p1 – p 2 = 0.4 – 0.5 = 0.1
Difference
0.1
=
= 3.03
S. E.
0.033
Since the difference is more than 2.58 S.E. at 1% level of significance our hypothesis
is rejected. Hence there is significance difference between Maharashtra and Gujarat
so far as the proportion of vegetable oil consumers is concerned.
Ex.9 : Before an increase in excise duty on coffee 800 people out of a sample of
1000 persons were found to be coffee drinkers. After an increase in the duty, 800
persons were found to be coffee drinkers in a sample of 1200 people. Do you think
that there has been a significant decrease in the consumption of coffee after the
increase in the excise duty ?
Sol: Let us take the hypothesis that there has been no significant decrease in the
consumption of coffee after increase in the excise duty.
Computing the standard error of the difference of proportions
S. E. ( p1 – p2 ) =
1 1
pq 

 n1 n 2 
800
400
= 0.80; p 2 =
= 0.667
1000
1200
x + x2
800 + 800
8
p= 1
=
=
n1 + n 2 1000 + 1200 11
p1 =
q = 1–
S. E. ( p1 – p2 ) =
8
3
= .
11 11
8
3  1
1 
×
+

 = 0.027

11 11 500 600
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Engineering Mathematics-III
p1 – p 2
0.8 – 0.667
=
= 4.93
S. E .
0.027
Since the difference is more than 2.58 S.E. (at 1% level of significance) it could not
have arisen due to fluctuations of sampling.Hence there is a significant decrease in
the consumption of coffee after the increase in excise duty.
Ex. 10 : A machine produced 20 defective articles in a batch of 400. After overt
hauling, it produced 10 defective in a batch of 300. Has the machine improved ?
Sol : Let us take the hypothesis that the machine has not improved. Symbolically,
H 0 : p1 = p 2
20
i. e.,
p1 =
= 0.05
400
Proportion of defective articles after overhauling
10
i. e.,
p2 =
= 0.033
300
p1 – p 2 = 0.4 – 0.5 = –0.1. However, the conclusion remains the same irrespective
of the fact whether the difference is positive or negative.
Pooled estimate of actual proportion in the population is given by :
x + x2
20 + 10
3
p= 1
=
=
n1 + n 2
400 + 300 70
q = 1– p = 1–
S. E.( p1 – p2 ) =
3
67
=
70 70
1
1
pq
+
 =
 n1 n 2 
3 67  1
1 
×
+

 = 0.0155
70 70  400 300
Difference p1 – p 2
0.05 – 0.033
=
=
= 1.1
S. E.
0.0155
0.0
Since the difference is less than 1.96 S.E. (at 5% level) our hypothesis is true, i. e.,
machine has not improved significantly.
Ex.11 : Two groups A and B consist of 100 people each who have a disease. A serum
is given to group A but not to group B. It is found that is groups A and B, 75 and 65
people respectively recover from the disease. Test the hypothesis that the serum
helps to curve the disease.
(I.C.W.A.1984)
Sol : Let us take the hypothesis that there is no significane difference in the effect of
the serum, i. e., p1 = p 2
75
65
p1 =
= 0.75 p 2 ;
= 0.65
100
100
x + x2
75 + 65
140
p= 1
=
=
= 0.7
n1 + n 2 100 + 100 200
q = 1 – p = 0.3
S. E.( p1 – p2 ) =
1
1
pq 
+

 n1 n 2 
Sampling Theory
375
=
1 
 1
0.7 × 0.31
+
 = 0.065
 100 100
Difference 0.75 – 0.65
0.1
=
=
= 1.54
S. E.
0.065
0.065
Since the difference is less than 1.96 S.E., the hypothesis is accepted. Hence the
scrum does hot help in curing the disease. At times we may be interested in
comparing the proportion of persons possessing an attribute in a sample with the
proportion given by the population. In such a case the following formula is
applicable :
n2
S. E. ( p1 – p 2 ) = p 0 q 0 ×
n1 ( n1 + n 2 )
where
p 0 = Population proportion
q0 = 1 – p0
n1 = number of observation in the sample
n1 + n 2 = Size of population
Ex. 12:. There are 1000 students in a college. Out of 20000 in the whole university
In a study 200 were found smokers in the college and 1,000 in whole university. It
there a significant difference between the proportion of smokers in the college and
university ?
Sol : Let us take the hypothesis that there is no significant difference in the
proportion of smokers in the college.
200
=
= 0.20
1000
Proportion of smokers in the university
1000
=
= 0.05
20000
Difference between the two proportions
= 0.20 – 0.05 = 0.15
p 0 , i.e., proportion of smokers in the university = 0.05
q 0 = 1 – p 0 = 0.95, n1 + n 2 = 20000
n 2 = 20000 – 1000 = 19000
n2
S.E. of difference
= p0 q0 ×
n1 ( n1 + n 2 )
=
19000


0.05 × 0.95
 =
 1000 + 20000
0.000045 = 0.0067
Difference
0.15
=
= 22.39
S. E .
0.0067
Since the difference is more than 2.58 S.E. (1% level of significance), it could not
have arisen due to fluctuations of sampling. Hence there is a significant difference
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Engineering Mathematics-III
between the proportion of smokers in the college and university–the proportion
begin significantly higher in the college.
Ex. 13 :
(i)
In some dice-throwing experiment, Weldon threw dice 42,152 times, and of
these 25,145 yielded 4 or 5 or 6. Is this consistent with the hypothesis that the
dice were unbiased ?
(ii) A certain cubical die was thrown 9000 times and a 5 or 6 was obtained 3240 times.
On assumption of the random throwing do the data indicate an unbiased die ?
(Meerut 1983)
Sol : (i) The total number of throws = 49,152.
The chance of throwing 4 or 5 or 6 with one die =
∴
1
2
The expected value of the number of successes =
1
× 49152 = 24576
2
and the observed value of success = 25145.
Thus the excess of the observed value over the expected value
= 25145 – 24576 = 569.
The standard deviation of simple sampling
= √ ( npq) =
Hence
z=
1 1

 49152 × ×  = 110 ⋅ 9

2 2
x − np
569
=
= 5 ⋅ 13
√ ( npq) 110 ⋅ 9
Since the observed deviation is 5 ⋅ 13 times the standard error, it is therefore highly
improbable that it arose as a sampling fluctuation. We must therefore seek some
other reason for this deviation. Hence it seems reasonable to suspect dice were
biased.
(ii) Proceed as in (i).
1
Here expected number of successes = × 9000 = 3000
3
3240 − 3000
and
z=
= 16 ⋅ 9. So die is biased.
1 2
√ ( 9000) × ×
3 3
Ex. 14: A sample of 900 days is taken from meteorological records of a certain
district, and 100 of them are found to be foggy. What are the probable limits to the
percentage of foggy days in the district ?
100 1
8
Sol : Here
p=
= and q =
900 9
9
∴ The standard error of the proportion of foggy days in the district
Sampling Theory
377
 pq
  =
 n
1 
1 8
 × ×
 = 0 ⋅ 0105 = 1 ⋅ 05 percent.
 9 9 900
1
to be the estimate of the number of foggy days, we have that limits to
9
100

the proportion of foggy days in the district are 
± 3 × 1 ⋅ 05 percent, i.e. 8
 9

Hence taking
percent and 13 ⋅ 25 percent approximately.
Ex. 15: Certain crosses of pea gave 5321 yellow and 1804 green seeds. The
expectation is 25 percent of green seeds on a Mendelian hypothesis. Can the
divergence from the expected value have arisen from the fluctuation of simple
sampling only ?
Sol : The total number of pea seeds examined = 5321 + 1804 = 7125
1
The expected value of green seed = 25% of the total = × 7125 = 1781.
4
And the observed value = 1804 green seeds.
∴ The difference between the observed and the expected value
= 1804 − 1781 = 23.
Also the standard error of green seeds = ( npq) =
Hence
z=
1 3

7125 × ×  = 36 ⋅ 6

4 4
23
= ⋅ 6.
36 ⋅ 6
Since the divergence of the observed from the expected value is only ⋅6 times the
standard error, we may very well say that this divergence is due to fluctuations of
simple sampling.
Ex. 16: A random sample of 500 pineapples was taken from a large consignment
and 65 were found to be bad. Estimate the proportion of bad pineapples in the
consignment, as well as the standard error of the estimate. Deduce that the
percentages of bad pineapples in the consignment almost certainly lies between 8 ⋅ 5
and 17 ⋅ 5.
65
13
87
Sol : Here
.
p=
=
and q =
500 100
100
The standard error of the proportion of bad pineapples in the consignment
∴
=
87
1 
 13
×
×

 = ⋅ 015 = 1 ⋅ 5 percent.
 100 100 500
13
= ⋅ 13 to be the estimate of the number of bad pineapples in the
100
consignment we have that the proportion of bad pineapples in the consignment are
(13 ± 3 × 1 ⋅ 5) percent, i. e. 8 ⋅ 5 percent and 17 ⋅ 5 percent approximately.
Hence taking
378
Engineering Mathematics-III
Ex. 17: Show that for a random sample of 0 drawn with replacement, the standard
error of sample proportion cannot exceed ⋅05.
Sol : Since p + q = 1 (i. e. constant) the product pq will be maximum, when
1
p = q = . Hence the maximum value of S.E. of proportion
2
=
 pq
  =
 n
1 
1 1
 × ×
 = 0 ⋅ 05.
 2 2 100
Ex. 18: In breeding certain stocks, 408 hairy and 126 glabrous plants were
obtained. If the expectation is one-fourth glabrous, is the divergence significant, or
might it have occurred as a fluctuation of sampling ?
Sol : Total number of plants = 408 + 126 = 534.
1
Expected number of glabrous plants = × 534 = 133 ⋅ 5.
4
Observed number of glabrous plants = 126.
Difference between expected and observed number of glabrous plants
= 133 ⋅ 5 − 126 = 7 ⋅ 5.
1
3
Hence
p = , q = and n = 534.
4
4
Hence the standard error of the number of glabrous plants
=
∴
z=
1 3

 534 × ×  = 10 ⋅ 0 approx.

4 4
7⋅5
= ⋅ 75
10
which shows that the difference is very insignificant and might have occurred as a
fluctuation of sampling.
Ex. 19: Balls were drawn from a bag containing equal number of black and white
balls each ball being returned before drawing another. The records were then
grouped by counting the number of black balls in consecutive 2’s, 3’s, 4’s, 5’s. The
following are the distributions derived for grouping by 5’s, 6’s and 7’s. Compare
actual with theoretical means and standard deviations.
Sol :
Successes
(a) Grouping by
fives
(b) Grouping by
sixes
(c) Grouping by
sevens
0
30
17
9
1
125
65
34
2
227
166
104
3
224
192
151
4
136
1
148
Sampling Theory
379
5
27
69
95
6
–
8
40
7
–
–
4
Total
819
683
585
Since there are equal number of black and white balls, the chance of drawing a black
1
ball is , so that we have
2
1
1
p = and q = .
2
2
5
1 1
Hence the theoretical distribution for group (a) is given by 819  +  .
 2 2
1
∴ Therefore mean = np = 5 × = 2 ⋅ 5
2
and theoretical standard deviation
=
1 1

 5 × ×  = 1 ⋅ 118.

2 2
Actual mean and standard deviation are found as follows :
f
d
fd
d2
fd 2
0
30
–2
– 60
4
120
1
125
–1
– 125
1
125
2
277
0
0
0
0
3
224
1
224
1
224
4
136
2
272
4
544
5
27
3
81
9
243
Total
819
392
Assumed origin
A = 2.
∴
M = A+
1256
1
392
Σ fd = 2 +
= 2 + ⋅ 48 = 2 ⋅ 48,
N
819
2
1
 Σ fd 
σ =  Σ fd 2 − 
 
 N  
 N

1
2
=
1256
2
 819 − (⋅48)  = 1 ⋅ 14.
Similarly theoretical and actual means and standard deviations for group (b) and
(c) can be found. This is left as an exercise for the students.
For group (b), Theoretically M = 3, σ = 1 ⋅ 225
Actually
M = 2 ⋅ 97, σ = 1 ⋅ 26,
380
Engineering Mathematics-III
For group (c), Theoretically M = 3 ⋅ 5, σ = 1 ⋅ 323,
and
Actually
M = 3 ⋅ 57, σ = 1 ⋅ 40.
Ex. 20: Out of 200 individuals 40 percent show a certain trait, and that the number
expected on a certain theory is 50 percent, find whether the number observed differs
significantly from expectation ?
1
1 1
Sol : Here
p = ⋅ 5 = and q = 1 − = , n = 200
2
2 2
The standard error of p
p=
z=
 pq
  =
 n
1 
1 1
 × ×
 = ⋅ 035
 2 2 200
⋅5 − ⋅ 4
= 2⋅ 8
⋅035
Hence the difference is significant.
Ex. 21: Experience has shown that 20 percent of a manufactured product is of top
quality. In one day’s production of 400 articles, there were only 50 of the top quality.
Does this contradict our hypothesis of 20 percent ? [You are given that if X is
normal with mean µ and variance σ 2 , then P {| x − µ | ≥ 1 ⋅ 96 σ } = 0 ⋅ 05.
Sol : Here
P=
20
1
1 4
50
1
= , q = 1 − = ; observed proportions =
= .
100 5
5 5
400 8
Difference between the expected and observed proportions
1 1
3
= − =
= 0 ⋅ 075.
5 8 40
Standard error of
p=
∴
z=
 pq
  =
 n
1 
1 4
 × ×
 = ⋅ 02.
 5 5 400
⋅075
= 3 ⋅ 75 > 1 ⋅ 96.
⋅02
Here the difference is significant and the hypothesis of 20 percent is contradicted.
Ex. 22: In a locality containing 18000 families, a sample of 840 families was
selected at random. Of these 840 families, 206 families were found to have a
monthly income of Rs. 50 or less. It is desired to estimate how many out of the 18000
families have a monthly income of Rs. 50 or less. Within what limits would you place
your estimate ?
206 103
317
Sol : Here
.
p=
=
and q =
840 420
420
∴The standard error of the proportion of families having a monthly income of Rs. 50
or less
=
 pq
  =
 n
1 
 103 317
×
×

 = ⋅ 015 = 1 ⋅ 5 percent.
 420 420 840
Sampling Theory
381
103
(or 24 ⋅ 5%) to be the estimate of families having a monthly
420
income of Rs. 50 or less in the locality, we have that the limits are
( 24 ⋅ 5 ± 3 × 1 ⋅ 5) percent i. e. 20 percent and 29 percent approximately.
Ex. 23: A dealer takes 100 samples from a consignment of 10000 items of a certain
goods and finds that there are 50 items of grade I worth Rs. 5 per thousand, 30 items
of grade II worth Rs. 4 per thousand and twenty items of grade. III worth Rs. 3 per
thousand. Within what limits should the value of the consignment be fixed ?
Sol : For items of grade I, we have
50
1
1 1
p=
= = 0 ⋅ 5 and q = 1 − = = 0 ⋅ 5.
100 2
2 2
Hence taking
For items of grade II, we have
30
p=
= 0 ⋅ 3 and q = 0 ⋅ 7,
100
And for items of grade III, we get
p = 20 / 100 = 0 ⋅ 2 and q = 0 ⋅ 8.
Total number of items in the sample = 100.
Hence the standard error of the simple sampling is
for grade
I = √ ( 0 ⋅ 5 × 0 ⋅ 5 × 100) = 5 ⋅ 0,
for grade
II = √ ( 0 ⋅ 3 × 0 ⋅ 7 × 100) = 4 ⋅ 6,
and for grade
III = √ ( 0 ⋅ 2 × 0 ⋅ 8 × 100) = 4 ⋅ 0.
Hence we have the lower and upper limits for the percentages of three grades as
follows :
Grade
Limits
I
Lower
50 – 3 × 5 = 35%
Upper
50 + 3 × 5 = 65%
II
30 – 3 × 4.6 = 16.2%
III
20 – 3 × 4 = 8%
30 + 3 × 4.6 = 43.8% 20 + 3 × 4 = 32%
Now the highest value that can be placed upon the consignment is the value for
which grade I is the highest and grade III is the lowest, so that we get
grade I = 65%
and
grade III = 8%.
The
grade II = {100 − ( 65 + 8)} = 27%
Hence the highest value of the consignment
= 65% of Rs. 50 + 27 % of Rs. 40 + 8 % of Rs. 30
= Rs. 32 ⋅ 5 + Rs. 10 ⋅ 8 + Rs. 2 ⋅ 4 = Rs. 45 ⋅ 7.
Similarly, the lowest value that can be given to the consignment is that value for
which grade I is the lowest and grade III the highest i. e.
Grade I = 35%
382
Engineering Mathematics-III
Grade III = 32%
so that grade
II = {100 − ( 35 + 32)} = 33%.
Hence the least value of the consignment
= 35% of Rs. 50 + 33% of 40 + 32% of Rs. 30
= Rs. 17 ⋅ 5 + Rs. 13 ⋅ 2 + Rs. 9 ⋅ 6 = Rs. 40 ⋅ 3.
Thus the value of the consignment almost certainly lies within the limits of Rs.
40 ⋅ 3 and 45 ⋅ 7.
9.9 Comparison of Large Samples
Two samples from distinct materials or different population give proportions of A' s,
as p1 and p 2 , the numbers of observations in the samples being n1 and n 2
respectively.
Two cases arise :
(a)
If the two populations are really similar as regards the proportions of A' s, can
the difference between the two proportions have arisen merely as a
fluctuation of simple sampling? In this case, we have no theoretical
expectation as to the proportion of A' s in the populations from which either
sample has been drawn. The best guide in such cases would be to take the
mean proportion in the two samples together for the proportion of A' s in the
populations. Hence the proportion of A' s in the population can be given by
p n + p2 n 2
p0 = 1 1
n1 + n 2
Let E1 and E 2 be the standard errors in the two samples; then
p q
p q
E12 = 0 0 and E 22 = 0 0 .
n1
n2
If E be the standard error of the difference between p1 and p 2 , then
1
1
E 2 = E12 + E 22 = p 0 q 0 
+

n
n
 1
2
...(9.1)
p1 ~ p 2
.
E
If z > 3, the difference between p1 and p 2 is real one and is not merely due to
fluctuations of simple sampling. If z < 3, the difference may be due to fluctuations of
simple sampling.
(b)
If the proportions of A' s are not the same in the two populations from which
the samples are drawn, but p1 and p 2 are the true values of the proportions,
the standard error E of the difference in this case is given by
p q
p q
E2 = 1 1 + 2 2
n1
n2
z=
p1 ~ p 2
< 3, the difference might have arisen due to fluctuations of simple
E
sampling only and may vanish on taking fresh samples in the same way from the
same material.
If z =
Sampling Theory
383
Ex. 24: In a large city A, 20 percent of a random sample of 900 school boys had a
certain slight physical defect. In another large city B,18 ⋅ 5 percent of a random
sample of 1,600 school boys had the same defect. Is the difference between the
proportions significant ?
Sol :
20
1
18 ⋅ 5
37
Here
p1 =
= , n1 = 900 and p 2 =
=
, n 2 = 1600.
100 5
100
200
n p + n 2 p2
180 + 296
Hence
p0 = 1 1
=
= ⋅ 19.
n1 + n 2
900 + 1000
q 0 = 1 − ⋅ 19 = ⋅ 81.
1
1
1 
 1
E 2 = p0 q0 
+
+
 = ⋅ 19 × ⋅ 81 
 = ⋅ 0017.

n2 
900 1600
 n1
Now
∴
Again
Then
E = ⋅ 04 approximately.
1⋅ 5
p1 − p 2 =
= ⋅ 015.
100
p − p 2 ⋅015
z= 1
=
= ⋅ 37.
E
⋅04
Since, z < 1, the difference between the proportions is not significant and might
vanish on taking fresh samples.
Ex. 25: In a random sample of 500 persons from town A, 200 are found to be
consumers of cheese. In a sample of 400 from town B, 200 are found to be
consumers of cheese. Discuss the question whether the data reveal significant
difference between A and B, so far as the proportion of cheese consumers is
concerned.
Sol :
200 2
200 1
Here
p1 =
= = ⋅ 4; p 2 =
= = ⋅ 5,
500 5
400 2
p n + p2 n 2
200 + 200 4
4 5
so that
p0 = 1 1
=
= and q 0 = 1 − = .
n1 + n 2
500 + 400 9
9 9
Hence
This gives
Now
E 02 =
4 5 1
1 
× 
+
 = ⋅ 001111.
9 9  500 400
E = ⋅ 033.
p − p 2 ⋅4 − ⋅ 5
z= 1
=
= − 3 ⋅ 03
E
⋅033
Since |z| > 3, the difference is significant.
Ex. 26: If for one half of n events, the chance of success is p and chance of failure is
q, whilst for the other half, the chance of success is q and the chance of failure is p.
Show that the standard deviation of the number of successes is the same as if the
chance of successes were p in all the cases i.e. √ ( npq) but the mean of the number of
successes is n/2 and not np.
384
Engineering Mathematics-III
Sol : Let the number of successes in first and second half be denoted by x and y
respectively.
np
nq
E ( x) =
, E (y) =
2
2
1
1
var ( x ) = npq, var ( y ) = npq
2
2
n
1
E ( x + y ) = E ( x ) + E ( y ) = ( q + p) = n
2
2
1
1
var ( x + y ) = var ( x ) + var ( y ) = npq + npq = npq.
2
2
Ex. 27: In two large populations there are 30 and 25 percent respectively of
fair-haired people. Is this difference likely to be hidden in samples of 1200 and 900
respectively from the two populations ?
30
25
Sol : Here
p1 =
= ⋅ 30, p 2 =
= ⋅ 25
100
100
so that
p1 − p 2 = ⋅ 05.
p q
p q
⋅30 × ⋅ 70 ⋅25 × ⋅ 75
.
E2 = 1 1 + 2 2 =
+
n1
n2
1200
900
This given on calculation, E = ⋅ 0195.
p − p2
⋅05
∴
z= 1
=
= 2 ⋅ 56 nearly
E
⋅0195
Hence it is unlikely that real difference will be hidden.
Ex. 28: In a random sample of 500 men from a particular district of U.P., 300 are
found to be smokers. In one of 1000 men from another district, 550 are smokers. Do
the data indicate that the two districts are significantly different with respect to the
prevalence of smoking among men ?
300 3
550
11
Sol : Here
.
p1 =
= , p2 =
=
500 5
1000 20
p n + p2 n 2
300 + 550 17
,
p0 = 1 1
=
=
n1 + n 2
500 + 1000 30
q0 = 1 −
E =
and
∴
17 13
,
=
30 30
17 13
 30 × 30

1 
 1
+

 = ⋅ 0271,
 500 1000 
p1 − p 2 = ⋅ 6 − ⋅ 55 = ⋅ 05,
p − p2
⋅05
z= 1
=
= 1 ⋅ 9 approx.
E
⋅0271
Hence the difference is not significant i. e. the data do not indicate that the two
districts are significantly different with respect to the prevalence of smoking among
men.
Sampling Theory
385
Ex. 29: The subject under investigation is the measure of dependence in Tamil on
words of Sanskrit origin. One newspaper article reporting the proceeding of the
constituent Assembly contained 2,025 words of which 729 words were declared by a
literary critic to be of Sanskrit origin. A second article by the same author describing
atomic research contained 1,600 words of which 640 words were declared by the
same critic to be of Sanskrit origin. Assuming that simple sampling conditions held,
estimate the limits for the proportion of Sanskrit terms in the writer’s vocabulary
and examine whether there is any significant difference in the independence of this
writer on words of Sanskrit origin in writing on the two subjects.
Sol : If p 0 denote the proportion of Sanskrit terms in the writer’s vocabulary in both
the articles taken together, then
729 + 640
p0 =
= ⋅ 3777 = 37 ⋅ 77%
2025 + 1600
and
q 0 = ⋅ 6223 = 62 ⋅ 23%
Now the proportion of Sanskrit terms in the first article =
proportion in the second article =
729
= ⋅ 36 = 36% and the
2025
640
= ⋅ 40 = 40%.
1000
The difference = 40 − 36 = 4%.
The standard error of the difference between these two proportions is given by
E =

1
1
+

 p0 q0 
n2  
 n1

1 
 1
= √ ( 37 ⋅ 77 × 62 ⋅ 23) 
+
 percent
 2025 1600
∴
= 1 ⋅ 16 percent.
4
z=
= 3 ⋅ 4 > 3.
1 ⋅ 16
And so the difference is a real one and could not have arisen from fluctuations of
simple sampling. Hence there is significant difference in the dependence of the
writer on words of Sanskrit origin in writing on the two given articles.
To estimate the limits for the proportion of Sanskrit terms in the writer’s vocabulary,
we first find the standard error of the proportion of Sanskrit terms in the writer’s
vocabulary. This is given by
σ=
=
 p0 q0 


 n 
where n = 2025 + 1600 = 3625
 37 ⋅ 77 × 62 ⋅ 23

 = ⋅ 81 percent.


3625
Hence taking p 0 = 37 ⋅ 77 percent to be the estimate to the Sanskrit words in the
writer’s vocabulary, we have that the limits are 37 ⋅ 77 ± 3 × ⋅ 81 percent i. e.
35 ⋅ 44% and 40 ⋅ 20% approximately.
386
Engineering Mathematics-III
Ex. 30: If for one half of n events the chance of success is p and the chance of the
failure is q, whilst for the other half the chance of success is q and the chance of
failure is p. Show that the standard deviation of the number of success is the same as
if the chance of success were p in all the cases i.e. √ ( npq) but that the mean of the
number of success is n / 2 and not np.
Sol : Let σ1 and σ 2 denote the standard deviation of first and second halves of n
events.
1
1
Thus,
σ12 = npq and σ 22 = nqp.
2
2
Hence, the standard deviation of the number of successes
= √ (σ12 + σ 22 ) =
n 
n
 pq + qp = √ ( npq).
2
2 
If p 0 denotes the proportion of success in the n-events, then
n
n
p + ⋅q
1
1
2
2
p0 =
= ( p + q) = .
n n
2
2
+
2 2
1
Hence the mean of the number of success = np 0 = n.
2
Ex. 31: In a certain association table, the following frequencies were obtained :
(AB) = 309, (Aβ ) = 214, ( αB) = 132, ( αβ ) = 119.
Can the association of the table have arisen as a fluctuation of simple sampling, the
true association being zero ?
Sol : We have
( A ) = ( AB) + ( Aβ ) = 309 + 214 = 523,
( B) = ( AB) + (αB) = 309 + 132 = 441 = n1
(β ) = ( Aβ ) + (αβ ) = 214 + 119 = 333 = n 2 ,
N = ( B) + (β ) = 441 + 333 = 774.
Proportion of A' s in
B' s =
( Aβ ) 309
=
= ⋅ 701.
( B)
441
A' s in β' s =
( AB) 214
=
= ⋅ 643.
(β )
333
Proportion of
The difference of the proportions = ⋅ 701 − ⋅ 643 = ⋅ 058.
( A ) 523
The proportion of A' s in the universe =
=
= ⋅ 676 = p 0 , say.
N
774
Then
q 0 = 1 − ⋅ 676 = ⋅ 324.
Hence, the standard error of the difference between the two proportions is given by
1
1
1 
 1
E 2 = p0 q0 
+
+
 = ⋅ 676 × ⋅ 324 
,
 441 333
n2 
 n1
Sampling Theory
whence
387
E 2 = ⋅ 034
z=
difference ⋅358
=
= 1 ⋅ 7 < 3.
E
⋅034
Hence, the association between A and B is not a real one and might have arisen as a
fluctuation of sampling.
Ex. 32: If a series of random samples of different size is taken from the same
material, show that the standard deviation of the observed proportion of successes
pq
in such sets is s, where s 2 =
and H is the harmonic mean of the numbers in the
H
samples.
Let there be f1 samples of n1 individuals each, f 2 samples of n 2 individuals each,
f 3 of n 3 and so on. Let p be the chance of success and q that of failure. The variance
of the observed proportion of success in f1 samples on n1 individuals each
pq f1
pq pq
etc.
=
+
+ … to f1 terms
n1
n1
n1
Hence the standard deviation s of the observed proportion of successes in all the sets
is given by
f
f
f

Ns 2 = pq  1 + 2 + 3 + …
n2
n3
 n1

..(9.1)
where N is the total number of samples.
Now the harmonic mean H of the numbers in the sample is given by
f
f

1
1  f1
=
+ 2 + 3 + ….

H N  n1
n2
n3

pq
Then (9.1) gives s 2 =
, as required.
H
9.10 The Sampling of Variables (Large Samples)
Already, we have In sampling of attributes, discussed the sampling of attributes. We
classified each member of a sample under one of two heads, success or failure. In the
case of sampling of variable such a classification is no longer possible. Each
individual member of the sample provides a value of the variate and these values are
generally spread over a range, which may be limited or unlimited. It is often
convenient to think of the variate as always having an infinite range. When the
range is actually finite, we may take the frequency to be zero outside the range. The
number of possible values of a continuous variable is infinite in any finite range,
however small the interval may be. In other words, the population of variate value
will always be infinite. Hence the drawing of a finite random sample does not affect
the drawing of any other sample from the population of variate values and
consequently the sampling is always simple. The examples for the sampling of
variables are provided by statures of men, ages of persons at death, prices of a
commodity etc.
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Engineering Mathematics-III
The aims of the study of the sampling of variables are the same as those of the
sampling of attributes. These are
(a) To compare the observed value with the expected value and to find how far the
difference between the two values can be attributed to fluctuations of simple
sampling.
(b) To estimate the parameters of the parent population from the sample, such as
means of a variate.
(c) To see how reliable our estimates are when they are obtained. The problems set
forth in (b) and (c) above are called the problems of estimation and the problem
of testing hypothesis.
9.10.1 Statistical Estimation
It is the technique of estimating the parameters of the population from that of
sample. Thus in order to test the amount of dust in a bag of grain a sample of the
grain is taken carefully from the middle portion of the bag, weighed, and after
cleaning it is weighed again so that the amount of dust in the sample is found out.
The weight of dust in the sample multiplied by the ratio of the weight of grain in the
bag to the weight of the sample gives an approximation to the weight of dust in the
whole bag.
9.10.2 Testing a Hypothesis
By testing a carefully drawn sample, it is possible to verify a hypothesis regarding
the population. Thus, to test the efficiancy of a drug against a disease, we can find
from a sample, the number of persons attacked even after using the drug and the
number attacked who have not used that drug. A similar example is that of a rope
manufacturer who would like to adopt a new process if the strength of the ropes is
increased. From past-experience he knows the breaking strength of the ropes in a
normal population with mean 100 lbs wt. He will thus want to test the hypothesis
that the new process gives ropes with breaking strength distributed according to
normal law with a mean more than 100 lbs. weight. If this mean is less that 100 lbs.,
the hypothesis is rejected.
9.10.3 Errors in testing hypothesis
The first type of error in testing a hypothesis is that a correct hypothesis is rejected
while the second type of error lies in the fact of accepting a wrong hypothesis. The
statistical testing of hypothesis aims at limiting the risk of the first type of error to a
pre-assigned value, say 1% or 5% and to minimize the second type of error.
9.10.4 Null hypothesis
Suppose we are given a sample from which a certain statistic such as mean is
calculated. We assumed that this sample is drawn from a population of known form
for which the corresponding parameter is tentatively specified. We call this tentative
specification a null hypothesis. For example, in a coin-tossing experiment, we want
to test whether the coin is biased. Therefore the null hypothesis in this case is that
1
the coin is unbiased, i.e. p = , where p is the probability of a head (or tail). If our
2
Sampling Theory
389
experiment gives a value of the statistics which deviates significantly from the value
of the parameter (i.e. 1 / 2), the null hypothesis is contradicted and we conclude that
the coin is biased. If, however, this deviation is not significant, the hypothesis is
accepted and the deviation may be attributed to sampling fluctuations. As another
example, in accepting a consignment of jute bags the user wants to know whether
the average warp strength of the batch is 100 lbs. The suitable null hypothesis is that
the mean batch warp strength is 100 lbs. and a sampling experiment will decide
whether the hypothesis is to be rejected. Again in sampling a certain drug for
immunization against a disease, the null hypothesis should be that the action of the
drug and attack of the disease are independent. Neither is it assumed that the drug
prevents the discease nor that it has bad effects and accelerates the spread of disease.
Thus a null hypothesis is a hypothesis which is tested for possible rejection under the
assumption that it is true.
By accepting a null hypothesis, we do not mean that it is proved to be true. This only
implies that on the basis of the statistic calculated from the sample, we find no
reason to question the validity of the hypothesis. The only way in which hypothesis
can be accepted with certainty is for us to know the population parameter which we
do not have.
9.10.5 Parameter and Statistics
In order to avoid verbal confusion with the statistical constants of the population,
viz., mean (µ ), variance (σ 2 ), etc., which are usually referred to as parameters,
statistical measures computed from the sample observations alone, e. g ., mean ( x ),
variance ( s 2 ) etc., have been termed by Professor R.A. Fisher as statistics. In
practice, parameter values are not known and the estimates based on the sample
values are generally used. Thus statistic which may be regarded as an estimate of
parameter, obtained from the sample, is a function of the sample values only. It may
be pointed out that a statistic, as it is based on sample values and as there are
multiple choices of the samples that can be drawn from a population, varies from
sample to sample. The determination or the characterisation of the variation (in the
values of the statistics obtained from different samples) that may be attributed to
chance or fluctuations of sampling is one of the fundamental problems of the
sampling theory.
9.11 Sampling Distribution of Large Sample
If a large number of samples are taken from a population and some statistics such as
mean, median, S.D. etc. of these samples are calculated for each sample, they will
form a frequency distribution. Thus the distribution given by the means of these
samples gives sampling distribution of the means, that given by the medians the
sampling distribution of medians and one given by standard deviations gives the
sampling distribution of the S.D.
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Engineering Mathematics-III
9.11.1 Utility of Sampling Distribution
Let the sampling distribution of a statistic be represented by the continuous curve
y = f ( x ), where x is the variate (statistic) and y is the corresponding frequency.
The total frequency of all the samples which give a value of x greater than a given
value x 0 will be given by the area to the right of the ordinate at x 0 . It follows that the
probability, that a sample chosen at random from all possible samples will give a
value of x greater than x 0 , is given by the area to the right of the ordinate at x 0
divided by the total area of the curve.
If P denotes this probability, we have
P=
∞
∫x
f ( x ) dx ∴
0
∞
∫− ∞
f ( x ) dx.
Similarly, the chance that a sample would give a value of x lying between x1 and x 2
is given by
P=
x2
∫x
f ( x ) dx ∴
1
∞
∫− ∞
f ( x ) dx
If the units are so chosen that the total area under the curve is unity, these
probabilities are given by
x1 =
∞
∫x
0
f ( x ) dx and
∫x
x2
f ( x ) dx respectively.
1
Now if we take a sample and find that it gives a very low value of P we are faced with
three possibilities.
(i) The hypothesis is not correct.
(ii) The sampling is not simple.
(iii) Some improbable event has happened.
Generally we are led to suspect our hypothesis provided we have tested our
sampling technique or on other grounds have no reason to suspect it; but it is always
a matter of choice, which of the above three explanations we should adopt under the
given circumstances.
9.11.2 Standard Error
In § 9.11 we have defined sampling distribution of a statistic as the distribution
given by the statistics of a large number of samples. The standard deviation of the
sampling distribution is known as Standard Error. Thus the standard deviation of
the sampling distribution of means is called standard error of means, the S.D. of
sampling distribution of medians as S.E. of medians etc. If the sampling size is large,
the sampling distribution is assumed to be normal and we are justified that the
parameter given by the sample lies within M ± 3 S.E.
Thus the standard error is used to gauge the precision of an estimate and to pass
judgements on the divergence between expected and observed values. Hence it is
necessary to know the standard error of various parameters which we have to
estimate. The most important of all the standard errors is that of the mean and we
proceed to find it in the following sections. The standard errors of various statistics
are given in § 9.11.
Sampling Theory
391
9.11.3 Mean and standard error of the sampling distribution of
the means of samples
Theroem : The mean value of the means of all positive random samples of size n
σ
from a population is the mean of the population and the standard error is
where
√n
σ is the standard deviation of the population.
9.11.4 Distribution of the sample Mean (Simple sampling)
Suppose from a population of whatever type, we draw a number of samples each
containing n variates, essentially under constant conditions so that each draw is
independent of the others. This is possible only when either the population is infinite
or sampling is by replacement. Then such samples are called random samples
with sample size n. Let x1 , x 2 , .....be the means of these samples. These means will
be randomly distributed variates, and it is required to be proved that
σ
E ( x ) = µ; σ x =
√n
where µ and σ are the mean and standard deviation respectively of the population.
Thus by knowing the means of a number of random samples, it is possible to
estimate the mean and variance of the population.
Let x be mean of sample of size n having variates x1 , x 2 ...., x n . If we take a number
of random samples each of size n, we can show that the expected value of these
means is the mean of the population µ. Now x is a statistic, it has a probability
distribution. Hence
 x + x2 + … + xn  1
E (x)= E  1
 = E ( x1 + x 2 + … + x n )

 n
n
1
{ E ( x1 ) + E ( x 2 ) + .... + E ( x n )} ( § 7 ⋅ 13)
n
Since the sample is random, it is fair to assume that
=
E ( x1 ) = E ( x 2 ) = … = E ( x n ) = µ
i.e. the expected value of any item in the sample is the same as in the population.
Hence
E ( x ) = µ.
We can therefore take the mean of the means of a large number of samples to be the
estimate of the mean of the population. We express it by saying that x is an unbiased
estimate of µ. Also variance of x is
 x + x2 + … + xn 
σ 2x = var  1



n
1
var ( x1 + x 2 + … + x n )
n2
Since the variance of sum of independent random variables is the sum of the
variances of the variables, we have
1
var ( x ) =
{var ( x1 ) + var ( x 2 ) + .... + var ( x n )}.( § 10 ⋅ 8)
n2
=
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Engineering Mathematics-III
Again since the sample is random, we can assume that
var ( x1 ) = var ( x 2 ) = … = var ( x n ) = var ( x ) = σ 2
∴
var ( x ) =
σ2
n
where σ 2 is the variance of probability distribution being sampled.
σ
Hence, σ x =
which is the standard error of the mean.
√n
It is clear that as n increases, the distribution of x is more concentrated about the
mean µ.
The above was the case when each draw was independent of other or the sampling
was simple or sampling was with replacement. In case the sampling is without
replacement, then,
 N − n σ 2
σ 2x = 

 N − 1 n
where N is the size of the population and n that of the sample with n ≤ N .
Note : It is interesting to note here that there is no restriction upon the distribution
of the population and nothing has been said about the distribution of the means of
the samples except finding a relation between the means and standard deviation of
the means of the samples and those of the parent population. However if the
population is normal, then it has been shown that :
The mean of the samples of size n from a normal population (µ, σ ) is
itself normally distributed about µ as mean and standard deviation σ
σ
or standard error
so that the sampling distribution of the means of random
√n
samples each of size n from a normal population with mean µ and S.D. σ is normal
σ 
i.e. N µ,

 √ n
or
y =
2
√n
 n ( x − µ ) 
exp  −

σ √ ( 2π )


2σ 2
There is no restriction on the size of the sample in this case.
Thus the standard deviation of the means of the samples of size n varies inversely as
the square root of the sample size. The larger the sample size, the more closely to the
values of the sample mean shall cluster to the mean of the population.
When the population is not normal, A modified form of the central limit theorem is :
Let X1 , X 2 , … , X n be independent random variables with the same distribution,
mean µ and S.D. σ ≠ 0, then the random variable
X + X2 + … + Xn
Zn = 1
n
σ
is normally distributed with mean µ and S.D.
as n→∞,
√n
Sampling Theory
393
Hence, we see that if a random sample of large size is drawn from a population (not
necessarily normal) of mean µ and variance σ 2 ; then
Lt
n→ ∞
σ 

x ∼ N µ,

 √ n
i.e. the distribution of sample mean (in large samples) is normal with mean µ and
σ
S.D.
√n
Thus, if the population is normal, the sampling distribution of the mean is normal
σ
with mean µ and S.D.
(irrespective of sample size), while for large sample the
√n
same result holds irrespective of the distribution of the population.
9.12 Distribution of the Difference Between Two Sample Means
Let there be two populations with mean m x and m y and standard deviation σ x and
σ y respectively. Further let N be the pairs of random sample that are formed from
each of two populations each having n1 variates from the first and n 2 from the
second population. Let the means of these samples be x1, x 2 , …, x n from the first
population and y1 , y 2 , ...., y n from the second. Then consider the difference of the
means i.e. x1 − y1 , x 2 − y 2 , …, x n − y n .
Then
Also
E (x − y )=
1 N
Σ ( xi − y i ) = E ( x ) − E ( y ) = m x − m y .
N i =1
var ( x − y ) = var ( x ) + var ( y ) (§ 10 ⋅ 8)
=
2
σ 2x σ y
+
n1
n2
as the two samples are independent and
σy
σ
σx = x , σy =
√ n1
√ n2
Hence the standard error of the difference between the means of samples from
two independent populations is √ (σ 2x + σ 2y ) where σ x and σ y are the strandard
errors of the means of two samples.
Note. In practical problems of statistics, we do not know the mean and S.D. of the
whole population. We have to estimate these values from samples. If the sample is a
fairly large one, we have to assume that its S.D. is the same as that of the whole
population.
9.13 Levels of Significance
As we have already seen that the probability of a variate in normal distribution lying
outside Mean ± 3σ is ⋅3 percent or only once in 300 trials which is a very small quantity.
Thus if a manufacturer with past experience knows that a normal worker in the factory
produces on the average 400 pieces per day with standard deviation 10 and by a new
process 360 pieces only are manufactured in a day we feel that the new process is highly
394
Engineering Mathematics-III
unsatisfactory, since this value deviates from the mean by 40 i.e. 4σ. The region in
which a sample point falling is rejected is known as the critical region or region of
rejection. Generally, we take two critical regions, which cover 5% and 1% areas of the
normal curve. When a hypothesis is rejected with
the variate deviating from the mean by more than
1 ⋅ 96σ on either side, it is known as a 5 percent
level of significance and it is rejected if lying
outside 2 ⋅ 58σ, it is 1 percent level of significance.
The probability of the value of the variate falling in
2.5%
2.5%
the critical region is known as the level of
M
significance. If the level of significance is taken at 5
–1.96 σ
1.96 σ
percent, 2 ⋅ 5 percent shall lie on the right hand side
Fig. 9.2
of x = 1 ⋅ 96σ and 2 ⋅ 5 percent on the left hand side
of x = − 1 ⋅ 96 σ (See the figure).
The choice of the level of significance will depend upon the nature of the problem and
is a matter of judgement for those who carry out experiment. Their judgement should
naturally be guided by the degree of confidence they have in the null hypothesis. If
they have the firm belief that the null hypothesis must be true, it will require very
improbable result before they can reject the hypothesis. On the other hand, if they
have no very strong feeling about the validity of the hypothesis, they may reject it on a
less improbable result. For example, suppose in a coin-tossing experiment the null
hypothesis is that the coin is unbiased, the observed result might have to have a
probability of 0 ⋅ 1 per cent, or even less, before he can reject the hypothesis of no bias.
If the experimenter is an expert to detect the property of bias in coins, and feels that
particular coin is biased he would reject the hypothesis if the observed result only
had a probability of 5 percent or even 10 per cent.
9.13.1 Nature of the double tail and single tail tests
It generally depends on the nature of the problem
whether we have to use a double tail or single tail
test in gauging the significance of a result. In the two
tail test, we take into consideration the areas of both
the tails of the curve represented by the sampling
distribution whereas in a single tail test the areas on
the right of an ordinate say at x = x 0 , is taken into
–X0
O
X0
account. For example, in a coin tossing experiment,
double tail test should be used to test whether a coin
Fig. 9.3
is biased, since a coin will be biased if either it gives
significantly more number of heads than tails (this gives right tail only) or it gives more
number of tails than heads (this gives left tail only). Most of the tests are of this two tail
type. In some cases, however, a single tail is desirable. For example if the manufacturer of
electric bulbs claims that the mean life of his bulbs is 1000 hours, the customer’s anxiety is
only if the mean life is shorter than this period. His sales go down only if the test shows the
mean life to be smaller and not if it is greater than 1000. Thus the customer is interested
only in one-tailed test.
Sampling Theory
395
Underlying assumption. A very important assumption in carrying out the
interpretation of the results is that the sample averages are distributed normally.
Even when the distribution of individuals is not normal it is nearly always true that
average and to a lesser extent, standard deviations of random samples from the
population of individuals are approximately normally distributed if the sample size
is not too small. The proof of the above statement is beyond the scope of the book.
9.13.2 A Choice of Significance Level and Distribution
There is no universal rule for choice of significance level. While in same cases 5%
level is used, others prefer 1% level. The higher the level of significance for testing a
hypothesis, the greater is the risk of rejecting a null hypothesis when it is true.
Regarding choice of distribution for testing the hypothesis, the rule is as given below
for using the normal distribution (area table IV) and `t ' distribution (table VI).
Sample Size
Population S.D. is known
Population S.D. unknown
n > 30
Normal
Normal
n ≤ 30 Population being
assumed normal
Normal
t distribution
9.14 Means of the Samples
If a number of random samples of size n is taken, we have seen that
σ
,
X = µ and σ x =
√n
where x and σ x are the mean and standard deviations of the means of the sample. If
the population is normally distributed, we have also seen that the mean of the
samples is normally distributed with mean µ and standard deviation σ / √ n. If we
consider the variate
X −µ
,
z=
σ / √n
then z is a normal variate with mean as zero and S.D. unity. On the null
hypothesis that the mean of the means of the samples is equal to µ, the value
of z should be zero, but it is not always so. We have to see whether the
difference between the observation and hypothesis is significant or merely
due to fluctuations of sampling, i.e. whether the value of z fall inside the
critical region or not. As, already observed, if |z| > 1 ⋅ 96, the difference is
significant on 5 percent level of significance and |z| > 2 ⋅ 58, the difference is
significant on 1 percent level. In some case where the rejection of the
hypothesis may mean serious implications, sometimes 1 percent level of
significance is also adopted.
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Engineering Mathematics-III
9.15 Fiducial or Confidence Limits
Since z =
or
or
X −µ
√ n is a standard normal variate, we have
σ
P {|z| < 1 ⋅ 96} = 0 ⋅ 95
σ


P |X - µ| <
1 ⋅ 96
√n


1 ⋅ 96σ
1 ⋅ 96σ 

P X −
<µ< X +
 = 0 ⋅ 95
√n
√n 

The relation shows that in repeated sampling, the probability that the interval
1 ⋅ 96σ
1 ⋅ 96σ 

,X +
X −
 will include µ, is 0 ⋅ 95. This means that if a very large
√n
√n 

number of samples, each of size n, is taken from the population and if we determine
the above interval for each sample, then in about 95 percent of the cases, the interval
will include µ. While in the remaining 5 per cent cases it will not do so. It is possible
that a particular sample gives that limits between which the population mean does
not lie. What the above statement means, is that under the same conditions repeated
sample will produce 95 percent results for which µ will lie in the above interval. The
1 ⋅ 96σ
1 ⋅ 96σ
values X −
and X +
are called 95 percent fiducial or confidence limits
√n
√n
of µ corresponding to the observed sample. The quantity 0 ⋅ 95 is called the
confidence coefficient. Similarly 99 percent confidence limits for µ are
2 ⋅ 58σ
2 ⋅ 58σ
and X +
X −
√n
√n
list of the formulae: The following is a list of the formulae for obtaining standard
error for different statistics :
1. Standard Error of Mean
(i) When standard deviation of the population is known
σp
S. E. X =
n
where S.E . X refers to the standard error of the mean
σ p = Standard deviation of the population
n = Number of observations in the sample
(ii) When standard deviation of population is not known, we have to use standard
deviation of the sample in calculating standard error of mean. Consequently, the
formula for calculating standard error of mean. Consequently, the formula for
calculating standard error is
σ( sample)
S. E. X =
n
where σ denotes standard deviation of the sample.
It should be noted that if standard deviations of both sample as well as population
are available then we should prefer standard deviation of the population for
calculating standard error of mean.
Sampling Theory
397
95% fiducial limits of population mean are
σ
X ± 1.96
n
99% fiducial limits of population mean are
σ
X ± 2.58
n
σ
2.S.E. of Median or S.E. Med = 1.25331
n
σ
3. S.E. of Quartile or S. E. q = 1.36263
n
4. S.E. of Quartile Deviation or S. E. QD = 0.78672
5. S.E. of Mean Deviation or S. E. MD = 0.6028
6. S.E. of Standard Deviation or S. E. σ =
7. S.E. of Variance or S. E. σ 2 = σ 2
σ
n
σ
n
σ
2n
2
n
8. S.E. of Coefficient of Variation of S. E. v
V
2n
9. S.E. of Coefficient of Skewness or S. E. sk =
10. S.E. of Coefficient of Correlation of S. E. r =
1+
2V
10 4
3
2n
1 – r2
n
11. S.E. of Regression Coefficient of Y on X, i. e., S.E. b xy
=
σ y 1 – r2
σx n
12. S.E. of Regression Coefficient of X on Y, i. e., S.E. b xy
=
σ x 1 – r2
σy n
13. S.E. of Regression Estimate of y on X or S. E. b xy = σ x 1 – r 2
14. S.E. of Regression Estimate of X on Y or S. E. yz = σ 2 1 – r 2
15. S.E. pf Coefficient of Association, i. e., S. E.Q
1 – Q2
1
1
1
1
=
+
+
+
2
( AB) ( Aβ ) αB (αβ )
16. S.E. of Rank Correlation Coefficient, i. e., S. E. rk
1
=
n–1
398
Engineering Mathematics-III
The following examples will illustrate how standard error of some of the statistics is
calculated :
Ex.33 : Calculate standard error of mean from the following data showing the
amount paid by 100 firms in Calcutta on the occasion of Durga Puja :
Mid Value (Rs.)
39
49
59
69
79
89
99
No of firms
2
3
11
20
32
25
7
Sol: S. E. X =
σ
n
Calculation of Standard Deviation
Mid value
m
f
(m – 69)/ 10
d
fd
fd 2
39
2
–3
–6
18
49
3
–2
–6
12
59
11
–1
–11
11
69
20
0
0
0
79
32
+1
+32
32
89
25
+2
+50
100
99
7
+3
+21
63
Σfd = 80
Σfd 2 = 236
N = 100
σ=
Σfd 2  Σfd
–
 ×i =
 N 
N
236  80 
–

100  100
2
× 10
2.36 – 0.64 × 10 = 1.311 × 10 = 13.11
13.11 13.11
=
=
= 1.311
10
100
=
S. E. X
Ex. 34. The mean height obtained from a random sample of size 100 is 64 inches.
The standard deviation of the height distribution of the population is known to be 3
inches. Test the statement that the mean height of the population is 67 inches at 5%
level of significance. Also set up 99% limits of the mean height of the population.
Sol : Let us take the hypothesis that there is no significant difference between the
sample mean and the population mean.
σ
σ
S. E. X =
=
= 0.3
n
100
Diff
67 – 64
=
= 10
S. E .
0.3
Since the difference is more than 1.96 S.E. (5% level) the hypothesis is rejected.
Hence the mean height of the population could not be 67 inches.
99% probable limits of the mean height of the population
= X ± 2.58 S. E.= 64 ± 2.58 (.3)
= 64 ± 0.774 or 63.2 to 64.8
Sampling Theory
399
Ex.35: A random sample of 200 tins of coconut oil gave an average weight of 4.95
kg with a standard deviation 0.21 kg. Do we accept the hypothesis of net weight of 5
kg. per tin, at 1 per cent level ?
Sol : We have to test the hypothesis that the net weight is 5 kg. per tin.
σ
S. E. X =
n
σ = 0.21, n = 200
0.21
0.21
S. E. X =
=
= 0.01485
14
.142
200
Diff
5 – 4.95
=
= 3.37
S. E . 0.01485
The difference is more than 2.58 S.E. (at 1% level of significance). The hypothesis is
rejected. Hence the net wt. cannot be 5 kg. per tin.
Ex.36 : A sample of 5,000 students from Bombay University was taken and the
average weight was found to be 112 pound with a standard deviation of 25 pound.
What can you infer about the average weight of the students in the entire university?
σ
25
25
Sol : S. E. X =
=
=
= 0.354
n
5000 70.711
The average weight of the student in the entire university shall be :
X ± 3 S. E. i. e., 112 ± 3 ( 0.354) or 110.938 to 113.062
Ex.37: If it costs a rupee to draw one number of a sample, how much would it cost
in sampling from a universe with mean 100 and standard deviation 10 to take
sufficient number as to ensure that the mean of a sample would in a 5 per cent
probability be within 0.01 per cent of the true value ? Find the extra cost necessary
to double this precision.
Sol : X = 100, σ = 10
Difference between universe mean and sample means = 0.01 per cent
σ
S. E. X =
n
For 95% confidence, difference of sample mean and population mean should be
10
equal to 1.96 S.E., i. e.,
.
n
1.96 × 10
∴
= 0.01
n
n × 0.01 = 19.6
19.6
n =
= 1960
0.01
n = (1960) 2 = 38,41,600
Doubling the precision would imply that the difference between sample mean and
population mean should be = 0.005 only.
1.96 × 10
19.6
= 0.005 or n =
= 3920
0
.005
n
∴
∴
n = ( 3920) 2 = 15,366,400
Extra cost = 15,366,400 – 3,841,600
400
Engineering Mathematics-III
= Rs.11,524,800
Hence in order to double the precision the extra cost required is Rs. 11,524,800.
Ex. 38: A sample of 400 male students is found to have a mean height of 171.38
cm. Can it be reasonably regarded as sample from a large population with mean
height 171.17 cm and standard deviation 3.30 cm ?
Sol : Let is take the hypothesis that there is no significant difference in the mean
height of the sample and the given population value.
σ
3.3
3.3
S. E. X =
=
=
= 0.165
20
n
400
X –µ
171.38 – 171.17
0.21
=
=
= 1.27
S. E. X
0.165
0.165
Since the difference is less than 1.96 S.E. (at 5% level of significance) the hypothesis
holds true. Hence the sample could have come from a population with mean height
171.77 and standard deviation 3.3.
Ex. 39: To study the correlation between the stature of father and the stature of
son, a sample of 1,600 is taken from the universe of fathers and sons. The sample
study gives the correlation between the two to be 0.80. Within what limits does it
hold true for the inverse?
Sol : The standard error of the correlation coefficient between the stature of the
father and son is :
S. E. r =
1 – r2
n
r = 0.8, n = 1,600
S. E. r =
1 – ( 0.8) 2
1600
=
1 – 0.64 0.36
=
= 0.009
40
40
If the sampling was simple random sampling the correlation in the universe cannot
deviate from the correlation in the sample by more than thrice the standard error.
i.e., 0.027. Hence correlation in the universe most probably lies between r ± 3 S. E.
or 0.8 + 0.027 or 0.773 and 0.827.
Ex. 40: A sample of 400 items is taken from a normal population whose mean is 4
and whose variance is 4. If the sample mean is 4 ⋅ 45 , can the sample be regarded as
a truely random sample.
Sol :
Mean of the population µ = 4.
S.D. of the population σ = 2.
S.D. of the mean of the sample
σ
2
=
=
= ⋅1
√ n √ ( 400)
Hence
z=
X −µ
4 ⋅ 45 − 4
=
= 4 ⋅ 5.
⋅1
σ / √n
Hence the deviation of the mean of the sample from the mean of the population is
4 ⋅ 5 S.E. is highly significant. Therefore the sample cannot be regarded as a random
sample.
Sampling Theory
401
Ex. 41: The mean of certain normal population is equal to the standard error of the
mean of the samples of 100 from that distribution. Find the probability that the
mean of the sample of 25 from the distribution will be negative.
Sol : If the mean of the distribution is µ and that of the sample is X, then
σ
σ
µ=
= , where σ is the S.D. of the distribution.
10
√ (100)
For a sample of , we have
X −µ
z=
=
σ
5
1
Since X is negative, z < − .
2
σ
10 = X − 1.
σ
σ
2
5
5
X −
The probability that z, a normal variate, is negative and < −
1
−
√ ( 2π )
− 1/ 2
∫− ∞
e
−
1 2
z
2
1
is given by
2
1
dz =
√ ( 2π )
∞
∫1/ 2
e
−
1 2
z
2
dz
= ⋅ 3085 (from the tables)
Ex. 42: The guaranteed average life of a certain type of electric light bulbs is 1,000
hours with a standard deviation of 125 hours. It is decided to sample the output so as
to ensure that 90% of the bulbs do not fall short of the guaranteed average by more
than 2 ⋅ 5%. What must be the minimum sample size ?
Sol : Let n be the size of the sample. Since the guaranteed mean is 1,000 we do not
want the mean of the sample to be less than 2 ⋅ 5% of 1000 (i.e. 25) from 1000 so that
it should not be below 1000 − 25 = 975. Hence X > 975. It follows that

 

X − µ 975 − 1000
√n
.
|z|=
>
=
 σ  

125
5

 

 √n  

√n
From the given condition, the area of the probability normal curve to the left of
√ n / 5 should be ⋅9 or the area between 0 and √ n / 5 is ⋅4. As we are not worried
about the bulbs which have life above the guaranteed life, it is a one-tailed problem.
From the table of areas, the area between t = 0 and t = 1 ⋅ 281 is ⋅4.
∴
√ n / 5 = 1 ⋅ 281 or n = 41 approx.
The sample should not consist of less than 41 bulbs.
Ex. 43: A sample of 400 male students is found to have a mean height of 67 ⋅ 47
inches. Can it be reasonably regarded as sample from a large population with mean
height 67 ⋅ 39 inches and S.D. 1 ⋅ 30 inches ?
Sol :
Here
X = 67 ⋅ 47 inches,
µ = 67 ⋅ 39 inches and σ = 1 ⋅ 30 inches, n = 400
x − µ 67 ⋅ 47 − 67 ⋅ 39
Here
z=
=
= 1 ⋅ 23
σ
1 ⋅ 30
20
√n
402
Engineering Mathematics-III
Hence the deviation of the mean of sample from the mean of the population is 1 ⋅ 23
S.E. which is not significant. Therefore the sample can reasonably be regarded as
drawn from a large population with mean height 67 ⋅ 39 inches and S.D. 1 ⋅ 30 inches.
Ex. 44: It is known that the mean and standard deviation of a variable are
respectively 100 and 10 in the universe. It is however considered sufficient to draw a
sample of sufficient size but such as to ensure that the mean of the sample would be
in all probability within 0 ⋅ 10% of the true value. How much would be the cost
(exclusive of overhead charges) if the charges for drawing 100 members of a sample
be one rupee ?Find the extra cost necessary to double the precision.
Sol : Assuming the conditions of simple sampling, the sample mean should not
differ from the true mean by 0 ⋅ 01% or by 0 ⋅ 01 since the true mean here is 100.
σ
10
The S.E. of the mean of the sample =
if the sample size is n and σ is the S.D.
=
√n √n
of the universe.
We know that in a normal distribution µ ± 3σ contains almost all the values of
variate, µ being the mean of the distribution.
∴
or


X − µ
should be equal to 3.
z=
 σ 


 √n 
⋅01
=3
10 / √ n
or
√ n = 3,000 i. e. n = 9,000,000.
The sample size is therefore 9,000,000. The sampling charges are thus Rs. 90,000.
⋅005
To double the precision, we should have
= 3.
10 / √ n
This gives n = 36,000,000.
Hence the extra cost = 360,000 − 90,000 = 270,000 rupees.
Ex. 45: To know the mean weight of all 10-year old boys in the state of Rajasthan, a
sample of 225 is taken. The mean weight of this sample is found to be 67 pounds
with a standard deviation of 12 pounds. Can you draw any inference from it about
the mean weight of the universe ?
Sol : Here S.D. of the universe is not given but we can use in its place the S.D. of the
sample which is given to be 12.
σ
12
∴ Standard error of the mean =
=
= ⋅ 8 pound.
√ n √ ( 225)
Assuming simple sampling conditions, the weight of the universe would in all
probability lie within three times this S.D. to the mean of the sample. Hence the
mean weight of all 10 year old boys in the state lies between
67 lbs. ± 2 ⋅ 4 lbs. i.e. 64 ⋅ 6 lbs. and 69 ⋅ 4 lbs.
Ex. 46: An industry desires to make a survey of the mean weekly wages of 10,000 of
its workers. The mean weekly wages are rupees thirty with standard deviation of
rupees 2 ⋅ 5. Since study of all the workers is impossible, a representative sample of
400 workers is selected; by how much would the results differ from the above
sample ?
Sampling Theory
Sol : Standard error of the mean
403
σ
2⋅ 5
=
= ⋅ 125 rupee.
√ n √ ( 400)
If fresh samples were taken their mean would not differ by the mean weekly wage of
this sample by more than three times the S.E. of the mean of the sample, that is, the
mean weekly wages of all the fresh samples would lie between Rs. 30 ± ⋅ 375 or
between Rs. 29 ⋅ 725 and Rs. 30 ⋅ 375.
Ex. 47: Suppose that the standard deviation of stature in mean is 2 ⋅ 48 inches. One
hundred male students in a large university are measured and their average height
is found to be 68 ⋅ 52 inches. Determine the 98 percent confidence limits for the
mean height of the mean of the university.
Sol : Standard error of the mean height of 100 male students
σ
2 ⋅ 48
=
=
= ⋅ 248 inches.
√ n √ (100)
Now 98% confidence limits for the mean height of the mean of the university means
that 49% of the total area under the normal curve lies on each side of the mean i.e.
⋅5 + ⋅ 49 = ⋅ 99 area under the standard normal curve should lie to the left of the
critical value of the variable z. From the table, we find that the critical value of z is
2 ⋅ 32.
Hence for the required confidence limits,
µ − 68 ⋅ 52
 < 2 ⋅ 32
we have
=
 ⋅248 
i.e.,
68 ⋅ 52 − 2 ⋅ 32 × ⋅ 248 < µ < 68 ⋅ 52 + 2 ⋅ 32 × ⋅ 248
67 ⋅ 945 < µ < 69 ⋅ 095.
Hence the 98% confidence limits for the mean height of the mean of the university
are 67 ⋅ 945 inches and 69 ⋅ 095 inches.
Ex. 48: The data concerning height measurement for a random sample of
individuals from a given population are as follows :
mean = 172, S. D. = 12, n = 65
If a large number of the samples of the same size were selected at random from the
given population, what would be the limits of 2% confidence interal for the true
mean ?
σ
12
Sol :
S.E. of the mean =
=
= 1 ⋅ 5 nearly.
√ n √ ( 65)
Now the limits of 2% confidence interval for the true mean, means the same thing as
98% confidence limits for the true mean. Hence as in the previous exercise, we have
the required confidence limits for the mean as
172 ± 1 ⋅ 5 × 2 ⋅ 32 = 172 ± 3 ⋅ 48.
i.e. the limits are 168 ⋅ 52 and 175 ⋅ 48.
Ex. 49: A research worker wishes to estimate mean of a population by using
sufficiently large sample. The probability is 95 percent that sample mean will not
differ from the true mean by more than 25 percent of the standard deviation. How
large a sample should be taken ?
(Jhansi M.Sc. 2002)
404
Engineering Mathematics-III
We know that the area under the curve between the values − 1 ⋅ 96 and 1 ⋅ 96 of the
standard normal variate is 0 ⋅ 95. Hence, we have
X − µ


