Where can I find an estimate to the $L^{\infty}(\mathbb{R})$-norms of $\dfrac{d^n}{dx^n}\exp(-x^2), ~n\in\mathbb{N}$? Where can I find an estimate to the $L^{\infty}(\mathbb{R})$-norms of $\dfrac{d^n}{dx^n}\exp(-x^2), ~n\in\mathbb{N}$ ?
Edit:
I am seeking an inequality of the form
$$\sup_{x\in\mathbb{R}}\left| \dfrac{d^n}{dx^n}\exp(-x^2)\right|\leq F(n),\quad n\in \mathbb{N}$$
and want to know how rapidly $F(n)$ grows as $n\to\infty$. In particular, what is the radius of convergence of the power series 
$$\sum_{n=0}^\infty F(n)\frac{t^n}{n!}$$
in $t$?
ps: I remember an equation that says
$$\dfrac{d^n}{dx^n}\exp(-x^2)=(-1)^n H_n(x)\exp(-x^2)$$
and the $H_n(x)$ is something called Hermite Polynomial or so, but I cannot tell whether this memory of mine is accurate.
 A: A few tests seem to indicate that the norms go to infinity as $n$ increases. This is not an answer per se, but perhaps it can be of help. Using Faà di Bruno's formula, we may write:
$$\frac{d^n}{dx^n}\exp\left(-x^2\right)=\exp\left(-x^2\right)\cdot n!\cdot \\\sum\limits_{\substack{m_1,\dots,m_n \geq 0\\\sum i\cdot m_i=n}}\frac{1}{{m_1}!{1!}^{m_1}{m_2}!{2!}^{m_2}\dots {m_n}!{n!}^{m_n}}\cdot\prod\limits_{j=1}^n\left(\frac{d^j}{dx^j}\left(-x^2\right)\right)^{m_j}$$
Now, this sum looks ugly, but observe the product inside it. If $m_j > 0$ for any $j\geq 3$, the product is $0$, so we might as well disregard them and rewrite the sum as:
$$\sum\limits_{\substack{m_1,m_2 \geq 0\\m_1+2m_2=n}}\frac{1}{{m_1}!{m_2}!2^{m_2}}\cdot(-2x)^{m_1}\cdot(-2)^{m_2}$$
So, all in all:
$$\frac{d^n}{dx^n}\exp\left(-x^2\right)=\exp\left(-x^2\right)\cdot n!\cdot \left(\sum\limits_{\substack{m_1,m_2 \geq 0\\m_1+2m_2=n}}\frac{(-1)^{m_1+m_2}\cdot 2^{m_1}}{{m_1}!{m_2}!}x^{m_1}\right)$$
Moreover, these functions are all bounded in $\mathbb{R}$, and in fact go to zero as $|x| \to \infty$ (the decaying exponential dominates any polynomial). Hence, the norm of any such function is its maximum, which is necessarily attained away from infinity. This means that at least one of the roots of the polynomial given by the sum in parantheses is the maximum of the corresponding function (think derivative equals zero).
I have to sleep now, but it seems the maximum always occurs at the root closest to $x=0$ (including $x=0$ itself, for odd $n$). It shouldn'tbe too hard to show this, and perhaps this can be used to make the estimate?
