MATH 371 -- Numerical Analysis

October 5, 2001

Error of Interpolation

Consider the degree 2 interpolating polynomial for

constructed using , , ,

on the interval .

> x[0]:=evalf(Pi/8): x[1]:=evalf(Pi/4): x[2]:=evalf(3*Pi/8):

> y[0]:=sin(x[0]): y[1]:=sin(x[1]): y[2]:=sin(x[2]):

The Lagrange polynomials

> L[2,0]:=expand((x-x[1])*(x-x[2])/((x[0]-x[1])*(x[0]-x[2])));

> L[2,1]:=expand((x-x[0])*(x-x[2])/((x[1]-x[0])*(x[1]-x[2])));

> L[2,2]:=expand((x-x[0])*(x-x[1])/((x[2]-x[0])*(x[2]-x[1])));

The interpolating polyomial

> p:=expand(y[0]*L[2,0]+y[1]*L[2,1]+y[2]*L[2,2]);

> plot({sin(x),p},x=0..Pi/2);

> plot(abs(sin(x)-p),x=0..Pi/2);

>

Note the pretty close agreement between the interpolating polynomial and the

sin graph (especially toward the middle of the interval). At x = 0.6, for instance:

> abs(sin(.6)-subs(x=.6,p));

Using the bound for the third derivative of sine, the error bound says

<=

> abs(.6-x[0])*abs(.6-x[1])*abs(.6-x[2])/6;

To get better approximations here we could try more interpolation points.