# Distribution of roots of Eulerian polynomials

Paul Melotti

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Distributed under a Creative Commons Attribution 4.0 International License
Distribution of roots of Eulerian polynomials

Paul MELOTTI*

July 23, 2025

Abstract

We show that the empirical measures of roots of Eulerian polynomials converge to a certain log-Cauchy distribution. To do so, we show that the moments of the roots of a related family of polynomials not only converge, but are in fact ultimately constant. These asymptotic moments are expressed in terms of Nörlund's numbers.

# 1 Introduction

For $1 \le k \le n$, let $A(n, k)$ denote the Eulerian number, that is the number of permutations of size $n$ with exactly $k-1$ descents. The Eulerian polynomials are

$$A_n(x) = \sum_{k=1}^{n} A(n, k)x^k.$$

It is a classical result that this polynomial has $n$ distinct real roots, see for instance [Bon22, Theorem 1.34]. Let us denote these roots by $x_{n,1} < x_{n,2} < \cdots < x_{n,n} = 0$. We are interested in their empirical measure, for which we consider $-x_{n,k}$ instead, so as to work with positive values:

$$\mu_n = \frac{1}{n} \sum_{k=1}^{n} \delta_{-x_{n,k}}.$$

Our main result is the following:

**Theorem 1.** As $n \to \infty$, the sequence of measures $\mu_n$ converges weakly to a probability measure $\mu$ with support $[0, \infty)$. This measure is the distribution of $\exp(\pi Z)$ where $Z$ is a standard Cauchy random variable. That is, $\mu$ has density

$$\frac{1}{t(\log^2 t + \pi^2)} 1_{t>0}.$$

Finding the asymptotic distribution of a family of (random or deterministic) zeros is a common problem in random matrix theory, see [AGZ09], from which we borrow the method of moments and Stieltjes transforms. In a recent preprint [JKM25], Jalowy, Kabluchko and Marynych develop another method to find the limiting distribution of zeros, based on the asymptotic behaviour of

<footnote>*Université Paris-Saclay.</footnote>
<figure><img src="image_2.png" /><figcaption>Figure 1: Cumulative distribution functions of $\mu_{10}$, $\mu_{100}$ and the limiting measure $\mu$.</figcaption></figure>

coefficients for many families of polynomials, of which Eulerian polynomial is an example that they will treat in a later paper. Our method differs from theirs in that we rely on exact computations of moments, as in Theorem 2 below, which we believe gives a short and interesting way to prove Theorem 1. Asymptotics of coefficients have also been used in the case of some Stirling polynomials by Elbert [Elb01], and for some orthogonal polynomials by Lubinsky and Sidi [LS94, LS08].

The method of moments consists in computing the moments of the sequence of measures, prove that they converge towards the moments of the limiting measure, and use a unicity argument. However, we cannot apply it directly in our case, as all moments diverge<fnref n="1" />. This problem can be avoided by studying instead

$$u_{n,k} = \frac{1}{1 - x_{n,k}} \in (0, 1]$$

and by computing the moments of the empirical measure of the $u_{n,k}$ instead. These moments (appropriately rescaled) happen to have a remarkable, closed expression. To state it, we denote by $N_p$ the p-th Nörlund number [How93], which can be defined as

$$N_p = \int_0^1 (x-1)(x-2)\dots(x-p)dx.$$

**Theorem 2.** For any $1 \le p \le n$,

$$\frac{1}{n+1} \sum_{k=1}^{n} u_{n,k}^{p} = \frac{(-1)^{p} N_p}{p!}.$$

We emphasize that the right-hand side does not depend on $n$. The choice of normalizing the moments by $n+1$ instead of $n$ comes from this remarkable property. The first values of this moment sequence are $\frac{1}{2}$, $\frac{5}{12}$, $\frac{3}{8}$, $\frac{251}{720}$, $\frac{95}{288}$, ...

