Analytically Solving Complex Expressions with Wolfram Alpha

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Discussion Overview

The discussion revolves around the analytical solving of complex expressions using Wolfram Alpha, specifically focusing on the infinite series \(\sum_{n=1}^{\infty} \frac{1}{n^2 + a^2}\) and its higher-order counterparts. Participants explore methods for evaluating these sums and express curiosity about their properties.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • Some participants note Wolfram Alpha's capability to analytically solve the series \(\sum_{n=1}^{\infty} \frac{1}{n^2 + a^2}\) and inquire about methods for higher-order expressions.
  • One participant asserts that the result of the series is a real number but admits uncertainty about how to calculate it.
  • A suggestion is made to split the series into partial sums for evaluation, with a hint involving the imaginary unit.
  • Another participant comments on the complexity of higher-order expressions, suggesting that while they may seem more complicated, the sums remain real due to the nature of \(n\) and \(a\).
  • One participant references a formula related to the Coth function as a source for the Wolfram Alpha result and mentions the need to series expand it to obtain the right-hand side.
  • Another participant expresses difficulty in finding a proof for the series expansion, indicating that the problem is not straightforward for them.

Areas of Agreement / Disagreement

Participants express varying levels of understanding and approaches to the problem, with no consensus on the ease of solving the series or the methods to be employed. Some participants find certain aspects straightforward, while others struggle with the complexity.

Contextual Notes

There are references to specific mathematical techniques and functions, but the discussion does not resolve the underlying assumptions or steps required for the series expansions or evaluations.

Office_Shredder
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I'm noticing wolfram alpha has the amazing ability to analytically solve
[tex]\sum_{n=1}^{\infty} \frac{1}{n^2 + a^2}[/tex]

Anyone know how to do this, and if it's also possible to deal with higher order guys (like it also can do 1/(n4+a2), but it's a way more complicated expression to the point where I'm staring at it wondering if it's actually a real number)
 
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It is a real number. However I have no idea how to calculate it.
 
Office_Shredder said:
I'm noticing wolfram alpha has the amazing ability to analytically solve
[tex]\sum_{n=1}^{\infty} \frac{1}{n^2 + a^2}[/tex]

Anyone know how to do this, and if it's also possible to deal with higher order guys (like it also can do 1/(n4+a2), but it's a way more complicated expression to the point where I'm staring at it wondering if it's actually a real number)

I'd split it into partial sums and then evaluate.

Hint:
##\frac{1}{n^2+a^2}=\frac{1}{(n+ai)(n-ai)}##, where i is the imaginary unit.

Higher order expressions will undoubtedly be a lot more complicated. Though, given n and a are real numbers, the sum will also be real, so they won't necessarily be more complex (heh. See what I did there? :-p).
 
Mandelbroth said:
Higher order expressions will undoubtedly be a lot more complicated. Though, given n and a are real numbers, the sum will also be real, so they won't necessarily be more complex (heh. See what I did there? :-p).

If you split it into two fractions of degree one neither of your series converge anymore.

By "wondering if it's a real number" I meant "gee wolfram alpha has a lot of 4th roots of -1 in that expression" not "I literally don't know if it's a real number"
 
dextercioby said:
This formula right here explains where the Wolframalpha result comes from

http://functions.wolfram.com/ElementaryFunctions/Coth/06/05/0001/

All you have to do now is to series expand the Coth function to get the RHS.

Oh wow that was easy. I was too busy approaching the problem from the wrong side of the equation. Thanks
 
It is actually not easy, not to me. I can't find a proof for the series expansion. :D
 

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