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Topic: Simpson Integration - a specific problem
Replies: 13   Last Post: Jun 15, 2006 10:36 AM

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 Toni Lassila Posts: 536 Registered: 12/13/04
Re: Simpson Integration - a specific problem
Posted: Jun 11, 2006 6:49 AM

On 11 Jun 2006 02:35:00 -0700, "MET" <Marcel.E.Tschudin@gmail.com>
wrote:

The correct place for this is sci.math.num-analysis.

>Normally Simpson is used for integrating between an interval a to b, by
>using the end points of the interval (a and b) as nodes. An advantage
>of this procedure is, that by doubling the number of nodes between two
>iteration loops the already calculated nodes can be reused, so that
>only about half of the nodes have to be calculated new.

>Due to a singularity at one of the end points of the interval, I try to
>apply the Simpson integration slightly different. I split the
>integration interval a to b in in equidistant subintervals and use the
>middle of these intervals as nodes. If, in this procedure, the number
>of nodes would be doubled one would loos the possibility of reusing
>the number of nodes from one iteration loop to the next, i.e. in an
>iteration loop one third of the nodes can be used from the previous
>loop and two thirds have to be calculated new. This adjusted procedure
>calculates the integral, but, contrary of what I would have expected,
>with a very poor "efficiency".
>
>Comparing the required number of iteration loops for a given
>Rel. Error = ABS( 1 - Result_new_Loop / Result_old_Loop )
>between the adjusted Simpson procedure and a "basic" numerical
>integration (sums of Fi*dF, doubling nodes without reusing already
>calculated ones) shows the following differences:
>
>Rel. Error Result Number of nodes
><1E-3 2062.2593 78732
><1E-4 2063.7296 4960116
><1E-5 2063.8892 389014812
>
>"Basic" iteration procedure
>Rel. Error Result Number of nodes
><1E-5 2063.9027 3072 (approx. * 2)
><1E-8 2063.9091 294912 (approx. * 2)
>
>Questions:
>I can somehow understand that the adjusted Simpson procedure requires
>more iteration loops since the relative error is calculated with twice
>the number of "new" nodes compared to the number of "old" nodes. But
>why is the result not more accurate with the much larger number of
>nodes? Or, are the Simpson factors for the sums (1, 4 and 2) not the
>same for this adjusted version? Is there a possibility to adjust the
>Simpson integration so that it becomes more efficient than the "basic"
>iteration procedure?
>
>
>Marcel

Date Subject Author
6/11/06 Toni Lassila
6/11/06 Julian V. Noble
6/11/06 MET
6/12/06 Julian V. Noble
6/11/06 Dave Dodson
6/11/06 MET
6/12/06 MET
6/12/06 Peter Spellucci
6/12/06 Martin Eisenberg
6/13/06 MET
6/14/06 Peter Spellucci
6/14/06 MET
6/15/06 MET