<figure><img src="image_1.png" /></figure>

HAL
open science

# Complete homogeneous symmetric polynomials and generalized Vieta formulas

Laura de Carli, Andrew Echezabal, Maurizio Laporta

► To cite this version:

Laura de Carli, Andrew Echezabal, Maurizio Laporta. Complete homogeneous symmetric polynomials and generalized Vieta formulas. 2025. hal-04938930v2

HAL Id: hal-04938930

https://hal.science/hal-04938930v2

Preprint submitted on 18 Jul 2025

HAL is a multi-disciplinary open access
archive for the deposit and dissemination of sci-
entific research documents, whether they are pub-
lished or not. The documents may come from
teaching and research institutions in France or
abroad, or from public or private research centers.

L'archive ouverte pluridisciplinaire HAL, est
destinée au dépôt et à la diffusion de documents
scientifiques de niveau recherche, publiés ou non,
émanant des établissements d'enseignement et de
recherche français ou étrangers, des laboratoires
publics ou privés.

Public Domain
Complete homogeneous symmetric polynomials
and generalized Vieta's formulas

Laura De Carli¹, Andrew Echezabal²†, Maurizio Laporta³*†

¹,²Department of Mathematics, Florida International University, Miami, 33199, Florida, USA.

³\*Dipartimento di Matematica e Applicazioni, Università degli Studi di Napoli "Federico II", Via Cinthia, Napoli, 80126, ITALY.

\*Corresponding author(s). E-mail(s): mlaporta@unina.it;

Contributing authors: decarlil@fiu.edu; ancheza@fiu.edu;

†These authors contributed equally to this work.

Abstract

We generalize the classical Vieta formulas that express the coefficients of a poly-
nomial in terms of all the roots. In particular, we focus on the case when just
some of the roots are known. Our formulas are established by exploiting some
properties of the complete homogeneous symmetric polynomials. Further, we pro-
vide new identities for these polynomials when the variables are in a geometric
progression and new identities involving the roots of unity.

**Keywords:** polynomials, complete homogeneous symmetric polynomials, Vieta's formulas, roots of unity

MSC Classification: 12D10, 05E05, 11A07, 11C08

# 1 Introduction

The homogeneous symmetric polynomials play a crucial role in various areas of math-
ematics, including algebraic geometry, combinatorics, and representation theory. A
complete homogeneous symmetric polynomial $\mathcal{H}_m(x_1, \ldots, x_n)$ is defined as the sum
of all monomials of total degree $m \in \mathbb{N}_0 := \mathbb{N} \cup \{0\}$ in the variables $x_1, \ldots, x_n \in \mathbb{C}$.
Specifically, we let $\mathcal{H}_0(x_1, \dots, x_n) := 1$ and for any $m \in \mathbb{N}$ one has

$$ \begin{aligned} \mathcal{H}_m(x_1, \dots, x_n) &:= \sum_{1 \le j_1 \le j_2 \le \dots \le j_m \le n} x_{j_1} x_{j_2} \cdots x_{j_m} \\ &= \sum_{\substack{m_1+m_2+\dots+m_n=m \\ m_i \in \mathbb{N}_0}} x_1^{m_1} x_2^{m_2} \cdots x_n^{m_n}. \end{aligned} $$

In particular, $\mathcal{H}_m(x) = x^m$ for any $m \in \mathbb{N}_0$.

The foundational work of MacDonald [8] provides a comprehensive treatment of the complete homogeneous symmetric polynomials within the more general context of the symmetric functions, whose theory has profound implications in the representation theory of the symmetric group ([4], [11], [12]). Other interesting applications of the complete homogeneous symmetric polynomials are presented in [1], [5], [9], [13].

Let us recall that the elementary symmetric polynomials of degree $m \in \mathbb{N}$ are defined as

$$ \mathcal{E}_m(x_1, \dots, x_n) := \sum_{1 \le j_1 < j_2 < \dots < j_m \le n} x_{j_1} \cdots x_{j_m} $$

for $1 \le m \le n$, while $\mathcal{E}_0(x_1, \dots, x_n) := 1$ for all $x_1, \dots, x_n$. The complete homogeneous symmetric polynomials and the elementary symmetric polynomials are related by the formula

$$ \sum_{j=0}^{m} (-1)^j \mathcal{E}_j(x_1, \dots, x_n) \mathcal{H}_{m-j}(x_1, \dots, x_n) = 0. $$

Both $\mathcal{H}_m(x_1, \dots, x_n)$ and $\mathcal{E}_m(x_1, \dots, x_n)$ are homogeneous functions and invariant under permutation of the variables $x_j$. We use such properties throughout the paper without mentioning them explicitly.