 = 1 ⋅ 96
σ
 / √ n
1 ⋅ 96σ
√n
σ
Also it is given that | X − µ| < .
4
From (1) and (2), we get
∴
1 ⋅ 95σ σ
< . or n > 16 × (1 ⋅ 96) 2 = 62 nearly.
4
√n
or
...(9.1)
| X − µ| =
Ex. 50: A normal population has a mean of 0 ⋅ 1 and a S.D. of 2 ⋅ 1. Find the
probability that the mean of simple sample of 900 members will be negative.
Sol :
σ
2⋅1
Here S.E. of the mean
=
=
= ⋅ 07
30
√n
X −µ
X − 0⋅1
X
∴
z=
=
=
− 1 ⋅ 43
⋅07
⋅07
σ / √n
Since X is negative, z < − 1 ⋅ 43.
∴ The probability that X is –ve i.e. z = − 1 ⋅ 43 is given by
1
√ ( 2π )
− 1⋅ 43
∫− ∞
e
−
1 2
z
2
dz =
1
√ ( 2π )
∞
∫1⋅43
e
−
1 2
z
2
dz
= ⋅ 0774 from the tables.
Ex. 51: The mean height of 9339 children of age 5 years is 41 ⋅ 26 inches, the
standard deviation is 2 ⋅ 238 inches. Find the odds against the possibility that the
mean of a random sample of 100 is greater than 41 ⋅ 70.
Sol :
σ
2 ⋅ 238
S.E. of the mean =
=
= ⋅ 2238.
10
√n
X − µ 41 ⋅ 70 − 41 ⋅ 26
∴
z=
=
= 1 ⋅ 96
σ
⋅2238
√n
From the table we see that for z = 1 ⋅ 96, the area to the left is 0 ⋅ 9750 and so the area
to the right of z = 1 ⋅ 96 is 1 − 0 ⋅ 9750 = 0 ⋅ 0250. Hence the probability that the
1
mean of random sample of 100 is greater than 41 ⋅ 70 is 0 ⋅ 0250 i.e.
or that the
40
odds against are as 39 to 1.
Ex. 52: Suppose light bulbs made by a standard process have an average life of
2000 hours with standard deviation of 250 hours. And suppose it is considered
worthwhile to replace the process if the mean life can be increased by at least 10
percent. An engineer wishes to test a proposed new process, and he is willing to
assume that the standard deviation of the distribution of bulbs is about the same as
for the standard process. How large a sample should he examine if he wishes the
Sampling Theory
405
probability to be about ⋅01 that he will fail to adopt the new process if in fact it
produces bulbs with a mean life of 2250 hours ?
Sol : Since there is to be an increase of 10 percent in the mean of the standard
process, the mean of the process
110
= 2000 ×
hours = 2200 hours = µ.
100
And S.D. of the new process = S.D. of the standard process = 250 hours = σ.
Mean of the sample = 2250 hours = X.
If n be the number in the sample, then
X −µ
2250 − 2200 √ n
.
z=
=
=
5
σ / √n
250 / √ n
Since the probability is to be about ⋅01 that we will fail to adopt the process, the
corresponding value of z for this probability is 2 ⋅ 58.
√n
Hence
= 2 ⋅ 58
5
or
n = 25 × ( 2 ⋅ 58) 2 = 166 approx.
Ex. 53: An unbiased coin is thrown n times. It is desired that the relative frequency
of the appearance of heads should lie between ⋅49 and ⋅ 51. Find the smallest value
of n that will ensure this result with 90% confidence.
Sol : Now 90% confidence means that ⋅45 of the total area under the standard
normal curve should lie on each side of the mean. From the tables, the
corresponding value of z is 1 ⋅ 645.
Also standard error of the proportion of head =
1
 1 1 1
.
 . .  =
 2 2 n 2 √ n
1
= ⋅ 49
[∵ p − 1 ⋅ 645σ = ⋅ 49]
2√ n
1
and
⋅5 + 1 ⋅ 645 ×
= ⋅ 51.
2√ n
1 ⋅ 645
These give
= ⋅ 01.
2√ n
1 ⋅ 645 329
or
√n =
=
⋅02
4
108241
or
n=
= 67 ⋅ 65 approx. = 68 nearly.
16
Ex. 54: Mean of 10 readings on the length of a given rod is 20 inches. The standard
deviation of errors of measurement is known to be 0 ⋅ 1 inch. Does the result
contradict the assumption that the length of the rod is 19 ⋅ 9 inches.
0⋅1
Sol : S.E. of mean =
√ 10
20 − 19 ⋅ 9
∴
z=
= √ 10 > 3.
0 ⋅ 1 / √ 10
Hence
⋅5 − 1 ⋅ 645 ×
Hence the difference is significant and so the length of the rod is not 19 ⋅ 9 inches.
406
Engineering Mathematics-III
Ex. 55:If the mean breaking strength of copper wire is 575 lbs. with a standard
deviation of 8 ⋅ 3 lbs., how large a sample must be used in order that there be one
chance in 100 that the mean breaking strength of the sample is less than 572 lbs.
Sol :
x −µ
572 − 575
Here
z=
√n =
√n
σ
8⋅ 3
3
or
...(9.1)
|z| =
√ n.
8⋅ 3
1
Now the probability that x < 572 is
= ⋅ 01. Hence we have to seek for that value
100
of z for which 0 ⋅ 01 = area to the right at the variate z,
so that the area to left = 0 ⋅ 99. From the tables of the areas of the normal curve, we
find that the corresponding value of z is 2 ⋅ 33. Hence we get from (1),
3
2 ⋅ 33 =
√ n, which gives n = 42 nearly
8⋅ 3
9.16 Test of Significance of the Means of Two Large Samples
(Agra M.Sc. 2006)
Suppose from a normal population, with standard deviation σ, two simple samples –
one of size n1 and mean x and the other of size n 2 and mean y are drawn. We wish to
test whether difference between mean x and y , the mean of the samples, is
significant or merely due to fluctuations of sampling. We know that the variance of
1
1
σ
σ
the difference of the means of two samples is σ 2 
+
and
 where
n2 
√ n1
√ n2
 n1
are the standard deviations of the means of the sample. Hence
x− y
z=
1
1
σ 
+

n2 
 n1
is normally distributed with mean zero and standard deviation unity. The
probability that |z| > 1 ⋅ 96 is 0 ⋅ 5 and hence if|z| > 1 ⋅ 96 the difference is significant
at 5% level of significance. That is, if the difference between x and y is not
significant|z| < 1 ⋅ 96. If|z| > 3, it is highly probable that either the samples have not
been taken from the sample population or the sampling is not simple.
If the samples are known to be taken from two normal populations with means µ1
and µ 2 and standard deviations σ1 and σ 2 , then x − y is normally distributed with
σ 2 σ 2 
mean µ1 − µ 2 and standard deviation  1 + 2  . We get
n
n 2 
 1
z=
( x − y ) − (µ1 − µ 2 )
 σ12 σ 22 


+
n
n 2 
 1
Sampling Theory
On the hypothesis that µ1 = µ 2 , we have z =
407
x− y
, and the same
 σ12 σ 22 


+
n
n 2 
 1
procedure of test of significance is applied.
Ex. 56: A random sample of 200 village was taken from Gorakhpur district and the
average population per village was found to be 485 with a standard deviation of 50.
Another random sample of 200 villages from the district gave an average population
of 510 per village with a standard deviation of 40. Is the difference between the
average of the two samples statistically significant ? Give reasons.
Sol : On the hypothesis that the samples have been taken from the same population,
we put
x− y
z=
 σ12 σ 22 


+
n
n 2 
 1
Now
so that
x = 485, σ1 = 50, n1 = 200, y = 510, σ 2 = 40, n 2 = 200.
485 − 510
− 25
z=
=
4 ⋅ 53
 ( 50) 2 ( 40) 2 
+


200 
 200
| z | = 5 ⋅ 5 nearly > 3.
Hence the difference between the means of the samples is highly significant and
could not have arisen from causes due to fluctuations of sampling.
Ex. 57: The mean of simple samples of 1000 and 2000 are 67 ⋅ 5 and 68 ⋅ 0 inches
respectively. Can the samples be regarded as drawn from the same population of
standard deviation 2 ⋅ 5 inches ?
Sol :
Here
x = 67 ⋅ 5, n1 = 1000, y = 68 ⋅ 0, n 2 = 2000.
The s.d. of the population σ = 2 ⋅ 5. On the hypothesis that the samples are drawn
from the same population of s.d. 2 ⋅ 5 inches, we get
x− y
67 ⋅ 5 − 68 ⋅ 0
⋅5
⋅5
z=
=
=
=
= 5⋅1
1  2 ⋅ 5 × ⋅ 0387 ⋅09675
1
 1
1
+
σ 
+



 1000 2000
n2 
 n1
Hence the difference between the means is more than three times the standard error
of the difference and so it is statistically significant. The samples cannot be regarded
as drawn from the same population of S.D. 2.5' '.
Ex. 58: A sample of heights of 6400 soldiers has a mean of 67 ⋅ 85 inches and a
standard deviation of 2 ⋅ 56 inches while a simple sample of heights of 1600 sailors
has a mean of 68 ⋅ 55 inches and a standard deviation of 2 ⋅ 52 inches. Do the data
indicate that the sailors are on the average taller than soldiers ?
Sol : Here x = 67 ⋅ 85, σ1 = 2 ⋅ 56, n1 = 6400, y = 68 ⋅ 55, σ 2 = 2 ⋅ 52, n 2 = 1600.
The standard error of the difference of the mean heights is given by
408
Engineering Mathematics-III
E =
 σ12 σ 22 

 =
+
n
n 2 
 1
 ( 2 ⋅ 56) 2 ( 2 ⋅ 52) 2 
+


1600 
 6400
= [(⋅001024) + (⋅003969)]1/ 2 = 0 ⋅ 07 nearly.
The difference between the means = y − x = ⋅ 7 inches which is nearly ten times the
standard error of the difference of the means and this is highly significant. Hence the
data indicates that the sailors with greater mean height are on the average taller
than the soldiers.
Two-tailed test for Difference between the Means of Two Samples
(i) If two independent random sample with n1 and n 2 numbers (both sample sizes
are greater than 30) respectively are drawn from the same population of standard
deviation σ, the standard error of the difference between the sample means is given
by the formula :
S.E. of the difference between sample means
1
1
= σ2
+

 n1 n 2 
If σ is unknown, sample standard deviation for combined samples must be
substituted.
(ii) If two random samples with X1 , σ1 , n1 , and X 2 , σ 2 , n 2 respectively are drawn
from different populations, then the S.E. of the difference between the sample
means is given by the formula:
S.E. of the difference between sample means
=
σ12 σ 22
+
n2 n2
and where σ1 and σ 2 are unknown
S.E. of the difference between means
=
S12 S 22
+
n1 n 2
where S1 and S 2 represent standard deviations of the two samples.
The null hypothesis to be tested is that there is no significant difference in the means
of the two samples, i. e.,
H 0 : µ1 = µ 2 ← null hypothesis, there is no difference
H1 : µ1 ≠ µ 2 ← alternative hypothesis, a difference exists.
Ex. 59 : Intelligence test on two groups of boys and girls gave the following results :
Mean
S.D.
N
Girls
75
15
150
Boys
70
20
250
Is there a significant difference in the mean scores obtained by boy and girls ?
Sol : Let us take the hypothesis that there is no significant difference in the mean
scores obtained by boys and girls.
Sampling Theory
S. E.( X
409
1– X2)
=
σ12 σ 22
+
n1 n 2
σ1 = 15, σ 2 = 20, n1 = 150, n 2 = 250
Substituting the values,
(15) 2 ( 20) 2
+
= 1.5 + 1.6 = 1.761
150
250
Difference 75 – 70
=
= 2.84
S. E .
1.761
Since the difference is more than 2.58 (1% level of significance), the hypothesis is
rejected. There seems to be a significant difference in the means score obtained by
boys and girls.
Ex. 60: A man buys 50 electric bulbs of Philips and 50 electric bulbs of HMT. He
finds that Philips bulbs give an average life of 1,500 hours with a standard deviation
of 60 hours and HMT bulbs give an average life of 1512 hour with a standard
deviation of 10 hours. Is there a significant difference in the mean life of the two
makes of bulbs ?
Sol : Let us set up the hypothesis that there is no significant difference in the mean
life of the two makes of bulbs. Calculating standard error of difference of means
S. E.( X
1– X2)
=
(15) 2 ( 20) 2
+
= 1.5 + 1.6 = 1.761
150
250
Difference 75 – 70
=
= 2.84
S. E .
1.761
Since the difference is more than 2.58 (1% level of significance), the hypothesis is
rejected. There seems to be a significant difference in the mean score obtained by
boys and girls.
Ex. 61: (a) A simple sample of the height of 6,400 Englishmen has a mean of 170
cm and a standard deviation of 6.4 cm, while a simple sample of heights of 1,600
Indians has a mean of 172 cm and a standard deviation of 5.3 cm. Do the data
indicate that the Indians are on the average talles than the Englishmen?
Sol : Let us take the hypothesis that the Indians are on an average not taller than the
Englishmen. Calculating standard error of the difference of means :
S. E.( X
S. E.( X
1– X2)
1–
X2
=
)=
σ12 σ 22
+
n
n2
σ1 = 6.4, n1 = 6400, σ 2 = 6.3, n 2 = 1600
( 64) 2 ( 6.3) 2
+
= 0.0064 + 0.0248 = 0.1767
6400
1600
Difference 6.4 – 6.3
=
= 0.566
S. E .
0.1767
Since the difference is less than 1.96 S.E. (5% level of significance) the hypothesis
holds true. Hence on the basis of given data we can not say that Indians are on an
average taller than the Englishmen.
Ex. 61: (b) In order to make a survey of the buying habits,two markets A and B are
chosen at two different parts of city. 400 women shoppers are chosen at random in
S. E.( X
1– X2)
=
410
Engineering Mathematics-III
super market 'A' located in a certain section of Bombay city. Their average monthly
food expenditure is Rs. 250 with a standard deviation of Rs. 40. For 400 women
shoppers chosen at random in super market 'B' in another section of the city, the
average monthly food expenditure is Rs. 220 with a standard deviation of Rs. 55.
Test at 1% level of significance whether the average food expenditure of the two
populations of shoppers from which the samples were obtained are equal.
(ICWA, June 1982)
Sol : Let us take the hypothesis that there is no difference in the average food
expenditure of the two populations of shoppers.
S.E. of the difference of means is given by
S. E.( X
1 − X2
)=
σ12 σ 22
+
n1 n 2
n1 = 400, X1 = 250, σ1 = 40, n 2 = 400, X 2 = 220, σ 2 = 55
Substituting the values
S. E.( X
Difference of
1 − X2
)=
( 40) 2 ( 55) 2
+
=
400
400
1600 3025
+
= 3 .4
400
400
Means = ( X 1 − X 2) = 250 − 220 = 30
Difference 30
=
= 8 .82
S. E.
3 .4
Since the difference is mor than 2.58 S.E. (1% level) the hypothesis is rejected.
Hence the average food expenditure of the populations of shoppers is not equal.
Ex. 62 : You are given the following information relating to purchase of bulbs from
two manufactures A and B :
Manufacturer
No. of Bulbs bought
Mean life
S.D.
A
100
2950 hours
100 hours
B
100
2970 hours
90 hours
Is there a significant difference in the mean life of two makes of bulbs ?
Sol : Let us take the hypothesis that there is no significant difference between the
mean life of the two makes of bulbs.
S. E.( X
1 − X2
)=
σ12 σ 22
+
n1 n 2
σ1 = 100, σ 2 = 90, n1 = 100, n 2 = 100
Substituting the values
10000 8100
∴
S. E.( X − X ) =
+
= 13 .45
1
2
100
100
Diff. 2970 − 2950
=
= 1 .49
S. E.
13 .45
Since the difference is less than 1.96 S.E. (at 5% level of significance), the
hypothesis holds true. Hence there is no significant difference in the mean life of the
two mak3es of bulbs.
Sampling Theory
411
Standard Error of the Difference between two Standard Deviations
In case of two large random samples, each drawn from a normally distributed
population, the S.E. of the difference between the standard deviations is given by :
S. E.( σ 1 − σ 2 ) =
σ12
σ2
+ 2
2n1 2n 2
When population standard deviations are not known
S. E.( S1 − S 2 ) =
S12
S2
+ 2 .
2n1 2n 2
Ex. 63: Intelligence test of two groups of boys and girls gave the following results :
Girls :
Mean score = 84.
S.D. =10,
n = 121
Boys :
Mean score = 81.
S.D. =12
n = 81
(a) Is the difference in mean scores significant ?
(b) Is the difference between standard deviations significant ?
Sol : (a) Let us take the hypothesis that there is no difference in mean scores.
S. E.( X
1 − X2
)=
σ12 σ 22
+
n1 n 2
σ1 = 10, σ 2 = 12, n1 = 121, n 2 = 81
S. E.( X
1 − X2)
=
(10) 2 (12) 2
+
=
121
81
100 144
+
121 81
= 2 .604 = 1 .61
Difference of
means = (84 – 81) =3
Difference
3
=
= 1 .86
S. E.
1 .61
Since the difference is less than 1.96 S.E. (at 5% level) the given facts support the
hypothesis. Hence the difference in mean scores of boys and girls is not significant.
(b) Let us take the hypothesis that there is no difference between the standard
deviations of the two samples.
S. E.( σ 1 − σ 2 ) =
σ12
σ2
+ 2
2n1 2n 2
σ1 = 10, σ 2 = 12, n1 = 121, n 2 = 81
S. E.( σ 1 − σ 2 ) =
=
(10) 2
(12) 2
+
2 × 121 2 × 81
100 144
+
= 1 .302 = 1 .14
242 162
Difference between the two standard deviations = (12 − 10) = 2
Difference
2
=
= 1 .754
S. E.
1 .14
412
Engineering Mathematics-III
Since the difference is less than 1.96 S.E. (at 5% level) the given data supports the
hypothesis. Hence the difference between standard deviations is not significant.
9.17 The Sampling of Variables (Small Samples)
So far we have discussed problems relating to large samples. When the size of
sample is small (less than 30) the above tests are not applicable because the samples.
In particular, it will no longer be possible for us to assume : (a) that that values given
by the sample data are sufficiently close to the population values and can be used in
their place for the calculation of the standard error of the estimate.
The removal of these assumptions makes it necessary to use entirely new techniques
to deal with the problems of small samples. The division between theories of large
and small samples is, therefore, a very real one, though it is not always easy to draw
a precise line of demarcation. It should be noted that as a rule, the methods and
the theory of small samples are applicable to large samples, though
the reverse is not true. While dealing with small samples our main interest is
not to estimate the population values as true as in large samples ; rather our interest
lies in testing a given hypothesis, i.e., in ascertaining whether observed values could
have arisen by sampling fluctuations from some value given in advance. For
example if a sample of 15 gives a correlation coefficient of +0.4, we shall be
interested not so much in the value of the correlation in the parent population, but
more generally whether this value could have arisen from an uncorrelated
population, i.e., whether it is significant of correlation in the parent population.
Again it will no longer be possible in small samples to assume that the random
sampling distribution of a statistic is approximately normal. A new technique is
therefore neccessary to deal with the theory of small samples. We had seen that the
mean of a random sample of size n from a normal population N (µ, σ ) is distributed
σ
normally with mean µ and S.D.
. However, if σ is not known, the distribution of x
√n
cannot be known as the S.D. of the sample cannot be regarded as S.D. of the
population. In order to find the distribution of sample mean x, we have to consider a
statistic which is a function of µ but whose distribution does not depend on σ. We
shall therefore discuss briefly the basis on which estimates of given parameters are
made. We shall confine our work to finding the estimates for the mean and the
standard deviations since these are two main parameters of interest.
It should be noted that the investigator who works with very small samples must
know that his estimates will vary widely from sample to sample. Moreover, he must
be satisfied with relatively wide confidence intervals. Precision of statement is less,
of course, the wider the intervals employed. Each inference drawn from large
sample results in far more precision in the limits it sets up than is an inference based
on smaller samples.
9.17.1 The Assumption of Normality
While dealing with small samples also an assumption is made that the parent
population is normal, unless otherwise stated. Strictly speaking, therefore, our
results will be true only for the normal population. However, as pointed out earlier,
the assumption of normality is not very much warranted in case of small samples.
Experiments have, therefore, been made to ascertain whether the results are true for
other types of population. Theoretical work confirms that the results remain true for
population. Theoretical work markedly from normality. However, if there is any
Sampling Theory
413
good reason to suspect that the parent population is markedly skew, i.e.,U or
J-shaped, the methods given below cannot be applied with confidence.
Since in many of the problems it becomes necessary to take a small size sample,
considerable attention has been paid in developing suitable tests for dealing with
problems of small samples. The greatest contribution to the theory of small samples
is that of Sir William Gossett and R.A. Fisher. Sir William Gossett published his work
in 1905 under the pen name 'Student'. He gave a test popularly known as 't test' and
Fisher gave another test known as 'z-test'. These tests are based on 't'-distribution
and 'z'-distribution.
9.18 Student's t-Distribution
Theoretical work on t-distribution was done by W.S. Gossett in early 1900. Gossett
was employed by the Guinness Brewery in Dublin, Ireland, which did not permit
employees to publish research findings under their own name. So Gossett adopted
the pen name 'Student' and published his findings under this name. Thereafter, the
t-distribution is commonly called Student's t-distribution or simply Student's
distribution. The t-distribution is used when sample size is 30 or less and the
population standard deviation is unknown. The ''t-statistic'' is defined as :
X −µ
t =
× n
S
where
S=
Σ ( X − X )2
n −1
The t-distribution has been derived mathematically under the assumption of a
normally distributed population. It has the following form

t2
f (t ) = C 1 + 
υ

−( υ +1)/ 2
(X − µ)
n
S
C = a constant required to make the area under the curve equal to unity.
υ = n –1, the number of degrees of freedom.
t =
9.18.1 Properties of t-distribution
1.
2.
3.
4.
The variable of t-distribution ranges from minus infinity to plus infinity.
The constant C is actually a function of υ , the distribution of f(t) is completely
specified. Thus, f(t) is a family of functions, one for each values of υ.
Like the standard normal distribution, the t distribution is symmetrical and has
a mean zero.
The variance of the t-distribution is greater than one, but approaches one as the
number of degrees of freedom and, therefore, the sample size becomes large.
Thus the variance of the t- distribution approaches the variance of the standard
normal distribution as the sample size increases. It can be demonstrated that
for an infinite number of degrees of freedom ( υ = ∞ ), the t-distribution and
normal distribution are exactly equal. Hence there is a widely practiced rule of
414
Engineering Mathematics-III
thumb that samples of size n > 30 may be considered large and the standard
distribution, where the latter is the theoretically correct functional form.
The following diagram compares one normal distribution with two t-distributions of
different sample sizes :
Normal distribution
t-Distribution for
sample size n=15
t-Distribution for
sample size n=2
Fig. 9.4
Normal distribution, t-distribution for sample size
n = 15, and t-distribution for sample size n = 2.
The above diagram shows two important characteristics of t-distribution. First, a
t-distribution is lower at the mean and higher at the tails than a normal distribution.
Second, the t-distribution has proportionately greater area in its tails than the
normal distribution. Interval widths from t-distribution are, therefore, wider than
those based on the normal distribution.
9.18.2 Probability Tables of t-distribution
The chance that on random sampling, we shall
get a value of t not greater than some value t 0 is
the area of the curve
t Curve
Normal
Curve
y0
y =

t2
1 + 

ν 

( ν + 1)/ 2
Fig. 9.5
to the left of the ordinate at the point t 0 and is
given by
t0
∫− ∞
y dt .Similarly, the probability that we get a value of t between the
limits t1 and t 2 is given by ∫
t2
t1
y dt . Student should himself prepared tables of the
integral
PS =
t
y dt
∫− ∞
for various degrees of freedom and different values of t which are generally required
in practical problems. The Fisher-Yates tables give the area given by
PF = 1 −
t
∫− t
y dt
Sampling Theory
415
Relation between PS and PF . From the above definition of PS and PF , it is clear
that PS denotes the probability that an observed value will not exceed t 0 whereas PF
denotes the probability that an observed value of t, regardless of sign, will exceed t 0 .
t
Now
PS =
and
PF = 1 −
y dt =
∫− ∞
t
∫− t
0
∫− ∞
y dt +
y dt = 1 − 2 ∫
0
t
∫0
t
y dt =
1
+
2
∫0
t
y dt
y dt ...(9.1)
...(9.2)
Multiplying (9.1) by 2 and adding the result to (9.2), we get
2PS + PF = 2 or PF = 2 (1 − PS ).
1
, the probability of a value of t lying between 0 and
π
1
= tan − 1 t 0
π
When n = 2, ν = 1, y 0 =
t 0 is t 0
1
π
∫0
t0
dt
1 + t2
If the level of significance is 20%, then
1
tan − 1 t 0 = ⋅ 4
π
or
t 0 = tan (⋅4π ) = tan 72 o .
For 10% level of significance, t 0 = tan (⋅45π ) = tan 81 o
For 5% level of significance, t 0 = tan (⋅475π ) = tan 85 ⋅ 5 o .
9.18.3 Use of t-distribution
We have seen that if the sample is large, the use is made of the tables of the normal
probability integral in interpreting the results of an experiment and on the basis of
that to reject or accept the null hypothesis. If however, the sample size n is small, the
normal probability tables will no longer be useful. It was the discovery of the
t-distribution by Student in 1908 that provided the answer to this difficulty. A rigorous
proof of student’s t-distribution was provided some years later by R.A. Fisher. New
tables giving the significance levels of observed values of t were prepared. There are
various uses of t-distribution. A few of them are mentioned below :
(a) To test the significance of the mean of a small random sample from a normal
population.
(b) To test the significance of the difference between the means of two samples
taken from a normal population.
(c) To test the significance of an observed coefficient of correlation including
partial and rank correlations.
(d) To test the significance of an observed regression coefficient.
We now discuss some of these tests in detail to emphasize the importance of
t-distribution.
416
Engineering Mathematics-III
9.19 Test of Significance of the Mean of Random Sample from
Normal Population
Suppose we have to test the hypothesis that the mean of a normal population from
( x − µ)
which a small sample has been drawn is µ. For this we first calculate t =
√n
s
and then from the tables, find the value of PS for the given degree of freedom. If
PF < ⋅ 05, we say that the difference between the population mean and the sample
mean is significant at 5 percent level of significance and the hypothesis is to be
rejected. If PF < ⋅ 01, the difference is said to be significant at 1 percent level of
significance. In this case the difference is said to be highly significant. On the other
hand, if PF > ⋅ 05, we say that the data are consistent with the hypothesis that µ is the
mean of the population.
Note : We shall specifically assume that the parent population is normal unless
otherwise stated.
Ex. 64: Find the Student’s t for the following variate values in a sample of eight
− 4, − 2, − 2, 0, 2, 2, 3, 3 taking the mean of the universe to be zero.
(Agra M.Sc. 2003; Kanpur M.Sc. 2007; M.U. 1990)
Sol :
x = mean of the sample =
2
= ⋅ 25 :
8
σ 2s = s.d. of the sample
=
1
[( − 4 − ⋅ 25) 2 + 2 ( − 2 − ⋅ 25) 2 + ( 0 − ⋅ 25) 2
8
+ 2 ( 2 − ⋅ 25) 2 + 2 ( 3 − ⋅ 25) 2 ]
∴
Hence
=
1
[( 4 ⋅ 25) 2 + 4 { 4 + (⋅25) 2 } + (⋅25) 2 + 2( 2 ⋅ 75) 2 ]
8
=
1  289 65
1
121
792
99
+
+
+
=
=
⋅


8  16
4
16
8  8 × 46 16
σs =
t =
√ ( 99)
= 2 ⋅ 487.
4
x −µ
0 ⋅ 25 − 0
√ ( n − 1) =
√ 7 = 0 ⋅ 27 nearly.
σs
2 ⋅ 487
Ex. 65: The nine items of a sample has the following values 45, 47, 50, 52, 48, 47,
49, 53, 51. Does the mean of the nine items differ significantly from the assumed
mean of 47 ⋅ 5 given that
 P = ⋅ 945 for t = 1 ⋅ 8
ν = 8
 P = ⋅ 953 for t = 1 ⋅ 9 ?
(Kanpur M.Sc. 2006; Agra M.Sc. 2002; Meerut 2000)
Sol : We find the mean and standard deviation of the sample as follows :
Sampling Theory
417
x
d
d2
45
–3
9
47
–1
1
50
2
4
52
4
16
48
0
0
47
–1
1
49
1
1
53
5
25
51
3
9
Total
10
66
Remarks
∴
i. e.
A = assumed mean = 48
1
10
µ1 ′ = Σ a =
= 1⋅1
9
9
x = mean = A = µ1 ′ = 49 ⋅ 1
1
66
µ 2 ′ = Σ d2 =
9
9
2
2
σ s = µ 2 ′ − µ1 ′
66 100
=
−
9
81
494
=
81
σ s = 2 ⋅ 47
x −µ
49 ⋅ 1 − 47 ⋅ 5
√ ( n − 1) =
√ 8 = 1 ⋅ 83
σs
2 ⋅ 47
Hence
t =
Here
ν = 9 − 1 = 8.
Now difference for ⋅1 = ⋅ 008,
⋅008
difference for⋅03 =
∴
× ⋅ 03 = ⋅ 0024.
⋅1
Hence when t = 1 ⋅ 83, P = ⋅ 945 + ⋅ 0024 = ⋅ 9474.
Here clearly P stands for PS so that we have
PF = 2 (1 − ⋅ 9474) = ⋅ 1052 > ⋅ 05.
It follows that the value of t is not significant at 5% level of significance and
therefore the test provides no evidence against the population mean being 47 ⋅ 5.
Ex. 66: Ten individuals are chosen at random from a population and their heights
are found to be in inches : 63, 63, 66, 67, 68, 69, 70, 70, 71, 71. In the light of these
data, discuss the suggestion that the mean height in the population is 66 inches
having given that for t = 1 ⋅ 8, P = 0 ⋅ 947 and for t = 1 ⋅ 9, P = 0 ⋅ 955, where P is the
area to the left of the ordinate at t.
Sol : We find mean and s.d. of the sample as follows :
418
Engineering Mathematics-III
x
f
d
fd
d2
fd 2
63
2
–4
–8
⋅6
32
66
1
1
–1
1
67
1
0
0
0
68
1
1
1
1
69
1
2
2
4
70
2
3
6
9
71
2
4
8
16
A = assumed mean = 67
8
µ1 ′ =
= ⋅8
1
10
x = mean = 67 ⋅ 8 inches
0 ∴
1
88
µ2′ =
Σ fd 2 =
= 8⋅ 8
1
10
10
88
64
204
4 ∴
σ 2s = µ 2 ′ − µ1 ′ 2 =
−
=
10 100
25
18
√ 2064
i.e. σ s =
= 2 ⋅ 86 inhes
5
32
Total 10
5
8
8
88
∴
t =
Ramarks
x −µ
67 ⋅ 8 − 66
5⋅ 4
√ ( n − 1) =
√9 =
= 1 ⋅ 89.
σs
2 ⋅ 86
2 ⋅ 86
Now difference for ⋅1 = ⋅ 008.
⋅008
∴ difference for ⋅09 =
× ⋅ 09 = ⋅ 0072
⋅1
Hence for
t = 1 ⋅ 89, P = 1 ⋅ 947 + 0 ⋅ 0072 = 0 ⋅ 954
∴
PF = 2 (1 − 0 ⋅ 954) = 0 ⋅ 092 > ⋅ 05.
It follows that the difference is not significant at 5% level of significance and the test
provides no evidence against the population mean being 66 inches.
Ex. 67: Ten individuals are chosen at random from a population and their heights
are found to be 63, 63, 66, 67, 68, 69, 70, 70, 71 and 71 inches respectively. Test
whether the mean height is 69 ⋅ 6 inches in the population given that for 9 degrees of
freedom
P {|t |≥ 2 ⋅ 262} = 0 ⋅ 05.
As in 64, above, we get
x = 67 ⋅ 8 inches and σ s = 2 ⋅ 86 inches.
x −µ
67 ⋅ 8 − 69 ⋅ 6
Hence
t =
√ ( n − 1) =
√ 9.
σs
2 ⋅ 86
5⋅ 4
∴
|t | =
= 1 ⋅ 89.
2 ⋅ 86
Since this value of|t | is less than 2 ⋅ 262, the difference is not significant at 5% level
of significance and the test provides no evidence against the population mean being
69 ⋅ 9 inches.
Ex. 68: Show that the 95 percent fiducial limits for the mean of the population are
s
x±
t 0⋅05 , where t 0⋅05 , stands for the value of t at 5 percent level of significance.
√n
Deduce that for a random sample of 16 values with mean 41 ⋅ 5 inches and the sum
of the squares of the deviations from the mean 135 (inches) 2 and drawn from a
normal population, 95 percent fiducial limits for the mean of the population are
39 ⋅ 9 and 43 ⋅ 1 inches.
Sampling Theory
419
Sol : The probabilities that|t |≤ t 0⋅05 is ⋅95. Hence the 95 percent confidence limits
for the mean µ of the population are given by
x −µ
s
| t | =
√ n ≤ t 0⋅05 or | x − µ | ≤
t 0⋅05
√n
 s

s
s
or
x−
t 0⋅05 ≤ µ ≤ x +
t 0⋅05 .
√n
√n
We can therefore say with a confidence coefficient ⋅95 that the confidence interval
s
x±
t 0⋅05 contains the population mean µ.
√n
For the given numerical example, we have
Σ ( x − x ) 2 145
so that s = 3.
n = 16, s 2 =
=
n−1
15
Also from the tables, t 0⋅05 = 2 ⋅ 13
s
3
∴
t 0⋅05 = × 2 ⋅ 13 = 1 ⋅ 6 approx.
4
√n
Hence the required fiducial limits are 41 ⋅ 5 ± 1 ⋅ 6 i.e. 39 ⋅ 9 and 43 ⋅ 1 inches.
Ex. 69: A mechinist is making engine parts with axle diameter of 0 ⋅ 700 inch. A
random sample of 10 parts shows a mean diameter of ⋅742 inch with a standard
deviation of 0 ⋅ 040 inch. On the basis of this sample, would you say that the work is
inferior ?
Sol :
Here we have
µ = ⋅ 700, x = 0 ⋅ 742, σ s = 0 ⋅ 040, n = 10.
x −µ
0 ⋅ 742 − 0 ⋅ 700
∴
t =
( n − 1) =
√ (10 − 1)
σs
0 ⋅ 040
0 ⋅ 126
=
= 3 ⋅ 16
0 ⋅ 040
The table value of t for 9 degrees of freedom at 5% level of significance is 2 ⋅ 262. The
calculated value of t is greater than this value.
Hence the difference is significant at 5% level of significance. Hence the work is
inferior. As a matter of fact, the work is inferior even if we make the level of
significance only 2 percent.
Ex. 70: A certain type of thread has a mean breaking strength of 1 ⋅ 32 ounce. A
change in its manufacture is recommended with a view to increasing its breaking
strength, but the cost of the proposed change will be substantial. A sample of 50
pieces of the new thread is tested and found to have a mean strength of 1 ⋅ 39 ounces
with a variance 0 ⋅ 0424. Discuss whether the new sample fulfils expectations to the
extent that the change should be made first by using the normal curve and secondly
by using t-distribution. To be quite sure of correct conclusion at 1 percent level of
significance is to be used. You may make use of the following tables.
(a) The area A under the standard normal curve lying to the left of the ordinate at
the standard normal variable z.
420
Engineering Mathematics-III
z
2 ⋅ 38
2 ⋅ 39
2 ⋅ 40
2 ⋅ 41
2 ⋅ 42
A
0 ⋅ 9943
0 ⋅ 9916
0 ⋅ 9918
0 ⋅ 9920
0 ⋅ 9922
(b) At 1% level of significance, t = 2 ⋅ 70 for ν = 40 and t = 2 ⋅ 66 for ν = 60 where ν
denote the number of degrees of freedom.
Sol : Firstly, if we regard this as a large sample, we have
x −µ
1 ⋅ 39 − 1 ⋅ 32
z=
√n =
√ ( 50) = 2 ⋅ 40.
σs
√ ( 0 ⋅ 0424)
From table (a) we see that for z = 2 ⋅ 40, A = 0 ⋅ 9918, so that the probability that the
sample value of z exceeds 2 ⋅ 40 in absolute value is (1 − 0 ⋅ 9918) = 0 ⋅ 0082 which
differs very slightly from ⋅01. It appears, then we have a border line case so far as the
1 percent level of significance is concerned and that the result is hardly significant.
Again, by using the t-distribution, we find from table (b) that for 49 degrees of
freedom t = 2 ⋅ 68, and the value calculated from the sample is given by
x −µ
1 ⋅ 39 − 1 ⋅ 32
t =
√ ( n − 1) =
√ 49 = 2 ⋅ 38,
σs
√ ( 0 ⋅ 0424)
which is less than 2 ⋅ 68. Hence the difference is not quite significant at 1 percent
level of significance.
This is an example of the occassional need for a careful study of the variable t or z to
be used and the usefulness of a more complete table of t.
Ex. 71: An observer made the following observations on the vertical diameter of the
planet Venus ( in second of arc) :
42 ⋅ 70, 42 ⋅ 56, 43 ⋅ 01, 43 ⋅ 48, 42 ⋅ 76, 43 ⋅ 06, 43 ⋅ 63, 42 ⋅ 87, 41 ⋅ 60, 42 ⋅ 78, 42 ⋅ 95, 43 ⋅ 20,
43 ⋅ 18, 43 ⋅ 39, 43 ⋅ 10
Assuming that the population of reading is normally distributed about a true value
which is estimated by the arithmetic mean, calculate 95 percent confidence limits
for the vertical diameter of Venus.
Sol : We first calculate mean and S.D. as follows :
x
( x − 43)
( x − 43) 2
42 ⋅ 70
–0.30
0 ⋅ 0900
42 ⋅ 56
–0.44
0 ⋅ 1936
43 ⋅ 01
0.01
43 ⋅ 48
0.48
0 ⋅ 2304
42 ⋅ 76
–0.24
0 ⋅ 0576
43 ⋅ 06
0.06
0 ⋅ 0036
43 ⋅ 63
0.63
0 ⋅ 3969
42 ⋅ 87
–0.13
0 ⋅ 0169
41 ⋅ 60
–1.14
1 ⋅ 9600
42 ⋅ 78
–0.22
0 ⋅ 0484
Sampling Theory
421
42 ⋅ 95
–0.05
0 ⋅ 0025
43 ⋅ 20
0.20
0 ⋅ 0400
43 ⋅ 18
0.18
0 ⋅ 0324
43 ⋅ 39
0.39
0 ⋅ 1521
43 ⋅ 10
0.10
0 ⋅ 0100
0.73
3 ⋅ 2345
Assumed mean A = 43
∴
and
x = 43 −
σ 2x =
0 ⋅ 73
= 43 − 0 ⋅ 0487 = 42 ⋅ 95 approx.
15
3 ⋅ 2345  0 ⋅ 73
−

 15 
15
2
= 0 ⋅ 2156 − 0 ⋅ 0024 = 0 ⋅ 2132
∴
σ x = 0 ⋅ 46 approx.
Now the five percent value of t for 14 degrees of freedom is 2 ⋅ 145 i.e. t 0⋅05 = 2 ⋅ 145.
x −µ
Also
t =
√ ( n − 1).
σx
Hence the 95 percent fiducial limits for µ are given by

x − µ √ ( n − 1)
 < t 0⋅05
 σx

or
or
x−
σx
t 0⋅05 ≤ µ ≤ x +
σx
t 0⋅05
√ ( n − 1)
√ ( n − 1)
0 ⋅ 46
0 ⋅ 46
42 ⋅ 95 −
× 2 ⋅ 145 ≤ µ ≤ 42 ⋅ 95 +
× 2 ⋅ 145
√ (14)
√ (14)
or
42 ⋅ 95 − 0 ⋅ 26 ≤ µ ≤ 42 ⋅ 95 + 0 ⋅ 26
or
42 ⋅ 69 ≤ µ ≤ 43 ⋅ 21.
Hence 95 percent confidence limits for the vertical diameter of Venus are 42 ⋅ 69 and
43 ⋅ 21.
9.20 Test of Significance of the Difference Between the Sample Means
Let there be two independent samples x1 , x 2 , x 3 , ..., x n and y1 , y 2 , y 3 ,… y n
with mean x and y and standard deviations σ x and σ y from normal population
with the same variance. We have to test two hypothesis.
(i) The population mean µ1 and µ 2 are different.
(ii) The population means µ1 and µ 2 are the same.
For (i) variate t is defined by the relation
( x − y ) − (µ1 − µ 2 )
...(9.1)
t =
1/ 2
1
1
s
+

n2 
 n1
where
s2 =
1
{ n1σ 2x + n 2σ 2y }
( n1 + n 2 − 2)
422
Engineering Mathematics-III
=
1
{ Σ ( x1 − x ) 2 + Σ ( y1 − y ) 2 }
( n1 + n 2 − 2)
If A1 and A 2 are the assumed means for two series, then
 Σ ( x1 − A1 ) 2 + Σ ( x 2 − A 2 ) 2 − n1 ( x1 − A1 ) 2 − n 2 ( x 2 − A 2 ) 2 
s2 = 

n1 + n 2 − 2


1/ 2
It can be shown the variate t defined by (9.1) has Fisher’s t–distribution with
n1 + n 2 − 2 degrees of freedom since both x and y are calculated from the data.
x− y
And for (ii),
...(9.2)
t =
1/ 2
1

1
s
+

n2 
 n1
having the same distribution with the same number of degrees of freedom.
Paired data. If the two samples are of the same size and the data are paired, then
we define t by the relation
d −µ
...(9.3)
t =
√ n,
s
1
where
s2 =
Σ ( dr − d ) 2
n−1
dr = difference of rth members of the samples; d = means of difference, i.e.,
1
d = Σ dr ; and µ = population mean of the differences.
n
If
...(9.4)
µ = 0, t = d / s √ n.
The number of degrees of freedom in this case = n − 1.
Ex. 72: Eleven school boys were given a test in drawing. They were given a month’s
further tuition and a second test of equal difficulty was held at the end of it. Do the
marks give evidence that the students have benefitted by extra coaching ?
Boys
1
Marks - Ist test
23
Marks - 2nd test
24
2
20
19
3
19
22
4
21
18
5
18
20
6
20
22
7
18
20
8
17
20
9
23
23
10
16
20
11
19
17
Sampling Theory
423
Sol : We calculate the mean and the standard deviation of difference between the
marks of the two tests as follows :
(d − d )2
Boys
x
y
d= y − x
1
23
24
1
0
0
2
20
19
–1
–2
4
3
19
22
3
2
4
4
21
18
–3
–4
16
5
18
20
2
1
1
6
20
22
2
1
1
7
18
20
2
1
1
8
17
20
3
2
4
9
23
23
0
–1
1
10
16
20
4
3
9
11
19
17
–2
–3
9
d−d
Σ d = 11
Total
Mean of the difference = d =
s2 =
11
= 1,
11
1
50
Σ (d − d )2 =
= 5,
n−1
10
so that s = 2 ⋅ 24.
Now we take the hypothesis that the students have not been benefitted by the extra
coaching. This implies that the mean of the difference between the marks of the two
tests is zero i.e. µ = 0.
−0
1− 0
and
√n =
√ 11 = 1 ⋅ 482
ν = 11 − 1 = 10.
s
2 ⋅ 24
The value of t for 10 degrees of freedom at 5 per cent level of significance is 2 ⋅ 228.
The calculated value is very much less than this value and so the difference is not
significant. That is, the test provides no evidence that the students have benefitted
by the extra coaching.
Ex. 73: The following table shows the mean number of bacterial colonies per plate
obtainable by four slightly different methods from soil samples taken at 4 P.M. and 8
P.M. respectively :
Then
t =
d
Method A
Method B
Method C
Method D
4 P.M.
29 ⋅ 75
27 ⋅ 50
30 ⋅ 25
27 ⋅ 80
8 P.M.
39 ⋅ 20
40 ⋅ 60
36 ⋅ 20
42 ⋅ 40
Are there significantly more bacteria at 8 P.M. than at 4 P.M. Calculations of the
mean and standard deviation of the difference between the number of bacterial
colonies per plate at 4 P.M. and 8 P.M.
424
Engineering Mathematics-III
Sol :
Method
4 P.M.
x
8 P.M.
y
d= y − x
d−d
(d − d )2
A
29 ⋅ 75
39 ⋅ 20
9 ⋅ 45
− 1 ⋅ 325
1 ⋅ 756
B
27 ⋅ 50
40 ⋅ 60
13 ⋅ 10
2 ⋅ 325
5 ⋅ 021
C
30 ⋅ 25
36 ⋅ 20
5 ⋅ 95
− 4 ⋅ 825
23 ⋅ 281
D
27 ⋅ 80
42 ⋅ 40
14 ⋅ 60
3 ⋅ 825
14 ⋅ 631
43 ⋅ 10
Total
Mean
d =
45 ⋅ 689
43 ⋅ 1
45 ⋅ 689
= 10 ⋅ 775; s 2 =
= 15 ⋅ 2296.
4
4−1
We take our hypothesis that there is no difference between the bacteria at 4 P.M.
and 8 P.M. i.e. µ = 0
Hence
t =
d
−0
10 ⋅ 75
21 ⋅ 550
√n =
√4 =
= 5 ⋅ 53.
s
√ (15 ⋅ 2296)
√ (15 ⋅ 2296)
Now for 3 degrees of freedom t ⋅05 = 3 ⋅ 182 from the tables. The calculated value of t
is much more than this value.
Hence the difference is significant and the hypothesis is refuted. It follows that there
are significantly more bacteria at 8 P.M. than at 4 P.M.
Ex. 74: The yields of two types ‘Type 17’ and ‘Type 51’ of grains in pounds per acre
in 6 replications are given below. What comment would you make on the difference
in the mean yields ?
You may assume that if there be 5 degrees of freedom and P = ⋅ 02, t is
approximately 1 ⋅ 476.
Replication
Yield in pounds
Type 17
Yields in pounds
Type 51
1
20 ⋅ 50
24 ⋅ 86
2
24 ⋅ 60
26 ⋅ 39
3
23 ⋅ 06
28 ⋅ 19
4
29 ⋅ 98
30 ⋅ 75
5
30 ⋅ 37
29 ⋅ 97
6
23 ⋅ 83
22 ⋅ 04
Sol : Calculate the mean and S.D. of the differences in the yields of the two types of
grains :
Replication
x
y
d= y − x
(d − d )
(d − d )2
1
20 ⋅ 50
24 ⋅ 86
4 ⋅ 34
2 ⋅ 717
7 ⋅ 38
2
24 ⋅ 60
26 ⋅ 39
1 ⋅ 79
0 ⋅ 147
0 ⋅ 02
Sampling Theory
425
3
23 ⋅ 06
28 ⋅ 19
5 ⋅ 13
3 ⋅ 487
12 ⋅ 15
4
29 ⋅ 98
30 ⋅ 75
0 ⋅ 77
− 0 ⋅ 873
0 ⋅ 76
5
30 ⋅ 37
29 ⋅ 97
− 0 ⋅ 40
− 2 ⋅ 043
4 ⋅ 17
6
23 ⋅ 83
22 ⋅ 04
− ⋅ 1 ⋅ 79
− 3 ⋅ 433
11 ⋅ 79
9 ⋅ 86
Mean of the difference of the yields d =
s2 =
36 ⋅ 27
9 ⋅ 86
= 1 ⋅ 643 lbs.
6
Σ (d − d )2
36 ⋅ 27
=
= 7 ⋅ 254 so that s = 2 ⋅ 69 lbs.
n−1
5
We adopt the null hypothesis that the difference in types has no effect on yields, i.e.
µ = 0.
∴
t =
d
−0
1 ⋅ 643
√n =
√ 5 = 1 ⋅ 489.
s
2 ⋅ 69
Now for the given value of t = 1 ⋅ 476, we have PF = 0 ⋅ 2 for 5 degrees of freedom.
The calculated value is greater than this value so that the difference is significant at
2% level of significance although it is not significant at 5% conventional level of
significance.
Ex. 75: Ten pairs of maize plants were grown in parallel boxes, and one member of
each pair was treated by receiving a small electric current. The differences in height
between the treated and untreated in m.m. were :
(treated)–(untreated) : 6 ⋅ 0, 1 ⋅ 3, 10 ⋅ 2, 23 ⋅ 9, 3 ⋅ 1, 6 ⋅ 8, − 1 ⋅ 5, − 14 ⋅ 7, − 3 ⋅ 3 and
11 ⋅ 1. Test whether the small electric current effected the growth of maize seedlings.
Sol : We set up the null hypothesis that the electric current does not affect the
growth, so that the observed differences are random observations from a population
with zero mean. By actual calculations, we find
42 ⋅ 9
x=
= 4 ⋅ 29
10
1
s2 =
Σ ( x − x ) 2 = 104 ⋅ 13
n−1
[this is left as an exercise for the students].
x −µ
4 ⋅ 29 − 0
4 ⋅ 29
∴
t =
√ (10) =
√ (10) =
= 1 ⋅ 33.
s
3 ⋅ 227
√ (104 ⋅ 13)
For 9 degrees of freedom, the value of t at 5 percent level of significance is 2 ⋅ 262
which is greater than the calculated value. Hence the hypothesis is not rejected. In
other words, the sample does not provide satisfactory evidence that the electric
treatment has made any difference to the growth of maize seedlings.
Ex. 76: For a random sample of 10 pigs, fed on a diet A, the increases in weight in a
certain period were 10, 6, 16, 17, 13, 12, 8, 14, 15, 9 lbs. For another random
sample of 12 pigs, fed on diet B, the increases in the same period were, 7, 13, 22, 15,
12, 14, 18, 8, 21, 23, 10, 17 lbs.
426
Engineering Mathematics-III
Find if the two samples are significantly different regarding the effect of diet, given
that for (d.f.) ν = 20, 21, 22, the five percent values of t are respectively 2 ⋅ 09,
2 ⋅ 07, 2 ⋅ 06.
(Rajasthan M.Sc. 2000)
Sol : Calculation of mean and S.D. of the two series is shown below :
Diet A
Diet B
y
y − y
( y − y )2
4
7
−8
64
−6
36
13
−2
4
16
4
16
22
7
49
17
5
25
15
0
0
13
1
1
12
−3
9
12
0
0
14
−1
1
8
−4
16
18
3
9
14
2
4
8
−7
49
15
3
9
21
6
36
9
−3
9
23
8
64
10
−5
25
17
2
4
x
x− x
10
−2
6
120
(x − x )
2
120
180
314
120
= 12, n1 = 10, Σ ( x − x ) 2 = 120;
10
180
y =
= 15, n 2 = 12, Σ ( y − y ) 2 = 314;
12
ν = n1 + n 2 − 2 = 10 + 12 − 2 = 20.
We assume that the samples do not differ in weight so far as the two diets are
concerned i.e., µ1 − µ 2 = 0.
x=
Hence
Now
so that
t =
(y − x)− 0
s
 n1 n 2 

.
 n1 + n 2 
1
− [Σ
n1 + n 2 − 2
1
=
[120 + 314] =
20
15 − 12  120
t =


√ ( 21 ⋅ 7 )  22 
s2 =
( x − x )2 + Σ ( y − y )2 ]
434
= 21 ⋅ 7
20
3 × √ ( 5 ⋅ 4545)
=
√ ( 21 ⋅ 7 )
3 × 2 ⋅ 335 7 ⋅ 005
=
=
= 1 ⋅ 5 nearly.
4 ⋅ 658
4 ⋅ 658
Sampling Theory
427
For ν = 20 the five percent value of t is 2 ⋅ 09. Since the calculated value of t is less
than this value, the two diets do not differ significantly as regards the increase in
weight.
Ex. 77: Two horses A and B were tested according to the time (in seconds) to run a
particulat track with the following result :
Horse A
28
30
32
33
33
29
Horse B
29
30
30
24
27
29
34
Test whether you can discriminate between two horses. You can use the fact that 5
percent value of t for 11 degrees of freedom is 2 ⋅ 20.
Sol : Calculation of the mean and S.D. of the two samples is given below :
Horse A
Horse B
x
( x − 32)
( x − 32) 2
y
( y − 29)
( y − 29) 2
28
−4
16
29
0
0
30
−2
4
30
1
1
32
0
0
30
1
1
33
1
1
24
−5
25
33
1
1
27
−2
4
29
−3
9
29
0
0
34
2
4
–
–
–
219
−5
35
169
−5
31
219
= 31 ⋅ 3, n1 = 7, ;
7
169
y =
= 28 ⋅ 1, n 2 = 6,
6
ν = n1 + n 2 − 2 = 11.
x=
1
35  5
Σ ( x − x )2 =
− − 
 7
n1
7
31
= 35 −
= 31 ⋅ 43
7
σ 2x =
n1 σ 2x
2
2
=
35 25
−
7
49
1
31  5
Σ ( y − y )2 =
− 
 6
n2
6
25
or
n 2σ 2y = 31 −
= 26 ⋅ 83.
6
On the null hypothesis that the horses do not differ, we get
and
σ 2y =
t =
x− y
s
 n1 n 2 

 ,
 n1 + n 2 
428
where
so that
Engineering Mathematics-III
31 ⋅ 43 − 28 ⋅ 1
11
31 ⋅ 3 − 28 ⋅ 1
t =
√ ( 5 ⋅ 3)
s2 =
= 5 ⋅ 3 approx.
 42 = 3 ⋅ 2 √ ( 3 ⋅ 2) = 5 ⋅ 76 = 2 ⋅ 5 approx.
 
 13 √ ( 5 ⋅ 3)
2⋅ 3
The five percent value of t for 11 degrees of freedom is given to be 2 ⋅ 20. The
calculated value of t is greater than this value and so the difference is significant.
Hence we can discriminate between the two horses. We can discriminate between
the horses only with 75 degrees of confidence.
Ex. 78: A certain stimulus administered to each of 12 patients resulted in the
following change in blood pressure :
5, 2, 8, − 1, 3, 0, − 2, 1, 5, 0, 4, 6.
Can it be concluded that the stimulus will in general be accompanied by an increase
in blood pressure ?
(Meerut M.Sc. 2003; Delhi University 2003)
d = 5, 2, 8, − 1, 3, 0, − 2, 1, 5, 0, 4, 6.
d 2 = 25, 4, 64, 1, 9, 0, 4, 1, 25, 0, 16, 36
31
Σ d = 31, Σ d 2 = 185, d =
= 2 ⋅ 583
12
s=
=
 1

Σ (d − d )2  =

n − 1

 1
{ Σ d 2 − nd

n − 1
2

}

1
{185 − 12 × ( 2 ⋅ 58) 2 } = 3 ⋅ 09
11
2 ⋅ 58 √ 12
d
√n =
= 2⋅ 9
s
3 ⋅ 09
ν = 12 − 1 = 11.
The table of t at 5% level = 2 ⋅ 201 The observed value of t is greater than the table
value and hence the difference is significant i.e., the stimulus is accompanied by
increase in blood pressure.
Ex. 79: Two independent samples of 8 and 7 items respectively had the following
values :
∴
t =
Sample 1
9
11
13
11
15
9
12
Sample 2
0
12
10
14
9
8
10
14
Is the difference between the means of samples significant ? Given that
 P = 0 ⋅ 875 for t = 1 ⋅ 2.
y = 3

 P = 0 ⋅ 892 for t = 1 ⋅ 3.
(Kanpur M.Sc. 2005; Purvanchal B.Sc. 2006; Agra M.Sc. 2004)
Sol : Calculation of mean; S.D. is given next :
Sampling Theory
429
Sample I
Sample II
y
( y − 10)
( y − 10) 2
9
10
0
0
−1
1
12
2
4
13
1
1
10
0
0
11
−1
1
14
4
16
15
3
9
9
−1
1
9
−3
9
8
−2
4
12
0
0
10
0
0
14
2
4
–
–
–
Total : 94
−2
34
73
3
25
x
( x − 12)
9
−3
11
We then have
n1 = 8, x =
σ 2x =
34  2
− 
 8
8
n 2 = 7, y =
σ 2y =
∴
( x − 12)
t =
2
94
= 11 ⋅ 75.
8
2
or
8σ 2x = 33 ⋅ 5.
73
= 10 ⋅ 43.
7
25  3
− 
 7
7
2
or
11 ⋅ 75 − 10 ⋅ 43
 33 ⋅ 5 − 23 ⋅ 7 




12
1/ 2
7σ 2y = 25 −
9
= 23 ⋅ 7
7
1 ⋅ 32
 8 × 7
√ ( 3 ⋅ 73).

 =
 15  √ ( 4 ⋅ 4)
t = 1 ⋅ 2144.
Now for 13 degrees of freedom, we have
for t = 1 ⋅ 2,
P = 0 ⋅ 874 ;
for t = 1 ⋅ 3,
P = 0 ⋅ 892.
Hence for t = 1 ⋅ 2144 , P = 0 ⋅ 8765.
∴
PF = 2 (1 − P ) = 2 (1 − 0 ⋅ 8765) = ⋅ 2470 > 0 ⋅ 05.
Since the value of PF is considerably greater than 0 ⋅ 05, the difference between the
mean of the two samples is not significant.
Ex. 80: In three samples of 50 lines each from Shakespeare’s “Romeo and Julliet”
(an early play), the following numbers of weak endings were observed; 7, 9, 10. In
three similar samples from “Cymbeline” (late), the number of weak endings were
15, 11, 12. Discuss the suggestion that Shakespeare’s prosody, as judged by the
number of weak endings, changed with advancing years.
25
Sol : Here
Σx = 7 + 9 + 10 = 26, n1 = 3 so that
x=
= 8 ⋅ 666.
3
430
Engineering Mathematics-III
and
Σx 2 = 49 + 81 + 100 = 230.
∴
σ 2x
or
Similarly
∴
and
∴
Σx 2  Σx 
=
− 
n1
 n1 
2
=
230  26
− 
 3
3
676
= 230 − 225 ⋅ 333 = 4 ⋅ 667.
3
Σy = 15 + 11 + 12 = 38, n 2 = 3.
38
Σy =
= 12 ⋅ 666
3
Σy 2 = 225 + 121 + 144 = 490.
3σ 2x = Σ ( x − x ) 2 = 230 −
σ 2y =
Σy 2  Σy 
− 
n2
 n2 
2
=
490  38
− 
 3
3
3σ 2y = Σ ( y − y ) 2 = 490 −
Hence
2
t =
2
1444
= 490 − 481 ⋅ 333 = 8 ⋅ 667.
3
x− y
2  1/ 2
 Σ ( x − x )2 + Σ ( y − y )


n1 + n 2 − 2


=
or
8 ⋅ 666 − 12 ⋅ 666
1/ 2
 n1 n 2 


 n1 + n 2 
 3
 
 2
 4 ⋅ 667 + 8 ⋅ 667 


4


8
9 ⋅ 7979
|t | =
× 1 ⋅ 2247 =
= 2 ⋅ 683.
3 ⋅ 651
√ (13 ⋅ 334)
Number of d.f. = 3 + 3 − 2 = 4.
From the tables, we get for v = 4, t 0⋅05 = 3 ⋅ 2776.
Since the observed value of t is nearly equal to t 0⋅05 , the difference of the means is
likely to be significant which supports the suggestion.
Ex. 81 : A sample of two observations x1 , x 2 has been drawn from a normal
population. It is required to test the hypothesis that the population mean is 0. Prove
that on the basis of t-test with equal test at 20% level of significance, the hypothesis
will be rejected if :
| x1 + x 2 | ≥ | x1 − x 2 | tan 72 o
Sol : There are two items in the sample so that
x + x2
1
x= 1
, σ s = | x1 − x 2 |
2
2
x1 + x 2
−0
| x + x2 |
2
Hence
t =
√ ( 2 − 1) = 1
1
| x1 − x 2 |
| x1 − x 2 |
2
For two tailed 20% test the table value for 1 d.f. is 3 ⋅ 078 = tan 72 o . The hypothesis
is rejected if t ≥ 3 ⋅ 078. Hence the result.
Sampling Theory
431
For 10% level of significance, the table value is 6 ⋅ 314 = tan 81 o .
9.18.4 Limitation of t-test
While drawing inferences on the basis of t-test, it should be remembered that the
conclusions arrived at on the basis of the 't-test' are justified only if the assumptions
upon which the test is based are true. If the actual distribution is not normally
distributed then, strictly speaking, the t-test is not justified for small samples. If is
not a random sample, then the assumption that the observations are statistically
independent is not justified and the conclusions based on the t-test may not be
correct. The effect of violating the normality assumption is slight when making
inference about means provided that the sampling is fairly large when dealing with
sample. However, it is a good idea to check the normality assumption, if possible. A
review of similar samples or related research may provide evidence as to whether or
not the population is normally distributed.
9.19 Limitations of Tests of Significance or Sampling
In testing statistical significance the following points must be noted :
1.
They should not be used mechanically. Tests of significance are simply the raw
materials from which to make decisions, not decisions in themselves. There
may be situations where real differences exist but do not produce evidence
that they are statistically significant or the other way round. In each case it is
absolutely necessary to exercise great care before taking a decision.
2.
Conclusions are to be given in terms of probabilities and not certainties. When
a test shows that differences was statistically significant it suggests that the
observed difference is probably not due to chance. Thus statements are not
made with certainty but with a knowledge of probability. Unusual events do
happen once in a while.
3.
They do not tell us why the difference exists. Though tests can indicate that the
difference has statistical significance, they do not telly us why the difference
exists. However, they do suggest the need for further investigation in order to
reach definite answers.
4.
If we have confidence in a hypothesis it must have support beyond the
statistical evidence. It must have a rational basis. This phrase suggests two
conditions : first, the hypothesis must be 'reasonable' in the sense of
concordance with a prior expectation. Secondly, the hypothesis must fit
logically into the relevant body of established knowledge.
The above points clearly show that in problems of statistical significance as in other
statistical problems, technique must be combined with good judgment and
knowledge of the subject-matter.
432
Engineering Mathematics-III
Objective Type Questions
Multiple Choice Questions
Tick the correct alternative :
1.
Standard error of number of successes is given by :
(a)
pq
n
(b) npq
(c) npq
(d)
np
qn
(e) n 2 p 2
2.
3.
95% fiducial limits of population mean are :
(a) X ± 3S. E.
(b) X ± 2 .51 S. E.
(c) X ± 2 .85 S. E.
(d) X ± 1 .96 S. E.
99% fiducial limits of population mean are :
(a) X ± 2 .58 S. E.
(b) X ± 1 .96 S. E.
(c) X ± 3 S. E.
(d) X ± 2 S. E.
(d) none of these.
4.
Large sample theory is applicable when :
(a) N is > 30
(b) N is > 30,
(c) N= 30
(d) N is at least 100
(e) N is at least 1,000.
5.
Student's 't' distribution was discovered by :
(a) Karl Pearson,
(b) Laplace
(c) Fisher
(d) Gosset,
(e) None of them,
6.
The difference of two means in case of small samples is tested by the formula :
(a) t =
X1 − X 2
S
(b) t =
(c) t =
X1 − X 2
S
n1 n 2
(d) t =
n1 + n 2
(e) t =
X1 − X 2
S
n1 − n 2
n1 n 2
X1 − X 2
S
n1 n 2
n1 + n 2
n1 + n 2
n1 n 2
Sampling Theory
433
State True or False
1. H a represents the null hypothesis and H o alternative hypothesis.
2. When the hypothesis is true and our test accepts it this is called Type I error.
3. The reciprocal of standard error is a measure of reliability or precision of the
sample.
4. An estimator is said to be biased if its expected value is identical with the
population parameter being estimated.
5. A point estimate is a single number which is used as an estimate of the
unknown population parameter.
6. An interval estimate should generally be preferred to point estimate.
7. The term standard error of the mean is used to refer to the standard deviation
of the distribution of sample means.
8. The sampling distribution has a standard deviation equal to the population
standard deviation divided by the sample size.
9. The sample mean X is the best estimator to the population mean µ.
10. A t-distribution is lower is the mean and higher at the tails than a normal
distribution.
Fill in the Blanks
1. The null hypothesis asserts that there is no true difference in the ...................
and the .................... in the particular matter under consideration.
2. Type I error is committed when the hypothesis is true but our test ................ it.
3. Type II errors are made when we accept a null hypothesis which is ..............
4. In ................. tail test the rejection region is located in one tail.
5. The standard deviation of sampling distribution is called ....................
6. The distribution formed of all possible values of a statistic is called the ............
7. The mean of sampling distribution of mean is equal to the .................
8. Standard error provides an idea about the ............... of sample.
9. An estimator is said to be .................. if it covers as much information as
possible about the parameter which is contained in the sample.
10. An .................. estimate of a population parameter provides two values,
between which it is estimated that parameter lies.
434
Engineering Mathematics-III
Short and Long Answer Type Questions
1. (a) What is test of significance ? Explain the concept of standard error related to it.
(b) Explain the concepts of Sampling Distribution and standard error. Discuss
the role of standard error in the large sample theory.
2. Write a note on either 'Sampling distribution of a statistic' or 'random sampling'.
3. Distinguish clearly between (i) Point estimation, and (ii) Interval estimation.
4. What is a null-hypothesis ? State the null-hypothesis which leads to
development of t-test. Give the situations where you will apply :
(i) Lift-tail test,
(ii) Right-tail test, and (iii) Two-tail test.
5. (a) Explain briefly the terms sampling distribution, standard error and level of significance.
(ICWA, Dec. 1985)
(b) Explain the concept of sampling distribution of sample statistics and points
out its role in statistical analysis.
6. Give some important application of the t-test and explain how it helps in taking
business decisions.
7. (a) Stating clearly the assumptions explain how will you proceed to test the
significance of the difference between (i) mean from two samples (ii) two
observed correlation coefficients.
(b) Explain how t-test is applied to test the difference of two population means.
8. (a) Define null hypothesis. State the hypothesis when the tests employed are
(i) t-test, and (ii) z-test.
(b) What do you mean by confidence interval ? How will you determine the
confidence interval for normal population means ?
9. Differentiate the following parts of concepts :
(i) Statistic and parameter,
(ii) Critical region of acceptance.
(iii) Null and alternative hypothesis.
10. (a) Explain the concept of statistical significance. What is meant by the
statement "the sample mean is significant at the 5% Level of Significance" ?
(b) Explain the notion of sampling distribution of an estimator. Why is the
knowledge of sampling distribution important in statistical theory ?
11. (a) Write down the formulae for standard error of sample mean, where
sampling is done from a finite population with replacement and without
replacement.
(b) Define : (i) Unbiased estimates, (ii) Efficient estimate, and (iii) interval
estimate.
12. (a) Why should there be different formulae of testing a significance of the
difference between means, when the samples are : (i) sample and small sample
test of significance. Explain the assumptions in large sample theory.
13. (a) What is standard error ? What are its uses ?
(b) Explain the concept of standard error in sampling distribution.
Sampling Theory
435
14. Explain how the Student's 't' distribution may be used to :
(i) Test the significance of the sample correlation coefficient in a sample drawn
from a bivariate normal population.
(ii) Test the significance of the difference between the yield of two varieties in
an agricultural experiment.
(iii) Explain briefly different applications of the t-test.
15. (a) Explain the uses of t-test and under what assumptions the test will be valid.
(b) Explain clearly the procedure generally followed in testing of a hypothesis.
(ICWA, June 1987)
16. (a) What is 'null hypothesis' ? How is it different from alternative hypothesis ?
(b) What is the significance of standard error in statistical work ?
(c) What is the difference between large sample and small sample ? List the
factors which affect the validity of a small sample.
(d) Explain how the Student's 't' test is a landmark in the development of
statistical method.
17. (a) Describe briefly the design of a statistical test of significance.
(b) Explain the meaning and importance of sampling distribution.
(c) How does small sampling theory differ from large sampling theory
18. What are the factors which lead to type I and type II errors in the process of
testing of hypothesis ? Explain with illustrations.
19. (a) Explain clearly the term sampling distribution and point out its role in
statistical analysis.
(b) What is t-test ? How is this test used to examine the significance of the
difference between two sample means ?
20. (a) State the principles of large sample theory. Describe the procedures for a
one latted large sample test of hypothesis.
(b) How does small sampling theory differ from large sampling theory ?
Explain the meaning and importance of sampling distribution.
21. (a) What are two types of errors in testing of hypothesis ? Which type is used in
test of significance ?
(b) For testing whether A has committed murder and is consequently to be
hanged to death, frame the null hypothesis. Is type I error to be assigned a low
or high level of significance ?
(c) Explain the criteria of a good estimator.
(d) Explain the terms 'Efficiency' and 'unbiasedness' of an estimate.
22. (a) Define the term 'Standard Error' Discuss its importance in large sample theory.
(b) Discuss the standard error of mean of mean as a parameter to assess
variability.
(c) What are the relative merits of point and interval estimates ?
436
Engineering Mathematics-III
(d) What is a "Point estimator" ? Give examples. How will you proceed to
construct an interval estimate of the mean of a normal population with known
variance ?
(e) Explain clearly the properties of a good estimator.
23. (a) Explain how you would test the equality of means populations when you
know that their standard deviations are equal but unknown and the
populations are independent, using small samples.
(b) In a correlation study 18 pairs of values had a correlation of – 0.75, test
whether this value is significant, using an appropriate 't' test.
24. (a) What is meant by a research hypothesis ? How is it formulated ? Mention
the characteristics of a hypothesis with suitable illustrations.
(b) what is test of significance ? Discuss various tests of significance for the
cases when the size of the sample is large.
(c) What are various tests of significance generally used in sampling of
attributes ? Discuss briefly.
25. (a) Explain the method used for testing the difference between means of two
large samples.
(b) What are the various tests of significance generally used in sampling of
attributes ? Discuss briefly.
(c) Why should those be difference in means when samples are : (a) small, and
(b) large ?
26. (a) What are the assumptions and applications of a t-test ? Explain fully
and give its utility in samples.
(b) What is t-test ? Explain its uses.
(c) Describe the procedure followed in testing a hypothesis.
(d) What is 'Test of Significance ' ? Explain the concept of standard error
related to that.
27. (a) Explain briefly the procedure of testing a hypothesis.
(b) Define Student's t-test and explain some of its applications.
28. (a) Explain the term Standard error and show how standard error is used in
large sample test of Statistical testing hypothesis.
(b) Explain the concept of 'significance'. Give the general procedure adopted
for testing the significance or testing hypothesis.
29. (a) What do you mean by 'level of significance' in testing a hypothesis ?
(b) Point out the characteristics of a good usable hypothesis.
(c) What do you understand by 'Standard Error' ? What are it importances ?
30. (a) Explain the terms : (i) Type I and Type II errors ; (ii) Critical region ; and
(iii) Level of Significance associated with testing hypothesis.
(b) Define a "Simple Hypothesis". Outline a method of testing a simple
hypothesis against a simple alternative.
Sampling Theory
437
(c) What do you mean by random sampling ? Define standard error of a
statistic Point out the importance of standard error in large sample theory.
31. (a) When and for what purposes 't' test of significance is used ?
(b) Illustrative the importance of level of significance in testing a hypothesis.
(c) Explain the concept of sampling distribution in general and also explain the
sampling distribution of mean in particular. What are its properties ?
(d) What is a point estimator ? Explain any one method of point estimation.
Give the form of an interval estimator of an unknown population mean of a
normal population with known variance σ 2 .
(e) Explain Type I and Type II error associated with testing of hypothesis. How
will you proceed to test difference between the mans in two populations, based
on large samples drawn from the populations.
Unsolved Numerical Problems
1.
A coin is tossed 400 times and it turns up head 216 times. Discuss whether the
coin may be an unbiased one, and explain briefly the theoretical principles you
would use for this purpose.
[z = 1 ⋅ 6 and so the coin can be said to be an unbiased one.]
2.
A biased penny is tossed 100 times and comes down heads 70 times. What are
the probable limits to the probability of getting a head in a single trial ?
[p lies between 0 ⋅ 55 and 0 ⋅ 85]
3.
In tossing a hundred pennies a student gets 66 heads. Do you think that he had
used sufficient care to obtain a random toss each time ?
[z = 3 ⋅ 2 and so the toss cannot be said to be random].
4.
A certain cubical die was thrown 9,000 times, and a 5 or 6 was obtained 3,240
times. On the assumption of random throwing, do the data indicate an
unbiased dice ?
[Die is biased]
5.
Twelve dice are thrown 3086 times and a throw of a 2, 3, 4 is reckoned as a
success. Suppose that 19142 throws of a 2, 3 or 4 have been made out. Do you
think that this observed value deviates from the expected value ? If so, can the
deviation from the expected value be due to fluctuations of simple sampling ?
[z = 6 ⋅ 5 and so the deviation is most unlikely to have risen due to
fluctuations of simple sampling].
6.
In a sample of 100 in a district, 60 are found to be wheat–eaters and 40
rice-eaters. Can we assume that both the food article are equally popular ?
[z = 2 and so the difference may be due to fluctuations of simple sampling, i. e.
the two food articles may be considered as equally popular].
438
7.
Engineering Mathematics-III
Balls are drawn from a bag containing equal numbers of black and white balls,
each being returned before drawing another. In 2250 drawings, 1018 black
and 1232 white balls have been drawn. do you suspect some bias on the part of
the drawer ?
[z = 4 ⋅ 5 so that it seems reasonable to suspect that the draw was biased].
8.
A man buys 1000 sacks of potatoes. He finds that from 1,000 potatoes, chosen
from the sacks at random, 400 are of class A, worth Rs. 10 a sack, 250 are of
class B, worth Rs. 7 per sack, 200 are of class C, worth Rs. 5 a sack, and 150 are
of class D, worth Rs. 4 per sack. What are the upper and lower bounds for the
value of the potatoes ?
[Hint. Find the upper and lower values for the four classes of potatoes by
adding and subtracting three times the standard error of the proportions from
actual proportions and express it in percentages. Then the maximum value of
the potatoes is the value for which class A has the maximum value and classes
C and D the minimum value. This gives on calculation Rs. 7667 ⋅ 10.
Similarly the minimum value of the potatoes is that value for which classes
C and D have the maximum value and class A the minimum. This gives Rs.
7032 ⋅ 9. Thus the upper and lower bounds for the value of the potatoes are Rs.
7677 ⋅ 10 and Rs. 7032 ⋅ 9].
9.
Balls are drawn from a bag containing equal number of black and white balls,
each ball being returned before drawing another. Out of 4,096 drawing 2,030
balls were black and 2,066 white. Is this divergence probably significant of bias
9
? [z =
so that the difference is not significant]
10
10. In a simple sample of 600 men from a certain large city, 400 are found smokers.
In one of 900 from another city, 450 are found smokers. Do the data indicate
that the cities are significantly different with respect to the prevalence of
smoking among men ?
[Here z = 6 ⋅ 56 nearly, so that the difference is significant.]
11. In a random sample of 1000 persons from town A, 400 are found to be consumers
of rice. In a sample of 800 from town B, 400 are found to be consumers of rice.
Discuss the question whether the data reveal a significant difference between
A and B so far as the proportion of rice-consumers is concerned.
[z = 4 ⋅ 2, so that difference is significant.]
12. One thousand articles from a factory are examined and found to be 3 percent
defective. Fifteen hundred similar articles from a second factory are found to be
only 2 percent defective. Can it reasonably be concluded that the product of the
first factory is inferior to that of the second ?
[z = 1 ⋅ 6, so that the difference is not significant and hence we cannot
reasonably say that the product of the first factory is inferior to that of the
second].
Sampling Theory
439
13. In a town A, 19400 persons were observed and 27 percent of them were found
to be short-sighted. In town B, 20750 persons were observed and 30 percent
were found to be short-sighted. Can the difference observed in the percentage
of short-sighted persons be attributed solely to the fluctuations of sampling ?
[z = 7 ⋅ 1 and so the difference is a real one.]
14. The sex ratio at birth is sometimes given by the ratio of male to female births,
p
intead of the proportion of male to total births. If z is the ratio, i. e. z = , show
q
1
z
that the standard error of z is approximately
√   , n being large, so
(1 + z )  n 
that deviations are small compared with mean.

p
z
1
,∴π =
,q=
, S. E. =
z =
1
−
p
1
+
z
1
+
z


 pq
  etc.
 n

15. Two large samples of sizes n1 , n 2 indicate proportions p1 , p 2 of an attribute.
Obtain the relevant formulae to test whether the samples belong to
populations.
(1) having the same proportion of that attribute.
(2) having different proportions and in the reverse order of magnitude to that
indicated by the sample proportion.
16. In a random sampling of 800 adults from the population of a certain large city,
600 are found to have dark hair. In a random sample of 1000 adults from the
inhabitants of another large city, 700 are dark-haired. Show that the difference
of proportions of dark-haired people is nearly 24 times the S.E. of this
difference for sample of above sizes.
17. In a sample of 500 people in Kerela, 280 are tea drinkers and the rest are coffee
drinkers. Can we assume that both coffee and tea are equally popular in the
state at 5% level of significance ?[No]
18. In drawing a sample of attributes of n members, the chance of success changes
at each drawing and if pi be the probability of success at the ith drawing, find
the variance E 2 of the number of success in the sample.
19. A simple sample of 1000 members is found to have a mean 3 ⋅ 42. Could it be
reasonable regarded as a simple sample from a large population whose mean is
3 ⋅ 3 cm. and standard deviation 2 ⋅ 6 cm. ?
Yes, since z = ⋅12 < 1.


