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Numerical Analysis Lecture 23.

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Presentation on theme: "Numerical Analysis Lecture 23."— Presentation transcript:

1 Numerical Analysis Lecture 23

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

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 The Newton’s forward difference formula for interpolation, which gives the value of f (x0 + ph) in terms of f (x0) and its leading differences.

13 This is also known as Newton-Gregory forward difference interpolation formula.
Here p=(x-x0)/h. An alternate expression is

14 NEWTON’S BACKWARD DIFFERENCE INTERPOLATION FORMULA

15 The formula,

16 Alternatively, this formula can also be written as
Here

17 LAGRANGE’S INTERPOLATION FORMULA

18 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.

19 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.

20 Let y = f (x) be a function which takes the values, y0 , y1 ,…yn corresponding to x0 , x1, …xn . 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.

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

22 Here, the coefficients ak are so chosen as to satisfy this equation by the (n + 1) pairs (xi, yi). Thus we get Therefore,

23 Similarly, we obtain and

24 Substituting the values of a0, a1, …, an we get
The Lagrange’s formula for interpolation

25 This formula can be used whether the values x0, x2, …, xn are equally spaced or not. Alternatively, this can also be written in compact form as

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

27 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.

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

29 Hence,

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

31 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).

32 Solution The given data can be arranged as

33 Using Lagrange’s interpolation formula, we have

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

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

36 Solution Using Lagrange’s formula,

37 Therefore

38 DIVIDED DIFFERENCES

39 Numerical Analysis Lecture 23


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