2
$\begingroup$

Question. Let $J_{\nu}$ be a standart Bessel function of the first order. What is the asymptotic of the sum $\sum_{n\ge 0}|J_n(t)|$ as $t\to\infty$? An upper bounds stronger than $O(t)$ are also interesting.

Thoughts. For $t > 0$ we have the asymptotic relation $$ J_{\nu}(t)\sim\frac{1}{\sqrt{2\pi\nu}}\left(\frac{et}{2\nu}\right)^{\nu},\quad \nu\to\infty. $$ Moreover, there is an inequality $$ \left|J_{\nu}(t)\right|\le \frac{(t/2)^\nu}{\Gamma(\nu + 1)}, \quad t > 0, \quad \nu > 0. $$ These formulas are 10.19.1 and 10.14.4 in DLMF library. We can use the latter inequality for all $n \ge 2t$ and $\left|J_{n}(t)\right|\le 1$ for $n < 2t$ to establish $$ \sum_{n\ge 0}|J_n(t)| = O(t), \quad t\to\infty, $$ however I think that the bound can be made much stronger.

$\endgroup$
1
  • 1
    $\begingroup$ Some Google-fu led me to a paper by Larry Landau (not to be confused with Lev Landau) "Monotonicity and bounds on Bessel functions" which claims a uniform bound $|J_\nu(t)|\le \frac{1}{x^{1/3}}$ which would yeild $O(t^{2/3})$ bound I believe. But take this paper at your own risk, I didn't read it and don't know if it's correct. $\endgroup$ Commented Oct 19, 2023 at 13:03

1 Answer 1

2
$\begingroup$

Since $\sum_{n\ge0}J_n(t)^2=1/2$ and $\sum_{n\ge2t}|J_n(t)|$ is negligible, applying Cauchy-Schwartz gives $S(t)=O(t^{1/2})$. Experimentally, I would guess that $S(t)=O(\log(t)^{5/2})$ or so.

Also, independently of this question, it seems that $\sum_{t<n\le 2t}|J_n(t)|$ tends to a limit around $1/3$. Is this true ?

$\endgroup$
1
  • $\begingroup$ The Cauchy argument is nice, thank you! Looks like the bounds of the functions $J_{\nu}(\nu x)$ are nice only when $x = 1 + O(1/\nu)$ and the conjecture about the sum over $[t, 2t]$ deserves its own question $\endgroup$ Commented Oct 19, 2023 at 16:49

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .