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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,115

To use the asymptotic form, the integrals need to be from something to infinity, do they not?

The first answer that M was able to give was about 446, this is not a good sign for convergence. Still could converge though.

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Not necessarily I don't think -- I think the series expansions hold for any positive argument, but I'll need to check.

Which parameters did you use to get 446?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,115

I am playing with this:

`NSum[(NIntegrate[((x^2 + y^2)^(-3/4))*(((b - x)^2 + (c - y)^2)^(-3/4)), {x, 0, 2*Pi}, {y, 0, 2*Pi}]^2), {b, 1, 10}, {c, 1, 10}]`

I would not get too excited over that answer it will be wildly erratic for small changes.

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Thanks. Hmm, 446 for just up to (b,c) = (10,10) is quite bad. It is probable that by bounding the Bessel functions we have lost too much.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,115

The error messages are more informative, M thinks there is a singularity or multiple singularities lurking inside the region of integration. Anyway, he is spending most of his time trying to do the integrals.

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Also, how long did it take you to get 446? I remember trying something similar and it would not let me do any higher than N = 24 without giving me some sort of error message.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,115

Hi;

Not too long, your workstation is probably a bit faster than my desktop.

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I got an answer for (b,c) = (1000,1000) after 32 seconds but it doesn't square the summand. Trying with the square now, going to leave it running for a few hours.

*Last edited by zetafunc (2016-10-09 19:18:45)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,115

Does it look like it converges?

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I'm not sure. I've replaced the sum with an integral instead (over the range [1, infinity)). Trying with (b,c) = (100,100) at the moment, might reduce that if it's taking too long.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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If there is convergence you should definitely see it happening with various (b,c).

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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Is it a bad idea to have NIntegrate nested inside NSum?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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That is what I have been doing. But the integral in post #28 may not converge. Numerical techniques like NIntegrate have to obey the rules too.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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I ran this:

bobbym wrote:

I am playing with this:

`NSum[(NIntegrate[((x^2 + y^2)^(-3/4))*(((b - x)^2 + (c - y)^2)^(-3/4)), {x, 0, 2*Pi}, {y, 0, 2*Pi}]^2), {b, 1, 10}, {c, 1, 10}]`

I would not get too excited over that answer it will be wildly erratic for small changes.

But after 4 hours I have no result yet.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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The integral as I said is consuming all the time.

For (b,c)=(n,m) as n and m approach infinity the expression seems to get smaller and smaller. So, at least it has a chance to converge. But the integral for 0 to 2 pi has that singularity at 0. This is where the problem is.

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I agree. Due to that singularity, the convergence of the integral (after bounding the Bessel functions) will depend highly on the powers of the polynomials of the integrand. (For instance, the integral of x^n over [0,1] is convergent only for n > -1, and I wouldn't be surprised if there were something similar going on here.)

I have been thinking about offering a bounty on this question on MathSE or perhaps MO.

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I managed to prove convergence for d = 1. I don't have time to write it up but I can bound it in terms of zeta(3), interestingly.

Unfortunately the problem is for d = 2 or higher but at least it is something. I think it is possible to get a similar result for d = 2.

*Last edited by zetafunc (2016-10-10 05:43:48)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,115

Are you sure about that integral?

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I think so. If d = 1 then with some substitutions you can reduce it to the integral of |1-t|^(-1) over the interval [0, 2*pi/b], which is just -log|1 - 2pi/b|.

EDIT: Whoops, I was wrong -- forgot a term.

*Last edited by zetafunc (2016-10-10 05:56:08)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Okay, post it over there if you think that is the way to go, I am going to get some rest. Have not slept in a while. See you later and leave a link about where you post it.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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OK, see you later.

Managed to get the 446 answer in about 38.94 seconds for (b,c) = (10,10). Trying (b,c) = (100,100).

EDIT: Managed to get 457.856 in 1668.29 seconds for (b,c) = (100,100).

I am now running the same calculation for all integers b,c on (-100, 100), not including 0.

*Last edited by zetafunc (2016-10-10 09:51:07)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Hi;

Why bother with -100 to -1? I thought your double sum was from 1 to infinity?

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No, the sum in post #1 varies over all integers apart from (0,0).

I left it on overnight and after 3 hours and 14 minutes, with b,c varying from -100 to 100, I got 482.362. But to be convinced I would need to use much larger numbers, and that could take a very long time to compute...

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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It will not add much so maybe it is converging.

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Hopefully. Of course, there are lots of series which diverge, but very very slowly.

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