# Expanding Dirac delta function with Hermite polynomial

My question is related to a formula in this paper

In that paper, they try to expand Dirac delta function $\delta(x)$, which has the property $$\int \delta(x)f(x) \, dx = f(0),$$ using Hermite polynomial. So they write

$$\delta(x) = \sum_{n=0}^{\infty}A_n H_{2n}(x)e^{-x^2}$$

and get the coefficient $A_n$ by

\begin{align} \int H_{2m}(x) \delta(x) \, dx &= \int H_{2m}(x) \sum_{n=0}^{\infty}A_n H_{2n}(x)e^{-x^2} \\ \Rightarrow H_{2m}(0) &= A_m \sqrt {\pi}4^m (2m)! \\ \Rightarrow A_m &= \frac{(-1)^m}{m! 4^m \sqrt{\pi}} ~~~~~~~~(H_{2n}(0)=\frac{(2n)!(-1)^n}{n!}) \end{align}

Usual $\delta(x)$ function has property that it equals to zero for $x\neq 0$, but $\delta(x) \rightarrow \infty$ for $x=0$

Now following above expansion, if we plug $x=0$ to the formula, we get

\begin{align} \delta(0) & = \sum_{n=0}^{\infty}A_n H_{2n}(0) \\ & = \sum_{n=0}^{\infty} \frac{(2n)!}{n!n!4^n\sqrt{\pi}} \end{align}

But this series converges, so the usual property of $\delta(x)$ is not recovered. So my question is, is this expansion for $\delta(x)$ valid?

• The series you've written diverges - use the explicit bound in the intro of the wiki article on Stirling's approximation to show that there exist a constant $C$ such that the $n$th term in your series is $\ge C/\sqrt{n},$ at which point divergence to $+\infty$ is immediate. – stochasticboy321 May 31 '18 at 3:06

The value for $$N$$-th approximation to the delta function at zero turns out to be
$$\frac{1}{\sqrt{\pi}} \frac{(2N+1)!!}{2^N N!}$$
This can be verified by induction over $$N$$. The values are quite small, climbing up in a leasurely fashion. For $$N=50$$ the value is $$\approx 4.5$$.