Simpson revival
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10-02-2016, 03:29 PM
(This post was last modified: 10-02-2016 03:37 PM by Dieter.)
Post: #11
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RE: Simpson revival
(09-30-2016 06:00 AM)I Wrote: The test should compare the last and previous improved approximations, cf. the posted VBA code. In your program that's the current and previous value of Y. It looks like even this remaining error can be estimated and an even more accurate result can be calculated from the last two improved values. Here is a more structured experimental version. Code: Function Simpson(a, b, Optional errlimit = 0.000001, Optional nmax = 3000) What do you think? The following diagram may give an impression for the integral of f(x) = 1/x from a=1 to b=2. Here X is the number of iterations (i.e. n=2x intervals) and Y is the accuracy, i.e. the number of valid digits. The red line is the classic Simpson method, the blue one represents the "improved" method, and finally the green graph shows the further extrapolated result as shown in the last program. Dieter |
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Messages In This Thread |
Simpson revival - Pekis - 09-28-2016, 08:04 AM
RE: Simpson revival - Namir - 09-28-2016, 10:40 AM
RE: Simpson revival - Dieter - 09-28-2016, 12:25 PM
RE: Simpson revival - Namir - 09-28-2016, 02:00 PM
RE: Simpson revival - Namir - 09-28-2016, 03:50 PM
RE: Simpson revival - Dieter - 09-28-2016, 10:00 PM
RE: Simpson revival - Albert Chan - 07-31-2018, 02:57 PM
RE: Simpson revival - Namir - 09-29-2016, 04:44 PM
RE: Simpson revival - Dieter - 09-29-2016, 06:23 PM
RE: Simpson revival - Namir - 09-29-2016, 10:27 PM
RE: Simpson revival - Dieter - 09-30-2016, 06:00 AM
RE: Simpson revival - Dieter - 10-02-2016 03:29 PM
RE: Simpson revival - Namir - 10-02-2016, 04:48 PM
RE: Simpson revival - Namir - 10-09-2016, 03:14 AM
RE: Simpson revival - Albert Chan - 08-04-2018, 05:29 PM
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