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12 votes
Accepted

Poisson Summation Formula appears to fail when applied to Hermite Functions (why?)

Your mistake is that you are not using the correct form of the Poisson summation formula based on your choice of how to define the Fourier transform. There are multiple conventions on how to define ...
KConrad's user avatar
  • 50.6k
6 votes

On Mehler's formula for Hermite polynomials

The formula is actually symmetric in $x\leftrightarrow y$. You can find a proof here: A combinatorial proof of the Mehler formula.
Carlo Beenakker's user avatar
5 votes

Complex Hermite polynomial orthogonality on weighted space

Fix $\xi$ and assume $k>n$ (the other case is similar). Integrate over $x$. You'll get $\int_{\mathbb R+i\xi}H_k(z)P_n(z-i\xi,\xi)e^{-z^2/2-\xi^2}\,dz$ where $P_n$ is some polynomial of 2 variables ...
fedja's user avatar
  • 61.9k
5 votes

On Mehler's formula for Hermite polynomials

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)}{...
user69642's user avatar
  • 778
5 votes

Generating function for products of complex Hermite polynomials

The OP's question has been addressed By Richard Stanley. So, we make attempt at Stanley's question on the complex counterpart to $$\sum_{n\geq0}\frac{H_n(x)H_n(y)}{n!}\left(\frac{u}2\right)^2 =\...
T. Amdeberhan's user avatar
5 votes

Generating function for products of complex Hermite polynomials

The formula (*) is trivial since it breaks up into a product of a sum over $m$ with a sum over $n$. The same reasoning works for the complex analog as a sum over $m,n,k,l$, using $$ \sum_{m,n\geq 0}...
Richard Stanley's user avatar
4 votes
Accepted

Closed formula for Hermite polynomials

Up to some normalization, the harmonic oscillator $H$ is self-adjoint such that $$ \langle Hu, u\rangle=\sum_{k\ge 0}(\frac12+k) \vert u_k\vert^2, $$ and thus defining a self-adjoint $A$ by the ...
Bazin's user avatar
  • 16.2k
4 votes

How to integrate the multinomial over a ball in $\mathbb{R}^{n}$?

$\newcommand\R{\mathbb R}\newcommand{\Ga}{\Gamma}$The integral in question is \begin{equation*} \begin{aligned} \int_0^1 dr\,\int_{rS_{n-1}}dx\,f(x) &=\int_0^1 dr\,r^{2m+n-1}\int_{S_{n-1}}...
Iosif Pinelis's user avatar
4 votes

How to integrate the multinomial over a ball in $\mathbb{R}^{n}$?

Let me separate the radial integration from the angular integration, $$\int_{|\mathbf{x}|\leq 1}f(\mathbf{x})d\mathbf{x}=\frac{2\pi^{n/2}}{\Gamma(n/2)}\int_0^1 r^{n-1}\bar{f}(r)\,dr,$$ where $\bar{f}(...
Carlo Beenakker's user avatar
4 votes
Accepted

In what sense does the Hermite expansion of a bounded smooth function converge?

According to the conclusion at the bottom of p. 603 of (say) Uspensky's paper, the Hermite expansion of $f$ converges pointwise to $f$ (under conditions much less restrictive that yours). If $f$ is (...
Iosif Pinelis's user avatar
3 votes

In what sense does the Hermite expansion of a bounded smooth function converge?

Here's a simple and far-from-optimal condition guaranteeing uniform convergence. Suppose for that there exists $C>0$ such that for any positive integer $n$ $\newcommand{\bR}{{\mathbb{R}}}$ $$ \int_\...
Liviu Nicolaescu's user avatar
3 votes
Accepted

Asymptotic form of $L^1$-norm of Hermite functions

Results for the $L^p$ norms of general orthogonal polynomials have been given in Ref. [1]. In the case of the orthonormal Hermite polynomials $H_n$, it was given in [1] for $L^{2p}$ norms, $0<p&...
user111's user avatar
  • 4,034
3 votes
Accepted

Hermite Transform of Tanh

$$|a_n|\approx\frac{\Gamma(n/2+1)}{\Gamma(n+1)}\exp\left(-\frac{\pi\sqrt{2n}}{2}\right).$$ Ref. Szego, Orthogonal polynomials, AMS. 1959, formula (8.23.4), see also (9.2.9).
Alexandre Eremenko's user avatar
2 votes
Accepted

Integral of product of Hermite polynomials w.r.t marginal distribution of first two-coordinate of random vector on unit-sphere

To find the dependence of $s_{nm}$ on $t=a\cdot b$, we take $a=(t,\sqrt{1-t^2},0,0,\ldots 0)$, $b=(1,0,0,0,\ldots 0)$, so that $$s_{nm} = \mathbb E[H_n(X^\top a)H_m(X^\top b)]=\mathbb E[H_n(X_1 t+X_2\...
Carlo Beenakker's user avatar
2 votes
Accepted

Integrating a B-Spline basis function with respect to the standard normal PDF

Since a B-spline is a piecewise polynomial function, the question is whether there exists an exact equality formula for the integral $\int_{-a}^{b}u^pe^{-u^2/2}du$. This integral equals an elementary ...
Carlo Beenakker's user avatar
1 vote
Accepted

Asymptotic behavior of the integral of Hermite functions/polynomials on half-lines

Your conjecture is not true. Moreover, $$J_m(y):=\int_y^\infty dx\, e^{-x^2}H_m(x)^2\sim c_m:=\frac12\,\pi^{1/2}\,2^m m! \tag{1}\label{1}$$ whenever $$0\le y=o(m^{1/2}).$$ Indeed, recalling that $H_m^...
Iosif Pinelis's user avatar
1 vote

Asymptotic behavior of the integral of Hermite functions/polynomials on half-lines

Let me first consider the case $x_0=0$, when the integral has a closed form expression: $$\int^{\infty}_{0}|H_m(x)|^2 e^{-x^2}\,dx=\sqrt{\pi } \,2^{m-1} m!\;\;(m\in\mathbb{N}).$$ See, for example, ...
Carlo Beenakker's user avatar
1 vote

A question regarding Hermite polynomials and exponential operators $\exp[e^{x^2/2}p(\frac{d}{dx})e^{-x^2/2}]f(x)$

You can of course write $$ k(x,t) = \exp \left[ e^{-t^2 /2} p\left( -\frac{d}{dt} \right) e^{t^2 /2} \right] \delta (x-t) $$ Not clear whether that affords you any sort of simplification you may be ...
Michael Engelhardt's user avatar
1 vote

Closed formula for Hermite polynomials

It was already written above that the required sum is the kernel of the operator $$(H+a−1/2)^{−1}$$ For your problem, this will be the Green's function for a second-order linear ordinary differential ...
ANDREY PUPASOV MAKSIMOV's user avatar
1 vote

A special solution to the Hermite Differential Equation

$$y(x)=a_1 M(-\frac{\lambda}{2},\frac{1}{2},x^2)+a_2 H(\lambda,x)\Rightarrow$$ $$y'(x)=-2a_1\lambda x M(1-\tfrac{1}{2}{\lambda},\tfrac{3}{2},x^2)+2 {a_2} {\lambda} H({\lambda}-1,x)$$ The asymptotics ...
Carlo Beenakker's user avatar

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