⋅82
20. A sample of 900 members is found to have mean of 3 ⋅ 4 cm. Can it be
reasonably regarded as a simple sample from a large population with the mean
3 ⋅ 25 cm. and S.D. 2 ⋅ 61 cm.(Meerut 1983)
[No, at 5% level of significance since z = 2 ⋅ 8]
440
Engineering Mathematics-III
21. Suppose that it has been determined that the average pulse rate of males in the
20 − 25 year age-group is 72 beats per minute and that the standard deviation
is 9 ⋅ 5 beats per minute. If a group of 55 distance runners all in the given
age-group, were examined and found to have an average pulse rate of 65,
should this be regarded as a significant deviation from the general average ?
[Yes : z = 5 ⋅ 47]
22. The average of 400 cases is 30 and the standard deviation is 16. (a) Find the
standard error of the mean and the probable error of the mean. Find also the
probability that the average of the population from which the sample is drawn
(b) is greater than 32, (c) is less than 27 ⋅ 5, (d) lies between 29 and 31, and (e)
does not differ from 30 by more than 2
[ (a) ⋅8, ⋅ 54; ( b) ⋅ 0062; ( c ) ⋅ 001; ( d) ⋅ 789; ( e) ⋅ 988 ]
23. If p is the observed proportion of success in n independent Bernoulian trials,
prove that the 99% fiducial limits for the proportion p′ for large samples, are
the roots of the quadratic
( p − p 2 ) ( 2 ⋅ 58) 2 = n ( p′ − p) 2 . (Agra M.Sc. 1980)
24. The grades of students in a certain course averaged 77 over a period of years. A
class of 40 has a mean grade of 70 with a standard deviation of 9. Can this
lower mean be attributed to ordinary sampling variation ?
[No, z = 4 ⋅ 03]
25. The numerical grades of graduates of large college have a mean of 2 ⋅ 83 with a
standard deviation of 0 ⋅ 538. If the mean grade of a group of 36 graduates who
majored in history is found to be 2 ⋅ 97, should this group be considered
different from the general run of graduates ?
[No; z = 1 ⋅ 56]
26. Ten individuals are chosen at random from a population and heights are found
to be in inches : 63, 63, 64, 55, 66, 69, 69, 70, 70, 71. Discuss the suggestion
that the mean height in the universe is 65 inches, given that for 9 degrees of
freedom the value of student’s t at 5 percent level of significance is 2 ⋅ 262.
[t = 2 ⋅ 02 < 2 ⋅ 262, so that it can be said that the mean
height in the universe is 65 inches].
27. Find Student’s t for the following variate values in a sample of 10 :
− 6, − 4, − 3, − 2, − 2, 0, 1, 1, 3, 5.
Taking µ to be zero, find from the tables the probability of getting a value of t as
great or greater on random sampling from a normal population.
[ t = − 0 ⋅ 662, ν = 9, PS = 0 ⋅ 738, so that PF = 2 (1 − 0 ⋅ 728) = 0 ⋅ 524.
∴ the probability that we should get a value of t greater in absolute value is
0 ⋅ 524].
Sampling Theory
441
28. A certain stimulus administered to each of 12 patients resulted in the following
increases of blood pressures :
5, 2, 8, − 1, 3, 0, 6, − 2, 1, 5, 0, 4.
Can it be concluded that the stimulus will be in general, accompanied by an
increase in blood pressure ?
[Hint. x = 2 ⋅ 58, σ s = 2 ⋅ 97, n = 12.
Taking the hypothesis that the stimulus does not result in an increase in blood
pressure, that is µ = 0, we get
t =
x −µ
2 ⋅ 58 − 0
√ ( n − 1) =
√ 11 = 2 ⋅ 747.
σx
2 ⋅ 97
The value of t for 11 degrees of freedom at 5 percent level of significance is
2 ⋅ 201, which is less than the calculated value of t. Hence the hypothesis is
refuted. It can therefore be said that the stimulus will in general be
accompanied by an increase in blood pressure].
29. A machine which produces mica insulating washers for use in electric devices is
set to turn out washers having a thickness of 10 miles ( 1 mile = 0 ⋅ 001 inch.). A
sample of washers has an average thickness of 9 ⋅ 52 miles with a standard
deviation of 0 ⋅ 60 mile. Find out the significance of such a deviation.
[t = 2 ⋅ 528, deviation is significant]
30. A sample of 11 rats from a central population has an average blood of viscosity
of 3 ⋅ 92 with s.d. of 0 ⋅ 61. On the basis of this sample, establish 95 per cent
limits of µ, the mean blood viscosity of central population.
[ 3 ⋅ 51 and 4 ⋅ 33]
31. If the mean I.Q. of 24 delinquent boys is 89 ⋅ 4 with a standard deviation of
11 ⋅ 0, find the 95% confidence limits for the population mean µ.
[ 84 ⋅ 6 < µ < 94 ⋅ 2]
32. The scores of 10 students in test were as follows :
65, 70, 86, 74, 90, 94, 57, 65, 76.
If the mean score of students in general in this test be 69, with standard
deviation 9 ⋅ 3, would you consider the group different from the general run of
students.
[Hint. x = 76, µ = 69, σ = 9 ⋅ 3.
∴
t =
x −µ
76 − 69
√n =
√ (10) = 2 ⋅ 4.
σ
9⋅ 3
For 9 degrees of freedom we get, corresponding to this value of t, PS = 0 ⋅ 980
∴
PF = 2 (1 − 0 ⋅ 98) = ⋅ 04 > 0 ⋅ 01 and ≤ 0 ⋅ 05
Hence the difference is not significant at 1 percent level of significance and is
significant to some extent at 5 percent level of significance].
442
Engineering Mathematics-III
33. The average breaking strength of steel rods is specified to be 18 ⋅ 5 thousand
pounds. To test this a sample of 14 rods was tested and gave the following
results (in units of 1,000 lbs.)
15, 18, 16, 21, 19, 21, 17 , 17, 15, 17, 20, 27, 18.
Is the result of the experiment significant ?
[No. t = 1 ⋅ 23]
34. The weights at birth of 15 babies born in Calcutta hospital are given below.
Each figure is correct to the nearest tenth of a pound.
6 ⋅ 2, 6 ⋅ 7, 7 ⋅ 1, 6 ⋅ 9, 7 ⋅ 5, 5 ⋅ 7, 4 ⋅ 8, 6 ⋅ 8, 7 ⋅ 6, 7 ⋅ 8, 8 ⋅ 1, 5 ⋅ 0, 5 ⋅ 8, 8 ⋅ 9, 8 ⋅ 5.
Find the 1 per cent fiducial limits for the mean weight at birth of all such babies,
given that for 14 degrees of freedom t = 2 ⋅ 77 at 1 percent level of significance.
[ 5 ⋅ 931 < µ < 7 ⋅ 723, x = 6 ⋅ 827 lbs., s = 1 ⋅ 126]
35. A certain colliery is supposed to supply coal of ash content about 15. To test
this, twenty random samples of the colliery’s coal are selected and tested. The
null hypothesis is that the ash content is in fact 15. The results of the twenty
tests gave an average ash content of 16 ⋅ 8 with a standard deviation of 3 ⋅ 6. Is
this sufficient evidence on which to reject the hypothesis ? You are given that 5
percent value of t for 19 degrees of freedom is 2 ⋅ 09. [Yes; t = 2 ⋅ 2]
36. The mean yield per plant for 11 tomato plants of a particular variety was found
to be 1 ⋅ 2873 gms with a standard deviation of 96 ⋅ 41 gms. Set up 99 percent
fiducial limits for the mean yield of all plants of this variety.
[1169 ⋅ 61 and
37. The following are 12 determinations (in degrees centigrade) of the melting
point of a compound made by an analysis, the true melting point being 165 o C.
Would you conclude from these data that his determinations are free from bias
164 ⋅ 4, 169 ⋅ 7, 163 ⋅ 9, 162 ⋅ 1, 160 ⋅ 9, 160 ⋅ 8, 161 ⋅ 4,162 ⋅ 2, 168 ⋅ 5, 163 ⋅ 4,
162 ⋅ 9, 167 ⋅ 7
[Yes; t = 1 ⋅ 49]
38. Discuss t-distribution and determine the moments of this distribution.
[Hint. See § 18 ⋅ 4 and 18 ⋅ 4 (b)].
39. Calculate the value of `t ' in the case of two characteristics A and B whose
corresponding frequencies are
A
16
10
8
9
9
8
B
8
4
5
9
12
4
[t = 1 ⋅ 8 nearly]
Sampling Theory
443
40. In certain experiment to compare two types of pig-food A and B, the following
results of increase in weights were observed in pigs :
Pig Number
1
Increase in Food A 49
Wt. in lbs
Food B 52
2
3
4
5
6
7
8
Total
53
51
52
47
50
52
53
407
55
52
53
50
54
54
53
423
Assuming that the two samples of pigs are independent, can we conclude that
food B is better than food A ? Examine the case when the same set of eight pigs
were used in both the foods. [Food B is better]
41. Mitchell conducted a paired feeding experiment with pigs on the relative value
of limestone and bone-meal for bone development. The results are :
Ash content in percentage of scapulus of pairs fed on limestone and bone-meal.
Pair
Limestone
Bone-meal
1
49 ⋅ 2
51 ⋅ 5
2
53 ⋅ 3
54 ⋅ 9
3
50 ⋅ 6
52 ⋅ 2
4
52 ⋅ 0
53 ⋅ 3
5
46 ⋅ 8
51 ⋅ 6
6
50 ⋅ 5
54 ⋅ 1
7
52 ⋅ 1
54 ⋅ 2
8
53 ⋅ 0
54 ⋅ 3
Determine the significance of the difference between the means in two ways –
(1) by assuming that the values are paired and
(2) by assuming that the values are not paired.
[The difference is significant ; t = 4 ⋅ 444]
42. The marks obtained by 20 students of college A and 15 students of college B in a
mathematics test are given below :
College A 89, 76, 63, 69, 55, 71, 84, 87, 88, 52, 47, 81, 32, 43, 86, 29, 49,73,
80,44.
College B 79, 61, 36, 42, 50, 12, 55, 81, 35, 73, 12, 91, 76, 67, 62.
Do yout think students of colleges A are more proficient in mathematics than
students of college B ? Given that for v ( d. f . ) = 33, t 0⋅05 = 2 ⋅ 36.
[No. t = 1 ⋅ 275]
43. (i) In the manufacture of steel by the open hearth process 63 casts charged with
refined iron showed an average charge to tap time of 10 ⋅ 36 hours with a
variance of 1 ⋅ 07. Then the same number of casts charged with basic iron
showed a corresponding mean and variance of 10 ⋅ 63 and 0 ⋅ 75. Does the
quality of iron used in charging affect the charge time ?
444
Engineering Mathematics-III
(ii) A random sample of 16 values from a normal population showed a men of
41 ⋅ 5" and the sum of the squares of deviations from this mean equal to 135
square inches. Show that the assumption of a mean of 43 ⋅ 5" for the population
is not reasonable. Given :
t15 ( 0 ⋅ 05) = 2 ⋅ 131 and t15 ( 0 ⋅ 01) = 2 ⋅ 947
44. Two samples, one from each of the two normal populations gave the following
results :
Sample
Size
Mean
Variance
I
10
3⋅ 4
3⋅ 0
II
12
4⋅ 0
3⋅ 6
Test whether the two populations are identical.
[v = 20, t = 3 ⋅ 27; t 0⋅05 for v = 20 is 2 ⋅ 086. The difference is significant]
45. The means of two random samples of sizes 9 and 7 respectively are 196 ⋅ 42 and
198 ⋅ 82 respectively. The sum of the squares of the deviations from the means
are 26 ⋅ 94 and 18 ⋅ 73 respectively. Can the samples be considered to have been
drawn from the same normal population, it being given that the value of t for
14 d.f. at 5 percent level of significance is 2 ⋅ 145 and at 1 percent level of
significance is 2 ⋅ 977. [No., t = 3 ⋅ 175]
46. How can `t ' test be applied for testing the significance of the difference between
two sample means ? Calculate the value of t in the case of two characters
A and B whose corresponding values are
A
41, 49, 34, 36, 86, 50, 36, 20, 18.
B
46, 44, 30, 35, 26, 29.
[t = 0 ⋅ 57]
47. Two random samples drawn from two normal populations are :
Sample I
20,16, 26, 27, 23, 22, 18,24, 25, 19.
Sample II
27, 33, 42,35,32, 24, 38, 28,41, 43, 30, 37.
How would you test hypothesis (a) µ1 = µ 2 ( b) σ12 = σ 22 to indicate that the
samples are drawn from the same population
[(a) t, (b) F tests]
48. The following figures show the mean breaking strength index of two types of
cotton fibre. Taking the figures in pairs, determine whether the mean
difference is significant.
Index of Breaking Strengths
Fibre A
7⋅2
7 ⋅7
7⋅8
7⋅6
7⋅3
7⋅8
7 ⋅7
7⋅6
7⋅4
Fibre B
7⋅2
7⋅2
7 ⋅1
7⋅3
7⋅3
7⋅3
7⋅6
7⋅2
7⋅9
[Not significant : t = 0 ⋅ 79]
Sampling Theory
445
49. The yield of wax extracted from peat depends on the nature of the solvent used
to extract the wax. Two solvents were tested on five sample of peat. The table
shows yield of crude wax (percent dry part). Is the effect significant ?
Sample
1
2
3
4
5
Solvent A
2⋅ 3
10 ⋅ 4
2⋅ 4
6⋅ 8
4⋅1
Solvent B
3⋅ 2
11 ⋅ 8
2⋅7
8⋅ 6
5⋅ 2
[t = 4 ⋅ 4, significant on 5 percent and not on 1 percent level of significance.]
50. The nine items of a sample had the following values 45, 47, 50, 52, 48, 47, 49,
53, 51. Does the mean of the nine items differ significantly from the assumed
mean 47 ⋅ 5, given that.
 p = ⋅ 945 for t = 1 ⋅ 8
v = 8
 p = ⋅ 953 for t = 1 ⋅ 9
51. Samples of two types of electric light bulbs were tested for length of life and the
following data were obtained.
Type I
Type II
Sample size
n1 = 8
n2 = 7
Sample mean
x1 = 1,234 hrs
x 2 = 1,036 hrs.
Sample S.D.
s1 = 36 hrs.
s 2 = 40 hrs.
Is the difference in the means sufficient to warrant that Type I is superior to
Type II regarding length of life ?
[Given : Tabulated t 0⋅05 for d.f. 13 = 2 ⋅ 160].
52. Bring out clearly the importance of t-distribution in the small sample theory.
Nine patients, to whom a drug was administered, registered the following
increments in blood pressure :
7, 3, – 1, 4, – 3, 5, 6, – 4, 1.
Compute the statistic which you would use to test whether the data indicate
that the drug was responsible for these increments. Also state how you would
proceed further.
446
Engineering Mathematics-III
ANSWERS
Multiple Choice Questions
1. (b)
2. (d)
5. (d)
6. (c)
3. (a)
4. (a)
Fill in the Blanks
1. Sample, populations
2. Rejects
3. Not true
4. one
5. Standard Error
6. Sampling distribution
7. Population mean
8. Unreliability
9. Sufficient
10. Interval
State True/False
1. False
2. False
3. True
4. False
5. True
6. True
7. True
8. False
9. True
10. True
❑❑❑
Unit-3
Chapter
10
Test of
Significance and Applications
10.1 Test of Significance
Already we have discussed that sampling theory deals with two types of problems as
Estimation and Testing of hypothesis. In this chapter, we shall deal with the
hypoblem of Testing of hypothesis. It was initiated by J. Neyman and E.S. Pearson. It
employs statistical techniques to arrive at decisions in certain situations where there
is an element of uncertainty on the basis of sample, whose size is fixed in advance.
Modern theory of probability plays an important role in decision making and the
branch of statistics which helps us in arriving at the criterion for such decisions is
known as 'Testing of Hypothesis'.
Hypothesis : A hypothesis is a statement about the population parameter. It can
be defined also as 'a hypothesis is a conclusion which is tentatively drawn on logical
basis'.
10.2 Statistical Hypothesis
Statistical hypothesis is a assumption which may or may not be true, about a
population or about the probability distribution characterising of the given
population which we want to test on the basis of the evidence from a random
sample.
Testing of Hypothesis :The testing of hypotheses is a procedure that helps us to
ascertain the likelihood of hypothesized population parameter being correct by
making use of the sample statistic. In testing of hypothesis, a statistic is computed
from a sample driven from the parent population and on the basis of this statistic, it
is observed whether the sample so drawn has come from the population with certain
specified characteristic.
10.3 Procedure for Testing a Hypothesis
The following are the steps involved in hypothesis testing.
1. Setting up of Hypothesis : There are two types of hypothesis :
(i) Null Hypothesis
(ii) Alternative Hypothesis
Null Hypothesis : The statistical hypothesis that is set up for testing a hypothesis
is known as null Hypothesis. It is set up in testing a statistical hypothesis only to
decide whether to accept or reject the null hypothesis. It is usually denoted by H 0 .
448
Engineering Mathematics-III
Alternative Hypothesis : The negation of Null hypothesis is called the
alternative hypothesis. It is set in such a way that the rejection of null hypothesis
implies the acceptance of alternative hypothesis. It is always denoted by N. For
example, if we want to find the null hypothesis that the average height of the
students of a college is 165 (m i.e., µ 0 = 165 cm then the null hypothesis is
H 0 :µ = 165 cm. Then the alternative hypothesis could be
(i) N1 : µ ≠ µ 0
(ii) N1 : µ > µ 0
(iii) N1 : µ < µ 0
Types of Errors in Hypothesis Testing :
It is possible that a decision regarding a null hypothesis may be correct or may not be
correct. There are two types of errors.
(a) Type I Error : It is the Error of rejecting null hypothesis N 0 , when it is true.
The probability of making a type I error is denoted by α. In order to control the type I
error, the probability of type I error is fixed at a certain level of significance α. The
probability of making a correct decision is then (1 – α ).
(b) Type II Error : It is the error of accepting the null hypothesis while it is not
true. The probability of making a type II error is denoted by β.
10.4 Level of Significance
Level of significance is the probability of type I error which are ready to write in
making a decision about H 0 . In other words, level. of significance is the maximum
probability of making a type I error and is denoted by α. The test value for fixing the
level of significance depends on the seriousness of the results of the types of error.
The commonly used level of significance in practice are 5% (.05) and 1% (.01). If we
use 5% level of significance (α =.05) we shall mean that the probability of making
type I error is .05 or 5% It is important to note that if no level of significance is given
then we always take α =.05.For more details, see (9.13) of chapter 9.
10.5 Critical Region or Rejection Region
The region of the standard normal curve corresponding to a predetermined level of
significance is known as the critical region or rejection region. the region under the
normal curve which is not covered by the rejection region is known as acceptance
region. The value of the test statistic computed to test the null hypothesis H 0 is
known as the critical value. For more details, see .................. .
10.6 Student 't' test
This test is used for small sample i.e., if the sample size n < 30 . It is defined as
x–µ
Difference of sample and population means
t =
=
S. E.( x )
Standard error of mean
with ( n – 1) degrees of freedom. The number of independent observation in a set is
called degree of freedom, in other words, degree of freedom may be defined as the
member of observation in a set minus the number of restrictions imposed on them.
where n is the size of sample.
Test of Significance and Applications
449
x–µ
It can also be defined as t =
where n = number of observations in the sample,
s/ n
x = sample mean, µ= The mean of the population from which the sample has been
drawn.
1
s=
∑( xi – x ) 2 is the standard deviation of the sample.
n–1
For more detail on t-test see 9.18 of chapter-9.
Illustrative Examples
Ex.1: A sample of 900 members has a mean 3.4 cms and s.d. 2.61 cms. can the
sample be regarded as one drawn from a population with mean 3.25 cms. Using the
level of significance as .05 is the claim acceptable. Make a conclusion.
Sol : n = 90,
x = 3.4 cm. s.d. of s = 2.61cm and µ = 3.25
Null Hypothesis : The sample has been drawn from a population with mean 3.25
cm, i.e.,
H 0 : µ = 3.25
Alternative hypothesis : N1 : µ ≠ 3.25
x–µ
Z =
S. E . of x
S.E. of
x=
s
2.61
=.087
n
900
3.4 – 3.25
Z =
= 1.72
.087
=
Level of significance
Critical value
α =.05
Z α = 1.96
|Z|< Z α
Hence null hypothesis is accepted.
Thus, the sample has been drawn from a population mean 3.25 cm.
Ex . 2: A manufacture of dry cells claimed that the life of their cell is 24.0 hours. A
sample of 10 cells observed the mean life of 22.5 hours with the standard duration of
3 hours. On the basis of available information, test weather the claim of the
manufacture is correct at 5% level of significance.
Sol: µ = 24.0 hours,
x = 10, x = 22.5 hours standard deviation S = 3 hours
Null hypothesis : The cleaim of manufacture is correct i.e., µ = 24.0 hours
Alternative hypothesis : µ ≠ 24
Ex. 3: A College conducts both day and night class intended to be identical. A
sample of 100 day students yields examination results as under :
x1 = 72 .4 and σ1 = 14 .8
A sample of 200 night students yields examination results as under :
450
Engineering Mathematics-III
x 2 = 73 .9 and σ 2 = 17 .9.
Are the two means statistically equal at 10% level ?
Sol : We have
n1 = 100, x1 = 72 .4, σ1 = 14 .8
n 2 = 200, x 2 = 73 .9, σ 2 = 17 .9
Null hypothesis :
M 0 : u1 = u 2
Calculation of Test statistic :
S.E.
σ 21
( x1 − x 2 ) =
Z =
n1
+
σ 22
(14 .8) 2 (17 .9) 2
=
+
n2
100
200
x1 − x 2
72 .4 − 73 .9
=
S. E .( x1 − x 2 )
1 .95
= − .769
Z = .769
at
α = . 10 Z α = 1 .645
Z < Z α = Hence No. is accepted. Therefore, conclude that the two means are
statistically equal. For more illustration, read chapter-9.
10.7 CHI–SQUARE (χ 2 ) Test for Goodness of fit
χ 2 −test is a measure of probabilities of association between the attributes. It gives
us an idea about the divergence between the observed and expected frequencies.
Thus the test is also described as the test of 'Goodness of fit". If the curves of these
two distributions, when superimposed do not coincide or appear to diverge much
We say that the fit is poor. On the other hand if they do not diverge much, then the fit
is less poor.
Conditions for Applying the Chi Square Test
1. Each of the observations constituting the sample for this test should be
independent of other.
2. The expected frequency of any item or cell. Should not be less than 5.
3. The total number of observations used in this test must be large i.e. n > 30
4. This test is used only for drawing inferences by testing hypothesis. It cannot be
used for estimation of parameter or any other value.
5. It is wholly dependent on the degree of freedom.
6. The frequencies used in χ 2 test should be absolute and not relative in terms.
Working Rule for χ 2 Test
Step 1 : Set up the hypothesis that no association exists between the attributes.
Step 2 : Calculate χ 2 by the following formula where Oi is observed frequency and
E i is expected frequency
χ2 =
∑ (Oi
− E i )2
Ei
Test of Significance and Applications
451
Step 3 : Find from the table the value of χ 2 for a given value of the level of
significant α and for the degree of freedom ν.
Step 4 : Compare the value χ 2 with the tabulated value of χ 2 .
(i) If χ 2cal < χ 2tab , then accept the null hypothesis H 0 .
(ii) If χ 2cal > χ 2tab , then reject the null hypothesis H 0 .
Illustrative Examples
Ex. 4: A survey of 320 families with 5 children each revealed :
No. of boys :-
5
4
5
2
1
0
No. of girls :-
0
1
2
3
4
5
No. of families :-
14
56
110
88
40
12
Is this result consistent with the hypothesis that the male and female births are
equally probable ?
Sol : The chances of success is 1/2 and that of failure is also 1/2. Hence the
theoretical frequencies of 0,1,2,3, ... success are the successive terms of binomial
5
 1 1
 1 5 10 5 1 
expansion of 320  +  = 320 ,
for
, , ,
 2 2
 32 32 32 32 32
= 10, 50, 100, 50, 10
On the basis of null hypothesis the expected frequencies are to be determent.
 1 1
320  + 
 2 2
5
 1 5 10 5 1 
= 320  , , , , 
 32 32 32 32 32
= 10, 50, 100, 50, 10
Table for χ
2
Observed
frequencies
' Oi '
Expected
frequencies
Ei
Oi − E i
(Oi − E i ) 2
(Oi − E i ) 2
Ei
14
10
4
16
1.60
56
50
6
36
1.72
110
100
10
100
1.00
88
100
–12
144
1.44
40
50
10
100
2.00
12
10
2
4
.40
Total
Level of Significance : Take α = 0 .5
7.16
452
Engineering Mathematics-III
Critical value : The table value of χ 2 at
α = 0 .5 for ( 6 − 1) = 5 degrees
∑ (Oi − E i ) = 7.16
χ2 =
Ei
χ 2tab = 11 .07 by table
Hence
Thus, χ 2cal = 7 .16 < χ 2tab = 11 .07 ⇒ χ 2cal < χ 2tab therefore the null hypothesis is accepted.
i.e. The male and female births are equally probable.
Ex. 5 : In 60 throws of a dice, face one turned up 6 times, face two or three 18 times,
face, four or five 24 times and face six 12 times. Test at 10% Significance level, if the
dice is honest it being given that p ( χ 2 > 6 .25) = .1 for 3 degrees of freedom.
Sol : Let us Consider that null hypothesis ( H 0 ) is that the dice is honest.
Alternative Hypothesis ( H1 ) is that the dice is not honest. if dice is honest the
1
expected frequency for each face is 60 × = 10
6
Table for χ 2
Face of the
die
Observed
(0)
frequency
Expected
frequency
(Ei )
Oi − Ei
(Oi − Ei )2
Ei
1
6
10
–4
1.6
2
18
10
–2
.2
4
.8
2
.4
3
4
10
24
10
5
10
6
12
10
Total
60
60
χ2 =
20
∑ (Oi − E i )2
Ei
20
3.0
= 3 .0
For 3 degree of freedom χ 2 is 6 .25 by table. Thus χ 2cal = 3 .0 < χ 2tab = 6 .25 Therefore,
the null hypothesis that the die is honest is accepted.
Ex. 6 : One thousand girls in a college were graded according to their I.Q. and the
economic conditions of their homes. Use χ 2 test to find out whether there is any
association between economic conditions at home and I.Q. of girls.
Test of Significance and Applications
453
Economic Condition
I.Q.
Total
High
Low
100
300
Poor
350
250
600
Total
450
550
1000
Rich
400
v = 1, χ .205 = 3 .84
Given for
Sol : Null Hypothesis H 0 : There is no association between economic Condition at
home and I.Q. Now,
Row Total ×Column total
E=
Grand total
Oi
Ei
(Oi − Ei )2
(Oi − Ei )2
Ei
100
180
6400
35.5
350
270
6400
23.7
300
220
6400
29.1
250
330
6400
19.4
Total
χ2 =
It is given that
∑ (Oi − E i )2
Ei
107.7
= 107 .7
χ .205 = 3 .84
χ 2cal > χ 2tab . So the null hypothesis is rejected. It means there is a
relation between economic condition at home and τ.θ.
Ex.7: A Sample analysis of examination results of 500 students was made.It was
found that 180 students has failed, 170 had secured third class, 110 were placed in
second class and 40 got first class. Are these figures commensurate with the general
examination result which is in the ratio of 4:3:2:1 for the various Categories. Table
value of Chi Square at α = . 05 2degree freedom (d.f.), 3 d.f.and 4. d.f. are 5.99, 7.81
and 9.49 respectively.
Sol : Null Hypothesis : ( H 0 ): The given data is commensurate with the general
examination result.
Thus,
454
Engineering Mathematics-III
Table of Chi square
Category
Observed
frequency
Oi
Expected
( ON − E N )2 ( ON − E N )2
frequency Ei
Ei
Failed
180
4
× 500 = 200
10
400
2.000
Third Class
170
3
× 500 = 150
10
400
2.666
Second Class
110
2
× 500 = 100
10
400
1.000
First Class
40
4
× 500 = 50
10
400
2.000
Total
500
χ2 =
It is given that
7.6667
∑ (Oi − E i )2
Ei
= 7 .666
χ 2 for 3 d.f. =7.81
Thus, χ 2tab > χ 2cal . Therefore, SQ we accept our null Hypothesis H 0 , i.e. the given
data is commensurate with the general examination results.
Ex.8: Mindilian theory indicates that the shape and colour of a certain variety of pie
ought to be grouped into four groups "round and yellow" " round and green",
"angular and yellow" and "angular and green" according to the ratios 9:3:3:1. For
556 peas, the following results were obtained : round and yellow =315, round and
green =108, angular and yellow =101, and angular and green =32. 20 is the
experimental results support theory (Given χ 20.05 = 7 .81 3 degree of freedom (d.f.)).
Sol : Null hypothesis : H 0 is that The experimental results support theory.
Theoretical frequencies are computed as :
Total number of peas =315+108+101+32=554. Since pecs are divided in the ratio
=9:3:3:1
9
3
Hence
E1 =
× 554 = 312 .75, E 2 =
× 554 = 104 .25,
16
16
3
1
E3 =
× 554 = 104 .25, E 4 =
× 554 = 34 .75 .
16
16
Test of Significance and Applications
455
Table for χ 2
Variety of pecs
Observed
frequency (Oi )
Expected
frequency
(Ei )
Round and yellow
315
312.75
Round and green
108
104.25
3.75
0.1348
Angular and
yellow
101
104.25
–3.25
0.1013
Angular and green
32
34.75
–2.75
0.2176
Total
Oi − Ei
2.75
554
(Oi − Ei )2 / Ei
0.01618
0.470
χ 2 (Computed) =0.47
The degree of freedom (d.f.) = n –1=4 –1=3
But, tabulated χ 20.05 for 3 d. f.= 7.81. Hence χ 2 (calculated) < χ 2 (tabulated). This
suggest that the experimental results support the the theory.
10.8 Analysis of Variance
Analysis of Variance is a method of splitting the total variation of a data into
constituent parts which measures different source of variations. It is developed by
Prof. R.A. Fisher in 1920's. With the help of this technique, we can test the
significance difference among three or more sample means. The total variation is
split up into the following two components :
(a) Variation within the subgroups of samples.
(b) Variation between the subgroups of the samples.
It is based on F–statistic which can be defined as the ratio of variance between
samples, and variance within sample, as,
Variance between samples
F=
Variance within samples
Some Assumptions :
1. Each one of the samples is independent of the other samples.
2. Each of the sample is a simple random sample.
3. Each one of the population has the same variance (σ12 = σ 22 = ...... µ n ).
4. The effect of various components are additive.
5. Populations from which the samples are selected is normally distributed.
Classification of Analysis of Variance : It is mainly carried on under the
following two classifications.
(i) One way Analysis of Variance or one way classification.
(ii) Two way Analysis of Variance or Manifold classification.
One Way Classification :- Under this classification only one factor is considered.
456
Engineering Mathematics-III
1. Direct Method :
Null Hypothesis : ( H 0 ) : µ1 = µ 2 ...... µ p i.e.the means of the population from
which p sample are drawn are equal to one another.
Alternative Hypothesis (H1 ) :- At least two of the means of the populations are
unequal.
(i) Sum of the Squares of Variations Amongst the Columns : It is the sum
of the squares of deviation between the columns or group means and the grand
mean, i.e.
2
S SC = r ∑ ( x j − x ) ,
where
x j = Mean of the j th sample
x = Mean of the sample mean
r = Number of rows
SSC
Variance amongst Columns : MSC =
, where n is the number of Column.
n−1
Sum of the Squares of Variations within Column : (SSE)
It is the sum of the squares of variations between individual items and the column
mean.
SSE =
where
∑ i ∑ j ( xij
− x j )2
xij = i th observation in
x j = Mean of j th Column
Mean of the Square of Column Error
SSE
MSE =
n ( r −1)
Where n no. of Column and r no. of rows.
(iii) Total Sum of Square of Variation :- It is denoted by SST and is calculated
by the relation
SST= SSC+SSE
SST
Totol Variance
=
M−1
Where M = n × r = Total no. of observations in all the samples.
ANOVA Table : The technique of analysis of Variance is referred to as ANOVA . A
table showing the source of variation, the sum of squares, degrees of freedom, mean
squares and the formula for the f–ratio is known as ANOVA Table.
Test of Significance and Applications
457
Table-A
ANOVA Table for One way Classification
Course of
Variation
Sum of Degree of
Square
freedom
Between samples
(Column means)
SSC
n –1
Within Samples
(Error)
SSE
n (r–1)
Total
SST=
Mean
Square
MSC ,
i
j
2. Compute the correction factor C, Where C =
MSC
MSE
SSE
n( r −1)
MSC =
Short Cut Method :
1. Find T = ∑ ∑ xij
SSC
n −1
F
T2
, N is the total number of
N
all the observations of all the samples
3. Find sum of squares of all items i.e. ∑ ∑ x 2ij
i
j
4. Find thd total sum of the squares SST, where SST=∑ ∑ xij2 − C
i
5. Calculate P =
6. SSC = P −
(∑ x1 )
2
n1
+
(∑ x 2 )
2
n2
+
(∑ x 3 )
j
2
n3
T2
N
7. Find the sum of the squares with in samples SSE = SST – SSC
8. Set up the ANOVA table and calculate F. Which is test statistic.
Ex. 9: A farmer applied three types of seeds on 4 separate plots. The observation on
per acre are given by the following table.
Seeds
A
B
C
D
Total
P
6
4
8
6
24
Q
7
6
6
9
28
R
8
5
10
9
32
Total
21
15
24
24
84
Thus, Fcalcuted < FTabulated , therefore, Hence we accept the hypothesis that the plots
are equally effective and the seeds have the same effect.
Ex.10: The three samples have been obtained from the normal population with
equal variances. Test the hypothesis at 5% level that the population means are
equal.
458
Engineering Mathematics-III
8
7
12
27
10
5
9
24
7
10
13
30
14
9
12
35
11
9
14
34
50
40
60
Sol : Null Hypothesis H 0 :- All the population means are equal.
Correction Factor :-
C =
=
T2
=
N
∑
x1 + x 2 + x 3
N
( 27 + 24 + 30 + 35 + 34) 2 150 × 150
=
= 1500
15
15
2
2
2
x1 )
x2)
x3)
(
(
(
∑
∑
∑
SSC =
+
+
−C
n1
2
n2
2
n3
2
( 50)
( 40)
( 60)
+
+
− 1500
5
5
5
2500 + 1600 + 3600
7700
=
− 1500 =
− 1500
5
5
=
= 1540 − 1500 = 40
Ex. 11 : In the previous example also find out if the plots are materially different in
fertility as also,if the three fertilizers make any material difference in plots.
Sol :
C =
SST =
T 2 ( 84) 2
=
= 588
N
12
∑ ∑ xij2 − C
= ( 36 + 49 + 64) + (16 + 36 + 25) + ( 64 + 36 + 100)
+( 36 + 81 + 81) − 588
= ( 349 + 275) − 588 = 624 − 588 = 36
Now
T2
SSC = P −
=
N
(∑ x1 ) 2 + (∑ x 2 ) 2 + (∑ x 3 ) 2 + (∑ x 4 ) 2
n
n
n
n
( 21) 2 (15) 2 ( 24) 2 ( 24) 2
+
+
+
− 588
3
3
3
3
441 + 225 + 576 + 576
1818
=
− 588 =
− 588
3
3
=
SSE = 36 − 18 = 18 = 606 − 588 = 18
−C
Test of Significance and Applications
459
ANOVA Table
Source
of
Variation
S.S
d.o.f
Between Plots
18
MSC=18/3=36
F=
MSC
= 3 .6
MSE
Between Seeds
8
MSR=0/2=4
F=
MSR
= 2 .4
MSE
Error
18
MSE=
18
= 1 .66
11
f tab = 4 .76 & 5 .14
ST =
∑ x12 + ∑ x 22 + ∑ x 32 − C
= 530 + 336 + 734 − 1500
= 1600 − 1500 = 100
SSE = SST – SSC =60
ANOVA TABLE
Source
Variation
of
S.S
D.o.f.
Between Samples
40
2
Within Samples
60
12
Mean Squares
F
40
= 20
2
60
MSE=
=5
12
MSC
=4
MSE
MSC=
We have that
f. 05 = 3 .88
Thus
f cal > f tab
Hence Null Hypotheses is rejected and the alternative Hypotheses is accepted.
Hence, all the population means are not equal.
Problem Set
Chi-Square test
1. The fallowing table occurs in the number of Karl -Pearson.
Eye colour in sons
Eye Colour in father
Not light
Not light
light
light
230
148
151
471
Test whether the sons eyes associated with that of the father.
460
2.
Engineering Mathematics-III
In a biochemical experiment 20 insects were put into each of 100 jars and were
subjected to a fumigant After three hours, the number of living insects is shown
below. Fit a binomial distribution to the data to test the goodness of fit.
No. of
alive
inscets
No. of jars
3.
4.
0
1
2
3
4
5
6
7
8
9
Total
3
8
11
15
16
14
12
11
9
1
100
A Company is considering five possible names for its new product. Before
choosing the name, the company. decide to test whether all five names are
equally appealing. A random sample of 100 people is choosen, and each person
is asked to state his or her choice of the best name among the five possibilities .
The number of people who choose each one of the names are as follows :
Name
A
B
C
D
E
No. of Choices
4
12
34
40
10
A list of wars of modern civilization proved the following data for the year A.D.
1500 to 1931.
No. of Out breaks in the
year
No. of such years
0
1
2
3
4
5
Total
233
142
48
15
4
0
432
Fit a Poisson distribution to the data and test the goodness of fit.
∑ fx = 299 = 0.69 app,
Hint : m =
∑ f 432
Theoretical frequencies are given by
( 0 .69) r
, r = 0, 1, 2, 3, 4, 5.
r!
The number of parts for a particular spare part in a factory was found to vary
from day to day. In a sample study, the following information was obtained.
432 i −0.69
5.
Days :
Mon
Tues
Wed
Thur
Fri
Sat
No. of parts
demanded :
1124
1125
1110
1120
1126
1115
6.
7.
Test the hypothesis that the number of parts demanded does not depend on the
day of the week.
The number of scooter accidents per month in a certain town were as follows :
12, 8, 20, 2, 14, 10, 15, 6, 9, 4
Use chi-square test to determine if these frequencies are in agreement with the
belief that accident conditions were same during 10 months period.
A manufacturer of refills claimed that the life of their refills is 24 hours. A
sample of 10 refills has mean life of 22.5 hours with a standard deviation of 3.0
hours. On the basis of available information test whether the claim of the
manufacturer is correct. (Apply t-test)
Test of Significance and Applications
8.
461
The measurements of diameter of 8 cylindrical rods by Vernier calliper and
micrometer were a follows :
Vernier
reading
(x1 )
2.265
2.267
2.264
2.267
2.268
2.263
2.264
2.268
Micrometer
reading
(x 2 )
2.270
2.268
2.269
2.273
2.270
2.270
2.268
2.268
9.
7.
Test whether the difference between measurements of diameter by vernier and
micrometer is significant or not. [Given t 7 .05 = 2 .365]
A tobacco company wishes to test whether its three salesmen A,B and C tend to
make sales of the same size or whether they differ in their selling ability as
measured by the average size of their sales. During the last week there have
been 14 sales calls – A made 5 calls, B made 4 calls and C made 5 calls.
Following are the weekly sales record of the three salesmen :
A
B
C
Rs.
Rs.
Rs.
300
600
700
400
300
300
300
300
400
500
400
600
0
–
500
Perform the analysis of variance and draw your conclusions.
Four experiments determine the moisture contents of samples of soil. Each
man taking a sample from each of six consignments. Their as sessments are
given below :
Observers
Consignments
1
2
3
4
5
6
1
9
10
9
10
11
11
2
12
11
9
11
10
10
3
11
10
10
12
11
10
4
12
13
11
14
12
10
Analysis the data and discuss whether there is any significant difference
between consignments or between observers.
11. The following data gives the retail price of commodity in some shops selected
at random in four different cities.
462
Engineering Mathematics-III
City
Prices (Rs. per kg)
A
4.2
5.3
6.8
4.0
3.5
B
3.3
4.4
5.2
6.1
3.1
C
3.9
3.8
4.4
5.2
6.2
D
4.3
4.7
7.2
6.8
4.5
Do the prices differ significantly in the four cities.
10.9 Application of Statistics to Engineering, Agriculture,
Medicine and other Sciences
Statistics is the mathematical science involving the collection of data, analyzing and
interpreting it to get an appropriate result. The discipline of statistics has got usages
in several fields and by applying the statistical investigation; a breakthrough can be
successfully achieved in respect of frauds and scams like fake currency and stamp
papers. The science of statistics had got several advantages and asserted that it
would be possible to clearly identify the genuineness of stamp papers and also
currency notes. The various fields, including agriculture, had been benefited by
statistics. Through statistical quality control and operational methods, a
manufacturer can know whether his product had got approval from buyers, more
particularly in the foreign market. In fact. There was no field which had not
benefited by the application of statistics. For instance, a layman can take lifetime
decisions, besides coming to a conclusion on wise investments. For the Government,
proper application of statistics would be useful in respect of policy decisions and
long range planning, for business it is useful in market research and planning and in
the field of medicine, statistics would be useful in respect of diagnosis, prognosis
and clinical trials. Even the parentage aspect also can be decided by statistical study.
Prof. C. R. Rao, who was conferred with the US President’s National Medal of
Science Laureate by President Bush, said the application of statistics was useful in
the study of literature to determine the originality of an author. There is no field
which is complete without statistics. Statistics is involved in every science. A
number of specialties have evolved to apply statistical theory and methods to
various disciplines.
♠
Actuarial science is the discipline that applies mathematical and statistical
methods to assess risk in the insurance and finance industries.
♠
Biostatistics is a branch of biology that studies biological phenomena and
observations by means of statistical analysis, and includes medical statistics.
♠
Chemometrics is the science of relating measurements made on a chemical
system or process to the state of the system via application of mathematical or
statistical methods.
♠
Demography is the statistical study of all populations. It can be a very general
science that can be applied to any kind of dynamic population, that is, one that
changes over time or space.
♠
Econometrics is a branch of economics that applies statistical methods to the
empirical study of economic theories and relationships.
Test of Significance and Applications
463
♠
Epidemiology is the study of factors affecting the health and illness of
populations, and serves as the foundation and logic of interventions made in the
interest of public health and preventive medicine.
♠
Geostatistics is a branch of geography that deals with the analysis of data from
disciplines such as petroleum geology, hydrogeology, hydrology, meteorology,
oceanography, geochemistry, geography.
♠
Operations research is an interdisciplinary branch of applied mathematics and
formal science that uses methods such as mathematical modeling, statistics, and
algorithms to arrive at optimal or near optimal solutions to complex problems.
♠
Population ecology is a sub-field of ecology that deals with the dynamics of
species populations and how these populations interact with the environment.
This study is handled by the application of data analysis and various statistical
tests.
♠
Psychometrics is the theory and techniques of educational and psychological
measurement of knowledge, abilities, attitudes, and personality traits. These
studies of these techniques are handled by the application of data analysis and
various statistical tests.
♠
Statistical finance, sometimes called econophysics, is an empirical attempt to
shift finance from its normative roots to a positivist framework using exemplars
from statistical physics with an emphasis on emergent or collective properties of
financial markets.
♠
Statistical mechanics is the application of probability theory, which includes
mathematical tools for dealing with large populations, to the field of mechanics,
which is concerned with the motion of particles or objects when subjected to a
force.
♠
Statistical physics is one of the fundamental theories of physics, and uses
methods of statistics in solving physical problems.
♠
Statistical thermodynamics is the study of the microscopic behaviors of
thermodynamic systems using probability theory and provides a molecular level
interpretation of thermodynamic quantities such as work, heat, free energy, and
entropy.
10.9.1 Application of Statistics in Agriculture
We know that about 72 percent of our population depends on agricultural for their
living. Agricultural contributes almost 50 percent of the national income. The main
objective of five years plans to increase the production of adequate quantities of
food in order to meet the demand of the rapidly growing population in India. To
fulfill these requirements, the role of comprehensive and reliable statistics for
realistic and detailed planning of agricultural development and implementation of
the plans cannot be ignored. Statistics are used in the scientific study of agriculture
as a tool to determine if the differences in variables are real or due to chance. This
translates to the farmer to let him know with confidence which varieties are better
than other varieties or which fertilizer treatments will give better yields than others.
464
Engineering Mathematics-III
10.9.2 Application of Statistics in the Business Field
You know that in the casino game a small ball falls at random into one of the
thirty-seven numbered compartments of a revolving wheel. Bets can be placed on a
single number or a set of numbers, including numbers having common attributes.
The payout of a winning bet on a single number is thirty-six for one but the chance of
winning is actually one out of thirty-seven. The payout of a winning bet on the odd
or even attribute of the drawn number is one for one but the chance of winning is
eighteen out of thirty-seven. Obviously, these payout rates are lower than the
probability of the outcomes. In this case, statistics is useful but not helpful. As each
round of the game is statistically independent of the previous rounds, its outcome is
a random variable. The only certainty is that the mathematical expectation of the
payoff of the game in the long run is zero, simply because the payout rates are
always in favor of the casino. This is also true of all other casino games. Markets and
people are the concerns of business. Their behaviors often display patterns and
trends, and therefore are not totally unpredictable. It is this partial predictability
that makes the application of statistics more useful and meaningful in the business
world than in the casinos. The higher the predictability of the outcome, the more
statistics is applicable. For instance, time series analysis is more usefully applied to
the consumer price index than the stock price index because retail prices have
repetitive seasonal patterns whereas stock prices have erratic cycles.
10.9.3 Application of Statistics in Decision Making
Business decisions are much more complex than gambling bets. They aim at the
maximization of output, or minimization of input, or optimization of the outcome in
the light of the changing market environment, changing consumer preferences and
intensifying competition. As the science of inference, statistics is helpful in these
respects. Sometimes decisions are made under certainty where the outcomes of the
strategic options are fully predictable. They essentially deal with the management of
resources, such as inventory control, where the issue in question is to optimize the
available resources to achieve the best possible outcome. These are classic
mathematical programming problems. Sometimes decisions are made under partial
uncertainty where knowledge of the outcomes under different strategic options is
incomplete. A classical case is acceptance testing in quality control where the
decision is to accept or reject the batch in question, and this is a problem for
pre-posterior analysis. Sometimes decisions are made under risk where only the
likelihood of the outcomes of various strategic options is known. They are typical
problems of risk and return in investment, where the issue in question is to optimize
the payoff. The usefulness of statistical methods depends very much on the validity
of the quantification of risk with the “variance”. Sometimes decisions are made
under conflicts where competitors are both rational and sophisticated. The issue in
question is to outperform the rivals. The development of game theory is to
generalize the actions and reactions of competitors and the related payoffs with
mathematical models. Although the real life situation is much more complex,
hypothetical games help managers to systematically analyze the possible interactive
moves of competitors against their marketing or product strategies. Someone says
Test of Significance and Applications
465
that you should know yourself and your opponents in order to win each and every
battle.
10.9.4 Application of Statistics in Market Research
Market research is perhaps the most common area of statistics application in
business. It aims at identifying any gap between customer needs and the products
and services being provided in the mass market. There are two broad approaches.
The quantitative approach deals more with customer profile, behavior and
preference while the qualitative approach deals more with customer attitude and
perception. No matter which approach is employed, the representative ness of the
data or information collected is vital to the reliability of the survey results. As it is
mostly impractical to survey the entire target population, a representative sample
will have to be drawn up. The fundamentals of survey sampling are randomness and
sufficiency, so that any inference from the sample may be projected to the
population as a whole and the possible deviation from the true picture arising from
the use of a sample may be objectively assessed. As the magnitude of the sampling
error is inversely related to the sample size but the survey cost increases with the
sample size, the benefit of increasing the sample size is subject to the law of
diminishing returns. The dilemma is the trade-off between cost and precision under
time and budget constraints. However, the main cause for concern in practice is
non-sampling errors. Substitutions for non-contacts or refusals are often necessary.
Systematic selection rather than random selection of respondents is often inevitable
in practice. Moreover, interviewer bias cannot be totally eliminated in data
collection.
Non-sampling errors cannot be avoided but they can be managed. The handling of
non-responses in household expenditure survey is an illustration. The household
expenditure survey is for deriving the weights of the components of the consumer
price index. As participating households had to keep a detailed expenditure diary for
four weeks, the response rate was around 50% only. This gave rise to two questions.
Firstly, how could we improve the response rate and in turn the accuracy of the
results? Secondly, would the results be distorted by non-responses? The answer to
the first question was to double the sample size and reduce the survey cycle to two
weeks. The response rate improved to over 60%. The answer to the second question
was to identify the demographic and economic characteristics of the non-responses
and make comparisons with those of the participants. Statistical tests showed that
any distortion would be insignificant. Apart from classical hypothesis testing like
chi-square and F-statistic, more sophisticated quantitative methods have been
developed to measure and analyze consumer satisfaction, perception and
preferences. Multiple regression analysis is used to identify the contributing factors
to consumer demand and the elasticity of demand. Factor analysis is used to deduce
the conceptual and beneficial dimensions underlying the expressed measures of
product perception and preference. Cluster analysis is used to deduce homogenous
groups from the data set for identification of potential market segments as well as
competitors. Conjoint analysis deals with the trade-off in consumer preferences, and
identifies the determinant attributes of brand choices. As human behavior is abstract
in nature, the use of mathematical models is at best a generalization and an
approximation of the real life situation. The reliability of these models in practice
466
Engineering Mathematics-III
depends much on the quality of the data input and the validity of the underlying
assumptions at the time of application.
10.9.5 Statistics in Forecasting
Theories in multiple regression and time series analysis have provided a well
developed mathematical framework for business forecasting. In practice, a
mathematical model consisting of a set of multiple regression equations is
developed from history. Apart from classic technical assumptions on the variables
used, such as randomness and linear relationships, there are a number of practical
presumptions for statistical forecasting to be effectively applied. Firstly, both the
target variables and explanatory variables are discretely measurable. Secondly,
causal relationships in variable sets are logical in a real life situation. Thirdly, such
relationships will persist into the future. In real life, however, the economic
environment is changing. So are consumer tastes and behaviors. These could upset
the demand-supply relationships built up over the years. As such, causal relations
between variables shift continuously over time. The result is that the future usually
bears a closer relationship with the immediate past than the distant past. As
forecasting models are often built on rather long time series, their prediction ability
is often impaired.
The econometric model of the Indian economy is a case in point. All the models in
use by the Government, academic institutions or financial institutions are trade
based and use the expenditure approach to Gross Domestic Product (GDP). A typical
model is constructed on the basis of an exporting city-state economy, with the bulk
of the exports being domestically manufactured from imported raw materials. The
model starts with the exports of Indian manufactures, which are determined by the
import demands of major markets and the price competitiveness of the goods. Price
competitiveness depends in turn on the exchange value of the Indian Rs, domestic
labour costs and imported material costs. Domestic labour costs are affected by
labour supply and inflation. The net income from exports will generate demands for
capital investments or reinvestment as well as consumption throughout the
economy. In turn, these domestic demands will determine the retained imports of
capital and consumer goods. Statistically, all these models have strong explanatory
relevance in relation to the past performance of the Indian economy. They meet all
the stringent statistical tests for a theoretically perfect regression analysis for time
series. However, their predictive power has not been impressive. The past 20 years
witnessed two phases of structural changes in the Indian economy. With the
embarkation of economic reforms in 1990, India reopened its doors to foreign trade
and investment. The decade also saw the massive relocation of Indian
manufacturing plants to the nearby Pearl River delta area on cost and resources
considerations, thereby effectively paving the way for Indian’s future reintegration
with the other economy.
Test of Significance and Applications
467
10.9.6 Some commonly used statistical tests
Parametric
test
Example
of Purpose of test
non-parametric
Two-sample
(unpaired) t test
Mann-Whitney
test
Example
U Compares
two To compare the height
independent samples of girls with boys
drawn from the same
population
One
sample Wilcoxon matched Compares two sets of To compare weight of
pairs test
observations on a infants before and after
(paired) t test
single sample
a feed
One way analysis Kruskal-Wallis
of variance (F analysis of variance
test) using total by ranks
sum of squares
Effectively,
a
generalization of the
paired t or Wilcoxon
matched pairs test
where three or more
sets of observations
are made on a single
sample
To determine whether
plasma glucose is higher
one, two, or three hours
after a meal
Two way analysis Two way analysis of As above, but tests the
variance by ranks
influence
(and
of variance
interaction) of two
different covariates
In the above example, to
determine whether the
results differ in male
and female subjects
χ 2 – chi square Fisher’s exact test
less 2 test
Tests
the
null
hypotheses that the
distribution
of
a
discontinuous
variable is the same in
two
(or
more)
independent samples
To determine whether
acceptance into medical
school is more likely if
the applicant was born
in Britain
Product moment Spearman’s
rank
correlation
coefficient (r2)
coefficient
(Pearson’s r)
Assesses the strength
of the straight line
association between
two
continuous
variables
To assess whether and
to what extent plasma
HbA1 concentration is
related
to
plasma
triglyceride
concentration
in
diabetic patients
Regression
by Non-parametric
Describes
the To see how peak
least
squares regression (various numerical
relation expiratory flow rate
tests)
between
two varies with height
method
quantitative variables,
allowing one value to
be predicted from the
other
468
Engineering Mathematics-III
Multiple
Non-parametric
regression
by regression (various
least
squares tests)
method
Describes
the
numerical
relation
between a dependent
variable and several
predictor
variables
(covariates)
To determine whether
and to what extent a
person’s age, body fat,
and
sodium
intake
determine their blood
pressure
Like any other science subject, the theory of statistics is meant for a perfect or ideal
world, which hardly exists in reality. In the application of statistics in the above field,
there is always a gap between theory and practice
Objective Type Questions
Multiple Choice Questions
Tick mark the correct alternative :
1. A hypothesis may be classified as :
(a) simple
(b) composite
(c) null
(d) all of these.
2. Whether a test is one sided or two sided depends on :
(a) alterrative hypothesis
(b) composite hypothesis
(c) null hypothesis
(d) simple hypothesis.
3. A hypothesis under test is:
(a) simple hypothesis
(b) alterative hypothesis
(c) null hypothesis
(d) all of these.
4. Level of significance is the probability of :
(a) type I error
(b) type II error
(c) not committing error
(d) any of these.
5. Degree of freedom is related to :
(a) no. of observations in a set
(b) hypothesis under test
(c) no. of independent observations in a set
(d) none of these.
6. Area of critical region depends on :
(a) size of type I error
(b) size of type II error
(c) value of the statistic
(d) number of observations.
7. Students t-test is applicable only when :
(a) the variate values are independent (b) the variable is distributed normally
(c) the sample is not large
(d) all of these.
8. Testing H 0 :µ = 1500 against µ < 1500 leads to :
(a) one -sided lower tailed test
(b) one-sided upper tailed test
(c) two-tailed test
(d) none of these.
Test of Significance and Applications
9. The range of statistics-t is
(a) –1 to 1
(c) 0 to α
10. The range of statistics χ 2 is :
469
(b) − α to α
(d) 0 to 1.
(a) − 1 to + 1
(b) − α to α
(c) 0 to α
(d) 0 to 1.
11. If all frequencies of classes are same, the value of χ 2 is :
(a) 1
(c) zero
(b) α
(d) none of these.
Fill in the Blanks
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
A hypothesis contrary to null hypothesis is known as ..... hypothesis.
The hypothesis which is under test for possible rejection is called ..... hypothesis.
A hypothesis is an .... about the parameter of a population.
There can be only .... types of errors in taking a decision about H 0 .
Level of significance lies between ...........and .......... .
The test statistic ............. in case of one-tailed and two-tailed test.
Probability of first kind of error is called the ...... of the test.
Critical region is also known as .......... .
A null hypothesis is rejected if the value of a test statistic lies in the ............. .
t has ( n − 1) d.f. when all the n observations in the sample are .......... .
11. The formula for student's -t statistic is ............... .
12. The value of chi-square varies from .......... to ............ .
13. The value of chi-square statistics depends on the difference between .............
and .......... frequencies .
14. On the basis of a sample of farm workers, the hypothesis that 50 per cent of
workers are farm owners can be tested by .......... .
State True or False
1.
2.
3.
4.
5.
6.
Type II error is more severe than type I error.
The number of independent values in a set of values is known as degree of freedom.
Student's t-test is applicable in case of big samples.
Student's t-test is a valid in case the variable x follows any distribution
Student's t-test is a robust test.
Critical value of t dereases as the sample size increases.
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Engineering Mathematics-III
ANSWERS
Problem Set
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
Binomial distribution gives poor fit.
Gives satisfactory fit.
All five names are equally appealing.
Gives satisfactory fit.
Does not depend upon the day of the week
Accident conditions were not some.
Correct.
Significant difference.
do not differ.
Significant difference.
Significant difference
Multiple Choice Questions
1.
(d)
2.
(a)
3.
(c)
4.
(a)
5.
(c)
6.
(a)
7.
(d)
8.
(a)
9.
(b)
10.
(c)
11.
(c)
Fill in the Blanks
1. altrnative
2. null
3. assertion
4.
5. 0, 1
6. remains same
8. rejion of rejection
9. critical region
two
7. size
10.
observed,
expected
11. Chi-square
13. ( x − µ ) n / s
12. independent
14. 0, α
True/False
1.
True
2.
True
3.
False
4.
False
5.
True
6.
True
❑❑❑
Time Unit-3
Series and Forecasting
Chapter
11
471
Time Series and
Forecasting
Introduction
I hope the reader will be familiar with the typical terms include stock prices,
currency exchange rates, the volume of product sales, biomedical measurements,
weather data, etc, which are collected over monotonically increasing time.
Arrangement of these types of statistical data in chronological order i.e., in
accordance with occurrence of time, is known as Time Series. Such series have a
unique important place in the field of economic and business statistics. The interest
of an economist always to estimates the likely population in the coming year so that
proper planning can be carried out with regard to food supply, job for the people etc.
Similarly, a business man guesses likely sales in the coming future, so that he could
adjust his production accordingly and avoid the possibility of inadequate production
to meet the demand. To solve this problem, business man usually deals with
statistical data, which are collected, observed or recorded at successive intervals of
time. In this chapter, we shall discuss time series and it analysis to the answer the
above question. Mathematically, we always interested to find out the causal effect
on a situation (variable Y) of a change in other situation (variable X) over time.
11.1 Time Series
Time series is an ordered sequence of values of a variable at equally spaced time
intervals. In other word, we can say that a time series is a sequence of data points,
measured typically at successive times, spaced at (often uniform) time intervals.
According to Mooris Hamburg - A time series is a set of statistical observations
arranged in chronological order. Ya-Lun- Chou defining the time series as a time
series may be defined as a collection of readings belonging to different time periods,
of some economic variable or composite of variables. It is clear that a time series is a
set of observations of a variable usually at equal intervals of time. Here time may be
yearly, monthly, weekly, daily or even hourly usually at equal intervals of time.
Hourly temperature reading, daily sales, monthly production are examples of time
series. Time series occur frequently when looking at industrial data. Number of
factors affects the observations of time series continuously, some with equal
intervals of time and others are erratic studying and interpreting analyzing the
factors is called analysis of time series. The Primary purpose of the analysis of time
series is to discover and measure all types of variations which characterize a time
series. The central objective is to decompose the various elements present in a time
series and to use them in business decision making.
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Engineering Mathematics-III
The usage of time series models is twofold. Time series provides an understanding of
the underlying forces and structure that produced the observed data and with the
help of time series, we can fit a model and proceed to forecasting, monitoring or
even feedback and feed forward control. Time series analysis is the analysis of an
economic series over time. Time series analysis consists methods that attempt to
understand such time series, often either to understand the underlying context of
the data points (where did they come from? what generated them?), or to make
forecasts (predictions). Time series forecasting is the use of a model to forecast
future events based on known past events: to forecast future data points before they
are measured. A standard example in econometrics is the opening price of a share of
stock based on its past performance. Also, it allows you to the best forecast the value
of some variable at a future date. Forecasting models do not need to be causal.
Example seeing individuals carrying umbrella make you forecast rain in the future,
but umbrella do not cause rain. Regression models can also produce reliable
forecasts even if their coefficients have no causal interpretation.
11.2 Components of time series
The components of a time series are the various elements which can be segregated
from the observed data. There are four components of time series .The following are
the broad classification of these components.
Long Term
♠
Secular Trend
♠
Cyclical (Regular)
Short Term
♠
Seasonal
♠ Irregular (Erratic)
In time series analysis, it is assumed that there is a multiplicative relationship
between these four components. Symbolically, Y = T S C I, Where Y denotes the
result of the four elements; T = Trend; S = Seasonal component; C = Cyclical
components; I = Irregular component. In the multiplicative model it is assumed that
the four components are due to different causes but they are not necessarily
independent and they can affect one another. Another approach is to treat each
observation of a time series as the sum of these four components. Symbolically Y = T
+ S+ C+ I. The additive model assumes that all the components of the time series
are independent of one another.
11.2.1 Secular trend
A time series data may show upward trend or downward trend for a period of years
and this may be due to factors like increase in population, change in technological
progress, large scale shift in consumer demands etc. For example, population
increases over a period of time, price increases over a period of years, production of
goods on the capital market of the country increases over a period of years. These
are the examples of upward trend. The sales of a commodity may decrease over a
period of time because of better products coming to the market. This is an example
Time Series and Forecasting
473
of declining trend or downward trend. The increase or decrease in the movements of
a time series is called secular trend.
11.2.1.1 Methods of Measuring Trend
Trend is measured by Graphical method, Method of Semi-averages, Method of
moving averages. These methods are described below.
11.2.1.1.1 Graphical Method
This is the easiest and simplest method of measuring trend. In this method, given
data must be plotted on the graph, taking time on the horizontal axis and values on
the vertical axis. Draw a smooth curve which will show the direction of the trend.
While fitting a trend line the following important points should be noted to get a
perfect trend line.
1. The curve should be smooth.
2. As far as possible there must be equal number of points above and below the
trend line.
3. The sum of the squares of the vertical deviations from the trend should be as
small as possible.
4. If there are cycles, equal number of cycles should be above or below the trend line.
5. In case of cyclical data, the area of the cycles above and below should be nearly equal.
Illustrative Examples
Ex 1: Fit a trend line to the following data by graphical method.
Year
1996
1997
1998
1999
2000
2001
2002
Sales
60000
72000
75000
65000
80000
85000
95000
(in Rs )
Hundreds
Sol : The graph refers the trend line as below.
1200
1000
800
600
Sales (in Rs )
Year
400
200
0
Merits of Graphical Method
♠
It is the simplest and easiest method. It saves time and labour.
♠
It can be used to describe all kinds of trends.
♠
This can be used widely in application.
♠
It helps to understand the character of time series and to select appropriate
trend.
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Engineering Mathematics-III
Demerits of Graphical Method
♠
It is highly subjective. Different trend curves will be obtained by different
persons for the same set of data.
♠
It is dangerous to use freehand trend for forecasting purposes.
♠
It does not enable us to measure trend in precise quantitative terms.
11.2.1.1.2 Method of semi averages
In this method, the given data is divided into two parts, preferably with the same
number of years. For example, if we are given data from 1981 to 1998 i.e., over a
period of 18 years, the two equal parts will be first nine years, i.e., 1981 to 1989 and
from 1990 to 1998. In case of odd number of years like 5, 7,9,11 etc, two equal parts
can be made simply by omitting the middle year. For example, if the data are given
for 7 years from 1991 to 1997, the two equal parts would be from 1991 to 1993 and
from 1995 to 1997, the middle year 1994 will be omitted. After the data have been
divided into two parts, an average of each part is obtained. Thus we get two points.
Each point is plotted at the mid-point of the class interval covered by respective part
and then the two points are joined by a straight line which gives us the required
trend line. The line can be extended downwards and upwards to get
intermediate values or to predict future values.
Ex 2 : Draw a trend by the method of semi-averages.
Year
2001
2002
2003
2004
2005
2006
Sales (in Rs )
30000
42000
65000
72000
70000
82000
Sol : Divide the two parts by taking 3 values in each part
Year
Sales (Rs)
Semi total
Semi average
Trend values
2001
2002
2003
2004
2005
2006
30
42
65
72
70
82
137
45.666
24.889
45.666
361
120.333
70.555
Difference in middle periods = 2005 –2002 = 3 years
Difference in semi averages = 120.333 –45.666 = 74.667
Annual increase in trend = 74.667/3 = 24.889
Trend of 2001 = Trend of 2002 -24.889
= 45.666-24.889 = 20.777
Trend of 2003 = Trend of 2002 + 24.889
= 45.666+24.889 = 70.555
Trend of 2004 = Trend of 2005 -24.889
95.444
120.333
145.222
Time Series and Forecasting
475
= 120.333-24.889 = 95.444
Trend of 2006 = Trend of 2005 + 24.889
= 120.333+24.889 = 145.222
The following graph will show clearly the trend line.
Ex 3: Calculate the trend value to the following data by the method of semiaverages.
Year
2001
2002
2003
2004
2005
2006
2007
Expenditure
1.5
1.8
2.0
2.3
2.4
2.6
3.0
(Rs in Lakhs)
Sol:
Year
2001
Sales (Rs)
Semi total
Semi average
1.5
Trend values
1.545
5.3
1.77
2002
1.8
1.770
2003
2.0
1.995
2004
2.3
2.220
2005
2.4
2.445
2006
2.6
2.670
2007
3.0
2.895
8.0
2.67
Difference between middle periods = 2006 – 2002 = 4 years
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Engineering Mathematics-III
Difference between semi-averages = 2.67 - 1.77 = 0.9
Annual trend values = 0.9/4= 0.225
Trend of 2001 = Trend of 2002 – 0.225
= 1.77 – 0.225 = 1.545
Trend of 2002 = 1.77
Trend of 2003 = Trend of 2002 + 0.225
= 1.77 + 0.225 = 1.995
Similarly we can find all the trend values
Merits of Method of semi averages
♠
It is simple and easy to calculate
♠
By this method every one getting same trend line.
♠
Since the line can be extended in both ways, we can find the later and earlier
estimates.
Demerits of Method of semi averages
♠
This method assumes the presence of linear trend to the values of time series
which may not exist.
♠
The trend values and the predicted values obtained by this method are not very
reliable.
11.2.1.1.3 Method of Moving Averages
This method is very simple. It is based on Arithmetic mean. Theses means are
calculated from overlapping groups of successive time series data. Each moving
average is based on values covering a fixed time interval, called period of moving
average and is shown against the center of the interval. The moving averages for
a+b+c b+c+d c+d+e
three years is
etc. The formula for five yearly moving
,
,
3
3
3
a+b+c+d+e b+c+d+e+ f c+d+e+ f +g
averages is
etc.
,
,
3
3
3
Steps for calculating the moving averages for odd number of years.
1.
2.
3.
4.
Find the value of three years total, place the value against the second year.
Leave the first value and add the next three years value (ie 2nd, 3rd and 4th
years value) and put it against 3rd year.
Continue this process until the last year’s value taken.
Each total is divided by three and placed in the next column.
