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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?

Michael Hardy
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user42298
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  • 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 03:06

2 Answers2

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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.

whpowell96
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  • 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! – user42298 May 31 '18 at 03:10
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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 $\approx 4.5$.

MSM
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