UNIT-IV

INTEPOLATION, NUMERICAL DIFFERENTIATION

AND NUMERICAL INTEGRATION
PART-A
1. State Lagrange’s interpolation formula.
Solution:
Let y=f(x) be a function which takes the values y0, y1 y2, y3…………….yn
corresponding to x= x0, x1 x2, x3…………….xn. Then Lagrange’s interpolation formula is
( x  x1 )( x  x2 ).......................( x  xn )
y  f ( x)  y0
( x0  x1 )( x0  x2 )....................( x0  xn )
( x  x0 )( x  x2 ).......................( x  xn )
 y1
( x1  x0 )( x1  x2 )....................( x1  xn )
+…………………………………………….
( x  x1 )( x  x2 ).......................( x  xn1 )
 yn
( xn  x0 )( xn  x1 )....................( xn  xn1 ) .
2. Give the inverse of Lagranges interpolation formula.
Solution:
( y  y1 )( y  y 2 ).......................( y  y n )
x x0
( y0  y1 )( y0  y 2 )....................( y0  y n )
( y  y0 )( y  y 2 ).......................( y  y n )
 x1
( y1  y0 )( y1  y 2 )....................( y1  y n )
+…………………………………………….
( y  y1 )( y  y 2 ).......................( y  y n1 )
 xn
( y n  y0 )( y n  y1 )....................( y n  y n1 )
3. What is the need of Newton’s & Lagrange’s interpolation formulae?
Solution:
To develop interpolation formula for unequally space3d values of x.
4. Find the parabola of the form y=ax2+bx+c passing through the points (0,0),
(1,1) & (2,20).
Solution: We use Lagrange’s interpolation formula
( x  x1 )( x  x2 ). ( x  x0 )( x  x2 ). ( x  x0 )( x  x1 ).
y  f ( x)  y0  y1  y2
( x0  x1 )( x0  x2 ). ( x1  x0 )( x1  x2 ). ( x2  x0 )( x2  x1 ).
( x  1)( x  2). ( x  0)( x  2). ( x  0)( x  1).
 0 1 20
(0  1)(0  2). (1  0)(1  2). (2  0)( 2  1).
y = 9x2-8x.

1
 1 1
  
3
5. Show that .
bcd a abcd
Solution:
1 1
If f ( x)  , f (a) 
x a
1 1

1 b a 1
f ( a, b)      ,
b a ba ab
1 1
 
f (b, c)  f (a, b) bc ab  1  c  a   1
f (a, b, c)    
ca ca abc  c  a  abc
1 1

f (b, c, d )  f (a, b, c) bcd abc 1 ad  1
f (a, b, c)     
d a d a abcd  d  a  abcd
1 1
  
3
Therefore, .
bcd a abcd
dy d2y
6. Write down the expressions for and at x  x0 by Newton’s forward
dx dx 2
difference formula. (or)
State the formula to find the first and second order derivative using the forward
differences.
Solution:
dy 1 1 1 1 
 y 0  2 y 0  3 y 0  4 y 0  ...........
dx x  x0 h  2 3 4 

d2y 1  2 11 4 
  y 0   y 0  12  y 0  ......................... .
3

dx 2 x  x0
h2

dy d2y
7. Write down the expressions for and at x  x n by Newton’s backward
dx dx 2
difference formula.
Solution:
dy 1 1 1 1 
 y 0   2 y 0   3 y 0   4 y 0  ...........
dx x  xn h  2 3 4 
d2y 1  2 11 4 
  y 0   y 0  12  y 0  .........................
3

dx 2 x  xn
h2

2
8. Create a forward difference table for the following data and state the degree of
polynomial for the same.
x 0 1 2 3
y -1 0 3 8
Solution:
The forward difference table is as follows

x: y ∆y ∆2y ∆3y
0 -1
1
1 0
2
3
2 3 0
2
5
3 8

Since ∆2y is having constant terms, it will have a polynomial of degree 2.