Ex 4: Calculate the three yearly averages of the following data.
Year
2000
2001
2002
2003
2004
2005
2006
2007
2008
2009
Producti
50
36
43
45
39
38
33
42
41
34
on in
Time Series and Forecasting
477
Sol :
Year
Production
in ( Metric tones)
3 years
moving total
3 years moving
average as
Trend values
50
36
43
45
39
38
33
42
41
-——129
124
127
122
110
113
116
117
-——
43.0
41.3
42.3
40.7
36.7
37.7
38.7
39.0
34
-——-
-——-
2000
2001
2002
2003
2004
2005
2006
2007
2008
2009
Even Period of Moving Averages:
When the moving period is even, the middle period of each set of values lies between
the two time points. So we must center the moving averages. The steps are
♠
Find the total for first 4 years and place it against the middle of the 2nd and 3rd
year in the third column.
♠
Leave the first year value, and find the total of next four-year and place it
between the 3rd and 4th year.
♠
Continue this process until the last value is taken.
♠
Next, compute the total of the first two four year totals and place it against the
3rd year in the fourth column.
♠
Leave the first four years total and find the total of the next two four years’ totals
and place it against the fourth year.
♠
This process is continued till the last two four years’ total is taken into account.
♠
Divide this total by 8 (Since it is the total of 8 years) and put it in the fifth
column.
Ex 5 : The production of a drink is given as below. Calculate the Four-yearly moving
averages
Year
2000
2001
2002
2003
2004
2005
2006
2007
2008
2009
Producti
on in
464
515
518
467
502
540
557
571
586
612
478
Engineering Mathematics-III
Sol:
Year
Production
in ( Metric tones)
000
464
2001
515
4 years
Moving
total
-
Total of Two 4
years
4 years moving
average as
Trend values
-
-
3966
495.8
4029
503.6
4093
511.6
4236
529.5
4424
553.0
4580
572.5
1964
2002
518
2003
467
2002
2027
2004
502
2005
540
2066
2170
2006
557
2254
2007
571
2008
586
2009
612
2326
Merits of Moving Averages
♠
The method is simple to understand and easy to adopt as compared to other
methods.
♠
It is very flexible in the sense that the addition of a few more figures to the data,
the entire calculations is not changed. We only get some more trend values.
♠
Regular cyclical variations can be completely eliminated by a period of moving
average equal to the period of cycles.
♠
It is particularly effective if the trend of a series is very irregular.
Demerits of Moving Averages
♠
It cannot be used for forecasting or predicting future trend, which is the main
objective of trend analysis.
♠
The choice of the period of moving average is sometimes subjective.
♠
Moving averages are generally affected by extreme values of items.
Time Series and Forecasting
♠
479
It cannot eliminate irregular variations completely.
11.2.2 Seasonal Variations
Seasonal variations are short-term fluctuation in a time series which occur
periodically in a year. This continues to repeat year after year. The major factors that
are responsible for the repetitive pattern of seasonal variations are weather
conditions and customs of people. In other words we can say that, seasonal
variations are fluctuations within a year during the season. For example, more
woolen clothes are sold in winter than in the season of summer .Regardless of the
trend we can observe that in each year more ice creams are sold in summer and very
little in winter season. The sales in the departmental stores are more during festive
seasons that in the normal days. The umbrella sales increase in rainy season and
agricultural production depends upon the monsoon etc., Secondly in marriage
season the price of gold will increase, sale of crackers and new clothes increase in
festival times. So seasonal variations are of great importance to businessmen,
producers and sellers for planning the future. The main objective of the
measurement of seasonal variations is to study their effect and isolates them from
the trend.
Measurement of seasonal variation
The following are some of the methods more popularly used for measuring the
seasonal variations.
1. Method of simple averages.
2. Ratio to trend method.
3. Ratio to moving average method.
4. Link relative method
Among the above four methods the method of simple averages is easy to compute
seasonal variations. The steps for calculations:
1. Arrange the data season wise
2. Compute the average for each season.
3. Calculate the grand average, which is the average of seasonal averages.
4. Obtain the seasonal indices by expressing each season as percentage of Grand
average
The total of these indices would be 100n where ‘n’ is the number of seasons in the
year.
Ex 6: Find the seasonal variations by simple average method for the data given
below.
Year
I
II
III
IV
2000
30
40
36
34
2001
34
52
50
44
2002
40
58
54
48
2003
54
76
68
62
2004
80
92
86
82
480
Engineering Mathematics-III
Sol :
Year
I
II
III
IV
2000
2001
2002
2003
2004
30
34
40
54
40
52
58
76
36
50
54
68
34
44
48
62
80
92
86
82
Total
238
318
294
270
Average
47.6
63.6
58.8
54
Seasonal
Indices
85
113.6
105
96.4
47.6 + 63.6 + 58.8 + 54
= 56
4
First quarterly Average
47.6
Seasonal Index for I quarter =
× 100 =
× 100 = 85
Grand Average
56
Grand average =
Seasonal Index for II quarter =
Second quarterly Average
63. 6
× 100 =
× 100 = 113. 6
Grand Average
56
Third quarterly Average
58.8
× 100 =
× 100 = 105
Grand Average
56
Fourth quarterly Average
54
Seasonal Index for IV quarter =
× 100 =
× 100 = 96.4
Grand Average
56
Seasonal Index for III quarter =
It ca be verified that the total of seasonal indices calculated must be equal to 400.
Here in this case, we have Total index is equal 85 + 113.6 + 105 + 96.4 = 400,
hence verified.
Ex 7 : Calculate the seasonal indices from the following data using simple average
method.
Year
I
II
III
IV
2000
72
68
80
70
2001
76
70
82
74
2002
74
66
84
80
2003
76
74
84
78
2004
74
74
86
82
Time Series and Forecasting
481
Sol :
Year
I
II
III
IV
2000
2001
2002
2003
2004
72
76
74
76
68
70
66
74
80
82
84
84
70
74
80
78
74
74
86
82
Total
372
352
416
384
Average
74.4
70.4
83.2
76.8
Seasonal Indices
97.6
92.4
109.2
100.8
74.4 + 70.4 + 83.2 + 76.8
= 76.2
4
First quarterly Average
74.4
Seasonal Index for I quarter =
× 100 =
× 100 = 97.6
Grand Average
76.2
Grand average =
Seasonal Index for II quarter =
Second quarterly Average
70. 4
× 100 =
× 100 = 92. 4
Grand Average
76. 4
Seasonal Index for III quarter =
Third quarterly Average
83. 2
× 100 =
× 100 = 100. 8
Grand Average
76. 2
Seasonal Index for IV quarter =
Fourth quarterly Average
76. 8
× 100 =
× 100 = 100. 8
Grand Average
76. 2
It ca be verified that the total of seasonal indices calculated must be equal to 400.
Here in this case, we have Total index is equal 97.6 + 92.4 + 109.2 +100.8 = 400,
hence verified.
11.2.3 Cyclical variations
Cyclical variations are recurrent upward or downward movements in a time series
but the period of cycle is greater than a year. The term cycle refers to the recurrent
variations in time series that extend over longer period of time, usually two or more
years. Also these variations are not regular as seasonal variation. There are different
types of cycles of varying in length and size. The ups and downs in business activities
are the effects of cyclical variation. A business cycle showing these oscillatory
movements has to pass through four phases-prosperity, recession, depression and
recovery. In a business, these four phases are completed by passing one to another in
this order. Most of the time series relating to economic and business show some kind
of cyclic variation. A business cycle consists of the recurrence of the up and down
movement of business activity. It is a four-phase cycle namely.
♠
Prosperity
♠
Decline
♠
Depression
♠
Recovery
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Engineering Mathematics-III
Each phase changes gradually into the following phase. The study of cyclical
variation is extremely useful in framing suitable policies for stabilizing the level of
business activities. Businessmen can take timely steps in maintaining business
during booms and depression.
11.2.4 Irregular variation
Irregular variations are fluctuations in time series that are short in duration, erratic
in nature and follow no regularity in the occurrence pattern. These variations are
also referred to as residual variations since by definition they represent what is left
out in a time series after trend, cyclical and seasonal variations. Irregular variations
are also called erratic. These variations are not regular and which do not repeat in a
definite pattern. These variations are caused by war, earthquakes, strikes flood,
revolution etc. This variation is short-term one, but it affects all the components of
series. There is no statistical technique for measuring or isolating erratic fluctuation.
Therefore the residual that remains after eliminating systematic components is
taken as representing irregular variations. Irregular fluctuations results due to the
occurrence of unforeseen events like floods, earthquakes, wars, famines, etc.
11.3 Utility of Time Series Analysis
1.
2.
3.
4.
5.
The analysis of time series has a great important in the every field of science.
It helps in understanding the past behaviour of any event. We know that the
plans can not be made without forecasting events and relationship they will
have. Time series analysis helps in planning the future operations.
Since the actual performance can be compared with the expected performance
and the cause of variation can be analysed, therefore the time series analysis
will enable us to apply the scientific procedure of “holding other things
constant” as we examine one variable at a time., For example, if we know how
much is the effect of seasonal influence on business, me may devise ways and
means of ironing out the seasonal influence.
Time series analysis facilitates comparison. Thus, by comparing two different
time series, some important conclusions can draw for further reading.
Time Series Analysis is also used for many applications such as: Budgetary
Analysis, Census Analysis, Economic Forecasting, Inventory Studies, Process
and Quality Control, Sales Forecasting, Stock Market Analysis, Utility Studies,
Workload Projections, Yield Projections etc.
11.4 Forecasting
A very important use of time series data is towards forecasting the likely value of
variable in future. In most cases, it is the projection of trend fitted into the values
regarding a variable over a sufficiently long period by any of the methods discussed
latter. Adjustments for seasonal and cyclical character introduce further
improvement in the forecasts based on the simple projection of the trend. The
importance of forecasting in business and economic fields lies on account of its role
in planning and evaluation. If suitably interpreted, after consideration of other
forces, say political, social governmental policies etc., this statistical technique can
be of immense help in decision making. The success of any business depends on its
future estimates. On the basis of these estimates a business man plans his
Time Series and Forecasting
483
production stocks, selling market, arrangement of additional funds etc. Forecasting
is different from predictions and projections. Regression analysis, time series
analysis, Index numbers are some of the techniques through which the predictions
and projections are made. Where as forecasting is a method of foretelling the course
of business activity based on the analysis of past and present data mixed with the
consideration of ensuring economic policies and circumstances. In particularly
forecasting means fore-warning. Forecasts based on statistical analysis are much
reliable than a guess work. According to T. S. Levis and R.A. Fox, “Forecasting is
using the knowledge we have at one time to estimate what will happen
at some future movement of time”.
Methods of Business forecasting: There are three methods of forecasting
1. Naive method
2. Barometric methods
3. Analytical Methods
1. Naive method: It contains only the economic rhythm theory. In this method
the manufactures analysis the time-series data of his own firm and forecasts on the
basis of projections so obtained. This method is applicable only for the individual
firm for which the data are analyzed; the forecasts under this method are not very
reliable as no subjective matters are being considered.
2. Barometric methods: It covers Specific historical analogy, Lead- Lag
relationship, Diffusion method, Action –reaction theory. The diffusion index method
is based on the principle that different factors, affecting business, do not attain their
peaks and troughs simultaneously. There is always time-log between them. This
method has the convenience that one has not to identify which series has a lead and
which has a lag. The diffusion index depicts the movement of broad group of series
as a whole without bothering about the individual series. The diffusion index shows
the percentage of a given set of series as expanding in a time period. It should be
carefully noted that the peaks and troughs of diffusion index are not the peaks
troughs of the business cycles. All series do not expand or contract concurrently.
Hence if more than 50% are expanding at a given time, it is taken that the business is
in the process of booming and vice - versa. The graphic method is usually employed
to work out the Diffusion index. The diffusion index can be constructed for a group
of business variables like prices, investments, profits etc.
3. Analytical Methods: It contains the factor listing method, Cross-cut analysis
theory, Exponential smoothing, and Econometric methods. In this method a
thorough analysis of all the factors under present situations has to be done and an
estimate of the composite effect of all the factors is being made. This method takes
into account the views of managerial staff, economists, consumers etc. prior to the
forecasting. The forecasts about the future state of the business are made on the
basis of over all assessment of the effect of all the factors.
11.5 Forecasting in an organization
In modern world, the competition among the companies is very tough. To take a proper
decision one should be aware of changes in the circumstances and the trends. The
decisions of a successive businessman are not based on guesses rather on a careful
analysis of data to decide the future course of action. For a better future it is important
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Engineering Mathematics-III
to analyze the past and we should try to find out what is likely to happen in future. The
solution of the future course of action always lies in the facts and figures of the past.
Thus when the future is decided by the systematic analysis of past, this process is
called as forecasting and the obtained statement is called as forecast. Forecasting
provides the best solution the future plans and helps to understand the implications
for the firm’s future of the alternative courses of action to them at present. In other
word we can say the future is not known with certainty, virtually every decision rest
upon a forecast. The successes or failures of a plan would depend upon the ability to
forecast the future course of events. Since future is not certain, therefore we can not
formulate the plans once for all without the need of subsequent revisions
Forecasts are needed throughout an organization — and they should certainly not be
produced by an isolated group of forecasters. Neither is forecasting ever “finished”.
Forecasts are needed continually, and as time moves on, the impact of the forecasts on
actual performance is measured; original forecasts are updated; and decisions are
modified, and so on. For example, many inventory systems cater for uncertain demand.
The inventory parameters in these systems require estimates of the demand and forecast
error distributions. The two stages of these systems, forecasting and inventory control, are
often examined independently. Most studies tend to look at demand forecasting as if this
were an end in itself or at stock control models as if there were no preceding stages of
computation. Nevertheless, it is important to understand the interaction between
demand forecasting and inventory control since this influences the performance of the
inventory system. This integrated process is shown in the following figure:
The decision-maker uses forecasting models to assist him or her in decision-making
process. The decision-making often uses the modeling process to investigate the
impact of different courses of action retrospectively; that is, “as if” the decision has
already been made under a course of action. That is why the sequence of steps in the
modeling process, in the above figure must be considered in reverse order. For
example, the output (which is the result of the action) must be considered first.
It is helpful to break the components of decision making into three groups:
Uncontrollable, Controllable, and Resources (that defines the problem situation). As
indicated in the above activity chart, forecasting has the following components:
1. Performance measure (or indicator, or objective): Measuring business
performance is the top priority for managers. Management by objective works if you
know the objectives. Unfortunately, most business managers do not know explicitly
what it is. The development of effective performance measures is seen as
increasingly important in almost all organizations. However, the challenges of
achieving this in the public and for non-profit sectors are arguably considerable.
Time Series and Forecasting
485
Performance measure provides the desirable level of outcome, i.e., objective of your
decision. Objective is important in identifying the forecasting activity. The following
table provides a few examples of performance measures for different levels of
management:
Level
Performance Measure
Strategic
Return of Investment, Growth, and Innovations
Tactical
Cost, Quantity, and Customer satisfaction
Operational
Target setting, and Conformance with standard
Clearly, if you are seeking to improve a system’s performance, an operational view is
really what you are after. Such a view gets at how a forecasting system really works;
for example, by what correlation its past output behaviors have generated. It is
essential to understand how a forecast system currently is working if you want to
change how it will work in the future. Forecasting activity is an iterative process. It
starts with effective and efficient planning and ends in compensation of other
forecasts for their performance
1. What is a System? Systems are formed with parts put together in a
particular manner in order to pursue an objective. The relationship between
the parts determines what the system does and how it functions as a whole.
Therefore, the relationships in a system are often more important than the
individual parts. In general, systems that are building blocks for other systems
are called subsystems.
2. The Dynamics of a System: A system that does not change is a static
system. Many of the business systems are dynamic systems, which mean their
states change over time. We refer to the way a system changes over time as the
system’s behavior. And when the system’s development follows a typical
pattern, we say the system has a behavior pattern. Whether a system is static or
dynamic depends on which time horizon you choose and on which variables
you concentrate. The time horizon is the time period within which you study
the system. The variables are changeable values on the system.
3. Resources: Resources are the constant elements that do not change during
the time horizon of the forecast. Resources are the factors that define the
decision problem. Strategic decisions usually have longer time horizons than
both the Tactical and the Operational decisions.
4. Forecasts: Forecasts input come from the decision maker’s environment.
Uncontrollable inputs must be forecasted or predicted.
5. Decisions: Decisions inputs ate the known collection of all possible courses of
action you might take.
6. Interaction: Interactions among the above decision components are the
logical, mathematical functions representing the cause-and-effect relationships
among inputs, resources, forecasts, and the outcome. Interactions are the most
important type of relationship involved in the forecasting. When the outcome
of a decision depends on the course of action, we change one or more aspects of
the problematic situation with the intention of bringing about a desirable
486
7.
Engineering Mathematics-III
change in some other aspect of it. We succeed if we have knowledge about the
interaction among the components of the problem. There may have also sets of
constraints which apply to each of these components. Therefore, they do not
need to be treated separately.
Actions: Action is the ultimate decision and is the best course of strategy to
achieve the desirable goal. Decision-making involves the selection of a course
of action (means) in pursue of the decision maker’s objective (ends). The way
that our course of action affects the outcome of a decision depends on how the
forecasts and other inputs are interrelated and how they relate to the outcome.
11.6 Controlling the Forecasting
Forecasting is a prediction of what will occur in the future, and it is an uncertain
process. Because of the uncertainty, the accuracy of a forecast is as important as the
outcome predicted by the forecast. This site presents a general overview of business
forecasting techniques as classified in the following figure:
Progressive Approach to Modeling: Modeling for decision making involves
two distinct parties, one is the decision-maker and the other is the model-builder
known as the analyst. The analyst is to assist the decision-maker in his/her
decision-making process. Therefore, the analyst must be equipped with more than a
set of analytical methods.
Integrating External Risks and Uncertainties: The mechanisms of thought
are often distributed over brain, body and world. At the heart of this view is the fact
that where the causal contribution of certain internal elements and the causal
contribution of certain external elements are equal in governing behavior, there is
no good reason to count the internal elements as proper parts of a cognitive system
while denying that status to the external elements.
In improving the decision process, it is critical issue to translating environmental
information into the process and action. Climate can no longer be taken for granted:
♠
Societies are becoming increasingly interdependent.
♠
The climate system is changing.
♠ Losses associated with climatic hazards are rising.
These facts must be purposeful taken into account in adaptation to climate
conditions and management of climate-related risks.
The decision process is a platform for both the modeler and the decision maker to
engage with human-made climate change. This includes ontological, ethical, and
historical aspects of climate change, as well as relevant questions such as:
Time Series and Forecasting
487
♠
Does climate change shed light on the foundational dynamics of reality
structures?
♠
Does it indicate a looming bankruptcy of traditional conceptions of
human-nature interplays?
♠
Does it indicate the need for utilizing nonwestern approaches, and if so, how?
♠
Does the imperative of sustainable development entail a new groundwork for
decision maker?
♠
How will human-made climate change affect academic modelers — and how
can they contribute positively to the global science and policy of climate change?
11.7 Effective Modeling for Forecasting
A Model is an external and explicit representation of a part of reality, as it is seen by
individuals who wish to use this model to understand, change, manage and control
that part of reality. “Why are so many models designed and so few used?” is a
question often discussed within the Quantitative Modeling (QM) community. The
formulation of the question seems simple, but the concepts and theories that must
be mobilized to give it an answer are far more sophisticated. Would there be a
selection process from “many models designed” to “few models used” and, if so,
which particular properties do the “happy few” have? This site first analyzes the
various definitions of “models” presented in the QM literature and proposes a
synthesis of the functions a model can handle. Then, the concept of
“implementation” is defined, and we progressively shift from a traditional “design
then implementation” standpoint to a more general theory of a model
design/implementation, seen as a cross-construction process between the model
and the organization in which it is implemented. Consequently, the organization is
considered not as a simple context, but as an active component in the design of
models. This leads logically to six models of model implementation: the technocratic
model, the political model, the managerial model, the self-learning model, the
conquest model and the experimental model. In order that an analyst succeeds in
implementing a model that could be both valid and legitimate, here are some
guidelines:
1. Be ready to work in close co-operation with the strategic stakeholders in order
to acquire a sound understanding of the organizational context. In addition,
the QM should constantly try to discern the kernel of organizational values
from its more contingent part.
2. The QM should attempt to strike a balance between the level of model
sophistication/complexity and the competence level of stakeholders. The
model must be adapted both to the task at hand and to the cognitive capacity of
the stakeholders.
3. The QM should attempt to become familiar with the various preferences
prevailing in the organization. This is important since the interpretation and
the use of the model will vary according to the dominant preferences of the
various organizational actors.
4. The QM should make sure that the possible instrumental uses of the model are
well documented and that the strategic stakeholders of the decision making
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Engineering Mathematics-III
process are quite knowledgeable about and comfortable with the contents and
the working of the model.
5. The QM should be prepared to modify or develop a new version of the model,
or even a completely new model, if needed, that allows an adequate
exploration of heretofore unforeseen problem formulation and solution
alternatives.
6. The QM should make sure that the model developed provides a buffer or leaves
room for the stakeholders to adjust and readjust themselves to the situation
created by the use of the model and
7. The QM should be aware of the pre-conceived ideas and concepts of the
stakeholders regarding problem definition and likely solutions; many decisions
in this respect might have been taken implicitly long before they become
explicit.
In model-based decision-making, we are particularly interested in the idea that a
model is designed with a view to action.
11.8 Main functions of Forecasting
Forecasting provides the best basis available for the formulation of intelligent
managerial expectations and handles the uncertainty about the future. In other
words we can say that forecasting can be viewed as part of management information
system. Forecasting is applied to capital investment decision, planning related to
production, sell stock control, budgetary control and finance. The main function of
forecasting are:
1. Forecasting provides a plan of action.
2. Forecasting is used to monitoring the continuing process of plans
3. Forecasting can be used as a warning system for the factors of a plan.
In this way, we can say that forecasting provides the facts and figures in advance and
then an executive takes the advantages of the future conditions. Since, it is not
always possible to predict the future precisely; therefore we must ready to allow
some errors in the forecast.
Exercise
Objective Type Questions
Multiple Choice Questions
Tick the correct alternative :
1. In forecasting :
(a) only future course of events is important
(b) only past is important
(c) neither future nor past is important
(d) both future and past are important.
2. Statistical forecasts torn out to be :
(a) 100 per cent accurate
Time Series and Forecasting
489
(b) 90 per cent accurate
(c) 50 per cent accurate
(d) depends on each individual case
(e) none of these.
3. While making forecasting it is desirable for a firm to :
(a) forecast the level of the industry first
(b) not to bother about the industry
(c) first make prediction for itself and then think of industry
(d) none of these.
4. The best forecasts are obtained with the help of :
(a) interpolation
(b) time series analysis
(c) regression analysis
(d) econometric model.
5. The statistical forecasting is done :
(a) without assumptions
(b) on the assumption ''other things being equal.''
(c) on one assumption that what has happened in the past will not happen in
future
(d) none of these.
6. Statistical forecasts :
(a) need not be supplemented with second judgement
(b) should be supplemented with sound judgement
(c) makes little difference in either case
(d) none of these.
7. 0 = :
(a) T × S + C + I
(b) T – S + C – I
(c) T – S × C × I
(d) T × S × C × I
(e) all fo the above
8. The most important factors causing seasonably variations are :
(a) growth of population
(b) technological improvements
(c) weather and social customs
(d) change in fashions
(e) none of these.
9. The most widely used method of measuring seasonal variations :
(a) ratio-to-moving average method
(b) ratio-to-trend method
(c) link relative method
(d) method of simple averages
(e) all of these.
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Engineering Mathematics-III
10. In the least square linear trend equation Y = a + bX , if b is positive it indicates :
(a) declining trend
(b) rising trend
(c) no trend at all
(d) all of these.
11. To reduce and annual trend equation to a monthly trend equation when the
original data are given as annual total :
(a) a is kept as it is and b is divided by 12
(b) a is divided by 12 and b by 144
(c) a is divided by 12 and b is kept as it is.
12. Cyclical fluctuations are caused by :
(a) wars
(b) earthquakes
(c) floods
(d) strikes
(e) all of these.
13. A time series consists of :
(a) two components
(b) three components
(c) four components
(d) five components .
14. Salient features responsible for the seasonal nations are :
(a) weather
(b) social Costumers
(c) festivals
(d) all of these.
15. Simple average method is used to calculate :
(a) trend Values
(b) cyclic Variations
(c) seasonal indices
(d) none of these.
16. Irregular variations are :
(a) regular
(b) cyclic
(c) episodic
(d) none of the above.
17. If the slope of the trend line is positive it show is :
(a) rising Trend
(b) declinin trend
(c) stagnation
(d) none these.
18. The sales of a departmental store on Diwali is associated with the compoment
of time-series :
(a) secular trend
(b) seasonal variation
(c) irregular variation
(d) all the above.
19. The component of time series attached to long-term variation is termed as :
(a) secular Trend
(b) seasonal Variation
(c) irregular fvariation
(d) cyclic variation.
20. Business forecasts are made on the basis of :
(a) present data
(b) past data
(c) polices and circumstances
(d) all of these.
21. Econometric methods involve :
(a) economics and mathematics
(d) economics and statistics
(c) economics, statistics and mathematics
Time Series and Forecasting
491
(d) none of these.
22. The economic rhythm theory comes under the category of :
(a) analytical methods
(b) naive method
(c) barometric methods
(d) none of these.
Fill in the Blanks
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
Forecasting is using the knowledge we have at one time to estimate what will
happen at some .........................moment in time.
Forecasting can be both................term and .............term.
The risk of error generally ................ with extension of the time horizon.
Despite all advance that have been made in the techniques of forecasting.
forecasting remains more an ...................than a............... .
All forecasts are based on certain ................. .
A time series consists of data arranged .............. .
Comparable monthly data may be obtained by multiplying each of the values
by ....................... and in a leap year by .............. .
The additive model of a time series is expressed as ...................... .
When the differences between successive observations of a series are constant
(or nearly so), the ................ may be an appropriate representation of trend
equation.
A polynomial of the form Y = a + bX + cX 2 is called a ................ .
In the trend equation Y = a + bX , a is the ............... and b represents ........ .
The line obtained by method of least squares is known as the line of ............... .
The equation of the Gompertz curve is of the form ....................... .
A time series in a set of values arranged in .................... order.
Quarterly fluctuations observed in a time series represent ..............variation.
Periodic changes in a business time series are called.......... .
A complete cycle passes through.............stages of phenomenon.
An overall tendency of rise and fall in a time series represents...................
The trend line obtained by the method of least square is known as the ............. .
Forecasting is different from ...................and ................... .
No statistical techniques measuring or isolating ................... .
State True/False
1.
2.
3.
4.
Secular trend refers to the long-term movement.
The most widely applied trend curve in practice is the second degree parabola.
When we shift the trend origin the value of b remains the same.
The semi-logarithmic trend curve is appropriate for those series on which
period to period changes are constant in absolute amount.
5. The Gompertz and Logistic curves are J-shaped.
6. The multiplicative model assumes that the value of original data is the product
of the values of the four components.
7. A second degree parabolic trend is expressed by the equation :
Y = a + bX + cX 2 + DX 3 .
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Engineering Mathematics-III
8. In the second degree curve Y = a + bX + CX 2 , C is the constant.
9. Forecasting business change involves more than analysis of statistical data.
10. Forecasting aims at reducing the areas of uncertainty that surround
management decision making.
11. Qualitative methods of forecasting are more commonly used in practices as
compared to the quantitative methods.
12. Time lag theory is by far the most important theory of business forecasting.
13. Specific historical analog theory is based on the assumption that all business
cycles are uniform in amplitude or duration.
Short/Long Answer Type Questions
1. (a)
2.
3.
4.
5.
6.
7.
8.
9.
Explain briefly the various methods of determining trend in a time series.
Explain merits and demerits of each method.
(b) Briefly explain the different methods of measuring seasonal variations.
(c) What do you understand by seasonal indices ? What methods are used to
determine them ?
Example the nature of cyclical variations in 9 time series. How do seasonal
variations differ form them ? Give an outline of the moving average method of
measuring seasonal variations?
Describe briefly the different components of a time series. With which
components of the time series would you associate each of the following ? Why
?
(i) The rainfall in Kolkatta that occurred for four days in March, 1992.
(ii) A decline in ice-cream sales during November to March.
(iii) An era of prosperity.
(iv) Increase in garment sales in October.
(a) What is business forecasting ?What is the importance of business
forecasting to :
(i) an administrator,
(ii) the society ? Bring out is limitations, if any.
(a) Exaplain briefly the various techniques of business forecasting.
(b) What are the limitations of business forecasting theories in general?
Discuss in detail and illustrate your answers.
(a) Critically examine the various methods that are used for business
forecasting. Why is time series considered to be an effective tool for
forecasting analysis ? Explain.
(b) What are the various theories of business forecasting ? Discuss any two of
them in detail.
Describe the techniques of forecasting that are commonly employed by a big
business house.
Discuss the important theories of business forecasting. Also explain the
practical importance of forecasting in business.
Distinguish between the 'historical analysis of past condition' and the 'crosssection analysis of current events' as methods of business forecasting.
(a) Examine 'long term forecasting' and 'short term forecasting'. Examine the
place of time series analysis in forecasting.
Time Series and Forecasting
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
493
(b) What is business forecasting ? What are the assumptions on which
business forecasts are made?
(a) ''Despite great limitations in statistical forecasting, it is invaluable to the
business.'' Examine this statement.
(b) Critically examine the various methods that are used for business
forecasting. What precautions will you keep in mind while using forecast
techniques ?
(c) ''Forecasting is much more than projecting a series mechanically into the
future.'' Elucidate.
(a) Discuss the importance and scope of business forecasting. Discuss the
internal and external factors affecting sales and profits of a large
organization.
(b) Point out the role and limitations of business forecasting as a tool of
analysing business data. Also explain any four methods of forecasting
known to you.
(a) Discuss the importance of business forecasting. Describe one of the
methods of business forecasting.
(b) ''The past can never be a perfect guide to the future''. Elucidate.
(a) ''Much of the success in business depends on the accuracy of the forecasts.''
Explain. Also critically examine any two theories of forecasting known to
you.
(b) What are the main types of techniques in use for solving business
forecasting problems ? Describe them very briefly.
(a) Explain the difference between forecasting and decision-making. How are
the two concepts related ?
(b) Write a short note on business forecasting.
[ICWA, Dec. 1983]
(a) What is meant by business forecasting ? Explain briefly the main methods
used in business forecasting.
(a) Explain the term ''Business Forecasting''. Name the different methods used
in Business Forecasting. Exaplain one method which you consider most
important in detail.
(b) Point out the role and limitations of business forecasting as a tool of
analysing business data. Also explain critically any two methods of
forecasting known to you.
(a) What is business forecasting ? Why is it done ?
(b) Mention the various methods of forecasting usually adopted by a business
house, in modern times.
[ICWA, May 1985]
(a) What is time series ? What is the need for the analysis of time series ?
(b) Explain briefly the different components of time series. Give some
illustrations of the use of time series in economic analysis.
Explain clearly the concept of Time Series Analysis. Indicate the
importance of such analysis in business.
(a) Describe briefly the various characteristic movements of time series.
Discuss briefly any one procedure for estimating secular trend.
494
22.
23.
24.
25.
26.
27.
28.
Engineering Mathematics-III
(b) What are the main component of time series ? How will you determine
them ? Discuss how the time series data of agricultural prices are useful
for preparing the agricultural developmental programme.
Describe the various components of time series. With which component of time
series would you mainly associate each of the following and why :
(a) an era of prosperity
(b) heavy sales on the occasion of Deepawali
(c) constantly rising demand for sugar in India and
(d) price hike in Petroleum products due to Iran-Iraq war.
Explain the meaning of time series. What are its main components? How would
you study seasonal variations in time series ?
(a) Explain briefly the additive and multiplicative models of times series.
Which of these models is more popular in practice and why ?
(b) Distinguish between ratio-to-trend and ratio-to-moving average as
methods of measuring seasonal variations. Which is better and why ?
(a) What is a time series ? What are its main components ? Discuss the various
methods of studying seasonal variations in a time series.
(b) Explain the utility of time-series. How can you compare one time series with
another ? What is 'lag' in time series and indicate how it is calculated ?
(a) What is economic time series ? Explain the meaning and importance of
trend in economic time series.
(b) What are the various influences on a time series. Explain briefly the
different methods of measuring trend.
(c) Distinguish between 'moving average' and 'least squares' as methods of
measuring trend in given time series. Which method is better and why ?
(a) How would you analyse a time series of records extending over 30 years ?
Describe in detail.
(b) Explain how would you fit a line of the form Y = a + b X to a given set of
data by the method of least squares.
[ICWA, July 1985]
(c) Describe any method for the determination of secular trend in a time
series data.
[AIMA, July 1983]
(d) Discuss the merits and limitations of the moving average method for
determining the trend in the analysis of time series.
(a) What is meant by trend ? How would you fit a straight line trend by the
method of least squares ?
(b) Are all 'periodical movements' necessarily seasonal? Give reasons for your
answer with appropriate illustrations.
Time Series and Forecasting
495
Unsolved Numerical Problems
1.
With the help of graph paper obtain the triend values .
Year
1996
1997
1998
1999
2000
2001
2002
Value
65
85
95
75
100
80
130
2.
Using graphical method, fit a trend-line to the following data.
Year
1982
1983
1984
1985
1986
1987
Value
24
22
25
26
27
26
3.
Draw a trend line by the method of semi-avereges.
Year
1993
1994
1995
1996
1997
1998
1999
2000
Sales
210
200
215
205
220
235
210
235
4.
The following figures are given relating to the output in a factory. Draw a
trend- line with the help of method of semi-averages.
Year
1996
1997
1998
1999
2000
2001
2002
Output
600
800
1000
800
1200
1000
1400
5.
Calculate three yearly moving average of the following data.
Year
91
92
93
94
95
96
97
98
99
00
Output
15
18
17
20
23
25
29
23
36
40
6.
7.
The following figures relating to the profit of a commercial concern for 8 years.
Find the 3-yearly moving averages.
Years
Profits
Years
Profits
1995
15,420
1999
26,120
1996
14,470
2000
31,950
1997
15,520
2001
35,370
1998
21,020
2002
35,670
Construct a four yearly centered moving average from the following data.
Year
1940
1950
1960
1970
1980
1990
2000
Imported cotton
consumption
(1000)
129
131
106
91
95
84
93
496
8.
Engineering Mathematics-III
From the following data calculate the 4-yearly moving average and determine
the trend values. Find the short-term fluctuations. Plot the original data and
the trend on a graph.
Year
93
94
95
96
97
98
99
Value
50
36.5
43
44.5
38.9
38.1
32.6
9.
00
01
02
41.7
41.1
33.8
Calculate trend value by taking 5 yearly period of moving average from the
data given below
Year
1987
1988
1989
1990
1991
1992
1993
1994
4
5
6
7
9
6
5
7
1995
1996
1997
1998
1999
2000
2001
2002
8
7
6
8
9
10
7
9
Production
in tones
Year
Production
in tones
10. Fit a straight line trend to the following data and calculate trend values.
Year
1996
1997
1998
1999
2000
4
6
7
8
10
Sales of
TV Sets
(Rs.'000)
Estimate the sales for the year 2005.
11. Below are given the figures of production in '000 quintals of a sugar factory. Fit
a straight line trend and calculate the trend values.
Year
Production
in tones
1994
1995
1996
1997
1998
1999
80
90
92
83
94
99
2000
92
12. Fit a straight line trend to the following data and calculate trend values.
Year
1996
1997
1998
1999
2000
2001
Profit
300
700
600
800
900
700
13. Fit a straight line trend to the following data. Estimate the earnings for the year
2002 and calculate trend values.
Year
1993
1994
1995
1996
1997
1998
1999
2000
Profit
38
40
65
72
69
60
87
95
Time Series and Forecasting
497
14. Compute the average seasonal movement for the following series.
Year
1st qiarter
IInd quarter
IIIrd quarter
IVthquarter
1999
3.5
3.9
3.4
3.6
2000
3.5
4.1
3.7
4.0
2001
3.5
3.9
3.7
4.2
2002
4.0
4.6
3.8
4.5
2003
4.1
4.4
4.2
4.5
15. Obtain seasonal fluctuations from the following time series Quarterly output of
coal for four years.
Year
2000
2001
2002
2003
I
6.5
58
70
60
II
5.8
63
59
55
III
56
63
56
51
IV
61
67
52
58
ANSWERS
Multiple Choice Questions
1
(d)
2.
(d)
3.
(a)
4.
(e)
5.
(b)
6.
(a)
7.
(d)
8.
(e)
9.
(e)
10.
(b)
11.
(a)
12.
(e)
13.
(c)
14.
(d)
15.
(c)
16.
(c)
17.
(a)
18.
(b)
19.
(a)
20.
(d)
22.
(b)
21.
(c)
Fill in the Blanks
1.
Future
2.
Short, long
3.
Increase
4.
Art, science
5.
Conditions and assumptions
6.
Chrono logical
7.
30.4167, 30.5
8.
Y = T + S +C + I
9.
Straight line
10.
Second degree equation
11.
intercept, slope of the trend line
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Engineering Mathematics-III
Y = Ka bx
12.
Best fit
13.
14.
Chronical
15.
Seasonal
16.
Cycles
17.
Four
18.
Secular trend
19.
line of best fit
20.
Prediction, projection
21.
Erratic Fluctuation
True/False
1
True
2.
False
3.
True
4.
False
5.
False
6.
True
7.
False
8.
True
9.
True
10.
True
11.
False
12.
True
13.
False
14.
False
15.
False
Unsolved Numerical Problems
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
Trend values are 200.94, 205.31, 209.69, 214.06, 218.43,
222.80, 227.19, 231.56
700, 800, 900, 1000, 1100, 1200, 1300
16.7, 18.3, 20, 22.7, 25.7, 29, 32.7, 36.3
15137, 17003, 20363, 26363, 31.147, 34330
110.0, 99.88, 92.38
42.1, 40.9, 39.8, 38.2. 38.1, 37.8
6.2, 6.6, 6.6, 6.8, 7.0, 6.6, 6.6, 7.2, 7.6, 8.0, 8.0
Trend values are 4.2, 5.6, 7; 8.4, 9.8
84, 86, 88, 90, 92, 94, 96
446.67, 546.67, 626.67, 706.67, 786.67, 866.67
40.06, 47.40, 54.74, 62.08, 69.42, 76.76, 84.10, 19.91.44
94.18, 105.82, 95.19, 105.32
106.4, 98.7, 94.9, 100
❑❑❑
Statistical Quality Control
499
Unit-3
Chapter
12
Statistical
Quality Control
Introduction
The word quality is not new for you. Quality means some countable property of a
product to satisfy the customer. Statistical quality control (SQC) is related to a
well planned collection and its better use for deciding the reason of variation in the
quality of a product, procedures, material, machines for a limited time. The study of
SQC is based upon the theory of probability and sampling. To improve the quality of
a product, SQC plays an important role in the industry. It covers all the factors and
related techniques of productions. Simply, we can say, SQC is useful to improve the
quality of material, manpower, machines and management.
12.1 History of statistical quality control
In 1924, Walter Shewhart developed the control chart. In 1931, Walter A. Shewhart
of bell Laboratories introduced statistical quality control in his book Economic
Control of Quality of Manufactured Products. In 1940, W. Edward Deming assisted
the U.S. Bureau of the Census in applying statistical sampling techniques. In 1941,
W. Edwards Deming joined the U.S. War Department to tech quality-control
techniques. In 1950, W. Edwards Deming addressed Japanese scientists, engineers,
and corporate executive on the subject of quality. In 1951, Joseph M. Juran
published the Quality Control Handbook. In 1954, Joseph M Juran addressed the
Union of Japanese Scientists and Engineers (JUSE). In 1968, Kaoru Ishikawa
outlined the elements of Total Quality Control (TQC). In 1970, Philip Crosby
introduces the concept of zero defects. In 1979, Philip Crosby publishes Quality is
Free. In 1980, Western industry began to import the concept of TQC under the name
Total Quality Management (TQM). In 1980, American electric giant, Motorola,
pioneered the concept of Six Sigma. In 1982, W. Edwards Deming published
Quality, Productivity, and Competitive Position. In 1984, Philip Crosby publishes
Quality Without Tears : The Art of Hassle-Free Management. In 1986, W. Edwards
Deming published Out of Crisis. In 1987, U.A. Congress created the Malcolm
Baldrige National Quality Award. In 1988, Secretary of Defense Frank Carlucci
direct the U.S. Department of Defense to adopt total quality. In 1993, the
total-quality approach began to be widely taught in the colleges and universities,
SQC is now an important part of the of any management studies.
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Engineering Mathematics-III
12.1.1 Statistical Quality Control: Evolution or Revolution?
Statistical quality control is an integral part of the paradigm of quality. For SQC to
be assessed either as evolutionary or as revolutionary, one has to understand the
paradigms of quality. Dooley indicates there have been three paradigms in the
discipline of quality :
♠
a pre-industrial paradigm of caveat of emptor,
♠
an industrial paradigm of quality control and
♠ post-industrial paradigm of total quality management.
However, since both industrial and post-industrial paradigms of quality emphasize
on the statistical methods to quality control, these two paradigms are labeled as the
paradigm of statistical quality control. Under the pre-industrial paradigm of caveat
emptor, individuals produced goods of certain quality and it was up to the consumer
to evaluate the quality of these goods. However, individual producers were at stake
for goods for which quality characteristics could be ascertained in a manifest way.
Dooley posits that trademarks, guilds, and punitive measures were used to such
situations.
On the contary, the paradigm of statistical quality brought about a broader set of
changes in quality control including the development of statistical quality practices,
the use of statistical methods within a framework of scientific management, the
improvement of quality, the total quality management, etc. We now turn to the
discussion of these two paradigms of quality – the pre-industrial paradigm of caveat
emptor, and the paradigm of statistical quality control – in relation to Kuhn's normal
versus revolutionary science. The pre-industrial era of quality revolves around the
production of goods and services produced by a single individual or a small group of
individuals. The individual or the small group producing the product or the service
determined the standard of quality and was responsible for conforming to those
standards. Feigenbaum calls this quality control the Operator Quality Control.
However, as the industrialization took place. the focus of production changed from
family-owned businesses to mass production by the industries. Complexity of
products and services grew exponentially along with industrialization. The operator
quality control became useless and more sophisticated measures of quality control
were warranted.
Quality philosophers such as Shewhart, Deming, Juran and others developed
statistical methods to address the unprecedented quality issues resulting form mass
production. The science of statistics replaced the traditional method of quality
control. The emergence of statistical quality control is a noncumulative
developmental episode in the history of quality control. The new phenomenon of
mass production demanded the rejection of the old methodology of operator quality
control. The paradigm of statistical quality control permitted the prediction of
quality that was unheard of in the pre-industrial paradigm of caveat emptor.
Therefore, the paradigm of statistical quality control emerged as a scientific
revolution in the arena of quality control.
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501
12.1.3 Application of statistical quality control
In those days, SQC plays an important role for an industry to grow. The technique of
SQC is applied in all industries. Following are the main objectives of SQC.
1. In this method, the quality of the product is checked without checking the every
unit involved in the production process.
2. SQC identifies the problems of an industry related to the production and then
helps to eliminate them.
3. SQC stops the wastage of material in an industry.
4. SQC helps to improve the quality of the product.
5. SQC helps an industry to boost its profit on the one side and provide the
guidance to improve the quality of the product on the other side.
6. SQC finds out the cause of concern for an industry.
7. With the help of SQC, an industry can determine the remedial measures for any
defect in the manufacturing process.
8. SQC provides good quality product in the market.
9. SQC provides a better environment in the industry and alertness in the
company for their long run.
12.1.3 Benefits of SQC
Since, SQC is scientific tool of manage men, therefore it has several benefits, which
are descries as:
1. SQC reduce the cost of inspection : In this method, since only a part of
the production is inspection, hence it reduce the cost of inspection.
2. SQC is an efficient method : SQC not only reduce the cost of impaction but
also an efficient method of inspection the output.
3. SQC is easy to apply : It is very easy to apply SQC to cheek the quality of an
output. This method can be applied by a person who has not haw the special
training or the extensive knowledge of mathematics.
4. SQC detects the early foults : Since in SQC a part of the output is always
checked, therefore once a sample part or output falls outside the control limits
at the same time the necessary corrective action has been taken.
5. SQC is better than 100% inspection : In 100% inspection method, the
variation in quality of the product may be detection at a stage, when a large
amount of faulty products have been already products. Thus, there would be a
big wastage. On the other side, SQC provides a graphical picture of how the
production is proceeding and to tell management where not to look for trouble.
In certain cases 100% inspection can not be applied without destroying all the
products inspected. for example if we want to apply the 100% inspection to
testing the breaking strength of chalks, then all the items inspected will be
spoiled. Thus SQC is superior than 100% inspection.
6. SQC determines effect of changed process : SQC helps to detect
whether or not a change in the production process results in a significant
change in quality.
502
7.
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Engineering Mathematics-III
SQC ensures overall co-ordination : SQC provides an overall
co-ordination among the staff of an organisation, for example, once the
production engineer may set specification that are so tight that operating staff
cannot meet them economically, consequently there will be an unnecessary
high scrapping rate. In case of loose specifications, the quality of the product
will be sacrificed. To avoid. these problem, SQC set up a reasonable control
limits for the workers and as well as ensures the quality of the product.
SQC has special role for India : SQC has a special role for our country
because there are so many variations here in row materials and in machine.
Since, we are in the age of development therefore we need to apply SQC to earn
more foreign exchange by supplying quality goods to compete in the global
market.
12.1.3 Limitations of SQC
No doubt, SQC has a lot of significance to ensure the quality of the product, but it is
not a complete solution to provide the best quality product. SQC has the following
limitations :
1. SQC can not be applied mechanically.
2. It is dangerous to apply the SQC without adequate study of the process.
3. Since SQC is applied as part of a generally quality awareness they may only
lead to false sense of security.
4. SQC is depends upper statistical methods and does not reduce the manager's
responsibility.
12.1.4 Cause of variation in the quality
Since change is the law of nature, therefore in spite of the modern technologies the
variation in the quality of a product is also inevitable and inherent. These variations
are classifies as follows:
12.1.4.1 Controlled Variation or Non random variation
There are bound to be many small, unobservable, chance effects that influence the
outcome. This kind of variation is said to be "control," not because the process
operator is able to control the factors absolutely, but rather because the variation is
the result of normal disturbances, called common causes, within the process.
This type of variation can be predicted. In other words, given the limitations of the
process, each of these common causes is controlled to the greatest extent possible.
12.1.4.2 Uncontrolled variation or random variation
Uncontrolled Variation or non random variation are the natural variations and
beyond the control of human being. For example, variation in the temperature,
pressure, humidity etc. These are the variations that you can never eliminate it
totally. Variation that arise periodically and for reasons outside the normally
functioning process, induced by a special cause. Special causes include differences
between machines, different skill or concentration levels of workers, changes in
atmospheric conditions, and variation in the quality of inputs. Unlike controlled
variation, uncontrolled variation can be reduced by eliminating its special cause.
Statistical Quality Control
503
12.2 Control Charts
The development and use of statistical tools known as control charts for monitoring
and managing a wide variety of processes has historically been credited to Dr.
Walter Shewart. During the 1920s, Dr. Shewart identified two types of variation
that prevailed among all processes:
♠
chance cause variation and
♠ assignable cause variation.
Any process will inherently contain chance cause variation, as this is normal,
random variation present in all systems. However, a process that contains anything
other than chance cause or random variation is exhibiting assignable cause variation
and is referred to as "out of statistical control." Control charts may be used to monitor
a process to determine whether or not it is in statistical control, to evaluate a process
and determine normal statistical control parameters, and to identify areas of
improvement in processes. Thus, control chart helps to check the quality standard of
the product as well as provide the ways to improve the level of standard of the
products and also makes them cost effective.
Control charts may be used to monitor variables that are continuous in nature, or
non-continuous attributes. Continuous variables include quantifiable or measurable
values that can be calculated across a continuous range, such as averages,
dimensions (length, height and width), weight, and temperature. Attributes are not
inherently quantifiable, but can be counted, such as number of parts defected, or
number of defects per unit inspected. The purpose of any control chart is to help
determine if variations in measurements of a product are caused by small, normal
variations that cannot be acted upon ("common causes"), or by some larger "special
cause" that can be acted upon or fixed. The type of chart to be used is based on the
nature of the data. The component of a control chart are CL, UCL and LCL, which are
described below.
♠
CL - In a control chart, CL (Control Line) is a straight line, which represents the
average quality of the product at which the related process should be performed.
♠
UCL - In a control chart, UCL (Upper Control Line) is a straight line, which is chosen
in such a way that any value falling above this line is treated as lack of control.
♠
LCL - In a control chart, LCL (Lower Control Line) is a straight line, which is chosen
in such a way that any value falling below this line is treated as lack of control.
However, when values are lying in between UCL and LCL, it means the related
product has a possibility of improvement. While values are beyond them are treated
as out of control.
12.2.1 Basic Model of Control Charts
Let w be a sample statistic that measures some quality characteristic of interest, and
suppose that the mean of w is µ w and the standard deviation of w is σ w . Then the
center line, the upper control limit, and the lower control limit become
UCL = µ w + k σ w ,Cl = µ w , LCL = µ w − k σ w
where k is the distance of the control limits from the center line, expressed in
standard deviation units
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Engineering Mathematics-III
12.2.2 How to Assessing the Results of a control chart
With respect to control charts, recording the observation, calculating the control
values, and plotting the observed values on the x and S charts is only the beginning
of the process. Once the chart is constructed, and data points are plotted, the chart
needs to be evaluated to determine whether or not the process is operating in
statistical control. There are several guidelines for evaluating control charts, all with
a prevailing philosophy: the plotted points should only exhibit random variation
that cannot be fitted with an identifiable or quantifiable pattern. Additionally, the
following features, in accordance with the parameters of a normal distribution and
the Central Limit Theorem, should also apply, which is described below.
12.2.2.1 Central Limit Theorem
Central Limit Theorem provides a statistical argument for control charts. The central
limit theorem suggests that even if the underlying population from which a series of
observations are gathered is not normally distributed, the resulting distribution of
averages from the sample observations will be normally distributed around an
average value, and that 99 ⋅ 73 percent of the values will be contained within three
standard deviations of that average value. Using this theorem, a control chart can be
constructed that utilizes the statistical mean as the reference value or centerline on
the chart, and an upper and lower control limit equivalent to ± 3 σ (three standard
deviations) from the statistical mean. If an observed value falls beyond an upper or
lower control limit, it can be concluded that this value is in an extreme tail of the
distribution (only 0 ⋅ 27 percent of values should fall in this region), and is therefore
out of control. Additionally, a series of values above or below the reference line can
be evaluated to determine process stability. In brief, we can say that
1. 68 percent of the points should be within ± 1σ of the reference line.
2. 4 ⋅ 27 percent of the points should be between ± 2σ and ± 3σ of the reference line.
3. No more than 0 ⋅ 27 percent of the points should exceed ± 3σ of the reference line.
In addition to the above general philosophy of evaluating the control chart, an
additional set of guidelines, as described in the AT&T Statistical Quality Control
Handbook. The AT&T Rules divide the control chart into three zones, mirrored
across the center line. This is illustrated in Figure 1.
Upper Control Limit
Zone A
Zone B
Zone C
Center Line
Zone C
Zone B
Zone A
Lower Control Limit
Fig .1 : AT&T Zone Chart
Statistical Quality Control
505
According to these guidelines, a process can be ruled "out of statistical control" is any
of the following apply:
1.
Any point falls outside of the upper or lower control limit (beyond ± 3σ).
2.
Two of three successive points in zone A or beyond..
3.
Four out of five successive points in zone B or beyond.
4.
Eight successive points fall in zone C or beyond, on one side of the center line.
5.
Trends: a series of points without any appreciable or random change in
direction, or values moving continuously up and down, or across the centerline
in pattern.
6.
Cycles: short trends in which the data may repeat in a pattern.
7.
Shifts: a sudden change in level in one direction or another.
8.
Stratification: a pattern of "unnatural consistency" within a single zone, or near
the centerline.
9.
Systematic Variables: a predictable pattern, where a high point is always
followed by a low point, or a low point by a high point.
12.2.3 Some Example of Control chart
For better understanding, some examples are given.
♠
As long as the points remain between the lower and upper control limits, we
assume that the observed variation is controlled variation and that the process is
in control. (Fig.2)
Fig. 2 :
♠
The process is out of control. Both the fourth and the twelfth observations lie
outside of the control limits, leading us to believe that their values are the result
of uncontrolled variation.(Fig.3)
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Engineering Mathematics-III
Fig. 3 :
♠
Even control charts in which all points lie between the control limits might
suggest that a process is out of control. In particular, the existence of a pattern in
eight or more consecutive points indicates a process out of control, because
an obvious pattern violates the assumption of random variability.(Fig.4.)
Fig. 4 :
♠
The first eight observations are below the center line, whereas the second seven
observations all lie above the center line. Because of prolonged periods where
values are either small or large, this process is out of control. (Fig.5)
Statistical Quality Control
507
Fig. 5 :
Other suspicious patterns could appear in control charts. Unfortunately, we cannot
discuss them all here. Control chart makes it very easy for you to identify visually
points and processes that are out of control without using complicated statistical
test. This makes the control chart an ideal tool for the shop floor, where quick and
easy methods are needed.
12.3 Types of control charts
Control charts can be divided in two categories: Variable Charts and Attribute
Charts
Variable Charts
Variable Charts monitor variables and display continuous measures, such as
weight of the product, diameter of a screw, thickness of the metal plate, purity of the
matter used in the product, specific resistance of the wire, life of an electric bulb. Its
statistical analysis focuses on the mean values of such measures. The variable charts
are given below.
♠
X-bar and R chart
♠
X-bar and S chart
12.3.1 X-bar and R chart
X-bar and R chart is an important number of a family of control charts. This
control chart is a plot of measurements of a product on two special scales, usually
located and below each other and running horizontally. X-Bar/R charts consist of
two charts, both with the same horizontal axis denoting the sample
number. The vertical axis on the top chart depicts the sample means
(X-Bar) for a series of lots or subgroup samples. It has a centerline represented by X
=
double bar ( X ), which is simply the overall process average, as well as two horizontal
lines, one above and one below the centerline, known as the upper control limit or
UCL and lower control limit or LCL, respectively. These lines are drawn at a
distance of plus and minus three standard deviations (that is, standard deviations of the
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Engineering Mathematics-III
sampling distribution of sample means) from the process average. In practice,
tabulated constants are available to determine the control limits, or they are
automatically calculated by the SQC software used.
R chart has the range (R) of each subgroup plotted on the vertical axis. Like an X-Bar
chart, R charts have a centerline and two control limits. However, for sample sizes
below 7, the LCL is zero. The X-bar/R chart is normally used for numerical data that
is captured in subgroups in some logical manner, for example three production parts
measured every hour. A special cause such as a broken tool will then show up as an
abnormal pattern of points on the chart.
12.3 Construction of Control Chart for X-bar and R, when µ, σ
are known
If the mean µ and the standard deviation σ are known, the LCL, CL and UCL for X-bar
and R chart are given below:
For X-bar Chart the control limits are given by
σ
µ ± 3σ x ⇒ µ ± 3
⇒ µ ± Aσ
n
Hence, the LCL, CL and UCL for X – bar chart are given as,
3
UCL = µ + A σ, LC = µ, LCL = µ − A σ, where A =
n
For R Chart the control limits are given by
µ R ± 3 σ R ⇒ d2σ ± 3d3 σ ⇒ ( d2 ± 3d3 ) σ
Hence, the LCL, CL and UCL for R chart are given as,
UCL = D2σ, CL = d2σ, LCL = D1σ
Where,
D1 = d2 − 3d3 , D2 = d2 + 3d3
The values A, d 2 , D1 and D 2 are determined from the table of the control chart
factors. The values of these factors depend on the size of the sample.
12.4 Construction of Control Chart for X-bar and R, when µ and
σ are unknown
In this case, when µ and σ are unknown, there is a need to calculate µ and σ for X-bar
and R chart. For this purpose compute the following parameters.
For X-bar Chart, we have to calculate,
m
m
µx = X =
i =1
m
n
m
∑ ∑ xij
∑ xi
=
i =1 j =1
mn
;
R
σ$ =
;
d2
∑ Ri
R =
i =1
Where, Range Ri = max ( X ij ) - min ( X ij ) for j = 1,..n
For X-bar Chart the control limits are given by
R / d2
σ$
µ$ x ± 3σ$ s ⇒ X ± 3
⇒ X ±3
⇒ X ± A2R
n
n
Hence, the LCL, CL and UCL for X – bar chart are given as,
m
Statistical Quality Control
509
UCL = X + A 2 R , CL = X , LCL = X − A 2 R where,
A2 =
3
d2 n
R
For R Chart, we have to calculate, µ R and σ R where σ$ =
d2
m
∑ Ri
µ$ R = R =
i =1
m
; σ R = d3 σ = d3 =
R
d2
For R chart the control limits are given by
d R
d
µ$ R ± σ$ R ⇒ R ± 3 3 ⇒ (1 ± 3 3 ) R
d2
d2
Hence, the LCL, CL and UCL for R chart are given, as
UCL = D4 R , CL = R , LCL = D3 R
3d
3d
Where,
D3 = 1 − 3 , D4 = 1 + 3
d2
d2
The values A 2 , D 3 and D 4 are determined from the table of the control chart
factors. The values of these factors depend on the size of the sample.
n
B3
B4
C4
A3
A2
D3
D4
2
0.000
3.267
0.7979
2.659
1.881
0
3.269
3
0.000
2.568
0.8862
1.954
1.023
0
2.574
4
0.000
2.666
0.9213
1.628
0.729
0
2.282
5
0.030
2.089
0.9400
1.427
0.577
0
2.114
6
0.118
1.970
0.9515
1.287
0.483
0
2.004
Table: 1 Control chart factors
Illustrative Examples
Ex. 1 : The data shown here are x-bar and R values for 24 samples of size n=5 taken
from a process producing bearings.
(a) Set up x-bar and R charts on this process, Does the process seem to be in
statistical control?
It necessary, revise the trial control limits.
(b) Estimate the process standard deviation
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Engineering Mathematics-III
Sol :
Calculation of X and R
Σ xi 
1

