There is a well-known bound on Hermite polynomials: $$ \left|\cfrac{H_n(x)}{\sqrt{2^n n!}}\right|\le e^{0.5x^2}, $$ where $x$ is real. I am trying to find a bound of the following form: $$ \left|\cfrac{H_n(ix)}{\sqrt{2^n n!}}\right|\le e^{\lambda n}f(x), $$ for the case of imaginary arguments, where $f(x)$ does not depend on $n$. I would be very grateful for suggestions. Numerical analysis indicates that such bounds might be possible. However, all bounds that I found in the literature are for real arguments.
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Can you access the paper referenced at https://www.rand.org/pubs/research_memoranda/RM97.html ? – Claude Leibovici Oct 10 '23 at 03:40
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Thank you for the reference. Unfortunately, I do not have access to this publication. I also searched many articles mentioning Hermite polynomials of imaginary arguments but I still cannot find any meaningful bounds... – george_ch Oct 10 '23 at 09:37
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1Check this question and the answer, which is valid for all complex numbers. Take $\lambda(n) = \log{\sqrt{2\pi n!}}$ and $f(x)=e^{\sqrt{2n} |x|}$. You cannot have $f$ independent of $n$ as the higher the value of $n$, the faster the polynomial grows on the imaginary axis. You will find no such bound. This objection come from looking at asymptotic estimates of Hermite plynomials (see work by Olver, and NIST/DMLF). – user12030145 Jan 28 '24 at 21:02