# 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?

• Plan of action: $D^nf=P_n(z)e^{-z^2}$ for some polynomial $P_n$. Calculate the first few, and look up the coefficients at oeis.org. – Gerry Myerson Jan 19 '17 at 22:06
• I calculated the first 200 and they all have only real roots. $P_0=1$ and $P_{n+1}=-2zP_n+P'_n$ (derivative wrt $z$ of course). But how do you look up a sequence of polynomials at oeis? – Kevin Buzzard Jan 19 '17 at 22:17
• The idea I guess is to recognise the polys as e.g. Euler-Jacobi polynomials of the 12th kind or whatever, and then appeal to the classical result that they only have real roots. But I don't know how to look for a list of polys on oeis as this is a 2-d array. – Kevin Buzzard Jan 19 '17 at 22:19
• You can look up two-dimensional arrays in oeis listing them by rows, e.g., 1,1,1,1,2,1,1,3,3,1 will get you oeis.org/A007318 "Pascal's triangle read by rows". oeis.org/A060821 gives coefficients of Hermite polynomials. – Gerry Myerson Jan 19 '17 at 22:50
• Oh many thanks @GerryMyerson -- I did not know that trick – Kevin Buzzard Jan 19 '17 at 22:54

## 3 Answers

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.

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.

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.