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I know that it exists for the Hermite polynomials an exponential generating function:

$$\sum_{n=0}^\infty H_n(x)\frac{t^n}{n!} = e^{2xt-t^2}$$

however, what happens if the sequence plugged into the formula is not $H_n(x)$ but $H_{4n}(x)$:

$$\sum_{n=0}^\infty H_{4n}(x)\frac{t^n}{n!} = ? $$

Does any generating (exponential or not) function exists which can represent this sum ? Any help is appreciated.

Lagrangian
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    The sum diverges. Given that Hermite numbers $H_{4n}(0)=2^{2n}(4n-1)!!$ (https://en.wikipedia.org/wiki/Hermite_polynomials ), you get a diverging sum at least for $x=0$ – Svyatoslav Sep 07 '24 at 21:33
  • To underline @Svyatoslav's observation, the culprit is that $n!$ doesn't grow fast enough to suppress the growth of the numerator. By contrast, using $(4n)!$ instead is enough to not only converge at $x=0$ but to guarantee convergence everywhere. – Semiclassical Sep 07 '24 at 22:26
  • Thank for the answer and the comments. I have to check up some formulas which I can only tomorrow to give a more qualified comment – Lagrangian Sep 08 '24 at 07:55
  • @Svyatoslav I read something similar twice. But $\exp(-1/8g)K_{1/4}(1/8g)(4g)^{-1/2} =\int \exp(-x^2-gx^4)dx =\sum\frac{(-g)^n}{n!} \int x^{4n} e^{-x^2} dx = \sqrt{\pi}\sum\frac{(-g)^n}{ 16^n n!}H_{4n}(0)$. So here $t=-g/16$. First equality is based on Gradshteyn-Ryzhik (GR) entry 3.469 #1 and the last one on GR 3.462 #4. For $g>0$ the integral is finite, the function looks like bell-shaped curve ($\neq$ Gaussian of course), and the area below this bell curve is certainly finite. All integrals go from $-\infty$ to $\infty$. Discrete sum goes from $n=0$ to $\infty$. – Lagrangian Sep 09 '24 at 08:47
  • IMHO, numeric check is a must for checking every hypothesis. In your case this is straightforward - please, see the results ($t=\frac14$): https://www.wolframalpha.com/input?i=%5Csum_%7Bn%3D1%7D%5E%7B10%7D+%284%5En%284n-1%29%21%21%2Fn%21%281%2F4%29%5En%29 and https://www.wolframalpha.com/input?i=%5Csum_%7Bn%3D1%7D%5E%7B100%7D+%284%5En%284n-1%29%21%21%2Fn%21%281%2F4%29%5En%29 and https://www.wolframalpha.com/input?i=%5Csum_%7Bn%3D1%7D%5E%7B1000%7D+%284%5En%284n-1%29%21%21%2Fn%21%281%2F4%29%5En%29 – Svyatoslav Sep 09 '24 at 11:28
  • @Svyatoslav there must be a reason for this. Does the summation and integration not commute ? – Lagrangian Sep 09 '24 at 11:49
  • This is a good point for investigation – Svyatoslav Sep 09 '24 at 12:31

1 Answers1

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This won't be quite what you want, but it's not too hard to get a power series of the form

$$G(t;x)=\sum_{n=0}^\infty H_{4n}(x)\frac{t^{4n}}{(4n)!}$$

using a root of unity filter. In this case, writing $$F(t;x)=\sum_{n=0}^\infty H_n(x)\frac{t^n}{n!} = e^{2xt-t^2}$$ we have \begin{align} G(t;x) &:=\frac14\Big[F(t;x)+F(it;x)+F(-t;x)+F(-it;x)\Big]\\ &=\sum_{n=0}^\infty \frac{1}{4}\left[t^n+(it)^n+(-t)^n+(-it)^n\right]H_n(x)\frac{1}{n!}\\ &=\sum_{n=0}^\infty \frac{1}{4}\left[1+i^n+(-1)^n+(-i)^n\right]H_n(x)\frac{t^n}{n!} \end{align} The term in brackets sums to zero unless $n$ is a multiple of 4. Hence

$$G(t;x)=\sum_{n=0}^\infty H_{4n}(x)\frac{t^{4n}}{(4n)!}$$

Additionally, note that we may group terms to obtain the simpler form

\begin{align} G(t;x) &=\frac14 [F(t;x)+F(-t;x)]+\frac14[F(it;x)+\frac14 F(-it;x)]\\ &=\frac14[e^{2xt-t^2}+e^{-2xt-t^2}]+\frac14[e^{2ixt+t^2}+e^{-2ixt+t^2}]\\ &=\frac12 e^{-t^2}\cosh(2xt)+\frac12 e^{t^2}\cos(2x t) \end{align} A lingering issue is that this is a series in powers $t^4$, not $t$. But this is remedied by replacing $t\to t^{1/4}$:

$$G(t^{1/4};x)=\sum_{n=0}^\infty H_{4n}(x)\frac{t^{n}}{(4n)!}=\frac12 e^{-t^{1/2}}\cosh(2xt^{1/4})+\frac12 e^{t^{1/2}}\cos(2x t^{1/4})$$

Semiclassical
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