A zeta-like sum

zetafunc · 2016-11-22 07:19:15

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.

bobbym · 2016-11-22 07:19:58

Just not zero?

zetafunc · 2016-11-22 07:32:51

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

bobbym · 2016-11-22 11:19:42

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

zetafunc · 2016-11-22 12:52:31

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

bobbym · 2016-11-22 13:17:39

Do you have one to see?

zetafunc · 2016-11-25 08:38:29

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)

bobbym · 2016-11-25 21:17:48

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

zetafunc · 2016-11-26 04:02:35

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.

bobbym · 2016-11-26 08:37:54

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

zetafunc · 2016-11-26 09:00:48

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)

bobbym · 2016-11-26 09:01:46

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}]

zetafunc · 2016-11-26 09:29:26

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

bobbym · 2016-11-26 09:31:28

A couple of seconds.

bobbym · 2016-11-26 09:34:36

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}]

zetafunc · 2016-11-26 09:36:12

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)

bobbym · 2016-11-26 09:38:06

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

zetafunc · 2016-11-26 09:38:50

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

zetafunc · 2016-11-26 09:39:53

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)

bobbym · 2016-11-26 09:51:18

For values that cause forms like /0 there will be big problems.

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

zetafunc · 2016-11-26 09:53:55

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)

bobbym · 2016-11-26 10:17:51

It might be alternating.

zetafunc · 2016-11-26 10:24:52

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).

bobbym · 2016-11-26 10:26:11

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

zetafunc · 2016-11-26 10:28:44

Leibniz property?

Math Is Fun Forum

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

Re: A zeta-like sum

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

Re: A zeta-like sum

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

Re: A zeta-like sum

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

Re: A zeta-like sum

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

Re: A zeta-like sum

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

Re: A zeta-like sum

#82 2016-11-25 08:38:29

Re: A zeta-like sum

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

Re: A zeta-like sum

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

Re: A zeta-like sum

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

Re: A zeta-like sum

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

Re: A zeta-like sum

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

Re: A zeta-like sum

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

Re: A zeta-like sum

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

Re: A zeta-like sum

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

Re: A zeta-like sum

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

Re: A zeta-like sum

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

Re: A zeta-like sum

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

Re: A zeta-like sum

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

Re: A zeta-like sum

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

Re: A zeta-like sum

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

Re: A zeta-like sum

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

Re: A zeta-like sum

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

Re: A zeta-like sum

#99 2016-11-26 10:26:11

Re: A zeta-like sum

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

Re: A zeta-like sum

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