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For almost esthetical reasons, I always have been fascinated by infinite sums of series and I always wonder if theyhave a closed form. A couple of days ago, I found on MSE a question related to the convergence of $$S=\Sigma_{n=0}^{\infty}{\frac{n^2 x}{1+n^4 x^2}}$$ where $x$ is a real number. So, I gave the problem to one of my former colleagues who ran Mathematica for this summation.

He sent me back the following (surprizing result) $$\frac{\sqrt[4]{-1} \pi \left(\cot \left(\frac{(-1)^{3/4} \pi }{\sqrt{x}}\right)+i \cot \left(\frac{\sqrt[4]{-1} \pi }{\sqrt{x}}\right)\right)}{4 \sqrt{x}}$$ telling that he did not find any way to make it simpler. For sure, the result is a real. I tried to simplify manually, expanding the $cot$ functions and ... arrived nowhere after pages of manual calculations (being almost blind does not make things easier).

Could any one tell me what should be the appropriate syntax to be used in order to remove all the complex numbers from this expression ? Any help will be really appreciated.

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@b.gatessucks. I asked that and it seems that the answer is just awful (very long formula with a lot of $Arg[x]$ arguments. –  Claude Leibovici Feb 17 at 13:15
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1 Answer

up vote 5 down vote accepted

You can do :

expr = (-1)^(1/4) π (Cot[(-1)^(3/4) π / Sqrt[x]] + 
        I Cot[(-1)^(1/4) π / Sqrt[x]])/(4 Sqrt[x]);

FullSimplify[ComplexExpand[expr, TargetFunctions -> {Re, Im}], Assumptions -> {x > 0}]
(* (π (Sin[(Sqrt[2] π)/Sqrt[x]] - Sinh[(Sqrt[2] π)/Sqrt[x]])) / 
   (2 Sqrt[2] Sqrt[x] (Cos[(Sqrt[2] π)/Sqrt[x]] - Cosh[(Sqrt[2] π)/Sqrt[x]])) *)
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Thank you very much ! Cheers. –  Claude Leibovici Feb 17 at 13:25
3  
FWIW I found that FullSimplify[TrigToExp[expr]] gets you to the same answer more quickly (I have 9.0.1 on OS X). –  Stephen Luttrell Feb 17 at 14:25
    
@StephenLuttrell Great, why not posting another answer ? –  b.gatessucks Feb 17 at 14:53
1  
It's so closely related to the already-accepted answer that I judged it to be more of a comment than an independent answer. –  Stephen Luttrell Feb 17 at 15:07
    
@StephenLuttrell. Thank you very much. Your help is really appreciated. Cheers. –  Claude Leibovici Feb 17 at 15:25
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