It is well-known that the coefficients of any monic polynomial can be expressed in terms of its roots by means of the elementary symmetric polynomials, namely (see e.g. [2, Ex. 4.6.6])

$$ \prod_{j=1}^{n} (z - x_j) = \sum_{k=0}^{n} (-1)^k \mathcal{E}_k(x_1, x_2, \dots, x_n) z^{n-k}. \quad (1.1) $$

More generally, if $x_1, \dots, x_m$ are the roots (not necessarily distinct) of

$$ \mathcal{P}(z) := a_0 z^m + a_1 z^{m-1} + \dots + a_{m-1} z + a_m, $$

with $a_0, \dots, a_m \in \mathbb{C}$ and $a_0 \neq 0$, then (1.1) yields

$$ (-1)^k \frac{a_k}{a_0} = \mathcal{E}_k(x_1, \dots, x_m), \quad \forall k \in \{0, 1, \dots, m\}. \quad (1.2) $$

This formula is usually named after François Viète (more commonly referred to by the Latinized form of his name Franciscus Vieta). He proved this relation in 1646 for
polynomials with real coefficients and with only positive roots [14]. The general form was established by Girard [6].

In our first theorem we generalize (1.2) by providing with the necessary condition that the coefficients of $\mathcal{P}(z)$ need to satisfy when not all the roots are given.

**Theorem 1.1** If $x_0, x_1, \dots, x_n$ are roots of the polynomial

$$ \mathcal{P}(z) := a_0 z^m + a_1 z^{m-1} + \dots + a_{m-1} z + a_m, \quad \text{with } a_j \in \mathbb{C}, a_0 \neq 0, $$

then

$$ a_{m-s} = - \sum_{j=0}^{m-s-1} a_j \mathcal{H}_{m-s-j}(x_0, \dots, x_s) \quad (1.3) $$

for any $s \in \{0, \dots, n\}$. Further,

$$ \mathcal{P}(z) = \left( \sum_{k=0}^{m-n-1} c_k z^{m-n-k-1} \right) \prod_{j=0}^{n} (z - x_j), $$

where

$$ c_k = \sum_{j=0}^{k} a_j \mathcal{H}_{k-j}(x_0, \dots, x_n) \quad (1.4) $$

for any $k \in \{0, \dots, m-n-1\}$.

In the case $n = m - 1$, by taking $s = m - 1$ in (1.3) one has

$$ a_1 = -a_0 \mathcal{H}_1(x_0, \dots, x_{m-1}) = -a_0 \mathcal{E}_1(x_0, \dots, x_{m-1}), $$

which is (1.2) for $k = 1$.

The formulas obtained in Theorem 1.1 sparked our interest in the properties of complete homogeneous symmetric polynomials. In this paper we recall some known properties of these polynomials and we establish new identities that, to the best of our knowledge, are not in the literature.

One of the most intriguing identities that we derive, a formula that connects the nth primitive roots of unity with the nth non-primitive root via $\mathcal{H}_k$ and $\mathcal{E}_k$, is a consequence of Theorem 1.1.

Recall that by definition the nth roots of unity are the zeros of $z^n - 1$, i.e., they are $\zeta_n^j, j \in \mathbb{N} \cup [1, n]$, with $\zeta_n := e^{2\pi i/n}$. In particular, the root $\zeta_n^j$ is primitive when $(n, j) = 1$. For brevity, here and in what follows, we let $(n, j) = \gcd(n, j)$.

Before stating our theorem, let us introduce the following notation.

