Math Forum / Mathematics / Undergraduate Math / December 2006

There should be a derivation in any book on numerical analysis.
It's not hard, using Taylor's Theorem.

Working it out just now, I found that the midpoint
rule, approximating int_a^b f(t) dt by

 (b-a) f((a+b)/2),

actually gives a smaller constant in the error than
the trapezoidal rule.

Let's assume that a = -b, to simplify the typing below.
Note that then b-a = 2b. Taylor's Theorem shows that

(*)  f(t) = f(0) + t f'(0) + R(t),

where

 |R(t)| <= m t^2/3.

So

 int_{-b}^b f(t) = 2b f(0) + int_{-b}^b R(t) dt
                 = (b-a) f(0) + int_{-b}^b R(t),

so the absolute value of the error in the midpoint
rule is

 |E_mid| <= int_{-b}^b m t^2/2 dt

          = m b^3/3 = m (b-a)^3/24.

Applying (*) with t = b and t = -b shows that the
difference between the midpoint rule and the
trapezoidal rule is

 b(f(b) + f(-b)) - 2b f(0) = b(R(b) + R(-b)),

so the absolute value of the difference is

 |mid - trap| <= bm(b^2/2 + b^2/2)
               = m b^3 = m (b-a)^3/8.

So the absolute value of the error in the trapezoidal
rule is

 |E_trap| <= m (b-a)^3/24 + m (b-a)^3/8 = m (b-a)^3/6.

You can use a similar argument to get an error estimate
for Simpson's Rule, with a constant slightly worse than
what you cite above:

(**) f(t) = f(0) + t f'(0) + t^2/2 f''(0) + R(t).

where now

 |R(t)| <= t^3/6 m_3,

where m_3 is the maximum of the third derivative.

So

 int_{-b}^b f = 2b f(0) + b^3/3 f''(0) + int R(t),

where

 int |R(t)| <= int_{-b}^b t^3/6 m_3 = b^4 m_3/12.

Now we don't want that f''(0) in the approximation
to the integral. Applying (**) with t = b and t = -b
and rearranging gives

 b^3/3 f''(0) = b(f(b) + f(-b))/3 - 2b/3 f(0)
                  + b/3 (R(b) + R(-b)),

so that

 2b f(0) + b^3/3 f''(0) = 4b/3 f(0) + b/3 (f(b)+f(-b))
             +  b/3 (R(b) + R(-b))

         = (b-a)/6 (4 f(0) + f(b) + f(a))
            +  b/3 (R(b) + R(-b)),

exactly the Simpson thing plus another error term.
The absolute value of the sum of the two error terms
is

  <= b^4 m_3 (1/12 + 1/9) = b^4/m_3 (7/36)
       
      = (b-1)^4 m_3 (7/288)

Of course 7/288 > 1/90.

On the other hand that's not the "right" error anyway!
According to http://en.wikipedia.org/wiki/Simpson's_rule
the error in Simpson's rule is actually bounded by a
constant times (b-a)^5, which is much better.
Oops, I just realized that you cited a bound in
terms of (b-a)^5 above...)

If I work through the above with a third-degree Taylor
polynoial in place of (**) I get an error of
(b-a)^5 m_4/560 (which is worse than the constant
at wikipedia, but better than the one you cite above),
where m_4 is the maximum of the fourth derivative.

         

************************

David C. Ullrich