<footnote n="1">¹This can be seen by noting that $\sum_{k=1}^{n} x_{n,k} = -A_{n,n-1} = -2^n + n + 1$, which does not scale like $n$.</footnote>
In Section 2, we proceed with the proof of Theorem 2, which relies on computing the symmetric functions of the roots $u_{n,k}$ in terms of Stirling numbers of the second kind, and a relation between Stirling numbers and Nörlund numbers. In Section 3 we find the asymptotic distribution of the $u_{n,k}$ via its Stieljes transform, and deduce that of the $x_{n,k}$, proving Theorem 1.

## 2 Symmetric functions and moments of the $u_{n,k}$

For $0 \le p \le n$, we denote the elementary symmetric functions of the $(u_{n,k})_{k=1}^n$ by

$$e_{n,p} = \sum_{1 \le i_1 < \dots < i_p \le n} u_{n,i_1} \cdots u_{n,i_p}$$

and the convention $e_{n,0} = 1$.

Let $S(n, k)$ denote the Stirling numbers of the second kind, that is the number of set partitions of $\{1, \dots, n\}$ into $k$ blocks.

**Lemma 3.** For any $0 \le p \le n$,

$$e_{n,p} = \frac{(n-p)!}{n!} S(n+1, n-p+1). \qquad (1)$$

*Proof.* Since $A_n$ has roots at $x_{n,k}$ and is monic,

$$\begin{aligned} A_n(1-x) &= \prod_{k=1}^{n} (1-x-x_{n,k}) \\ &= \prod_{k=1}^{n} \left( \frac{1}{u_{n,k}} - x \right) \\ &= n! \prod_{k=1}^{n} (1-x u_{n,k}) \\ &= n! \sum_{p=0}^{n} (-1)^p e_{n,p} x^p \end{aligned}$$

where we used that the constant term is $A_n(1) = n!$. On the other hand,

$$\begin{aligned} A_n(1-x) &= \sum_{k=1}^{n} \sum_{p=0}^{k} (-1)^p \binom{k}{p} A(n,k) x^p \\ &= \sum_{p=0}^{n} (-1)^p x^p \sum_{k=p}^{n} \binom{k}{p} A(n,k) \\ &= \sum_{p=0}^{n} (-1)^p \frac{1}{(n-p+1)!} S(n+1, n-p+1) x^p \end{aligned}$$

where in the last line, we used a well-known relation between Eulerian numbers and Stirling numbers of the second kind, see for instance [Bon22, Theorem 1.18].

Identifying the term in $x^p$ in both expressions yields the lemma. □
The proof of Theorem 2 relies on Newton identities, Lemma 3, and a new relation between Nörlund numbers and Stirling numbers of the second kind, which we state now. We use the common convention $N_0 = 1$.

**Lemma 4.** For any $1 \le p \le n$,

$$ \sum_{i=0}^{n-p} \binom{p+i-1}{i} S(n, p+i) N_i = \frac{p}{n} S(n, p). $$

*Proof.* Let us start with the mixed bivariate generating function of the Stirling numbers of the second kind [Sta11, (1.94b)]:

$$ \sum_{p \ge 0} \sum_{n \ge p} S(n, p) \frac{x^n}{n!} y^p = \exp(y(e^x - 1)) \quad (2) $$

with the convention that $S(n, 0) = 0$ for $n \ge 1$, and $S(0, 0) = 1$. Differentiating with respect to $y$ and then multiplying by $y$, we get

$$ \sum_{p \ge 1} \sum_{n \ge p} \frac{p}{n} S(n, p) \frac{x^n}{(n-1)!} y^p = y(e^x - 1) \exp(y(e^x - 1)) \quad (3) $$

which is a mixed generating function of the right-hand side of the lemma. Let us compute the same function of the left-hand side, and change index $i$ into $j = p + i$:

$$ \begin{aligned} & \sum_{p \ge 1} \sum_{n \ge p} \sum_{i=0}^{n-p} \binom{p+i-1}{i} S(n, p+i) N_i \frac{x^n}{(n-1)!} y^p \\ & = \sum_{p \ge 1} \sum_{j \ge p} \binom{j-1}{j-p} N_{j-p} y^p \sum_{n \ge j} S(n, j) \frac{x^n}{(n-1)!}. \end{aligned} $$