𝒅𝒚
9. Find 𝒅𝒙 at x=1 from the following table.

x: 1 2 3 4
y: 1 8 27 64

Solution:
The forward difference table is as follows

x: y ∆y ∆2y ∆3y
1 1
7
2 8 12
19
3 27 6
18
37
4 64

dy 1 1 1 1 
  y 0  2 y 0  3 y 0  4 y 0  ...........
dx x  x0 h 2 3 4 
Here h=1, x0=1, ∆y0=7, ∆2y0=12 and ∆3y0=6.
dy 1  12 6 
Therefore  7     3.
dx x 1 1  2 3

3
10. Find the area under the curve passing through the points (0,0), (1,2), (2,2.5), (3,2.3),
(4,2), (5,1.7) & (6,1.5).
Solution: Given x: 0 1 2 3 4 5 6
y: 0 2 2.5 2.3 2 1.7 1.5
y0 y1 y2 y3 y4 y5 y6

By Trapezoidal rule,
𝑏 ℎ
∫𝑎 𝑓(𝑥)𝑑𝑥 = 2{(sum of the first and last ordinates)+2(sum of remaining ordinates}
6 ℎ
Area = ∫0 𝑦𝑑𝑥 = 2{( y0+ y6)+2(y1+ y2+ y3+ y4+ y5)}.
1
= {(0+1.5)+2(2+2.5+2.3+2+1.7)}
2
22.5
= = 11.25.
2
1
1 by trapezoidal rule dividing the range into 4 equal parts.
11. Evaluate

1/ 2
x
dx
1
1− 1
2
Solution: Here ℎ = = 0.125: 𝑦=𝑥
4
x: ½=0.5 0.625 0.75 0.875 1
y: 2 1.6 1.3333 1.1429 1

By Trapezoidal rule,
𝑏 ℎ
∫𝑎 𝑓(𝑥)𝑑𝑥 = 2{(sum of the first and last ordinates)+2(sum of remaining ordinates}
1 0.125
∫1/2 1/𝑥𝑑𝑥 = 2
{(2+1)+2(1.6+1.3333+1.1429)}.
= 0.6970.
1
dx
12. Using Trapezoidal rule, evaluate
 1 x
0
2
with h=0.2. Hence obtain an approximate

falue of π.
Solution: Given h=0.2
x: 0 0.2 0.4 0.6 0.8 1
1 1 0.96154 0.886207 0.73529 0.60976 0.5
𝑦=1+𝑥 2:
1 1 ℎ
∫0 𝑑𝑥 = 2{( y0+ yn)+2(y1+ y2+ y3+ y4+ …………..yn-1)}
1+𝑥 2
1
= {(0+0.5)+2(0.96154+0.886207+0.73529+0.60976+0.5)}
2
=0.783732……………………………………………(1)
By actual integration
1 1 𝜋
∫0 𝑑𝑥 = (𝑡𝑎𝑛−1 𝑥)10 = 4 …………………………………..(2)
1+𝑥 2
𝜋
From (1) & (2) = 0.783732
4
π = 3.13493 (approximately).

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 𝒙
13. State the Trapezoidal rule to evaluate ∫𝒙 𝒏 𝒇(𝒙)𝒅𝒙.
𝟎
Solution:
𝑥𝑛 ℎ
∫𝑥 𝑓(𝑥)𝑑𝑥 = 2 ⌊(𝑦0 + 𝑦𝑛 ) + 2(𝑦1 + 𝑦2 + ⋯ 𝑦𝑛−1 )⌋
0
𝑥 ℎ
(i.e) ∫𝑥 𝑛 𝑓(𝑥)𝑑𝑥 = 2{(sum of the first and last ordinates)+2(sum of remaining ordinates)}
0

14. Write down the Simpson,s 1/3-Rule in numerical integration.

Solution: Simpson’s one-third rule is given by
𝑥0 +𝑛ℎ ℎ
∫𝑥 𝑓(𝑥)𝑑𝑥 = 3 ⌊(𝑦0 + 𝑦𝑛 ) + 2(𝑦2 + 𝑦4 + ⋯ 𝑦𝑛−2 ) + 4(𝑦1 + 𝑦3 + ⋯ 𝑦𝑛−1 )⌋
0
𝑥0 +𝑛ℎ ℎ
(i.e) ∫𝑥 𝑓(𝑥)𝑑𝑥 = {(sum of the first and last ordinates)+2(sum of remaining odd
0 3
ordinates+4(sum of even ordinates)}