* $\vec{\zeta}_n := (\zeta_n, \dots, \zeta_n^j, \dots, \zeta_n^{n-1}, \zeta_n^n) = (\zeta_n, \dots, \zeta_n^j, \dots, \zeta_n^{n-1}, 1)$

* $\vec{\zeta}_n^* = (\zeta_{1,n}^*, \dots, \zeta_{j,n}^*, \dots, \zeta_{n,n}^*)$, where

$$ \zeta_{j,n}^* := \begin{cases} \zeta_n^j & \text{if } (j,n) = 1, \\ 0 & \text{otherwise.} \end{cases} $$
• $\vec{\zeta}_n^{**} := (\zeta_{1,n}^{**}, \dots, \zeta_{j,n}^{**}, \dots, \zeta_{n,n}^{**})$, where

$$ \zeta_{j,n}^{**} := \begin{cases} \zeta_n^j & \text{if } (j,n) > 1, \\ 0 & \text{otherwise.} \end{cases} $$

Note that the nonzero components of $\vec{\zeta}_n^*$ are the $n$th primitive roots and $\vec{\zeta}_n = \vec{\zeta}_n^* + \vec{\zeta}_n^{**}$.

For any $f: \mathbb{C}^n \to \mathbb{C}$ and every $\vec{v} = (v_1, \dots, v_n) \in \mathbb{C}^n$ we let $f(\vec{v}) := f(v_1, \dots, v_n)$.

Since $\mathcal{H}_k(x_1, x_2, \dots, x_h, 0, \dots, 0) = \mathcal{H}_k(x_1, x_2, \dots, x_h)$, it turns out that $\mathcal{H}_k(\vec{\zeta}_n^*)$ is evaluated only at the $n$th primitive roots of unity. Analogous considerations hold for $\mathcal{E}_k(\vec{\zeta}_n^*)$, $\mathcal{H}_k(\vec{\zeta}_n^{**})$, and $\mathcal{E}_k(\vec{\zeta}_n^{**})$.

Our result on the roots of unity is the following

**Corollary 1.2** Let $n \in \mathbb{N}$. We have that

$$ \mathcal{E}_k(\vec{\zeta}_n^*) = (-1)^k \mathcal{H}_k(\vec{\zeta}_n^{**}), \quad \forall k \in \{0, \dots, \varphi(n)\}, $$

$$ \mathcal{E}_k(\vec{\zeta}_n^{**}) = (-1)^k \mathcal{H}_k(\vec{\zeta}_n^*), \quad \forall k \in \{0, \dots, n - \varphi(n)\}, $$

where $\varphi(n) := \#\{s \in \mathbb{N} \cap [1, n] : (s, n) = 1\}$ is the so-called Euler totient function.

In this paper we provide identities for the complete homogeneous symmetric polynomials $\mathcal{H}_n(x_0, \dots, x_m)$ when the variables $x_j$ are in a geometric progression.

**Theorem 1.3** For every $x \in \mathbb{C}$ and all $n, m \in \mathbb{N}_0$ we have that

$$ \mathcal{H}_n(1, x, \dots, x^m) = \mathcal{H}_m(1, x, \dots, x^n). \qquad (1.5) $$

It is known that

$$ \mathcal{H}_n(\underbrace{x, \dots, x}_{m+1 \text{ times}}) = x^n \binom{n+m}{m} $$

for any $m, n \in \mathbb{N}$ and every $x \in \mathbb{C} \setminus \{0\}$ (see e.g. [7]). In particular, this yields

$$ \mathcal{H}_n(\underbrace{1, \dots, 1}_{m+1 \text{ times}}) = \binom{m+n}{m} = \binom{m+n}{n}, \qquad (1.6) $$

that is (1.5) when $x = 1$.

The next theorem gives an explicit formula for $\mathcal{H}_n(1, x, \dots, x^m)$ when $x^s \neq 1$ for every $s \in \mathbb{N} \cap [1, n]$.

**Theorem 1.4** Let $n, m \in \mathbb{N}$. If $x^s \neq 1$ for every $s \in \mathbb{N} \cap [1, n]$, then

$$ \mathcal{H}_n(1, x, \dots, x^m) = \prod_{s=1}^{n} \frac{1 - x^{m+s}}{1 - x^s}. \qquad (1.7) $$
By Theorem 1.3, it follows immediately that if $x^s \neq 1$ for every $s \in \mathbb{N} \cap [1, m]$, then

$$ \mathcal{H}_n(1, x, \dots, x^m) = \prod_{s=1}^{m} \frac{1 - x^{n+s}}{1 - x^s}. $$

In particular, taking $x = \zeta_n$, $m = n - 1$ and after recalling that $\zeta_n^n = 1$, we recover

$$ \mathcal{H}_n(\vec{\zeta}_n) = \prod_{s=1}^{n-1} \frac{1 - \zeta_n^{n+s}}{1 - \zeta_n^s} = 1. $$

