Numerical Analysis Lecture 25

Chapter 5 Interpolation

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

Newton’s Forward Difference Interpolation Formula

An alternate expression is

NEWTON’S BACKWARD DIFFERENCE INTERPOLATION FORMULA

The formula is,

Alternatively form Here

LAGRANGE’S INTERPOLATION FORMULA

The Lagrange Formula for Interpolation

Alternatively compact form

Also

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

DIVIDED DIFFERENCES

Let us assume that the function y = f (x) is known for several values of x, (xi, yi), for i=0,1,..n. The divided differences of orders 0, 1, 2, …, n are now defined recursively as:

The zero-th order divided difference

The first order divided difference is defined as

Second order divided difference

Generally

Standard format of the Divided Differences

NEWTON’S DIVIDED DIFFERENCE INTERPOLATION FORMULA

Let y = f (x) be a function which takes values y0, y1, …, yn corresponding to x = xi, i = 0, 1,…, n. We choose an interpolating polynomial, interpolating at x = xi, i = 0, 1, …, n in the following form

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

The first equation gives The second equation gives

Third equation yields which can be rewritten as that is

Thus, in terms of second order divided differences, we have Similarly, we can show that

Newton’s divided difference interpolation formula

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

Assuming equi-spaced values of abscissa, we have

By induction, we can in general arrive at the result

Similarly

In general, we have

Also, in terms of central differences, we have

In general, we have the following pattern

Example Find the interpolating polynomial by (i) Lagrange’s formula and (ii) Newton’s divided difference formula for the following data. Hence show that they represent the same interpolating polynomial. X 1 2 4 Y 5

Solution The divided difference table for the given data is constructed as follows: X Y 1st divided difference 2nd divided difference 3rd divided difference 1 2 1/2 -1/2 4 5 3/2 1/6

(i) Lagrange’s interpolation formula gives

(ii) Newton’s divided difference formula gives We observe that the interpolating polynomial by both Lagrange’s and Newton’s divided difference formulae is one and the same.

Note! Newton’s formula involves less number of arithmetic operations than that of Lagrange’s.

Example Using Newton’s divided difference formula, find the quadratic equation for the following data. Hence find y (2). X 1 4 y 2

Solution: The divided difference table for the given data is constructed as: x y 1st divided difference 2nd divided difference 2 1 -1 1/2 4

Now, using Newton’s divided difference formula, we have Hence, y (2) = 1.

Example A function y = f (x) is given at the sample points x = x0, x1 and x2. Show that the Newton’s divided difference interpolation formula and the corresponding Lagrange’s interpolation formula are identical.

Solution For the function y = f (x), we have the data

The interpolation polynomial using Newton’ divided difference formula is given as

Using the definition of divided differences, we can rewrite the equation in the form

On simplification, it reduces to which is the Lagrange’s form of interpolation polynomial. Hence two forms are identical.

Newton’s Divided Difference Formula with Error Term

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

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

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

Numerical Analysis Lecture 25