[ 34 ⋅ 5 + 34 ⋅ 2 + 31 ⋅ 6 + 31 ⋅ 5 + 35 ⋅ 0 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ 34 ⋅ 2] = 34  ∵x =

24
24 

Σ Ri 
1

R=
[ 3 + 4 + 4 + 4 + 5 + 6 ⋅ ⋅ ⋅ ⋅ ⋅ 2] = 4 ⋅ 7
R =

24
24 

=
X =
LCL, CL and UCL for X bar chart
As described earlier, when the mean and standard deviation are unknown, the LCL,
CL and UCL for X bar chart are given, as,
UCL = x + A 2 R , CL = X , LCL = X − A 2 R
Here X = 34 and R = 4 ⋅ 7, Since in this problem, n = 5, therefore, from the table of
control chart, as given in the table ......1 the value of A 2 = 0 ⋅ 577, we get
UCL = 34 + [( 0 ⋅ 577 )( 4 ⋅ 7 )] = 36 ⋅ 7, CL = 34, LCL = 34 − [( 0 ⋅ 577 )( 4 ⋅ 7 )] = 31 ⋅ 3
LCL, CL and UCL for R chart
As described earlier, when the mean and standard deviation are unknown, the LCL,
CL and UCL for S chart are given, as,
UCL = D4 R , CL = R , LCL = D3 R
Here R = 4 ⋅ 7. Since, in this problem, n = 5, therefore, from the table of control
chart, as given in the table .....1... the value of D 3 = 0 and D 4 = 2 ⋅ 114, thus, we get
UCL = ( 2 ⋅ 114) ( 4 ⋅ 7 ) = 9 ⋅ 96, CL = 4 ⋅ 7, LCL = ( 0) ( 4 ⋅ 7 ) = 0
Construction of X and R charts
Typically, the construction of the X and R charts is done together, so that the
plotted values can be compared at even intervals. Below, the two charts are plotted.
Fig. 6 : X and R chart
Revised X and R charts
Observe the above X bar chart. Since the sample 12 and 15 out of control therefore
removing these two out of control samples and recalculate control limits for both
charts
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511
Calculation of Revised X and R
In this case
X = 33 ⋅ 7
and
R = 4⋅ 5
Revised LCL, CL and UCL for X bar chart
Since the revised value of X = 33 ⋅ 7 and R = 4 ⋅ 5, therefore
UCL = 33 ⋅ 7 + [( 0 ⋅ 577 )( 4 ⋅ 5)] = 36 ⋅ 3, CL = 34, LCL = 33 ⋅ 7 − [( 0 ⋅ 577 )( 4 ⋅ 5)] = 31
Revised LCL, CL and UCL for R chart
As described earlier, when the mean and standard deviation are unknown, the LCL,
CL and UCL for S chart are given, as,
UCL = D4 R , CL = R , LCL = D3 R
Since the revised value R = 4 ⋅ 5, therefore
UCL = ( 2 ⋅ 114) ( 4 ⋅ 7 ) = 9 ⋅ 96, CL = 4 ⋅ 7, LCL = ( 0) ( 4 ⋅ 7 ) = 0
Constructing the Revised X and R charts
Typically, the construction of the revised X and R charts is done together, so that the
plotted values can be compared at even intervals, Below, the two charts are plotted.
Fig. 7
12.5 Interpretation of X bar R Chart
♠
First check the R chart and eliminate the assignable causes from R chart, and
then check the X bar chart
♠ Check non-random pattern
– Cyclic pattern due to temperature, regular rotation of operators or machines,
maintenance schedules, tool wear (Fig. 8,)
– Mixture pattern when the plotted points tend to fall near or slightly outside the
control limits. Two overlapping distributions are resulted from too often
process adjustment (Fig 9,).
– Shift in process level due to introduction of new workers, methods, materials,
or inspection standard (Fig. 10,)
– Trend pattern due to gradual tool wear (Fig. 11,)
– Stratification pattern for the points to cluster around the center line due to
incorrect calculation of Control limits or inappropriate reasonable sampling
group (Fig. 12,)
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Engineering Mathematics-III
Fig. 8 : Cycles on a control chart
Fig. 10 : A shift in process level
Fig. 9 : A mixture pattern
Fig. 11 : A trend in process level
Fig. 12 : Stratification Pattern
12.6 Application Conditions of X bar and R chart
♠
Underlying distribution of the quality characteristics is normal
– X bar chart is more robust to nonnormality than R chart
Statistical Quality Control
513
– samples of 4 or 5 are sufficient to ensure reasonable robustness to the
normality assumption for X bar chart
♠
Calculation accuracy of Type I error is dependent on distribution
♠
X bar chart (n = 4, 5, 6) is not effective to detect a small mean shift (less than
1 ⋅ 5 σ) on the first sample following the shift
♠
R chart is insensitive to small or moderate shifts (σ 1 / σ 0 < 2 ⋅ 5) for the sample
size of n=4, 5, or 6. If n>10. an S chart should be used instead of an R chart
12.7 X bar and S chart
The X chart relays information about the central tendency of the data, or the
tendency of the measurements to accumulate in a normal distribution around an
average value, or x (x-bar). While this is an important statistic to know and
understand, it is also critical that the amount of variation around a particular
average is also known. The s chart, recording the standard deviation from the average
value in a given sample is used for measuring and monitoring variation, and together
x and s charts can provide insight into the stability of a process or system. For
example, a sample of five measurements might have an average value ( x ) of 100 with
a standard deviation (s) of 10; while another sample of five measurements might have
an average value ( x ) of 100 with a standard deviation (s) of 30. Although the average
values are the same in booth samples, the sample with s equal to 30 exhibits much
more variation. In some processes, variation within broad control limits might be
acceptable, but in others, this kind of variation might be undesirable. In some
applications, such is in high-tech manufacturing or machine part manufacturing;
tolerances for variation are often very limited. The use of the x and s charts can be
valuable in monitoring these types of processes. X bar-S charts are generally employed
for plotting variability of sub-groups with sizes greater than 10. X-bar/R charts are
used for plotting variability when sub-group sizes are less than 10.
Both the x bar and S chart must be seen together to interpret the stability of the
process. The S chart must be examined first as the control limits of the X bar chart
are determined considering both the process spread and center. Process variation,
which is a characteristic of spread, must be in control to correctly interpret the X bar
chart. If data-points in the S chart are outside control limits, then the limits on the X
bar chart may be inaccurate and may falsely indicate an out-of-control condition. As
in other types of control charts, data-points outside of control limits in a X bar-S
chart indicate special causes.
12.7.1 Construction of Control Chart for X-bar and S, when µ and
σ are known
If the mean µ and the standard deviation σ are known, For X-bar Chart the control
limits are same as that in X-bar and R chart the LCL, CL and UCL for S chart are given
below:
For S Chart the control limits are given by
µ s ± 3σ s ⇒ c 4σ ± 3σ 1 − c 42 ⇒ ( c 4 ± 3 1 − c 42 )σ
Hence, the LCL, CL and UCL for R chart are given, as,
UCL = B6σ, CL = µ s , LCL = B5σ
where B5 = c 4 − 3 1 − c 42 ,B6 = c 4 + 3 1 − c 42
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Engineering Mathematics-III
The values B 5 and B 6 are determined from the table of the control chart factors.
The values of these factors depend on the size of the sample.
12.3.2.2 Construction of Control Chart for X-bar and S, when µ
and σ are unknown
In this case when µ and σ are unknown, there is a need to calculate µ and σ for
X –bar and S chart. For this purpose compute the following parameters. For X-bar
Chart, we have to calculate,
m
m
∑
∑ xi
µ = X i =1
m
S =
1
m
=
n
n
∑ xij
i =1 j =1
m
∑ Si , σ$ =
i =1
mn
µ$ s
c4
∑ ( xij
;
=
− xi ) 2
j =1
Si =
;
n −1
S
3
; ( based on µ = c 4σ ) A 3 =
c4
c4 n
For X-bar Chart, the control limits are given by,
3 S
σ$ ± 3σ$ x ⇒ x ±
⇒ x ± A3S
n c4
Hence, the LCL, CL and UCL for R chart are given, as,
UCL = X + A 3 S , CL = X , LCL = X − A 3 S
For S Chart, we have to calculate µs and σs, as,
n
m
1
µ$ s = S = ∑ Si ; Si =
m i =1
∑ ( xij
− xi ) 2
i =1
n −1
σ =
S
;
c4
For S Chart, the control limits are given by,
1 − c 42
S
1 − c 42 ⇒ (1 ± 3
)S
c4
c4
Hence, the LCL, CL and UCL for R chart are given, as,
UCL = B4 S , CL = S , LCL = B3 S
3
3
where
B3 = 1 −
1 − c 42 ,B4 = 1 +
1 − c 42
c4
c4
µ s ± 3σ s ⇒ S ± 3
The values A 3 , B 3 and B 4 are determined from the table of the control chart
factors. The values of these factors depend on the size of the sample.
Ex. 2 : Construct the x and s charts from a sample size, n = 5 and the number of
samples taken, m = 10. The observations are recorded, and average values (x-bar)
and standard deviations are calculated for each sample. Additionally, average s and
x-bar values are calculated for the all of the sample averages and standard
deviations. The resulting measurements and calculations can be recorded in a grid,
in the following table.
Statistical Quality Control
515
Sample m
1
2
3
4
5
6
7
8
9
10
1
0.015
0.018
0.017
0.018
0.014
0.017
0.012
0.014
0.015
0.015
2
0.019
0.017
0.018
0.016
0.013
0.015
0.013
0.015
0.018
0.015
3
0.022
0.013
0.019
0.012
0.014
0.018
0.015
0.014
0.016
0.018
4
0.016
0.014
0.014
0.020
0.015
0.019
0.015
0.013
0.014
0.016
5
0.013
0.015
0.015
0.019
0.017
0.014
0.016
0.012
0.013
0.017
x-bar
0.017
0.015
0.017
0.017
0.015
0.017
0.014
0.014
0.015
0.016
s
0.004
0.002
0.002
0.003
0.002
0.002
0.002
0.001
0.002
0.001
Average x-bar = 0.016,.....
Average's
=
0.002
Table 2 : Completed grid with measurements and calculations
Calculation of X and S
Σ xi
⋅017 + 0 ⋅ 015 + 0 ⋅ 017 + 0 ⋅ 017 + 0 ⋅ 015 + ...0 ⋅ 016
X =
=
= 0 ⋅ 016
10
10
ΣSi
0 ⋅ 004 + 0 ⋅ 002 + 0 ⋅ 002 + ... 0 ⋅ 001
R =
=
= 0 ⋅ 002
10
10
UCL, CL and UCL for X bar chart
As described earlier, when the mean and standard deviation are unknown, the LCL,
CL and UCL for X bar chart are given, as,
UCL = X + A 3S , CL = X , LCL = X − A 3S
In this problem, n = 5, therefore, from the table of control chart, as given in the
table..1.....the value of A 3 = 1 ⋅ 427, we get
UCL = 0.016 + [(1.427 )(0.002)] = 0.0189, CL = 0.016, LCL = 0.016 −[(1.427 )(0.002)]
= 0.0131
LCL, CL and UCL for S chart
As described earlier, when the mean and standard deviation are unknown, the LCL,
CL and UCL for S chart are given, as,
UCL = B4 S , CL = S , LCL = B3 S
In this problem, n = 5, therefore, from the table of control chart, as given in the
table...1...the value of B3 = 0.030 and B 4 = 0.089. we get
UCL = ( 2.089)( 0.002) = 0.00418, CL = 0.002, LCL = ( 0.030)(.002) = 0.00006
Constructing the X and S charts
Typically, the construction of the X and s charts is done together, so that the plotted
values can be compared at even intervals. Below, the two charts are plotted in
Figures 1 and 2:
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Engineering Mathematics-III
Fig. 13 : x-bar Chart
Fig. 14 : S - Chart
Evaluating the above Example
As the rules and guidelines are applied to the example provided in the
section.12.2.2., it is apparent that the s chart is exhibiting signs of instability, or that
the underlying process is out of statistical control. No values ever exceed control
limits, however, there appears to be a sustained trend in variations from the positive
side of the centerline to the negative or lower side of the control line. When checked
against the x chart, it is less apparent that the system is out of control. The x and s
chart together suggest that while the process output is consistently clustered around
the mean value, the variation is out of control and needs to be addressed.
Statistical Quality Control
517
12.8 Comparison of R Chart and S Chart
♠
R Chart
– Simple for hand calculation;
– good for small sample size;
– lose information between x min and x max ;
– not used for variable sample size.
♠
S Chart
– when the sample size is large (n>10);
– Used for variable sample size ;
– Computation complexity can be simplified by using a computer.
12.9 Attribute Charts
Attribute charts monitor attributes and differ from variable charts in a way that they
describe a feature of the process rather than a continuous variable such as a weight
or volume. Attributes can be either discrete quantities, such as the number of defects
in a sample, or proportions, such as the percentage of defects per lot. There are three
control charts for attributes :
♠
p-Chart - control chart for fraction defective.
♠
c-Chart - control chart for number of defects.
♠
np-Chart- control chart for number of defective.
12.9.1 p-Chart
In industrial statistics, the p-chart is a type of control chart that monitors the
proportion of nonconforming units in a sample. The appropriate data for p-charts
are attribute data (conform or non-conform, yes or no, etc.). The subgroup size
should ideally be equal, although unequal sample sizes can be accommodated. Let n
be the sample size taken from the production process at different interval of time, if
d is the number of defectives in this sample size, then the sample proportion
defective p is given by
p = total number of defective units/ Total numbers of units
Control limits for the p-chart are calculated on the basis of the binomial distribution
and an approximation based on the central limit theorem.
The UCL and LCL for p- chart are determined by the formula:
p (1 − p )
p±3
n
Hence, UCL, CL and LCL are given, as
p (1 − p )
p(1 − p )
UCL = p + 3
, CL = p, LCL = p − 3
n
n
12.9.2 c-Chart
In industrial statistics, the c-chart is a type of control chart used to monitor "count"
-type data, typically total number of nonconformities per unit. It is also occasionally
used to monitor the total number of events occurring in a given unit of time. The
c-chart differs from the p-chart in that is accounts for the possibility of more than
518
Engineering Mathematics-III
one nonconformity per inspection unit. The p-chart models "pass"/"fail" -type
inspection only. Nonconformities may also be tracked by type or location which can
prove helpful in tracking down assignable causes. Examples of processes suitable for
monitoring with a c-chart include:
♠
Monitoring the number of voids per inspection unit in injection molding or
casting processes
♠
Monitoring the number of discrete components that must be re-soldered per
printed circuit board
♠ Monitoring the number of product returns per day
The Poisson distribution is the basis for the chart and requires the following
assumptions:
♠
The number of opportunities or potential locations for nonconformities is very
large
♠
The probability of nonconformity at any location is small and constant
♠
The inspection procedure is same for each sample and is carried out consistently
from sample to sample
The UCL and LCL for c- chart type are determined by the formula
c ±3 c
where c is the estimate of the long-term process mean established during
control-chart setup. Hence UCL, CL and LCL are given as
UCL = c − 3 c, CL = c, LCL = c − 3 c
12.5.3 np-Chart
In industrial statistics, the np-chart us a type of control chart that is very similar to
the p-chart except that the statistic being plotted as a number count rather than a
sample proportion of items. For example, an np-chart often shows the number of
nonconforming items in each sample. Since we are counting failures or successes,
clearly the appropriate data for np-charts need to be attribute data. The subgroup
size must be constant, as comparisons of counts would otherwise be meaningless. In
np-chart
1. The "np" stands for the number of conforming items, which can be expressed as
n (sample size) times p (proportion of nonconforming items)
2. Need a good definition of nonconforming items – usually a categorical
definition
3. Subgroup size must be constant
4. Normally need large subgroups – can even be up to total for the period
Control limits for the np-chart are calculated on the basis of the binomial
distribution and an approximation based on the central limit theorem.
The UCL and LCL for np-chart type ares determined by the formula:
np ∓ 3 np (1 − p )
Hence, UCL, CL and LCU are given as
UCL = np + 3 np(1 − p), CL = np, LCL = np − 3 np − (1 − p) 2
Statistical Quality Control
519
Illustrative Examples
Ex. 3 : Construction a control chart (x and R chart) from the following data on
basis of fuses and also comment on the basis of chart. Also find σ .
Sol :
Sample observations
Σ xi
xi =
Σ xi
n
R i = R max R min
1
42
65
75
78
347
347
69.4
45
2
42
45
68
72
317
317
63.4
48
3
19
24
80
81
285
285
57.0
62
4
36
54
69
77
320
320
64.0
48
5
42
51
59
59
287
287
57.4
36
6
51
74
75
78
410
410
82.0
81
7
60
60
72
95
425
425
85.0
78
8
18
20
27
42
167
167
33.4
42
9
15
30
39
62
230
230
46.0
69
10
69
109
113
118
562
562
112.4
84
11
64
90
93
109
468
468
93.6
48
12
61
78
94
109
478
478
95.6
75
Σxi = 859.2
Σ R i = 716
Now
x=
Σ xi
859 ⋅ 2
=
= 71 ⋅ 6
n
12
R=
ΣR i
716
=
= 52 ⋅ 67
n
12
Construction of x-bar Chart
In this case n = 5 so by control chart table 1, A 2 = 0 ⋅ 58, hence the UCL, CL And LCL
are given as,
UCL = x + A 2 R = 71 ⋅ 60+ 0 ⋅ 58 × 52 ⋅ 67 = 106 ⋅ 21
CL = x = 71 ⋅ 60
520
Engineering Mathematics-III
LCL = x − A 2 R = 71 ⋅ 60 − 0 ⋅ 58 × 52 ⋅ 67 = 36 ⋅ 99
the x is given below
Sample number
Fig. 15 :
Construction of R chart
In this case n = 5 so D 3 = 0 and D 4 = 2 ⋅ 11, hence the UCL, CL and LCL for R chart
are given below
UCL = D 4 R = 2 ⋅ 11 × 59 ⋅ 67 = 125 ⋅ 904
CL = R = 59 ⋅ 67
LCL = D 3 R = 0 × 59 ⋅ 67 = 0
the R chart is given below
Sample number
Fig. 16 :
Comment on x chart and R chart
(i) Process is out of control because the points corresponding to sample 8 and 10
out of the limits.
Statistical Quality Control
521
(ii) R chart shows the process is in control.
From x and R chart we can say that there are some assignable causes due to which
process is not is statistical control.
In this case, n = 5 therefore from control chart table d 2 = 2 ⋅ 326
R
52 ⋅ 67
we know that σ =
=
= 22.6440
d2
2 ⋅ 326
Ex. 4 :
Machine A
Machine B
Sub group No.
Mean x
Range 'R'
Sub group No.
Mean x
Range 'R'
1
2.77
0.66
1
2.53
0.12
2
2.70
0.29
2
2.67
0.30
3
2.78
0.19
3
2.66
0.17
4
2.67
0.12
4
2.57
0.25
5
2.75
0.34
5
2.60
0.24
6
2.77
0.23
6
2.60
0.05
7
2.75
0.17
7
2.70
0.30
8
2.73
0.06
8
2.56
0.04
9
2.76
0.23
9
2.70
0.19
10
2.63
0.20
10
2.67
0.08
11
2.73
0.17
11
2.80
0.11
12
2.73
0.28
12
2.63
0.14
13
2.74
0.26
13
2.71
0.24
14
2.72
0.13
14
2.63
0.31
15
2.73
0.13
15
2.75
0.17
The above data (pertaining to two subgroups of size 4) is from two different
machines, which are supposed to be alike. Plot the necessary charts to show whether
their products would support this assumption. If they do not support this
assumption, does this prove that machines are not essentially alike?
Sol :
Control limits for X − chart, for machine-1 is given by
∑ xi = 40 ⋅ 96 = 2.73
X =
15
15
522
Engineering Mathematics-III
∑R
2.86
= 0 ⋅ 191
15
15
R
0 ⋅ 191
σ=
=
= 0 ⋅ 0927
d2
2 ⋅ 059
R =
and
=
CL x = X = 2 ⋅ 73
UCL x = X + A 2 R = 2 ⋅ 73 + 0 ⋅ 73 × 0 ⋅ 191 = 2 ⋅ 87
LCL x = X − A 2 R = 2 ⋅ 73 − 0 ⋅ 73 × 0 ⋅ 196 = 2 ⋅ 59
For machine No. 2
X =
∑ xi
15
∑ Ri
=
39 ⋅ 60
= 2 ⋅ 64
15
2 ⋅ 71
=
= 0 ⋅ 181
15
15
R
0 ⋅ 181
σ=
=
= 0 ⋅ 088
d2
2 ⋅ 059
R =
CL x = X = 2 ⋅ 64
UCL x = X + A 2 R = 2 ⋅ 64 + 0 ⋅ 73 × 0 ⋅ 181 = 2 ⋅ 77
LCL x = X − A 2 R = 2 ⋅ 64 − 0 ⋅ 73 × 0 ⋅ 181 = 2 ⋅ 51
It is clearly evident that the two machines are not alike (readers are advised to draw
X-chart). Further standard divination for first machine is 0.027 while that of second
is 0.088, which are different. Therefore given two machines are not essentially alike.
Ex. 5 : Control charts X and R are maintained on a certain dirneonsion of a
manufactured part measured in inches. The subgroup size is 4. The value of X and R
are computed for each subgroup, After 20 subgroups, Σ X = 41 ⋅ 340 and
ΣR = 0 ⋅ 320. Compute the value of the 3-sigma limits for the X and R-charts, and
estimate the values σ on the assumption that the process is in statistical control.
Sol :
Given, Σ X = 41 ⋅ 340, n = 20
41 ⋅ 340
X =
= 2 ⋅ 067
20
ΣR = 0 ⋅ 320
∑ R = 0 ⋅ 320 = 0 ⋅ 016
hence
R =
20
20
For subgroup size 4, from the control chart table 1.1 A 2 = 0 ⋅ 73, D3 = 0, and
D4 = 2 ⋅ 28
CL x = X = 2 ⋅ 067
UCL x = X + A 2 R = 2 ⋅ 067 + 0 ⋅ 730 × 0 ⋅ 016 = 2 ⋅ 055
LCL x = X = − A 2 R = 2 ⋅ 067 − 0 ⋅ 73 × 0 ⋅ 016 = 2 ⋅ 079
CLR = R = 0 ⋅ 016
UCLR = D4 R = 2 ⋅ 28 × 0 ⋅ 016 = 0
Statistical Quality Control
523
LCLR = D3 R = 0 × 0 ⋅ 016 = 0.
For a subgroup size of 4, from Table 7.1. d2 = 2 ⋅ 059
R
0 ⋅ 016
σ=
=
= 0 ⋅ 0078
d2
2 ⋅ 059
Ex. 6 :
Sample No.
Sample Size
No. of defectives (d)
1
2,000
425
2
1,500
430
3
1,400
216
4
1,350
341
5
1,250
225
6
1,760
322
7
1,875
280
8
1,955
306
9
3,125
337
10
1,575
305
The above data give the number of defectives in 10 independent sample of varying
sizes from a production process. Construct p-chart for the data.
Sol :
In the given problem, we have variable sample size.
Here, the total number of defectives Σdi = 3187
And the total number of sample size Σni = 17790.
∑ di
∑ ni
3187
= 0 ⋅ 1791
17790
So that
p=
and
q = 1 − p = 1 − 0 ⋅ 1791 = 0 ⋅ 8209
=
q = 1 − p = 1 − 0 ⋅ 1791 = 0 ⋅ 8209
CL = p = 0 ⋅ 1791
UCL = p + 3 pq / ni
and
LCL = p − 3, pq / ni
524
Engineering Mathematics-III
Let us form the following table for the computation of p-chart :
d
S.No.
n
d
UCL
LCL
1
2000
425
0.2125
0.0000735
0.025719
0.2048
0.1534
2
1,500
430
0.2867
0.000098
0.029698
0.2088
0.1494
3
1,400
216
0.1543
0.000105
0.030741
0.2098
0.1484
4
1350
341
0.2526
0.000109
0.031321
0.2104
0.1478
5
1250
225
0.1800
0.000118
0.32588
0.2117
0.1465
6
1760
322
0.1829
0.000084
0.027413
0.2065
0.1517
7
1875
280
0.1495
0.000078
0.026562
0.2057
0.1525
8
1955
306
0.1565
0.000075
0.026015
0.2041
0.1531
9
3125
337
0.1078
0.000047
0.020567
0.1997
0.1585
10
1575
305
0.1937
0.000093
0.029877
0.2080
0.1502
Total
17790
3487
p=
n
pq / b
3×
pq / n
on the basis of above table let us form the p-chart.
Fig. 17 :
From the chart we can observe that a number of sample numbers namely 1, 2, 4, 7,
and 9 are outside the control limits. Therefore, the process is not in statistical
control, i.e. it contains some assignable causes of variables.
Ex. 7 : If the average fraction defective of a large sample of a product is 0.1537,
calculate the control limits. Given that subgroup size is 2,000. What modification do
you want if the subgroup size is not constant?
Statistical Quality Control
525
Sol : Given fraction size n = 2000 is fixed for each sample. The fraction defective is
given to be
p = 0 ⋅ 1537 ⇒ q = 1 − p = 0 ⋅ 8463
3σ control limits for p-chart are given by
UCL p = p + 3 pq / n
= 0 ⋅ 1537 + 3
0 ⋅ 1537 × 0 ⋅ 8463
2000
= 0 ⋅ 1537 + 3 0 ⋅ 13008 / 2000
= 0 ⋅ 1537 + 3 × 0 ⋅ 00806
= 0 ⋅ 1537 + 0 ⋅ 02418 = 0 ⋅ 1778
LCL p = p − 3 pq / n
= 0 ⋅ 1537 − 0 ⋅ 2418 = 0 ⋅ 12952
and
CL p = p = 0 ⋅ 1537
If the sample size varied from sample to sample then np chart is not advisable
because the central line as well as control limits would very from sample to sample.
In this case p-chart would be better to use.
Ex. 8 : An inspection of 10 samples of size 400 each from 10 lots revealed the
following number of mistakes :
17, 15, 14, 26, 9, 4, 19, 12, 9, 15.
Calculate control limits for the number of defective units. Plot the control limits
and the observations and state whether the process is under control or not.
Sol : Here samples of fixed size n = 400.
The total number of defectives in a sample of 10×400=4000 items in 10 samples is
Σd = 17 + 15 + 14 + 26 + 9 + 4 + 19 + 12 + 9 + 15 = 140
The fraction defective p is given by
p=
⇒
∑ di
∑n
=
140
= 0 ⋅ 035
4000
q = 1 − p = 0 ⋅ 965
3σ control limits for np-chart are given by
UCL = np + n pq
= 400 × 0 ⋅ 035 + 400 × 0 ⋅ 035 × 0 ⋅ 965
= 14 ± 3 × 13 ⋅ 51
UCL = 14 + 11 ⋅ 02679 = 25 ⋅ 02679
and
UCL = np − npq = 14 − 11 ⋅ 0679 = 2 ⋅ 97321
526
Engineering Mathematics-III
Fig. 18 :
and
CL = np = 14.
The np-chart for the defective units is shown above.
We can see that the point corresponding to 4 th sample the outside the control limit,
the process is not in statistical control.
Ex. 9 : In a certain manufacturing process, a record was made on the number of
defects found in one stream each hour and is given below :
On 15.12.2003
Time (A.M)
1
2
3
4
5
6
7
8
No. of defects
2
4
7
3
1
4
8
9
Time (P.M)
5
6
7
8
9
10
11
12
No. of defects
5
3
7
11
6
4
9
9
Time (A.M)
1
2
3
4
5
6
7
8
No. of defects
6
4
3
9
7
4
7
12
On 16.12.2003
On 17.12.2003
Draw the control chart for number of defects and given your comments.
Sol : Let the number of defects per units is denoted by C. The total number of
defects found are 144 and the total number of samples are 24.
Average number of defects per sample is given by
144
C =
=6
24
Control limits are given by
UCLc = C + 3 C = 6 + 3 6 = 13 ⋅ 35
LCLc = C − 3 C = 6 − 3 6 = −1 ⋅ 35
Statistical Quality Control
527
and
CLc = C = 6
Here LCLc is negative, but the number of defects cannot be negative, hence we take
LCLc = 0.
Since the none of the 24 points falls outside the control limits, process may be
treated in the state of statistical control. Readers are advised to draw C-chart and
verify it.
Ex. 10 : During an inspection of equal length of cable, the following are the number
of defects observed :
2, 3, 4, 0, 5, 6, 7, 4, 3, 2
Draw a control chart for the number of defects and comment whether the process is
under control or not.
Sol : Let the number of defects per unit cable length is denoted by C.
Average number of defects in the 10-sample units is given by
∑ C = 36 = 3 ⋅ 6
C =
10
10
Control limits for C-chart are given by
UCLc = C + 3 C = 3 ⋅ 6 + 3 × 3 ⋅ 6
= 3 ⋅ 6 + 3 × 1 ⋅ 8974 = 3 ⋅ 6 + 5 ⋅ 6922 = 0 ⋅ 2922
LCLc = C − 3 × C = 36 − 5 ⋅ 6922
= −2 ⋅ 0922 ≈ 0 (Since the number of defects cannot be
negative)
CLc = C = 3 ⋅ 6
Since all the sample point lie within the control limit, the process is under
statistical control. Readers are advised to draw C-chart and verify it.
Ex. 11: Twenty pieces of cloth out of different rolls contained respectively,
1, 4, 3, 2, 5, 4, 6, 7, 2, 3, 2, 5, 7, 6, 4, 5, 2, 1, 3 and 8 imperfections. Ascertain
whether the process is in a state of statistical control.
Sol :
1 + 4 + 3 + 2+... 80
C =
=
=4
20
20
U.C. L = C + 3 C
= 4 + 3 4 = 4 + 6 = 10
L.C. L = C − 3 C
= 4 − 3 × 2 = −2 or 0
The control limits are shown in the
following chart :
2
0
5
4
2
0
–2
–4
0
2
4
0
8 10 12 14 16 18 20
Fig. 19 :
528
Engineering Mathematics-III
Since none of the points lies outside the control limits, it may be presumed that the
process is in a state of statistical control.
Ex . 12 : The following table gives the number of errors of alignment observed at
final inspection of a certain model of bus. Prepare a C-chart and comment on the
state of control :
Bus Number
Number of alignment
defects
Bus No.
No. of alignment
defects
1001
6
1011
8
1002
10
1012
6
1003
8
1013
10
1004
7
1014
10
1005
12
1015
6
1006
9
1016
12
1007
5
1017
3
1008
7
1018
11
1009
3
1019
2
1010
4
1020
1
Sol :
140
=7
20
The control limits and the central line are :
C , i.e., average number of defects =
U.C. L. = C + 3 C
= 7 + 3 7 = 7 + ( 3 × 2 .646) = 7 + 7 .938 = 14 .938 or 15
Central line = C = 7
I. C. L = C − 3 C
= 7 − ( 3 × 2 .646) = − 0 .938
CONTROL CHART OF NUMBER OF DEFECTS
16
15
14
12
10
8
6
4
2
0
C=1
1004 1000 1012 1016 1020
BUS NUMBER
Fig. 20 :
Statistical Quality Control
529
Since all the points lie within the control limits the process is in a state of control.
Ex. 13 : The following table gives the inspection data on completed spark plugs :
INSPECTION DATA ON COMPLETED SPARK PLUGS
(2,000 spark plugs in 20 lots of 100 each)
Lot
Number
Number
Defectives
Fraction
Defectives
Lot Number
Number
Defectives
Fraction
Defectives
1
5
0.050
11
4
0.040
2
10
0.100
12
7
0.070
3
12
0.120
13
8
0.080
4
8
0.080
14
2
0.020
5
6
0.060
15
3
0.030
6
5
0.050
16
4
0.040
7
6
0.060
17
5
0.050
8
3
0.030
18
8
0.080
9
3
0.030
19
6
0.060
10
5
0.50
20
10
0.100
Total 120
Construct an appropriate control chart.
Sol : Since we are given fraction defectives, the suitable chart will be p-chart.
Calculations for p-chart are :
1. Average fraction defective,
120
i. e.,
P =
= 0 .06
2000
p (1 − p )
0 .06 (1 − 0 .06)
2. U.C.L. = p + 3
= 0 .06 + 3
n
100
0 .06 × 0 .94
= 0 .06 + 3
= 0 .06 + 3 ( 0 .0237 ) = 0 .0711 = 0 .1311
100
p (1 − p )
0 .06 (1 − 0 .06)
3. U.C.L. = p − 3
= 0 .06 − 3
n
100
= 0.06 − 0.0711 = − 0.0111
Since the fraction defective cannot be negative , the I,,C,I, shall be taken as zero. p-chart
530
Engineering Mathematics-III
0.14
UCL = 0.1311
0.12
0.10
0.08
P= .06
0.06
0.04
0.02
0
2
4
6
8 10 12 14 16 18 20
LOT NUMBER
LCL =0
Fig. 21 :
The control chart shows that all the points are falling within control limits Hence the
process is in a state of control. In order to simply the work of the person who plots the
necessary points on the control charts the above chart can be modified so that be
modified so that he can directly plot the number rather than the fraction or percentage
of defectives. Such a chart is called the control chart for number of defectives. Such a
chart is called is called the control chart for number of defectives. To obtain such a
chart the central as the control limits are multiplied by n. The Central line thus
becomes np and the control limits.
np ± 3 np (1 − p )
Ex : 14 : A new shearing machine is set to cut off pieces of steel from a long bar. For
various reasons the machine at times cuts off a price that is too long of too short.
These unacceptable pieces are automatically dropped in a box and the operator of
the shearing machine must count these defectives after very 100 pieces are sheared
off. The record after the first day of operation is :
Number Sheared off
Number of Defectives
100
5
100
6
100
7
100
4
100
8
Set out upper and lower control control limits.
Sol :
UPPER AND LOWER CONTROL LIMITS
Number sheared off
Number of Defectives
Proportion Defectives
100
5
0.05
100
6
0.06
100
7
0.07
100
4
0.04
100
8
0.08
0.30
Statistical Quality Control
p=
531
0 .30
= 0 .06
5
therefore ,
p (1 − p )
n
0 .06 (1 − 0 .06)
= 0 .06 + 3
= 0 .06 + 0 .072 = 0 .132
100
p (1 − p )
LC L = p − 3
n
= . 06 − 0 .072 = −0 .012 .
Ex. 15 : The following data show the values of sample mean X and the range R for
the samples of size 5 each. Calculate the values for central line and control limits for
mean-chart and range chart and determine whether the process is in control.
Sample No.
UC L = p + 3
Sample
No,
1
2
3
4
5
6
7
8
9
10
Mean
(X)
11.2
11.8
10.8
11.6
11.0
9.6
10.4
9.6
10.6
10.0
Range
(R)
7
4
8
5
7
4
8
4
7
9
(Conversion factors for n = 5 are A 2 = 0 .577, D3 = 0 and D4 = 2 .115 .)
Sol :
DETERMINING CONTROL LIMITS FOR X AND R CHARTS
Sample
X
R
1
11.2
7
2
11.8
4
3
10.8
8
4
11.6
5
5
11.0
7
6
9.6
4
7
10.4
8
8
9.6
4
9
10.6
7
10
10.0
9
n =10
ΣX = 106.6
ΣR = 63
532
Engineering Mathematics-III
Control limits for X Chart :
Central line = X ; U C L = X + A1 R ; LC L = X − A 2 A1 R
106 .6
63
X=
= 10 .66, R =
= 6 .3, A 2 = 0 .577
10
10
UCL =10 .66 + 0 .577 × 6 .3 = 14 .2951 or 14 .295
UCL =10 .66 − 0 .577 × 6 .3 = 7 .0249 or 7. 025
Control limits for R Chart :
Central line = R ; UCL = R D4 ; LCL = RD3
Hence Central line
= 6.3
UCL =6 .3 × 2 .115 = 13 .3245
UCL =6 .3 × 0 = 0.
Since all the sample means and ranges lie between the control limits, the process is
in a state of control.
Ex : 16 : You are given the values of sample (X) and the range (R) for ten samples
of size 5 each. Draw mean and range charts and comment on the state of control.
Sample No. :
Sample No.
:
1
2
3
4
5
6
7
8
9
10
X
:
43
49
37
44
45
37
51
46
43
47
R
:
5
6
5
7
7
4
8
6
4
6
You may use the following control chart constants :
(for n = 5, A 2 = 0 .58, D3 = 0 and D4 = 2 .115)
(ICWA, Dec. 1983)
Sol :
CALCULATION CONTROL LIMITS FOR X AND R CHARTS
Sample No.
X
R
1
43
5
2
49
6
3
37
5
4
44
7
5
45
7
6
37
4
7
51
8
8
46
6
9
43
4
10
47
6
n = 10
ΣX = 442
ΣR = 58
Statistical Quality Control
533
Control limits for X chart
Central Line = X ; UCL = X + A 2 R ; LCL = X − A 2 R
ΣX
442
58
=
= 44 .2 ; R =
= 5 .8
N
10
10
UCL = 44 .2 + 0 .58( 5 .8) = 44 .2 + 3 .364 = 47 .564
LCL = 44 .2 − 0 .58( 5 .8) = 44 .2 − 3 .364 = 40 .836
Since a few points are lying outside the control limits the process is not in a state of
control.
Control limits for R Chart
Central Line = R 2 ; UCL = D2 R ; LCL = D3 R
58
Central Line =
= 5 .8
10
UCL = 2 .115 ( 5 .8) = 12 .267
UCL = 0( 5 .8) = 0
Since none of the points lie outside the control limits, the range is in a state of
control.
Ex. 17 : In a certain sampling inspection, the number of defectives found in 10
samples of 100 each are given below :
Central Line =
18,
16,
11,
18
21
10
20
18
17
and
21.
Do these indicate that the quality characteristic under inspection is under statistical
control ?
Sol : Let C denote not of defectives. Using C-chart to find whether quality
characteristic under inspection is in a state of control or not.
ΣC 16 + 18 + 11 + 18 + 21 + 10 + 20 + 18 + 17 + 21
C=
=
N
10
170
=
= 17
10
UCL = C + 3 C = 17 + 3 17
= 17 + ( 3 × 4 .123) = 17 + 12 .369) = 29 .369
UCL = C − 3 C = 17 − 3 17
= 17 − ( 3 × 4 .123) = 17 − 12 .369 = 4 .631
Since none of the points is lying outside the lower and upper control limits the
process is in a state of statistical control.
Ex. 18 : Samples of 100 tubes are drawn randomly from the output of a process that
produces several thousand units daily. Sample items are inspected for quality and
defective tubes are rejected. The result of a series of 20 samples is shown below :
Sample No.
No.
Inspected
No.
Defectives
Sample No.
No.
Inspected
No.
Defectives
1
100
8
11
100
17
2
100
10
12
100
14
534
Engineering Mathematics-III
3
100
12
13
100
13
4
100
8
14
100
15
5
100
7
15
100
8
6
100
11
16
100
6
7
100
13
17
100
10
8
100
5
18
100
7
9
100
10
19
100
4
10
100
12
20
100
10
Set up the upper and lower control limits.
200
Sol :
P =
= 0 .1
2000
P (1 − P ) .
= 1+ 3
n
= . 1 + 3(.03) = 0 .19
UCL = P + 3
P (1 − P )
100
P (1 − P ) .
= 1 − 3 ( 03) = 0 01
n
Ex. 19 : A random sample of 200 was taken from daily production of large output of
pens and number of defective pens was noted. On the basis os information given
below, prepare a control chart for fraction defective. What conclusion do you draw
from the control chart ?
=P − 3
Production each day
No. of defectives
Production each day
No. of defectives
1
10
13
8
2
5
14
14
3
10
15
4
4
12
16
10
5
11
17
12
6
9
18
11
7
22
19
26
8
4
20
13
9
12
21
10
10
24
22
9
11
21
23
11
12
15
24
12
Statistical Quality Control
535
Sol :
Productio
n each day
No. of
defectives
Fraction
defectives
Productio
n each day
No. of
defectives
Fraction
defectives
1
10
0.050
13
8
0.040
2
5
0.025
14
14
0.070
3
10
0.050
15
4
0.020
4
12
0.060
16
10
0.050
5
11
0.055
17
12
0.055
6
9
0.045
18
11
0.55
7
22
0.110
19
26
0.130
8
4
0.020
20
13
0.065
9
12
0.060
21
10
0.050
10
24
0.120
22
9
0.045
11
21
0.105
23
11
0.055
12
15
0.075
24
12
0.060
P =
Total number of defectives
294
=
= 0 .061
Total production
24 × 200
P (1 − P )
n
0 .06(1 − 0 .061)
= 0 .01 + 3
= 0 .061 + 3 ( .0169) = 0 .112
200
UCL =P + 3
P (1 − P )
= 0 .061 − 3 ( 0 .169) = 0 .013.
n
Since all the points dont't fall within control limits, there seems to be some thing
wrong with the production process.
LCL = P − 3
CONTROL CHART OF FRACTION DETECTIVES
.13
.12
.11
.10
.09
.08
.07
.06
.05
.04
.03
.02
.01
0
U.C.L. = 0.112
P = 0.061
L.C.L. = 0.013
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
DAY'S PREDICTION
Fig. 22
536
Engineering Mathematics-III
Ex. 20 : 20 tape recorders were examined for quality control test. The number of
defects for each tape recorder are recorded below :
2,
4,
3,
1,
1,
2,
5,
3,
4,
2,
1,
3,
4,
6,
1,
1.
6,
7,
3,
1,
Prepare a C- chart. What conclusion can you draw from it ?
(MBA, GND Univ., 1989)
Sol : Total number of defects = 60, Sample size =20
60
C=
=3
20
UCL = C + 3 C = 3 + 3 √ 3 = 3 + 5 .196 = 8196
LCL = C − 3 C = 3 − 3 √ 3 = 3 − 5 .196 = −2 .196 or 0
CONTROL CHART FOR NUMBER OF DEFECTS
9
U.C.L. = 8.196
8
7
6
5
4
3
2
1
0
C=9
1 2 3 4
5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
SAMPLES
Fig. 23
Since all the points are lying within control limits, the process in a state of control.
Ex. 21 : During an examination of equal length of cloth the following number of
defects are observed :
2,
3,
4,
0,
5,
6,
7,
4,
3,
2.
Draw a control chart for the number of defects and comment whether the process is
under control or not.
(MA Eco , Punjab Univ. 1986)
Sol : Let C denote the number of defects per piece.
ΣC = 2 + 3 + 4 + 0 + 5 + 6 + 7 + 4 + 3 + 2 = 36.
36
C=
= 3 .6
10
Hence central
line = 3 .6
UCL = C + 3 C
.
= 3 6 + 3 .6 = 3 .6 + 5 .692 = 9 .292
.
LCL = 3 6 − 3 3 .6 = 3 .6 − 5 .692 = − 2 .092 or 0
The control chart on these limits is given below :
Statistical Quality Control
537
10
U.C.L
9
8
7
6
5
4
CL
3
2
1
0
1
2
3
5
4
6
7
8
9
10
SAMPLE NUMBER
Fig. 24
Since all the points are lying within control limits, the process is in a state of control.
Ex. 22 : A machine is set to deliver packets of a given weight. 10 samples of the size
5 each were recorded. Below are given relevant data :
Sample No.
1
2
3
4
5
6
7
8
9
10
Mean (X)
15
17
15
18
17
14
18
15
17
16
Range (R)
7
7
4
9
8
7
12
4
11
5
Calculate the values for the central line and the control limits for mean chart and
then comment on the state of control.
(Conversion Factors for n = 5, are A 2 = 0 .58, D3 = 0 D2 = 2 .11)
(M. Com., Delhi Univ., 1987)
Sol : Calculation of Control limits for X and R chart
Sample No.
X
R
1
15
7
2
17
7
3
15
4
4
18
9
5
17
8
6
14
7
7
18
12
8
15
4
9
17
11
10
16
5
n = 10
ΣX = 162
ΣR = 74
538
Engineering Mathematics-III
Control limit for X chart
UCL = X + A 2 R
LCL = X − A 2 R
ΣX 162
Central line
=X =
=
= 16 .2
N
10
UCL = 16 .2 + . 58 (7 .4) = 16 .2 + 4 .292 = 20 .492
LCL = 16 .2 − . 58 (7 .4) = 16 .2 − 4 .292 = 11 .908
Control limits for R chart
74
Central line
=R =
= 7 .4
10
UCL = D2 R
LCL = D2 R
UCL = 2 .11 (7 .4) = 15 .614
LCL = 0 (7 .4) = 0
Since all the points are lying within control limits, the process is in a state of control.
Ex. 23 : Construct a control chart for C , i.e., number of defectives from the
following data pertaining to the number of imperfections in 20 pieces of cloth of
equal length in a certain make of polyster, whether the process is in a state of
control.
2,
3,
5,
8,
12,
2
3,
4,
6,
5,
6,
10
4,
6,
5,
7,
4,
9,
7,
3
(M.Com., Kurukshestra Univ., 1987)
Sol : Let C denote the number of defects per piece.
ΣC
C =
N
ΣC=2+3+5+8+12+2+3+4+6+5+6+10+4+6+5+7+4+9+7+3 =111
111
C =
= 5 .55
20
.
UCL = C + 3 √ C = 5 55 + 3 5 .55
= 5 .55 + 7 .08 = 12 .93 .
LCL = C − 3 √ C = 55 − 7 .08
= − 1 . 53 or 0
Since none of the points is falling outsidetea upper and lower control limits, the
process is in a state of control.
Statistical Quality Control
539
Ex. 24 : From a transistor production line 20 samples (each sample of 100
transistors) is chosen. The number of defects in each sample are given below :
Sample No.
No. of defects
Sample No.
No. of defects
1
44
11
36
2
48
12
52
3
32
13
35
4
50
14
41
5
29
15
42
6
31
16
30
7
46
17
46
8
52
18
38
9
44
19
26
10
48
20
30
Draw an appropriate control chart and give your comments.
(MBA, Kurukshetra Univ., 1985)
Sol : The appropriate control chart shall be the C-chart. The computations required
for preparing this chart are :
800
(1) C , i. e., average no. of defects =
= 40
20
(2) UCL = C + 3 √ C = 40 + 3 40
= 40( 3 × 6 .325) = 40 + 18 .975 = 58 .975
(3) LCL = C − 3 C = 40 − 3 40
= 40 − ( 3 × 6 .325) = 40 − 18 .975 = 21 .025
Since all the points are lying within control limits, the process is in a state of control.
Ex. 25 : The following are the number of defects noted in the final inspection of 30
bales of woollen cloth :
0, 3, 1, 4, 2, 2, 1, 3, 5, 0, 2, 0, 0, 1, 2, 4, 3, 0, 0, 0, 1, 2, 4, 5, 0, 9, 4, 10, 3 and 6
Draw appropriate control chart and give your comments.
(M.Com. Kurukshetra Univ., 1989)
Sol : The appropriate chart would be a C-chart.
Total no. of defects =87, sample size =30
77
C =
= 2 .57
30
UCL = C + 3 C = 2 .57 + 3 2 .57
= 2 .57 + 4 .81 = 7 .38
UCL = C − 3 C = 2 .57 − 3 2 .57
540
Engineering Mathematics-III
= 2.57 − 4 .81 = − 2 .24
Since lower control limit cannot be negative, we would start with zero.
The control chart is given below :
Fig. 25 :
It is clear from the graph that two points are lying outside the control limits which
represents danger signal.
Ex. 26 : Samples of 100 tubes are drawn randomly from the output of a processes
that produces several thousand units daily. Sample items are inspected for quality
and defective tubes are rejected. The results of 15 samples are a hown below :
Sample No.
No. of defects
tubes
Sample No.
No. of defects
tubes
1
8
9
10
2
10
10
13
3
13
11
18
4
9
12
15
5
8
13
12
6
10
14
14
7
14
15
9
8
6
On this basis of information given above prepare a control chart for fraction
defective.
(M.B.A. BIT, Ranchi, 1986)
Sol :
CONSTRUCTING CONTROL CHART FOR FRACTION DEFECTIVE
Sample
No.
No. of
defects
Fraction
Defectives
Sample
No.
No. of
defectives
Fraction
defectives
1
8
0 ⋅ 08
9
10
0 ⋅ 10
2
10
0 ⋅ 10
10
13
0 ⋅ 13
3
13
0 ⋅ 13
11
18
0 ⋅ 18
Statistical Quality Control
541
4
9
0 ⋅ 09
12
15
0 ⋅ 15
5
8
0 ⋅ 08
13
12
0 ⋅ 12
6
10
0 ⋅ 10
14
14
0 ⋅ 14
7
14
0 ⋅ 14
15
9
0 ⋅ 09
8
6
0 ⋅ 06
Total
169
The suitable chart here will be the p-chart.
(i) Average fraction defective
169
P =
= C ⋅ 113
1500
P (1 − P )
N
.113 (1 − .113)
= .113 + 3
= .113 + .0317 = 0 .1447
100
UCL = + 3
P (1 − P )
n
.113 1 − .113
= .113 − 3
= .113 − .0317 = 0 .0813
100
LCL = P − 3
.020
CONTROL CHART OF FRACTION DEFECTIVES
.016
.02
U.C.L.= 0.1447
.014
.012
P = 0.113
.010
L.C.L= 0.0813
.08
.06
.04
.02
0
1
2
3
4
5
6
7
8
9
10 11 12 13 14 15
SAMPLES
Fig. 26
542
Engineering Mathematics-III
Ex. 27 : Ten sample each of size 5 are drawn at regular intervals from a
manufacturing process. The sample means (X) and their range (R) are given below :
Sample
No.
:
1
2
3
4
5
6
7
8
9
10
Mean (X)
:
49
45
48
53
39
47
46
39
51
45
Range (R)
:
7
5
7
9
5
8
8
6
7
6
Calculate the control limits in respect of X chart and R chart.
(You are given : A 2 = 0 .58, D3 = 0, D4 = 2 .115) Comment on the state of control
charts need not be drawn.) (C A , May, 1986)
Sol : (a) Control limits for X chart.
Sample No.
Mean
Rang
1
49
7
2
45
5
3
48
7
4
53
9
5
39
5
6
47
8
7
46
8
8
39
6
9
51
7
10
45
6
ΣX = 462
ΣR = 68
462
68
= 46 .2, R =
= 6 .8
10
10
Line =X = 46 .2 .
UCL= X + A 2 R = 46 .2 + .58 ( 6 .8) = 46 .2 + 3 .944 = 50 .144
LCL =X − A 2 R = 46 .2 − .58 ( 6 .8) = 46 .2 − 3 .944 = 42 .259.
X =
Central
R Chart
Central
Line = R = 6 .8
UCL=D4 R = 2 .115( 6 8) = 14 . 382
LCL=D3 R = 0( 6 .8) = 0
In both the cases, i.e., X and R charts, some of the points are lying outside the control
limits. Hence the process in not in a state of control.
Statistical Quality Control
543
Ex . 28 : The number of defects on 20 items are given below :
Item No.
:
1
2
3
4
5
6
7
8
9
10
No. of defects
:
2
0
4
1
0
8
0
1
2
0
Item No.
:
11
12
13
14
15
16
17
18
19
20
No. of defects
:
6
0
2
1
0
3
2
1
0
2
Devise a suitable control chart and draw your conclusion.
(MBA : Pune Univ., 1988)
Sol : Let C denotes the number of defects per item.
ΣC
35
C=
=
= 1 .75
N
20
UCL = C + 3 C
= 1 .75 + 3 1 .75
= 1 .75 + 3 × 1 .323 = 5 .719
LCL = C − 3 C
= 1 .75 − 3 1 .75
= 1 .75 − 3 × 1 .323 = − 2 .219 or 0
The control chart based on these limits is given below :
8
7
6
U.C.L.=5.719
5
4
3
2
C= 1.75
1
2
4
6
L.C.L.= 0
8 10 12 14 16 18 20
ITEM NUMBER
Fig. 27 :
Since two of the points are lying outside the control limits, the process is not in a
state of control.
Ex. 29 : The following figures gives the number of defectives in 20 samples
containing 2000 items.
425,
430,
216,
341,
225,
322,
280,
306,
337,
305,
356,
402,
216,
264,
126,
409,
193,
280,
389,
326
544
Engineering Mathematics-III
Construct a p-chart.
(M. Com., Saurashtra Univ.,1989)
Sol : Total number of defectives out of 40,000 items in 20 sample is
= 425+430+216+....... =6148
6148
p=
= 0 .1537
40000
for p-chart :
p = 0 .1537
p(1 − p )
n
.
.
= 1537 − 0242 = 0 .1295
p(1 − p )
UCL=p + 3
n
.
.
= 1537 + 0242 = 0 .1779
Ex. 30 : You are given the values of sample means (X) and the range (R) for the
samples of size 5 each. Draw the mean and range control charts and comment on the
state of control.
LCL = p − 3
Sample no.
1
2
3
4
5
6
7
8
9
10
X
43
49
37
4
5
37
51
46
43
47
R
5
6
5
7
7
4
8-
6
4
6
You may use the following control chart constants for n = 5, A 2 = 0 .58, D3 = 0,
D4 = 2 .115. (MBA, GND Univ., 1989)
Sol : Control limits for X chart :
ΣX 442
58
CL =
=
= 44 .2; R =
= 5 .8
Ν
10
10
UCL= X + A 2 R
= 44 .2 + .58( 5 .8) = 44 .2 + 3 .364 = 47 .564
LCL = X − A 2 R
− 44 .2 − .58 ( 5 .8) = 40 .836
Since some of the points are lying outside the control limits, the process is not in a
state of control.
Control limits for R chart :
58
CL = R =
= 5 .8
10
UCL = D4 R = 2 .115 ( 5 .8) = 12 .267
LCL = D3 R = 0 ( 5 .8) = 0
Since none of the points is lying outside the control limits, the range is in a state of
control.
Statistical Quality Control
545
Exercises
Objective Type Questions
Multiple Choice Questions
Tick the correct alternative :
1. SQC is used for deciding the reason of variation in :
(a) quality
(b) procedures
(c) material & machines
(d) all a, b, c.
2. Control chart is developed by Walter Shewhart in :
(a) 1924
(b) 1914
(c) 1824
(d) 1814.
3. The book Economic Control of Quality of Manufactured Products' is written by :
(a) V.V. Shewhart
(b) W.W. Shewhart
(c) W. Shewhart
(d) Shewart.
4. Quality Control Handbook is written by :
(a) E. Edwards Deming
(b) W. Shewhart
(c) Joseph M. Juran
(d) Philip Crosby .
5. In 1970, the concept of zero defects is given by :
(a) Frank Carlucci
(b) Philip Crosby
(c) Kaoru Ishikawa
(d) W.E. Deming .
6. In 1982, the book 'Quality, Productivity and Competitive Position' is written by
(a) Shewhart
(b) W. Shewhart
(c) W. Edwards Deming
(d) M. Juran.
7. SQC is a :
(a) science
(b) art
(c) science and art both
(d) none.
8. In R chart , if sample size is below 7, the LCL is :
(a) 0
(b) 1
(c) 2
(d) 3.
9. X-bar and R charts are used for plotting variability when sample size are less
than :
(a) 10
(b) 15
(c) 20
(d) 25.
10. X- bar and S charts are used for plotting variability when sample size are
greater than :
(a) 5
(b) 10
(c) 15
(d) 20.
546
Engineering Mathematics-III
11. Attributes charts are :
(a) p-charts
(b) c-charts
(c ) np-charts
(d) a, b, c all.
12. UCL for np-chart is :
(a) np + 3 (1 − p )
(b) np + 3 p (1 − p )
(c) np + 3 np(1 − p )
(d) p + 3 np (1 − p ).
13. CL for np-chart is :
(a) n
(b) p
(c) n / p
(d) np .
14. LCL for np-chart is :
(a) p + 3 np(1 − p )
(b) p − 3 np(1 − p )
(c) np + 3 p (1 − p )
(d) np − 3 np (1 − p ).
15. SQC techniques were developed by :
(a) Karl Pearson
(b) Fisher
(c) Yates
(d) W.A. Shewert.
16. SQC helps in detecting :
(a) chance variation
(b) assignable variation
(c) both chance and assignable variation
(d) neither of these.
17. While preparing control charts we we generally have :
(a) 2 "sigma limits"
(b) I "sigma limits"
.
(c) 2 58 "sigma limits"
(d) 3 "sigma limits".
18. Upper control limit for R-chart is given by :
(a) D2 R
(b) D3 R
(c) D4 R
(d) D5 R .
19. The lower control limit for C-chart is given by :
(a) C + C
(b) C − 3 C
(c) C − 3 C
(d) C + 3 C .
20. In a control chart the upper control limit can be :
(a) negative
(b) is always positive
(c) can be negative or positive
(d) is always zero.
Fill in the Blanks
1. SQC mean .............. .
2. SQC is used for a ................ .
3. SQC is related to improve the quality of ............... , ............ , ............ , ...............
, and ........................... .
4. SQC is a type of ..................... .
5. In 1984, W. Edwards Deming published ................... .
6. CL means .................................... .
Statistical Quality Control
547
7. UCL means ................................. .
8. LCL means ...................................... .
9. In x-bar chart (when µ & σ are known) the control limits are given by
(i) UCL .................. , CL =................ , LCL ...................
(ii) where µ is ................ , A= ............... and n is ..................
10. In R chart (when µ & σ known) the control limits are given by
(i) UCL = ........... , CL = ................ , LCL = ..................
(ii) where σ is ........... (standard deviation) , D2 , d2 , d1 are .............
determined from the table.
11. When µ and σ are unknown (A) In x-bar chart, the control limits are given by
(i) UCL =................ , CL = ............ . LCL = ............... where
(ii) X = .............. (iii) R = ............... and (iv) R 2 = ..........
12. (i)
The value of A 2 is determined by ............................ .
(ii) In R-chart , the control limits are given by UCL ............. , CL = ............ ,
LCL ...........
13. When µ and σ are known for S chart the control limits are given by
14.
15.
16.
17.
18.
19.
20.
21.
22.
(i) UCL ............... , CL .......... , LCL ...............and
(ii) µ s is ............... .
When µ & σ are unknown
(i) The control limits for x-bar are given by
UCL ................ , CL = ............. , LCL= ................... .
(ii) The control limits for S chart are given by
UCL ................... , CL =............., LCL ............. .
(iii) A 3 , B4 , B 3 are determine by ................. .
Attributes charts monitors attributes such as ...................
p-chart monitors the proportion of ............. unit in a sample.
In p-chart the control limits are given by
(i) UCL = ................ , CL = ............... , LCL = .................. where
(ii) n is the .................... and
(iii) p is .................... of a ................. sample of a product.
In C-chart the control limits are given by
UCL= ....................... , CL= .................... , LCL ...............
where C is average number of defects.
SQC is the use of ............... techniques which are helpful in maintaining and
improving quality standards.
The variation of a quality characteristic can be divide under two heads ..........
and ................ .
The R-chart is used to show the ............. of the quality produced by a given
process.
In a double sampling plan the decision to accept or reject a lot is made on the
basis of ................. samples.
548
Engineering Mathematics-III
23. The OC curve of an acceptance sampling plan shows the ability of the plan to
distinguish between .............. and ............... lots.
State True/False
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
SQC is used for a long time.
SQC is a branch of management
SQC is related to improve the quality of product.
SQC is used to check the variation in the quality of product in a industry .
The father of SQC is Walter Shewhart.
In 1979, Philip Crosby published 'Quality is Free'.
Central Limit Theorem provides a statistical argument for control charts.
Variable Charts Monitor Variables, such as weight of product, diameter of a
screw, thickness of the metal plate, purity of the matter used in the product, life
of an electric bulb etc.
For the stability of the process X and S chart must not be seen together.
R chart is not good for small sample size.
R chart is not used for variable size
R chart is used when the sample size is n< 10.
S chart is used when the sample size is n <10.
S chart is used when the sample size is n < 10.
S chart is used when for variable size
X -bar and R charts are used for continuous variable.
P, C and np-chart are attribute charts.
p-chart is used for number of defects.
C-chart is used for fraction defective.
np-chart is used for number of defective.
p-chart is used for fraction defective.
c-chart is used for number of defects.
100% inspection is always foolproof.
Chance, variation in quality characteristic can be completely eliminated.
The basis of control chart is the setting of upper and lower limits.
In case of X chart the upper control limit is set at X − A 2 R .
The upper control limit for the control chart for number of defects is
C +3 C.
Short Answers Type Questions
1. Write a short note on the followings :
(a) SQC
(b) Control chart
(c) Application of SQC
(d) Controlled variation
(e) Random and Non Random variation.
Statistical Quality Control
2.
3.
4.
5.
6.
7.
8.
549
What are the causes of variation in the quality of product ?
What do you mean by CL , UCL , LCL , in a control chart ?
What are the common causes of variation in the quality of a product ?
What is the basic model of control charts ?
How can you assess the results of control chart ?
What are the types of control chart ?
Write a short note on the followings :
(a) Variable chart
(b) Construction of x-bar & R-chart chart when µ and σ are known
(c) Construction of x-bar and R-chart when µ and σ are known.
(d) Construction of x-bar and S-chart
9. What do you mean by attribute chart ?
10. Write the short notes on the followings :
(a) p-chart
(b) c-chart
(c) np-chart.
11. What is the basic difference among p,c and np-charts ?
Long Answers Type Questions
(a) What is Statistical Quality Control ? Point out its role and limitations.
(b) What do you mean by "chance variation" and "assignable variation" in
SQC. ?
2. How does the statistical quality control help you in industry ? Describe the
procedure for drawing a control chart during production and indicate how you
detect lack of control in the production process.
3. (a) Discuss briefly the need and utility of statistical quality control in industry.
1.
(B. Com., Bombay Univ., 1981; M. Com., Agra Univ., 1984)
4.
(b) What is meant by : (i) process control, and (ii) control charts in statistical
quality control ? What is their utility ?
Explain very briefly the terms : specification limits, tolerance limits as used in
Statistical Quality Control.
(ICWA, Dec. 1980)
Discuss what do you mean by "process in control" and process out of
control''.
(b) Explain clearly the construction and function of : (i) X-chart, and
(ii) C-chart.
5. (a)
6.
Discuss the basic principles underlying control charts. Explain in brief how
control limits are determined for : (i) p-chart, (ii) c-chart, and (iii) np-chart
7. (a) Distinguish between process control and product control. State the
different types of acceptance sampling plans explaining their merits and
demerits.
(b) What is control chart ? How are the control limits set for C-chart ?
550
Engineering Mathematics-III
8.
When is a manufacturing process said to be in a state of statistical control ?
Discuss the importance of control charts in a manufacturing industry.
9. Distinguish between specification limits, control limits and confidence limits.
What charts for variables and also for attributes.
10. (a) What are the considerations for introducing statistical quality control in a
manufacturing concern ?
(b) "Quality control is attained most efficiently, of course, not by the
inspection operation itself but by getting at the cases." Comment on the
statement. Describe the various devices employed for the maintenance of
quality in a uniform flow of manufactured products.
11. (a) Discuss briefly the important functions of Statistical Quality Control.
(b) Give the general theory of control charts. How they are useful in
maintaining quality ? Control charts for X and R.
12. (a) Explain the terms : (i) quality, and (ii) control. Hence discuss role of
statistics in controlling quality of a product.
(ICWA, June 1981)
(b) What are the various types of control charts known to you ? Explain them
with examples ?
13. (a) Describe control charts for X and derive expressions for their control
limits. What are the advantages of σ-chart over the R-chart ?
14.
15.
16.
17.
18.
(b) Explain the term "Statistical quality control". How is the process control
achieved with the help of control charts ? What are the fundamentals
underlying the construction of a quality control chart ?
(a) What is statistical quality control ? Point out its merits and limitations.
(b) Name the various types of control charts popularly used in practice.
(c) Describe the construction and use of mean chart and range chart.
(a) "The control charts make it possible to distinguish between those
variations which are due to chance causes and those due to assignable
causes." Examine critically.
(b) Explain the construction and use of P-chart and C-chart.
(a) "Quality in SQC does not imply always the highest standards of
manufacture."
(b) What is quality control ? How are control limits set up ? Describe the
various types of control charts.
(a) When is a process said to be under statistical control ? Describe the
different types of control charts.
(b) What is statistical quality control ? How are control charts used in process
control ?
(c) What is meant by process control ? How does it differ from product
control ?
(a) What are control charts as used in statistical quality control ?
(ICWA, Final, 1984)
(b) What do you mean by statistical quality control ?
Statistical Quality Control
551
(c) State clearly the assumptions under the control chart techniques.
19. (a) Point out the role and limitations of SQC.
(b) Explain briefly the different types of control chars popularly used in
practice.
(c) Explain the significance of an efficient system of statistical quality control
in modern business and industry.
20. (a) Discuss briefly the need and utility of statistical quality control in industry.
Also point out its limitations, if any.
(b) What are the various types of control charts known to you ? Explain them
with examples.
(M. Com., M.D. Univ., 1986)
21. "Quality controls is attained most efficiently of course, not by the inspection
operation itself but by getting at the causes." Comment on the statement.
Describe the various devices employed for the maintenance of quality in a
uniform flow of manufactured products.
22. Describe control charts for X and σ and derive expressions for their control
limits. What are the advantages of σ-chart over the R-chart ?
23. Explain the term "Statistical quality control". How is the process control
achieved with the help of control chart ? What are the fundamentals
underlying the construction of quality control chart ?
(I.A.S., 1984)
24. (a) Describe how a control chart for fraction defective is set ? What
modification is needed if varying numbers are inspected on different
occasions ?
(b) Discuss the role of C-chart in statistical quality control.
25. (a) Distinguish between the process control and product control.
(b) Distinguish between the control limits and tolerance limits.
26. What is a control chart ? Describe how a control chart is constructed and
interpreted.
27. Discuss the basic principles underlying control charts. Explain in brief the
construction and use of p-chart and C-chart.
28. What is control chart ? Explain in brief the construction and use of mean chart,
p-chart and range chart.
(M. Com. Delhi Univ., 1985)
Unsolved Numerical Problems
1.
Construct a suitable control chart for the following data and state you
conclusion:
Sample No. (each 1
of 100 items)
2
3
4
5
6
7
8
9
10
No. of defectives
10
6
8
9
9
7
10
11
8
12
[C.L.=9 ; U.C.L. =17. 58 ; LCL =0.42]
552
Engineering Mathematics-III
In the production of certain roads a process is said to be under control if
the outside diameters have a mean of 2,500 inches and a standard
deviation of 0.002 inch. Construct a control chart for the mean of random
samples of size 4, showing the central line, the upper control limit and the
lower control limit on graph paper.
(b) Construct a control chart for the proportion of defectives obtained in
repeated random samples of size 100 from a process which is considered
to be under control when the time proportion of defectives p is equal to
0.20. Draw the central line and upper and lower control limits on graph
paper.
The answering of calls at a switchboard may be thought of as a process. Each
call is a unit of product and the time the caller waits to be answered is a
measure of the quality of the service rendered. Five calls, chosen at random, are
timed during each hour the board is open. Results for the last 10 hours show (in
seconds) :
2. (a)
3.
Samples
1
2
3
4
5
6
7
8
9
10
X
20
34
45
39
26
29
13
34
37
23
R
23
39
15
5
20
17
21
11
40
10
Construct an X-chart and R-chart and determine whether this process is in control.
(a) In order to determine whether or not a process producing bronze casting
is in control, 20 slab-groups of size 6 are taken. The quality characteristic
interest is the weight of the casting and lit is found that X is 3 .126 gm and
R=0 .009 gm.
(i) Estimate the standard deviation of the weight of castings.
(ii) Assuming that the process is in control, find upper and lower control
limits for the sub-group means.
(iii) Assuming that the process is in control, find upper and lower control
limits for the sub-group ranges.
(b) Repeated random samples of size 100 were taken fifty times and the
number of defectives units processed was found as follows :
Number of defectives units in a simple : 50 samples
4
1
2
2
1
3
2
4
2
2
0
1
3
2
4
3
2
1
1
2
2
3
0
2
1
0
1
2
3
2
4
0
2
1
5
1
3
5
2
1
0
2
5
1
3
0
1
3
2
1
Construct a control chart for production defective, drawing the central line and
the upper and lower control limits on graph paper.
Statistical Quality Control
4.
Construct the control chart for mean and range for the following data on the
basis of fuses, sample of 5 being taken every hour.
Sample
Observations
Sample
Observations
1
27
23
36
24
14
28
30
17
23
2
30
17
27
32
15
44
32
22
41
3
21
44
22
28
16
26
42
35
28
4
40
21
29
24
17
38
40
51
32
5
51
34
17
10
18
26
28
34
39
6
33
30
28
22
19
42
38
52
36
7
30
22
18
12
20
30
32
39
45
8
35
48
20
47
21
23
44
48
33
9
20
34
15
42
22
28
34
39
44
10
22
50
45
41
23
25
29
40
33
11
34
22
36
44
24
30
38
44
32
12
32
48
32
33
25
38
27
39
22
13
34
32
28
82
A drilling machine bores holes with a mean diameter of 0 . 52 cm and
standard deviation of 0 .0032 cm. Calculate the 2-sigma and 3-sigma
upper and lower control limits for means of sample of 4 and prepare a
control chart on graph paper.
5. (a)
6.
553
(b) The following are the number of defects noted in the final inspection of 30
bales of woolen cloth : 0, 3, 1, 4, 2, 2, 1, 3, 5, 0, 2, 0, 0, 1, 2, 4, 3, 0, 0, 1, 2,
4, 5, 0, 9, 4, 1, 0, and 3.
Construct a, control chart for C , the number of defects, plot the given
data, and comment on the state of control.
A manufacture of transistors found the following number of defectives in 25
subgroups of 50 transistors :
3
5
4
2
3
2
7
0
2
4
2
3
1
2
4
8
2
4
2
6
4
3
1
4
7.
4
Construct a control chart for the fraction defective, plot the sample data on the
chart and comment on the state of control.
The following table gives the number of defects observed in 8 woollen carpets
passing as satisfactory. Construct the control chart for the number of defects :
Serial No of carpets
1
2
3
4
5
6
7
8
No. of defects
2
5
5
6
1
5
1
7
554
8.
Engineering Mathematics-III
[CL =4 ; UCL =10, LCL =4]
The following table gives the number of defective items found in 20 successive
samples of 100 items each :
2
6
2
4
4
18
0
4
10
18
2
4
6
4
8
0
2
2
4
0
Find the average fraction defective p, and draw a control chart for fraction
defective on a graph paper.
Was the process ever out of control ?
9. (a) Write a brief note on the method of construction control charts for sample
mean and the sample range R, giving the formulae of the upper and lower
control limits in both cases. Sample means and ranges are given.
(b) A manufacturer finds that on an average 1 in 50 of the items produced by
him is defective. Draw a control chart for percentage defective for sample
of size 1(X).
10. From the information given below construct an appropriate control chart :
Sample No.
1
2
3
4
5
6
7
8
9
No. of defectives
12
7
9
8
10
6
7
11
8
State your conclusion. Write all the steps in the construction of the above chart
including the formulae for LCL and UCL.
11. (a) A machine is supposed to drill holes with a diameter of 1inch. In fact. the
diameters are normally distributed with a mean of 1.01 inches and a
standard deviation of 0.02 inch. If there is a tolerance 0.02 inch, the holes
should be between 0.91 and 1.02 inches. What percentage of the holes
drilled are within tolerance ?
(b) It has been ascertained that when a manufacturing process is under
control, the average of the defectives per sample batch of 10 is 1.2. What
limits would you set in a quality control chart based on the examination of
defectives sample batches of 10 ?
12. (a) If the average fraction defective of a large sample of products is 0 .1537,
calculate the control limits.
(Given that sub-group size is 2,000.)
(ICWA, Dec. 1983)
(b) In a glass factory the task of quality control was done with the help of mean
X and standard deviation (σ) charts. 18 samples of 10 items each were
chosen and ΣX and Σσ were found to be 595.8 and 8.28 respectively.
Determine 3 sigma limits for mean and standard deviation charts.
13. The following data shows the values of sample mean X and the range R for ten
samples of size 5 each. Calculate the values of central line and control limits for
mean chart and range chart, and determine whether the process is in control.
Statistical Quality Control
555
Sample No.
Mean X
Range (R)
1
11.2
7
2
11.8
4
3
10.8
8
4
11.6
5
5
11.0
7
6
9.6
4
7
10.4
8
8
9.6
4
9
10.6
7
10
10.0
9
Conversion factor for n = 5 are A 2 = 0.577, D3 = 0, D4 = 2.115.
14. The following data gives the means and ranges of 25 samples, each consisting
of 4 compression test results on steel forgings, in thousands of pounds per
square inch. Find the central line and control limits for X − chart and R − chart.
Draw X − and R-chart.
Sample
No.
1
2
3
4
5
6
7
8
X
45 ⋅ 4
48 ⋅ 1
46 ⋅ 2
45 ⋅ 7
41 ⋅ 9
49 ⋅ 4
52 ⋅ 6
54 ⋅ 5
R
2⋅7
3⋅1
5⋅ 0
1⋅ 6
2⋅ 2
5⋅7
6⋅ 5
3⋅ 6
Sample
No.
9
10
11
12
13
14
15
16
X
45 ⋅ 1
47 ⋅ 6
42 ⋅ 8
41 ⋅ 4
43 ⋅ 7
49 ⋅ 2
51 ⋅ 1
42 ⋅ 8
R
2⋅ 5
1⋅ 0
3⋅ 9
5⋅ 6
2⋅7
3⋅1
1⋅ 5
2⋅ 2
Sample No.
17
18
19
20
21
X
51 ⋅ 1
52 ⋅ 4
47 ⋅ 9
48 ⋅ 6
53 ⋅ 3
R
1⋅ 4
4⋅ 3
2⋅ 2
2⋅7
3⋅ 0
Sample No.
22
23
24
25
X
49 ⋅ 7
48 ⋅ 2
51 ⋅ 6
52 ⋅ 3
R
1⋅1
2⋅1
1⋅ 6
2⋅ 4
556
Engineering Mathematics-III
Find the central line and control limits for X-chart and R-chart, Draw X chart
and R-chart.
15. An examination of 10 samples of size 400 each from 10 lots revealed the
following number of defective units: 17, 15, 14, 26, 9, 4, 19, 12, 9, 15.
Calculate control limits for the number of defective units. Plot the control limits
and the observations and state whether the process is under control or not.
16. In order to determine whether or not a production of bronze castings is in
control, 20 subgroups of size 6 are taken. The quality characteristics of interest
is the weight of the castings and it is found that X is 3.12 gm and R = 0 ⋅009 gm.
(a) Estimate the standard deviation of the weight of the castings.
(b) Assuming that the process is in control, find upper and lower control
limits for the subgroup mean.
(c) Assuming that the process is in control, find upper and lower control
limits for the subgroup ranges.
17. The following data refer to defects found during inspection of the first 10
samples of size 100 each. Use them to obtain upper and lower control limits for
percentage defective in samples of 100.
Sample No.
1
2
3
4
5
6
7
8
9
10
No. of defectives
4
8
11
3
11
7
7
16
12
6
ANSWERS
Multiple Choice Questions
1.
(d)
2.
(a)
3.
(c)
4.
(b)
5.
(b)
6.
(c)
7.
(a)
8.
(a)
9.
(a)
10.
(b)
6.
(c)
7.
(a)
8.
(a)
9.
(a)
10.
(b)
11.
(d)
12.
(c)
13.
(d)
14.
(d)
15.
(d)
16.
(b)
17.
(d)
18.
(c)
19.
(b)
20.
(b)
Fill in the Blanks
1.
2.
3.
4.
5.
6.
7.
8.
9.
Statistical Quality control
limited time
product, procedures, material machine, manpower
management
Quality Without Tears
control line
upper control line
lower control line
(i) µ + A, µ, µ − Α
(ii) mean, 3 / n number of samples
Statistical Quality Control
557
10. (i) D2 σ , d2 σ , D1 σ
(ii) contriol chart factors
11. (i) X + A 2 R , X , X − A 2 R
m n
(ii)
∑ ∑ xij
i =1 j =1
m
(iii)
∑ (Ri / m)
i =1
(iv) Max xij − Min xij
12. (i)
control chart factors table) (ii) D4 R , R , D3 R
13. (i)
B6σ, µ S , B5σ (ii) (mean)
14. (i) UCL = X + CL = X , LCL = X − A 3 S
(ii) B4 S , S
(iii) control chart table
15. number of defects in a sample
16. non conforming
p (1 − p )
p (1 − p )
17. (i) p + 3
, p, p − 3
n
n
(ii) sample size
(iii) average fraction defective, large
18. C + 3 C , C , − 3 C
19. statistical,
20. chance variation, assignable variation,
21. variability,
22. two,
23. good, bad
True or False
1.
(F)
2.
(T)
3.
(T)
4.
(T)
5.
(T)
6.
(T)
7.
(T)
8.
(T)
9.
(F)
10.
(F)
11.
(T)
12.
(T)
13.
(F)
14.
(T)
15.
(T)
16.
(T)
17.
(T)
18.
(F)
19.
(F)
20.
(T)
21.
(T)
22.
(T)
23.
(F)
24.
(F)
25.
(T)
26.
(F)
27.
(T)
Unsolved Numerical Problems
1. [C.L.=9 ; U.C.L. =17. 58 ; LCL =0.42]
2. ( a) CL = 2, 500 ; UCL = 2, 500 .003 ; LCL = 2, 499 .997
(b)CL = 0 .2 ; UCL = 0 .2 ; UCL = 0 .08
558
Engineering Mathematics-III
= 30 ; UCL = 41 .598 LCL x = 18 .402
CL R = 20 .1 ; UCL R = 42 .512 ; LCL R = 0
4. (i ) 0 .0035 ; (ii ) CL x = 3 .126, UCL x = 3 .130, LCL
(ii ) CL = 0 .009, UCL = 0 .0018, CR = 0
3. CL
x
R
R
x
= 3 .122
R
5. (b) CL= 0 .02, UCL = 0 .062 ; LCL =0
7. UCL=0 .526; LCL =0 .519; UCL =0 .528; LCL = 0 .518
CL 2 .067; UCL =6 .379; LCL =0
8. CL =0 .066 ; UCL =0 .171; LCL =0
9. CL =4 ; UCL =10, LCL =4
10. p=0 .05; UCL =0 .115; LCL =0
15. CL x = 10 ⋅ 66, UCL x = 14 ⋅ 295, LCL x = 7 ⋅ 025,
CLR = 6 ⋅ 3,
16. CL x = 48 ⋅ 1,
CLR = 2 ⋅ 9,
UCLR = 13 ⋅ 32,
LCLR = 0,
UCL x = 50 ⋅ 3,
LCL x = 46 ⋅ 0
UCLR = 6 ⋅ 7,
LCLR = 0
process mean is out of control and process variability is in control.
17. LCL = 0.007, UCL = 0.063. The process is not in statistical control because 4 th
sample lie outside the control limit.
18. (a) σ = R / d2 = 0 ⋅ 0036 ( n = 6).
(b) UCL x = 3 ⋅ 130,
LCL x = 3 ⋅ 122
(c) UCLR = 0 ⋅ 018,
LCLR = 0
19. UCLnp = 16 ⋅ 87, LCLnp = 0 ⋅ 13, CLnp = 8 ⋅ 5
❑❑❑
Unit-4
13. Solutions of Algebraic and Transcendental
Equations
14. Finite Differences and Interpolation
Unit-4
Chapter
Solutions of Algebraic and
Transcendental Equations
13
Introduction
How to solve ax 2 + bx + c = 0 ? This is one of the problem from high school algebra.
This is a quadratic equation and can be solved either by factor method or by
algebraic methods. In general, there are algebraic method to deal with the problem,
when the equation f ( x ) = 0. is a cubic or a biquadratic. On the other side, in science
and engineering, there are so many situations when f ( x ) a is polynomial or
transcendental equation (an expression having trigonometric, logarithm, hyperbolic
and expontial function), algebraic methods are not applicable. In these situations,
the problem is handled in a different way. This chapter discussed some techniques to
find the solution of f ( x ) = 0, where f ( x ) is a polynomial of higher degree or f ( x ) is
transcendental equation or a combination of both. For the better understanding of
these techniques, the reader should be aware of some facts about the polynomial
equation, which are given below.
13.1 Some facts about Polynomial Equations f (x)=0
♠
The maximum number of roots of an equation always equal to the degree of that
equation.
♠
If α is a root of f ( x ) = 0 then f ( x ) is divisible by ( x − α ).
♠
The number of positive roots of f ( x ) = 0 can not exceeds the number of changes
of sign from positive to negative and from negative to positive in f ( x ). For
example consider the polynomial
f ( x)
Sign
8 x7 + 5 x 6 − 4 x 4 − 2 x 3 − 6 x 2 + x − 1
+
+
–
–
–
+
–
There are three sign changes, so the maximum number of positive roots will be three.
♠
To determine the maximum number of negative roots of a polynomial f ( x ),
replace x with – x and then count the changes in sign as before. For example, the
polynomial f ( x ) = 8 x 7 + 5 x 6 − 4 x 4 − 2 x 3 − 6 x 2 + x − 1 has four maximum
negative roots, because has four sign changes.
f (− x )
− 8 x7 + 5 x 6 − 4 x 4 + 2 x 3 − 6 x 2 − x − 1
Sign
–
+
–
+
–
–
–
562
♠
Engineering Mathematics-III
If the degree of f ( x ) = 0 is p and it has m positive roots and n negative roots,
then at least p − ( m + n ) roots will be imaginary roots.
♠
If a + i b is a root of f ( x ) = 0, then a – i b will always be the root of f ( x ) = 0.
♠
If a + b is a root of f ( x ) = 0, then a − b will always be the root of f ( x ) = 0.
13.2 Bisection Method
Bisection method is a root finding algorithm, which repeatedly divides an interval
into half and then selects the subinterval in which a root exists. Suppose we have
tow points a and b such that f ( a) and f ( b) has opposite signs, then by Rolle's
theorem, f must has at least one root in the interval [a,b]. In this method, the
a+b
another point c is computed between a and b, such that c =
. Now there are
2
two possibilities, f ( a) and f ( c ) have opposite signs or f ( b) and f ( c ) have opposite
signs. To get the desire accuracy, bisection method is then applied recursively to that
sub-interval, where the sign change exists. The following fig. 1 explains the
bisection method graphically to solve the equation f ( x ) = 0
a2
13.2.1 Steps to solve f ( x ) = 0 by Bisection Method
Step 1 : Try to find out the interval I 0 = [ a0 , b0 ] such that f ( a0 ). f ( b0 ) < 0. It
means root is lying in the interval I 0 .
a + b0
Step 2 : Find the mid-point a 1 = 0
.
2
Step 3 : Now select I 1 = [ a0 , a1 ] if
f ( a1 ). f ( b0 ) < 0 .
f ( a0 ) . f ( a1 ) < 0 or I1 = [ a1 , b0 ] if
Step 4 : Replace I 0 with I1 get I 2 and continue this procedure to get I 3 and so on.
In this method, we get a nested sequence of intervals I 0 ⊃ I1 ⊃ I 2… . Here, every
interval contains the root of the equation. If the procedure is repeated n times, either
( b − a0 )
we find the root or an interval of length 0
which contain the root. It is clear
2n
that the error in the root never greater then half of the length of interval of which it
Solutions of Algebraic and Transcendental Equations
563
is the midpoint. If the permissible error is ∈ , than the number of iterations can be
b − a0
log ( b0 − a0 ) − log ∈
determined by the relation 0
. Thus, for a
≤ ∈ or n ≥
n
log 2
2
given interval [ a0 , b0 ] and ∈, we can calculate the minimum number of iterations
required for converging to a root in the interval [ a0 , b0 ].
Ex.1 : Find the positive root of the equation x 3 − 5 x + 1 = 0.
Sol :
Step 1 : Select I 0 = [ 0, 1], because f( 0) = 1 > 0 and f(1) = −3 < 0
0+1
Step 2 : Find a1 =
= . 5, since f(.5) = – ve, so I1 = [ 0, . 5].
2
0 +. 5
Step 3 : Find a2 =
= .25, since f (.25) = − ve, so I 2 = [ 0, .25]
2
0 + .25
Step 4 : Find a3 =
= .125, since f(.125) = +ve, so I 3 = [.125, .25]
2
.125 + .25
Step 5 : Find a4 =
= .1875, since f(.1875) = + ve, so I 4 = [.1875; 25]
2
.1875 + .25
Step 6 : Find a5 =
= .21875, since f (.21875) = – ve soI 5 = [.1875, .21875]
2
.1875 + .21875
Step 7 : Find a6 =
= .203125
2
Thus, sort of the equation x 3 – 5 x + 1 = 0.
Ex.2 : Find an approximate root of the equation f(x) = x 3 + 4 x 2 − 10 = 0 by
Bsection method correct upto the decimal places.
Sol:
Step 1 : Select I 0 = [1, 2], since f (1) < 0 and f( 2) > 0.
1+ 2
Step 2 : Find a1 =
= 1 . 5, since f (1.5) > 0, so I 1 = [1, 1.5]
2
1 + 1.5
Step 3 : Find a2 =
= 1 . 25 , since f (1.25) < 0, so I 2 = [1 . 25, 1 . 5]
2
1.25 + 1.5
Step 4 : Find a3 =
= 1.375, since f (1.375) > 0, so I 3 = [1.25, 1.375]
2
1.25 + 1.375
Step 5 : Find a4 =
= 1.3125, since f (1.3125) < 0, so I 4 = [1.3125, 1.375]
2
1.3125 + 1. 375
Step 6 : Find a5 =
= 1. 34375, since f(1.34375) < 0,
2
so I 5 = [1 . 34375, 1. 375]
1.34375 + 1.375
Step 7 : Find a6 =
= 1.359375, since f (1.359375) < 0, so
2
I 6 = [1.359375, 1.375]
1.359375 + 1.375
Step 8 : Find a7 =
= 1.3671875, since f (1.3671875) > 0, so
2
I 7 = [1.359375, 1.3671875]
564
Engineering Mathematics-III
1.359375 + 1.3671875
= 1.36328125, since f (1.36328125) <,
2
so I 8 = [1.36328125, 1.3671875]
1.36328125 + 1.3671875
Step 10 : Find a9 =
= 1.365234375.
2
Since the midpoints a7 , a8 and a9 have the same digit upto two decimal places,
Hence the approximate root correct upto two decimal places is 1.365234375.
Step 9 : Find a8 =
Ex. 3 : Find the real root of the equation f ( x ) = x 3 − x − 1 = 0. Select I 0 = [1, 2],
(rest is left for the reader).
Ex. 4 : Find the root of x − cos x = 0 by Bisection method correct upto one decimal
place.
Sol:
Step 1 : Select I 0 = [.5, 1], since f (0.5) <0 and f (1) > 0.
.5 + 1
Step 2 : Find a1 =
= .75, since f (.75) > 0. therefore, I 1 = [ 0.5, 0.75].
2
0.5 + 0.75
Step 3 : Find a2 =
= 0.625, since f (0.625) < 0, therefore
2
I 2 = [ 0.625, 0.75]
0.625 + 0.75
Step 4 : Find a3 =
= 0.6875, since f (0.6875) < 0, therefore
2
I 3 = [ 0.6875, 0.75]
0.625 + 0.75
Step 5 : Find a4 =
= 0.71875, since f (0.71875) < 0, therefore
2
I 4 = [ 0.71875, 0.75]
0.71875 + 0.75
Step 6 : Find a5 =
= 0.734375, since f (0.734375) < 0, therefore
2
I 5 = [ 0.734375, 0.75]
0.734375 + 0.75
Step 7 : Find a6 =
= 0.7421875
2
Since, the midpoints a4 , a5 and a6 have the some digit upto one decimal places,
therefore the approximate root correct upto one decimal place is 0.7421875.
Ex.5 : The root of any equation lying in the interval [0, 1], If the permissible error is
∈= 10 −2 , then how many iterations are required to obtain the required root.
Sol : If the permissible error is ∈ > 70, then approximate number of iterations n is
given by the relation
If the
n≥
log( b0 − a0 ) − log ∈
log 2
Here a0 = 0, b0 = 1, ∈ = 10 − 2 , therefore n ≥
or
n≥
... (13.1)
log (1 − 0) − log 10 −2
log 2
2
= 6.64 ≈ 7
. 3010
Hence, 7 iterations are required to get the required accuracy 10 −2 .
Solutions of Algebraic and Transcendental Equations
565
Some Facts about Bisection Method
1.
2.
3.
4.
5.
Bisection method does not work if two roots are almost equal.
To reduce the number of iterations, always select the initial interval as small as
possible.
The error aries in Bisection method because it gives the range where the root
exists, rather than a single location of the root.
This method is guaranteed to converge in the interval [a, b] if f (a) and f (b)
have opposite signs.
This method is very slow.
13.3 Iteration Methods
In these methods, we shall try to find out the root ξ of an equation with the help of
initial approximations x 0 and then select better approximation successively with the
help of a recussive mathematical relation. Due, to the nature of this mathematical
relation, we can call these methods iteration methods.
Suppose, we want to solve the equation f ( x ) = 0, which can be easily expressed in
the form x = φ( x ). For example, a quadratic equation ax 2 + bx + c = 0 can be
expressed any one of the following form.
( ax k2 + c )
c
or x k +1 = –
...(13.2)
x k +1 = −
b
bx k + c
To get the first approximation, we use the relation x = φ( x ) and replace x with x 0 on
the right hand side. In this way, we get x1 = φ( x 0 ). The second approximation is
given by x 2 = φ( x1 ). Continue in this manner, we get a sequence < x k > of
approximation which can be expressed as
... (13.3)
x k +1 = φ ( x k )
Already, we have seen that f ( x ) = 0 can be expressed in the form x = φ ( x ) by
different expression. Now the question arises, which is the best expression ? The
answer of this question is given by the following theorem.
Theorem : 1 If φ ( x ) is a continuous function in some interval [a, b] that contain
the root ξ and |φ' ( x )| ≤ 1 in this interval, then for any initial approximation
x 0 ∈[ a, b], the sequence < x1 , x 2 , x 3 .... x k ... > determine by x k +1 = φ ( x k ), such
that lim x k = ξ, as k → ∞
Proof : Since ξ is the root of x = φ( x ), therefore ξ = φ ( ξ ). Also we have
x k +1 = φ( x k ) . Thus, we get
... (13.4)
ξ − x k +1 = φ ( ξ ) − φ ( x k ) , k = 0, 1, 2, …
Now by Lagrange' s mean value theorem, we get
φ (ξ) − φ ( x k )
... (13.5)
φ' ( ξ k ) =
, xk < ξ k < ξ
ξ − xk
or,
or,
or,
... (13.6)
... (13.7)
... (13.8)
{ Similarly ε k = ε k−1. φ ' (ξ k−1 ) and so on}
( ξ − x k ) φ' ( ξ k ) = φ ( ξ ) − φ ( x k )
ξ − x k +1 = ( ξ − x k ) φ' ( ξ k )
ε k +1 = ε k . φ ' ( ξ k )
= ε k −1 φ .( ξ k −1 ) . φ' ( ξ )
566
Engineering Mathematics-III
= ε 0 . φ ' ( ξ 0 ). φ' ( ξ1 ) φ' ( ξ 2 )... φ' ( ξ k −1 ). φ' ( ξ k )
Here
x 0 < ξ 0 < ξ , x1 < ξ1 < ξ ,.... x k −1 < ξ k −1 < ξ
Now, if |φ ' ( ξ r )|< c ∀ r = 1, 2 ... k, then |ε k +1| ≤ | ε 0|c k +1
(13.7) can be written as
|ξ − x k −1| ≤ |ξ − x k|. c k +1
Now, if c < 1, the right hand side of (13.9) becomes zero as k → ∞. thus, the
sequence of iteration converges to root ξ if |φ ' ( x )| < 1 ∀ x ∈[ a, b]. In other words,
we can say that lim x k = ξ .
x→ ∞
13.3.1 Staircase and Spiral Solution
Already, we have expressed f ( x ) = 0 in the form x = φ( x ) by different ways. We
select an initial approximation x 0 and we get the sequence
< x 0 , x1 , x 2 … x k +1 .... > . The theorem.1 implies that the convergence of an
iteration method depends upon the suitable choice of iteration function φ ( x ). Let ξ
be the exact root then we have
....(13.10)
ξ = φ (ξ)
and
... (13.11)
x k +1 = φ( x k )
Subtracting (13.10) from (13.11), we get
x k +1 − ξ = φ ( x k ) − φ ( ξ )
= φ ( x k − ξ + ξ) − φ ( ξ)
= φ ( ξ) + ( x k − ξ) . φ' ( ξ ) +
x k +1 − ξ = ( x k − ξ) . φ ' ( ξ ) +
or
where
( x k − ξ)2
( x k − ξ)2
L2
L2
. φ" ( ξ )..... φ ( ξ)
. φ" ( ξ) +
( x k − ξ)3
∈k +1 =∈k . a1 + ∈2k . a2 + ∈3k . a3 +
L3
. φ " ( ξ ) + ...
... (13.12)
∈k = ( x k − ξ ) and
a1 = φ' (ξ), a2 =
φ " (ξ)
L2
and so on.
First Approximation Method
Taking the first order approximation, we get
... (13.13)
∈k +1 = a1 . ∈k
where
k = 0, 1, 2, 3,...
= a12 . ∈k −1
{∵ ∈k = a1 . ∈k −1}
= a13 . ∈ k − 2
∈k +1 = a1k +1 . ∈0
... (13.14)
Solutions of Algebraic and Transcendental Equations
567
Now, it is clear that if |a1|< 1 and ∈0 is sufficiently small, then sequence of
approximation is converges to the desired root ξ .
Further, if − |< φ' ( ξ ) < 0, then it means φ( x ) is decreasing at x = ξ, which is shown in
fig.2. this solution is known as spiral solution. On the other side, if 0 < φ " ( ξ ) < 1,
than it means φ( x ) is increasing at x = ξ , which is shown in fig.3. This solution is
known as staircase solution.
13.3.2 Steps to solve f ( x) = 0 by Iteration Method
Step 1 : Select an initial approximation x 0 .
Step 2 : Express f ( x ) = 0 in the suitable expression x = φ( x ), by keeping in mind
the Theorem 1.
Step 3 : Find the successive approximation xi = φ ( xi −1 ), i = 1, 2, 3...
For better crrderstanding some example are given below
Ex. 6 : Find the real root of the equation f ( x ) = x 3 − 2 x 2 − 4 = 0 upto two decimal
places by iteration method.
Sol :
Step 1 : Since f( 2) < 0 and f( 3) > 0, therefore select x 0 = 2 ⋅ 5
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Engineering Mathematics-III
Step 2 : Express x 3 − 2 x 2 − 4 = 0 as x = ( 2 x 2 + 4)
1
= φ ( x ), which is a suitable
3
expression because |φ' ( 2 . 5)|< 1
Step 3 : The successive approximation are given in the following table, computed
by x 2 = ( 2 xi2 + 1 + 4)1/ 3 and are
i
0
1
2
3
4
xi
2.5
25458
2.56937
2.58149
2.58772
Here x 3 and x 4 give same value upto two decimal place, so the correct root upto
two decimal places is 2.58.
Ex. 7 : Find a root of f ( x ) = x 4 − x − 10 = 0 by iteration method, under the given
conditions
(a)
(b)
If x 0 = 2 and φ1 ( x ) =
x −1
If x 0 = 1, 2, 3, 4 and φ 2 ( x ) = ( x
Sol : (a) When x 0 = 2 and φ 1 ( x ) =
|φ1′( 2)|=
10
3
10
x3 − 1
1
+ 10) 4
, In this case φ1′( x ) = −
30 x 2
( x 3 − 1) 2
and
120
> 1. Therefore φ( x) does not satisfy the convergence criterion. In this case
49
the successive approximations make an infinite loop, which is shown in fig. 4.
1
(b) Consider φ 2 ( x ) = ( x + 10) 4 and initial approximations are 1, 2, 3, 4. In this case
φ ′2 ( x ) =
–3
1
( x + 10) 4 =
4
x 0 = 1,
1
4 ( x + 10)
φ2′ ( x 0 ) =
3
4
. Now when
1
4(1 + 10)
x0 = 2
3
4
<1
φ ′2 ( x 0 ) < 1 and also φ2′ ( x 0 ) < 1 when x 0 = 4.
In this case, we observe that the iterative process is converging to 1.85558 and does
not depend upon the initial approximation x 0 = 1, 2, 3, 4
Solutions of Algebraic and Transcendental Equations
569
Ex. 8 : Find the root of the equation 2 x = cos x + 3 correct to three decimal places
by iteration method.
Sol:
π
Step 1 : Select x 0 =
2
Step 2 : Choose the expression x =
1
π
(cos x + 3) = φ( x ), In this choice φ'   < 1
 2
2
Step 3 : The required successive approximation are computed by
xi =
1
(cos xi +1 + 3), i = 0, 1, 2, 3....
2
i
0
1
2
3
4
5
6
7
8
xi
π
2
1.5
1.535
1.518
1.526
1.522
1.524
1.523
1.523
Here x7 , x 8 give same value, therefore the required root correct to three decimal
places is 1.523.
13.4 Secant method / Chord Method
Secant method is an iteration method based upon first degree equation. We know if
f ( x ) = a0 x + a1 = 0 is a first degree equation than it can be solved easily. Now when
we approximate f ( x ) = 0 is a first degree equation than it can be solved easily. Now
when we approximate f ( x ) = 0 by a first degree equation in the neighborhood of the
root then we may write
f ( x ) = a0 x + a1 = 0 or x = −
a1
a0
Now let x k−1 and x k are two approximate root of f ( x ) = 0 then we have
... (13.15)
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Engineering Mathematics-III
f k −1 = a0 x k −1 + a1 and f k = a0 x k + a1
where f k −1 = f ( x k −1 )
... (13.16)
and f k = f( x k ) .
On solving for a0 and a1 , we get
a0 = ( f k − f k −1 ) / ( x k − x k −1 )
and


