Timeline for Errors in `NIntegrate` - or in coding?
Current License: CC BY-SA 4.0
18 events
| when toggle format | what | by | license | comment | |
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| Jun 6 at 9:59 | comment | added | Richard Burke-Ward | Thanks to you all | |
| Jun 5 at 18:39 | comment | added | Michael E2 | @yarchik No, I was agreeing with you. I suggested PV just looking at the denominators, and took it back in my 2nd comment after realizing the effect of floor[] on the asymptotes at integers. As you said, there's no point going into it much. | |
| Jun 5 at 17:50 | comment | added | yarchik | @MichaelE2 Probably I misunderstood your comment, I thought you seek a way to fix the f[3] case. | |
| Jun 5 at 16:06 | comment | added | yarchik |
@MichaelE2 The whole approach does not work if it does not work in some cases. It is sufficient to Plot[f[2] + g[2] - h[2], {u, 1, 2}] to see two non-integrable singularities at x=1 and at x=2. Maybe there are also integrable singularities in other cases. But what is the point of such in depth analysis when already the simplest case diverges?
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| Jun 5 at 15:08 | comment | added | Michael E2 |
I take that back about principal value. The Floor[u] means some PVs do not exist in your problem.
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| Jun 5 at 14:53 | comment | added | Michael E2 |
I think @yarchik means, for example that FunctionSingularities[f[3], u] shows a pole at u == 3/2, which is within the interval of integration. The principal value is an approach, if valid.
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| Jun 5 at 10:31 | vote | accept | Richard Burke-Ward | ||
| Jun 5 at 9:49 | review | Close votes | |||
| Jun 24 at 3:05 | |||||
| Jun 5 at 9:32 | answer | added | Roland F | timeline score: 1 | |
| Jun 5 at 9:29 | comment | added | yarchik | I am voting to close this question because it is just a simple mistake. There is nothing to fix here with PrincipalValue, just set x=2, and plot the functions in the integration interval. | |
| Jun 5 at 8:30 | comment | added | Richard Burke-Ward |
Hi @yarchick. OK, thank you. There are certainly discontinuities at integer x because of the floor function... But since each interval [1,2], [2,3], [3,4] etc are separately integrable, my understanding was that the functions still counted as Riemann integrable over the entire [1,x] interval. Perhaps I am wrong, or perhaps there is some other issue? Is it something I could get around by using PrincipalValue, for example?
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| Jun 5 at 8:27 | comment | added | yarchik | Your math is wrong, the functions have singularities and are not integrable, at least in the case x=2, which I checked. | |
| Jun 5 at 8:05 | comment | added | Richard Burke-Ward |
I agree about the errors, @Michael. They make me suspicious too. But I am at a loss for alternatives, because plain old Integrate just won't work with Floor...
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| Jun 5 at 7:56 | comment | added | Richard Burke-Ward |
Thanks @Daniel. Just FYI, the x vs u thing is deliberate; this is part of a bigger function with many variables, and the required result is its value for a given x.
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| Jun 5 at 7:28 | comment | added | Daniel Huber | You integrate over "u" but you function are given as "f[x]". | |
| Jun 5 at 6:59 | comment | added | Michael E2 |
I don't know about you, but NIntegrate gives me a bunch of "failed to converge" errors. It makes me think that is where to look for the problem.
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| Jun 5 at 6:47 | history | edited | Richard Burke-Ward | CC BY-SA 4.0 |
edited title
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| Jun 5 at 6:40 | history | asked | Richard Burke-Ward | CC BY-SA 4.0 |