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?

  • $\begingroup$ 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. $\endgroup$ – stochasticboy321 May 31 '18 at 3:06

Plugging the series into Wolfram Alpha gives us that it diverges, see here. So the expansions seems to be valid. Just a minor thing: it is usually a trap to actually evaluate distributions like the Dirac function, as they are not actually functions and only really make sense under an integral sign. Viewing it as a functional with the property you listed at the top is much more clear and avoids function having infinite values and other weird stuff.

  • $\begingroup$ I've just noticed that I mistyped (2n)! to (2n!) in Wolfram when I did convergence test, so it said that the series converges. Thanks! $\endgroup$ – user42298 May 31 '18 at 3:10

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 $\approx 4.5$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.