1. Lecture 22 Numerical Analysis

2. Chapter 5 Interpolation

3. Finite Difference Operators Newton’s Forward Difference Interpolation Formula Newton’s Backward Difference Interpolation Formula Lagrange’s Interpolation Formula Divided Differences Interpolation in Two Dimensions Cubic Spline Interpolation

5. Thus Similarly

6. Shift operator, E

7. The inverse operator E -1 is defined as Similarly,

8. Average Operator,

9. Differential Operator, D

10. Important Results

11. Newton’s Forward Difference Interpolation Formula

12. Let y = f (x) be a function which takes values f (x 0 ), f (x 0 + h), f (x 0 +2h), …, corresponding to various equi- spaced values of x with spacing h, say x 0, x 0 + h, x 0 + 2h, …. Suppose, we wish to evaluate the function f (x) for a value x 0 + ph, where p is any real number, then for any real number p, we have the operator E such that

14. 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.

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

16. Exercise Find a cubic polynomial in x which takes on the values -3, 3, 11, 27, 57 and 107, when x = 0, 1, 2, 3, 4 and 5 respectively.

17. Solution Here, the observations are given at equal intervals of unit width. To determine the required polynomial, we first construct the difference table

18. Difference Table

19. Since the 4 th and higher order differences are zero, the required Newton’s interpolation formula

20. Here,

21. Substituting these values into the formula, we have The required cubic polynomial.

22. NEWTON’S BACKWARD DIFFERENCE INTERPOLATION FORMULA

23. For interpolating the value of the function y = f (x) near the end of table of values, and to extrapolate value of the function a short distance forward from y n, Newton’s backward interpolation formula is used

24. Derivation 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),

25. where p is any real number, then we have the shift operator E, such that Binomial expansion yields,

26. That is,

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

28. If we retain (r + 1)terms, we obtain a polynomial of degree r agreeing with f (x) at x n, x n-1, …, x n-r. Alternatively, this formula can also be written as Here

29. Example For the following table of values, estimate f (7.5).

30. Solution The value to be interpolated is at the end of the table. Hence, it is appropriate to use Newton’s backward interpolation formula. Let us first construct the backward difference table for the given data

31. Difference Table

32. Since the 4 th and higher order differences are zero, the required Newton’s backward interpolation formula is

33. In this problem,

34. Example The sales for the last five years is given in the table below. Estimate the sales for the year 1979

35. Solution Newton’s backward difference table for the given data as -3

36. In this example, and

37. Newton’s interpolation formula gives Therefore,

38. LAGRANGE’S INTERPOLATION FORMULA

39. Newton’s interpolation formulae developed earlier can be used only when the values of the independent variable x are equally spaced. Also the differences of y must ultimately become small.

40. If the values of the independent variable are not given at equidistant intervals, then we have the basic formula associated with the name of Lagrange which will be derived now.

41. Lecture 22 Numerical Analysis

42. 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.

43. Suppose we write this polynomial in the form. or in the form

44. 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,

45. Similarly, we obtain and

46. Substituting the values of a 0, a 1, …, a n we get The Lagrange’s formula for interpolation

47. This formula can be used whetherthe values x 0, x 2, …, x n are equally spaced or not. Alternatively, this can also be written in compact form as This formula can be used whether the values x 0, x 2, …, x n are equally spaced or not. Alternatively, this can also be written in compact form as

48. Where, We can easily observe that, and Thus introducing Kronecker delta notation

49. Further, if we introduce the notation That is is a product of (n + 1) factors. Clearly, its derivative contains a sum of (n + 1) terms in each of which one of the factors of will be absent.

50. We also define, which is same as except that the factor (x–x k ) is absent. Then But, when x = x k, all terms in the above sum vanishes except P k (x k )

51. Hence,

52. Finally, the Lagrange’s interpolation polynomial of degree n can be written as

53. 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).

54. Solution The given data can be arranged as

55. Using Lagrange’s interpolation formula, we have

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

57. Example Given the following data, evaluate f (3) using Lagrange’s interpolating polynomial.

58. Solution Using Lagrange’s formula,

59. Therefore