[racket] Implementation of Simpson's Rule (SICP Exercise 1.29)

Thanks, Jens and Daniel.

There's no problem with Simpson's Rule. I got another way to implement
Simpson's Rule from Eli
Bendersky<http://eli.thegreenplace.net/2007/07/11/sicp-section-131/>.
And re-wrote it in Racket:

    (define (simpson-integral2 f a b n)
      (define h (/ (- b a) n))
      (define (next counter) (+ counter 1))
      (define (term counter)
        (* (f (+ a (* h counter))) (cond
                                     ((= counter 0) 1)
                                     ((= counter n) 1)
                                     ((odd? counter) 4)
                                     ((even? counter) 2))))
      (* (/ h 3) (sum term 0 next n)))

The result is very accurate:

    > (simpson-integral2 cube 0 1.0 10)
    0.25
    > (simpson-integral2 cube 0 1.0 100)
    0.24999999999999992
    > (simpson-integral2 cube 0 1.0 1000)
    0.2500000000000003
    > (simpson-integral2 cube 0 1.0 10000)
    0.2500000000000011

Eli's code and mine are two implementations for the same Rule, but mine is
far less accurate. So I think there's some wrong with my implementation.
But I still can't find it out.


On Tue, Nov 5, 2013 at 1:59 AM, Jens Axel Søgaard <jensaxel at soegaard.net>wrote:

> 2013/11/4 Ben Duan <yfefyf at gmail.com>:
> > I've asked the question on Stack Overflow [1]. Óscar López gave me an
> > answer. But I still don't understand it. So I re-post it here:
> >
> > Following is my code for SICP exercise 1.29 [2]. The exercise asks us to
> > implement Simpson's Rule using higher order procedure `sum`. It's
> supposed
> > to be more accurate than the original `integral` procedure. But I don't
> know why
> > it's not the case in my code:
>
> A couple of points:
>
>  - to compare to methods M1 and M2 we need to specify the class of
> functions,
>    we will use the methods on
>  - when comparing the results of the two methods on a single function f,
>    we must use the same number of evaluations of f
>  - the most accurate method for a given number of evaluations n,
>    is the method which has the best worst case behaviour
>
> Thus when SICP says:
> "Simpson's Rule is a more accurate method of numerical integration
> than the method illustrated above."
> it doesn't mean that the Simpson rule will work better than, say, the
> midpoint method for all
> function. Some methods give very good results for polynomials of a small
> degree.
> Try your methods on some non-polynomials: say 1/x , ln(x), sqrt(x) and
> sin(x).
>
> See also some interesting examples here:
>
> http://www.math.uconn.edu/~heffernan/math1132s13/files/trapezoid_rule.pdf
>
>
> --
> Jens Axel Søgaard
>
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