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## Roots of the Bessel Function

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### Roots of the Bessel Function

Tue Nov 29, 2016 5:41 pm

Posts: 10
I had seen this post on the digamma root sum and began to wonder about extending these zeta type sums to the roots of other functions. Playing around with the Bessel function of the first kind I obtained a few nice results that you guys might enjoy deriving or proving.

Probelm 1:

Let $m=2n,\;n\in\left\{1,\;2,\;3,\dots\right\}$

Show that

$$\Omega_\alpha(m)=\sum_{j\;\in\; U_\alpha}\frac{1}{j^m}=\frac{3^{\frac{m}{2}-1}\zeta(m)\Gamma(\alpha+2)\Gamma\left(\frac{m+3}{2}\right)}{2^{\frac{m}{2}-2}\pi^{m+\frac{1}{2}}(\alpha+1)^{\frac{m}{2}}\Gamma\left(\alpha+\frac{m}{2}+1\right)}$$

where

$$U_\alpha=\left\{j\;|\;J_\alpha(j)=0,\;j>0\right\}.$$

The above can be simplified* quite a bit since $m$ is a positive even integer, most notably all the $\pi$s cancel out. Also, the values that $\alpha$ is valid for are dependent on $m$ and it's fairly easy to see what the invalid values are. I was never able to make a solid proof for this formula but it was obtained from a pattern I saw from doing repeated long division of Taylor series and using the result from the case with $\alpha=1/2$, so I admit there is still a chance this (and everything besides the contents of problem 2) could be wrong.

*It simplifies to $\Omega_\alpha(2n)=\left(\frac{3}{2}\right)^{n-1}\frac{(-1)^{n+1}(2n+1)B_{2n}}{2\cdot n!(\alpha+1)^{n-1}}\prod_{k=1}^n(\alpha+k)^{-1}$ where $B_{2n}$ is a Bernoulli number.

Problem 2:

In addition to the above sum an easier (much easier) pair of related sums to consider are

$$\sum_{j\;\in\; U_\frac{1}{2}}\frac{1}{j^s}=\frac{\zeta(s)}{\pi^s}$$

and

$$\sum_{j\;\in\;U_{-\frac{1}{2}}}\frac{1}{j^s}=\frac{\left(2^s-1\right)\zeta(s)}{\pi^s}$$

where $\Re[s]>1$ in each sum.

Another nice result:

We have

$$\frac{J_{\alpha+1}(z)}{J_\alpha(z)}=2\sum_{n=0}^\infty\Omega_\alpha\left(2(n+1)\right)z^{2n+1}$$

where the radius of convergence is determined by the root of $J_{\alpha}(z)$ with smallest absolute value. Setting $\alpha=-\frac{1}{2}$ should give the series for $\tan(z)$.
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