# Explicit forms for the roots of Eulerian polynomials

Let $E_n(z)$ be the Eulerian polynomial $$E_n(z) = \sum_{\tau \in \mathfrak{S}_n} z^{\operatorname{des}(\tau)}$$ where $\mathfrak{S}_n$ denotes the set of all permutations of $\{1,\ldots,n\}$ and $\operatorname{des}(\tau)$ is the number of descents $\tau(i) > \tau(i+1)$ in the permutation $\tau$. These are studied in great detail, see its OEIS wiki page for other definitions and several properties. In particular, it is known the Eulerian polynomial has only negative and simple roots, $$E_n(z) = \prod_{i=1}^n (z+q^{(n)}_i)$$ for different positive numbers $q^{(n)}_i$.

My question now is

What is known about the $q^{(n)}_i$'s? Do they have an explicit description?

(It is known that the roots of $E_n$ separate the roots of $E_{n+1}$. This is, $$q_1^{(n+1)} < q_1^{(n)} < \cdots < q_{n}^{(n+1)}< q_n^{(n)} < q_{n+1}^{(n+1)}$$ when they come in sorted order. That's not the type of property I am looking for, but only properties towards their explicit values.)

Here are the first two examples and the type of property I would like to have answered from a description I search for:

\begin{align*} E_2(z) &= z^2 + 4z + 1 = (z+2-\sqrt{3})(z+2+\sqrt{3}) \\ E_3(z) &= z^3 + 11z^2 + 11z + 1 = (z+5+2\sqrt{6})(z+5-2\sqrt{6})(z+1) \end{align*}

(The roots become more complicated than $a\pm b\sqrt{c}$ for bigger $n$'s.)

It is known that the mean value of the discrete distribution given by $E_n$ is $n/2$ and the variance is $(n+2)/12$. This can be used to show that \begin{align*} \sum_i \frac{1}{1+q^{(n)}_i} &= \frac{n}{2} \\ \sum_i \frac{q^{(n)}_i}{\big(1+q^{(n)}_i\big)^2} &= \frac{n+2}{12} \end{align*}

Doing this computation in the first example yields $$\frac{1}{1+2+\sqrt{3}} + \frac{1}{1+2-\sqrt{3}} = 1$$ and $$\frac{2+\sqrt{3}}{(1+2+\sqrt{3})^2} + \frac{2-\sqrt{3}}{(1+2-\sqrt{3})^2} = \frac{1}{3}$$

Is there any known immediate property of the roots that can be used to get these identities?

• As for the roots having an explicit description, if the Galois group of $E_n(z)$ were abelian then there would be a trigonometric formula for the roots. However, this is not the case. It is known that $E_n(z)$ is irreducible. By the symmetry of the coefficients, the largest possible Galois group has order $2^m m!$, where $m=\lfloor n/2\rfloor$ (and is the hyperoctahedral group $B_m$) . It has been checked for $n\leq 9$ that $B_m$ is indeed the Galois group of $E_n(z)$, so one could conjecture that this is the case for all $n$. – Richard Stanley Dec 4 '17 at 17:02
• I was mistaken when I said that it is known that $E_n(z)$ is irreducible. First, $E_{2m}(z)$ is divisible by $1+z$. It is only conjectured that $E_{2m+1}(z)$ and $E_{2m}(z)/(1+z)$ are irreducible. For some instances of irreducibility see ac.els-cdn.com/0022314X84900507/…. – Richard Stanley Dec 6 '17 at 3:43

## 1 Answer

Here is some partial information. If $E_n(z)=\sum A_n(i)x^i$, then $A_n(i)$ grows roughly like $(i+1)^n$ for fixed $i$. Now $A_n(i)$ is (up to sign) the $i$th elementary symmetric function of the roots. Hence $|q_1^{(n)}|$ is about $2^n$. Then $|q_2^{(n)}|$ is about $(3/2)^n$, $|q_3^{(n)}|$ is about $(4/3)^n$, etc. It is not hard to show in fact that for fixed $i$, $$\lim_{n\to\infty} |q_i^{(n)}|^{1/n} = \frac{i+1}{i}.$$ Note also that there is an explicit formula for $A_n(i)$ as a sum with $i+2$ terms (https://en.wikipedia.org/wiki/Eulerian_number). Thus any rational symmetric function of the roots can be written in terms of these sums.