I want to expand $e^{-a^2 x^2}$ to a series with the hermite polynomials $H_{2n}(x)$ but i can't find out how. I've already expanded $x^{2n}$,$x^{2n+1}$ and $e^{ax}$ to a series of$H_n(x)$ but i can't relate them. Any help?
Hermite polynomial expression
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$\begingroup$
hermite-polynomials
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0What makes you think that there is a relation ? How did you get the expansions ? – 2017-02-13
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0they are in the same book and usually this author connect's the exercises. for the first 2 i used a given formula for $x^n$ and for the last one i used the Hermite polynomial generator function. – 2017-02-13
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It is known that $$\sum_{n=0}^\infty \frac{H_n(x)H_n(y)}{n!}\left(\frac u 2\right)^n= \frac{1}{\sqrt{1-u^2}} e^{\frac{2u}{1+u}x y-\frac{u^2}{1-u^2}(x-y)^2}$$ If we let $y=0$ we find that $$\sqrt{1-u^2}\sum_{n=0}^\infty \frac{H_n(x)H_n(0)}{n!}\left(\frac u 2\right)^n= e^{-\frac{u^2}{1-u^2}x^2}$$ If we run the substitution $a^2 = \frac{u^2}{1-u^2}$ we can solve for $u$ and subsitute in to get a closed form for your series. Wikipedia cites this source, which is in German but provides the needed documents for free as far as I could tell. Note that you can probably clean this up a bit to be in terms of $H_{2n}(x)$, since the Hermite Numbers $H_n(0)$ are defined to be $0$ for odd $n$.
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0[+1] Excellent. Is there a connection with Gram-Charlier expansions ? – 2017-02-13