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That is the sum of the modulus of the old integral squared, which unfortunately is now wrong, according to my supervisor. The new integral is in the newer thread (though the old thread was not made in vain as we still learned some things from it).

My supervisor also says he is sceptical that I can get the integral to converge for d > 2, because he says he remembers trying something similar himself and finding that the integral diverged.

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

I would daresay that your supervisor and yourself are among the world's leading authorities on either double integral. I can not touch it with anything I know.

**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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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi;

You can post it and accept 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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I've just sent my solution to my supervisor -- about 3 pages long. I'll wait and see what he has to say about it.

The case for d = 2 certainly seems much more difficult. If I can't get an answer I might consider putting a bounty up on there, although it might hurt to lose a third of my reputation.

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

How long will it take before you know?

**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'm not sure. He could respond over the weekend. It's a fairly simple proof, though -- I can post it in Members Only if you would like to have a look.

On the other hand I get the feeling the result might not be correct because of a more recent result.

*Last edited by zetafunc (2016-11-03 21:45:21)*

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

On the other hand I get the feeling the result might not be correct because of a more recent result.

What is that result?

**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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There is a result which seems to suggest that the -norm is bounded below by . My results are saying they are bounded above by which is obviously impossible. Then again, that more recent result would also seem to contradict some older results about the and -norms of the remainder, which suggests perhaps we are misinterpreting the results of that paper.

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Did you ask him about that?

**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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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

That is a good idea. Start to set up a meeting to see him.

**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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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Does your integral look like Jacky thinks?

**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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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

That ought to be doable with M, did you try it just to see?

**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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Yeah -- but I already posted this integral in the other thread. Couldn't get an answer with M, but it doesn't matter, because he proved it converges for d > 2. d = 2 is what I'm waiting on, though someone on MO is claiming it converges after bounding the Bessel functions. I don't believe their claim, however.

Some interesting observations: if b = c = 1, then the integral below grows very large. If b and c are large, then the integral grows more slowly and has a much smaller value when integrated over a large region (say, [-10000,10000]). This is similar to the observations my supervisor made on the annulus problem (that the lattice vectors had to be bounded away from zero).

*Last edited by zetafunc (2016-11-05 03:48:14)*

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

though someone on MO is claiming it converges after bounding the Bessel functions.

Which link?

**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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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

I would not go with his answer just yet for the practical reason that he has not been able to deal with your comment yet. I would really like to see some other form for the integral than the one given in post 92.

**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 agree. I'm mostly sceptical about this change to polar co-ordinates, because there is no way that integral he is talking about converges, unless I've misunderstood what he says.

He's just given an answer but I still don't understand it, I'll need to think about it for a while.

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Yea, I see it too. See you in a bit. Got a chore to do.

**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.

M suggests that larger values of b tend to result in something very close to zero. Using M on the integral

we have:

```
NIntegrate[((x^2 + y^2)^(-1/2)) (((100 - x)^2 + (100 - y)^2)^(-1/2))*
BesselJ[1, (x^2 + y^2)^(1/2)]*
BesselJ[1, (((100 - x)^2 + (100 - y)^2)^(1/2))], {x, -10^5,
10^5}, {y, -10^5, 10^5}]
```

which produces -0.000411696 with an error of 0.002530016802142993. The output and error get much bigger though for a range higher than [-10^6, 10^6]. I've tried plotting this to see how NIntegrate behaves for varying b, or for a varying range, but I can't get either to work.

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi;

I am getting a lot of error messages with that so I would not be too confident in the answer or that estimate. Do you use v11?

**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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