II/IV B.Tech CSE&ECM Mathematical Methods for Computing (11BS203) Chapter-2
Chapter-2
Finite Differences and Interpolation
Suppose we are given the following values of y = f(x) for a set of values of x :
x :
x0
x1
x2
……
xn
y :
y0
y1
y2
……
yn
The process of finding the values of y corresponding to any value of x=xi between x0 and xn is
called interpolation.
 The technique of estimating the value of a function for any intermediate value of the
independent variable is called interpolation.
 The technique of estimating the value of a function outside the given range is called
extrapolation.
 The study of interpolation is based on the concept of differences of a function.

Suppose that the function y=f(x) is tabulated for the equally spaced values x = x0,
x1=x0+h, x2=x0+2h, …, xn=x0+nh giving y = y0, y1, y2, …, yn. To determine the values of
f(x) and f '(x) for some intermediate values of x, we use the following three types of
differences
1. Forward differences
2. Backward differences
3. Central differences
Forward differences: The forward differences are defined and denoted by ∆f(x)=f(x+h)-f(x),
∆y0 = y1 – y0
∆y1 = y2 – y1
∆y2 = y3 – y2
…………….
∆yr = yr+1 – yr
…………….
∆yn-1 = yn – yn-1
These are called the first forward differences and ∆ is the forward difference operator.
Similarly the second forward differences are defined by
∆2 yr = ∆ yr+1 – ∆yr.
In general
∆p yr = ∆p-1 yr+1 – ∆p-1yr ,
th
p forward differences.
The forward differences systematically set out in a table called forward difference table.
Value of Value of 1st diff.
2nd diff.
3rd diff.
4th diff.
5thdiff.
x
y
∆
∆2
∆3
∆4
∆5
x0
y0
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II/IV B.Tech CSE&ECM Mathematical Methods for Computing (11BS203) Chapter-2
∆y0
x1
∆2y0
y1
∆y1
x2
∆3y0
∆2y1
y2
∆4y0
∆y2
x3
∆3y1
∆2y2
y3
∆4y1
∆y3
x4
∆5y0
∆3y2
∆2y3
y4
∆y4
x5
y5
Backward Differences: The backward differences are defined and denoted by f(x)= f(x)-f(x-h),
y1 = y1 – y0
y2 = y2 – y1
y3 = y3 – y2
…………….
yr = yr – yr-1
…………….
yn = yn – yn-1.
These are called the first backward differences and is the backward difference operator.
Similarly the second backward differences are defined by
2
yr =
yr –
yr-1.
In general
p
p-1
yr= p-1yr –
yr-1 ,
th
p backward differences. The backward differences systematically set out in a table called
backward difference table.
Value of
x
Value of
y
x0
y0
1st diff.
2nd diff.
2
3rd diff.
3
4th diff.
4
5thdiff.
5
y1
x1
y1
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2
y2
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II/IV B.Tech CSE&ECM Mathematical Methods for Computing (11BS203) Chapter-2
3
y2
x2
2
y2
4
y3
3
y3
x3
2
y3
x4
2
y4
y4
5
y4
4
y4
3
y4
y3
y5
y5
y5
y5
y5
x5
y5
Central differences: The central differences are defined and denoted by δf(x)=f(x+h/2)-f(x-h/2),
δy1/2 = y1 – y0
δy3/2 = y2 – y1
δy5/2 = y3 – y2
…………….
δyr+1/2 = yr+1 – yr
…………….
δyn-1/2 = yn – yn-1
These are called the first central differences and δ is the central difference operator.
Similarly the second central differences are defined by
δ2 yr = δyr+1/2 – δyr-1/2.
In general
δp yr = δp-1yr+1/2 – δp-1yr-1/2 , (if p is even)
δp yr+1/2 = δp-1yr+1 – δp-1yr , (if p is odd)
pth central differences.
Note : It is observe that, the differences are not change only notation is changed.
y1-y0 = ∆y0= y1= δy1/2 .
Example#1. Evaluate
(i) ∆ tan-1x
(ii) ∆ (ex log 2x)
(iii) ∆2 cos 2x
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II/IV B.Tech CSE&ECM Mathematical Methods for Computing (11BS203) Chapter-2
Sol. From the definition of forward differences ∆f(x) = f(x+h) – f(x).
(i) Let f(x) = tan-1x, then
∆ tan-1x = tan-1(x+h) - tan-1x
 xhx 
h


  tan 1 
= tan 1 
.
2 
 1  hx  x 
 1  ( x  h) x 
(ii)
 (e x log 2 x)  e x  h log 2( x  h)  e x log 2 x
 e x  h log 2( x  h)  e x  h log 2 x  e x  h log 2 x  e x log 2 x
 xh
xh
x
 e x  h log 
  (e  e ) log 2 x
x


 

