1. 1 Chapter 3 INTERPOLATION

2. 2 WHAT IS INTERPOLATION? Given (x 0,y 0 ), (x 1,y 1 ), …, (x n,y n ), finding the value of ‘y’ at a value of ‘x’ in ( x 0, x n ) is called interpolation.

3. 3 INTERPOLANTS Polynomials are the most common choice of interpolants because they are easy to: Evaluate, Differentiate, and Integrate.

4. 4 LAGRANGIAN INTERPOLATION Lagrangian interpolating polynomial is given by    n i iin xfxLxf 0 )()()( where ‘n’ in)(xf n stands for the th n order polynomial that approximates the function)(xfy  given at)1(  n data points as      nnnn yxyxyxyx,,,,...,,,, 111100 , and       n ij j ji j i xx xx xL 0 )( )(xL i is a weighting function that includes a product of)1(  n terms with terms ofij  omitted.

5. 5 Example A manufacturer of thermistors makes the following observations on a thermistor. Determine the temperature corresponding to 754.8 ohms using the Lagrangian method for interpolation. R T Ohm C  1101.0 911.3 636.0 451.1 25.113 30.131 40.120 50.128

6. 6 Solution

7. 7

8. 8

9. 9 NEWTONS DIVIDED DIFFERENCE What is divided difference? f[x 0,x 1,x 2 ] = f[x 0,x 1 ] = f[x 1,x 2 ] – f[x 0,x 1 ] x 2 – x 1 f[x 0, x 1, …, x k-1, x k ] = for k = 3, 4, ….. n. These I st, II nd... and k th order differences are denoted by f,  2 f, …,  k f. f[x 1 ] – f[x 0 ] x 1 – x 0

10. 10 INTERPOLATION USING DIVIDED DIFFERENCE The divided difference interpolation polynomial is: P(x) = f(x 0 ) + (x – x 0 ) f [x 0, x 1 ] + L + (x – x 0 ) L (x – x n-1 ) f[x 0, x 1, …, x n ]

11. 11 Example For the data x:–1025 f(x) :71022235 Find the divided difference polynomial and estimate f(1).

12. 12 Solution

13. 13 INTERPOLATION FOR EQUALLY SPACED POINTS Let (X 0,,Y 0 ), (X 1,,Y 1 ), …, (X n,,Y n ) be the given points with X i+1 = X i +h,i = 0,1,2,…, (n-1). Finite Difference Operators ● Forward difference operator  f(x i ) = f (x i + h) – f (x i ) ● Backward difference operator  f(x i ) = f (x i ) – f (x i – h) ● Central difference operator  f(x i ) = f - f

14. 14 INTERPOLATION FOR EQUALLY SPACED POINTS ● Shift operators ● Averaging operator

15. 15 RELATION BETWEEN OPERATORS  f i =  f i+1 = df i+ ½  = E – 1,  = 1 – E -1  = E ½ – E -½  = ( E ½ + E - ½ )

16. 16 NEWTON GREGORY FORWARD INTERPOLATION For convenience we put p = and f 0 = y 0. Then we have

17. 17 Example Estimate f (3.17)from the data using Newton Forward Interpolation. x: 3.13.23.33.43.5 f(x): 00.61.01.21.3

18. 18 Solution

19. 19 Solution

20. 20 NEWTON GREGORY BACKWARD INTERPOLATION FORMULA

21. 21 Estimate f(42) from the following data using newton backward interpolation. x:20 25 30 35 40 45 f(x):354332291260231204 Example

22. 22 Solution Here x n = 45, h = 5, x = 42 and p = - 0.6

23. 23 Solution

24. 24 INTERPOLATION USING CENTRAL DIFFERENCES Suppose the values of the function f (x) are known at the points a -3h, a – 2h, a – h, a, a +h, a + 2h, a + 3h,... etc. Let these values be y -3, y -2, y -1, y 0, y 1, y 2, y 3..., and so on. Then we can form the central difference table as: We can relate the central difference operator  with  and E using the operator relation  = E½.

25. 25 GAUSS FORWARD INTERPOLATION FORMULA

26. 26 GAUSS FORWARD INTERPOLATION FORMULA The value p is measured forwardly from the origin and 0<p<1. The above formula involves odd differences below the central horizontal line and even differences on the line. This is explained in the following figure.

27. 27 Example Find f(30) from the following table values using Gauss forward difference formula: x:2125293337 F(x):18.470817.814417.107016.343215.5154

28. 28 Solution f (30) = 16.9217

29. 29 GAUSS BACKWARD INTERPOLATION FORMULA This formula is used to interpolate the value of the function for a negative value of p and–1<p<0. By substituting for  y 0,  2 y 0,  3 y 0,....... in terms of central difference in the Newton Forward Difference formula, we get P(x) = y 0 +         1 p  y -1 +          2 1p  2 y -1 +          3 1p  3 y -2 +          4 2p  4 y -2 +…

30. 30 GAUSS BACKWARD INTERPOLATION FORMULA This formula involves odd differences above the central horizontal line and even differences on the central line.

31. 31 Example Estimate cos 51  42' by Gauss backward interpolation from the following data: x: 50  51  52  53  54  cos x:0.64280.62930.61570.60180.5878

32. 32 Solution

33. 33 Solution P(51.7) = 0.6198

34. 34 STIRLING’S FORMULA This formula gives the average of the values obtained by Gauss forward and backward interpolation formulae. For using this formula we should have – ½ < p< ½. We can get very good estimates if - ¼ < p < ¼. The formula is:

35. 35 Example Using Sterling Formula estimate f (1.63) from the following table: x: 1.50 1.60 1.70 1.80 1.90 f(x):17.609 20.412 23.045 25.527 27.875 Solution X 0 = 1.60, x = 1.63, h = 0.1

36. 36 Solution Difference table

37. 37 BESSELS’ INTERPOLATION FORMULA This formula involves means of even difference on and below the central line and odd difference below the line. The formula is

38. 38 Note When p = 1/2, the terms containing odd differences vanish. Then we get the formula in a more simple form: BESSELS’ INTERPOLATION FORMULA

39. 39 Example Use Bessels’ Formula to find (46.24) 1/3 from the following table of x 1/3.

40. 40 Solution x 0 = 45,x = 46.24, h = 4 p = 0.31 Difference table is: Applying Bessels’ formula, we get (46.24) 1/3 = 3.5893