Note that from Vieta's formula (1.2) applied to the polynomial $z^n - 1$ it follows immediately that

$$ \mathcal{E}_n(\vec{\zeta}_n) = (-1)^{n+1}. $$

This is generalized in the next corollary (see also (1.8) below). Before stating it, let us introduce the following notation: if $n \in \mathbb{N}$ and $m \in \mathbb{N}_0$, then we let

$$ m = nm_n + m'_n, \quad \text{i.e., } m_n := \lfloor m/n \rfloor \text{ and } 0 \le m'_n < n, $$

where $\lfloor t \rfloor$ denotes the integer part of the real number $t$. When there is no risk of confusion, we often write $m'$ instead of $m'_n$.

**Corollary 1.5** Let $n \in \mathbb{N}$, $n \ge 2$. For all $k \in \mathbb{N}$ and $m \in \mathbb{N}_0$ we have that

$$ \mathcal{H}_m(1, \zeta_n, \dots, \zeta_n^k) = \mathcal{H}_{m'}(1, \zeta_n, \dots, \zeta_n^{k'}) = \begin{cases} 1 & \text{if } k' = 0 \text{ or } m' = 0, \\ \prod_{s=1}^{k'} \frac{1 - \zeta_n^{m'+s}}{1 - \zeta_n^s} & \text{if } 1 \le k' + m' < n, \\ 0 & \text{if } k' + m' \ge n, \end{cases} $$

where $m' := m'_n$ and $k' := k'_n$.

The particular instance $k = n - 1$ gives the identity

$$ \mathcal{H}_m(\vec{\zeta}_n) = \begin{cases} 1 & \text{if } n \text{ divides } m, \\ 0 & \text{otherwise.} \end{cases} \qquad (1.8) $$

In view of Theorem 1.3, the formula in Corollary 1.5 can also be written as

$$ \mathcal{H}_m(1, \zeta_n, \dots, \zeta_n^k) = \mathcal{H}_{k'}(1, \zeta_n, \dots, \zeta_n^{k'}) = \begin{cases} 1 & \text{if } m' = 0 \text{ or } k' = 0, \\ \prod_{s=1}^{m'} \frac{1 - \zeta_n^{k'+s}}{1 - \zeta_n^s} & \text{if } 1 \le m' + k' < n', \\ 0 & \text{if } m' + k' \ge n. \end{cases} $$

A further consequence of Theorem 1.4 is the following
**Corollary 1.6** Let $n \in \mathbb{N}$ and $m, r \in \mathbb{N}_0$, with $n > m$. If $x^s \neq 1$ for every $s \in \mathbb{N} \cap [1, n]$, then

$$ \frac{\mathcal{H}_n(1, x, \dots, x^r)}{\mathcal{H}_m(1, x, \dots, x^r)} = \frac{\mathcal{H}_{n-m}(1, x, \dots, x^{r+m})}{\mathcal{H}_{n-m}(1, x, \dots, x^m)}. \qquad (1.9) $$

The proofs of all theorems and corollaries stated above are given in the next section. In Sect. 3 we provide new and alternate proofs of formulas involving complete homogeneous symmetric polynomials including some remarkable identities established by Cornelius [3].

## 2 Proofs of the main results

We need to apply some properties of the complete homogeneous symmetric polynomials that we summarize in the following lemma.

**Lemma 2.1**

* Given $n \in \mathbb{N}$, we have that

$$ \mathcal{H}_n(x_0, x_1, \dots, x_m) = x_0 \mathcal{H}_{n-1}(x_0, x_1, \dots, x_m) + \mathcal{H}_n(x_1, \dots, x_m) \quad (2.1) $$

$$ = \sum_{s=0}^{n} x_0^s \mathcal{H}_{n-s}(x_1, \dots, x_m). \qquad (2.2) $$

* Given $n \in \mathbb{N}_0$, we have that

$$ \mathcal{H}_n(x_0, \dots, x_{m_1}, y_0, \dots, y_{m_2}) = \sum_{s=0}^{n} \mathcal{H}_s(x_0, \dots, x_{m_1}) \mathcal{H}_{n-s}(y_0, \dots, y_{m_2}). \quad (2.3) $$

