# On Mehler's formula for Hermite polynomials

In the reference article of Richard Askey and George Gasper published in the American Journal of Mathematics, Autumn, 1976, Vol. 98, No. 3 (Autumn,1976), pp. 709-737, they attribute on page 731 the following formula to Mehler (quoting a book by Erdelyi): $$\sum_{n=0}^{\infty}\frac{r^n H_n(x) H_n(y)}{2^n n!}=(1-r^2)^{-1/2} \exp\bigl\{ x^2-\frac{(x-ry)^2}{(1-r^2)} \bigr\},$$ where $$H_n$$ is the $$n$$th Hermite polynomial.

Question: I do not believe that formula, since the lhs is symmetric in $$x,y$$ whereas the rhs fails to be symmetric in $$x,y$$. I am also puzzled since, as said above, this article is a reference material for numerous articles on Laguerre and Hermite polynomials.

The formula is actually symmetric in $$x\leftrightarrow y$$. You can find a proof here: A combinatorial proof of the Mehler formula.
• @Bazin I'm a bit lost. If you literally expand everything in the exponent in the rhs from your post the resulting expression will be symmetric in $x$ and $y$ (and would symbol-by-symbol coincide with the one from the answer by user69642). Jun 10, 2020 at 19:09
• $x^2-\frac{(y-r x)^2}{1-r^2}=y^2-\frac{(x-r y)^2}{1-r^2}$ Jun 10, 2020 at 19:25
• @Carlo Beenakker Take $r=0$ in your equality, you get $x^2-y^2=y^2-x^2$. Nevertheless you are right for my post since $x^2-\frac{(x-ry)^2}{1-r^2}$ is indeed symmetric. Sorry for the absurd question. Jun 10, 2020 at 20:51
• @Aleksei Kulikov You are right for my post since $x^2-\frac{(x-ry)^2}{1-r^2}$ is indeed symmetric. Sorry for the absurd question. Jun 10, 2020 at 20:52
I think that the formula is equal to, for all $$-1, $$\sum_{\ell \geq 0} \dfrac{r^\ell H_{\ell}(x)H_{\ell}(y)}{2^{\ell}\ell!} = (1-r^2)^{-\frac{1}{2}} \exp \left(\dfrac{2xyr -r^2(x^2+y^2)}{1-r^2}\right),$$