Simpson's rule, was Re: Integrating a Taylor series?

You're quite correct.  I just did a small experiment and your method works
perfectly well (as indeed it should, as the mathematics of it is sound).
Thank you very much!


On Tue, May 14, 2013 at 1:50 AM, Leo Butler
<l_butler at users.sourceforge.net>wrote:

>    >From mailnull  Sun May 12 23:52:31 2013
>    Received-SPF: pass (sog-mx-2.v43.ch3.sourceforge.com: domain of
> math.utexas.edu designates 146.6.25.7 as permitted sender)
> client-ip=146.6.25.7; envelope-from=maxima-bounces at math.utexas.edu; helo=
> ironclad.mail.utexas.edu;
>    Date: Mon, 13 May 2013 09:51:04 +1000
>    From: Alasdair McAndrew <amca01 at gmail.com>
>    Content-Type: multipart/mixed;
>            boundary="===============6772570308007447215=="
>
>    --===============6772570308007447215==
>    Content-Type: multipart/alternative;
> boundary="20cf307cfc7eb8d1a404dc8e1275"
>
>    --20cf307cfc7eb8d1a404dc8e1275
>    Content-Type: text/plain; charset="ISO-8859-1"
>
>    Yes, I know that in general Newton-Cotes rules of order 2n are exact for
>    polynomials of order 2n+1.  However, I was interested in some more
> general
>    rules (including Newton-Cotes with derivative end-corrections), some of
>    which have errors of a very high order of h, and working using a Taylor
>    series seemed the best way to go.
>
>    Thanks,
>    Alasdair
>
> No, I think you are mistaken. Let me expand upon RJF's remark. Let's
> look at the problem from a more abstract perspective. You are looking
> to approximate a linear map I : P -> P, where I is the operator
>
> I(f) := integrate(f,x,a,a+h);
>
> and P is the linear space of polynomials in a single variable (f is a
> polynomial in x, g:I(f) is a polynomial in h). You are using a linear
> operator J : P -> P to approximate I. Your notion of approximation is
> that
>
> I == J mod P_n  ---(*)
>
> where P_n is the subspace of degree at most n polynomials. Then
>
> R[n] : I(x^{n+1}) - J(x^{n+1}) ;
>
> is the lowest-order remainder term in your approximation. At the point
> where you find n, you can employ Taylor's theorem, etc. to
> characterize the remainder term for your algorithm in terms of the
> n+1-th derivative, etc.
>
> But, and this simplifies things greatly, you never need to work
> directly with a Taylor polynomial to do any of this analysis. For
> example, to check (*), you can compute R[0], ..., R[n-1].
>
> Leo
>



-- 
Blog: http://amca01.wordpress.com
Web:  http://sites.google.com/site/amca01/
Facebook: http://www.facebook.com/alasdair.mcandrew