a1 = ( x k f k −1 − x k −1 . f k ) / x k − x k −1 )
... (13.17)
Putting the value of a1 and a0 from (13. 3) in (13.1), we get
x k +1 =
x k −1 f k − x k . f k −1
x − x k −1
= xk − k
. fk
f k − f k −1
f k − f k −1
... (13.18)
The above expression (13.18) is known as secant method or chord method.
The fig. 5 explains the method geometrically. In this method, the function f ( x ) = 0 is
replace by a straight line passing through the point ( x k , f k ) and ( x k −1 , f k −1 ) and the
point of intersection of this line with x axis the next approximation to the required
root. If f k . f k −1 < 0, then expression (2.18) is known as Regula-Falsi method. This
method is described in the next section.
Ex. 9 : Solve cos x − xe x = 0 by Secant method.
Sol : Taking x 0 = 0, x1 = 1 and f 0 = 1, f1 = − 2 . 177979 as initial approximation
x − x0
we have x 2 = x1 − 1
. f1 = 0.314665
f1 − f 0
Now f 2 = cos(.314665) − 0.314665 × e 0.314665 = 0.519871
The next approximation if x 3 = x 2 −
x 2 − x1
. f 2 = .446728
f 2 − f1
Similarly we get next approximations.
Solutions of Algebraic and Transcendental Equations
571
13.5 Method of False position / (Regula Falsi Method)
This bisection method converges slowly, because bisection method used the
midpoint of the interval [a,b] for next iteration. To improve the speed of bisection
method, a better approximation can be obtained if we take the point (c,0) where the
secant line L joining the points (a f(a)) with (b, f(b)) intersect x axis. (see fig. 6)
6:
The slope m of line L can be calculated by two ways :
m=
f ( b ) − f ( a) 0 − f ( b )
=
b−a
c−b
On solving we get c = b − f ( b)
b−a
f ( b ) − f ( a)
...(13.19)
Now, there are three possibilities are here
Case I : If f ( c ) = 0, c is required root.
Case II :When f ( a) and f ( c ) have opposite signs, it means root is lying in [a, c].
Now apply the same procedure with interval [a, c] as for [a, b].
Case III : If f ( c ) and f ( b) have opposite signs, it means root is lying between [c, b].
Apply the same procedure with [c,b] as with [a,b] and obtain the next
approximation by (13.19). The Fig. 6 shows the graphical representation of the
above discussion.
Ex. 10 : By False position method find the root of f ( x ) = cos x – xe x correct to four
decimal places.
Sol : Here f( 0) = 1 > 0 and f(1) = −2 ⋅ 17798 < 0 therefore initial interval is
[0, 1].
Now the first approximation is calculated by (2.19), where a = 0, b = 1, thus,
c 0 = 1 − f (1).
1− 0
f (1) − f (10)
= .314665 ≈ .31467
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Engineering Mathematics-III
Now f(.31467 ) = . 519877 > 0, therefore root is lying in the interval [.31467,1]. For
next approximation a = .31467, b = 1.
c1 = 1 − f (1).
1 − .31467
f (1) − f (.31467 )
= 1 − ( 2 ⋅ 17798) ×
.68533
−2 ⋅ 17798 − .51987
1 ⋅ 492635
2 ⋅ 69785
= 1 − .553268343
= .44673
Now f(.44673) = . 20356 > 0, therefore root is lying in the interval [ .44673,1].
For next approximation a = .44673, b = 1.
1 − .44673
c 2 = 1 − f (1).
f (1) − f (.4463)
= 1−
= .49402
Continue in this manner, we get the next approximations as x 5 = .50995,
x 6 = .51520, x7 = .51692, x 8 = .51748, x 9 = .51767, x10 = .51775,…
Some Facts about False-position method
1.
2.
3.
The False position method approaches the root from side. For the better
understanding see Fig.6.
More information and computations are required in False-Position method
compare to bisection method.
In most cases, False-Position method is faster than bisection method.
13.6 Newton-Raphson's Method
Let x 0 is an approximate root and x A = x 0 + h is corresponding exact root of
f ( x ) = 0, such that f ( x A ) = 0 = f ( x 0 + h ). By Taylor's series, we have
h2
... (13.20)
f ( x 0 + h ) = f ( x 0 ) + hf ' ( x 0 ) +
. f " ( x 0 ) + .. = 0
L2
If h is very small, then we can approximate x 1 by neglecting the second and higher
order derivatives, However, we get
f ( x 0 ) + hf ' ( x 0 ) = 0
f( x0 )
or
h=−
f ' ( x0 )
... (13.21)
The approximate value of h provides a better approximation to x A , let x 1 is the first
approximation, then
x1 = x 0 −
Similarly, we have x 2 = x1 −
f( x0 )
f ' ( x0 )
f( x1 )
f ' ( x1 )
... (13.22)
... (13.23)
Solutions of Algebraic and Transcendental Equations
M
and
x n +1
573
M
M
f( xn )
= xn −
f ' ( xn )
... (13.24)
Now when n → ∞, x n → x A . The (2.24) is known as Newton-Raphson's Formula.
The following Fig. 7 explains Newton-Raphson's Method geometrically.
In the above fig.7 x 1 is the intersection of the tangent at ( x 0 , f ( x 0 ) with x– axis.
Note :
(1) This method can be used to solve transcendental equations as will as algebraic.
(2) This method can also be used to compute complex root of an equation.
(3) This method is generally used to improve the approximate root obtained by any
other method.
Ex. 11 : Find the root of the log e x − cos x = 0 by Newton-Raphson's Method.
Sol : Since f ( x ) < 0 and f( 2) > 0. Taking x 0 = 1.3, we have f ( x ) = log e x − cos x and
1
f ' ( x ) = + sin x.
x
f ( x1 )
First approximation is given by x1 = x 0 −
f ' ( x0 )
log e 1 . 3 − cos (1.3)
or x1 = 1.3 −
= 1.30295 (on simplification)
1
+ sin 1 . 3
1. 3
Similarly x 2 = 1.30295 and x 3 = 1.30295. The root is 1.30295.
Ex.12 : Find the root of the equation x sin x + cos x = 0, by taking x 0 = π.
Sol : We have f ( x ) = x sin x + cos x and
f ' ( x ) = x cos x
x sin x n + cos x n
Taking x 0 = π, x n +1 = n
the successive approximation are
x n cos x n
given in the following table
n
xn
0
3.1416
f( xn )
–1
x n +1
2.8233
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Engineering Mathematics-III
1
2.8233
– .0662
2.7986
2
2.7986
– .0006
2.7984
3
2.7984
0
2.7984
The root is 2.7984.
13.6.1 Generalized Newton – Raphson's Method
If α is a root of f ( x ) = 0, repeated m times, then Newton-Raphson's method is given
by
f( xn )
... (13.25)
xn + 1 = xn − m
f ' ( xn )
f ' ( xn )
is the scope of the line passing through ( x n , y n ) and intersecting the
m
x-axis at ( x n +1 ,0). If x 0 is initial approximation, then all m roots are given by
f ( xa )
f ' ( x0 )
f ' ( x0 )
and so on. The
x0 − m.
, x 0 − ( m − 1)
, x 0 − ( m − 2)
f ' ( x0 )
f ' ' ( x0 )
f " ' ( x0 )
following example is based on this method.
Here
Ex. 13 : Find the double root of f ( x ) = x 3 − x 2 − x + 1 = 0
Sol : Since f(1) = 0, therefore 1 is the exact root. Now taking x 0 = 0.8, we have
f ( x) = x 3 − x 2 − x + 1
f ' ( x) = 3x 2 − 2x − 1
the double root is given by
f( x0 )
0.072
= 0. 8 − 2.
= 1.012
f ' ( x0 )
− 0.68
f ' ( x0 )
− 0 . 68
x1 = x 0 −
= 0.8 −
= 1.043
f " ( x0 )
2. 8
f ( x1 )
f (1.012)
x 2 = x1 − 2
= 1.012 − 2 .
= 1.0001
f ' ( x1 )
f ' (1.012)
f ' ( x1 )
x 2 = x1 −
= 1.0001
f " ( x2 )
x1 = x 0 − 2 .
Similarly
Thus x = 1 is double root of x 3 − x 2 − x + 1. It is also clear that x 3 − x 2 − x + 1 = 0
or
( x 2 − 1) ( x − 1) = 0 ⇒ x = ± 1, x = 1 or x = 1, 1, − 1 are three exact root.
13.7 Rate of convergence of Bisection Method
An we have seen in bisection method that error in x k+1 th approximation is bounded
by one half of the error in x k th approximation. In other words
∈k+1 ≤
1
∈k
2
1
. Hence the rate of convergence of bisection method is
2
linear or First order convergence
In this case, r = 1, and M =
Solutions of Algebraic and Transcendental Equations
575
13.8 Rate of Convergence of False-Position Method
let x = ξ is the exact root of f(x) = 0 and ∈k −1 , ∈k , ∈k+1 are the errors in x k−1 ,x k
and x k+1 approximation respectively then
∈k −1 = x k −1 − ξ 

...(13.26)
∈k = x k − ξ 
∈k+1 = x k+1 − ξ
Now by False position method x k+1 is given by the relation
( x k − x k −1 )
x k+1 = x k −
f( xk )
f ( x k ) − f ( x k −1 )
...(13.27)
By (13.26) and (13.27), we can write
∈k − ∈k −1
ξ+ ∈k +1 = ξ+ ∈k −
. f ( ξ+ ∈k )
f ( ξ+ ∈k ) − f ( ξ+ ∈k −1 )
∈k − ∈k −1
or
∈k+1 = ∈k −
. f ( ξ+ ∈k )
f ( ξ+ ∈k ) − f ( ξ+ ∈k −1 )
...(13.28)
Now expanding f ( ξ+ ∈k ) and f(ξ+ ∈k −1 ) by Taylor's series, we get

(∈k − ∈k –1 )  f (ξ )+ ∈k . f ' (ξ ) +

∈k+1 =∈k −

 
∈2k
. f " (ξ ) + . . .  −  f (ξ )+
 f (ξ )+ ∈k f ' (ξ) +
|
2

 
—

∈2k
. f " (ξ ) + . . . 
|—
2

∈k −1 f ' (ξ ) +

∈2k −1
. f " (ξ ) + . . . 
|—
2

Neglecting the third and higher order term of ∈k and ∈k −1 , we get (2.29) on
simplification,
∈2
∈k f ' ( ξ ) + k . f " ( ξ )
2
...(13.29)
∈k+1 = ∈k –
, since f(ξ) = 0
 ∈k + ∈k −1 
f ' (ξ) + 
 f " (ξ)


2
On dividing the numerator and denomenelor by f ' (ξ), we get
−1

∈2
f " ( ξ )   ∈k + ∈k −1 f " ( ξ )
∈k+1 =∈k − ∈k + k .
1
+

|—
2
f ' ( ξ )  
2
f ' ( ξ ) 


∈2 f " ( ξ )   ∈k + ∈k −1  f " ( ξ ) 
∈k+1 =∈k − ∈k + k .
....
.
 1− 
 f ' (ξ)
|—
2 f ' ( ξ )   
2


On neglecting second and higher order term in second bracket, we get

∈2
f " ( ξ )    ∈k + ∈k+1  f " ( ξ ) 
or
∈k+1 =∈k − ∈k + k .
.
 1− 
 f ' ( ξ ) 
|—
2
f ' ( ξ )  
2

or
or

∈ . (∈k + ∈k+1 ) f " ( ξ ) ∈2k f " ( ξ ) ∈2k (∈k + ∈k+1 ) f " ( ξ )
∈k+1 =∈k − ∈k − k
.
+
.
.

2
f ' (ξ)
2 f ' (ξ)
4
f ' ( ξ ) 

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Engineering Mathematics-III
or

∈ (∈ + ∈k+1 ) f " (ξ ) ∈2k
f " (ξ ) ∈2k (∈k + ∈k +1 ) f " (ξ )
∈k+1 =∈k − ∈k − k k
.
+
.
−
.

2
f ' (ξ ) 2
f ' (ξ )
4
f ' (ξ ) 

or
∈k+1 = ∈k .∈k −1 =
1 f " (ξ)
+ 0 (∈2k )
2 f ' (ξ)
On neglecting 0 (∈2k ), we get
...(13.30)
∈k+1 =∈k .∈k −1 . D
1 f " (ξ)
where D =
is a constant
2 f ' (ξ)
r
Since we have ∈k+1 = M ∈k , therefore ∈k = M ∈k −1
r
1
or
∈  r
∈k −1 =  k 
M
...(13.31)
By above expressions, we get
1
 ∈k  r
r
M.∈k =∈k .   .
M
or
∈rk =
D. M
 1
− 1+ 
 r
D

1
. ∈k 1+ r 
Comparing the power of ∈k on both side, we get
1
or
r = 1+
r2 − r − 1 = 0
r
On solving we get r = 1⋅618, and we have
∈k+1 = M. ∈1k.618
Therefore, we rate/order of convergence of False position method is 1 .618.
13.10.3 Rate of Convergence of Newton-Raphson's Method
In this method, we have
x k+1 = x k −
f( xk )
f ' ( xk )
...(13.32)
We also have
∈k = x k − ξ or x k = ∈k + ξ and ∈k+1 – ξ = x k +1 – ξ
or
x k+1 = ∈k +1 + ξ
By (13.63) and (13.64), we have
f (∈k + ξ )
f (∈k + ξ )
or ∈k+1 =∈k −
∈k+1 + ξ = ∈k +ξ −
f ' (∈k + ξ )
f ' (∈k + ξ )
On expanding by Taylor's series, we get
f ( ξ )+ ∈k . f ' ( ξ ) +
∈k+1 = ∈k −
f ' ( ξ )+ ∈k . f " ( ξ ) +
∈2k
. f " ( ξ ) +...
|—
2
∈2k
. f " ' ( ξ ) +...
|—
2
Neglecting the third and higher order, we get
...(13.33)
Solutions of Algebraic and Transcendental Equations
577
{ ∵ f (ξ ) = 0
∈k . f ' ( ξ )
∈k+1 =∈k −
f ' ( ξ ) + ∈k . f " ( ξ )
=
∈2k . f " ( ξ )
f ' ( ξ )+ ∈k . f " ( ξ )
=
∈2k . f " ( ξ ) 
f " ( ξ )
. 1+ ∈k .
f ' (ξ) 
f ' ( ξ ) 
=
∈2k f " ( ξ ) 
f " (ξ) 
. 1− ∈k .
...
f ' ( ξ ) 
f ' ( ξ ) 
∈k+1 ≈
or
−1
∈2k . f " ( ξ )
f ' (ξ)
∈k+1 = M .∈2k where M =
f " (ξ)
f ' (ξ)
It is clear that rate / order of convergence of Newton-Raphson's method is quadratic.
It means subsequent error at each approximation is proportional to the square of the
previous error.
Illustrative Examples
Ex.19 : Find a root of the equation x 3 – x – 11 = 0 correct to four decimals using
bisection method.
Sol: Let
f ( x ) = x 3 – x – 11.
Since f( 2) = –5 < 0 and f( 3) = 13 > 0, a root lies in between 2 and 3.
Hence, the first approximation to the root is
x1 = ( 2 + 3) / 2 = 2.5
Now
f( 2.5) = ( 2.5) 3 – 2.5 – 11 = 2.215( + ) ve.
Therefore, the root lies between 2 and 2.5 ( = x1 ).
Thus the second approximation to the root is
x 2 = ( 2 + 2.5) / 2 = 2.25
Now
f( 2.25) = ( 2.25) 3 – 2.25 – 11 = –1.859375(–)ve
Therefore, the root lies between x1 and x 2 .
Thus the third approximation to the root is
x 3 = ( x1 + x 2 ) / 2 = ( 2.5 + 2.25) / 2 = 2.375
Now
f ( 2.375) = ( 2.375) 3 – 2.375 – 11 = 0.0214843( + )ve
Therefore, the root lies in between x 2 and x 3 .
Thus the fourth approximation to the root is
x 4 = ( x 2 + x 3 ) / 2 = ( 2.25 + 2.375) / 2 = 2.3125
f( 2.3125) = ( 2.3125) 3 – 2.3125 – 11 = –0.9460449(–)ve
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Engineering Mathematics-III
which means that the root lies in between x 3 and x 4 .
Thus the fifth approximation to the root is
x 5 = x 4 / 2 = ( 2.375 + 2.3125) / 2.34375
f( 2.34375) = ( 2.34375) 3 – 2.34375 – 11 = – 0.4691467 (–)ve
Therefore, the root lies in between x 3 and x 4 .
Thus the sixth approximation to the root is
x 6 = ( x 3 + x 3 ) / 2 = ( 2.375 + 2.34375) / 2 = 2.359375
f( 2.359375) = ( 2.359375) 3 – 2.359375 – 11 = – 0.2255592 (–)ve
Therefore, the root lies in between x 3 and x 6 .
Thus the seventh approximation to the root is
x7 = ( x 3 + x 4 ) / 2 = ( 2.375 + 2.359375) / 2 = 2.3671875
f( 2.3671875) = ( 2.3671875) 3 – ( 2.3671875) – 11
= –0.1024708(–)ve
which means that the root lies in between x 3 and x7 .
Thus the eighth approximation to the root is
x 8 = ( x 3 + x7 ) / 2 = ( 2.375 + 2.3671875) / 2 = 2.3710938
f( 2.3710938) = ( 2.3710938) 3 – ( 2.3710938) – 11
= –0.040601(–)ve
Therefore, the root lies in between x 3 and x 8 .
Thus the ninth approximation to the root is
x 9 = ( x 3 + x 8 ) / 2 = ( 2.375 + 2.3710937 ) / 22.3730469
f( 2.3730469) = ( 2.3730469) 3 – ( 2.3730469) – 11
= 9.585468 × 10 –3 (–)ve
Therefore, the root lies in between x 3 and x 9 .
Thus the tenth approximation to the root is
x10 = x 3 + x 9 / 2 = ( 2.375 + 2.3730469) / 2 = 2.3740235
f( 2.3740235) = ( 2.3740235) 3 – ( 2.3740235) – 11
= 5.942661 × 10 –3 ( + )ve
Therefore, the root lies in between x 9 and x10 .
Thus the eleventh approximation to the root is
x11 = ( x 9 + x10) / 2 = ( 2.3730469 + 2.3740235) / 2 = 2.3135352
f( 2.3735352) = ( 2.3735352) 3 – 2.3735352 – 11
= 1.823101 × 10 –3 (–)ve
Therefore, the root lies in between x10 and x11 .
Thus the twelfth approximation to the root is
x12 = ( x10 + x11 ) / 2 = ( 2.3740235) + 2.3735352) / 2 = 2.3737794
f( 2.3736573) = ( 2.3737794) 3 – 2.377794 – 11
Solutions of Algebraic and Transcendental Equations
579
2.059952 × 10 –3 ( + )ve
Therefore, the root lies in between x11 and x12 .
Thus the thirteenth approximation to the root is
x13 = ( x11 + x12 ) / 2 = ( 2.3735352 + 2.3737794 = 2.3736573)
f( 2.3736573) = ( 2.3736573) 3 – 2.3736573 – 11
= 1.18815 × 10 – 4 ( + )ve
Therefore, the root lies in between x11 and x13 .
Thus the fourteenth approximation to the root is
x14 = ( x11 + x13 ) / 2 = ( 2.3735352 + 2.3736573) / 2 = 2.3735963
f( 2.3735963) = ( 2.3735963) 3 – 2.3735963 – 11
= – 8.51921 × 10 – 4 (–)ve
Therefore, the root lies in between x13 and x14 .
Thus the fifteenth approximation to the root is
x15 = ( x13 + x14 ) / 2 = ( 2.3736573 + 3.3735963) / 2 = 2.3736268 and
f ( 2.3736268) = – 3.66112 × 10 –4 (–)ve.s
Therefore, form x14 and x15 we can see that f ( x14 ) and f ( x15 ) are nearly equal to
zero. Hence the root correct to four decimal places is = 2.3736 .
Ex. 20 : Using bisection method, find the negative root of
x 3 – x + 11 = 0
f ( x ) = x 3 – x + 11
Sol : Let
f (– x ) = – x 3 + x + 11
The negative root of f ( x ) = 0 is the positive root of f (– x ) = 0.
Therefore, we will find the positive root of f (– x ) = 0.
φ( x ) = x 3 – x – 11 = 0
Proceeding as explained in Example 3.1, we get x = 2.3736 and hence. the negative
root is –2.3736.
Ex. 21 : Find a real root of the equation x 3 + x 2 – 1 = 0 by iteration method.
f ( x) = x 3 + x 2 – 1
Sol : Let
Now f( 0) = –1 is negative and f(1) = 1 is positive. Hence a real root lies between 0
and 1 .
Now, x 3 + x 2 – 1 = 0 can be written as
x = 1 / ( x + 1) = φ( x )
φ' ( x ) = – 12 [1 / ( x + 1) 3/ 2 ]
Clearly, |φ' ( 0)|=
i.e..,
1
2<
1 and |φ' (1)|= 1 / 2 5/ 2 < 1
|φ' ( x )|< 1 for all x in (0,1)
Hence the iterative method can be applied. Let x 0 = 0.65 be the initial
approximation, then
580
Engineering Mathematics-III
x1
x2
x3
x4
x5
= φ( x 0 )
= 1 / (1 +
= 1 / (1 +
= 1 / (1 +
= 1 / (1 +
x1 )
x2 )
x3 )
x4 )
= 1/
= 1/
= 1/
= 1/
= 1/
(1 + x 0 ) = 1 / 1.65 = 0.7784989
1.7784989
= 0.7498479
1.7498479
= 0.7559617
1.7559617
= 0.7546446
1.7546446
= 0.7549278
x 6 = 1 / (1 + x 5 )
= 1 / 1.7549278
x7 = 1 / (1 + x 6 )
= 1 / 1.7548668
x 8 = 1 / (1 + x7 )
= 1 / 1.7548799
x 9 = 1 / (1 + x 8 )
= 1 / 1.7548771
x10 = 1 / (1 + x 9 )
= 1 / 1.758777
x11 = 1 / (1 + x10 )
= 1 / 1.7548776
Hence the root is 0.7548776.
Ex.22 : Find a real root of the equation cos x = 3 x – 1 correct
places by the method of successive approximation.
Sol: Let
f ( x ) = cos x – 3 x – 1 = 0
f( 0) = –1(–) ve and f ( π / 2) = –3( π / 2) – 1(–)ve
Therefore, a real root lies in between 0 and π / 2.
Rewriting the equation as
=
=
=
=
=
=
to seven decimal
x = 1 / 3(cos x + 1) = φ( x )
we have φ' ( x ) = –1 / 3 sin x
Now |φ' ( x )|= 11 / 3 sin x|< 1 for all x in ( 0, π / 2).
Hence, the iteration method can be applied.
Let x 0 = 0.5 be the initial approximation, then
x1
x2
x3
x4
= φ( x 0 ) = 1 / 3[cos( 0.5)+] = 0.6258608
= 1 / 3[cos( 0.6258608) + 1] = 0.6034863
= 1 / 3[cos( 0.6034863) + 1] = 0.6077873
= 1 / 3[cos( 0.6077873) + 1] = 0.6069711
x 5 = 1 / 3[cos( 0.6069711) + 1] = 0.6071264
x 6 = 1 / 3[cos( 0.6071264) + 1] = 0.6070969
x7 = 1 / 3[cos( 0.6070969) + 1] = 0.6071025
x 8 = 1 / 3[cos( 0.6071025) + 1] = 0.6071014
x 9 = 1 / 3[cos( 0.6071014) + 1] = 0.6071016
x10 = 1 / 3[cos( 0.6071016) + 1] = 0.6071016
Hence the root is 0.6071016.
Ex. 23 : Find the smallest root of the equation.
x 3 x 5 x7
x9
x11
+
–
+
–
+.... = 0.4431135
3
10 42 216 1320
Sol: Writing the given equation as
x–
0.7548668
0.7548799
0.7548771
0.7548777
0.7548776
0.7548776
Solutions of Algebraic and Transcendental Equations
x = 0.4431135 +
581
x 3 x 5 x7
x9
x11
–
+
–
+
+.... = φ ( x )
3
10 42 216 1320
Neglecting powers of x 3 and higher powers of x in the above equation, we get
x = 0.4431135 (approximately). To solve it let the initial approximation be
x 0 = 0.44.
Then x1 = φ( x 0 )
= 0.4431135 +
( 0.44) 3 ( 0.44) 5 ( 0.44)7 ( 0.44) 9 ( 0.44)11
–
+
–
+
–....
3
10
42
216
1320
= 0.4699
Similarly,
x 2 = φ( x1 ) = φ( 0.4699) = 0.4755
x 3 = φ( x 2 ) = φ( 0.4755) = 0.47664
x 4 = φ( x 3 ) = φ( 0.47664) = 0.47686
x 5 = φ( x 4 ) = φ( 0.47686) = 0.47690
The values of x 4 and x 5 indicate that x = 0.4769, correct to four decimal places.
Ex. 24 : Find a real root of the equation x 3 – 2 x – 5 = 0 by the method of false
position correct to three decimal places.
Sol: Given : f ( x ) = x 3 – 2 x – 5; a = 2, b = 3
Now
f ( a) = f ( 2) = 2 3 – 2( 2) – 5 = –1(–)ve
and
f ( b) = f ( 3) = 3 3 – 2( 3) – 5 = –16( + )ve
Therefore, the root of f ( x ) = 0 lies in between 2 and 3.
The first approximation of the root is x1 and is given by
af ( b) – bf ( a) 2(16) – 3(–1)
x1 =
=
f ( b ) – f ( a)
16 – (–1)
= 35 / 17 = 2.0588 (approximately)
Now
f ( x1 ) = f ( 2.0588) = ( 2.0588) 3 – 2( 2.0588) – 5
= – 0.391(–)ve
Therefore, the root of f ( x ) = 0 lies in between x1 = 2.0588 and b = 3.
The second approximation to the root is given by
x f ( b) – bf ( x1 ) ( 2.0588)(16) – 3(–0.391)
x2 = 1
=
f ( b) – f ( x1 )
16 – (–0.391)
=
34.1138
= 2.08125
16.391
f ( x 2 ) = f ( 2.08125) = ( 2.08125) 3 – 2( 2.08125) – 5
= –0.147(–)ve
which means that the root of f ( x ) = 0 lies in between x 2 = 2.08125 and b = 3.
The third approximation to the root is given by
x3 =
x 2 f ( b) – bf ( x 2 ) ( 2.08125)(16) – 3(–0.147 )
=
f ( b) – f ( x 2 )
16 – (–0.147 )
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Engineering Mathematics-III
=
33.741
= 2.0896
16.147
f ( x 3 ) = f ( 2.0896) = ( 2.0896) 3 – 2( 2.0896) – 5
= –0.0551(–)ve
Therefore, the root of f ( x ) = 0 lies in between x 3 = 2.0896 and b = 3.
33.52418
=
= 2.0943
16.000726
Now
f ( x 6 ) = f ( 2.0943) = –0.0028
Therefore, the root lies in between x 6 = 2.0943 and b = 3.
The seventh approximation to the root is given by
af ( b) – bf ( a) 2(16) – 3(–1)
x1 =
=
f ( b ) – f ( a)
16 – (–1)
= 35 / 17 = 2.0588 (approximately)
Now
f ( x1 ) = f ( 2.0588) − ( 2.0588) 3– 2 ( 2.0588) – 5
= –0.391(–) ve
Therfore, the root of f ( x ) = 0 lies in between x1 = 2.0588 and b = 3.
The second approximation to the root is given by
x f ( b) – bf ( x1 ) ( 2.0588)(16) – 3(–0.391)
x2 = 1
=
f ( b) – f ( x1 )
16 – (–0.391)
=
34.1138
= 2.08125
16.391
f ( x 2 ) = f ( 2.08125) = ( 2.08125) 3 – 2( 2.08125) – 5
= –0.147(–) ve
which means that the root of f ( x ) = 0 lies in between x 2 = 2.08125 and b = 3,
The third approximation to the root is given by
x f ( b) – bf ( x 2 ) ( 2.08125)(16) – 3(–0.147 )
x3 = 2
=
f ( b) – f ( x 2 )
16 – (–0.147 )
=
33.741
= 2.0896
16.147
f ( x 3 ) = f ( 2.0896) = ( 2.08125) 3 – 2( 2.0896) – 5
= –0.0551(–) ve
Therefore, the root of f ( x ) = 0 lies in between x 3 = 2.0896 and b = 3.
33.52418
=
= 2.0943
16.00726
Now
f ( x 6 ) = f ( 2.0943) = –0.0028
Therefore, the root lies in between x 6 = 2.0943 and b = 3.
The seventh approximation to the root is given by
x f ( b) − b f ( x 6 ) ( 2 .0943)(16) − 3 ( −0 .0028)
x7 = 6
=
f ( b) − f ( x 6 )
16 − ( −0.0028)
Solutions of Algebraic and Transcendental Equations
=
583
33 .5172
= 2 .0944
16 .0028
Therefore, the root is 2.094, correct to three decimal places.
Ex. 25 : Find the root of xe x = 3 by Regula Falsi Method correct to three decimal
places.
Sol : Given : f ( x ) = xe x − 3
Now
and
f (1) = e − 3 = −0 .28172 ( − ) ve
f (1 .5) = 1 .5 e1.5 − 3 = 3 .72253 ( + )ve
Therefore, the root lies in between 1 land 1.5 Let a=1 and b=1.5, them
f ( a) = −0 .28172 and f ( b) = 3 .72253.
The first approximation to the root is given by
a f ( b) − b f ( a) 1( 3 .72253) − 1 .5( −0 .28172)
x1 =
=
f ( b ) − f ( a)
3 .72253 − ( −0 .28172)
=
Now
4 .14511
= 1 .035
4 .00425
f ( x1 ) = f (1 .035) = 1 .035 e1.035 − 3
= −0 .0864 ( − )ve
Therefore, the root of f(x)=0 lies in between x1 = 1 .035 and b = 1 .5.
The second approximation to the root is given by
x f ( b) − b f ( x1 ) (1 .035)( 3 .72253) − 15( −0 .0286485)
x2 = 1
=
f ( b) − f ( x1 )
3 .72253 − ( −0 .00286485)
=
Now
3 .98242
= 1 .045
3 .80893
f ( x 2 ) = f (1 .045) = 1 .045 e1.045 − 3 = −0 .0286485 ( − )ve
with implies that the root lies in between x 2 = 1 .045 and b = 1 .5.
The third approximation to the root is given by
x f ( b) − b f ( x 2 ) (1 .045)( 3 .72253) − 15( −0 .0286485)
x3 = 2
=
f ( b) − f ( x 2 )
3 .72253 − ( −0 .00286485)
=
3 .9330166
= 1 .048
3 .7511785
Now
f ( x 3 ) = f (1 .048) = −0 .0111652 ( − ) ve
Therefore, the root lies in between x 3 = 1 .048 and b = 1 .5.
The fourth approximation to the root is given by
x f ( b) − b f ( x 3 ) (1 .048)( 3 .72253) − 15( −0 .0111652)
x4 = 3
=
f ( b) − f ( x 3 )
3 .72253 − ( −0 .0111652)
=
Now
3 .9179594
= 1 .049
3 .7336953
f ( x 4 ) = f (1 .049) = −5 .320155 × 10 −3 ( − )ve
584
Engineering Mathematics-III
Therefore, the root lies in between x 4 = 1 .049 and b = 1 .5.
The sixth approximation to the root is given by
x f ( b) − b f ( x 4 ) 3 .9129142
x5 = 4
=
= 1 .0496
f ( b) − f ( x 4 )
3 .7278502
Now f ( x 5 ) = f (1 .0496) = −1 .808903 × 10 −3 which means that the root lies in
between
x 5 = 1 .0496 and b = 1 .5.
Ex. 26 : Find an iterative formula to find √ N, where N is a positive number and
hence, find √ 12 correct to four decimal places. (M.U,B.E., 1992)
Sol : Let x = √ N ∴ x 2 − N = 0
If
f ( x ) = x 2 − N then f ′ ( x ) = 2 x
Now, from Newton-Raphson formula,
x n +1 = x n −
f( xn )
x2 − N
= xn − n
f ′ ( xn )
2x n
∴
x n +1 = 1/ 2 [ x n + ( N / x n )]
Eqn (13.1) is the required iterative formula.
....(13.1)
Putting N=12 in f(x), we have f(x)=x 2 − 12.
Now, f(3)<0 and f(4)>0. Therefore, the root lies in between 3 and 4. Let the initial
approximation x 0 be 3 .1.
Then, from Eqn (i) the first approximation to the root.
x1 = 1/ 2[ x 0 + 12/ x 0 ] = 1/ 2[ 3 .1 + 12/ 3 .1] = 3 .4854839
The second approximation is

