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

wrote:

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

- Hide quoted text — Show quoted text -

>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

>already calculated nodes. This disadvantage can be overcame by trebling

>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:

>Adjusted Simpson Procedure:

>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?

>Thank you for your help.

>Marcel