From (2), we can also extract $\sum_{n \ge j} S(n, j) \frac{x^n}{n!} = \frac{(e^x - 1)^j}{j!}$, which by differenting $x$ and multiplying by $x$ gives the value of the inner sum. Thus the previous expression becomes

$$ \begin{aligned} & xe^x \sum_{p \ge 1} \sum_{j \ge p} \binom{j-1}{j-p} N_{j-p} y^p \frac{(e^x - 1)^{j-1}}{(j-1)!} \\ & = xye^x \left( \sum_{q \ge 0} N_q \frac{(e^x - 1)^q}{q!} \right) \left( \sum_{r \ge 0} \frac{(y(e^x - 1))^r}{r!} \right) \end{aligned} $$

where we have set $q = j - p$ and $r = p - 1$. The exponential generating function of Nörlund numbers is known to be [AD10]

$$ \sum_{q \ge 0} N_q \frac{t^q}{q!} = \frac{t}{(1+t) \log(1+t)}. \quad (4) $$

Injecting into the previous expression, we get

$$ xye^x \frac{e^x - 1}{xe^x} \exp(y(e^x - 1)) = y(e^x - 1) \exp(y(e^x - 1)) $$

which is the same as (3).

<figure><img src="image_3.png" /></figure>
We now have all the elements to prove Theorem 2.

*Proof of Theorem 2.* For any fixed $n \ge 1$, we proceed by induction over $p$. Throughout the proof let us drop the $n$ from the subscript notations $e_{n,p}$, $u_{n,p}$ etc., and also denote $m_p = \sum_{k=1}^n u_{n,k}^p$.

For $p = 1$, first note that the Eulerian numbers satisfy $A(n, n+1-k) = A(n, k)$, which implies that for any nonzero root $x_k$, $\frac{1}{x_k}$ is also a root of $A_n$. Using this involution,

$$m_1 = 1 + \sum_{k=1}^{n-1} \frac{1}{1 - x_k} = 1 + \sum_{k=1}^{n-1} \frac{1}{1 - \frac{1}{x_k}} = 1 + \sum_{k=1}^{n-1} \frac{-x_k}{1 - x_k}.$$

Summing the second and last expressions, we obtain $2m_1 = n+1$ and therefore $\frac{1}{n+1}m_1 = \frac{1}{2} = -N_1$.

Now for $2 \le p \le n$, suppose that the formula holds for all values $1, \dots, p-1$. By Newton's identities, then Lemma 3 and the induction hypothesis,

$$ \begin{aligned} m_p &= (-1)^{p-1} p e_p + \sum_{i=1}^{p-1} (-1)^{p-1+i} e_{p-i} m_i \\ &= (-1)^{p-1} p \frac{(n-p)!}{n!} S(n+1, n-p+1) \\ &\quad + (-1)^{p-1} (n+1) \sum_{i=1}^{p-1} \frac{(n-p+i)!}{n!} S(n+1, n-p+i+1) \frac{N_i}{i!} \end{aligned} \quad (5) $$

For the last sum, we use Lemma 4 with $n$ replaced by $n+1$ and $p$ by $n-p+1$, which gives

$$\sum_{i=0}^{p} \binom{n-p+i}{i} S(n+1, n-p+i+1) N_i = \frac{n-p+1}{n+1} S(n+1, n-p+1)$$

or, removing the two extremal indices,

$$\sum_{i=1}^{p-1} \binom{n-p+i}{i} S(n+1, n-p+i+1) N_i = -\frac{p}{n+1} S(n+1, n-p+1) - \binom{n}{p} N_p.$$

