# Differentiation of generating function of Hermite's polynomials

The generating function of Hermite's polynomials is given by $G(x,t)=e^{2xt-t^2}$ for $x, t \in \mathbf{R}$. It is known that $\displaystyle G(x,t)=\sum_{n=0}^\infty H_n(x) \frac{t^n}{n!}$ for $x, t \in \mathbf{R}$, where $H_n(x)=(-1)^n e^{x^2} \dfrac{d^n}{dx^n}e^{-x^2}$ for $x \in \mathbf{R}$ -is the $n$-th Hermite's polynomial. By properties of power series, for fixed $x\in \mathbf{R}$ we have that $$\frac{\partial G}{\partial t}= \sum_{n=1}^\infty \frac{\partial}{\partial t}\left[H_n(x) \frac{t^n}{n!}\right]=\sum_{n=1}^\infty H_n(x) \frac{t^{n-1}}{(n-1)!}\mbox{ for }t \in \mathbf{R}.$$

I have seen in some books that also

$$\frac{\partial G}{\partial x}= \sum_{n=0}^\infty \frac{\partial}{\partial x}\left[H_n(x) \frac{t^n}{n!}\right]= \sum_{n=0}^\infty H_n'(x) \frac{t^n}{n!} \mbox{ for }x \in \mathbf{R}.$$

I don't understand, what is the reason that we can differentiate this series with respect to $x$ (maybe series $\sum_{n=0}^\infty \frac{\partial}{\partial x} (...)$ is uniformly convergent or locally uniformly convergent, but I don't see why).

Thanks.

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We can show that we have the normal convergence of the series $f_1(x):=\sum_{n=0}^{+\infty}H_n(x)\frac{t^n}{n!}$ and $f_2(x):=\sum_{n=0}^{+\infty}H_n'(x)\frac{t^n}{n!}$ on each compact $\left[-a,a\right]$ for each $t\neq 0$ fixed (it's clear for $t=0$). Thanks to that, we can differentiate under the sum. For each $x\in\mathbb R$, we have, applying the Cauchy formula to the holomorphic function $e^{2zx-z^2}$ (for $z\in\mathbb C$, and we can extend the formula $G(x,t)=e^{2tx-t^2}$ it since these functions are equals for $t\in \mathbb R$, which is not discrete) on the circle $C(0,r), r>0$: $$\left|\frac{H_n(x)}{n!}\right|=\left|\frac 1{2\pi i}\int_{C(0,r)}\frac{e^{2xz-z^2}}{z^{n+1}}dz\right|\leq \frac 1{2\pi}\int_{C(0,r)}\frac{e^{|2xz-z^2|}}{r^{n+1}}dz\leq \frac{e^{2xr+r^2}}{r^n}.$$ Now, we fix $t\neq 0$ and we choose $r$ such that $0< r<|t|$. Then we get $$\sup_{-a\leq x\leq a}\left|H_n(x)\frac{t^n}{n!}\right|\leq \sup_{-a\leq x\leq a}\left(\frac{|t|}r\right)^ne^{2r|x|+r^2}=\left(\frac{|t|}r\right)^ne^{2ra+r^2},$$ and since $|t|<r$, the series whose general term is the RHS is convergent.
For $f_2$, note that for $n\geq 1$ we have $H_n'(x)=2nH_{n-1}(x)$, therefore $$\sup_{-a\leq x\leq a}\left|H_n'(x)\frac{t^n}{n!}\right|=2n\sup_{-a\leq x\leq a}\left|H_{n-1}(x)\frac{t^n}{n!}\right|\leq 2n\left(\frac{|t|}r\right)^ne^{2ra+r^2},$$ and the ratio test show that the series whose general terme is the RS is convergent.
And we can generalize the asked result, since by induction we get $H_{n}^{(d)}(x)=2^dn(n-1)\ldots (n-d+1)H_{n-d}(x)$ for $n\geq d$, hence a polynomial of $n$ times a Hermite polynomial, and the ratio test and the previous bound gives the normal convergence on compacts of the series $\sum_{n=0}^{+\infty}H_n^{(d)}(x)\frac{t^n}{n!}$ for all $t$.