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 ...
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.
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 ...
5
votes
On Mehler's formula for Hermite polynomials
I think that the formula is equal to, for all $-1<r<1$,
$$
\sum_{\ell \geq 0} \dfrac{r^\ell H_{\ell}(x)H_{\ell}(y)}{2^{\ell}\ell!} = (1-r^2)^{-\frac{1}{2}} \exp \left(\dfrac{2xyr -r^2(x^2+y^2)}{...
5
votes
Generating function for products of complex Hermite polynomials
The OP's question has been addressed By Richard Stanley.
So, we make attempt at Stanley's question on the complex counterpart to
$$\sum_{n\geq0}\frac{H_n(x)H_n(y)}{n!}\left(\frac{u}2\right)^2
=\...
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
$$ \sum_{m,n\geq 0}...
4
votes
Accepted
Closed formula for Hermite polynomials
Up to some normalization, the harmonic oscillator $H$ is self-adjoint such that
$$
\langle Hu, u\rangle=\sum_{k\ge 0}(\frac12+k) \vert u_k\vert^2,
$$
and thus defining a self-adjoint $A$ by the ...
4
votes
How to integrate the multinomial over a ball in $\mathbb{R}^{n}$?
$\newcommand\R{\mathbb R}\newcommand{\Ga}{\Gamma}$The integral in question is
\begin{equation*}
\begin{aligned}
\int_0^1 dr\,\int_{rS_{n-1}}dx\,f(x)
&=\int_0^1 dr\,r^{2m+n-1}\int_{S_{n-1}}...
4
votes
How to integrate the multinomial over a ball in $\mathbb{R}^{n}$?
Let me separate the radial integration from the angular integration,
$$\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,$$
where $\bar{f}(...
4
votes
Accepted
In what sense does the Hermite expansion of a bounded smooth function converge?
According to the conclusion at the bottom of p. 603 of (say) Uspensky's paper, the Hermite expansion of $f$ converges pointwise to $f$ (under conditions much less restrictive that yours).
If $f$ is (...
3
votes
In what sense does the Hermite expansion of a bounded smooth function converge?
Here's a simple and far-from-optimal condition guaranteeing uniform convergence.
Suppose for that there exists $C>0$ such that for any positive integer $n$ $\newcommand{\bR}{{\mathbb{R}}}$
$$
\int_\...
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&...
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).
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
$$s_{nm} = \mathbb E[H_n(X^\top a)H_m(X^\top b)]=\mathbb E[H_n(X_1 t+X_2\...
2
votes
Accepted
Integrating a B-Spline basis function with respect to the standard normal PDF
Since a B-spline is a piecewise polynomial function, the question is whether there exists an exact equality formula for the integral $\int_{-a}^{b}u^pe^{-u^2/2}du$. This integral equals an elementary ...
1
vote
Accepted
Asymptotic behavior of the integral of Hermite functions/polynomials on half-lines
Your conjecture is not true. Moreover,
$$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}).$$
Indeed, recalling that $H_m^...
1
vote
Asymptotic behavior of the integral of Hermite functions/polynomials on half-lines
Let me first consider the case $x_0=0$, when the integral has a closed form expression:
$$\int^{\infty}_{0}|H_m(x)|^2 e^{-x^2}\,dx=\sqrt{\pi } \,2^{m-1} m!\;\;(m\in\mathbb{N}).$$
See, for example, ...
1
vote
A question regarding Hermite polynomials and exponential operators $\exp[e^{x^2/2}p(\frac{d}{dx})e^{-x^2/2}]f(x)$
You can of course write
$$
k(x,t) = \exp \left[ e^{-t^2 /2} p\left( -\frac{d}{dt} \right) e^{t^2 /2} \right] \delta (x-t)
$$
Not clear whether that affords you any sort of simplification you may be ...
1
vote
Closed formula for Hermite polynomials
It was already written above that the required sum is the kernel of the operator
$$(H+a−1/2)^{−1}$$
For your problem, this will be the Green's function for a second-order linear ordinary differential ...
1
vote
A special solution to the Hermite Differential Equation
$$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 ...
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