h 
 e x e h  log( 1  )   (e h  1) log 2 x .
x 
 

2
(iii) ∆ cos 2x = ∆ [∆ cos 2x]
= ∆ [ cos 2(x+h) – cos 2x]
= ∆ cos 2(x+h) – ∆ cos 2x
= cos 2(x+2h) – cos 2(x+h) – [cos 2(x+h) ) – cos 2x]
= – 2 cos (2x+3h) sin h + 2 sin(2x+h) sin h
= – 2 sin h [sin(2x+3h) – sin(2x+h) ]
= – 2 sin h [2 cos(2x+2h)sin h]
= – 2 sin2 h cos 2(x+h).
Example#2. Evaluate the following, with interval of difference being unity
(i) ∆2 (abx)
(ii) ∆n ex
Sol. From the definition of forward differences ∆f(x) = f(x+h) – f(x).
(i) ∆(abx) = a ∆bx = a(bx+1 – bx) = abx(b – 1)
∆2 (abx) = ∆ [∆ (abx)]
= ∆ abx(b – 1) = a(b – 1) ∆(bx)
= a(b – 1) (bx+1 – bx)
= a(b – 1)2 bx.
x
(ii) ∆e = ex+1 – ex = ex(e – 1)
∆2ex = ∆[∆ex] = ∆ [ex+1 – ex ]= (e – 1)∆ex
= (e – 1) ex(e – 1) = (e – 1)2 ex.
Similarly ∆2ex = (e – 1)2 ex, ∆3ex = (e – 1)3 ex, … and ∆nex = (e – 1)n ex.
Differences of a Polynomial:
Let f(x) = a0 xn + a1 xn-1 + a2 xn-2 + … + an-1x + an be an nth degree polynomial in x, then
∆f(x) = f(x+h) – f(x)
= a0 [(x+h)n – xn] + a1 [(x+h)n-1 – xn-1 ]+ a2 [(x+h)n-2 – xn-2 ]+ … + an-1 [x+h-x]
= a0 n h xn-1 + a11xn-2 + a21xn-2 + … + an-1 h,
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II/IV B.Tech CSE&ECM Mathematical Methods for Computing (11BS203) Chapter-2
where a11, a12, … are new constant coefficients. Thus, the first difference of a polynomial of nth
degree is a polynomial of degree n-1.
Similarly
∆2f(x) = ∆[∆f(x)]
= a0 n h [(x+h)n-1 – xn] + a11 [(x+h)n-2 – xn-2 ]+ a21 [(x+h)n-3 – xn-3 ]+ … + an-2, 1 [x+h-x]
= a0 n(n – 1) h2 xn-2 + a12xn-3 + a22xn-4 + … + an-2,1 h,
where a12, a12, … are new constant coefficients. Thus, the second difference of a polynomial of
nth degree is a polynomial of degree n-2.
Continuing this process, for the nth difference, we get a polynomial of degree zero i.e.
∆nf(x) = a0 n(n – 1) (n – 2) … 3.2.1. hn = a0 n! hn
which is a constant.
From the above discussion, we have the following results:
 The differences of a polynomial of the nth degree are constant and all higher order
differences are zero.
 If the nth differences of a function tabulated at equally spaced intervals are constant, then
the function is a polynomial of degree n (It is important in numerical analysis as it
enables as to approximate a function by a polynomial).
Example#3. Evaluate ∆10 [(1-ax)(1-bx2)(1-cx3)(1-dx4)].
Sol. Taking interval of difference h = 1.
∆10 [(1-ax)(1-bx2)(1-cx3)(1-dx4)] = ∆10 [abcd x10 + k1 x9 + k2 x8 + … + 1]
= abcd ∆10 (x10) + k1 ∆10 (x9)+ k2 ∆10 (x8) + … + ∆10(1),
where k1, k2, … are constant coefficients. Since ∆10 (xn) = 0 for n < 10, we have
∆10 [(1-ax)(1-bx2)(1-cx3)(1-dx4)] = abcd ∆10 (x10) = abcd 10!.
Factorial Notation: A product of the form x(x – 1) (x – 2) …(x – r +1) is denoted by [x]r and is
called a factorial. In particular,
[x] = x, [x]2 = x(x – 1) , [x]3 = x(x – 1) (x – 2), …[x]n = x(x – 1) (x – 2) …(x – n +1).
If the interval of difference is h, then
[x]n = x(x – h) (x – 2h) …(x – (n – 1)h).
The factorial notation is of special utility in the theory of finite differences. It helps in finding the
successive differences of a polynomial directly by simple rule of differentiation ([x]r as xr).
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II/IV B.Tech CSE&ECM Mathematical Methods for Computing (11BS203) Chapter-2
To express a polynomial of nth degree in the factorial notation, we use the following two steps
1. Arrange the coefficients of the powers of x in descending order, replacing missing
powers by zeros.
2. Using detached coefficients divide by x, x – 1, x – 2, … x – (n – 1) successively.
Example#4. Express f(x) = 2x3 – 3x2 + 3x – 10 in a factorial notation and hence find all
differences.
Sol. Let f(x) = A[x]3 + B[x]2 + C[x] + D. Then
x3
x2
x
1
2
_
-3
2
3
-1
2
2
_
-1
4
2=C
3
2
_
3=B
-10 = D
2=A
Hence f(x) = 2[x]3 + 3[x]2 + 2[x] – 10. Therefore,
∆f(x) = 6[x]2 + 6[x] + 2
∆2f(x) = 12 [x] + 6
∆3f(x) = 12.
Other Difference Operators:
(1) Shift operator : Shift operator E is the operation of increasing the argument x by h so that
E f(x) = f(x+h), E2 f(x) = f(x+2h), …
En f(x) = f(x+nh).
-1
The inverse operator E is defined by
E-1 f(x) = f(x-h).
Similarly
E-n f(x) = f(x-nh).
(2) Averaging operator : Averaging operator μ is defined by the equation
μ f(x) = ½ [f(x + h/2) + f(x - h/2)].
In the difference calculus, ∆ and E are regarded as the fundamental operators and , δ
and μ can be expressed in terms of these.
Relations Between the Operators :
1. ∆ = E – 1
2.  = 1 – E-1
3. δ = E1/2 – E-1/2
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II/IV B.Tech CSE&ECM Mathematical Methods for Computing (11BS203) Chapter-2
4. μ = ½ [E1/2 + E-1/2 ]
5. ∆ = E  =  E =δ E1/2
6. E = ehD.
Example#1. Determine the missing values in the following table:
x
45
50
55
60
65
y
3
?
2
?
-2.4
Sol. Let p and q be the missing values in the given table, then the difference table is as follows:
x
y
∆y
∆2y
∆3y
45
3
p–3
50
p
5 – 2p
2–p
3p + q – 9
55
2
p+q–4
q–2
3.6 – p – 3q
60
q
–0.4 – 2q
–2.4 – q
65
-2.4
Since three entries are given, the function y can be represented by a second degree polynomial.
Therefore, ∆3y0 = 0 and ∆3y1 = 0. Thus 3p + q – 9 = 0 and 3.6 – p – 3q = 0. Solving these
equations, we get p = 2.925 and q = 0.225.
Example#2. Determine the missing values in the following table without using difference table.
x
45
50
55
60
65
y
3
?
2
?
-2.4
Sol. Given that y0 = 3, y2 = 2 and y4 = -2.4 and missing values be taken as y1= p and y3 = q.
Since three entries are given, the function y can be represented by a second degree polynomial.
Therefore, ∆3y0 = 0 and ∆3y1 = 0.
(E – 1)3y0 = 0
(E – 1)3y1 = 0
3
2
(E – 3E + 3E – 1)y0 = 0
(E3 – 3E2 + 3E – 1)y1 = 0
y3 – 3y2 + 3y1 – y0 = 0
y4 – 3y3 + 3y2 – y1 = 0
q – 3(2)+ 3p – 3 = 0
-2.4 – 3q + 3(2) – p = 0
3p + q – 9 = 0
3.6 – p – 3q = 0.
Solving these equations, we get p = 2.925 and q = 0.225.
Newton’s Forward Interpolation Formulae:
Let the function y=f(x) take the values y0, y1, y2, … corresponding to the values x0, x1, x2, …
of x. Suppose it is required to evaluate f(x) for x=x0+ph, p is any real number.
