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.

  • $\begingroup$ Thanks for your answer and for the reference, which I need to study. However, I am still puzzled, since if user69642 is correct, the logarithms of the two different rhs must be the same. $\endgroup$ – Bazin Jun 10 '20 at 18:43
  • $\begingroup$ @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). $\endgroup$ – Aleksei Kulikov Jun 10 '20 at 19:09
  • $\begingroup$ $x^2-\frac{(y-r x)^2}{1-r^2}=y^2-\frac{(x-r y)^2}{1-r^2}$ $\endgroup$ – Carlo Beenakker Jun 10 '20 at 19:25
  • $\begingroup$ @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. $\endgroup$ – Bazin Jun 10 '20 at 20:51
  • $\begingroup$ @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. $\endgroup$ – Bazin Jun 10 '20 at 20:52

I think that the formula is equal to, for all $-1<r<1$, $$ \sum_{\ell \geq 0} \dfrac{r^\ell H_{\ell}(x)H_{\ell}(y)}{2^{\ell}\ell!} = (1-r^2)^{-\frac{1}{2}} \exp \left(\dfrac{2xyr -r^2(x^2+y^2)}{1-r^2}\right), $$


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