12
12


x 2 = 1/ 2  x1 +  = 1/ 2 3 .4854839 +
= 3 .4641672
x1 
3 .4854839


The third approximation is
12


x 3 = 1/ 2 3 .4641672 +
= 3 .4641016
3 .4641672

The fourth approximation is
12


x 4 = 1/ 2 3 .4641016 +
= 3 .4641016
3 .4641016

Thus, the value of √ 12 correct to four decimals is 3 .4641.
Ex. 27 : Solve x 3 + 2 x 2 + 10 x − 20 = 0 by Newton-Raphson Method.
Sol : Let f ( x ) = x 3 + 2 x 2 + 10 x − 20
∴
f ′ ( x ) = 3 x 2 + 4 x + 10
We have that in Newton-Ranspson's method
x n −1 = x n −
 x 3 + 2 x n2 + 10 x n − 20
f( xn )
= xn −  n

f ′ ( xn )
3 x n2 + 4 x n + 10 

Solutions of Algebraic and Transcendental Equations
=
585
2( x n3 + x n2 + 10)
....(13.1)
3 x n2 + 4 x n + 10
Now we can see that f (13.1)=–7<0 and f (13.2)=16>0.
Therefore, the root lies in between 1 and 2.
Let x 0 = 1 .2 be the initial approximation ( ∵ f (1 .2) < 0).
Putting n = 0 in Eqn (i), first approximation x1 is given by
x1 =
=
2( x 03 + x 02 + 10)
3 x 03 + 4 x 0 + 10
=
2[(1 .2) 3 + (1 .2) 2 + 10]
3(1 .2) 2 + 4(1 .2) + 10
26 .336
= 1 .3774059
19 .12
The second approximation x 2 is
2( x13 + x12 + 10)
x2 =
3 x12 + 4 x1 + 10
=
2[(1 .3774059) 3 + (1 .3774059) 2 + 10]
3 (1 .3774059) 2 + 4 (1 .3774059) + 10
29 .021052
=
= 1 .3688295
21 .201364
The third approximation x 3 is given by
2( x 23 + x 22 + 10)
x3 =
3 x 22 + 4 x 2 + 10
=
2[(1 .3688295) 3 + (1 .3688295) 2 + 10]
3 (1 .3688295) 2 + 4 (1 .3688295) + 10
28 .876924
=
= 1 .3688081
21 0964
The fourth approximation x 4 (to the root )is given by
2( x 33 + x 32 + 10)
x4 =
3 x 32 + 4 x 3 + 10
=
2[(1 .3688081) 3 + (1 .3688081) 2 + 10]
3 (1 .3688081) 2 + 4 (1 .3688081) + 10
28 .876567
=
= 1 .3688081
21 .09614
Hence the root is 1.3688081.
Ex. 28 : Using Newton-Raphson Method, find the root of the equation
x log10 x = 1 .2.
(M.U,B.E., 1991, 1994, 1995)
Sol : Let f ( x ) = x log10 x − 1 .2.
f ′ ( x ) = log10 x + x (log10 e / x )
= log10 x + 0 .4343
[∵
d
1
log a x = log a e ]
dx
x
586
Engineering Mathematics-III
From Newton-Raphson formula,
f( xn )
x log10 x n − 1 .2
x n +1 = x n −
= xn − n
f ′ ( xn )
log10 x n + 0 .4343
0 .4343 x n − 1 .2
∴
x n +1 =
log10 x n + 0 .4343
....(13.1)
Now f ( 2 .5) = −0 .2051499 < 0 and f ( 3) = 0 .2313637 > 0. Therefore, the real root
fo f(x) lies in (2.5,3). Let x 0 = 2 .7 be the initial approximation. Putting n=0 in Eqn
(i), the first approximation x1 is given by
0 .4343 x 0 + 1 .2
0 .4343( 2 .7 ) + 1 .2
x1 =
=
log10 x 0 + 0 .4343 log10 2 .7 + 0 .4343
2 .37261
=
= 2 .7407986
0 .8656637
The second approximation x 2 is
0 .4343 x1 + 1 .2
0 .4343( 2 .7407986) + 1 .2
x2 =
=
log10 x1 + 0 .4343 log10 ( 2 .7407986) + 0 .4343
2 .3903288
=
= 2 .7406461
0 .8721771
Similarly, the third approximation is
0 .4343 x 2 + 1 .2
0 .4343( 2 .7406961) + 1 .2
x3 =
=
log10 x 2 + 0 .4343 log10 ( 2 .7406461) + 0 .4343
2 .3902626
=
= 2 .7406461
0 .8721529
Hence, the root is 2 .7406461.
Ex. 29 : Solve sin x = 1 + x 3 using Newton-Raphson Method.
(M.U, B.E., 1992)
Sol : Let f ( x ) = sin x − 1 − x 3 ∴ f ′ ( x ) = cos x − 3 x 2
Then, from Newton-Raphson formula,
f( xn )
sin x n − 1 − x n3
x n +1 = x n −
= xn −
f ′ ( xn )
cos x n − 3 x n2
∴
x n +1 =
x n cos x n − sin x n − 2 x n3 + 1
cos x n − 3 x n2
....(13.1)
Now f( −1) = sin ( −1) − 1 − ( −1) 3 = −0 .8414709 < 0
and f( −2) = sin ( −2) − 1 − ( −2) 3 = 6 .0907026 > 0 which means that the root lies in
between – 1 and – 2. Let x 0 = −1 .1 be the initial approximation. Then, by putting
n = 0, 1, 2,... in Eqn (i), we obtain the successive approximations as
x cos x 0 − sin x 0 − 2 x 03 + 1
4 .0542516
x1 = 0
=
= 1 .2763653
2
−
3 .1764039
cos x 0 − 3 x 0
5 .7452469
= −1 .2497465
− 4 .5971297
5 .4584049
x3 =
= −1 .2490526
− 4 .370036
x2 =
Solutions of Algebraic and Transcendental Equations
5 .4510835
= −1 .2490522
− 4 .364176
5 .4510786
x5 =
= −1 .2490521
− 4 .3641722
5 .4510785
x6 =
= −1 .2490522
− 4 .3641721
x4 =
Hence the root is average of x 5 and x 6 , i. e. − 1 .24905215.
Exercises
Objective Type Questions
Multiple Choice Question
1. If α is a root of f(x)=0, then f(x) is divisible by :
(a) x + α
(b) x − α
(c) x − 2α
(d) x ± α.
2. Bisection method does not work if two roots are :
(a) equal
(b) unequal
(c) almost equal
(d) none of these.
3. In False position method , an initial interval is required :
(a) true
(b) false
(c) can not say
(d) sometimes.
4. Bisection method is based upon :
(a) Laurent's series
(b) Euler's series
(c) Moclaurin's series
(d) Taylor's series.
5. Newton-Raphson's method gives :
(a) exact solution
(b) approximate root
(c) real roots
(d) b and c both.
6. Newton-Raphson's method is used to solve :
(a) Transcendental equation
(b)Algebraic equation
(c) a and b both
(d) every equation
7. Newton-Raphson's method is used to find
(a) real roots
(b) negative root
(c) positive roots
(d) complex roots
(e) a, b, c, d all.
587
588
Engineering Mathematics-III
8. If f ( x ) = 8 x 7 + 5 x 6 − 4 x 4 − 2 x 3 − 6 x 2 + x − 1, then the maximum number
of positive roots for f ( x ) = 0 will be :
(a) 3
(b) 4
(c) 2
(d) 7.
3
9. The root of the equation x − 5 x + 1 = 0 lying in the interval :
(a) [0, 1]
(b) [0, –1]
(c) [– 2, –1]
(d) [1, 2].
10. If f ( x ) = cos x. e x , then initial interval is :
(a) [1, 2]
(b) [–1, –2]
(c) [1, 0]
(d) [0, 1].
Fill in the Blanks
1. The number of .................... roots of f(x)=0 can not ............ the number of
changes sign from ............... to ................... and from negative to positive.
2. Bisection method does not work if two roots are almost .................. .
3. Bisection method is very ............ .
4. Bisection method is always ................in the interval [a,b] if f(a) and f(b) have
........... sign.
5. Bisection method is always ..................... than Regula Falsi method .
6. An .................. solution is needed, to apply Newton-Raphson's method.
7. Newton-Raphson method is based upon ................... .
8. Newton's Method can be used to solve ................... as well as ...................
equations.
9. ..................... can be used to get an approximate root to start Newton-Raphson
method.
10. The first approximation for f(x)=log e x– cos x=0 is .................... . If we take
x 0 = 1. 3 in Newton-Raphson's method.
True/False
1. To determine the maximum number of negative roots of a polynomial f ( x ) = 0,
replace x with -x.
2. If the deg f ( x ) = p and it has m positive roots and n negative roots, then at least
p-m-n roots will be imaginary.
3. Bisection method always gives exact root of the equation.
4. A suitable choice of interval [a,b ] is required to apply bisection method.
Solutions of Algebraic and Transcendental Equations
589
5. Bisection method never fails.
6. Bisection method is a slow method and the speed depend upon the choice of
interval [a ,b ].
7. Bisection method is always convergent in the interval [a, b] if f ( a) and f ( b)
have same sign.
8. Regula False method is more faster than Bisection method in most cases.
9. Newton- Raphson's method is used to find the complex root of an equation.
10. Newton-Raphson's method can be applied directly to solve an equation.
Short Answer Type Questions
1.
What are the Descarte's rules of sign ?
2.
Describe the bisection method in brief.
3.
Discuss Regula False method in brief.
4.
Discuss Newton-Raphson's method.
Numerical Problems
1.
Find a root of the following equations correct to three decimal places, using the
Bisection method.
(i)
x3 − x2 + x − 7 = 0
(ii)
x 3 − 2x − 5 = 0
(iii)
x 3 − 3 x − 5 = 0 (Bangalore, B.E., 1989)
(iv)
x 3 − 4 x − 9 = 0 (Mysore, B.E., 1987 )
(v)
x 4 − x − 10 = 0 (S. Gujarat B.E., 1990)
(vi)
x − cos x = 0 (B.U. B.E., 1995)
(vii) 3 x − e x = 0
(viii) 3 x =√ (1 + sin x )
(ix)
2.
x log10 x − 1 .2 = 0
Using Bisection method find the negative root of x 3 − 4 x + 9 = 0, correct to
three decimal places.
3.
Find a root of the following equations correct to three decimal places, using
Iteration method.
(i)
x 3 + x 2 − 100 = 0
(ii)
x = 1/ 2 + sin x
(iii)
3 x − 6 = log10 x
(iv)
xe x − cos x = 0
590
Engineering Mathematics-III
(v)
sin x = e x − 3 x (vi) 2 x −7 − log10 x = 0
(vii) 1 − x +
4.
x2
( 2 !) 2
−
x3
( 3 !) 2
+
x4
( 4 !) 2
−
x5
( 5 !) 2
+... = 0
Find a negative root of x 3 − 2 x + 5 = 0, correct to three decimal places, using
Successive Approximation method.
5.
Find a root of the following equations correct to four decimal places using the
method of False Position (regula false method).
(i)
x 3 − 4x − 9 = 0
(ii)
x 3 + 2 x 2 + 10 x − 20 = 0
(iii)
x 3 − 4x − 1 = 0
(iv)
x6 − x4 − x3 − 1 = 0
(v)
xe x = 2
(vi)
e x sin x = 1
(vii)
x = cos x
(viii) x tan x = −1 in ( 2 .5, 3)
(ix) x log10 x = 1 .2
Using Newton-Raphson method, find a root correct to three decimal places of
the following :
6.
x 3 − 3x 2 + 7 x − 8 = 0
7.
x 3 − 3x − 5 = 0
(M.U. B.E., 1992)
(Kerala B.Tech 1989)
3
8.
x − 5x + 3 = 0
9.
x 4 − x − 10 = 0
(Gulbarga B.E. 1993)
10. x 4 − x − 13 = 0
11. e x = 1 + 2 x
12. xe x − cos x = 0
13. e x sin x = 1
(Gujarat B.E., 1990; B.U., B.E., 1995)
14. x x = 1000
15. 3 x − 1 = cos x
16. sin x = 1 − x
17. x 2 + 4 sin x = 0
18. 2 x tan x = 1
19. x(1 − log e x ) = 0 .5
20. 3 x − e x + sin x = 0
21. x sin x + cos x = 0 near x = π (Karnataka B.E., 1993)
(B.R., B.E., 1993)
(M.U., B.E., 1987)
Solutions of Algebraic and Transcendental Equations
591
22. Find the iterative formulae for finding 1/ √ N , 3√ N , 4√ N where N is a positive
real number, using Newton's method. Hence evaluate 1/ √ 17, 3 √ 10, 4√ 25.
23. Find a negative root of the following equations using Newton's method.
x 3 − x 2 + x + 100 = 0 (ii) x 3 − 21 x + 3500 = 0
(i)
24. Find the cube root of 24 by Newton-Raphson's method
[Hint : x 3 = 24 or x 3 − 24 taking x 0 = 2 ⋅ 7]
ANSWERS
Multiple Choice Questions
1. (b)
2. (c)
3. (a)
4. (d)
5. (d)
6. (c)
7. (e)
8. (a)
9. (a)
10. (d)
Fill in the Blanks
1
positive, exceeds, positive, negative
2.
equal
3.
slow
4.
convergent, opposite
5.
slow
6.
approximate
7.
Taylor's series
8.
transcendental, algebraic
9.
Bisection method
10.
1.30295
State True/False
1. True
2. True
3. False
4. True
5. False
6. True
7. False
8. True
9. True
10. False
Unsolved Numerical Problem
1.
(i) 2.105
(vi) 0.739
(ii) 2.095
(vii)
0.619
(iii) 2.280
(viii)
(iv) 2.706
0.392
(ix) 2.740
(iv) 0.518
(v) 1.813
2. – 2.706
3.
(i) 4.331
(ii) 1.497
(iii) 2.108
(v) 0.360
(vi) 3.789
(vii)
1.445
4. –2.095
5.
(i)
2.7065
(ii)
1.3688
(iii)
0.2541
(iv)
1.7365
(vi)
0.5885
(vii)
0.7391
(viii)
2.7981
(ix)
2.7406
6. 1.674
(v)
0.8526
592
Engineering Mathematics-III
7. 2.279
8. 1.834
9. 1.856
10. 1.961
11. 1.256
12. 0.518
13. 0.589
14. 3.592
15. 0.607
16. 0.511
17. – 1.934
18. 0.653
19. 0.187
20. 0.360
21. 2.798
22. 0.24246, 2.15466, 2.236
23. (i) – 4.264
(ii) – 16.56
❑❑❑
Unit-4
Chapter
14
Finite Differences and
Interpolation
Introduction
Consider, we have only some points ( x 0 , y 0 ), ( x1 , y 2 ), ( x 2 , y 2 )... ( x n , y n )
satisfy the relation y = f(x) and the explicit nature of f(x) is not given. In this
situation, we shall try to find another function φ ( x) such that φ ( x) always agree at
the given point. This process of approximating a polynomial is called interpolation
and φ ( x) is known as interpolating polynomial.
14.1 Finite Differences
Suppose, there are (n + 1) points ( x 0 , y 0 ), ( x1 , y1 ) … ( x n , y n ) corresponding to
any function y = f(x), which are equally spaced i.e.
x1 = x 0 + h, x 2 = x 0 + 2h,… xi = x 0 + i h, i = 0, 1, 2, 3,... n.
Now to solve the problem of interpolation, the concept of differences of a function is
discussed in next discussion. For this purpose, some operators are defined, which
are given below.
The Shift Operator (E)
E f ( xi ) = f ( xi + h ), means E f ( x1 ) = f ( x1 + h )
E n f ( xi ) = f ( xi + nh ) = f ( x n )
The forward difference operator (∆)
∆ f ( xi ) = f ( xi + h ) − f ( xi )
∆f ( x1 ) = f ( x1 + h ) − f ( x1) = f ( x 2 ) − f ( x1 )
∆ n f ( xi ) = ∆ n −1 f ( xi +1 ) − ∆ n −1 f ( xi )
The Backward difference operator (∇)
∇ f ( xi ) = f ( xi ) − f ( xi − h )
∇f ( x 2 ) = f ( x 2 ) − f ( x 2 − h ) = f ( x 2 ) − f ( x1 )
∇ n f ( xi ) = ∇ n −1 f ( xi ) − ∇ n −1 f ( xi −1 )
594
Engineering Mathematics-III
The Central-difference operator (δ)
h
h


δ f ( xi ) = f  xi +  − f  xi − 


2
2
1

δ n f ( xi ) = δ n −1 . f  xi +  − δ n −1 .

2
1

f  xi − 

2
The Averaging Operator (µ)
µf ( xi ) =
1
f
2 
h
h


 xi +  + f  xi −  


2
2 
To calculate the differences of various order, we should know how to make a
difference table. The forward, backward and central difference tables are given
below.
(A) Forward difference table, Here, fi = f( x i ) .
It is clear that all forward differences of f 0 , ∆f 0 , ∆ 2 f 0… ∆ 9 f 0 lies on a straight line
sloping downward to the right. Similarly, the differences of f1 , f 2 ... f 8 lies on the
straight line sloping downward. We should also observe that f 0 has 9 difference
while f1 has 8 and f 2 has 7, ... f 8 has only first order difference.
Finite Differences and Interpolation
595
Backward difference table, Here fi = f ( xi )
It is clear that all backward difference of f 9 lies on the straight line sloping upward.
f 9 has differences upto nine order while f1 has only first order difference.
Central difference table, Here fi = f ( xi )
596
Engineering Mathematics-III
It is clear that all differences of f 5 lies on a horizontal line and f 5 has higher
difference upto tenth order.
If we observe forward, backward and central difference table, the numerical value is
same at every position, thus we can say
∆f 0 = ∇f1 = δf1/ 2
∆ 2 f 0 = ∇ 2 f 2 = δ 2 f1
M
M
M
n
n
∆ f 0 = ∇ f n = δ n f n /2
It verifies that
∆f ( x ) = f ( x + h ) − f ( x )
= Ef ( x ) − f ( x )
= (E – 1) f(x)
or
∆ ≡ E −1
1 1/2
[E
+ E −1/2 ]
2
Relation among various difference operators ∆, ∇, δ, µ and ∆.
Relation between ∆ and E1
We know ∆f ( x ) = f ( x + h ) − f ( x ) = Ef ( x ) − f ( x ) = ( E − 1) f ( x )
so,
∆ ≡ E − 1 or E ≡ 1 + ∆
Relation between E and ∇
Similarly ∆ ≡ 1 − E −1
and δ = E 1/ 2 − E −1/ 2 and µ =
∇f ( x ) = f ( x ) − f ( x − h )
= f ( x ) − E −1 f ( x ) = (1 − E −1 ) f ( x )
∴
∇ = 1 − E −1 or E −1 = 1 − ∇
∴
E = (1 − ∇ ) −1 [∵ ( E −1 ) −1 = E ]
Relation between E and δ
δf ( x ) = f ( x + h / 2) − f ( x − h / 2)
= E 1/ 2 f ( x ) − E − 1/ 2 f ( x )
= ( E 1/ 2 − E − 1/ 2 ) f ( x )
∴
δ=E
–
1
2
1/ 2
−E
1/ 2
(1 − E −1 ) = E 1/ 2 ∇
Again,
δ=E
and
δ = E −1/ 2 ( E − 1) = E −1/ 2 ∆
∴
δ = E 1/ 2 ∇ = E − 1/ 2 ∇
Relation between E and µ
µf ( x ) = 1 / 2[ f ( x + h / 2) + f ( x − h / 2)]
= 1 / 2[ E 1/ 2 f ( x ) + E −1/ 2 f ( x )]
= 1 / 2 ( E 1/ 2 + E − 1/ 2 ) f ( x )
∴
µ = 1 / 2 ( E 1/ 2 + E − 1/ 2 )
Finite Differences and Interpolation
597
Relation of differential operator D with other Operators
d
We know that Df ( x ) =
f ( x ) = f ' ( x ) etc.
dx
By Taylor’s series
h
h2
h3
f ( x + h) = f ( x) + f ' ( x) +
f " ( x) +
f "' ( x ) + ...
1!
2!
3!
h2 2
h3 3
or
Ef ( x ) = f ( x ) + hD f ( x ) +
D f ( x) +
D f ( x ) + ...
2!
3!
h 2 D2
h 3 D3
= [1 + hD +
+
+ ...] f ( x )
2!
3!
= e hD f ( x )
∴
E = e hD
Taking logarithms on both sides, we get
hD = log E = log (1 + ∆ )

1
∆2
∆3
∆4
∴
D = ∆ −
+
−
+ ...
h 
2!
3!
4!

Also,
i.e.
∇ = 1 − E −1 , ∴ E −1 = 1 − ∇
e − hD = 1 − ∇
Taking logarithm on both sides,
− hD = log(1 − ∇ )

1
∇2 ∇3 ∇4
∴
D = ∇ −
+
−
+ ...
h 
2!
3!
4!

Again,
e hD − e − hD
E − E −1
=
2
2
 E 1/ 2 + E − 1/ 2   1/ 2

=
+ e −1/ 2  = µδ
 E
2




sin ( hD) =
hD = sin −1 (µδ )
1
or
D = sin −1 (µδ )
h
Theorem. It f(x) is a polynomial of degree n, then show that ∆ n f ( x ) is constant
∴
and ∆ n+1 f ( x ) is equal to zero.
.Proof : let f(x) = a0 x n + a1 x n −1 + a2 x n − 2 ...+an is a polynomial of degree n.
Now
∆f ( x ) = ∆a0 x n + ∆ a1 x n −1 + ∆a2 x n − 2 ...+ ∆an
= a0[( x + h ) n − x n ]+ a1 [( x + h ) n −1 − x n −1 ]... + ( an − an )
= a0[ x n + n c1 . x n −1 . h + n c 2 x n − 2 . h 2 + n c 3 x n − 3 . h 3 .+… − x n ]
+ a1[ x n −1 + n −1 C1 . x n − 2 h+....
Collecting the coefficient of x n −1 , x n − 2… we have
− x n −1 ] ...+0
598
Engineering Mathematics-III
∆f ( x ) = a0 n c1 . h. x n −1 + b1 . x n − 2 + b2 x n − 3+… bn −1
...(14.1)
where b1 , b2… bn −1 are new coefficient. The expression (14.1) show that ∆ f ( x ) is a
polynomial of ( n − 1) th degree. Similarly, second difference will be a polynomial of
degree (n – 2) and the coefficient of x n − 2 will be a0 n c 2 . h 2 . Proceed in the same
manner we get the n th difference of f(x) will be equal to a0 n c n .h n or a0|n
.h n ,
—
which is constant.
Since ∆ n f(x) = a0|n
.h n (a constant), therefore ∆ n+1 f(x) = 0
—
Note : By above discussion, we can say if n th difference of a tabulated function are
constant, then the tabulated function will be a polynomial of degree n. With the help
of this result, we can approximate a function of n degree in case its n th difference
are constant.
Ex.1 : Solve the following if interval of differencing being h:
(a) ∆ sin x
(b) ∆e x
(c) ∆ log x
(d) ∆ 2 cos x
(e) ∆ ( x 3 + 1)
(f) ∆ (sin x cos x )
(g) ∆ tan −1 x
 x2 

(h) ∆ 

 sin 2 x 
2x 
(i) ∆  
 x !
h
 x + h
Sol : (a) ∆ sin x = sin (x + h) – sin x = 2 cos 
 . sin
 x 
2
(b) ∆ e x
= e x+h − e x ( e h − 1)
(c) ∆ log x
 x + h

= log ( x + h ) − log x = log 
 = log 1 +
 x 

h

x
(d) ∆ 2 cos x = ∆ [ ∆ cos x] = ∆ [cos( x + h ) − cos x] = ∆ cos( x + h ) − ∆ cos x
= cos( x + 2h ) − cos( x + h ) − cos ( x + h ) + cos x
= cos( x + 2h ) − 2cos( x + h ) + cos x
3
(e) ∆ ( x + 1) = ∆x 3 + ∆1 = ( x + h ) 3 − x 3 + 1 − 1 = ( x + h ) 3 − x 3
1
1
 1
(f) ∆ sin cos x = ∆  sin 2 x = ∆ sin 2 x = [sin 2 ( x + h ) − sin 2 x]
2
2
 2
1
= [ 2 cos( 2 x + h ). sin h] = cos ( 2 x + h ) sin h
2
h
−1
(g) ∆ tan x = tan −1 ( x + h ) − tan −1 x = tan −1
1 + ( x + h) x
2
2
2
 x 
( x + h)
x
 =
(h) ∆ 
(Simplify it)
−

 sin 2 x  sin 2 ( x + h ) sin 2 x
2x 
2 x+1 − ( x + 1) 2 x
2 x +1
2x
2 x (1 − x )
(i) ∆   =
−
=
=
( x + 1)!
( x + 1)!
 x ! ( x + 1) x !
Finite Differences and Interpolation
599
Ex. 2 : Construct a forward difference table from the following data :
x
0
1
2
3
4
yx
1
1.5
2.2
3.1
4.6
Evaluate ∆ 3 y1 , y x and y 5 .
Sol : The forward difference table is as given below :
x
yx
x0 = 0
y0 = 1
∆2y
∆y
∆3y
∆4y
∆ y 0 = 0 .5
x1 = 1
∆ 2 y 0 = 0 .2
y1 = 1 .5
∆3y0 = 0
∆ y1 = 0 .7
x2 = 2
∆ 2 y1 = 0 .2
y 2 = 2 .2
∆ 3 y1 = 0.4
∆ y 2 = 0 .9
x3 = 3
∆ 4 y 0 = 0 .4
∆ 2 y 2 = 0 .6
y 3 = 3 .1
∆ y 3 = 1 .5
x4 = 4
Now,
y 4 = 4 .6
∆ 3 y1 = y 4 − 3 y 3 + 3 y 2 − y1
= 4 .6 − 3( 3 .1) + 3( 2 .2) − 1 .5 = 0 .4
we know that
C1 ∆y 0 + x C 2 ∆ 2 y 0 + x C 3 ∆ 3 y 0 + x C 4 ∆ 4 y 0
1
1
= 1 + x( 0 .5) + x( x − 1)( 0 .2) +
x( x − 1) ( x − 2) ( 0)
2
3!
1
+
x( x − 1)( x − 2)( x − 3)( 0 .4)
4!
1
1
1
=1+ x+
( x 2 − x) +
( x 4 − 6 x 3 + 11 x 2 − 6 x )
2
10
60
1
∴
yx =
( x 4 − 6 x 3 + 17 x 2 + 18 x + 60)
60
1
∴
y5 =
[ 5 4 − 6( 5) 3 + 17( 5) 2 + 18( 5) + 60] = 7 .5
60
Ex. 3 : Prove the following results :
(i) ∆ ∇ = ∇ ∆ = ∆ − ∇ = δ 2
(M.U, B.E., 1996)
∆ ∇
(ii) ∆ + ∇ = −
(B.R, B.E., 1996)
∇ ∆
(iii) ( E 1/ 2 + E −1/ 2 )(1 + ∆ )1/ 2 = 2 + ∆
(Madurai B.E., 1998)
y x = y0 +
x
600
Engineering Mathematics-III
(iv) 1 + µ 2δ 2 = (1 + 1/ 2 δ 2 ) 2
(v) ∆ = 1/ 2 δ 2 + δ √ (1 + δ 2 / 4)
(M.U, B.E., 1997, Madurai B.E., 1988)
1
3 4
5
(vi) µ −1 = 1 − δ 2 +
δ −
δ 6 + ...
8
128
1024
Sol : (i) We have,
∆ ∇ f ( x ) = ∆ [∇ f ( x )] = ∆ [ f ( x ) − f ( x − h )]
= ∆ f ( x) − ∆ f ( x − h)
= [ f ( x + h ) − f ( x )] − [ f ( x ) − f ( x − h )]
= ∆ f ( x ) − ∇ f ( x ) = (∆ − ∇) f ( x )
∴
∆∇ = ∆ − ∇
Similarly, ∇ ∆ f ( x ) = ∇ [ ∆ f ( x )] = ∇ [ f ( x + h ) − f ( x )]
= ∇ f ( x + h) − ∇ f ( x)
= [ f ( x + h ) − f ( x )] − [ f ( x ) − f ( x − h )]
= ∆ f ( x ) − ∇ f ( x ) = (∆ − ∇) f ( x )
∴
∇∆ = ∆ − ∇
Again, δ 2 f ( x ) = [ E 1/ 2 − E −1/ 2 ]2 f ( x )
= ( E + E −1 − 2) f ( x )
=
=
=
=
f ( x + h) + f ( x − h) − 2 f ( x)
f ( x + h ) − f ( x )] − [ f ( x ) − f ( x − h )]
∆ f ( x ) − ∇ f ( x ) = (∆ − ∇) f ( x )
∆ −∇
∴
δ2
Hence,
∆ ∇ = ∇ ∆ = ∆ − ∇ − δ2
(ii) RHS
∆ ∇ ∆2 − ∇2
− =
∇ ∆
∇∆
( ∆ + ∇ )( ∆ − ∇ )
=
= ∆ + ∇ = LHS
(∆ − ∇)
=
(iii) ( E 1/ 2 + E −1/ 2 ) (1 + ∆ )1/ 2
= ( E 1/ 2 + E − 1/ 2 ) E 1/ 2
= E +1=1+ ∆ +1= 2+ ∆
(iv) 1 + µ 2δ 2
 E 1/ 2 + E
=1+ 
2

− 1/ 2  2
 E − E −1 
=1+ 

2




E 1/ 2 − E −1/ 2 


2
4 + ( E − E −1 ) 2  E + E −1 
=
=

4
2


1 

Now, 1 + δ 2 
2 

2
2
2

1
= 1 + E 1/ 2 + E −1/ 2  

2 


2
2
...(1)
Finite Differences and Interpolation
601
1


= 1 + E + E −1 − 2


2


2
 E + E −1 
=

2


2
...(2)
Hence, from Eqns (1) and (2), we have
1 

1 + µ 2δ 2 = 1 + δ 2 
2 

(v) RHS
2
1 2
δ2
δ +δ 1+
2
4
1
= δ + δ + 4 + δ 2 


2
1
= δ + ( E 1/ 2 − E −1/ 2 ) + 4 + E 1/ 2 − E −1/ 2 ) 2 


2
1
= δ + ( E 1/ 2 − E −1/ 2 ) + ( E 1/ 2 + E −1/ 2 )
2


=
1 1/ 2
(E
− E −1/ 2 )( 2 E 1/ 2 )
2
= E − 1 = ∆ = LHS
(vi) By definition, we have
=
2
1

µ 2 =  ( E 1/ 2 + E −1/ 2 )
2

1  1/ 2
− 1/ 2 2
=
(E
−E
) + 4

4 
=
1/ 2
}
– E –1/ 2 = δ
1 2
δ2
(δ + 4) =
+1
4
4
1/ 2

δ2 
µ = 1 +

4 

∴
{∵ E
or

δ2 
µ −1 = 1 +

4 

2
− 1/ 2
3
2
2
1 δ2
1  1  1
1  1  1
 δ 
 1
 δ 
∴ µ
=1−
+
   − 1   −
   − 1  − 2   + ...
  4
 2
 4
2 4
2 !  2  2
3 !  2  2
 
 
1 2
3 4
5
6
=1− δ +
δ −
δ + ...
8
128
1024
Ex. 4 : Prove the following :
∆2
∆ 2u x
 ux ≠
(i) 
(M.U, B.E., 1996)

Eu x
 E 
−1
 ∆ 2  x E (e x )
e .
(ii) 
= ex

2 x
E


∆ e
Sol :
(i)
LHS = E −1 ( ∆2 u x ) = E −1 ( E − 1) 2 u x
= ( E − 2 + E −1 )u x
= u x − h − 2u x + u x − h
(Kerala, B.Tech. 1990)
602
Engineering Mathematics-III
( E − 1) 2 u x ( E 2 − 2E + 1) u x
=
U x+ h
u x+ h
u x + h − 2u x + h + u x
=
u x+ h
RHS =
∴
(ii)
LHS ≠ RHS
LHS = E −1 ( ∆2 e x )
e x+ h
e x ( e h − 1) 2
= E −1[ e x ( e h − 1) 2 ]
eh
( e h − 1) 2
= E −1 e x . e h
= e e − h e h = e x = RHS
Ex. 5: Using the method of seperation of symbols, prove that
(i) u1 x + u 2 x 2 + u 3 x 3 + ... =
2
3
x
 x 
 x  2
u1 + 
 ∆u1 + 
 ∆ u1 + ...
 1 − x
 1 − x
1− x
(Kerala, B. Tech.,1985)
(ii) u 0 + u1 + u 2 + ...+ =
n +1
C1u 0 +
n +1
C 2 ∆u 0 +
n +1
C 3 ∆u 0 +... + ∆n u 0 .
(Rourkela,B.Tech., 1985)
Sol :
(i) We know that u x + h = E h u x
∴ LHS = xu1 + x 2 Eu1 + x 3 E 2u1 +...
= x [1 + xE + x 2 E 2 +...]u1
1


 1 
= x
 u1 = x 
u
 1 − xE 
 1 − x(1 + ∆ ) 1
1
x 
x∆ 


= x
1−
 u1 =

 1 − x − x∆ 
1 − x  1 − x 
−1
u1
=

x 
x∆
x 2 ∆2
+
+... u1
1 +
2
1 − x  1 − x (1 − x )

=
x
x2
x3
u1 +
∆u1 +
∆2u1 ... = RHS
2
3
(1 − x )
(1 − x )
(1 − x )
(ii) LHS = u 0 + u1 + u 2 + ...+ u n
= u 0 + Eu 0 + E 2u 0 + ... + E n u 0
= ( 1 + E + E 2 + ... + E n ) u 0
 E n +1 − 1
(1 + ∆ ) n +1 − 1
=
 u0 = 
 u0
∆
 E − 1 


1
= [(1 + n +1 C1 ∆ + n +1 C 2 ∆2 + n +1 C 3 ∆3 + .... + ∆n +1 − 1]u 0
∆
1
= [ n +1 C1 ∆ + n +1 C 2 ∆2 + n +1 C 3 ∆3 + .... + ∆n +1 ]u 0
∆
Finite Differences and Interpolation
=
n +1
603
C1u 0 + n +1 C 2 ∆2u 0 + n +1 C 3 ∆2u 0 + .... + ∆n u 0 = RHS]
 f ( x )
Ex. 6: Evaluate (i) ∆ [ f ( x ) g ( x )] and (ii) ∆ 

 g ( x )
Sol : Let h be the interval of differencing.
(i) ∆ [ f ( x ) g ( x )] = f ( x + h ) g ( x + h ) − f ( x ) g ( x )
f ( x + h) g( x + h) − f ( x + h) g( x)
f ( x + h) g( x) − f ( x) g( x)
f ( x + h )[ g ( x + h ) − g ( x )] + g ( x )[ f ( x + h ) − f ( x )]
f ( x + h) ∆ g( x) + g( x) ∆ f ( x)
 f ( x ) f ( x + h ) f ( x )
∆
−
=
 g ( x ) g ( x + h ) g ( x )
f ( x + h) g( x) − f ( x) g( x + h)
=
g( x + h) g( x)
f ( x + h ) g ( x ) − f ( x ) g ( x ) + f ( x )g ( x ) − g ( x + h ) f ( x )
=
g( x + h) g( x)
g ( x )[ f ( x + h ) − f ( x )] − f ( x )[ g ( x + h ) − g ( x )]
=
g( x + h) g( x)
g ( x )∆f ( x ) − f ( x ) ∆ g ( x )]
=
g( x + h) g( x)
=
+
=
=
(ii)
Ex.7: If y = ( 3 x + 1)( 3 x + 4)...( 3 x + 22), prove
that
∆4 y = 136080 ( 3 x + 13)( 3 x + 16)( 3 x + 19)( 3 x + 22)
Sol : The given equation y = ( 3 x = 1)( 3 x + 4)...( 3 x + 22) contains eight factors :
1 
4 
22

∴
y = 3 8  x +   x +  ...  x + 





3
3
3

= 38  x +

∴
1

3
( 8)
22

∆ y = 38 . 8  x + 

3
(7 )
22

∆2 y = 3 8 . 8 . 7  x + 

3
( 6)
22

∆3 y = 3 8 . 8 . 7. 6  x + 

3
]
( 5)
( 4)
and
∴
22

∆4 y = 3 8 . 8 . 7. 6 . 5  x + 

3
22 
22  
22
22



∆4 y = 11022480  x +   x +
− 1  x +
− 2  x +
− 3








3
3
3
3
= 136080 ( 3 x + 22)( 3 x + 19)( 3 x + 16)( 3 x + 13)
604
Engineering Mathematics-III
1


Ex. 8 : Find ∆2 

x
(
x
+
3
)(
x
+
6
)


1
Sol : y =
= ( x − 3)( −3) where h = 3
x( x + 3) ( x + 6)
∆y = ( −3)( 3)( x − 3)( −4)
∆2 y = ( −3)( −4)( 3) 2 ( x − 3)( −5)


108
=

x
(
x
+
3
)(
x
+
6
)(
x
+
9
)(
x
+
12
)


Ex. 9: Evaluate ∆10 (1 − ax )(1 − bx 2 )(1 − cx 3 )(1 − dx 4 ) (B.R, B.E., 1996)
Sol : ∆10 (1 − ax )(1 − bx 2 )(1 − cx 3 )(1 − dx 4 )
= ∆10 [ abcd x10 + terms involving lesser degree]
= abcd ∆10 ( x10 ) = abcd (10 !)
Ex.10. Find the value of all forward differences of 2, 5, 10, 17, 10 and 8 if it is given.
x
1
2
3
4
5
6
y
2
5
10
17
10
8
Sol : All forward differences can be obtained with be help of difference table
It is clear that ∆( 2) = 3 ∆ 2 ( 2) = 2 ∆ 3 ( 2) = 0 ∆ 4 ( 2) = −2 ∆ 5 ( −2) = 13, similarly,
∆( 5) = 5 ∆ 2 ( 5) = 2, ∆ 3 ( 5) = −2, ∆ 4 ( 5) = 13, ∆(10) = 7, ∆ 2 (10) = 0 ∆ 2 (10) = 9
∆(17 ) = −7 ∆ 2 (17 ) = 9, ∆(10) = 2
Ex. 11 : Find the value of ∇ 2 y 5 , given
y1 = 2, y 2 = 5, y 3 = 10, y 4 = 17, y 5 = 26
Finite Differences and Interpolation
605
Sol : Method I - First we shall make the backward difference table
By difference table ∇ 2 y 5 = 2
Method II we know
∇ ≡ 1 − E −1 , therefore
∇ 2 y 5 = (1 − E −1 ) 2 y 5 = (1 + E −2 − 2E −1 ) y 5
= y 5 + E 2 y 5 − 2E −1 y 5
Hence
= y 5 + y 3 − 2y 4
= 26 + 10 − 2 × 17
=2
∇2 y 5 = 2
Ex.12 : Find y1 is y 2 = 8, y 3 = 3, y 4 = 0, y 5 = −1, y 6 = 0
Sol : First, we shall make the forward difference table
we know that
y1 = E −1 . y 2 = (1+ ∆ ) −1 . y 2
= (1 − ∆ + ∆ 2 − ∆ 3 ... ) y 2
= y 2 − ∆ y 2 − ∆ 2 y 2 − ∆ 3 y 2…
= 8 − ( −5) + 2 − 0 {Since ∆ 3 y 2 and other higher order equal to zero}
y1 = 15
606
Engineering Mathematics-III
14.2 Computation of Missing Terms
Consider, we have ( n +1) equidistant values of x. If out of ( n + 1) values, one or more
values are missing, then with the help of differences, we can find these missing
terms. For the better understanding, related examples are given below.
Ex. 13 : Find the missing term in the following table
x
1
2
3
4
5
6
7
y
1
4
–
16
25
36
49
Sol : In the above table one term is missing, which six terms are given Since, we
have six values, therefore ∆ 6 f 0 = 0, or ( E − 1) 6 f 0 = 0. By Binomial expansion
( E − 1) 6 f 0 = ( E 6 − 6 C1 . E 5 + 6 C2 E 4 − 6 C3 E 3 + 6 C4 E 2 − 6 C5 E − 6 C6 ) f 0 = 0
E 6 f 0 − 6 E 5 f 0 + 15E 4 f 0 − 20 E 3 f 0 + 15E 2 f 0 − 6E f 0 − f 0 = 0
(Since f 0 = 1, f1 = 4, f 2 = ?. f 3 = 16, f 4 = 25, f 5 = 36, f 6 = 49)
f 6 − 6 f 5 +15 f 4 − 20 f 3 − 15 f 2 − 6 f1 − f 0 = 0
putting the corresponding values, we get
49 − 6 × 36 + 15 × 25 − 20 × 16 − 15 f 2 − 6 × 4 − 1 = 0
on solving we get f 2 = 9
Ex.14: Find the missing term in the data, f(10) = 270, f(15) = ?, f(20) = 222,
f(25) = 200, f(30) = ?, f(35) = 164, f(40) = 148
Sol : Here five terms are given and two terms are missing, therefore
∆ 5 f 0 = 0 and ∆ 5 f1 = 0
It is also clear
f 0 = 270, f1 = ?, f 2 = 222, f 3 = 200, f 4 = ?
f 5 = 164, f 6 = 148, Now, we have
∆ 5 f 0 = ( E − 1) 5 f 0 = E 5 f 0 − 5E 4 f 0 +10E 3 f 0 − 10E 2 f 0 + 5E f 0 − f 0 = 0
or
or
or
or
f 5 − 5 f 4 +10 f 3 − 10 f 2 + 5 f1 − f 0 = 0
164 − 5 f 4 +10 × 200 − 10 × 222+ 5 f1 − 270 = 0
164 − 5 f 4 + 2000 − 2220+ 5 f1 − 270 = 0
5 f 4 − 5 f1 = 326
...(1)
5
Similarly, with the help of ∆ f1 = 0 we get
∆ 5 f1 = 0 = ( E − 1) 5 f1 = E 5 f1 − 5E 4 f1 +10E 3 f1 − 10E 2 f1 + 5E f1 − f1 = 0
or f 6 − 5 f 5 +10 f 4 − 10 f 3 + 5 f 2 − f1 = 0
or 148 − 5 × 164 + 10 f 4 − 10 × 200 + 2 × 222 − f1 = 0
or 148 − 820 + 10 f 4 − 2000 + 444 − f1 = 0
or 10 f 4 − f1 = 2228
Solving (1) of (2) we get
10814
f1 = 1576 / 9 and f 4 =
45
...(2)
Finite Differences and Interpolation
607
Ex. 15: find the missing value in the following table.
x
16
18
20
22
24
26
y
34
89
–
155
268
388
Sol : Since five values are given it is possible to express y as a polynomial of fourth
degree.
Hence, the fifth differences of y are zeros.
Taking the origin for x at 16, from the given data we have
y 0 = 43, y1 = 89, y 3 = 155, y 4 = 268, y 5 = 388 and we have to find y 2 . We know
that
∆5 y x = 0 for all values of x
∆5 y 0 = 0, i. e. ( E − 1) 5 y 0 = 0
∴
i.e.
( E 5 − 5C1 E 4 + 5 C 2 E 3 − 5 C 3 E 2 + 5 C 4 E − 1 ) y 0 = 0
or
( E 5 − 5E 4 + 10E 3 − 10E 2 + 5E − 1 ) y 0 = 0
i.e.
E 5 y 0 − 5E 4 y 0 + 10E 2 y 0 + 5Ey 0 − y 0 = 0
or
y 5 − 5 y 4 + 10 y 3 − 10 y 2 + 5 y1 − y 0 = 0
Substituting the given values,
388 – 5(268)+10 (155) – 10y2+ 5(89) – 43 =0
y 2 = 100
Ex. 16: Find the missing values in the following table of values of x and y:
x
0
1
2
3
4
5
6
y
–4
–2
-
-
220
546
1148
(M.U, B.E., 1997)
Sol : There being given five values and two missing figures, we may have
∴
∆5 y 0 = 0 and ∆6 y 0 = 0
Let y 0 = − 4, y1 = −2, y 4 = 220, y 5 = 546, y 6 = 1148 and we have to find out y 2 , y 3 .
∆5 y 0 = 0 and ∆6 y 0 = 0
i.e.,
E 5 y 0 − 5E 4 y 0 + 10E 3 y 0 − 10E 2 y 0 + 5Ey 0 − y 0 = 0
or
y 5 − 5 y 4 + 10 y 3 − 10 y 2 + 5 y1 − y 0 = 0
Substituting the values, we get
546 − 5( 220) + 10 y 3 − 10 y 2 + 5 ( −2) − ( −4) = 0
or
10 y 3 − 10 y 2 − 560 = 0
or
y 3 − y 2 = 56
Again
∆6 y 3 = ( E − 1) 6 y 0 = 0
i.e.,
E 6 y 0 − 6E 5 y 0 + 15E 4 y 0 − 20E 3 y 0 + 15E 2 y 0 6Ey 0 + y 0 = 0
or
or
or
y 6 − 6 y 5 + 15 y 4 − 20 y 3 + 15 y 2 − 6 y1 + y 0 = 0
1148 − 6( 546) + 15( 220) − 20 y 3 + 15 y 2 − 6 ( −2) − 4 = 0
−20 y 3 + 15 y 2 + 1180 = 0
...(1)
...(2)
608
Engineering Mathematics-III
Solving (14.1) and (14.2), we get
y 2 = 12 and y 3 = 68
14.3 Summation of Series
One of the very important applications of the calculus of finite differences is to
obtain a formula for summing a series of upto n terms. The method is best illustrated
by the following examples.
Ex.17: Find the sum to n terms of the series u 0 + u1 + u 2 + .... u n , where the general
term u x is the first difference of another function. Using this concept, find the sum to
n terms of the following series.
(i) 1.3.5.+2.4.6. +3.5.7+...
1
1
1
(ii)
...
+
+
2.3.4 3.4.5 4.5.6
Sol : It is given that the general term u x of the given series
u1 + u 2 + u 3 + ...+ u n is the first difference of another function.
Let u 0 = ∆y x = y x +1 − y x (∵ h = 1)
n
∴
∑ux
= u1 + u 2 + u 3 + ....+ u n
x =1
= ∆y1 + ∆y 2 + ∆y 3 +...+ ∆y n
= ( y 2 − y1 ) + ( y 3 − y 2 ) + ( y 4 − y 3 ) + ...+ ( y n +1 − y n )
= ( y n +1 − y1 ) = [ y x ]1n +1
[
= ∆ − 1u x
n +1
]
1
(∵ y x = ∆−1u x , where ∆−1 = 1/ ∆)
If the general term u x of the series can be expressed in factorial notation, then can
be found easily by integration and hence, the sum of the series.
(i) Consider the series 1.3.5 + 2.4.6 + 3.5.7 + ...
General term, u x ( x + 2)( x + 4) = x 3 + 6 x 3 + 8 x.
In factorial notation, u x = x ( 3) + 9 x ( 2) + 15 x (1) = ∆y x
1
9
15 ( 2)
∴
y x = ∆ − 1u x = x ( 4 ) + x ( 3 ) +
x { see 14.4 for factorial notation}
4
3
2
n
Hence
[
∑ u x = ∆ − 1u x
x =1
n +1
]
1
n +1
15 ( 2) 
1
=  x ( 4) + 3 x ( 3) +
x 
4
2

1
15
1

=  ( n + 1)( 4) + 3 ( n + 1)( 3) +
( n + 1)( 2) 
2
4

15 ( 2) 
1
−  (1)( 4) + 3 (1)( 3) +
(1) 
2
4

1
= ( n + 1)( n )( n − 1)( n − 2) + 3( n + 1)( n )( n − 1)
4
15
+ ( n + 1) n − 0
2
( n + 1) n 2
=
{ n + 9n + 20}
4
Finite Differences and Interpolation
609
n( n + 1)( n + 4)( n + 5)
4
(ii) Consider the series
1
1
1
=
+
+
+ ...
2.3.4 3.4.5 4.5.6
1
Here,
ux =
= x ( −3) (∵ h =1)
( x + 1)( x + 2)( x + 3)
1
But
u x = ∆y x
∴ y x = ∆ − 1u x = − x ( − 2 )
2
=
n
Hence,
[
∑ u x = ∆ − 1u x
x =1
n +1
]
1
n +1
 1

= − x ( −2) 
 2
1
n +1
1
 1

= −

 2 ( x + 1)( x + 2)1
1
1
1
=− 
−

2 ( n + 2)( n + 3) 2.3
1
1
1
=− 
− 
2 ( n + 2)( n + 3) 6
n
Ex. 18: Prove that
∑ u r = n C1u1 + n C 2 ∆u1 + n C 3 ∆2u1 +...+ ∆n −1u1 .
r =1
n
Hence or otherwise find the sum ∑ r 2 .
1
n
Sol :
∑ ur
= u1 + u 2 + u 3 + ...+ u n
r =1
=u1 + Eu1 + E 2u1 +... + E n −1u1
(
)
= 1 + E + E 2 +...+ E n −1 u1
n
n
 E − 1
(1 + ∆ ) − 1
=
u1 = 
u1
 E − 1 
 1 + ∆ − 1 
1
=
(1+ n C1 ∆ + n C 2 ∆2 +....+ ∆n ) − 1 u1
∆
= n C1u1 + n C 2 ∆u1 +...+ ∆n −1u1 ...(i)
[
]
To find the sum
n
∑ r2
= 1 2 + 2 2 + 3 2 +...+ n 2
r =1
Let us first construct the difference table for the coefficients of the series given in
Eqn (i) taking u1 = 1, u 2 = 4, u 3 = 9, u 4 = 16 and u 5 = 25.
610
Engineering Mathematics-III
∆ 2u
∆ 3u
u
∆u
1 = u1
3
4 = u2
5
2
9 = u3
7
2
0
16 = u 4
9
2
0
25 = u 5
Here, u1 = r 2 is a polynomial of second degree. Hence, the third and higher order
differences will be zero.
∴
∆u1 = 3, ∆2u 2 = 2
Hence from Eqn (i),
n
∑ r 2 = n C1 (1) + n C 2 (3) + n C 3 (2)
r =1
n ( n − 1)
n( n − 1)( n − 2)
( 3) +
( 2)
2!
3!
n ( 2n 2 + 3n + 1) n ( n + 1)( 2n + 1)
=
=
6
6
Ex. 19: Prove that
u
x∆u 0
x 2 ∆ 2u 0
u 0 + u1 x + u 2 x 2 +.... = 0 +
+
+ ...
1 − x (1 − x ) 2 (1 − x ) 3
= n+
and hence find the sum to infinity of the series 1 .2 + 2 .3 x + 3 .4 x 2 +.....
Sol : Consider u 0 + u1 x + u 2 x 2 + ......
= u 0 + xEu 0 + x 2 E 2u 0 +.....
= (1 + xE + x 2 E 2 +...... ) u 0
1


 1 
=
u0 = 
 u0

1
−
xE
1
−
x
(
1
+
∆
)




1
1


=
u0 =

x∆ 

1 − x − x∆ 
(1 + x )1 −
u0
(
1
− x )

−1
=
1 
x∆ 
1−
(1 − x )  1 − x 
=

1 
x∆
x 2 ∆2
+
+ .... u 0
1 +
(1 − x )  1 − x (1 − x ) 2

=
u0
x∆u 0
x 2 ∆ 2u 0
+
+
+....
1 − x (1 − x ) 2 (1 − x ) 3
= R.H.S
u0
...(1)
Finite Differences and Interpolation
611
To find the sum of the series
1 .2 + 2 .3 x + 3 .4 x 2 +....
Let us construct the difference table for the coefficients of the series given in Eqn
(14.1) by taking u 0 = 2, u1 = 6, u 2 = 12, u 3 = 20 and u 4 = 30.
∆2 u
∆3 u
u
∆u
2
4
6
6
2
12
8
2
0
20
10
2
0
30
Hence, from Eqn (i), the sum of the series
u0
x
x2
=
+
∆ u0 +
∆2 u 0
3
1 − x (1 − x ) 2
(1 − x )
=
=
u0
4x
2x 2
+
+
1 − x (1 − x ) 2 (1 − x ) 3
2(1 − x ) 2 + 4 x(1 − x ) + 2 x 2
(1 − x )
3
=
2
(1 − x ) 3
Ex. 20: Show that
u0 +


u1 x u 2 x 2
x2 2
+
+ ... = e x u 0 + x∆u 0 +
∆ u 0 +....
1!
2!
2!


and hence sum to infinity the series
23
33 2 43 3
1+
x+
x +
x +...
1!
2!
3!
u x u x2
Sol : Consider u 0 + 1 + 2
+ ...
1!
2!
x
x2 2
= u 0 + Eu 0 +
E u 0 + ....
1!
2!


xE x 2 E 2
= 1 +
+
+... u 0
1!
2!


= ( e xE )u 0 = [ e e(1+ ∆ ) ]u 0 = e x . e x∆ u 0


x 2 ∆2 x 3 ∆3
= e x 1 + x∆ +
+
+.... u 0
2!
3!



x2 2
x3 3
= e x u 0 + x ∆ u 0 +
∆ u0 +
∆ u 0 +....
2!
3!

= R.H.S
...(1)
612
Engineering Mathematics-III
To find the sum of the series
23
33 2 43 3
1+
x+
x +
x +...
1
2!
3!
let us construct the difference table for the coefficients of the series given in Eqn
(14.1) by taking u 0 = 1, u1 = 8, u 2 = 27, u 3 = 64, u 4 = 125, and u 5 = 216.
∆u
∆ 2u
∆ 3u
8
7
12
6
27
19
18
6
0
64
37
24
6
0
125
61
30
216
91
u
∆ 4u
1
From the table, u 0 = 1, ∆u 0 =7, ∆2u 0 = 12, ∆3u 0 = 6, ∆4u 0 = 0.
Hence, from Eqn (14.1), sum of the series is

x2 2
x3 3 
= e x u 0 + x ∆ u 0 +
∆ u0 +
∆ u0
2!
3!


2
3

12 x
6x 
= e x 1 + 7 x +
+

2!
3 ! 