*Proof* The equation (2.1) follows as

$$ \begin{aligned} \mathcal{H}_n(x_0, x_1, \dots, x_m) &= \sum_{\substack{n_0+n_1+\dots+n_m=n \\ n_i \in \mathbb{N}_0}} x_0^{n_0} x_1^{n_1} \dots x_m^{n_m} \\ &= \sum_{\substack{n_1+\dots+n_m=n \\ n_i \in \mathbb{N}_0}} x_1^{n_1} \dots x_m^{n_m} + \sum_{\substack{n_0+n_1+\dots+n_m=n \\ n_i \in \mathbb{N}_0, n_0 \ge 1}} x_0^{n_0} x_1^{n_1} \dots x_m^{n_m} \\ &= \mathcal{H}_n(x_1, \dots, x_m) + x_0 \sum_{\substack{n_0+n_1+\dots+n_m=n \\ n_i \in \mathbb{N}_0, n_0 \ge 1}} x_0^{n_0-1} x_1^{n_1} \dots x_m^{n_m} \\ &= \mathcal{H}_n(x_1, \dots, x_m) + x_0 \mathcal{H}_{n-1}(x_0, x_1, \dots, x_m). \end{aligned} $$

Similarly, after recalling that $\mathcal{H}_0(x_1, \dots, x_m) = \mathcal{H}_0(x_0, x_1, \dots, x_m) = 1$, the formula (2.2) is obtained by writing

$$ \mathcal{H}_n(x_0, x_1, \dots, x_m) = \sum_{s=0}^{n} x_0^s \sum_{\substack{n_1+\dots+n_m=n-s \\ n_i \in \mathbb{N}_0}} x_1^{n_1} \dots x_m^{n_m} = \sum_{s=0}^{n} x_0^s \mathcal{H}_{n-s}(x_1, x_1, \dots, x_m). $$
In order to prove (2.3), let us take $m_1 \in \mathbb{N}_0$ fixed and proceed by induction on $m_2 \in \mathbb{N}_0$. If $m_2 = 0$, then (2.3) reduces to (2.2). After assuming that (2.3) holds for any $m_2 \ge 0$, we prove it for $m_2 + 1$. To this end, we apply (2.2) to $\mathcal{H}_{n-s}$ and write

$$ 
\begin{aligned}
& \sum_{s=0}^{n} \mathcal{H}_s(x_0, \dots, x_{m_1}) \mathcal{H}_{n-s}(y_0, \dots, y_{m_2+1}) \\
&= \sum_{s=0}^{n} \mathcal{H}_s(x_0, \dots, x_{m_1}) \sum_{j=0}^{n-s} y_{m_2+1}^j \mathcal{H}_{n-s-j}(y_0, \dots, y_{m_2}) \\
&= \sum_{j=0}^{n} y_{m_2+1}^j \sum_{s=0}^{n-j} \mathcal{H}_s(x_0, \dots, x_{m_1}) \mathcal{H}_{n-s-j}(y_0, \dots, y_{m_2}).
\end{aligned}
$$

Now, using (2.2) again, from the induction hypothesis on (2.3) it follows that

$$ 
\begin{aligned}
\sum_{s=0}^{n} \mathcal{H}_s(x_0, \dots, x_{m_1}) \mathcal{H}_{n-s}(y_0, \dots, y_{m_2+1}) &= \sum_{j=0}^{n} y_{m_2+1}^j \mathcal{H}_{n-j}(x_0, \dots, x_{m_1}, y_0, \dots, y_{m_2}) \\
&= \mathcal{H}_n(x_0, \dots, x_{m_1}, y_0, \dots, y_{m_2+1}),
\end{aligned}
$$

as required.

□

**Remark 2.1** In view of (1.6), the formula (2.3) yields

$$ \mathcal{H}_n(\underbrace{x, \dots, x}_{m+1 \text{ times}}, y_0, y_1, \dots, y_r) = \sum_{s=0}^{n} \binom{s+m}{m} x^s \mathcal{H}_{n-s}(y_0, y_1, \dots, y_r). $$

*Proof of Theorem 1.1* First, note that in the case $n = 0$, the equation (1.3) reduces to

$$ a_m = - \sum_{j=0}^{m-1} a_j \mathcal{H}_{m-j}(x_0). $$

Since $\mathcal{H}_{m-j}(x_0) = x_0^{m-j}$, the equation above follows immediately from $\mathcal{P}(x_0) = 0$, which true by assumption.

