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Let's denote with $\{x_n\}_{n\in N}$ the sequence of the largest root of the (statistical) Hermite polynomial $h_n$. Much is known about upper and lower bounds of the $x_n$, see for example here. Clearly, the sequence is monotonically increasing (due to the interlacing property). Studying the numerical values \begin{eqnarray} x_1&=&0\\ x_2&=&1\\ x_3&=&1.7320\ldots\\ x_4&=&2.3344\ldots\\ x_5&=&2.8569\ldots\\ \end{eqnarray} one can observe that the difference $x_{n+1}-x_n$ is monotonically decreasing which I like to prove. A promising approach seems to be the via the facts $$h_n''-xh_n'+nh_n=0\quad \text{and}\quad h_n'=nh_{n-1},$$ but I failed. Help would be welcome!

Ben Hur
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1 Answers1

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You wish to show that the sequence of largest positive zeros form a concave sequence. This (and much more) is shown in the paper Inequalities and monotonicity properties for zeros of Hermite functions by Á. Elbert and M. E. Muldoon. See especially the end of Section $7$.

Gary
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  • Thanks for the source. Unfortunatelly, the paper is not public. Is it possible to sketch the main idea of the proof or is it to profound? – Ben Hur Oct 25 '22 at 14:35
  • It is highly non-trivial. They consider a continuous version $H_\lambda(x)$ of $H_n(x)$ and show that if $x_\lambda$ is any zero then $x'\lambda$ is completely monotonic. In particular they use $$ \frac{{{\rm d}x\lambda }}{{{\rm d}\lambda }} = \sqrt {\frac{\pi }{2}} \int_0^{ + \infty } {{\rm e}^{ - (2\lambda + 1)t} {\rm e}^{x_\lambda ^2 \tanh t} {\mathop{\rm erfc}\nolimits} (x_\lambda \sqrt {\tanh t} )\frac{{{\rm d}t}}{{\sqrt {\sinh (2t)} }}} . $$ – Gary Oct 25 '22 at 14:46