For any real number p, we have defined E such that
Ep f(x) = f(x0+ph)
yp = f(x0+ph) = Epf(x0)= (1+∆)py0
= [1+p∆+p(p-1)/2! ∆2 + p(p-1)(p-2)/3! ∆3 +…] y0
= y0 + p ∆y0 + p(p-1)/2! ∆2 y0+ p(p-1)(p-2)/3! ∆3 y0+…
It is called Newton’s forward interpolation formulae.
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II/IV B.Tech CSE&ECM Mathematical Methods for Computing (11BS203) Chapter-2
Newton’s Backward Interpolation Formulae:
Suppose it is required to evaluate f(x) for x=xn+ph, where p is any real number.
Ep f(x) = f(xn+ph)
yp = f(xn+ph) = Epf(xn)= (1-  )-p yn
= [1+p  +p(p+1)/2!  2 + p(p+1)(p+2)/3!  3 +…] yn
= yn + p  yn + p(p+1)/2!  2 yn+ p(p+1)(p+2)/3!  3 yn+…
It is called Newton’s backward interpolation formulae.
Choice of Newton’s Interpolation formulae:
 Newton’s forward interpolation formulae is used for interpolating the values of y near the
beginning of a set of tabulated values and extrapolating values of y a little backward of
y0.
 Newton’s backward interpolation formulae is used for interpolating the values of y near
the end of a set of tabulated values and also extrapolating values of y a little ahead of yn.
Example#1. The table gives the distances in nautical miles of the visible horizon for the given
heights in feet above the earth’s surface :
x=height
100
150
200
250
300
350
400
y=distance
10.63
13.03
15.04
16.81
18.42
19.90
21.27
Find the values of y when (i) x= 218 ft. (ii) x= 410 ft.
Sol. The difference table is
x
y
100
10.63
∆
∆2
∆3
∆4
2.4
150
13.03
-0.39
2.01
x0=200
0.15
-0.24
15.04
0.08
1.77
250
16.81
18.42
0.03
-0.13
1.48
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-0.05
-0.16
1.61
300
-0.07
-0.01
0.02
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II/IV B.Tech CSE&ECM Mathematical Methods for Computing (11BS203) Chapter-2
350
19.90
-0.11
1.37
xn=400
21.27
(i) If we take x0=200, then y0=15.04, ∆y0=1.77, ∆2y0=-0.16, ∆3y0=0.03, ∆4y0=-0.01.
Since x=218, step length h=50 and p=(x-x0)/h =18/50 = 0.36.
By Newton’s forward interpolation formula, we have
y(218) = y0 + p ∆y0 + p(p-1)/2! ∆2 y0+ p(p-1)(p-2)/3! ∆3 y0+ p(p-1)(p-2)(p-3)/4! ∆4 y0
= 15.04 + 0.36 (1.77) + 0.36(0.36-1)/2 (-0.16)+ 0.36(0.36-1)(0.36-2)/6 (0.03)
+ 0.36(0.36-1)(0.36-2)(0.36-3)/24 (-0.01)
=15.04+0.6372+0.0184+0.0018+ 0.00041 = 15.69741
≈ 15.7 nautical miles.
(ii) If we take xn=400, then yn=21.27,  yn=1.37,  2yn=-0.11,  3yn=0.02,  4yn=-0.01.
Since x=410, step length h=50 and p=(x-xn)/h =10/50 = 0.2.
By Newton’s backward interpolation formula, we have
y(410) = yn + p  yn + p(p+1)/2!  2 yn+ p(p+1)(p+2)/3!  3 yn+ p(p+1)(p+2)(p+3)/4!  4 yn
= 21.27 + 0.2 (1.37) + 0.2(0.2+1)/2 (-0.11) + 0.2(0.2+1)(0.2+2)/6 (0.02)
+ 0.2(0.2+1)(0.2+2)(0.2+3)/24 (-0.01)
=21.27+0.274-0.0132+0.0017- 0.0007 = 21.5318
≈ 21.53 nautical miles.
Central Difference Interpolation:
It is best suited for interpolation near the middle of the table. If x takes the values
x0-2h, x0-h, x0, x0+h, x0+2h
and the corresponding values of y=f(x) are
y-2, y-1, y0, y1,
y2 ,
then we can write the difference table in the two notations as follows:
Value
Value
1st diff.
2nd diff.
3rd diff.
4th diff.
of x
of y
x0-2h
y-2
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II/IV B.Tech CSE&ECM Mathematical Methods for Computing (11BS203) Chapter-2
∆y-2(=δy-3/2)
x0-h
∆2y-2(=δ2y-1)
y-1
∆y-1(=δy-1/2)
x0
∆3y-2(=δ3y-1/2)
∆2y-1(=δ2y0)
y0
∆y0(=δy1/2)
x0+h
∆4y-2(=δ4y0)
∆3y-1(=δ3y1/2)
∆2y0(=δ2y1)
y1
∆y1(=δy3/2)
x0+2h
y2
Gauss’s forward interpolation formula: The Newton’s forward interpolation formula is
y(x) = y0 + p ∆y0 + p(p-1)/2! ∆2 y0+ p(p-1)(p-2)/3! ∆3 y0+…
We have ∆2y0 - ∆2y-1 = ∆3 y-1
∆2 y0 = ∆2 y-1 + ∆3 y-1
Similarly ∆3 y0 = ∆3 y-1 + ∆4 y-1
∆4 y0 = ∆4 y-1 + ∆5 y-1
Also
∆3y-1 - ∆3y-2 = ∆4 y-2
∆3 y-1 = ∆3 y-2 + ∆4 y-2
Similarly ∆4 y-1 = ∆4 y-2 + ∆5 y-2
∆5 y-1 = ∆5 y-2 + ∆6 y-2
Substituting the above in Newton’s forward interpolation formula, we get
yp = y0 + p ∆y0 + p(p-1)/2! ∆2 y-1+ (p+1)p(p-1)/3! ∆3 y-1+ (p+1)p(p-1)(p-2)/4! ∆4 y-2+ …,
which is called Gauss forward interpolation formula.
In the central difference notation
y(x) = y0 + p δy1/2 + p(p-1)/2! δ2 y0+ (p+1)p(p-1)/3! δ3 y1/2+ (p+1)p(p-1)(p-2)/4! δ4 y0+ …
It employs odd differences just below the central line and even differences on the central line.
This formula is used to interpolate the values of y for p (0<p<1) measured forwardly from the
origin.
Value
of x
Value
of y
x0-2h
y-2
1st diff.
2nd diff.
3rd diff.
4th diff.
∆y-2(=δy-3/2)
x0-h
∆2y-2(=δ2y-1)
y-1
∆y-1(=δy-1/2)
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∆3y-2(=δ3y-1/2)
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II/IV B.Tech CSE&ECM Mathematical Methods for Computing (11BS203) Chapter-2
x0
∆2y-1(=δ2y0)
y0
∆y0(=δy1/2)
x0+h
∆4y-2(=δ4y0)
∆3y-1(=δ3y1/2)
∆2y0(=δ2y1)
y1
∆y1(=δy3/2)
x0+2h
y2
Example#1. Use Gauss’s forward formula to evaluate y30, given that y21=18.4708, y25=17.8144,
y29=17.1070, y33=16.3432, and y37=15.5154.
Sol. Taking x0 = 29. The interval of difference h=4. We require to find value of y for x=30.
Therefore, p=(x-x0)/h =(30-29)/4 = 0.25. The difference table is given below
x
y
21
18.4708
∆y
∆ 2y
∆ 3y
∆ 4y
-0.6564
25
17.8144
-0.0510
-0.7074
x0=29
17.1070
-0.0054
-0.0564
-0.7638
33
16.3432
-0.0022
-0.0076
-0.0640
-0.8278
37
15.5154
Gauss forward interpolation formula is
y(x) = y0 + p ∆y0 + p(p-1)/2! ∆2 y-1+ (p+1)p(p-1)/3! ∆3 y-1+ (p+1)p(p-1)(p-2)/4! ∆4 y-2+ …
y(30) = 17.1070+(0.25)(-0.7638) + (0.25)(-0.75)/2 (-0.0564) + (1.25)(0.25)(-0.75)/6 (-0.0076)
+ (1.25)(0.25)(-0.75)(-1.75)/24 (-0.0022)
y30 = 16.9216.
Gauss’s backward interpolation formula:
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II/IV B.Tech CSE&ECM Mathematical Methods for Computing (11BS203) Chapter-2
The Newton’s forward interpolation formula is
y(x) = y0 + p ∆y0 + p(p-1)/2! ∆2 y0 + p(p-1)(p-2)/3! ∆3 y0 +…
We have ∆y0 - ∆y-1 = ∆2 y-1,
∆y0 = ∆y-1 + ∆2 y-1.
Similarly ∆2 y0 = ∆2 y-1 + ∆3 y-1,
∆3 y0 = ∆3 y-1 + ∆4 y-1.
Also
∆3y-1 - ∆3y-2 = ∆4 y-2,
∆3 y-1 = ∆3 y-2 + ∆4 y-2.
Similarly ∆4 y-1 = ∆4 y-2 + ∆5 y-2,
∆5 y-1 = ∆5 y-2 + ∆6 y-2 .
