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I was wondering if there is a closed form expression for the zeros of the following equation:

$$\sum\limits_{n=1}^\infty\frac{1}{n^4 + x^2} \text{ where } x \in \rm I\!R$$

If it not exists, could you suggest a numerical method for calculate the approximate values of these zeros.

Thanks you in advance for every response!

Alessandro Palla
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  • Since for $x\in \mathbb{R}$ this is a series of positive terms, it is unclear how you expect there to be "zeros" of the expression (not an equation). – hardmath Mar 11 '16 at 18:24
  • Totally agree with you, I've made a mistake writing the equation. My problem is there: http://math.stackexchange.com/questions/1693376/closed-form-solution-for-the-zeros-of-an-infinite-sum – Alessandro Palla Mar 11 '16 at 18:36

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See this answer in which we determine that, for $a \gt 0$,

$$\sum_{n=-\infty}^{\infty} \frac{1}{n^4+a^4} = \frac{\pi}{\sqrt{2} \, a^3} \frac{\sinh{(\sqrt{2} \pi a)}+\sin{(\sqrt{2} \pi a)}}{\cosh{(\sqrt{2} \pi a)} -\cos{(\sqrt{2} \pi a)}}$$

Therefore,

$$\sum_{n=1}^{\infty} \frac{1}{n^4+x^2} = \frac12 \left [ \frac{\pi}{\sqrt{2 |x|^3} } \frac{\sinh{(\pi \sqrt{2 |x|} )}+\sin{((\pi \sqrt{2 |x|})}}{\cosh{((\pi \sqrt{2 |x|})} -\cos{((\pi \sqrt{2 |x|})}}- \frac1{x^2} \right ] $$

Not sure how you expect to derive zeroes from the sum.

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Ron Gordon
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