1. Numerical Analysis
Lecture 26

2. Chapter 5 Interpolation

3. Finite Difference Operators Newton’s Forward Difference
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

4. DIVIDED DIFFERENCES

5. The zero-th order divided difference

6. The first order divided difference is defined as

7. Second order divided difference

8. Generally

9. NEWTON’S DIVIDED DIFFERENCE INTERPOLATION FORMULA

10. Newton’s divided difference interpolation formula

11. Newton’s divided differences can also be expressed in terms of forward, backward and central differences.

12. In terms of forward differences

13. In terms of backward differences

14. In terms of central differences

15. Newton’s Divided Difference Formula with Error Term

16. Following the basic definition of divided differences, we have for any x

17. Multiplying the second Equation by (x – x0), third by (x – x0)(x – x1) and so on, and the last by (x – x0)(x – x1) … (x – xn-1) and adding the resulting equations, we obtain

18. Please note that for x = x0, x1, …, xn, the error term vanishes
Where
Please note that for x = x0, x1, …, xn,
the error term vanishes

19. Error Term in Interpolation Formulae
We know, if y (x) is approximated by a polynomial Pn (x) of degree n then the error is given by
where,

20. Alternatively it is also expressed as
Now, consider a function F (x), such that
Determine the constant K in such a way that F(x) vanishes for x = x0, x1, …, xn and also for an arbitrarily chosen point , which is different from the given (n + 1) points.

21. Let I denotes the closed interval spanned by the values
Let I denotes the closed interval spanned by the values Then F (x) vanishes (n + 2) times in the interval I. By Rolle’s theorem vanishes at least (n + 1) times in the interval I, at least n times, and so on.

22. Eventually, we can show that
Eventually, we can show that vanishes at least once in the interval I, say at Thus, we obtain
Since Pn(x) is a polynomial of degree n, its (n + 1)th derivative is zero. Also, from the definition of

23. Therefore, we get
Hence
for some in the interval I.

24. Further more, we observe that
Thus the error committed in replacing y (x) by either Newton’s divided difference formula or by an identical Lagrange’s formula is

25. INTERPOLATION IN TWO DIMENSIONS

26. Let u be a polynomial function in two variables, say x and y, in particular quadratic in x and cubic in y, which in general can be written as

27. This relation involves many terms
This relation involves many terms. If we have to write a relation involving three or more variables, even low degree polynomials give rise to long expressions. If necessary, we can certainly write, but such complications can be avoided by handling each variable separately.

28. If we let x, a constant, say x = c, the equation simplifies to the form
Now we adopt the following procedure to interpolate at a point (1, m) in a table of two variables, by treating one variable a constant say x = x1. The problem reduces to that of a single variable interpolation.

29. Any one of the methods discussed in preceding sections can then be applied to get f (x1, m). Then we repeat this procedure for various values of x say x = x2, x3, …, xn keeping y constant. Thus, we get a new table with y constant at the value y = m and with x varying. We can then interpolate at x = 1.

30. Example Tabulate the values of the function f (x) = x2+y2-y
for x = 0,1,2,3,4 and y = 0,1,2,3,4.
Using the table of the values, compute f(2.5, 3.5) by numerical double interpolation

31. Solution
The values of the function for the given values of the given values of x and y are given in the following table X ……
y…. 1 2 3 4 6 12 7 13 10 16 9 11 15 21 18 22 28

32. Cont!
Using quadratic interpolation in both x and y directions we need to consider three points in x and y directions. To start with we have to treat one variable as constant, say x. Keeping x = 2.5, y = 3.5 as near the center of the set, we choose the table of values corresponding to x = 1,2, 3 and y = 2, 3, 4.

33. Cont !
The region of fit for the construction of our interpolation polynomial is shown in different color in the table X ……
y…. 1 2 3 4 6 12 7 13 10 16 9 11 15 21 18 22 28

34. Cont ! Thus using Newton’s forward difference formula, we have
At x = 1 Y f 2 3 7 4 13 6

35. Cont !
Similarly
At x = 2 Y f 2 6 3 10 4 16

36. Cont !
Similarly
At x = 3 Y f 2 11 3 15 4 21 6

37. Cont !

38. Cont !

39. Cont ! Therefore we arrive at the following result At y=3.5 X f 1 9.752
12.75 3
17.75 5

40. Cont !

41. Cont ! And hence no error in interpolation !!!
From the functional relation, we also find that
And hence no error in interpolation !!!

42. Numerical Analysis
Lecture 26