Substituting the above in The Newton’s forward interpolation formula, we get
y(x) = y0 + p ∆y-1 + p(p+1)/2! ∆2 y-1+ (p+1)p(p-1)/3! ∆3 y-2+ (p+2)(p+1)p(p-1)/4! ∆4 y-2+ …
which is called Gauss backward interpolation formula.
In the central difference notation
y(x) = y0 + p δy-1/2 + (p+1)p/2! δ2 y0+ (p+1)p(p-1)/3! δ3 y-1/2+ (p+2)(p+1)p(p-1)/4! δ4 y0+ …
It employs odd differences just above the central line and even differences on the central line.
This formula is used to interpolate the values of y for p (-1<p<0).
Value
of x
Value
of y
x0-2h
y-2
1st diff.
2nd diff.
3rd diff.
4th diff.
∆y-2(=δy-3/2)
x0-h
∆2y-2(=δ2y-1)
y-1
∆y-1(=δy-1/2)
x0
∆2y-1(=δ2y0)
y0
∆y0(=δy1/2)
Drgsk@KLUniversity
∆3y-2(=δ3y-1/2)
∆4y-2(=δ4y0)
∆3y-1(=δ3y1/2)
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II/IV B.Tech CSE&ECM Mathematical Methods for Computing (11BS203) Chapter-2
x0+h
∆2y0(=δ2y1)
y1
∆y1(=δy3/2)
x0+2h
y2
Example#2. Interpolate by means of Gauss’s backward formula, the population of a town for the
year 1974, given that
Year
1939
1949
1959
1969
1979
1989
Population
12
15
20
27
39
52
(in thousands)
Sol. Let x be the year and y be the population. Taking x0 = 1969, h=10. We require to find value
of y for x=1974. Therefore,
p=(x-x0)/h =(1974-1969)/10 = 0.5.
The difference table is
x
y
1939
12
∆y
∆2y
∆3y
∆4y
∆5y
3
1949
15
2
5
1959
20
0
2
7
x0=1969
27
3
5
12
1979
3
39
-10
-7
-4
1
13
1989
52
Gauss’s backward formula is
y(x)= y0 + p ∆y-1 + p(p+1)/2! ∆2 y-1+ (p+1)p(p-1)/3! ∆3 y-2+ (p+2)(p+1)p(p-1)/4! ∆4 y-2+ …
y(1974) = 27+(0.5)(7)+(1.5)(0.5)/2 (5)+ (1.5)(0.5)(-0.5)/6 (3) + (2.5)(1.5)(0.5)(-0.5)/24 (-7)
+ (2.5)(1.5)(0.5)(-0.5)(-1.5)/120 (-10)
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II/IV B.Tech CSE&ECM Mathematical Methods for Computing (11BS203) Chapter-2
= 32.345 thousands.
Stirling’s interpolation formula:
The Gauss’s forward formula is
yp = y0 + p ∆y0 + p(p-1)/2! ∆2 y-1+ (p+1)p(p-1)/3! ∆3 y-1+ (p+1)p(p-1)(p-2)/4! ∆4 y-2+ …
(1)
Gauss’s backward formula is
yp = y0 + p ∆y-1 + p(p+1)/2! ∆2 y-1+ (p+1)p(p-1)/3! ∆3 y-2+ (p+2)(p+1)p(p-1)/4! ∆4 y-2+ … (2)
Taking the mean of (1) and (2), we obtain
yp = y0 + p (∆y0 + ∆y-1)/2+ p2/2! ∆2 y-1+ p(p2-1)/3! (∆3 y-1+ ∆3y-2)/2 + p2(p2-1)/4! ∆4 y-2
+ p(p2-1)(p2-22)/5! (∆5 y-2+ ∆5 y-3)/2+…
which is called Stirling’s formula.
This formula involves means of the odd differences just above and below the central line
and even differences on this line.
Value
of x
Value
of y
x0-2h
y-2
1st diff.
2nd diff.
3rd diff.
4th diff.
∆y-2(=δy-3/2)
x0-h
∆2y-2(=δ2y-1)
y-1
∆y-1(=δy-1/2)
x0
∆3y-2(=δ3y-1/2)
∆2y-1(=δ2y0)
y0
∆y0(=δy1/2)
x0+h
∆4y-2(=δ4y0)
∆3y-1(=δ3y1/2)
∆2y0(=δ2y1)
y1
∆y1(=δy3/2)
x0+2h
y2
Example#1. Employ Stirling’s formula to compute y12.2 from the following table
(yx=1+log10sinx):
xo
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10
11
12
13
14
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II/IV B.Tech CSE&ECM Mathematical Methods for Computing (11BS203) Chapter-2
105 yx
23,967
28,060
31,788
35,209
38,368
Sol. Taking x0=12, h=1, since x=12.2, we have
p=(x-x0)/h = (12.2-12)/1 = 0.2.
The difference table is
x
yx
10
0.23967
∆y
∆2y
∆3y
∆4y
0.04093
11
0.28060
-0.00365
0.03728
x0=12
0.31788
0.00058
-0.00307
0.03421
13
0.35209
-0.00013
0.00045
-0.00262
0.03159
14
0.38368
By Stirling’s formula
y12.2 = y0 + p (∆y0 + ∆y-1)/2+ p2/2! ∆2 y-1+ p(p2-1)/3! (∆3 y-1+ ∆3y-2)/2 + p2(p2-1)/4! ∆4 y-2
= 0.31788+0.2(0.03728+0.03421)/2 + (0.2)2/2 (-0.00307)+0.2[(0.2)2-1]/6
(0.00058 - 0.00045)/2 + (0.2)2[(0.2)2-1]/24 (-0.00013)
= 0.31788 + 0.00715 - 0.00006 - 0.000002 + 0.0000002 =0.32495.
Bessel’s Interpolation Formula:
Gauss forward interpolation formula is
yp = y0 + p∆y0 + p(p-1)/2! ∆2y-1 + (p+1)p(p-1)/3! ∆3y-1 + (p+1)p(p-1)(p-2)/4! ∆4y-2 + … (1)
We have
∆2y0 - ∆2y-1 = ∆3y-1
∆2y-1 = ∆2y0 - ∆3y-1.
Similarly
∆4y-2 = ∆4y-1 - ∆5y-2.
Now we write (1) as
yp = y0 + p∆y0 + p(p-1)/2! [(1/2)∆2y-1+(1/2) ∆2y-1] + (p+1)p(p-1)/3! ∆3y-1
+ (p+1)p(p-1)(p-2)/4! [(1/2)∆4y-2+(1/2) ∆4y-2] + …
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II/IV B.Tech CSE&ECM Mathematical Methods for Computing (11BS203) Chapter-2
= y0 + p∆y0 + p(p-1)/2! [(1/2)∆2y-1+(1/2) (∆2y0-∆3y-1)] +(p+1)p(p-1)/3! ∆3y-1
+ (p+1)p(p-1)(p-2)/4! [(1/2)∆4y-2+(1/2) (∆4y-1-∆5y-2)] +…
After simplification, we get
yp = y0 + p∆y0 + p(p-1)/2! [∆2y-1+∆2y0]/2 + (p-1/2)p(p-1)/3! ∆3y-1
+(p+1)p(p-1)(p-2)/4! [∆4y-2+∆4y-1]/2+ …,
which is known as Bessel’s formula. This is very useful formula for practical purposes. It
involves odd differences below central line and means of even differences of and below this line.
It is best suitable for value of p between ¼ and ¾ .
Value
of x
Value
of y
x0-2h
y-2
1st diff.
2nd diff.
3rd diff.
4th diff.
∆y-2(=δy-3/2)
x0-h
∆2y-2(=δ2y-1)
y-1
∆y-1(=δy-1/2)
x0
∆3y-2(=δ3y-1/2)
∆2y-1(=δ2y0)
y0
∆y0(=δy1/2)
x0+h
∆4y-2(=δ4y0)
∆3y-1(=δ3y1/2)
∆2y0(=δ2y1)
y1
∆y1(=δy3/2)
x0+2h
y2
Example#1. Apply Bessel’s formula to obtain y25, given that y20=2854, y24=3162, y28=3544 and
y32=3992.
Sol. Taking x0=24, h=4. Since x=25, we have
p= (x-x0)/h = (25-24)/4 = ¼ = 0.25.
The difference table is
x
y
20
2854
∆
∆2
∆3
308
x0=24
3162
74
382
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II/IV B.Tech CSE&ECM Mathematical Methods for Computing (11BS203) Chapter-2
28
3544
66
448
32
3992
Bessel’s formula is
yp = y0 + p∆y0 + p(p-1)/2! [∆2y-1+∆2y0]/2 + (p-1/2)p(p-1)/3! ∆3y-1
+ (p+1)p(p-1)(p-2)/4! [∆4y-2+∆4y-1]/2+ …
y25 = 3162 + (0.25)382 + (0.25)(0.25-1)/2 [74+66]/2
+ (0.25-1/2)0.25(0.25-1)/6 (-8)
= 3162 + 95.5 – 6 – 5625 – 0.0625
= 3250.875.
Interpolation with unequal intervals:
The disadvantage for the previous interpolation formulas is that, they are used only for
equal intervals. The following are the interpolation with unequal intervals;
1) Lagrange’s formula for unequal intervals,
2) Newton’s divided difference formula.
Lagrange’s interpolation formula: If y = f(x) takes the values y0, y1, y2, …, yn corresponding
to x0, x1, x2, …, xn, then
(x  x 0 )(x  x 2 ) (x  x n )
(x  x 1 )(x  x 2 ) (x  x n )
f(x) 
y0 
y1  
(x 0  x 1 )(x 0  x 2 ) (x 0  x n )
(x 1  x 0 )(x 1  x 2 ) (x 1  x n )

