To
make this discussion easier, we can translate the y-axis so that
it passes through the point x_{j+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 x_{j+1} as 0 and then label
x_{j} as -h and x_{j+2} as +h as shown. Now the
cubic polynomial f(x) has the general form
ax^{3}+bx^{2}+cx+d, and the quadratic polynomial
generated by Simpson's rule has the general form
ex^{2}+fx+g. The error function then is
E(x)=ax^{3}+bx^{2}+cx+d-(ex^{2}+fx+g) which
has the general form px^{3}+qx^{2}+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 px^{3}+qx^{2}+rx
Evaluating E(-h) and E(h), we get:

E(-h)=-ph^{3}+qh^{2}-rh=0

E(h)=ph^{3}+qh^{2}+rh=0 and adding these
two equations together gives

2qh^{2}=0

Since h is not zero, we must have q=0, so E(x) has the form
E(h)=px^{3}+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

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.

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Date last modified: 5/8/00

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