Further, we can write

$$ \mathcal{P}(z) = (z - x_0)(c_0 z^{m-1} + c_1 z^{m-2} + \dots + c_{m-2} z + c_{m-1}), $$

where it is readily seen that

$$ 
\left\{ 
\begin{array}{l}
a_0 = c_0 \\
a_1 = c_1 - x_0 c_0 \\
\vdots \\
a_{m-2} = c_{m-2} - x_0 c_{m-3} \\
a_{m-1} = c_{m-1} - x_0 c_{m-2}
\end{array}
\right.
$$

and $a_m = -c_{m-1}x_0$. Thus, for any $k \in \{0, \dots, m-1\}$ we easily obtain

$$ c_k = \sum_{j=0}^{k} a_j x_0^{k-j} = \sum_{j=0}^{k} a_j \mathcal{H}_{k-j}(x_0), $$
which is (1.4) when $n = 0$.

Now, we proceed by induction to show that (1.3) and (1.4) hold for any integer $n \in [0, m]$.
Let us assume that

$$ \mathcal{P}(z) = (z - x_0) \cdots (z - x_{n-1}) (b_0 z^{m-n} + b_1 z^{m-n-1} + \dots + b_{m-n+1} z + b_{m-n}) $$

with

$$ b_k = \sum_{j=0}^{k} a_j \mathcal{H}_{k-j}(x_0, \dots, x_{n-1}), \quad \forall k \in \{0, \dots, m-n\}. \qquad (2.4) $$

Further, we assume that (1.3) holds for every $0 \le s \le n-1 < m$.

We have to show that $\mathcal{P}(z) = (z - x_0) \cdots (z - x_n) \mathcal{F}(z)$, where

$$ \mathcal{F}(z) := c_0 z^{m-n-1} + c_1 z^{m-n-2} + \dots + c_{m-n} z + c_{m-n-1}, $$

with $c_k$ given by (1.4). Further, we need to prove that (1.3) holds also for $s = n$.

To this end, by using $\mathcal{P}(x_n) = 0$ we write

$$ \begin{aligned} & b_0 z^{m-n} + b_1 z^{m-n-1} + \dots + b_{m-n+1} z + b_{m-n} \\ & = (z - x_n)(c_0 z^{m-n-1} + c_1 z^{m-n-2} + \dots + c_{m-n-2} z + c_{m-n-1}). \end{aligned} $$

The proof of the case $n = 0$ shows that

$$ c_k = \sum_{j=0}^{k} b_j \mathcal{H}_{k-j}(x_n) $$

for any $k \in \{0, \dots, m-n-1\}$. Moreover,

$$ b_{m-n} = - \sum_{k=0}^{m-n-1} b_k \mathcal{H}_{m-n-k}(x_n). $$

On the other hand, (2.4) for $k = m-n$ gives

$$ b_{m-n} = \sum_{j=0}^{m-n} a_j \mathcal{H}_{m-n-j}(x_0, \dots, x_{n-1}). \qquad (2.5) $$

Further, (2.4) allows us to express each $b_{m-n}$ in terms of $a_j$, namely

$$ b_{m-n} = - \sum_{k=0}^{m-n-1} \mathcal{H}_{m-n-k}(x_n) \sum_{i=0}^{k} a_i \mathcal{H}_{k-i}(x_0, \dots, x_{n-1}). $$

Thus, after exchanging the order of the sums, making a change of variables, and using the identities (2.3) and (2.5), we obtain

$$ \begin{aligned} b_{m-n} &= - \sum_{i=0}^{m-n-1} a_i \sum_{k=i}^{m-n-1} \mathcal{H}_{k-i}(x_0, \dots, x_{n-1}) \mathcal{H}_{m-n-k}(x_n) \\ &= - \sum_{i=0}^{m-n-1} a_i \sum_{j=0}^{m-n-1-i} \mathcal{H}_j(x_0, \dots, x_{n-1}) \mathcal{H}_{m-n-i-j}(x_n) \\ &= \sum_{i=0}^{m-n-1} a_i \mathcal{H}_{m-n-i}(x_0, \dots, x_{n-1}) - \sum_{i=0}^{m-n-1} a_i \sum_{j=0}^{m-n-i} \mathcal{H}_j(x_0, \dots, x_{n-1}) \mathcal{H}_{m-n-i-j}(x_n) \\ &= \sum_{i=0}^{m-n-1} a_i \mathcal{H}_{m-n-i}(x_0, \dots, x_{n-1}) - \sum_{i=0}^{m-n-1} a_i \mathcal{H}_{m-n-i}(x_0, \dots, x_n) \end{aligned} $$
$$ = b_{m-n} - a_{m-n} - \sum_{i=0}^{m-n-1} a_i \mathcal{H}_{m-n-i}(x_0, \dots, x_n), $$

