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**Symmetry and Simpson's Rule**

Mathematicians appreciate that exploiting symmetry is a powerful tool.
We shall show that in Simpson's rule for estimating integrals, better error
bounds and a simpler derivation result from symmetry considerations. These
ideas may be applied to all Newton-Cotes formulae.

Suppose that *h* > 0. Recall that Simpson's rule says that if *f* is
continuous on [*a*,*a*+2*h*] and four times differentiable on (*a*,*a*+2*h*) then

| |
(1) |

Generally Simpson's rule is applied to integrals of the form
by letting *N* be a positive even integer, letting *h*= (*B*-*A*)/*N* and writing
to obtain the familiar formula
| |
(2) |

This bound on the error is clearly not optimum because it assumes that the
maximum value of |*f*^{(4)}| on all of (*A*,*B*) occurs on every subinterval, but
it is quick and easy (relatively speaking). However, one cannot find the
maximum of |*f*^{(4)}| on every subinterval before one knows how many
subintervals are to be used. Hence the cruder estimate in (2) is
used to get an upper bound on the number of subintervals needed.
Put

We shall show now that that symmetry gives a painless improvement of
(2) to
| |
(3) |

It is easy to check that
| |
(4) |

and that the Simpson's rule estimate of each of these integrals is given by
exactly the same expression:
Applying Simpson's rule to the integral in the right-hand side of
(4) then gives (3).
By way of examples, consider two standard calculations:

and
In the latter, using symmetry guarentees that the difference between and
the Simpson's rule estimate of the integral is no more than (42/180)*N*^{-4}
compared with the usual bound of (96/180)*N*^{-4}. Thus to obtain an error
tolerance of not more than 10^{-8} we need only take *N* = 70 versus 86
with the conventional estimate, or more generally, an improvement by a factor
of . In the former example, the difference
between and the Simpson's rule estimate of the integral is no more than
(24/180)*N*^{-4} by the usual estimate
compared with
when symmetry is taken
into account, allowing us to use a value of *N* about smaller to be
guarenteed the same accuracy of the estimate. Of course, one can produce
examples where there is no improvement and where the improvement is
overwhelming.
Considering symmetry also simplifies the derivation and proof of (1).
We will show that for an even function which is sufficiently smooth,

| |
(5) |

It is clear from the discussion above that (5) leads quickly to an
improved version of (1). As we shall now demonstrate, (5) is
easier to derive than (1).
The key to deriving and proving (5) is that even polynomials should be
used to interpolate even functions, and that it is easy to find such
polynomials if only two interpolation points are needed. In fact, all one has
to do is look at the form for linear interpolation as in the trapezoid rule and
one sees what to do.

If the trapezoid rule were to be employed to estimate

we would replace the integral of *g* with the integral of the affine function
*L*(*u*) = *g*(*h*)(*h*-*u*)/*h* + *g*(0)*u*/*h* since *y*=*L*(*x*) is an equation of the line which
passes through (0,*g*(0)) and (*h*,*g*(*h*)). It is then clear that since *g* is
even, we should try to change the variable *u* in *L*(*u*) to *u*^{2} to find a
quadratic polymomial with the same property. It is not hard to see that what
we want is *P*(*u*) = *g*(0)(*h*^{2}-*u*^{2})/*h*^{2} + *g*(*h*)*u*^{2}/*h*^{2}, and that *y*=*P*(*x*) passes
through all three points (0,*g*(0)) and . Note that we did
not have to solve any systems of equations to find *P*, nor did we need higher
order interpolation formulae, we simply proceeded by analogy with the linear
case. It is then easy to check that
| |
(6) |

We see that (6) tells us that to complete the proof of (5) we can
estimate |*g*(*u*)-*P*(*u*)| on [0,*h*]. We proceed in a well-known fashion. Since
*g* -*p* is an even function, we try to identify the constant *C* such that
*g*(*u*)-*P*(*u*) = *Cu*^{2}(*h*^{2}-*u*^{2}),

where *u* is a fixed element of (0,*h*). Our goal is to express *C* in terms
of the some derivative of *g*. For put *E*(*x*) =
*g*(*x*)-*P*(*x*)-*Cx*^{2}(*h*^{2}-*x*^{2}). Note that *E* continuous on [-*h*,*h*], four times differentiable on
(-*h*,*h*) and even. Therefore, not only do we have *E*(*u*) = *E*(0) = *E*(*h*) = 0,
but also *E*'(0) = *E*^{(3)}(0) = 0. Therefore we may apply Rolle's Theorem
successively to *E*, *E*', *E*'' and *E*^{(3)} and we find that for some
we have *E*^{(4)}(*z*) = 0. Of course, *E*^{(4)}(*z*) = *g*^{(4)}(*z*) +
24*C*, so . Take *C*=0 if *u*=0 or
*u*=*h*. This tells us that
which is (5).

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*Eric S Key*

*9/15/1998*