Series of inverse zeros of bessel functions

Question

I am interested in the numerical values for the series $\displaystyle\sum_{n=1}^\infty\frac{1}{j_{0,n}^4}$ and $\displaystyle\sum_{l=1}^\infty\sum_{m=1}^\infty\frac{1}{j_{l,m}^4}$. where $j_{k,m}$ is the $m-th$ positive zero of the Bessel function of order $k$. Are there known formulas for other exponents (rather than $4$)?

Answer 1

From 'Spectral Sum Rules for the Circular Aharonov-Bohm Quantum Billiard,' F. Steiner, Fortshcr. Phys. 35 1, p 87-114 (1987). The history of this problem is much older than this reference.

A table is given of the closed forms of the following, for $s=1,2,...10$. I'll stop a short list at s=4. $$\sum_{k=1}^\infty (j_{a,k})^{-2s}$$ $$s=1: \quad \frac{1}{4(a+1)}$$ $$s=2: \quad \frac{1}{16(a+1)^2(a+2)}$$ $$s=3: \quad \frac{1}{32(a+1)^3(a+2)(a+3)}$$ $$s=4: \quad \frac{5a+11}{256(a+1)^4(a+2)^2(a+3)(a+4)} $$

Answer 2

I do not know any formula for the computation of $$S_k=\sum _{n=1}^{\infty } \left(j_{0,n}\right){}^{-k}$$ but we can compute them to high accuracy (even if it takes a quite long time).

For $k=4$, the result, for twenty five significant figures is $0.03125000000000000$ and the number of trailing $0$'s is quite impressive (notice that, for $k=2$, the result is $0.25000000000000000$ ). I suppose that this hides something I totally ignore.

I suppose that this hides something I totally ignore. I tried (with no success at all) to use some of the approximations of the zeros.

Answer 3

For $k\in\mathbb{N}$ and $\nu\in\mathbb{C}\setminus\{0,-1,-2,\dotsc\}$, the sequence $$\zeta_{\nu}(2k)=\sum_{n=1}^{\infty}\frac{1}{j_{\nu,n}^{2k}}$$ can be determinantally represented as \begin{equation}\label{zeta=bessel-det-eq} \zeta_{\nu}(2k) =\frac{(-1)^{k+1}\nu}{(2k)!} \begin{vmatrix} \frac{1}{(\nu)_0} & \frac{1}{(\nu+1)_0} & 0 &\dotsm & 0 & 0\\ 0 & 0 & \frac{1}{(\nu+1)_0} &\dotsm& 0 & 0\\ \frac{1}{2(\nu)_1} & \frac{1}{2(\nu+1)_1} & 0 &\dotsm & 0 &0\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ \frac{(2k-3)!!}{2^{k-1}(\nu)_{k-1}} & \frac{(2k-3)!!}{2^{k-1}(\nu+1)_{k-1}} & 0 & \dotsm & \frac{1}{(\nu+1)_0} & 0\\ 0 & 0 & \binom{2k-1}{1}\frac{(2k-3)!!}{2^{k-1}(\nu+1)_{k-1}} & \dotsm & 0 & \frac{1}{(\nu+1)_0}\\ \frac{(2k-1)!!}{2^{k}(\nu)_k} & \frac{(2k-1)!!}{2^{k}(\nu+1)_k} & 0 & \dotsm & \binom{2k}{2k-2}\frac{1}{2(\nu+1)_1} & 0 \end{vmatrix}. \end{equation}

Y. Hong, B.-N. Guo, and F. Qi, Determinantal expressions and recursive relations for the Bessel zeta function and for a sequence originating from a series expansion of the power of modified Bessel function of the first kind, CMES Comput. Model. Eng. Sci. 129 (2021), no. 1, 409--423; available online at https://doi.org/10.32604/cmes.2021.016431.

Answer 4

This Rayleigh sums can be calculated using a recurrence formulas, see papers:

[1] Kishore, The Rayleigh function, Proc. Amer. Math. Soc. 14 (1963), 527-533.

[2] Sneddon I. N., On some infinite series involving the zeros of Bessel functions of the first kind, Glasgow Mathematical Journal , Volume 4 , Issue 3 , January 1960 , pp. 144 – 156

[3] Elizalde, E., Leseduarte, S., & Romeo, A. (1993). Sum rules for zeros of Bessel functions and an application to spherical Aharonov-Bohm quantum bags. Journal of Physics A: Mathematical and General, 26(10), 2409–2419. doi:10.1088/0305-4470/26/10/012

[4] J.L. deLyra, On the sums of inverse even powers of zeros of regular Bessel functions, arXiv:1305.0228 [math-ph]