roots of higher derivatives of exponential Consider the Gaussian function $f(z)=e^{-z^2}$ which has no zeros on the complex domain. Let $D$ denote derivative w.r.t. the variable $z$.

Question. Is it true that $D^nf(z)=0$ has only real roots that are simple?
  If so, any slick proof?

 A: There are several complete characterizations of real entire functions whose all derivatives have all roots real: a) this is a closure of polynomials with the same property,
and b) this class is represented by the formula
$$f(z)=cz^me^{-az^2+bz}\prod_{k}\left(1-\frac{z}{z_k}\right)e^{z/z_k},$$
where $a\geq0$, $b,c$ are real, $m\geq 0$ is an integer and $z_k$ real, with
$$\sum\frac{1}{|z_k|^2}<\infty.$$
This is a parametric description: each such function is represented by this formula,
and each function represented by this formula has the stated property.
This class of function has a standard name: Polya-Wiman class.
These results are due to Wiman, Polya, Hellerstein and Williamson.
For recent generalizations and survey, see
 arXiv:math/0510502.
A: The (physicists') Hermite polynomials are
$$ H_n(x) = (-1)^n e^{x^2} D^n e^{-x^2}$$
And their roots are real.  For that you don't need to know they are Hermite polynomials: just Rolle's theorem.  See this.
A: I've found a slightly different argument.
If $f_m(z)=\left(1-\frac{z^2}m\right)^m$ then $\lim_{m\rightarrow\infty}f_m(z)=e^{-z^2}$ uniformly on every compact subset of $\mathbb{C}$. Hence, the same holds for $\lim_{m\rightarrow\infty}D^nf_m(z)=D^ne^{-z^2}$. On the other hand, $D^nf_m(z)$ has only real zeros. Therefore $D^ne^{-z^2}$ can not have non-real zeros, by Hurwitz's Theorem.
A: There are several ways to skin this cat. Most are presented in the excellent survey "Zeros of entire Fourier transforms" by
Dimitar K. Dimitrov and Peter K. Rusev with an account of the historical development. Jensen's contributions are particularly interesting to me with the emphasis on Appell polynomial sequences--the Hermite polynomials are the only set of orthogonal Appell polynomials.
