Does there exist a closed form expression for the zeros of the following equation?

$$\sum\limits_{n=1}^\infty\frac{1}{n^4 - x^2} = 0 \text{ where } x \in \rm \mathbb R$$

Could you suggest a numerical method for calculate the approximate values of these zeros if that solution doesn't exist?

  • 2
    $\begingroup$ Strictly writing, you should add $=0$ in order to turn your expression into an EQUATION. .. $\endgroup$ – Jean Marie Mar 11 '16 at 18:57

If you decompose $\frac{1}{n^4 - x^2}$ into $\frac{1}{2x}(\frac{1}{n^2 - x} - \frac{1}{n^2 + x})$, and compute both series independently (everything converges absolutely, so it is OK to do so), you'd end up with an equation

$$\pi\sqrt{x}(\coth{\pi\sqrt{x}} + \cot{\pi\sqrt{x}}) = 2$$

which doesn't look promising for a closed form. It might be a good starting point for a numerical solution.

  • $\begingroup$ In that form Newton's method works reasonably well - it seems like convergence is quadratic. $\endgroup$ – feralin Mar 11 '16 at 20:47
  • $\begingroup$ Just a question: how do you compute the two series? Using the residue theorem or similar? $\endgroup$ – Alessandro Palla Mar 11 '16 at 21:31
  • $\begingroup$ @AlessandroPalla No, I just happen to know them. They follow from Euler's infinite product representation of $\sin x$. Another way is a direct demonstration - see www-m9.ma.tum.de/foswiki/pub/WS2010/IMCSeminar/Cotangent.pdf for an inspirational example. $\endgroup$ – user58697 Mar 11 '16 at 22:34
  • $\begingroup$ @AlessandroPalla It easy to understand that the equation $y(\coth y + \cot y) = 2$ has an infinite number of real solutions (for $k\in \mathbb Z_{\ge1}$, every interval $]\pi k,\pi(k+1)[$ will contain exactly one solution). To get their asymptotic form, we can note that for large $k$ we can replace $\coth y$ by $1$, so that $y_k\approx \pi k+\frac{3\pi}{4}$ with good precision. $\endgroup$ – Start wearing purple Mar 12 '16 at 9:47

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