I am trying to find an expansion of centered Gaussian - $\frac{1}{\sqrt{2\pi}\sigma}\exp({-\frac{x^2}{2\sigma^2})}$ in terms of Hermite polynomials.

Namely to calculate $a_n=\frac{1}{\sqrt{2\pi}\sigma}\int_{-\infty}^{\infty}{\exp({-\frac{x^2}{2\sigma^2}})}H_{n}(x)\exp({-\frac{x^2}{2})}dx$

Any comments are welcome.

Edited later:


Equivalently, I am looking for the value of -


for some arbitrary $\alpha$


  • $\begingroup$ The integrands of two parts of the question are not the same. In the first part, the integrand is: $exp(.)H[x] exp(.)$. In the second part, the integrand is $exp(.) H[x]$. $\endgroup$ – wolfies Jan 17 '15 at 5:00
  • $\begingroup$ If you add exponents you get the general form as in the second part. $\endgroup$ – them May 15 '16 at 5:05

If $\hat{H}_{n}$ are the normalized Hermite functions, and $-1 < r < 1$, then $$ \begin{align} \sum_{n=0}^{\infty}r^{n}\hat{H}_{n}(x)^{2} & =\frac{1}{\sqrt{\pi(1-r^{2})}}\exp\left(-\frac{1-r}{1+r}x^{2}+x^{2}\right) \\ & = \frac{1}{\sqrt{\pi(1-r^{2})}}\exp\left(\frac{2r}{1+r}x^{2}\right). \end{align} $$ The normalized $\hat{H}_{n}$ are chosen so that $\int_{-\infty}^{\infty}hat{H}_{n}(x)e^{-x^{2}}dx = 1$. By choosing $r$ appropriately, you can get what you want. Look for Mehler's kernel on this Wikipedia page: http://en.wikipedia.org/wiki/Hermite_polynomials#Hermite_functions . You'll find more information about how one derives the above special case of Mehler's kernel $\sum_{n=0}^{\infty}r^{n}H_{n}(x)H_{n}(y)$ where $x=y$.

  • $\begingroup$ Could you kindly explain how does one prove that the Mehler kernel expansion converges? $\endgroup$ – Anirbit Apr 26 '16 at 23:52
  • $\begingroup$ @Anirbit : Proving the Mehler expansion is not particularly easy. The first version I found was in Norbert Wiener's book The Fourier Integral and Certain of Its Applications (Chapter 1, Section 7, "The Generating Function of the Hermite Functions"). I'm sure there must be other versions out there, and ones that are simpler. $\endgroup$ – DisintegratingByParts Apr 27 '16 at 0:23
  • $\begingroup$ Thanks! Let me look up that reference! This is particularly confusing because the upper bound on the Hermite polynomials is exponentially diverging and hence I have no intuition for its convergence! $\endgroup$ – Anirbit Apr 28 '16 at 14:46
  • $\begingroup$ And is this a pointwise convergence (uniform on compact sets?) or is this a convergence in the sense of L^2 norm (which is like a convergence on average) ? $\endgroup$ – Anirbit Apr 28 '16 at 14:47
  • $\begingroup$ @Anirbit : Uniform on compact subsets. $\endgroup$ – DisintegratingByParts Apr 28 '16 at 14:48

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