HP71B Integral Questions
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02-05-2020, 07:52 PM
(This post was last modified: 02-05-2020 08:46 PM by Albert Chan.)
Post: #9
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RE: HP71B Integral Questions
(02-05-2020 06:20 PM)J-F Garnier Wrote: The IBOUND value is still not clear for me, both on the HP71 and HP75. With same number of samples and same extrapolation, why is the IBOUND value different and depends on the user-supplied target accuracy P? For instance in the cases above, IBOUND is exactly divided by 10 when P is changed from 1E-4 to 1E-5 - same samples, same extrapolation, same INTEGRAL result. I think P is maximum relative error allowed. Let E = | rough expected integral result | → Maximum absolute error ≈ IBOUND = E × P Then, it just walk across the romberg numbers, using the difference as a guide, pick the correct gap. (02-03-2020 01:58 PM)Albert Chan Wrote: Extending rows of 128 intervals, extrapolate to its limits, we have: Say, E = 62563e6 For P=1E-5, 63575 < (IBOUND = 625630) < 40399538 → return 62563050740 For P=1E-6, 1007 < (IBOUND = 62563) < 63575 → return 62563050648 This implied we need at least 4 romberg numbers, or minimum of 7 sample points. |
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