from which it follows that

$$ a_{m-n} = - \sum_{i=0}^{m-n-1} a_i \mathcal{H}_{m-n-i}(x_0, \dots, x_n), $$

which is (1.3) for $s = n$, as required. For any $k = 0, \dots, m-n-1$ the same argument yields

$$ \begin{aligned} c_k &= \sum_{j=0}^{k} \mathcal{H}_{k-j}(x_n) \sum_{i=0}^{j} a_i \mathcal{H}_{j-i}(x_0, \dots, x_{n-1}) = \sum_{i=0}^{k} a_i \sum_{j=i}^{k} \mathcal{H}_{k-j}(x_n) \mathcal{H}_{j-i}(x_0, \dots, x_{n-1}) \\ &= \sum_{i=0}^{k} a_i \sum_{j=0}^{k-i} \mathcal{H}_{k-i-j}(x_n) \mathcal{H}_{j}(x_0, \dots, x_{n-1}) = \sum_{i=0}^{k} a_i \mathcal{H}_{k-i}(x_0, \dots, x_{n-1}, x_n), \end{aligned} $$

that is (1.4). □

*Proof of Corollary 1.2.* Let us consider the polynomial

$$ z^n - 1 = \prod_{1 \le j \le n} (z - \zeta_n^j) = \Phi_n(z) \Psi_n(z), $$

where

$$ \Phi_n(z) := \prod_{\substack{1 \le j \le n \\ (j,n)=1}} (z - \zeta_n^j), $$

is the so-called nth cyclotomic polynomial of degree $\varphi(n)$. Clearly, the zeros of $\Psi_n$ are the nth non-primitive roots of unity. In view of Vieta's formula (1.1), we can write

$$ \begin{aligned} \Phi_n(z) &= \sum_{k=0}^{\varphi(n)} (-1)^k \mathcal{E}_k(\zeta_n^*) z^{\varphi(n)-k}, \\ \Psi_n(z) &= \sum_{k=0}^{n-\varphi(n)} (-1)^k \mathcal{E}_k(\zeta_n^{**}) z^{n-\varphi(n)-k}. \end{aligned} $$

Now, by applying (1.4) of Theorem 1.1 to $\mathcal{P}(z) = z^n - 1$ we see at once that the coefficient of $z^{\varphi(n)-k}$ in $\Phi_n(z)$ is given by $(-1)^k \mathcal{E}_k(\zeta_n^*) = \mathcal{H}_k(\zeta_n^{**})$.

The same argument shows that the coefficient of $z^{n-\varphi(n)-k}$ in $\Psi_n(z)$ is

$$ (-1)^k \mathcal{E}_k(\zeta_n^{**}) = \mathcal{H}_k(\zeta_n^*) $$

for $k = 0, \dots, n - \varphi(n)$. □

*Proof of Theorem 1.3.* We proceed by induction on $n+m$. If $n+m=0$, i.e. $n=m=0$, then (1.5) is trivially true. By assuming that (1.5) is true for $m+n$ we mean that

$$ \mathcal{H}_k(1, x, \dots, x^r) = \mathcal{H}_r(1, x, \dots, x^k) \qquad (2.6) $$

provided that $k+r=m+n$. Thus, we have to prove that (1.5) is also true for $m+n+1$, i.e., (2.6) holds for any $k,r$ such that $k+r=m+n+1$. By using (2.1) of Lemma 2.1 and the induction assumption for $k+r-1=m+n$, we can write

$$ \mathcal{H}_k(1, x, \dots, x^r) = \mathcal{H}_{k-1}(1, x, \dots, x^r) + \mathcal{H}_k(x, \dots, x^r) $$
$$
\begin{align*}
&= \mathcal{H}_r(1, x, \dots, x^{k-1}) + x^k \mathcal{H}_k(1, \dots, x^{r-1}) \\
&= \mathcal{H}_r(1, x, \dots, x^{k-1}) + x^k \mathcal{H}_{r-1}(1, x, \dots, x^k) \\
&= \mathcal{H}_r(1, x, \dots, x^{k-1}, x^k),
\end{align*}
$$

as required.