(x  x 0 )(x  x 1 ) (x  x n 1 )
yn ,
(x n  x 0 )(x n  x 1 ) (x n  x n 1 )
which is known as Lagrange’s formula.
Divided Differences: If (x0, y0), (x1, y1), …, (xn, yn) are given points, then the first divided
differences for the argument x0, x1 is defined by
y  y0
.
[x 0 , x 1 ]  1
x1  x 0
Similarly
y  y2
y  y1
y  y n 1
[x 1 , x 2 ]  2
, [x 2 , x 3 ]  3
,, [x n 1 , x n ]  n
.
x 2  x1
x3  x2
x n  x n 1
The second divided differences for x0, x1, x2 is
[x , x ]  [x 0 , x 1 ]
.
[x 0 , x 1 , x 2 ]  1 2
x2  x0
The third divided differences for x0, x1, x2 , x3 is
[x , x , x ]  [x 0 , x 1 , x 2 ]
.
[x 0 , x 1 , x 2 , x 3 ]  1 2 3
x3  x0
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II/IV B.Tech CSE&ECM Mathematical Methods for Computing (11BS203) Chapter-2
And so on, the nth divided differences for x0, x1, x2, …, xn is
[x , x ,, x n ]  [x 0 , x 1 ,, x n -1 ]
[x 0 , x 1 , x 2 ,, x n ]  1 2
.
xn  x0
All the divided differences systematically set out in a table called divided difference table.
Value
of x
Value
of y
x0
y0
1st
divided
difference
2nd
divided
difference
3rd divided
difference
4th divided
difference
5th divided
difference
[x0,x1]
x1
y1
[x0,x1,x2]
[x1,x2]
x2
y2
[x0,x1,x2,x3]
[x1,x2,x3]
[x2,x3]
x3
y3
[x1,x2,x3,x4]
[x2,x3,x4]
[x3,x4]
x4
[x0,x1,x2,x3,x4]
y4
[x0,x1,x2,x3,x4,x5]
[x1,x2,x3,x4,x5]
[x2,x3,x4,x5]
[x3,x4,x5]
[x4,x5]
x5
y5
Newton’s divided difference formula: If y = f(x) takes the values y0, y1, y2, …, yn
corresponding to x0, x1, x2, …, xn, then
f(x) = y0+(x-x0)[x0, x1] + (x-x0)(x-x1)[x0, x1, x2] + …+ (x-x0)(x-x1)…(x-xn-1)[x0, x1, x2, …, xn],
which is known as Newton’s general interpolation formula with divided differences.
Example#1. Given the values
x:
5
f(x):
150
7
11
13
17
392
1452
2366
5202
Evaluate f(9), using
(i) Lagranges formula
(ii) Newton’s divided difference formula.
Sol. Let y = f(x), then from the given data, we have
x0 = 5, x1 = 7, x2 = 11, x3 = 13, x4 = 17 and y0 = 150, y1 = 392, y2 = 1452, y3 = 2366, y4 = 5202.
(i) By Lagrnge’s interpolation formula
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II/IV B.Tech CSE&ECM Mathematical Methods for Computing (11BS203) Chapter-2
f(x) 
(x  x 1 )(x  x 2 )( x  x3 )(x  x 4 )
(x  x 0 )(x  x 2 )( x  x3 )(x  x 4 )
y0 
y1
(x 0  x 1 )(x 0  x 2 )( x0  x3 )(x 0  x n )
(x 1  x 0 )(x 1  x 2 )( x1  x3 )(x 1  x n )

