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Using conditional expression in summation

Michael Morami, tel aviv uni
Posted 3 years ago

Hi, I have the following sum to compute numerically using wolfram mathematica and i cant seem to succeed limiting my variables as intended. The sum is:

The sum i have to solve

I tried using ConditionalExpression as follows:

A simple code

But to no avail. Code for copying:

ConditionalExpression[\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = \(-\[Infinity]\)\), \(\
\[Infinity]\)]
\*FractionBox[\(1\), \(
\*SuperscriptBox[\(n\), \(2\)] 
\*SuperscriptBox[\((n - m)\), \(2\)]\)]\), {n != {0, m}, 
  m \[Element] Integers}]

Does anybody know how to write the conditions properly so mathematica will solve the summation?

edit: I have also tried the following way, but still the compiler doesn't recognize m,n from the conditions in the actual summation:

New try

code for copy:

Assuming[Element[m, Integers] && n != 0 && n != m, Simplify[\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = \(-\[Infinity]\)\), \(\
\[Infinity]\)]
\*FractionBox[\(1\), \(
\*SuperscriptBox[\(n\), \(2\)] 
\*SuperscriptBox[\((n - m)\), \(2\)]\)]\)]]

Thanks for helping! :>
POSTED BY: Michael Morami
12 Replies
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POSTED BY: Richard Frost

Wow that is amazing. Thank you!

When I asked how splitting the 'm' into cases is the same as 'm in Z' I meant that by splitting you are still allowing 'm' to any Real number within those cases.

I see what you did there and this is very clever. I guess the fact that 'm' is part of the SUM limit force it to be an integer? Am I right?
POSTED BY: Michael Morami

I took m to be an integer because it is stated in your problem definition. :-)
POSTED BY: Richard Frost

haha obviously :D.

But how does Wolfram Mathematica know it needs to limit m to be an integer? The way you wrote it does not imply such a restriction on the code, yet it outputs the correct solution.

To be clear: m>0 contains the case where m=1.59723621 which is not something i want to allow when im doing the summation
POSTED BY: Michael Morami

m is a parameter of the expression. Whether it is integer or real has no impact on the solution.

The sum looks familiar from decades ago. I've no recollection though of where it arises.
POSTED BY: Richard Frost

The sum can be split into 2 large pieces: n from -infinity to -1, plus n from 1 to +infinity. Then you have 3 cases for m: m < 0, m==0, and m>0.
POSTED BY: Richard Frost

I actually did the first one (splitting it into 2 summations), but i cant see how seperating it to m<0, m=0, m>0 is equivalent to 'm is an integer'. i have tried a new approach based on your answer:

New approach

code:

Assuming[m \[Element] Integers, \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = \(-\[Infinity]\)\), \(-1\)]
\*FractionBox[\(1\), \(
\*SuperscriptBox[\(n\), \(2\)] 
\*SuperscriptBox[\((n - m)\), \(2\)]\)]\) + \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 1\), \(\[Infinity]\)]
\*FractionBox[\(1\), \(
\*SuperscriptBox[\(n\), \(2\)] 
\*SuperscriptBox[\((n - m)\), \(2\)]\)]\)]

and arrived to the following solotioun:

An answer

This might be the intended solution, but i am not certain. We have never used PolyGamma functions before, and although we dont really need to study it to get this result, i suspect it is not what the exercise was aiming for. I guess i am expecting a nice convergance to a compact form. So am i just paranoid or wolfram really misinterpet my quary?
POSTED BY: Michael Morami

Do you have access to Gradshteyn and Ryzhik?

https://en.m.wikipedia.org/wiki/Gradshteyn_and_Ryzhik
POSTED BY: Richard Frost

I do, but couldnt find it there... At least not in the series section.
POSTED BY: Michael Morami

For the 3 cases of m, think about what happens to the denominator. For example, it is n^4 when m=0, so the two split summation terms are equal.
POSTED BY: Richard Frost
Rohit Namjoshi
Rohit Namjoshi
Posted 3 years ago

Hi Michael,

Try FullSimplify on the result

Assuming[Element[m, Integers], 
 Sum[1/(n^2*(n - m)^2), {n, -Infinity, -1}] + Sum[1/(n^2*(n - m)^2), {n, 1, Infinity}]]

(*
(12 EulerGamma + m \[Pi]^2 + 12 PolyGamma[0, 1 - m] + 
  6 m PolyGamma[1, 1 - m])/(
 6 m^3) + (-12 EulerGamma + m \[Pi]^2 - 12 PolyGamma[0, 1 + m] + 
  6 m PolyGamma[1, 1 + m])/(6 m^3)
*)

FullSimplify@%

(*
(-9 + m^2 \[Pi]^2 + 3 m \[Pi] Csc[m \[Pi]]^2 (m \[Pi] + Sin[2 m \[Pi]]))/(3 m^4)
*)
POSTED BY: Rohit Namjoshi

Cool this is helpful! Now it looks like a more sensable answer thanks. I guess the code was interperted correctly
POSTED BY: Michael Morami
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