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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?

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    $\begingroup$ 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. $\endgroup$ Jan 19, 2017 at 22:06
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    $\begingroup$ 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? $\endgroup$ Jan 19, 2017 at 22:17
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    $\begingroup$ 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. $\endgroup$ Jan 19, 2017 at 22:19
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    $\begingroup$ 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. $\endgroup$ Jan 19, 2017 at 22:50
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    $\begingroup$ Oh many thanks @GerryMyerson -- I did not know that trick $\endgroup$ Jan 19, 2017 at 22:54

4 Answers 4

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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.

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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.

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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.

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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.

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