(x  x 0 )(x  x 1 )( x  x3 )(x  x 4 )
(x  x 0 )(x  x 1 )( x  x 2 )(x  x 4 )
y2 
y3
(x 2  x 0 )(x 2  x 1 )( x 2  x3 )(x 2  x 4 )
(x 3  x 0 )(x 3  x 1 )( x3  x 2 )(x 3  x 4 )
(x  x 0 )(x  x 1 )( x  x 2 )(x  x 3 )
y4.
(x 4  x 0 )(x 4  x 1 )( x 4  x 2 )(x 4  x 3 )
(9  7)(9  11)(9  13)(9  17)
(9  5)(9  11)(9  13)(9  17)
f(9) 
 150 
 392
(5  7)(5  11)(5  13)(5  17)
(7  5)(7  11)(7  13)(7  17)
(9  5)(9  7)(9  13)(9  17)
(9  5)(9  7)(9  11)(9  17)

 1452 
 2366
(11  8)(11  7)(11  13)(11  17)
(13  5)(13  7)(13  11)(13  17)
(9  5)(9  7)(9  11)(9  13)
50 3136 3872 2366 578

 5202   



 810.
(17  5)(17  7)(17  11)(17  13)
3
15
3
3
5
(ii) The divided difference table is

Value
of x
Value
of y
5
150
1st
divided
difference
2nd
divided
difference
3rd divided
difference
4th divided
difference
121
7
392
24
265
11
1452
1
32
457
13
2366
0
1
42
709
17
5202
By Newton divided difference formula
f(x) = y0+(x-x0)[x0, x1] + (x-x0)(x-x1)[x0, x1, x2] +(x-x0)(x-x1)(x-x2)[x0, x1, x2,x3]
+ (x-x0)(x-x1)(x-x2)(x-x3)[x0, x1, x2,x3,x4].
f(9) = 150 + (9 – 5)×121 + (9 – 5) (9 – 7)×24 + (9 – 5)(9 – 7)(9 – 11)×1
+ (9 – 5)(9 – 7)(9 – 11)(9 – 13)×0
= 150 + 484 + 192 – 16 + 0
= 810.
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Numerical Differentiation:
Mathematically, the derivative represents the rate of change of a dependent variable with respect
to an independent variable. For example, if we are given a function y(t) that specifies an object’s
position as a function of time, differentiation provides a means to determine its velocity, as in:
As in following Figure, the derivative can be visualized as the slope of a function.
Numerical differentiation is used when the function y = f(x) is given in tabular form or it
is highly complex. The basic idea in numerical differentiation is to replace the given function y =
f(x) on the interval by an interpolating polynomial P(x) and set f '(x) = P'(x), f '' (x) = P'' (x) etc.
Numerical differentiation is less exact than interpolation.
Numerical differentiation using Newton’s forward formula: Suppose y = f(x) is specified in
an interval [a, b] at equally spaced points xi = x0 + ih (i = 0, 1, …, n) (x0=a, xn=b) by means of
values yi = f(xi). By Newton forward interpolation formula
p(p  1) 2
p(p  1)(p  2) 3
p(p  1)(p  2)(p  3) 4
y  f(x)  y 0  py 0 
Δ y0 
Δ y0 
Δ y 0  ... ,
2!
3!
4!
x  x0
dp 1
 .
where p 
and h = xi+1 – xi, for i = 1, 2, …, n. Here p is a function of x and
dx h
h
Rewriting the above equation, we have
p2  p 2
p 3  3p 2  2p 3
p 4  6p 3  11p 2  3p 4
y(x)  y 0  py 0 
Δ y0 
Δ y0 
Δ y 0  ...
2
6
24
Differentiating the above equation with respect to x, we have

dy dy dp 1 dy 1 
2p  1 2
3p 2  6p  2 3
4p 3  18p 2  22p  6 4


 Δy 0 
Δ y0 
Δ y0 
Δ y 0  ....
dx dp dx h dp h 
2
6
24

Again differentiating with respect to x, we get
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II/IV B.Tech CSE&ECM Mathematical Methods for Computing (11BS203) Chapter-2

d 2 y d  dy  d  dy  dp 1 d  dy  1  2
6p 2  18p  11 4
3




Δ
y

(
p

1
)
Δ
y

Δ y 0  ....






0
0
2
2 
dx  dx  dp  dx  dx h dp  dx  h 
12
dx

Special case: If the derivative is required to find at a basic tabulated point x = xi, then choose
x0= xi and the formulas become
1
1
1
1
1
 dy 

 Δy 0  Δ 2 y 0  Δ 3 y 0  Δ 4 y 0  Δ 5 y 0  
 
2
3
4
5
 dx  x  x 0 h 

And
 d2y 
1  2
11
5




Δ y 0  Δ 3 y 0  Δ 4 y 0  Δ 5 y 0  .
2 
 dx 2 
12
6

h

x x
0
Numerical differentiation using Newton’s backward formula:
In this case we replace y(x) by Newton’s backward interpolation formula
p(p  1) 2
p(p  1)(p  2) 3
p(p  1)(p  2)(p  3) 4
y(x)  y n  py n 
 yn 
 yn 
 y n  ...
2!
3!
4!
x  xn
dp 1
 .
and h = xi+1 – xi, for i = 1, 2, …, n. Here p is a function of x and
dx h
h
Rewriting the above equation, we have
where p 
p2  p 2
p 3  3p 2  2p 3
p 4  6p 3  11p 2  3p 4
y(x)  y n  py 0 
 yn 
 yn 
 y n  ...
2
6
24
Differentiating the above equation with respect to x, we have

dy dy dp 1 dy 1 
2p  1 2
3p 2  6p  2 3
4p 3  18p 2  22p  6 4


 y n 
 yn 
 yn 
 y n  ....
dx dp dx h dp h 
2
6
24

Again differentiating with respect to x, we get

d 2 y d  dy  d  dy  dp 1 d  dy  1  2
6p 2  18p  11 4




 y n  ( p  1) 3 y n 
 y n  ....







12
dx 2 dx  dx  dp  dx  dx h dp  dx  h 2 

Special case: If the derivative is required to find at a basic tabulated point x = xi, then choose
xn= xi and the formulas become
1
1
1
1
1
 dy 

 y 0   2 y n   3 y n   4 y n   5 y n  
 
2
3
4
5
 dx  x  x n h 

and
 d2y 
1  2
11
5




 y n   3 y n   4 y n   5 y n  .

2 
 dx 2 
12
6


x x n h
Example#1. Compute f'(x) and f''(x) at (i) x = 16 (ii) x = 15 (iii) x = 24 (iv) x = 25 from the
following table
x
15
17
19
21
23
25
f(x)
3.873
4.123
4.359
4.583
4.796
5.8
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II/IV B.Tech CSE&ECM Mathematical Methods for Computing (11BS203) Chapter-2
Sol. Let y = f(x), then from the given table x0 = 15, x1 = 17, x2 = 19, x3 = 21, x4 = 23, x5 = 25
and y0 = 3.873, y1 = 4.123, y2 = 4.359, y3 = 4.583, y4 = 4.796, y5 = 5.8. The finite difference table
is
x
y
∆y
∆2y
∆3y
∆4y
∆5y
15
3.873
0.25
17
4.123
-0.014
0.236
0.002
19
4.359
-0.012
-0.001
0.224
0.001
0.002
21
4.583
-0.011
0.001
0.213
0.002
23
4.796
-0.009
0.204
25
5
(i) Since x = 16 is nearer to the beginning of the table we use Newton forward formula. Here the
x  x0 16  15 1
step size h = 2. Taking x0 = 15, then p 

  0.5.
h
2
2
Newton forward formula to compute first derivative of y=f(x) is

dy 1 
2p  1 2
3p 2  6p  2 3
4p 3  18p 2  22p  6 4
 Δy 0 
Δ y0 
Δ y0 
Δ y 0  ....
dx h 
2
6
24

2
3
4
5
Substituting x = 16, p = 0.5, ∆y0 = 0.25, ∆ y0= – 0.014, ∆ y0 = 0.002, ∆ y0= – 0.001, ∆ y0=0.002,
we have
1
2(0.5)  1
3(0.5) 2  6(0.5)  2
 dy 
 0.25 
(-0.014) 
(0.002)
 
2
6
 dx  x 16 2 

4(0.5) 3  18(0.5) 2  22(0.5)  6

(-0.001).
24

fʹ(16) = 0.1249375.
Newton forward formula to compute second derivative of y=f(x) is

d2y
1  2
6p 2  18p  11 4

Δ y 0  ( p  1)Δ 3 y 0 
Δ y 0  ....

12
dx 2 h 2 

Substituting the values from the table, we have
 d2y 

1 
6(0.5) 2  18(0.5)  11



- 0.014  (0.5  1)(0.002) 
(-0.001).

