Download presentation

Presentation is loading. Please wait.

Published byRandall Tune Modified over 2 years ago

1
Numerical Analysis

2
TOPIC Interpolation TOPIC Interpolation

3
Interpolation is the process of estimating the value of function for any intermediate value of the variable with the help of its given set of values. Let us assume that the function y=f(x) is known for certain values of x say a for x 0,x 1,x 2,………x n. As f(x 0 ),f(x 1 ),……..f(x n ). The process of finding the value of f(x) corresponding to x=x i. Where x 0

4

5
1.Finite Difference Operators 2.Newton’s Forward Difference Interpolation Formula 3.Newton’s Backward Difference Interpolation Formula 4.Lagrange’s Interpolation Formula

6
rth forward difference kth backward difference

7
Thus Similarly

8
Shift operator, E

9
The inverse operator E -1 is defined as Similarly,

10
Average Operator,

11
Differential Operator, D

12
Important Results

13
Newton’s Forward Difference Interpolation Formula

14

15

16
This is known as Newton’s forward difference formula for interpolation, which gives the value of f (x 0 + ph) in terms of f (x 0 ) and its leading differences.

17
This formula is also known as Newton-Gregory forward difference interpolation formula. Here p=(x-x 0 )/h. An alternate expression is

18

19

20

21

22
Here,

23
The required cubic polynomial.

24

25

26
Let y = f (x) be a function which takes on values f (x n ), f (x n -h), f (x n -2h), …, f (x 0 ) corresponding to equispaced values x n, x n -h, x n -2h, …, x 0. Suppose, we wish to evaluate the function f (x) at (x n + ph),

27
Binomial expansion yields,

28

29
This formula is known as Newton’s backward interpolation formula. This formula is also known as Newton’s-Gregory backward difference interpolation formula.

30

31
Example:- For the following table of values, estimate f (7.5).

32

33
Difference Table

34

35
In this problem,

36

37
-3

38
and

39
Newton’s interpolation formula gives Therefore,

40

41

42

43
Let y = f (x) be a function which takes the values, y 0, y 1,…y n corresponding to x 0, x 1, …x n. Since there are (n + 1) values of y corresponding to (n + 1) values of x, we can represent the function f (x) by a polynomial of degree n. DERIVATION:-

44
or in the form

45
Here, the coefficients a k are so chosen as to satisfy this equation by the (n + 1) pairs (x i, y i ). Thus we get Therefore,

46
and

47
The Lagrange’s formula for interpolation

48

49
We can easily observe that, and Thus introducing Kronecker delta notation

50
Example:- Find Lagrange’s interpolation polynomial fitting the points y(1) = -3, y(3) = 0, y(4) = 30, y(6) = 132. Hence find y(5).

51

52
Using Lagrange’s interpolation formula, we have

53
On simplification, we get which is required Lagrange’s interpolation polynomial. Now, y(5) = 75.

54

55
Solution:- Using Lagrange’s formula,

56
Therefore

57
1. Evaluate : (x.log x) 2. Evaluate : (sin2x.cos4x) 3. Estimate the missing term: x01234 Y139_81

58
4.Derive newton forward formula for interpolation. 5. Derive newton backward formula for interpolation.

59

60
1.Evaluate (x 2 +sinx), the interval of difference being . 2. Find the lowest degree polynomial which satisfies the following table: X012345 F(x)038152435

61
3.The population of a town is as follow: YEAR196119711981199120012011 POPULATION202435404251

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google