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Polynomial Interpolation - Newton's method

(i) Use the tabulated data and Newton's method of divided differences to (a) calculate the degree three interpolating polynomial based on x_0, x_1, x_2, x_3 and (b) calculate the degree four interpolating polynomial based on x_0, x_1, x_2, x_3, x_4.

x_k || 0 | -1 | 1 | 3 | 2
y_k || 5 | 15 | 3 | 47 | 9

(Use exact arithmetic (with fractions, if necessary); do not re-order the given points.)

(ii) Check whether the function f(x) is a cubic spline, where

f(x) = {0.5x^3 - 2x^2 + 1.3x - 8 if 0 <= x <= 1
{0.5(x-1)^3 - 0.5(x-1)^2 - 1.2(x-1) - 8.2 if 1 <= x <= 3
{0.5(x-3)^3 + 2.4(x-3)^2 + 2.8(x-3) - 8.6 if 3 <= x <= 4

Solution Preview

Interpolating polynomial based on Newton’s method of divided differences
(i) Table of divided differences
x0 y0 = f[x0]
x1 y1 = f[x1] f[x0,x1]
x2 y2 = f[x2] f[x1,x2] f[x0,x1,x2]
x3 y3 = f[x3] f[x2,x3] f[x1,x2,x3] f[x0,x1,x2,x3]

Where
f[xi] = f(xi)
f[xi,xi+1] = (f[xi+1 ] - f[xi])/(xi+1 - xi) (1st divided difference)
f[xi,xi+1,xi+2] = (f[xi+1, xi+2] - f[xi, xi+1])/( xi+2- xi)
f[x0, x1, ..., xn] = (f[x1, x2, ..., xn] - f[x0, x1, ..., xn-1])/( xn - x0)

Interpolating Polynomials
P0(x) = f[x0]
P1(x) = f[x0] + (x - x0)f[x0,x1]
P2(x) = f[x0] ...

Solution Summary

This shows how to use Newton's method to calculate a degree three interpolating polynomial.

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