2
 dx 2 
12


 x16 2 
fʹʹ(16) = -0.0038229.
(ii) Since x = 15 is in the beginning of the table, we use Newton forward formula. Here the step
size h = 2, x = x0 = 15 and p = 0.
Newton forward formula to compute first derivative of y=f(x) at x = x0 is
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II/IV B.Tech CSE&ECM Mathematical Methods for Computing (11BS203) Chapter-2
1
1
1
1
1
 dy 

 Δy 0  Δ 2 y 0  Δ 3 y 0  Δ 4 y 0  Δ 5 y 0  
 
2
3
4
5
 dx  x  x 0 h 

Substituting the values from the table, we have
1
1
1
1
1
 dy 

 0.25  (-0.014)  (0.002)  (-0.001)  (0.002)
 
2
3
4
5
 dx  x 15 2 

f'(15) = 0.128958.
Newton forward formula to compute second derivative of y=f(x) at x = x0 is
 d2y 
1  2
11
5




Δ y 0  Δ 3 y 0  Δ 4 y 0  Δ 5 y 0  .
2 
 dx 2 
12
6

h

x x
0
Substituting the values from the table, we have
 d2y 
1 
11
5




- 0.014  0.002  (-0.001)  (0.002) .
2 
 dx 2 
12
6


 x 15 2
f''(15) = -0.004229.
(iii) Since x = 24 is nearer to the ending of the table, we use Newton backward formula. Here the
x  xn 24  25
1
step size h = 2. Taking xn = 25, then p 

   0.5.
h
2
2
Newton backward formula to compute first derivative of y=f(x) is

dy 1 
2p  1 2
3p 2  6p  2 3
4p 3  18p 2  22p  6 4
 y n 
 yn 
 yn 
 y n  ....
dx h 
2
6
24

Substituting x = 25, p = – 0.5,  yn = 0.204,  2yn= – 0.009,  3yn = 0.002,  4yn= 0.001,
 5yn=0.002, we have
2(-0.5)  1
3(-0.5) 2  6(-0.5)  2
1
 dy 
 0.204 
(0.009) 
(0.002)
 
2
6
 dx  x 25 2 


4(-0.5)3  18(-0.5) 2  22(-0.5)  6
(0.001).
24

f'(24) = 0.09727.
Newton backward formula to compute second derivative of y=f(x) is

d2y
1  2
6p 2  18p  11 4
3

 y n  ( p  1) y n 
 y n  ....

12
dx 2 h 2 

Substituting the values from the table, we have
 d2y 

1 
6(-0.5)2  18(-0.5)  11




0
.
009

(

0
.
5

1
)(
0
.
002
)

(0.001).

2
 dx 2 
12


 x24 2 
f''(24) = -0.00242708.
(iv) Since x = 25 is in the ending of the table, we use Newton forward formula. Here the step size
h = 2, x = x0 = 15 and p = 0.
Newton backward formula to compute first derivative of y=f(x) at x = xn is
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II/IV B.Tech CSE&ECM Mathematical Methods for Computing (11BS203) Chapter-2
1
1
1
1
1
 dy 

 y 0   2 y n   3 y n   4 y n   5 y n   .
 
2
3
4
5
 dx  x  x n h 

Substituting the values from the table, we have
1
1
1
1
1
 dy 

 0.204  (0.009)  (0.002)  (0.001)  (0.002).
 
2
3
4
5
 dx  x  25 2 

f '(25) = 0.10048
Newton backward formula to compute second derivative of y=f(x) at x = xn is
 d2y 
1  2
11
5




 y n   3 y n   4 y n   5 y n  .
2 
 dx 2 
12
6


x x n h
Substituting the values from the table, we have
 d2y 
1 
11
5





0
.
009

0
.
002

(
0
.
001
)

(0.002)  .
2 
 dx 2 
12
6


 x  25 2
f ''(25) = -0.001833.
Numerical Integration
Integration is the inverse of differentiation. Just as differentiation uses differences to quantify an
instantaneous process, integration involves summing instantaneous information to give a total
result over an interval. Thus, if we are provided with velocity as a function of time, integration
can be used to determine the distance traveled:
According to the dictionary definition, to integrate means “to bring together, as parts, into
a whole; to unite; to indicate the total amount” Mathematically, definite integration is
represented by
(1)
which stands for the integral of the function f (x) with respect to the independent variable x,
evaluated between the limits x = a to x = b. As suggested by the dictionary definition, the
“meaning” of Eq. (1) is the total value, or summation, of f (x)dx over the range x = a to b. In fact,
the symbol ∫ is actually a stylized capital S that is intended to signify the close connection
between integration and summation.
Geometrically, integration is just finding the area under a curve from one point to another. It is
b
represented by  f ( x)dx , where the numbers a and b are the lower and upper limits of
a
integration, respectively, the function f is the integrand of the integral, and x is the variable of
integration. Figure 1 represents a graphical demonstration of the concept.
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II/IV B.Tech CSE&ECM Mathematical Methods for Computing (11BS203) Chapter-2
Why are we interested in integration: because most equations in physics are differential
equations that must be integrated to find the solution(s). Furthermore, some physical quantities
can be obtained by integration (example: displacement from velocity).
The problem is that sometimes integrating analytically some functions can easily become
laborious. For this reason, a wide variety of numerical methods have been developed to find the
integral.
The process of evaluating a definite integral from a set of tabulated values of the integrand f(x) is
called numerical integration. This process when applied to a function of single variable, is known
as quadrature.
b
Let I   y(x)dx , where y(x) takes the values y0, y1, y2, …, yn for x = x0, x1, x2, …, xn. Let
a
us divide the interval (a, b) into n sub-intervals of width h so that x0 = a, x1 = x0 + h, x2 = x0 + 2h,
…, xn = x0 + nh = b.
xn
Trapezoidal rule :
 y(x)dx
x0

h
(y 0  y n )  2(y1  y 2  ...  y n -1 ).
2
Simpson’s 1/3rd rule :
xn
 y(x)dx

x0
h
(y 0  y n )  4(y1  y 3  ...  y n -1 )  2(y 2  y 4  ...  y n - 2 )
3
Simpson’s 3/8th rule :
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II/IV B.Tech CSE&ECM Mathematical Methods for Computing (11BS203) Chapter-2
xn
 y(x)dx

x0
3h
(y 0  y n )  3(y1  y 2  y 4  y 5  ...  y n -1  y n - 2 )  2(y 3  y 6  ...  y n -3 )
8
6
Problem#1. Compute the integral

dx
01 x
2
, using (i) Trapezoidal rule (ii) Simpson’s 1/3rd rule
(iii) Simpson’s 3/8 rule and also determine the relative true error.
1
Sol. Let y(x) 
and divide the interval (0, 6) into n = 6 subintervals each of length h = 1.
1 x2
Then, we have the following tabular values.
x
0
1
2
3
4
5
6
y(x)
1
0.5
0.2
0.1
0.0588
0.0385
0.027
(i) Trapezoidal rule
th
xn
 y(x)dx

h
(y 0  y n )  2(y1  y 2  ...  y n -1 ).
2

h
(y 0  y 6 )  2(y1  y 2  y 3  y 4  y 5 )
2
x0
6
 y(x)dx
0
1
(1  0.027)  2(0.5  0.2  0.1  0.0588  0.0385)
2
 1.4108.
(ii) Simpson’s 1/3rd rule

xn
 y(x)dx

x0
6
 y(x)dx

0
h
(y 0  y n )  4(y1  y 3  ...  y n -1 )  2(y 2  y 4  ...  y n - 2 )
3
h
(y 0  y 6 )  4(y1  y 3  y 5 )  2(y 2  y 4 )
3
1
(1  0.027)  4(0.5  0.1  0.0385)  2(0.2  0.0588)
3
 1.3662.
(iii) Simpson’s 3/8th rule

xn
 y(x)dx

x0
6
3h
(y 0  y n )  3(y1  y 2  y 4  y 5  ...  y n -1  y n - 2 )  2(y 3  y 6  ...  y n -3 )
8
 y(x)dx