= e x (1 + 7 x + 6 x 2 + x 3 )
14.4 Factorial Notation or Factorial Polynomial
Consider the continued product,
x( x − h )( x − 2h )...[ x − ( r − 1)h]
containing r factors of which x is the first one and the successsive factors are
decreased by a constant difference h. This is known as factorial polynomial and is
denoted x ( r ) .
∴
x ( r ) = x( x − h )( x − 2h )...[ x − ( r − 1)h]
Hence,
x (1) = x, x ( 2) = x( x − h ), x ( 3) = x( x − h )( x − 3h ) and so on.
Differences of x ( r )
∆ x ( r ) = [ x + h]( r ) − x r
= ( x + h ) x ( x − h )...[ x − ( r − 2)h] − x( x − h )...[ x − ( r − 1)h]
= x( x − h )...[ x − ( r − 1)h]{ x + h − x − ( r + 1)h}
= x ( r −1) rh = rhx ( r −1)
[
] [
Similarly,∆2 x ( 2) = ∆ ∆x ( r −1) = ∆ rh x ( r −1)
]
= rh ( r − 1) h x ( r − 2) = r ( r − 1)h 2 x ( r − 2)
Finite Differences and Interpolation
613
∆3 x ( r ) = r ( r − 1)( r − 2) h 3 x ( r − 2)
Proceeding on, we get
∆r x ( r ) = r ( r − 1)( r − 2)...1. h r x ( r − r ) = r ! h r
Note:
1. The result of differencing x ( r ) is analogous to that of differentiating x r .
2. In particular, if h=1 then
∆ x ( r ) = r x ( r −1) , ∆2 x ( r ) = r( r − 1) x ( r − 2) ,..., ∆r x ( r ) = r !
14.5 Reciprocal Factorial
The reciprocal factorial function, x ( − r ) , is defined as
1
x(− r ) =
( x + h )( x + 2h )...( x + rh )
where r is a positive integer.
Differences of x ( − r )
(i)
∆ x ( − r ) = ( x + h )( − r ) − x ( − r )
1
1
=
−
( x + 2h )( x + 3h )...[ x + ( r + h )h] ( x + h )( x + 2h )...( x + rh )
( x + h ) − [ x + ( r + 1)h]
=
( x + h )( x + 2h )...[ x + ( r + 1)h]
− rh
=
= ( − r ) h x[ −( r +1)]
( x + h )( x + 2h )...[ x + ( r + 1)h]
(ii)
∆2 x ( − r ) = ∆ ( ∆ x ( − r ) ) = ∆ [ − rhx[ −( r +1)] ]
= ( − r ) h [ −( r + 1)] h x[ −( r + 2)]
= ( −1) 2 r( r + 1) h 2 x[ −( r + 2)]
Proceeding on, we get
∆k x ( − r ) = ( −1) k r( r + 1)...( r + k − 1) x[ −( r + k )]
14.6 Polynomial in Factorial Notation
Let f ( x ) = a0 + a1 x + a2 x 2 + a3 x 3 +...+ an x n be the polynomial of nth degree to be
expressed in factorial notation. Since x ( n ) , x ( n −1) , x ( n − 2) etc., are respectively,
polynomial of nth degree, ( n − 1)th degree, ( n − 2)th degree etc. we can express f (x)
as
....(14.2)
f ( x ) = A 0 + A1 x (1) + A 2 x ( 2) + A 3 x ( 3) +...+ A n x ( n )
where A 0 , A1 , A 2 , A 3 ..., A n , are to be determined.
Now,
∆f ( x ) = A1 + 2 A 2 x (1) + 3 A 3 x ( 2) +...+ n A n x ( n −1)
∆2 f ( x ) = 2 A 2 + 6 A 3 x (1) +...+ n ( n − 1) A n x ( n − 2)
∆3 f ( x ) = 6 A 3 +...+ n ( n − 1)( n − 2) A n x ( n − 3)
L
L
L
∆ f ( x ) = n ( n − 1)( n − 2)...2 .1. A n x ( 0) = A n ( n !)
n
614
Engineering Mathematics-III
Putting x=0 in the above equations, we get
f ( 0) = A 0 ; ∆f ( 0) = A1 ; ∆2 f ( 0) = 2 A 2 ⇒ A 2 =
1 2
∆ f ( 0)
2!
1 3
∆ f ( 0);
3!
L L L L L
∆n f ( 0) = A n ( n !); ⇒ A n = 1/ 3 ! ∆n f ( 0)
∆3 f ( 0) = 6 A 3 ⇒ A 3
Substituting the above values in Eqn (14.2), we get
1
1
f ( x ) = f ( 0) + ∆f ( 0) x (1) + ∆2 f ( 0) x ( 2) + ∆3 f ( 0) x ( 3)
2!
3!
1 n
+ L + ∆ f ( 0) x ( n )
n!
Thus, any polynomial of degree n can be expressed as a factorial polynomial of the
same degree and vice versa.
Note : We have some other methods to express a polynomial in factorial notation.
They are best illustrated through examples.
Ex. 21: Represent the function f ( x ) = x 4 − 12 x 3 + 42 x 2 − 30 x + 9 and its successive
differences in factorial notation in which the interval of differencing is one.
(M.U, B.E., 1997)
Sol :
Method 1: The values of f ( x ) at x = 0, 1, 2, 3, and 4 are 9,10, 37, 54 and 49. The
forward difference table is as follows.
∆2 f ( x )
∆3 f ( x )
∆4 f ( x )
27
26 = ∆2 f( 0)
−36 = ∆3 f( 0)
24 = ∆4 f( 0)
37
17
−10
−12
3
54
−5
−22
4
49
x
f ( x)
∆f ( x )
0
9 = f( 0)
1 = ∆f( 0)
1
10
2
Therefore,
f ( x ) = x 4 − 12 x 3 + 42 x 2 − 30 x + 9
in factorial form is
∆f ( 0) (1) ∆2 f ( 0) ( 2) ∆3 f ( 0) ( 3) ∆4 f ( 0) ( 4)
x +
x +
x +
x
1!
2!
3!
4!
1
26 ( 2) −36 ( 3) 24 ( 4)
= 9 + x (1) +
x +
x
+
x
1!
2!
3!
4!
= 9 + x (1) + 13 x ( 2) − 6 x ( 3) + x ( 4)
f ( x ) = f ( 0) +
Now,
∆f ( x ) = 1 + 26 x (1) − 18 x ( 2) + 4 x ( 3)
∆2 f ( x ) = 26 − 36 x (1) + 12 x ( 2)
Finite Differences and Interpolation
615
∆3 f ( x ) = −36 + 24 x (1)
∆4 f ( x ) = 24
Method 2 : Let f ( x ) = x 4 − 12 x 3 + 42 x 2 − 30 x + 9
= ax ( x − 1)( x − 2)( x − 3) + bx ( x − 1)( x − 2) + cx ( x − 1) + dx + e
...(1)
Putting x=0 in Eqn (1), we get e=9.
Putting x=1 in Eqn (1), we get d+e=10. ∴ d = 1.
Putting x=2 in Eqn (1), we get 2c+2d+e=37 ∴ c = 13
Now comparing the coefficients of x 4 and x 3 on both sides of Eqn (1), we get
a = 1; − 6a + b = −12 ∴ b = −6
Hence, Eqn (1) becomes
f ( x ) = x ( 4) − 6 x ( 3) + 13 x ( 2) + x (1) + 9
Method 3 : (Synthetic division method) : Dividing the given polynomial
successively by x, x − 1, x − 2 and x − 3, we have
Now the required equation is
y x = x ( 4) − 6 x ( 3) + 13 x ( 2) + x (1) + 9.
Note : Unless otherwise stated, the interval of differencing may be taken as 1.
Ex.22: Find the second difference of the polynomial
x 4 − 12 x 3 + 42 x 2 − 30 x + 9
with interval of differencing h=2.
Sol :
Synthetic division method : Dividing the given polynomial successively by
x, x − 2, x − 4 and x − 6, we have
0
2
4
6
∴
1
–12
42
–30
9
0
0
0
0
0
1
–12
42
–30
9
0
2
–20
44
1
–10
22
14
0
4
–24
1
–6
–2
0
6
1
0
f ( x ) = x 4 − 12 x 3 + 42 x 2 − 30 x + 9
In factorial notation
616
Engineering Mathematics-III
f ( x ) = x ( 4) − 2 x ( 2) + 14 x (1) + 9
∆f ( x ) = 4 .2 x ( 3) − 2 .2 .2 x (1) + 14 .2 .1
(using ∆x r = rh x ( r −1) )
∆2 f ( x ) = 4( 2) 2 3 x ( 2) − 2 .2( 2) 2
∆2 f ( x ) = 48 x ( 2) – 16
14.7 Interpolation
There are many practical situation in our daily life, when it is required to find out the
value of the function at some intermediate stage. For example, if we have a table
concern with the sales of product in some particular years and we want to find out
the sales of any other year. This process is known as interpolation. Some important
interpolation results are given below.
14.7.1 Newton's Forward Interpolation Formula
Consider we have a set of (n+1) equidistant values, i.e. ( x 0 , f 0 ), ( x1 , f1 ) ( x 2 , f 2 )...
( x n , f n ). Now we shall find a polynomial Pn ( x ) of degree n such that Pn ( x ) agree at
the given set of point i.e. Pn ( xi ) = fi .
Since, Pn ( x ) is an approximate polynomial Pn ( x ) ≈ f ( x ) of degree n, therefore it can
be written as
Pn ( x ) = A 0 + A1 ( x − x 0 )+ A 2 ( x − x 0 )( x − x1 )+ A 3 ( x − x 0 )( x − x1 )( x − x 2 )
..........A n ( x − x 0 )( x − x1 ) … ( x − x n −1 )
...(14.3)
Now it is given that when x = x 0 , Pn ( x ) = f 0 = f ( x 0 ), we get A 0 = f 0 .
Similarly, when x = x1 , Pn ( x1 ) = f1 = f ( x1 ), thus we get
f − f0
∆ f0
f1 = f 0 + A1 ( x1 − x 0 )or A1 = 1
=
{ ∵ x1 = x 0 + h}
x1 − x 0
h
Similarly,
A2 =
∆ 2 f0
|—
2 h2
,... A n =
∆ n f0
|—
n hn
Since we have
x1 = x 0 + h, x 2 = x 0 + 2h, ... x n = x 0 + nh
keeping in mind all the above results, we have
Pn ( x ) = f 0 +( x − x 0 ).
∆f 0
∆ 2 f0
+ ( x − x 0 ) ( x − x 0 − h ).
+ ...
h
|2 h 2
—
n
...
∆ f0
|—
n .h n
.( x − x 0 )( x − x 0 − h )( x − x 0 − 2h ) + ... ( x − x 0 − n − 1 h )...(14.2)
x − x0
the expression (14.3) can be written as,
h
u (u − 1)(u − 2) … (u − n − 1) n
u(u − 1) 2
Pn ( x 0 + uh ) = f 0 + u. ∆f 0 +
∆ f 0 + ...
∆ f0
n
|—
2
|—
It we take u =
...(14.3)
Finite Differences and Interpolation
617
 ( x − x 0 ) ( x − x 0 − h)  x − x 0   x − x 0


=
− 1 = u(u − 1)
 
∵
2




h
h
h


Note : Newton forward interpolation formula is used when x is lying in the upper half
of the table.
Ex. 23 : Estimate the population in year 1985, if the population table is given below :
Year
1891
1901
1911
1921
1931
46
66
81
93
101
Population (in thousand)
Sol : The corresponding difference table is given below
x
Here
y
x = 1985, h = 10, x 0 = 1981 so u =
1985 − 1981
= 0 .4
10
By Newton-Forward formula, we have
f (1985) ≈ 46+ 0 .4 × 20+
0 .4 ( 0 .4 − 1)
0 .4 ( 0 .4 − 1)( 0 .4 − 2)
× ( − 5) +
× ( 2)
|—
2
|—
3
+
0 .4 ( 0 .4 − 1)( 0 .4 − 2)( 0 .4 − 3)
( −3)
|—4
On solving, we get f(1985) = 54 .85 thousands
Ex. 24 : Find f(x), which takes the following values
f(0) = 1,
f(1) = 0,
f(2) = 1,
f(3) = 10
Sol : The corresponding difference table is given below :
618
Here
Engineering Mathematics-III
x−0
= x, so
1
x ( x − 1)
x( x − 1)( x − 2)
f( x ) = 1+ x ( −1)+
× 2+
6 = x 3 − 2x 2 − 1
|—
2
|—
3
h = 1, x 0 = 0, so u =
Ex. 25 : Find the number of men getting the wages Rs. 10 and Rs. 15 from the
following table
Wages
0 – 10
10 – 20
20 – 30
30 – 40
9
30
35
42
Frequency
Sol : The given table can be written as
Wages in Rs.
less than 10
Number of men
9
,,
,, 20
9 + 30 = 39
,,
,, 30
39 + 35 = 74
,,
,, 40
74 + 42 = 116
It is clear that number of men getting wages between Rs. 10 and Rs. 15
= number of men getting wages below Rs. 15 – number of men getting wages
Below 10
= f (15) − f (10)
...(1)
= f(15) − 9
The corresponding difference table is
Finite Differences and Interpolation
619
5
= 0 .5, so
10
0 .5 ( 0 .5 − 1)
0 .5 ( 0 .5 − 1) ( 0 .5 − 2)
f(15) = 9+ 0.5 × 30+
× 5+
×2
2
6
= 24 (app)
...(2)
By (1) & (2), the number of men getting wages between Rs. 10 and Rs. 15 is
24 − 9 = 15 .
Here x 0 = 10, x = 15, h = 10, so u =
14.7.2 Newton Backward-Interpolation Formula
When x is lying on the lower half of the table, it is better to assume Pn ( x ) in the
following form
Pn( x )= A0 + A1( x − xn)+ A2( x − xn)( x − xn−1 ) + a3( x − xn)( x − xn−1 ) ( x − xn−2)
....(14.5)
...... A n ( x − x n ) ( x − x n −1 )....( x − x1 )
To compute the value of A 0 , A1… A n , we impose the given conditions i.e.
when x = x n , Pn ( x ) = f ( x n ) , we get A 0 = f n .
Similarly, when x = x n −1 , Pn ( x ) = f n −1 = f ( x n −1 ) , thus we get
∇f
f n −1 = f n + A1 ( x n −1 − x n )or A1 = n
{∵ x n − x n −1 = h}
h
Again, we get A 2 =
∇ 2 fn
|2
h2
—
,... A n =
∇ n fn
|n
.h n
—
.
{ f n − f n −1 = ∇f n
x − xn
or x = x n +uh
h
u (u +1) 2
u(u +1)(u + 2)... u + n − 1 n
Pn ( x ) = f n +u. ∇f n +
. ∇ f n +....
∇ f n ...(14.6)
|2
|n
—
—
Putting the value of A 0 , A1… A n and u =
Ex. 26 : Find the value of sin 54° it is given that
x
30°
35°
40°
45°
50°
55°
sin x
0 .5000
0 .5736
0 .6428
0 .7071
0 .7660
0 .8192
Sol : Since x = 54° is lying in the lower half. therefore we apply Newton's Backward
interpolation formula. The corresponding difference table is given below
620
Engineering Mathematics-III
x − xn
54 − 55
=
= − 0 .2
h
5
By Newton backward interpolation formula, we have
( − 0 .2)( − 0 .2 − 1)
sin 54° = f (54° ) = 0 .8192+(– 0 .2) × (0 .0532) +
.( − 0 .0057 )
|2
—
Here
x = 54°, x n = 55°, h= 5°, so u =
+
+
+
( − 0 .2) ( − 0 .2 − 1)( − 0 .2 − 2)
.( − 0 .0004)
|3
—
( − 0 .2)( − 0 .2 − 1)( − 0 .2 − 2)( − 0 .2 − 3)
× ( − 0 .0001)
|4
—
( − 0 .2) ( − 0 .2 − 1)( − 0 .2 − 2)( − 0 .2 − 3)( − 0 .2 − 4
⋅ ( − 0 .0001)
|5
—
≈ . 809016994
14.8 Central Difference Interpolation Formulae
In this section, we shall discuss the central difference formulae, which is suitable for
interpolation near middle of the table. Some important central difference formulae
are given below.
14.8.1 Gauss's Forward central difference formulae
To construct Gauss's forward formula, the central ordinate value is taken as ( x 0 , f 0 )
and we make the following difference table.
Finite Differences and Interpolation
621
Table : 3.1
The above table show the differences (∆f 0 , ∆ 2 f −1 , ∆ 3 f −1 , ∆ 4 f −2 , ∆ 5 f −2 , ∆ 6 f −3 .... )
used in this formula, therefore, we have
Pn( x )= f0 + A1∆f0 + A2∆ 2 f−1 + A3∆ 3 f−1 + A4∆ 4 f−2 + A5∆ 5 f−2 + A6. ∆ 6 f−3
2
3
...(14.7)
4
we can also express ∆ f −1 , ∆ f −1 , ∆ f −2 ... in the following way. i.e.
∆ 2 f −1 = ∆ 2 E −1 f 0 = ∆ 2 (1+ ∆) −1 f 0 = ∆ 2 (1 − ∆ + ∆ 2 − ∆ 3 ... ) f 0
= ∆ 2 f 0 − ∆ 3 f 0 + ∆ 4 f 0 − ∆ 5 f 0+…
Now, ∆ 3 f −1 = ∆ 3 f 0 − ∆ 4 f 0 + ∆ 5 f 0 − ∆ 6 f 0…
Similarly, ∆ 4 f −2 = ∆ 4 E −2 . f 0 = ∆ 4 (1+ ∆ ) −2 . f 0 = ∆ 4 (1 − 2∆ + 3∆ 2 − 4∆ 3 ... ) f 0
= ∆ 4 f 0 − 2∆ 5 f 0 + 3∆ 6 f 0 − 4∆ 7 f 0+...
Now (14.7) becomes, as
Pn( x )= f0 + A1∆f0 + A2( ∆ 2 f0 − ∆ 3 f0 + ∆ 4 f0. . . )+ A3 ( ∆ 3 f0 − ∆ 4 f0 + ∆ 5 f0 − ∆ 6 f0. . . )
+ A 4 ( ∆ 4 f 0 − 2∆ 5 f 0 + 3∆ 6 f 0 − 4∆ 7 f 0 ... )+...
...(14.8)
On the other side, Newton forward difference formula is
u(u − 1) 2
u(u − 1)(u − 2)
Pn ( x ) = f 0 + u ∆f 0 +
. ∆ f0 +
.
|—
|—3
2
∆ 3 f 0 ++
u(u − 1)(u − 2)(u − 3) 4
. ∆ f0 +
|—4
...(14.9)
Now on comparing (14.8) with (14.9), we get
A1 = u,
A2 =
u (u − 1)
(u + 1)u (u − 1)
(u + 1) u (u − 1) (u − 2)
, A3 =
, A4 =
, etc
|2
|3
|4
—
—
—
622
Engineering Mathematics-III
Putting the value of A1 , A 2 , A 3 ... in (14.7), we get
Pn( x ) ≈ f 0 + u. ∆f 0 +
+
u (u − 1) 2
(u +1) u (u − 1) 3
(u +1) u (u − 1)(u − 2) 4
. ∆ f −1 +
. ∆ f −1 +
∆ f −2
|2
|3
|4
—
—
—
(u + 2)(u +1) u (u − 1)(u − 2) 5
(u + 2)(u +1) u(u − 1)(u − 2)(u − 3) 6
∆ f −3
∆ f 3 +…
|5
|6
—
—
The above expression is known as Gauss-Forward difference formula.
14.8.2 Gauss's Backward Difference Interpolation Formula
The table 3.1 shows the difference ∆f −1 , ∆ 2 f −1 , ∆ 3 f −2 , ∆ 4 f −2 , ∆ 5 f −3 ... ) which is
used to construct Gauss backward formula. thus we have
Pn ( x ) = f 0 + A'1 ∆f −1 + A' 2 ∆ 2 f −1 + A' 3 ∆ 3 f −2 + A' 4 ∆ 4 f −2 + A' 5 ∆ 5 f −3 +
Now to determine the value of A'1 , A' 2 , A' 3 ..., we follow the same procedure
as in case of gauss forward difference formula. In this way, we get
u(u +1) 2
(u − 1) u (u +1) 3
Pn ( x ) = f 0 + u. ∆f −1 +
. ∆ f −1 +
∆ f −2
|2
|3
—
—
+
+
(u − 1) u (u +1)(u + 2) 4
. ∆ f −2
|4
—
(u − 2)(u − 1) u (u +1)(u + 2) 5
(u − 2)(u − 1) u (u +1)(u + 2)(u + 3) 6
. ∆ f−3 +
.∆ f−3 + …
|5
|6
—
—
...(14.10)
The above expression is known as Gauss backward difference interpolation formula.
Stirling's Formula
Stirling's formula is the mean of Gauss's forward and backward interpolation
formulae, therefore sterling's formula is given, as
Pn ( x ) = f 0 + u.
∆f 0 + ∆f −1 u 2
u(u 2 − 1) ∆ 3 f −1 + ∆ 3 f −2
+
. ∆ 2 f −1 +
.
2
2
|3
2
—
+
u 2 (u 2 − 1) 4
. ∆ f2 +
|4
—
...(14.11)
14.8.3 Bessel's Interpolation Formula
In Bessel's formula, the difference are used in a different way as shown in the table
3.2 By using this assumption, we may write
Pn( x ) =
f0 + f1
∆ 2 f0 + ∆ 2 f−1
∆ 4 f−2 + ∆ 4 f = 1
+ A1. ∆f0 + A2
+ A3. ∆ 3 f−1 + A4 .
+
2
2
2
...(14.12)
Finite Differences and Interpolation
623
Table : 3.2
Now to determine the value of constant term A1 , A 2 ..., we follow the same method
as in case of Gauss's Formula, thus we get,
1
u(u − 1)
A1 = u − , A2 =
, A3 =
2
|2
—
1

u(u − 1) u − 

(u + 1) u (u − 1) (u − 2)
2
, A4 =
|3
|4
—
—
Now putting the value of A1 , A 2 , A 3 ..., Bessel's interpolation formula is written, as
1

u(u − 1)u − 
2
2

u (u − 1) ∆ f −1 + ∆ f 0
2 3
Pn ( x ) = f 0 + u. ∆f 0 +
.
+
∆ f −1
|2
2
|3
—
—
+
(u +1) u (u − 1)(u − 2) ∆ 4 f −2 + ∆ 4 f −1
.
|4
2
—
...(14.13)
14.8.4 Everett's Interpolation Formula
In this formula only even order differences ( ∆ 2 y −1 , ∆ 4 y −2, ∆ 6 y −3 ... ) are used,
therefore Evertt's formula can be written as
Pn ( x ) = A 0 f 0 + A 2 ∆ 2 f −2 + A 4 ∆ 4 f −2 + A 6 . ∆ 6 f −3+...
+ A' 0 f1 + A' 2 ∆ 2 f 0 + A' 4 ∆ 4 f −1 + A' 6 . ∆ 6 F− 2 +...
+ A" 0 f 2 + A" 2 ∆ 2 f1 + A" 4 ∆ 4 f 0 + A' 6 . ∆ 6 f −1+…
Now to determine the value of A 0 , A' 0 , A" 0 ..., A1 , A'1 , A"1 ..., we follow the
same procedure as in Gauss's formula. Thus, at last. We get
Pn ( x ) = v. f 0 +
+ uf1 +
v (v 2 − 12 ) 2
v ( v 2 − 1 2 ).( v 2 − 2 2 ) 4
. ∆ f −1 +
. ∆ f −2 +…
|3
|5
—
—
u (u 2 − 1 2 ) 2
u (u 2 − 1 2 ).(u 2 − 2 2 ) 4
. ∆ f0 +
. ∆ f −1 +
|3
|5
—
—
Here u + v = 1.
...(14.14)
624
Engineering Mathematics-III
Ex. 27 : Find a polynomial, which takes the values as given in the table given below
by using Gauss's forward difference formula.
x
1
2
3
4
5
y
1
–1
1
–1
–1
Sol : Here h = 1, f 2 = 1, f −1 = −1, f 0 = 1, f1 = −1, f 2 = +1,
x−3
If we take n = 3 as origin u =
= x−3
1
Now, the corresponding difference table is given below
∆2 f
x
f
∆f
–2
1
1
–2
–1
2
–1
0
3
1
1
4
–1
2
5
1
2
–2
4
–4
∆3 f
∆4 f
–8
8
16
4
2
From the above table we have ∆f 0 = −2, ∆ 2 f −1 = − 4, ∆ 4 f −2 = 16 Putting all these
value in Gauss-forward difference formula, we have
u(u − 1) 2
(u +1) u (u − 1) 3
Pn ( x ) = f 0 + u . ∆f 0 +
. ∆ f −1 +
. ∆ f −1
|2
|3
—
—
+
(u +1) u (u − 1)(u − 2) 4
∆ f −2 ...
|4
—
= 1 + u .( − 2) +
+
u (u − 1)
(u +1) u (u − 1)
.( − 4) +
×8
|2
|3
—
—
(u +1) u (u − 1) (u − 2)
× 16.
|4
—
= 1 − 2 u − 2 u (u − 1)+
4 4

= 1 + u  −2+ 2 − + 

3 3
8
2
= 1 − u2 + u4
3
3
4 u (u 2 − 1) 2 (u 2
+
3
2  4

+ u 2  −2 −  +  −

3  3
− 1)(u 2 − 2u )
3
4 3 2 4
u + u
3
3
Finite Differences and Interpolation
=
625
1
[( x − 3) 4 − 8( x − 3) 2 + 3]
3
{ ∵ u = x − 3}
Ex. 28 : Find the value of e1.17 by using Gauss-forward difference formula if it is
given that
x
1 .00
1 .05
1 .10
1 .15
1 .20
1 .25
1 .30
ex
2 .7183
2 .8517
3 .0042
3 .1582
3 .3201
3 .4903
3 .6693
Sol : Here x = 1 .17, x 0 = 1 .15, h = 0 .05, so u =
1 .17 − 1 .15
= 0 .4,
. 05
The corresponding difference table is given below.
ex
x
If, we approximate Gauss forward formula, upto second order, then
u (u − 1) 2
Pn ( x ) ≈ f 0 + u . ∆f 0 +
. ∆ f −1
|2
—
= 3 .1582 + 0 .4 × 0 .1619 +
0 .4 ( − 0 .4 − 1)
× 0 .0079
2
≈ 3 .2220
Ex. 29 : If it is given that f ( 25) = 0 .2707, f ( 30) = 0 .3027, f ( 35) = 0 .3386,
f(40) = 0.3794, find the value f(32). by Gauss-forward difference method
Sol : If we select x 0 = 30, then f −1 = 0.2707, f 0 = 0 .3027, f1 = 0 .3386,
x − x0
32 − 30
f 2 = 0 .3794 Here h = 5, x = 32, so u =
=
= 0 .4
h
5
x
f ( x)
–1
25
0 .2707
0
30
0 .3027
1
35
0 .3386
2
40
0 .3794
∆
∆2
∆3
0 .0320
0 .0359
0 .0408
. 0039
. 0049
. 0010
626
Engineering Mathematics-III
From the above table, we have ∆f −1 = 0 .0359, ∆ 2 f −1 = . 0039, ∆ 3 f −2 = . 0010
putting all these values in the Gauss's forward formula, we get,
f(32) =. 3165
Ex 30 : Find the sales in the year 1946, if it is given that
Year
1911
1921
1931
1941
1951
1961
12
15
20
27
39
52
Sales in thousand
Sol : Here x = 1946, which is near to bottom, so use Gauss' backward difference
formula, select x 0 = 1941, we get u = 0 .5 . the corresponding difference is given
below
x
f( x )
–3
1911
12
–2
1921
15
–1
1931
20
0
1941
27
1
1951
39
2
1961
52
∆f
∆2 f
∆3 f
∆4 f
∆5 f
3
5
7
12
13
2
2
5
0
3
–4
3
– 10
–7
1
Putting
u = 0.5, f 0 = 27, ∆f 0 = 7, ∆ 2 f −1 = 5, ∆ 3 f −2 = 3, ∆ 4 f −2 = −7, ∆ 5 f −3 = −10, in the
Gauss backward formula
Pn ( x ) ≈ f 0 + u ∆f −1 +
+
u (u +1) 2
(u +1) u (u − 1) 3
. ∆ f −1 +
. ∆ f −2
|2
|3
—
—
(u+ 2)(u+1) u (u − 1) 4
(u+ 2)(u+1) u (u − 1) (u − 2) 5
. ∆ f −2 +
. ∆ f −3 ,
|4
|—5
—
We get f (1946) ≈ 32. 3437 thousand
Finite Differences and Interpolation
627
Ex.31 : Find tan 16° by using stirling formula, if is given that
θ°
0
5
10
15
20
25
30
tan θ
0
0 .0875
0 .1763
0 .2679
0 .3640
0 .4663
0 .5774
Sol : Here x = 16° , select x 0 = 15, h = 5, so u = 0.2 the corresponding difference
table is given below.
θ°
tan θ°
–3
0
0
–2
5
0 . 0875
–1
10
0 .1763
0
15
1
20
2
25
3
30
∆
∆2
∆3
∆4
∆5
∆6
0 . 0875
0 . 0013
0 . 888
0 . 0015
0 . 0028
0 . 916
0 . 2679
0 . 0002
0 . 0017
0 ⋅ 0045
0 . 961
0 . 3640
0 . 0000
0 . 0017
0 . 0062
0 .1023
0 . 4663
− 0 . 0002
0 . 0011
0 . 0009
0 . 0009
0 . 0026
0 . 0088
0 .1111
0 . 5774
Since the forth differences are almost equal to zero, therefore by approximating
sterling's formula upto forth order we get
tan 16° = f 0 +
+
u( ∆f 0 + ∆f −1 ) u 2
u (u 2 − 1 2 )  ∆ 3 f −1 + ∆ 3 f −2 
+
. ∆ 2 f −1 +


2
|2
|3
2


—
—
u 2 (u 2 − 1 2 ) 4
∆ f −2
|4
—
Putting
f 0 = 0 . 2679, ∆f 0 = 0 . 916, ∆f −1 = 0 . 916, ∆ 2 f −1 = 0 . 0045, ∆ 3 f −1 = 0 . 0017 = ∆ 3 f −2,
∆ 4 f −2 = 0 .0000. We get, tan 16° = 0 .2876
628
Engineering Mathematics-III
Ex. 32 : Find the value of e 0⋅644 by using central difference formulae from the table
given below
x
0 ⋅ 61
ex
0 ⋅ 62
0 ⋅ 63
0 ⋅ 64
0 ⋅ 65
0 ⋅ 66
0 ⋅ 67
1 ⋅ 840431 1 ⋅ 858928 1 ⋅ 877610 1 ⋅ 896481 1 ⋅ 915541 1 ⋅ 934792 1 ⋅ 954237
Sol : Here x = 0 ⋅ 644, select x 0 = 0 ⋅ 64, so u =
x − x0
= 0 ⋅ 4 (∵ h = 0 ⋅ 01) . the
h
corresponding difference table is given below
f 0 = 1 ⋅ 896481, ∆ f −1 = 0 ⋅ 019060, ∆ 2 f −1 = 0 ⋅ 000189, etc.
Here
Bessel's Formula gives
e 0⋅644 = 1 ⋅ 896481 + 0 ⋅ 4 ⋅ ( 0 ⋅ 019060) +
0 ⋅ 4 ( 0 ⋅ 4 − 1)
.0 ⋅ 000189
2
= 1 ⋅ 904082
Sterling's Formula gives
e 0⋅644 = 1 ⋅ 896481 + 0 ⋅ 4
0 ⋅ 018871 + 0 ⋅ 019060 0 ⋅ 16
+
( 0 ⋅ 000189)
2
2
= 1 ⋅ 904082
Everett's formula gives
Since
u = 0 ⋅ 4, so v = 0 ⋅ 6
e 0⋅ 644 = 0 ⋅ 6 (1 ⋅ 896481) +
{ ∵ u + v = 1}
0 ⋅ 6 ( 0 ⋅ 6 2 − 1)
( 0 ⋅ 000189)
2
Finite Differences and Interpolation
629
+ 0 ⋅ 4 (1 ⋅ 915541) +
0 ⋅ 4 (0 ⋅ 42 − 12 )
( 0 ⋅ 000191)
2
= 1 ⋅ 904082
Ex.33 : The amount A of a substance remaining lin a reacting system after an
interval of time t in a certain chemical experiment is tabulated below :
t (min)
2
5
8
11
A (gm)
94.8
87.9
81.3
75.1
Obtain the value of A where t=9 using Newton's backward interpolation formula.
Sol : Since the value t=9 is near the end of the table, to get the corresponding value
of t we use Newton's backward interpolation formula. The backward differences are
calculated and tabulated below.
Here,
∇2 A
∇3 A
–6.6
0.3
0.1
–6.2
0.4
t
A
∇A
2
94.8
–6.9
5
87.9
8
81.3
11
75.1
t n = 11, A n = 75 .1, ∇A n = −6 .2, ∇ 2 A n = 0 .4, ∇ 3 A n = 0 .1.
Hence, the interpolation polynomial will be of degree 3. That is,
p( p + 1) 2
p( p + 1)( P + 2) 3
A = A n + p ∇A n +
∇ An +
∇ An
2!
3!
Let A p be the value of A when t =9.
Then
∴
t − tn
9 − 11
2
=
=−
h
3
3
1  2  2 
 2
= 75 .1 +  −  ( −6 .2) +  −   − + 1 ( 0 .4)
 3
2 !  3  3 
p=
Ap
+
1  2  2   2

 −   − + 1  − + 2 ( 0 .1)

3 !  3  3   3
=75.1+4.1333333 – 0.0444444 – 4.9382716 ×10 −3
= 79.183951
Ex. 34: Find a cubic polynomial which takes the following set of values
( 0, 1),(1, 2),( 2, 1) and ( 3, 10).
(M.U, B.E., 1997)
630
Engineering Mathematics-III
Sol : The difference table is
∆2 f ( x ) ∆3 f ( x )
x
f ( x)
∆f ( x )
0
1
1
1
2
–1
–2
2
1
9
10
3
10
12
x − x0
x−0
=
= x
h
1
∴ Using Newton's forward interpolation forumulae, we get
x( x − 1) 2
x( x − 1)( x − 2) 3
f ( x ) = f ( 0) + x ∆f ( 0) +
∆ f ( 0) +
∆ f ( 0)
2!
3!
x( x − 1)
x( x − 1)( x − 2)
= 1 + x.1 +
( −2) +
(12)
2
6
= 2x 3 − 7 x 2 + 6x + 1
we take x 0 = 0 and p =
which is the required polynomial.
Ex. 35: The following data give I, the indicated HP and V, the speed in knots
developed by a ship.
V
8
10
12
14
16
I
1000
1900
3250
5400
8950
Find I when V =9, is near the beginning if the table. Hence, to get the corresponding
I, we use Newton's forward interpolation formula. The forward differences are
calculated and tabulated as below.
∆2 I
∆3 I
1350
450
350
3250
2150
800
600
14
5400
3550
1400
16
8950
V
I
∆I
8
1000
900
10
1900
12
∆4 I
250
Here, V0 = 8, I 0 = 1000, ∆I 0 = 900, ∆2 I 0 = 450, ∆3 I 0 = 350, ∆4 I 0 = 250.
Hence, the interpolation polynomial will be of degree 4. That is,
p( p − 1) 2
p( p − 1)( p − 2) 3
I = I0 + p∆ I0 +
∆ I0 +
∆ I0
2!
3!
p( p − 1)( p − 2)( p − 3) 4
+
∆ I0
4!
Let I p be the value of I when V=9
Finite Differences and Interpolation
631
V − V0
9− 8 1
=
= = 0 .5
h
2
2
0 .5( 0 .5 − 1)
∴
I p = 1000 + ( 0 .5)( 900) +
( 450)
2!
( 0 .5)( 0 .5 − 1)( 0 .5 − 2)
0 .5 ( 0 .5 − 1)( 0 .5 − 2)( 0 .5 − 3)
+
( 350) +
( 250)
3!
4!
=1000+450 –56.25+21.875 –9.765625
= 1405.8594
Ex. 36: From the following data, estimate the number of persons having income in
between (i) 1000 –1700 and (ii) 3500–4000.
Then
p=
Income
below 500
500–1000
1000–2000
2000–3000
3000–4000
No. of
persons
6000
4250
3600
1500
650
(M.U,B.E., 1991)
Sol : First we prepare the cumulative frequency table as follows ;
Person's income
less than
x:
1000
2000
3000
4000
No. of persons
y x:
10250
13850
15350
16000
Now the difference table is as follows:
∆2 y x
∆3 y x
1500
–2100
1250
650
–850
x
yx
∆y x
1000
10250
3600
2000
13850
3000
15350
4000
16000
(i) Here, we have to find the number of persons having income in between
1000–1700.
Taking x 0 = 1000, x = 1700, we have
x − x 0 1700 − 1000
p=
=
= 0 .7.
h
1000
Let
y p = y1700.
Using Newton's forward interpolation formula, we get
p( p − 1) 2
p( p − 1)( p − 2) 3
y p = y 0 + p∆y 0 +
∆ y0+
∆ y0
2!
3!
1
∴
y1700 = 10250 + ( 0 .7 )( 3600) + ( 0 .7 )( 0 .7 − 1)( −2100)
2!
1
= ( 0 .7 )( 0 .7 − 1)( 0 .7 − 2)(1250)
3!
632
Engineering Mathematics-III
=10250+2520+220.5+56.875
= 13047.375
∴ The number of persons having income less than 1700 is 13047.375, i.e. 13047
But the number of persons having income less than 1000 is 10250.
∴ The number of persons having income in between 1000–1700 is 13047
–10250 =2797.
(ii) Here, we have to find the number of persons having income in between
3500–4000.
Taking x n = 4000, x = 3500, we have
x − xn
3500 – 4000
p=
=
= −0 .5.
h
1000
Let
y p = y 3500 . Using Newton's backward interpolation formula, we get
p( p + 1) 2
p( p + 1)( p + 2) 3
y p = y n + p ∇y n +
∇ yn +
∇ yn
2!
3!
where y n =16000 + ∇y n = 650, ∇ 2 y n = − 850, ∇ 3 y n = 1250
1
∴
y 3500 = 16000 + ( −0 .5)( 650) + ( −0 .5)( −0 .5 + 1)( −850)
2!
1
+ ( −0 .5)( −0 .5 + 1)( −0 .5 + 2)(1250)
3!
= 1600 –325+106.25 – 78.125
= 15703.125
∴ The number of persons having income less than 3500 is 15703. 125, i.e. 15703.
But the number of persons having income less than 4000 is 16000.
∴ No. of persons whose income in between 3500–4000 is 16000–15703 =297.
Ex. 37: Find a polynomial which takes the following values
x
1
3
5
7
9
11
yx
3
14
19
21
23
28
and hence compute y x at x = 2 and x = 12.
Sol : The difference table is as follows :
x
yx
∆y x
∆2 y x
∆3 y x
∆4 y x
1
3
11
–6
3
14
5
–3
3
0
5
19
2
0
3
0
7
21
2
3
3
9
23
5
11
28
Finite Differences and Interpolation
633
x −1
2
Using Newton's forward interpolation formula, we get
p( p − 1) 2
p( p − 1)( p − 2) 3
y p = y 0 + p∆y 0 +
∆ y0+
∆ y0
2!
3!
1
1 1
1

= 3 + ( x − 1)(11) +
( x − 1)  ( x − 1) − 1 ( −6)
2
2! 2
2

1 1
1
1



+
( x − 1) ( x − 1) − 1  ( x − 1) − 2 ( 3)
3! 2
2
2

Take
x 0 = 1, y 0 = 3, p =
= 3 + 11/ 2( x − 1) − 3/ 4( x 2 − 4 x + 3) + 1/16 ( x 3 − 9 x 2 + 23 x − 15)
= 1/ 6 ( x 3 − 21 x 2 + 159 x − 91)
...(14.1)
Again take
x −11
.
2
Using Newton's backward interpolation formula,
p( p + 1) 2
p( p + 1)( p + 2) 3
y p = y n + p ∇y n +
∇ yn +
∆ yn
2!
3!
5
1 1
= 28 + ( x − 11) +
( x −11)( x − 9)( 3)
2
2! 22
1 1
+
( x −11)( x − 9)( x −7 )( 3)
3! 23
5
1
= 28 + ( x − 11) +
( x −11)( x − 9)( x − 1)
2
16
1
...(2)
=
( x 3 − 21 x 2 + 159 x − 91)
16
which is the same as Eqn (1) representing the data. So we can use any one of the
formula to find the polynomial.
1
∴
yn =
( x 3 − 21 x 2 + 159 x − 91)
16
1 3
Now
yn =
( 2 − 21 .2 2 + 159 .2 − 91) = 9 .4375
16
1
and
y12 =
[(12) 3 − 21(12) 2 + 159(12) − 91] = 32 .5625
16
x n = 11, y n = 28, p =
10
10
Ex. 38: Given ∑ f ( x ) = 500426, ∑ f ( x ) = 329240
1
4
10
∑ f ( x ) = 175212 and f (10) = 40365, find f (2).
7
(Karnataka, B.E., 1989)
Sol : Here, we are given the cumulative function f(x) and
10
F(1) =
∑ f ( x ) = 500426,
1
10
F( 4) =
∑ f ( x ) = 329240,
4
634
Engineering Mathematics-III
10
F(7 ) =
∑ f ( x ) = 175212,
and F(10) = 40365.
7
Here is the difference table.
∆F( x )
∆2 F ( x )
∆3 F ( x )
329240
–171186
17158
2023
7
175212
–154028
19181
10
40365
–134847
x
F( x )
1
500426
4
10
Now we shall find F( 2) =
∑ f ( x ).
2
Taking x 0 + ph = 2, we get
x − 2 1− 2 1
p= 0
=
=
h
3
3
∴ By Newton's forward interpolation formula, we get
1
F( 2) = F(1) + p∆F(1) + p( p − 1) ∆2 F(1)
2!
1
+ p( p − 1)( p − 2)∆3 F(1)
3!
=5000426+1/3 (–171186)+1/2 (1/3) (–2/3) (17158)
+1/6(1/3)(–2/3)(–5/3)(2023)
=500426–57062–1906.4444+124.8765
=441582.432
10
Hence,
f ( 2) = F(1) − F( 2) =
10
∑ f ( x) −∑ f ( x)
1
2
=500426–441582.432
=58843.568
Ex. 39: Given sin 45° = 0 .7071, sin 50° = 0 .8192 and sin 60° = 0 .8660, find sin 52°
using Newton's interpolation formula. Estimate the error.
(Punjab B.E., 1987)
Sol : Let y = sin x be the function. We construct the following difference table:
∆2 y
x
y = sin x
∆y
45°
0.7071
0.0589
50°
0.7660
0.0532
–0.0057
55°
0.8192
0.0468
–0.0064
60°
0.8660
∆ 3y
–0.0007
Finite Differences and Interpolation
Here,
and
635
x 0 = 45, y 0 = 0. 7071, ∆y 0 = 0 .0589 ∆2 y 0 = −0 .0057
∆3 y 0 =− 0 .0007
Using Newton's forward interpolation formula,
1
1
y = y 0 + p∆y 0 + p( p − 1) ∆2 y 0 + p( p − 1)( p − 2) ∆3 y 0
2!
3!
x − x0
where
p=
. Let y p be the value of y at x = 52°
h
∴
p = ( 52 − 45)/ 5= 7 / 5 = 1 .4
1
∴
y 52 = 0 .7071 + (1 .4)( 0 .0589) + (1 .4)(1 .4 − 1)( −0 .0057 )
2
1
+ (1 .4)(1 .4 − 1)(1 .4 − 2)( −0 .0007 )
6
= 0 .7071 + 0 .08246 − 0 .001596 + 0 .0000392
= 0 .7880032
∴
sin 52° = 0 .7880032
p( p − 1)…( p − n ) n +1
Error
=
∆
y (c )
3!
1 .4(1 .4 − 1)(1 .4 − 2) 3
=
∆ y ( c ) [by taking n=2]
6
1 .4( 0 .4)( −0 .6)
=
( − 0 .0007 ) = 0 .0000392
6
Ex. 40: The following table gives the values of density of saturated water for
various temperatures of saturated steam.
Temp°
C(=T)
100
150
200
250
300
Density hg/
m 3 (=d)
958
917
865
799
712
Find by interpolation, the densities when the temperatures are 130° C and 275° is
near the end of the table. So for the former, we use Newton's Backward interpolation
formula. The difference table is as given below:
∆d
∆2 d
917
–41
–11
200
865
–52
250
799
–66
300
712
–87
T
d
100
958
150
∆3 d
∆4 d
–14
–3
–4
–21
–7
Here T0 = 100, d0 = 958, ∆d0 = − 41, ∆2 d0 − 11, ∆3 d0 = − 3 and ∆4 d0 = − 4.
Hence the interpolation polynomial will be of degree 4.
636
Engineering Mathematics-III
Let du be the value of d at T=130°
T − T0 130 − 100
∴
u=
=
= 0 .6
h
50
u(u − 1) 2
u(u − 1)(u − 2) 3
∴
du = d0 + u∆d0 +
∆ d0 +
∆ d0
2!
3!
u(u − 1)(u − 2)(u − 3) 4
+
∆ d0
4!
( 0 .6)( 0 .6 − 1)
= 958 + ( 0 .6)( −41) +
( −11)
2!
( 0 .6)( 0 .6 − 1)( 0 .6 − 2)
( 0 .6)( 0 .6 − 1)( 0 .6 − 2)( 0 .6 − 3)
+
( −3) +
( −4)
3!
4!
= 658 + 24 .6 + 1 .32 − 0 .168 + 0 .1344
_ 935
= 934 .6864 ~
Tn = 300, dn = 712, ∆dn = −87, ∆2 dn = − 21, ∆2 dn = − 7 and ∆4 dn = −4.
Again
Let dv be the value of d at T=275°C
T − Tn
275 − 300
∴
v=
=
= − 0 .5
h
50
v( v + 1) 2
v( v + 1)( v + 2) 3
∴
dv = dn v ∇ dn
∇ dn +
∇ dn
2!
3!
v( v + 1)( v + 2)( v + 3) 4
+
∇ dn
4!
( −0 .5)( −0 .5 + 1)
= 712 + ( −0 .5)( −87 ) +
( −21)
2!
( −0 .5)( −0 .5 + 1)( −0 .5 + 2)( −0 .5 + 3)
+
( −4)
4!
= 712 + 43 .5 + 2 .625 + 0 .4375 + 0 .15625
= 758 .71875 ≈ 759
Ex. 41: The following are data from the steam table :
temp C°(t)
140
150
160
170
180
Pressure
kgf/cm2(P)
3.685
4.854
6.302
8.076
10.225
Using Newton's formula, find the pressure of the steam for temperatures 142° and
175°.
(M.U, B.E., 1996)
Sol : We note that t =142° is near the beginning of the table and t =175° is near the
end of the table. So for the former we use Newton's Forward interpolation formula
and for the later, we use Newton's Backward interpolation formula. The differences
are calculated and tabulated below.
t
P
∆P
140
3.685
1.169
∆2 P
∆3 P
∆4 P
Finite Differences and Interpolation
637
150
4.854
1.448
0.279
0.047
160
6.302
1.774
0.326
0.049
170
8.076
2.149
0.375
180
10.225
Here
and
0.002
t 0 = 140, P0 = 3 .685, ∆P0 = 1 .169, ∆2 P0 = 0 .279, ∆3 P0 = 0 .047
∆4 P0 = 0 .002
Hence the interpolation polynomial will be of degree 4.
Let Pu be the value of P when t =142°
t − t 0 142 − 140
∴
U=
=
= 0 .2
h
10
u(u − 1) 2
u(u − 1)(u − 2) 3
i.e.,
Pu = P0 + u ∆P0 +
∆ P0 +
∆ P0
2!
3!
u(u − 1)(u − 2)(u − 3) 4
+
∆ P0
4!
( 0 .2)( 0 .2 − 1)
∴
Pu = 3 .685 + ( 0 .2)(1 .169) +
( 0 .279)
2!
( 0 .2)( 0 .2 − 1)( 0 .2 − 2)
( 0 .2)( 0 .2 − 1)( 0 .2 − 2)( 0 .2 − 3)
+
( 0 .047 ) +
( 0 .002)
3!
4!
= 3 .685 + 0 .2338 − 0 .2332 + 0 .002256 − 0 .0000672
_ 3 .899
= 3 .8986688 ~
Again
∴
and
t n = 180°
Pn = 10 .225, ∇ Pn = 2 .149, ∇ 2 Pn = 0 .375, ∇ 3 Pn = 0 .049,
∇ 4 Pn = 0 .002
Let Pv be the value of P at t=176°.
t − tn
175 − 180
v=
=
= − 0 .5
h
10
v( v + 1) 2
v( v + 1)( v + 2) 3
Pv = Pn + v ∇Pn +
∇ Pn +
∇ Pn
2!
3!
v( v + 1)( v + 2)( v + 3) 4
+
∇ Pn
4!
( −0 .5)( −0 .5 + 1)
= 10 .225 + ( −0 .5)( 2 .149) +
( 0 .375)
2!
( −0 .5)( −0 .5 + 1)( −0 .5 + 2)
+
( 0 .049)
3!
( −0 .5)( −0 .5 + 1)( −0 .5 + 2)( −0 .5 + 3)
+
( 0 .002)
4!
= 10 .225 − 1 .745 − 0 .046875 − 0 .0030625 − 0 .000078125
_ 9 .1005
= 9 .10048438 ~
638
Engineering Mathematics-III
14.9 Interpolation with Unequal Intervals
All the interpolation formulae, which are described in our earlier discussions are
useful when the value of x at equal is equally spaced. In this section, we shall discuss
some interpolation formulae which are applied when x is not equally spaced.
14.9.1 Lagrange's Interpolation Formula for Unequal Interval
Consider, we have ( x 0 , f 0 ) , ( x1 , f1 )...( x n ,f n ) any (n + 1) points, of the function y =
f ( x) where fi =f ( xi ) = ∀ i = 0, 1, 2... n. Here the interval xi − xi −1 is not unique.
Now to approximate a polynomial Pn ( x ) of degree n such that Pn ( xi ) = fi , we can
assume Pn ( x ) has the following form. i.e.
Pn(x)=A0(x − x1 ) (x − x 2 ) ... (x − x n ) + A1(x − x 0 )(x − x 2 )...( x − x n )
...(14.15)
+ A2 (x − x 0 )(x − x1 )(x − x 3 ) … (x − x n )+. . . − An(x − x 0 )(x − x1 )(x − x 2 ) … (x − x n-1 )
where A 0 , A1… A n are constant such that Pn (xi )=fi , therefore when x=x 0 ,
A 0=
f0
(x 0 − x1 )(x 0 − x 2 ) − (x 0 − x n )
when x=x1 ,A1 =
f1
(x1 − x 0 )(x1 − x 2 ) − (x1 − x n )
we get
Similarly, we get the value of A 2 , A 3 , A 4… A n . Putting all these values in (14.1) we
get
Pn (x)=
+
... +
(x − x1 ) (x − x 2 )...(x − x n )
f0
(x 0 − x1 ) (x 0 − x 2 )...(x 0 − x n )
(x − x1 ) (x − x 2 ) ... (x − x n )
f1 + ...
(x1 − x 0 )(x1 − x 2 ) … (x1 − x n )
(x − x 0 ) (x − x1 ) (x − x 2 ) … (x − x n −1 )
fn
(x n − x 0 )(x n − x1 ) … (x n − x n −1 )
...(14.16)
The above expression is known as Lagrange's interpolation formula for unequal
interval
Ex. 42 : Find f(x) given by
x
0
1
2
3
4
f(x)
3
6
11
18
27
Sol : Here x 0 = 0 , x1 = 1 , x 2 = 2 , x 3 = 3 , x 4 = 4
f 0=3 ,
f1=6 ,
f 2=11 ,
f 3=18 ,
2
f 4=27
Putting all these value in (14.16) we get f ( x ) = x + 2 x + 3
Finite Differences and Interpolation
639
Ex. 43 : Find log10 301 from the table given below
x
300
304
305
307
log10 x
2 ⋅ 4771
2 ⋅ 4829
2 ⋅ 4843
2 ⋅ 4871
Sol : Here x 0 = 300,
x1 = 304,
x 2 = 305,
x 3 = 3 ⋅ 7, and x = 300
f 0=2 ⋅ 4771 , f1 = 2 ⋅ 4829 , f 2 = 2 ⋅ 4843 , f 3 = 2 ⋅ 4871 ,
By Lagrange's formula
log10 301 =
+
( −3) ( − 4) ( − 6)
(1) ( − 4)( − 6)
2 ⋅ 4771 +
2 ⋅ 4829
( − 4)( −5)( −7 )
( 4) ( −1)( −3)
(1) ( −3)( − 6)
(1)( −3)( − 4)
2 ⋅ 4843 +
2 ⋅ 4871
( 5) (1) ( −2)
(7 )( 3)( 2)
= 1 ⋅ 2739 + 4 ⋅ 9658 − 4 ⋅ 4714 + 0 ⋅ 7106.
= 2 ⋅ 4786
Ex. 44 : If f (1)= 4 , f ( 3)= 12 , f ( 4)= 19 and f(x)=7 Find x .
Sol : We have x 0 = 1, x1 = 3, x 2 = 4, x 3 = x
f 0 = 4, f1 = 12, f 2 = 19 f 3 = 7
Now interchanging xi ↔ fi in the expression (14.16) we get
x 0 = 4, x1 = 12, x 2 = 19, x = 7
f 0 = 1, x1 = 3, f 2 = 4, f x = ?
Putting all values we get x = 1 ⋅ 86, For which f(1 ⋅ 86) = 7.
Ex. 45 : Express
3 x 2 + x +1
as a sum of partial fractions.
( x − 1)( x − 2)( x − 3)
Sol : Let f ( x ) = 3 x 2 + x +1, and make the following table
x
1
2
3
f( x)
5
15
31
Now By Lagrange's Interpolation, for x 0 = 1, x1 = 2, x 2 = 3, f 0 = 5, f1 = 15,
f 2 = 31, we get
( x − 2)( x − 3)
( x − 1)( x − 3)
( x − 1)( x − 2)
f ( x) =
5+
15 +
31
( 5 − 2)( 5 − 3)
(15 − 5)(15 − 3)
( 31 − 5)( 31 − 15)
5
31
or, 3 x 2 + x +1 = ( x − 2)( x − 3)+15( x − 1)( x − 3)+ ( x − 1)( x − 2)
2
2
or
3 x 2 + x +1
5
15
31
=
−
+
( x − 1)( x − 2)( x − 3) 2( x − 1) x − 2 2 ( x − 3)
640
Engineering Mathematics-III
14.10 Inverse Interpolation
So, far, given a set of values of x and y, we were required to find the value of y
corresponding to a value of x. Sometimes, we may require to find the value of x
corresponding to a certain value of y. The process of finding such a value of x is
called inverse interpolation. In this section, we shall illustrate how lagrange's method
Can be applied for inverse interpolation.
14.11 Lagrange's Method
choosing y as an independent variable and x as a dependent variable in Lagrange's
interpolation formula, i.e.
( y − y1 )( y − y 2 )…( y − y n )
x = f(y ) =
x1
( y 0 − y1 )( y 0 − y 2 )…( y 0 − y n )
( y − y 0 )( y − y 2 )…( y − y n )
+
x2
( y1 − y 0 )( y1 − y 2 )…( y1 − y n )
… … … …
( y − y 0 )( y − y1 )…( y − y n −1 )
+
xn
( y n − y 0 )( y n − y1 )…( y n − y n −1 )
+…
...(14.17)
which is use for inverse interpolation.
Ex. 46: The following table gives the values of x and y.
x
30
35
40
45
50
y
15.9
14.9
14.1
13.3
12.5
Find the value of x corresponding to y =13.6
Sol : Here, y 0 = 15 .9, y1 = 14 .9, y 2 = 14 .1, y 3 = 13 .3, y 4 = 12 .5,
x 0 = 30, x1 = 35, x 2 = 40, x 3 = 45, and x 4 = 12 .5 ., Taking y = 13 .6, Eqn (14.17).
gives
(13 .6 − 14 .9)(13 .6 − 14 .1)(13 .6 − 13 .3)(13 .6 − 12 .5)
x=
( 30)
(15 .9 − 14 .9)(15 .9 − 14 .1)(15 .9 − 13 .3)(15 .9 − 12 .5)
(13 .6 − 15 .9)(13 .6 − 14 .1)(13 .6 − 13 .3)(13 .6 − 12 .5)
+
( 35)
(14 .9 − 15 .9)(14 .9 − 14 .1)(14 .9 − 13 .3)(14 .9 − 12 .5)
(13 .6 − 15 .9)(13 .6 − 14 .9)(13 .6 − 13 .3)(13 .6 − 12 .5)
+
( 40)
(14 .1 − 15 .9)(14 .1 − 14 .9)(14 .1 − 13 .3)(14 .1 − 12 .5)
(13 .6 − 15 .9)(13 .6 − 14 .9)(13 .6 − 14 .1)(13 .6 − 12 .5)
+
( 45)
(13 .3 − 15 .9)(13 .3 − 14 .9)(13 .3 − 14 .1)(13 .3 − 12 .5)
(13 .6 − 15 .9)(13 .6 − 14 .9)(13 .6 − 14 .1)(13 .6 − 13 .3)
+
( 50)
(12 .5 − 15 .9)(12 .5 − 14 .9)(12 .5 − 14 .1)(12 .5 − 13 .3)
= 0 .4044117 − 4 .3237305 + 21 .41276 + 27 .79541 − 2 .1470014
= 43 .14185 ≅ 43 .1
Finite Differences and Interpolation
641
14.12 Divided differences
If we are interested to have an interpolation formula which has the property that
polynomial of highest degree may be derived from it by simply adding a new term.
For this purpose, we need to discuss divided differences.
Consider we have ( x 0 , f 0 ), ( x1 , f1 )...( x n , f n ) any ( n +1) points of the function
y=f(x) , such that f(xi )=fi ∀ i = 0 , 1 , 2 , 3 ...n. Now, divided differences of under 1,
2...n are defined as below.
f − f0

[ x 0 , x1 ]= 1
= ∆| f ( x 0 )

x1 − x 0
x1

[ x1 , x 2 ] − [ x 0 , x1 ]
2

|
[ x 0 , x1 , x 2 ] =
= ∆ f( x0 )

x2 − x0
x1 x 2
...(14.18)

… … … … … … … … …


… … … … … … … … …
[ x1 , x 2 ... x n ] − [ x 0 , x1 , x 2 ,...x n −1 ]
[ x 0 , x1 , x 2… x n ] =

xn − x0

= ∆| n f( x 0 )
x1, x 2 .... x n
14.13 Properties of Divided Differences
1. The divided differences are symmetrical in all their arguments, i.e. the value of
any divided difference is independent of the order of the arguments.
f ( x1 ) − f ( x 0 ) f ( x 0 ) − f ( x1 )
f ( x 0 , x1 ) =
=
= f ( x1 , x 0 )
x1 − x 0
x 0 − x1
f( x0 )
f ( x1 )
Also,
f ( x 0 , x1 ) =
+
x 0 − x1 x1 − x 0
which shows that f ( x 0 , x1 ) is symmetric with respect to the arguments x 0 , x1 .
f ( x1 , x 2 ) − f ( x 0 , x1 )
Now f ( x 0 , x1 , x 2 ) =
x 2 − x1
 f ( x1 )
f( x2 )   f( x0 )
f ( x1 )  
+
+
−


 x1 − x 2 x 2 − x1   x 0 − x1 x1 − x 0  
f( x0 )
f ( x1 )
f( x2 )
...(14.19)
=
+
+
( x 0 − x1 )( x 0 − x 2 ) ( x1 − x 0 )( x1 − x 2 ) ( x 2 − x 0 )( x 2 − x1 )
RHS =
1
x2, x0
Eqn (14.19) shows that f ( x 0 , x1 , x 2 ) is symmetric with respect to the arguments
x 0 , x1 , x 2 .
Similarly, it can be shown that
f( x0 )
f ( x 0 , x1 , x 2 ,… x n ) =
( x 0 − x1 )( x 0 − x 2 )…( x 0 − x n )
f ( x1 )
f( x2 )
+
+
+…
( x1 − x 0 )( x1 − x 2 )…( x1 − x n ) ( x 2 − x 0 )( x 2 − x1 )…( x 2 − x n )
f( xn )
+
( x n − x 0 )( x n − x1 )…( x n − x n −1 )
642
Engineering Mathematics-III
That is, f ( x 0 , x1 , x 2… x n ) is symmetric with respect to the arguments
x 0 , x1 , x 2… x n .
2. The divided difference (of any order) of the sum or difference of two functions is
equal to the sum of difference of the corresponding separate divided differences;
that is the operators ∆, ∆2 ,…etc. are linear.
Let f(x) and g(x) be two functions with arguments x 0 , x1 . Then,
[ f ( x1 ) ± g ( x1 )] − [ f ( x 0 ) ± g ( x 0 )]
∆[ f ( x ) ± g ( x )] =
x1 − x 0
[ f ( x1 ) − f ( x 0 )] [ g ( x 0 ) − g ( x 0 )]
=
±
x1 − x 0
x1 − x 0
= ∆ f ( x) ± ∆ g( x)
Similarly, we can prove it for any higher order difference, i.e.
∆r [ f ( x ) ± g ( x )] = ∆r f ( x ) ± ∆r g ( x )]
3. The divided difference of the product of a constant and a function and a function
is equal to the product of the constant and the divided difference of the function, i.e.
∆{ cf ( x )} = c∆f ( x ).
cf ( x1 ) − cf ( x 0 )
By definition, ∆cf ( x ) =
x1 − x 0
 f ( x1 ) − f ( x 0 )
= c
 = c∆f ( x )
x1 − x 0


This is also true for any higher order difference.
4. The nth divided difference of a polynomial of nth degree are constants.
Let f ( x ) = x n be the polynomial, where n is a positive integer. Then,
RHS is a polynomial function of (n-1)th degree, symmetrical in x 0 and x1 with
leading coefficient 1.
[ f ( x1 , x 2 ) − f ( x 0 , x1 )]
Again, f ( x 0 , x1 , x 2 ) =
x2, x0
RHS=
=
[ x 2n −1 − x1 x 2n − 2 +L+ x1n −1 ] − [ x 0n −1 + x1 x 0n − 2 +L+ x1n −1 ]
[ x 2 − x1 ]
x 2n −1 − x 0n − 2
x ( x n − 2 − x 0n − 2 )
x n−2 ( x2 − x0 )
+ 1 2
+L+ 1
x2 − x0
x2 − x0
x2 − x0
= ( x 2n − 2 + x 0 x 2n − 3 +L+ x 0n − 2 ) + x1 ( x 2n − 3 + x 0 . x 2n − 4 +L+ x 0n − 3 ) +L+ x1n − 2
= a polynomial of ( n − 2)th degree, symmetrical in x 0 , x1 , and x 2 with leading
coefficient 1.
Proceeding in this way, the pth order divided difference of x n will be a polynomial of
( n − p)th degree, symmetrical in x 0 , x1 , x 2 ,… x p with leading coefficient 1.
Hence, the nth order divided difference of x n will be a polynomial of ( n − n )th
degree, i.e. of order 0 symmetrical in x 0 , x1 , x 2 ,… x n with leading coefficient 1.
i.e. it will be a constant = 1 ∴ ∆n x n = 1 and ∆n + i x n = 0 for i = 1, 2,…
Finite Differences and Interpolation
643
Hence ∆n { a0 x n + a1 x n −1 + az x n − 2 +L+ an }
= a0 .1 + a1 .0 + a2 .0 +L+ 0 = a0 .
14.14 Relation Between Divided Differences and Forward Differences
Let the arguments x 0 , x1 , x 2 ,…, x n be equally spaced.
i.e. xi +1 − xi = h ( say ) for i = 0, 2,…, n − 1
f ( x1 ) − f ( x 0 ) ∆f ( x 0 )
Then
f ( x 0 , x1 ) =
=
x1 − x 0
h
f ( x1 , x 2 ) − f ( x 0 , x1 )
f ( x 0 , x1 , x 2 ) =
=
x2 − x0
Similarly, f ( x 0 , x1 , x 2 , x 3 ) =
1
3!h
…
…
3
∆f ( x 1 ) ∆f ( x 0 )
−
1
h
h
=
∆2 f ( x 0 )
2
2h
2h
∆3 f ( x 0 )
…
…
…
…
…
…
f ( x 0 , x1 , x 2 ,… x n ) =
1
n!h
n
∆n f ( x 0 )
14.14.1 Newton's Divided Difference Formula
We know that [ x x 0 ] =
f x − f0
, or
x − x0
f x=f 0 + ( x − x 0 ).[ x, x 0 ]
Again, by second order divided difference, we have
[ x, x 0 ] − [ x 0 , x1 ]
[ x, x 0 , x1 ] =
x − x1
...(14.20)
...(14.21)
By (14.20) and (14.21) we get
Again
f x = f 0 + (x − x 0 )[x 0 , x1 ] + (x − x1 )(x − x1 ) [x, x 0 , x1 ] ...(14.22)
[ x, x 0 , x1 ] − [ x 0 , x1 , x 2 ]
...(14.23)
[ x, x 0 , x1 , x 2 ] =
x2 − x0
By (14.22) and (14.23), we get
f x=f 0+( x − x 0 )[ x 0 , x1 ]+( x − x 0 ) ( x − x1 ]+(x 0 , x1 , x 2 )
+ ( x − x 0 ) ( x − x1 )( x − x 2 )[ x,x 0 ,x1 ,x 2 ]
...(14.24)
Continue in this manner, we get
f x=f 0+( x − x 0 )[ x 0 , x1 ]+( x − x 0 )( x − x1 )[ x 0 , x1 ,x 2 ]
+ ( x − x 0 )( x − x1 )( x − x 2 )[ x 0 , x1 , x 2 , x 3 ]
.......................+............
+ ( x − x 0 ) ( x − x1 ) ( x − x 2 ) ... ( x − x n −1 ) [ x 0 , x1 x 2 … x n ] ...(14.25)
644
Engineering Mathematics-III
The expression (14.6) is also known Newton general interpolation formula with
divided differences.
Deduce Newton's forward enrerpolation formula from Newton's
divide differnce inderpolation formula for enequal interval
Deduction :
We know that if the argumetns x 0 , x1 ,....., x n are equally spaced, then
1
1
f ( x 0 x1 ) = ∆ f ( x 0 ); f ( x 0 , x1 , x 2 ) =
∆2 f ( x 0 )
n
2!h 2
1
f ( x 0 , x1, x 2 , x 3 ) =
∆3 f 3 ( x 0 )
3!h 3
......................
1
f ( x 0 , x1 , x 2 ,,........., x n ) =
∆n f ( x 0 )
n
h !h
Setting x – x 0 = ph, we get
x – x1 = ( x – x 0 ) – ( x1 – x 0 ) = ph – h = ( p – 1)h
( x – x 2 ) = ( p – 2)h and so on.
Substituting these values in Newton's divide difference interpolation formula we get
∆f ( x 0 ) ph( p – 1)h 2
f ( x ) = f ( x 0 + ph ) = f ( x 0 ) + ph
+
∆ f( x0 )
h
2!h 2
ph( p – 1)h( p – 2)h 3
+
∆ f ( x 0 )+....
3!h 3
ph( p – 1)h,{ p – ( n – 1)} h n
+
∆ f( x0 )
n !h n
p( p – 1) 2
p( p – 1)( p – 2) 3
= f ( x 0 ) + p∆f ( x 0 ) +
∆ f( x0 )
∆ f( x0 )
2!
3!
p( p – 1)…[ p – ( n – 1)] n
+......+
∆ f( x0 )
n!
which is Newton's forward interpolation formula.
Ex. 47 : Find f ( x ), using the following table
x
–1
0
3
6
7
f(x)
3
–6
39
822
1611
Sol : The corresponding divided difference is given below
Finite Differences and Interpolation
x
Here x 0 = −1 x1 = 0,
f 0 = 3,
645
f(x)
x 2 = 3 x 3 = 6,
x4 = 7
f1 = − 6, f 2 = 39, f 3 = 822, f 4 = 1611
[ x 0 , x1 ] = − 9, [ x 0 , x1 , x 2 ] = 6,[ x 0 , x1 , x 2 , x 3 ] = 5,[x 0 , x1 , x 2 , x 3 , x 4 ] = 1
Putting all these value in (3.21) we get
f x = 3+( x+1)( − 9)+ x( x +1)(6)+ x ( x +1)( x − 3)(5)+ x ( x +1)( x − 3)( x − 6) 1
= x 4 − 3x 3 + 5x 2 − 6
Ex. 48 : Find f (8) and f (15) from the following table
x
4
5
7
10
11
13
f( x )
48
100
294
900
1210
2028
Sol : The corresponding divided difference table is given below
646
Engineering Mathematics-III
We have x 0=4 ,
x1 = 5,
x 2 = 7,
x 3 = 10, x 4 = 11,
x 5 = 13
f 0=48 , f1=100 , f 2=294 , f 3 = 900 , f 4=1210 , f 5=2028
[ x 0 , x1 ] = 52, [ x 0, x1, x 2 ] = 15, [ x 0 , x1 , x 2 , x 3 ] = 1, [ x 0 , x1 , x 2 , x 3 , x 4 ] = 0
Putting all these values in (3.21) we get
f(8) = 48+(8 ⋅ 4), 52 + ( 8 − 4)( 8 − 5) × 15 + ( 8 ⋅ 4)( 8 − 5)( 8 − 7 ) ⋅ 1 = 448
Similarly f (15) = 3150
1
 1
| 2   =
Ex. 49 : Show that ∆
bc  a abc
1
1
1
1
Sol : Here f ( x )= , x 0=a, x1=b, x 2 = c, then f 0 = , f1 = , f 2 = , Now we
x
a
b
c
shall make the divided difference table.
1
 1
so D 2   =


a
abc
bc
14.15 Hermite's Interpolation
The interpolation formulae so for discussed uses only tabular points. Now consider
the situation, when both the value of the function and its first derivative are to be
assigned at each given point. Hermite interpolation is the problem in which the
interpolating polynomial satisfy the following conditions.
...(14.26)
P( xi )= f ( xi ) and P'(xi )=f '(xi ), i=0 , 1 , 2...n
Since we have 2n + 2 conditions therefore P( x ) must be a polynomial of degree
≤ 2n+1. P( x ) can be written as
n
P( x )=
∑
i= 0
n
Ai (x). f(x i ) +
∑ Bi (x) . f '(x )
i
...(14.27)
i= 0
Here Ai ( x ) and Bi ( x ) are polynomials of degree ≤ 2 n +1, which satisfy

0 i ≠ j
Ai (x j ) =
and Ai '(x j ) = 0 ∀ i and j
1 i=j

...(14.28)

0 i ≠ j

B ' 2(x j ) = 
and Bi (xi )=0 ∀ i and j

1 i=j

Finite Differences and Interpolation
647


x − xi
x − x0
 We know that in Lagrange interpolation, let x − x = l0(x) and x − x = li(x)
0
1
1
0


 then l0(n) + l1(x)=1 and l0(x0 ) = 1, l0( x1 )= 0 and l1(x0 ) = 0 and l1(x1 )=1.In



1 i = j
 other words, we can say li (x j ) = δ ij = 



0 i ≠ j
Now using the above result, due to Lagrange interpolation, we may write as
Ai ( x ) = α i ( x ). li2 ( x ) and Bi ( x ) = β i ( x ). li2 ( x )
...(14.29)
Since l 2i (x) is a polynomial of degree 2n, therefore α i (x) and β i (x) must be of degree
one. Let
ai ( x ) = ai x+b and β i ( x ) = bi x+di
Now using the condition (14.28), we can get
ai = − 2 li ' (xi ), bi = 1 + 2 xi l'i (x), c i = 1 and di = x i
Putting all value in (14.27), we get
...(14.30)
...(14.31)
n
P( x ) =
∑ [1 − 2( x − xi ) li ' ( xi )]. li2 ( x ). f ( xi )
i=0
n
+
∑
i =0
( x − xi ) . li2 ( x ) f ' ( xi )
...(14.32)
the above expression is known as Hermite interpolating polynomial.
Ex. 50 : Find the corresponding Hermite interpolating polynomial from the
following data
x
–1
0
1
f( x )
1
1
3
f '( x )
–5
1
7
Sol : Here n = 2, x 0 = −1, x1 = 0 and x 2 = 1
f( x 0 ) = 1, f( x1 ) = 1,
f( x 2 ) = 3
f ' ( x 0 ) = − 5, f ' ( x i ) = 1, f ' ( x 2 ) = 7
2
We know p ( x ) =
∑
i=0
2
Ai ( x ) f ( xi )+
∑ Bi ( x ). f ' ( x 2 ),
where
i=0
 A 0(x)= [ 1 − 2 (x − x 0 ) . l' 0(x 0 )] l02(x)

 B0(x) = (x − x 0 ) . l02(x)
 A1(x)= [ 1 − 2 (x − x1 ) . l1' (x1 )]. l 2(x)

 B1(x)= (x − x1 ) . l12(x)
 A 2(x)=[ 1 − 2 (x − x 2 ) . l22(x)

 B2(x)= (x − x 2 ) . l22(x)
648
Engineering Mathematics-III
where, l0(x), l' 0( − 1 ), l1(x), l'1( 0 ), l2(x), l2'( 1 ) are given below.
(x − 0 ) (x − 1 )
x (x − 1 )
l0(x)=
=
, l0'( − 1 )= − 3 / 2
( − 1 − 0) ( − 1 − 1)
2
l1(x)=
(x+1 ) (x − 1 )
= − (x 2 − 1 ), l1'( 0 ) = 0
( 0 + 1) ( 0 − 1)
l2(x)=
(x+1 ) (x − 0 )
x (x+1 )
3
=
, l2'( 1 )=
(1 + 1 ) (1 − 0 )
2
2
Now
3 x 2(x − 1 ) 2 1

A 0(x)=1 − 2 (x+1 ) × −  .
=
[ 3 x 5 − 2 x 4 − 5 x 3+4 x 2 ]
2
4
4

B0(x)=(x+1 ) .
x 2(x − 1 ) 2 1
= [x 5 − x 4 − x 3+x 2 ]
4
4
Similarly A1(x)=x 4 − 2 x 2+1 , B1(x)=x 5 − 2 x 3+x
A 2(x)=
1
1
[ − 3 x 5 − 2 x 4+5 x 3+4 x 2 ] , B2(x)= [x 5+x 4 − x 3 − x 2 ]
4
4
Putting the value of A 0(x), B0(x), A1(x), B1(x), A 2(x), B2(x) in P(x), we get
P(x)=2 x 4 − x 2+x+1
14.16 Approximation of Function
When a function is represented by a series and that series has infinite term and we want
to develop a software for a digital computer, we have to approximate that series. Since
interpolation techniques has been found costlier in comparison to approximation
techniques. This section provides two techniques of approximation : Approximation of
function by Taylor's series and appr