Injecting this into (5), we have

$$ \begin{aligned} m_p &= (-1)^{p-1} p \frac{(n-p)!}{n!} S(n+1, n-p+1) \\ &\quad + (-1)^{p-1} (n+1) \frac{(n-p)!}{n!} \left( -\frac{p}{n+1} S(n+1, n-p+1) - \binom{n}{p} N_p \right) \\ &= (-1)^p (n+1) \frac{N_p}{p!} \end{aligned} $$

which concludes the induction step. $\square$
### 3 Weak convergence and limiting distribution

We can now work towards the proof of Theorem 1.

Let $\nu_n$ be the probability measure

$$ \nu_n = \frac{1}{n} \sum_{k=1}^{n} \delta_{u_{n,k}}. $$

As a direct consequence of Theorem 2, for any $p \ge 1$ the p-th moments of $\nu_n$ converge as $n \to \infty$ towards $\frac{(-1)^p N_p}{p!}$. Since they are all supported in the compact set $[0, 1]$, they are tight and the only subsequential weak limit is the (unique) probability measure $\nu$ with moments

$$ \forall p \ge 1, \quad \int_0^1 u^p \nu(du) = \frac{(-1)^p N_p}{p!}. $$

As a result, $\nu_n$ converges weakly towards this measure $\nu$, see [Dur19, 3.3.5].

This allows for other characterizations of $\nu$, for instance via its Stieltjes transform, which can be directly computed from (4):

$$ \forall t \in \mathbb{C} \setminus [0, 1], \quad S_{\nu}(t) = \int_{0}^{1} \frac{1}{u-t} \nu(du) = \frac{1}{t(t-1)\log\left(1-\frac{1}{t}\right)}. \quad (6) $$

where we use the principal value of the logarithm, which gives an analytic function in the domain as the argument is never in $(-\infty, 0]$.

Moreover, since $x_{n,k} = 1 - \frac{1}{u_{n,k}}$, by the continuous mapping theorem we also get that the sequence of measures $\mu_n$ converges weakly towards $\mu$, where $\mu$ is the pushforward measure of $\nu$ by the map $u \mapsto 1 - \frac{1}{u}$. We can also find its Stieltjes transform by using (6) and changing variables, which leads to

$$ \forall t \in \mathbb{C} \setminus [0, \infty), \quad S_{\mu}(t) = \int_{-\infty}^{0} \frac{1}{x-t} \mu(dx) = -\frac{1}{1-t} + \frac{1}{t \log(-t)}. $$

From there, by the inverse Stieltjes transform procedure, see for instance [AGZ09, Theorem 2.4.3], we get that for any interval $I \subset [0, \infty)$,

$$ \mu(I) = \lim_{\epsilon \to 0} \frac{1}{\pi} \int_I \Re(S_{\mu}(\lambda + i\epsilon)) d\lambda. $$

For the chosen logarithmic branch, we have $\log(-\lambda - i\epsilon) = \log \lambda - i\pi + O(\epsilon)$, from which we get $S_{\mu}(\lambda + i\epsilon) = -\frac{1}{1+\lambda} + \frac{\log \lambda + i\pi}{\lambda(\log^2 \lambda + \pi^2)} + O(\epsilon)$, with a $O$ constant uniform in $\lambda$ for $\lambda$ bounded away from 0. Therefore, for $I = [a, b]$ with $0 < a < b$, we may switch the limit and the integral, which gives

$$ \mu([a, b]) = \int_a^b \frac{1}{\lambda (\log^2 \lambda + \pi^2)} d\lambda $$

and we can extract the density of $\mu$. The fact that is also the distribution of $\exp(\pi Z)$, where $Z$ is a standard Cauchy random variable, is a direct computation. This concludes the proof of Theorem 1.
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