3h
(y 0  y 6 )  3(y1  y 2  y 4  y 5 )  2(y 3 )
8

3
(0  0.027)  3(0.5  0.2  0.0588  0.0385)  2(0.1)  1.3571.
8
0
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II/IV B.Tech CSE&ECM Mathematical Methods for Computing (11BS203) Chapter-2
The relative true error 
True value :
6

dx
01 x
2

True value - Approximat e value
 100% .
True value

6
 tan 1 ( x) 0  tan 1 (6)  tan 1 (0)  1.4056  0  1.4056.
1.4056 - 1.4108
 100%  0.3699%.
1.4056
1.4056 - 1.3662
(ii) Simpson’s 1/3rd rule =
 100%  2.803%.
1.4056
1.4056 - 1.3571
(iii) Simpson’s 3/8th rule =
 100%  3.4504%.
1.4056
Relative true error for (i) Trapezoidal rule =
Problems
1. Estimate the missing values in the following table
x
45
50
55
60
y
3.0
?
2.0
?
65
-2.4
2. Express y=2x3-3x2+3x-10 in factorial notation and hence show that Δ3y=12.
3. Given Sin450= 0.7071, Sin500= 0.7660, Sin550= 0.8192, Sin 600= 0.8660, find Sin520, using
Newton’s forward formula.
4. From the following table, estimate the number of students who obtained marks between 40
and45
Marks
30-40
40-50
50-60
60-70
70-80
No. of
31
42
51
35
students
5. Construct a cubic polynomial which takes the following values:
x
0
1
2
f(x)
1
2
1
6. The area of a circle of diameter d is given for the following values:
d
80
85
90
95
A
5026
5674
6362
7088
Calculate the area of a circle of diameter 105.
31
3
10
100
7854
7. Use Gauss forward formula to evaluate y30, given that y21 = 18.4708, y25 = 17.8144,
y29 = 17.1070, y35 = 16.3432 and y37 = 15.5154.
8. Interpolate by means of Gauss’s backward formula, the population of a town for the year
1974, given that
Year
1939
1949
1959
1969
1979
1989
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II/IV B.Tech CSE&ECM Mathematical Methods for Computing (11BS203) Chapter-2
Population
(in thousands)
12
15
20
27
39
52
9. Employ Sterling’s formula to compute y12.2 from the following table:
x0
10
11
12
13
14
105ux
23,967
28,060
31,788
35,209
38,368
10. Apply Bessel’s formula to obtain y25, given y20 = 2854, y24 = 3162, y28 = 3544, y32 = 3992.
11. Given the values
x:
5
7
11
13
17
f(x):
150
392
1452
2366
5202
Evaluate f(9) by using (a) Lagrange’s formula (b) Newton’s divided difference formula.
12. Use Lagrange’s formula to find the form of f(x) when,
x:
f(x):
0
2
3
6
648
704
729
792
13. Determine f(x) as a polynomial in x for the following data using Newton’s divided difference
formula
x:
-4
-1
0
2
5
f(x):
1245
33
5
9
1335
14. Apply Lagrange’s formula inversely to obtain a root of the equation f(x) = 0 f(30)= -30, f(34)
= -13,f(38) = 3 and f(42)= 18.
15. Given that
x
1.0
1.1
1.2
1.3
1.4
1.5
1.6
y
7.989
8.403
8.781
9.129
9.451
9.750
10.031
find the values of dy/dx and d2y/dx2 at x = 1.1and at x 1.6.
16. Find the first and second derivatives of the function tabulated below, at the point x = 1.1
x:
1.0
1.2
1.4
1.6
1.8
2.0
f(x):
0
0.128
0.544
1.296
2.432
4.00
17. From the following table, find the values of dy/dx and d2y/dx2 at x = 2.03
x:
1.96
1.98
2.00
2.02
2.04
y:
0.7825 0.7739 0.7651 0.7563 0.7473
18. The following data gives the corresponding values of pressure and specific volume of a
superheated steam.
v
2
4
6
8
10
p
105
42.7
25.3
16.7
13
19. From the table below, for each value of x, y is minimum? Also find the value of y
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II/IV B.Tech CSE&ECM Mathematical Methods for Computing (11BS203) Chapter-2
x
y
3
0.205
4
0.240
5
0.259
6
0.262
7
0.250
8
0.224
20. Given that ,
x:
4.0
4.2
4.4
4.6
4.8
5.0
5.2
logx:
1.3863 1.4351 1.4816 1.5261 1.5686 1.6094 1.6484
Evaluate
by (a) Trapezoidal rule(b) Simpson’s 1/3 rule (c) Simpson’s
th
3/8 rule.
21. Use Simpson’s 1/3rd rule to find
dx by taking seven ordinates.
22. The velocity (km/min) of a moped which starts from rest, is given at fixed intervals of time t
(min) as follows:
T
2
4
6
8
10
12
14
16
18
20
V
10
18
25
29
32
20
11
5
2
0
Estimate approximately the distance covered in 20 minutes.
23. A river is 80 feet wide. The depth d in feet at a distance x feet from one bank is given by the
following table:
x
0
10
20
30
40
50
60
70
80
d
0
4
7
9
12
15
14
8
3
Find approximately the area of the cross-section.
24. A solid of revolution is formed by rotating about the X-axis, the area between the X-axis, the
lines x=0 and x=1 and a curve through the points with the following coordinate:
x
0.00
0.25
0.50
0.75
1.00
y
1.0000
0.9896
0.9589
0.9089
0.8415
Short Answer Questions
1. Evaluate Δ2 (abx), interval of differences being unity.
2. Show that Δ log f(x)= log{1+Δf(x)/f(x)}
3. Evaluate Δ10 [(1-x) (1-2x2) (1-3x3) (1-4x4)], if the interval of differencing is 2.
4. Obtain the function whose first difference is 2x3+3x2-5x+4.
5. Prove that y3 = y2 + Δy1 + Δ2y0 + Δ3y0.
6. Prove with usual notations that (E1/2 + E-1/2) (1 + Δ) ½ = 2 + Δ.
7. Prove with usual notations that Δ3y2 =  3y5.
8. Prove with usual notations that µ2 = 1 + δ2/4.
9. State Newton’s forward interpolative formula.
10. State Bessel’s formula.
11. Write the relation between Δ and E.
12. If f(x) = 3x3 – 2x2 + 1, then find Δ3f(x)
13. State Stirling’s formula.
14. Write the relation between Δ , E and 
15. Prove with usual notations that (1+Δ)(1-  ) = 1
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II/IV B.Tech CSE&ECM Mathematical Methods for Computing (11BS203) Chapter-2
16. State Lagrange’s Interpolation formula
17. State Newton’s divided difference formula
18. By Trapezoidal rule, write the value of
dx
19. State Simpson’s 3/8 th rule.
20. If f(x) is given by
x= 0
0.5
1
1.5
f(x) = 0
0.25 1
2.25
Then the value of
dx by Simpson’s 1/3 rule.
Drgsk@KLUniversity
2
4.
Page 30