<![CDATA[Math Is Fun Forum / A zeta-like sum]]> 2016-12-17T16:28:08Z FluxBB https://www.mathisfunforum.com/viewtopic.php?id=23540 <![CDATA[Re: A zeta-like sum]]> That integral looks tough too.  Have you applied M to it?

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https://www.mathisfunforum.com/profile.php?id=33790 2016-12-17T16:28:08Z https://www.mathisfunforum.com/viewtopic.php?pid=391590#p391590
<![CDATA[Re: A zeta-like sum]]> My supervisor says that it might be best to estimate the sum by an integral instead. In other words, instead of

we consider

There will be singularities, but he says that they won't lead to divergence in this case. The integral

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https://www.mathisfunforum.com/profile.php?id=206089 2016-12-17T11:29:06Z https://www.mathisfunforum.com/viewtopic.php?pid=391579#p391579
<![CDATA[Re: A zeta-like sum]]> Yes, the value of the error is less than the absolute value of the first neglected term.

I did a much easier one you can look at as an example over here:

http://math.stackexchange.com/questions … 25#1240125

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https://www.mathisfunforum.com/profile.php?id=33790 2016-11-26T22:35:01Z https://www.mathisfunforum.com/viewtopic.php?pid=390921#p390921
<![CDATA[Re: A zeta-like sum]]> Leibniz property?

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https://www.mathisfunforum.com/profile.php?id=206089 2016-11-26T22:28:44Z https://www.mathisfunforum.com/viewtopic.php?pid=390920#p390920
<![CDATA[Re: A zeta-like sum]]> Alternating series have the Leibniz property and they can be bounded.

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https://www.mathisfunforum.com/profile.php?id=33790 2016-11-26T22:26:11Z https://www.mathisfunforum.com/viewtopic.php?pid=390919#p390919
<![CDATA[Re: A zeta-like sum]]> It does seem that way, especially when varying rho. And that would make sense (this is how the remainder behaves in the Gauss circle problem).

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https://www.mathisfunforum.com/profile.php?id=206089 2016-11-26T22:24:52Z https://www.mathisfunforum.com/viewtopic.php?pid=390918#p390918
<![CDATA[Re: A zeta-like sum]]> It might be alternating.

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https://www.mathisfunforum.com/profile.php?id=33790 2016-11-26T22:17:51Z https://www.mathisfunforum.com/viewtopic.php?pid=390917#p390917
<![CDATA[Re: A zeta-like sum]]>
``````NSum[NIntegrate[(Cos[Sqrt[x^2 + y^2] - 3*\[Pi]/4]*
Cos[Sqrt[(b - x)^2 + (c - y)^2]])/((((x^2 + y^2)^(3/
4)))*((b - x)^2 + (c - y)^2)^(3/4)), {x, 1, 10}, {y, 1,
10}], {b, 1, 10}, {c, 1, 10}]``````

works a lot better and yielded 4.02468 in 16 seconds, surprisingly.

``````NSum[NIntegrate[(Cos[Sqrt[x^2 + y^2] - 3*\[Pi]/4]*
Cos[Sqrt[(b - x)^2 + (c - y)^2]])/((((x^2 + y^2)^(3/
4)))*((b - x)^2 + (c - y)^2)^(3/4)), {x, 1, 50}, {y, 1,
50}], {b, 1, 20}, {c, 1, 20}]``````

produced 3.72777 in 45.49 seconds.

``````NSum[NIntegrate[(Cos[Sqrt[x^2 + y^2] - 3*\[Pi]/4]*
Cos[Sqrt[(b - x)^2 + (c - y)^2]])/((((x^2 + y^2)^(3/
4)))*((b - x)^2 + (c - y)^2)^(3/4)), {x, 1, 500}, {y, 1,
500}], {b, 1, 20}, {c, 1, 20}]``````

produced 4.23585 in 198.92 seconds.

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https://www.mathisfunforum.com/profile.php?id=206089 2016-11-26T21:53:55Z https://www.mathisfunforum.com/viewtopic.php?pid=390916#p390916
<![CDATA[Re: A zeta-like sum]]> For values that cause forms like /0 there will be big problems.

My revised post only uses one NIntegrate and gets faster results.

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https://www.mathisfunforum.com/profile.php?id=33790 2016-11-26T21:51:18Z https://www.mathisfunforum.com/viewtopic.php?pid=390915#p390915
<![CDATA[Re: A zeta-like sum]]>
``````NIntegrate[
NIntegrate[(Cos[Sqrt[x^2 + y^2] - (3 \[Pi])/4] Cos[
Sqrt[(1 - x)^2 + (1 - y)^2]])/((x^2 + y^2)^(3/
4) ((1 - x)^2 + (1 - y)^2)^(3/4)), {x, 1, 1000}], {y, 1,
1000}]``````

produced 1.13796 in 89.72 seconds.

And here is what happens when trying a sum over one variable.

``````NSum[NIntegrate[
NIntegrate[(Cos[Sqrt[x^2 + y^2] - (3 \[Pi])/4] Cos[
Sqrt[(b - x)^2 + (10 - y)^2]])/((x^2 + y^2)^(3/
4) ((b - x)^2 + (10 - y)^2)^(3/4)), {x, 1, 1000}], {y, 1,
1000}], {b, 10, 12}]``````

produced 0.655201 in about 50 seconds. And that is with just 1 variable summed over 3 terms.

Currently running a double NSum with a double NIntegrate and that is producing lots and lots of errors, several of which include the words "catastrophic loss of precision".

EDIT:

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

produced 3.05475*10^16 in about 395 seconds.

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https://www.mathisfunforum.com/profile.php?id=206089 2016-11-26T21:39:53Z https://www.mathisfunforum.com/viewtopic.php?pid=390914#p390914
<![CDATA[Re: A zeta-like sum]]> It looks like Mathematica struggles more with values very close to b,c = 0.

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https://www.mathisfunforum.com/profile.php?id=206089 2016-11-26T21:38:50Z https://www.mathisfunforum.com/viewtopic.php?pid=390913#p390913
<![CDATA[Re: A zeta-like sum]]> That is going to be another problem, could get very ugly.

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https://www.mathisfunforum.com/profile.php?id=33790 2016-11-26T21:38:06Z https://www.mathisfunforum.com/viewtopic.php?pid=390912#p390912
<![CDATA[Re: A zeta-like sum]]> Thanks, this works very well. The question is how to get this to combine this with the sum. Going to try and see what NSum does.

EDIT: Sorry, didn't notice your post just now. I managed to get your original code (with the nested NIntegrates) to work after closing Mathematica fully.

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https://www.mathisfunforum.com/profile.php?id=206089 2016-11-26T21:36:12Z https://www.mathisfunforum.com/viewtopic.php?pid=390911#p390911
<![CDATA[Re: A zeta-like sum]]> Okay, try this instead:

``````NIntegrate[(Cos[Sqrt[x^2 + y^2] - 3*\[Pi]/4]*
Cos[Sqrt[(100 - x)^2 + (100 - y)^2]])/((((x^2 + y^2)^(3/
4)))*((100 - x)^2 + (100 - y)^2)^(3/4)), {x, 1, 5}, {y, 1,
5}]``````
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https://www.mathisfunforum.com/profile.php?id=33790 2016-11-26T21:34:36Z https://www.mathisfunforum.com/viewtopic.php?pid=390910#p390910
<![CDATA[Re: A zeta-like sum]]> A couple of seconds.

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https://www.mathisfunforum.com/profile.php?id=33790 2016-11-26T21:31:28Z https://www.mathisfunforum.com/viewtopic.php?pid=390908#p390908