change of summation, Hermite polynomial formula from generating functions

Question

Now given the exponential generating function of the Hermite polynomial $$e^{ux-(1/2)u^2} = 1+\sum_{n=1}^\infty \dfrac{u^n}{n!}h_n(x),$$how do I derive the closed form formula of the Hermite polynomial $h_n(x)$ given by $$ h_n(x) = n!\sum_{m=0}^{\left\lfloor\dfrac{n}{2}\right\rfloor}\dfrac{(-1)^m}{m!(n-2m)!}\dfrac{x^{n-2m}}{2^m}. $$

My attempt is to use the exponential expansion of $e^{ux-(1/2)u^2} = 1+\sum_{n=1}^\infty \dfrac{(ux-1/2u^2)^n}{n!}$ and further use the binomial formula $$(ux-\dfrac{1}{2}u^2)^n = \sum_{k=0}^n\binom{n}{k}\left(\dfrac{-1}{2}\right)^{n-k}u^{2n-k}x^k, $$which turns the problem into changing the double summation $$e^{ux-(1/2)u^2} = 1+\sum_{n=1}^\infty \dfrac{1}{n!}\sum_{k=0}^n\binom{n}{k}\left(\dfrac{-1}{2}\right)^{n-k}u^{2n-k}x^k $$ into the double summation$$ e^{ux-(1/2)u^2} = 1 + \sum_{\ell=1}^\infty \dfrac{u^\ell}{\ell!}\ell!\sum_{m=0}^{\left\lfloor\dfrac{\ell}{2}\right\rfloor}\dfrac{(-1)^m}{m!(\ell-2m)!}\dfrac{x^{\ell-2m}}{2^m}, $$ I'm struggling on the last step, and hope that a detailed explanation can be given on how a change of index can work. Thanks.

Answer 1

It is a simple convolution: $$e^{ux}=\sum_{a\geq 0}\frac{x^a}{a!}u^a,\qquad e^{-u^2/2}=\sum_{b\geq 0}\frac{(-1)^b}{2^b b!} u^{2b} $$ hence the coefficient of $u^n$ in the product between $e^{ux}$ and $e^{-u^2/2}$ (i.e. $n!h_n(x)$) is given by

$$ \sum_{a+2b=n} \frac{x^a(-1)^b}{a!b! 2^b} = \sum_{b=0}^{\lfloor n/2\rfloor}\frac{x^{n-2b}(-1)^b}{b!(n-2b)!2^b}.$$

Answer 2

Starting with $$ e^{ux-(1/2)u^2} = 1+\sum_{n=1}^\infty \dfrac{1}{n!}\sum_{k=0}^n\binom{n}{k}\left(\dfrac{-1}{2}\right)^{n-k}u^{2n-k}x^k, $$ we want pull $u$ out of the inner summation, allowing us to extract the inner summation as a coefficient of a power of $u$. The problem is, $u$ appears as $u^{2n-k}$, which involves both index variables, so it cannot be pulled out. We need to re-index.

To this end, let $$\ell = 2n-k.$$ As $k$ goes from $0$ to $n$, $\ell$ goes from $2n$ to $n$, or equivalently from $n$ to $2n$. Therefore, after replacing every instance of $k$ with $2n-\ell$, we get $$ 1+\sum_{n=1}^\infty \dfrac{1}{n!}\sum_{\ell=n}^{2n}\binom{n}{2n-\ell}\left(\dfrac{-1}{2}\right)^{\ell-n}u^{\ell}x^{2n-\ell} $$ Great. Next, we need to switch the order of summation. This is because we can cannot pull $u^\ell$ out of the $\ell$ summation. To make this easier, write this as $$ 1+\sum_{n=1}^\infty \dfrac{1}{n!}\sum_{\ell=\color{red}1}^{\color{red}{\infty}}\binom{n}{2n-\ell}\left(\dfrac{-1}{2}\right)^{\ell-n}u^{\ell}x^{2n-\ell} $$ We have changed the bounds, which results in adding a bunch of terms. This is legal, since the binomial coefficient $\binom{n}{2n-\ell}$ ensures these added terms are all zero. Now that the inner summation bounds do not depend on the outer summation index, we can switch the order of summation as follows: $$ \begin{align} \sum_{n=1}^\infty \dfrac{1}{n!}\sum_{\ell=1}^{\infty}\binom{n}{2n-\ell}\left(\dfrac{-1}{2}\right)^{\ell-n}u^{\ell}x^{2n-\ell} &= \sum_{n=1}^\infty \sum_{\ell=1}^{\infty}\dfrac{1}{n!}\binom{n}{2n-\ell}\left(\dfrac{-1}{2}\right)^{\ell-n}u^{\ell}x^{2n-\ell} \\&= \sum_{\ell=1}^{\infty}\sum_{n=1}^\infty \dfrac{1}{n!}\binom{n}{2n-\ell}\left(\dfrac{-1}{2}\right)^{\ell-n}u^{\ell}x^{2n-\ell} \\&= \sum_{\ell=1}^{\infty}u^\ell\sum_{n=1}^\infty \dfrac{1}{n!}\binom{n}{2n-\ell}\left(\dfrac{-1}{2}\right)^{\ell-n}x^{2n-\ell} \\&\stackrel{*}= \sum_{\ell=1}^{\infty}u^\ell \sum_{n=\color{blue}{\lceil \ell/2\rceil}}^{\color{blue}{\ell}} \dfrac{1}{n!}\binom{n}{2n-\ell}\left(\dfrac{-1}{2}\right)^{\ell-n}x^{2n-\ell} \\&= \sum_{\ell=1}^{\infty}u^\ell\sum_{n=\lceil \ell/2\rceil}^\ell \frac1{(\ell-n)!(2n-\ell)!}\left(\dfrac{-1}{2}\right)^{\ell-n}x^{2n-\ell} \end{align} $$ In the step labeled $\stackrel{*}=$, we tighted the bounds of the inner summation, since the binomial coefficient $\binom{n}{n-2\ell}$ is zero unless $0\le n-2\ell\le n$, which is equivalent to $\lceil \ell/2\rceil \le n\le \ell$.

We have at last rearranged the summation into $u^\ell$ times a summation, implying that $h_n(x)$ is that inner summation: $$ h_n(x)=\sum_{n=\lceil \ell/2\rceil}^\ell \frac1{(\ell-n)!(2n-\ell)!}\left(\dfrac{-1}{2}\right)^{\ell-n}x^{2n-\ell} $$ This does not quite look like what you want. In this expression, the index goes from $\lceil \ell/2\rceil$ to $\ell$, but we want it to go from $0$ to $\lfloor \ell/2\rfloor$. This can be fixed by reindexing, using the variable $$ m=\ell-n $$ I leave this last step to you, as practice with re-indexing.