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#76 2016-11-22 07:19:15

zetafunc
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Re: A zeta-like sum

We just want b and c to be non-zero. Maybe something like Boole[b!=0 AND c!=0], or define the integrand as a piecewise function taking 0 whenever b or c are 0.

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#77 2016-11-22 07:19:58

bobbym
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From: Bumpkinland
Registered: 2009-04-12
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Re: A zeta-like sum

Just not zero?


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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#78 2016-11-22 07:32:51

zetafunc
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Re: A zeta-like sum

Yes, b,c non-zero is the only restriction on b,c in this case.

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#79 2016-11-22 11:19:42

bobbym
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From: Bumpkinland
Registered: 2009-04-12
Posts: 108,547

Re: A zeta-like sum

I remember this, there is also the restriction of b ≠ x and c ≠ y


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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#80 2016-11-22 12:52:31

zetafunc
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Re: A zeta-like sum

That may be necessary. If the Bessel functions are too difficult it is possible to replace them with cosines.

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#81 2016-11-22 13:17:39

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 108,547

Re: A zeta-like sum

Do you have one 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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#82 2016-11-25 08:38:29

zetafunc
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Re: A zeta-like sum

The first term in the asymptotic expansion of
is:

so we can write the expression as looking like:

ignoring all the constants.

Unfortunately Mathematica has a lot of trouble just trying to evaluate this integral for given b,c, so the sum looks hopeless.

Last edited by zetafunc (2016-11-25 10:06:42)

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#83 2016-11-25 21:17:48

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 108,547

Re: A zeta-like sum

What did you use to get that asymptotic form for the Bessel function?


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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#84 2016-11-26 04:02:35

zetafunc
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Re: A zeta-like sum

I looked it up in my supervisor's copy of Gradshteyn and Ryzhik. There is a version involving sin as well

Unfortunately, neither version is workable. I cannot even compute the integral over [1,2], putting b = c = 10, say.

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#85 2016-11-26 08:37:54

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 108,547

Re: A zeta-like sum

Can you post the commands? My typing skills are notoriously bad.


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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#86 2016-11-26 09:00:48

zetafunc
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Re: A zeta-like sum

Here is what the integral looks like in Mathematica over [1,2]*[1,2], not including the sum and putting the values b = c = 100 in.

Integrate[(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, 2}, {y, 1, 
  2}]

I've been running that all day and have not got anything though.

Last edited by zetafunc (2016-11-26 09:02:02)

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#87 2016-11-26 09:01:46

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 108,547

Re: A zeta-like sum

Okay, thanks for posting that.

You can get a quick answer for this.

NIntegrate[
 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}]

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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#88 2016-11-26 09:29:26

zetafunc
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Posts: 1,823
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Re: A zeta-like sum

How long did it take you to get something with that?

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#89 2016-11-26 09:31:28

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 108,547

Re: A zeta-like sum

A couple of seconds.


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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#90 2016-11-26 09:34:36

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 108,547

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}]

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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#91 2016-11-26 09:36:12

zetafunc
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Posts: 1,823
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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.

Last edited by zetafunc (2016-11-26 09:38:02)

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#92 2016-11-26 09:38:06

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 108,547

Re: A zeta-like sum

That is going to be another problem, could get very ugly.


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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#93 2016-11-26 09:38:50

zetafunc
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Re: A zeta-like sum

It looks like Mathematica struggles more with values very close to b,c = 0.

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#94 2016-11-26 09:39:53

zetafunc
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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.

Last edited by zetafunc (2016-11-26 09:50:53)

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#95 2016-11-26 09:51:18

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 108,547

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.


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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#96 2016-11-26 09:53:55

zetafunc
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Registered: 2014-05-21
Posts: 1,823
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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.

Last edited by zetafunc (2016-11-26 10:02:02)

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#97 2016-11-26 10:17:51

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 108,547

Re: A zeta-like sum

It might be alternating.


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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#98 2016-11-26 10:24:52

zetafunc
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Registered: 2014-05-21
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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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#99 2016-11-26 10:26:11

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 108,547

Re: A zeta-like sum

Alternating series have the Leibniz property and they can be bounded.


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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#100 2016-11-26 10:28:44

zetafunc
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Registered: 2014-05-21
Posts: 1,823
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Re: A zeta-like sum

Leibniz property?

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