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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.numanalysis.
>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 ><1E3 2062.2593 78732 ><1E4 2063.7296 4960116 ><1E5 2063.8892 389014812 > >"Basic" iteration procedure >Rel. Error Result Number of nodes ><1E5 2063.9027 3072 (approx. * 2) ><1E8 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



