# Is a Bessel function larger than all other Bessel functions when evaluated at its first maximum?

Let $\mathcal{J}_{n+1/2}$ be the Bessel function of order $n+1/2$. Let $j'_{n+1/2,1}$ denote the first zero of its derivative, which is also the location of the first maximum of $\mathcal{J}_{n+1/2}$.

My question: Given a non-negative integer $n$, is it true that $|\mathcal{J}_{m+1/2}(j'_{n+1/2,1})|<|\mathcal{J}_{n+1/2}(j'_{n+1/2,1})|$ holds for all non-negative integers $m\neq n$?

EDIT: The answer is yes and the proof can be found in the appendix of the paper http://arxiv.org/abs/1504.07306 Thanks for the helpful comment @Jung Wen Chen!

• Have you written up the proof somewhere? I would like to read it. – username Oct 23 '14 at 8:53
• Yes, see the edit – M Lemm Apr 29 '15 at 18:12
• Jung Wen Chen isn't my name. I think MO is nicer if answers are answers judged for what they are worth without wondering whether the author is a graduate student or a tenured faculty or a field medalist, as it leads to automatically favouring answers from annointed mathematicians. If you really want to say something in your paper, a general acknowledgement to fruitful discussions on mathoverflow for example would be more fair to the process which lead to the tip. – username May 5 '15 at 16:48

We suppose $0<m<n$. It is known, from Landau's paper quoted above that $n\to n^{1/3}J_n(x_n)$ increases (to some universal constant $b$). Thus $$J_n(x_n)>\left(\frac{m}{n}\right)^{1/3}J_m(x_m)>\left(\frac{m}{x_n}\right)^{1/3}J_m(x_m).$$ It is also known that for $x>n$, $x\to \sqrt{x^2-n^2}((J_n(x))^2+(Y_n(x))^2)$ increases to $2/\pi$. Therefore, as $x_n>n>m$ $$|J_m(x_n)|^2\leq \frac{2}{\pi}\frac{1}{\sqrt{x_n^2 -m^2}}.$$ Therefore, the inequality holds at least for all $n,m$ such that $$\frac{2}{\pi m^{2/3}(J_m(x_m))^2}<\sqrt{x_n^{2} -m^{2}} \frac{1}{x_n^{2/3}},$$ and the left hand-side is between $1.39$ and $1.89$. If we write $$m=\frac{x_n}{1+\lambda x_n^{-2/3}}$$ we find that this inequality holds for all $x_n\geq40$, and $\lambda>11/10$ for example. The asymptotic of $x_n$ is quite well known, and it is precisely of that sort of magnitude, $x_n=n+\beta n^{1/3}+O(1)$. So, refining the argument and using $n$ instead of $x_n$, you would be very close. For $m=n-1$, you get $$\frac{2}{\pi m^{2/3}(J_m(x_m))^2}<\sqrt{\beta^2 n^{2/3} +2\beta n^{4/3}+2n-1} \frac{1}{n^{2/3}},$$ Asymptotically, this is not enough, because $\beta\approx 0.8$ and $\sqrt{1.6}=1.26<1.39$, but that does limit the range of $m$ to investigate, to $m\geq n- n^{1/3}/5$, so you are only have to consider the case when $J_m(x_n)>0$, $J_m^\prime(x_n)<0$, as the first root is further away ($m+ 1.8m^{1/3}$).
• @MLemm I am note sure what you mean by second line. If you mean second inequality, it is true for all $\nu\geq\frac{1}{2}$: $|H^{(1)}_{\nu}(x)|^2\sqrt{x^2-\nu^2}$ is increasing to its limit (for $x>\nu$), $2/\pi$. Page 447, 13.74 Watson's book. I used no asymptotic result which is not true for all $m$. The bound $x_{\nu}>\nu+0.8\nu^{1/3}$ is true for all $n>1$.Everything is for all $m$, of the form integer plus a half or not. – username Jun 9 '14 at 18:37
• Thanks for the clarification. Do you have a reference for $x_\nu>\nu +0.8 \nu^{1/3}$? I can only find an old paper by Gatteschi & Laforgia, which adresses uniform lower bounds on $x_\nu$, but does not quite yield this for all $\nu > 1$. – M Lemm Jun 12 '14 at 1:06