15. What are the errors in Trapezoidal and Simpson’s rules of numerical integration?
Solution:
(b  a) 2 2
Error in Trapezoidal rule E  h .M
12
(b  a) 2 4
Error in Simpson’s 1/3rd rule E  h .M
180

16. Comp[are Simpson’s 1/3rd rule with Trapezoidal rule.

Solution:

S. No. Trapezoidal rule Simpson’s 1/3rd rule

1. Any number of intervals Number of intervals must be
even
2. Least accuracy More accuracy
3. Here y is a linear function Here y is a polynomial of
of x degree two.

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 PART-B
Lagrangian method:
1. Using Lagrange’s interpolation formula find the value of f(3), from the following table
x: 0 1 2 5
f(x): 2 3 12 147
[Pg. No. 4.3: Eg: 4.1.1].
2. Using Lagrange’s interpolation formula, find y(10) from the following table
x: 5 6 9 11
y: 12 13 14 16
[Pg. No. 4.9: Eg: 4.1.6].

3. Use Lagrange’s formula to fit a polynomial to the following data hence find y(x=1) &
y(x=1.5)
x: -1 0 2 3
y: -8 3 1 12
[Question Bank Pg. No. 4.15: Q.No.: B9].
Inverse Lagrangian method:
1. Find the correspoding to the annuity value 13.6 given the table:
Age x: 30 35 40 45 50
Annuity value y: 15.9 14.9 14.1 13.3 12.5
[Pg. No. 4.16: Eg: 4.1.13].
Newton’s divided difference method:
1. Using Newton’s divided difference formula, find the value of f(2), f(8) and f(15) given
the following table
x: 4 5 7 10 11 13
f(x): 48 100 294 900 1210 2028
[Pg. No. 4.34: Eg: 4.2.5].
Newton’s forward and backward difference method:
1. Using Newton’s forward interpolation formula, find the polynomial f(x) satisfying the
following data. Hence evaluate at f(x) at x = 5.
x: 4 6 8 10
f(x) 1 3 8 10
[Pg. No. 4.47: Eg: 4.3.1].
2. From the following data find  at x=43 and x=84.
x: 40 50 60 70 80 90
: 184 204 226 250 276 304
Also  in terms of x. [Pg. No. 4.16: Eg: 4.1.13].
3. From the data given below, find the number of students whose wight is between 60 to to 70.
Weitht in Ibs 0-40 40-60 60-80 80-100 100-120
No. of students 250 120 100 70 50
[Pg. No. 4.59: Eg: 4.3.8].

6
Numerical differentiation:
1. Use the Newton divided difference formula to calculate f(3), f’(3) and f”(3) from the
following table
x: 0 1 2 4 5 6
f(x) 1 14 15 5 6 19
[Question Bank Pg. No. 19: 14 (a) (i)].
2. Compute f ‘(0) and f”(4) from the following data:
x: 0 1 2 3 4
f(x) 1 2.718 7.381 20.086 54.598
[Pg. No. 4.78: Eg: 4.4.3].
3. Given the following data, find y’(6) and the maximum value of y.
x: 0 2 3 4 7 9
f(x) 4 26 58 112 466 922

[Pg. No. 4.99: Eg: 4.4.17].

Numerical integration:
 /2
1. By dividing the range into ten equal parts, evaluate  sin x dx by trapezoidal &
0
Simpson’s rule. Verify your answer with actual integration. [Pg. No. 4.113: Eg: 4.5.4].

2. By dividing the range into ten equal parts, evaluate  sin x dx by trapezoidal & Simpson’s
0
rule. Verify your answer with actual integration. [Pg. No. 4.115: Eg: 4.5.8].
6
dx by (i) Trapezoidal rule (ii) Simpson’s rule. Aso check up the results by
3. Evaluate
 1 x
0
2

actual integration. [Pg. No. 4.119: Eg: 4.5.12].