Make your own free website on Tripod.com
This is an attempt to offer a somewhat intuitive explanation of why Simpson's rule is exact for cubic polynomials even though it only uses quadratic polynomials to approximate the integrand function. Recall that in each interval (xj, xj+2), Simpson's rule constructs a quadratic polynomial that agrees with the function being integrated at the points [xj, f(xj)], [xj+1, f(xj+1)], and [xj+2, f(xj+2)]. (See figure)

To make this discussion easier, we can translate the y-axis so that it passes through the point xj+1 without changing the shape of the curves involved, the width of the intervals, or the results of any of the computations. Since Simpson's rule uses equally spaced intervals, we can now label the point that was originally xj+1 as 0 and then label xj as -h and xj+2 as +h as shown. Now the cubic polynomial f(x) has the general form ax3+bx2+cx+d, and the quadratic polynomial generated by Simpson's rule has the general form ex2+fx+g. The error function then is E(x)=ax3+bx2+cx+d-(ex2+fx+g) which has the general form px3+qx2+rx+s, and the integral of E(x) from -h to h is the error in the Simpson method within this interval. Now since the quadratic polynomial agrees with f(x) at the endpoints and midpoints, we have E(-h)=E(0)=E(h)=0. Evaluating E(0) we have s=0, so E(x) actually has the form px3+qx2+rx   Evaluating E(-h) and E(h), we get:
E(-h)=-ph3+qh2-rh=0
E(h)=ph3+qh2+rh=0     and adding these two equations together gives
2qh2=0
Since h is not zero, we must have q=0, so E(x) has the form E(h)=px3+rx
But this means E(x) is an odd function [that is, E(-x)=-E(x), so E(x) is symmetric about the origin], and therefore the integral from -h to 0 of E(x) equal negative the integral from 0 
to h of E(x), so the integral from -h to h of E(x) is 0

In summary, this shows that when Simpson's rule is used to generate a quadratic approximation to a cubic polynomial over two subintervals, the quadratic polynomial underestimates the area under the cubic in one subinterval and overestimates it by the same amount in the other subinterval. This is one of those rare cases when two wrongs really do make a right!

It may still be a gift from the god of mathematics, but perhaps this gives a more intuitive feel for the phenomenon.

Return to Bob's Homepage
Aurora University Homepage

Date last modified: 5/8/00

Send me mail