☐

*Proof of Theorem 1.4.* First, note that

$$
\mathcal{H}_1(1, x, \dots, x^m) = 1 + x + \dots + x^m = \frac{1 - x^{m+1}}{1-x},
$$

i.e., (1.7) holds for $n = 1$. As for Theorem 1.3, we proceed by induction on $n+m$. Let us assume that (1.7) holds for any $k, r \in \mathbb{N}$ such that $k+r = m+n$, i.e.,

$$
\mathcal{H}_k(1, x, \dots, x^r) = \prod_{s=1}^{k} \frac{1 - x^{r+s}}{1 - x^s}.
$$

We have to prove that it holds for any $k, r \in \mathbb{N}$ such that $k+r = m+n+1$.

By using (2.1) and the induction assumption for $k+r-1=m+n$, we can write

$$
\begin{align*}
\mathcal{H}_k(1, x, \dots, x^r) &= \mathcal{H}_{k-1}(1, x, \dots, x^r) + \mathcal{H}_k(x, x^2, \dots, x^r) \\
&= \mathcal{H}_{k-1}(1, x, \dots, x^r) + x^k \mathcal{H}_k(1, x, \dots, x^{r-1}) \\
&= \prod_{s=1}^{k-1} \frac{1-x^{r+s}}{1-x^s} + x^k \prod_{s=1}^{k} \frac{1-x^{r-1+s}}{1-x^s} \\
&= \left( \prod_{s=1}^{k-1} (1-x^{r+s}) + \frac{x^k}{1-x^k} \prod_{s=0}^{k-1} (1-x^{r+s}) \right) \prod_{s=1}^{k-1} (1-x^s)^{-1} \\
&= \left( (1-x^k) \prod_{s=1}^{k-1} (1-x^{r+s}) + x^k \prod_{s=0}^{k-1} (1-x^{r+s}) \right) \prod_{s=1}^{k} (1-x^s)^{-1} \\
&= \left( 1-x^k + x^k(1-x^r) \right) \prod_{s=1}^{k-1} (1-x^{r+s}) \prod_{s=1}^{k} (1-x^s)^{-1} \\
&= \prod_{s=1}^{k} \frac{1-x^{r+s}}{1-x^s},
\end{align*}
$$

as required.

☐

*Proof of Corollary 1.5.* Since $\zeta_n^n = 1$ and $\zeta_n^s \neq 1$ for $1 \le s \le n-1$, one has that $\zeta_n^r = \zeta_n^m$ if and only if $r'_n = m'_n$.

Now, assume that $k = k' := k'_n$. Since the case $k = 0$ gives $\mathcal{H}_m(1) = 1$, let us take
$1 \le k = k' < n$. By Theorem 1.4

$$
\mathcal{H}_m(1, \zeta_n, \dots, \zeta_n^k) = \prod_{s=1}^{k} \frac{1 - \zeta_n^{m+s}}{1 - \zeta_n^s} = \prod_{s=1}^{k} \frac{1 - \zeta_n^{m'+s}}{1 - \zeta_n^s} = \mathcal{H}_{m'}(1, \zeta_n, \dots, \zeta_n^k).
$$

In particular, we have $\mathcal{H}_m(1, \zeta_n, \dots, \zeta_n^k) = 1$ when $m' = 0$. If $m' + k < n$, all terms of the product above are nonzero, while $m' + k = m' + k' \ge n$ yields $\mathcal{H}_m(1, \zeta_n, \dots, \zeta_n^k) = 0$ because the term corresponding to $s = n - m'$ is zero. Therefore, the claimed formula for $\mathcal{H}_m(1, \zeta_n, \dots, \zeta_n^k)$ remains proved when $k = k'$.
Let us turn our attention to the case $k > k'$, so that $k = rn + k'$ for some $r \in \mathbb{N}$. We have to show that $\mathcal{H}_m(1, \zeta_n, \dots, \zeta_n^k) = \mathcal{H}_{m'}(1, \zeta_n, \dots, \zeta_n^{k'})$. To this end, we consider only the case $k = n + k'$ because the proof for $r \ge 2$ is similar. Further, in view of the fact that $\mathcal{H}_m(1, \zeta_n, \dots, \zeta_n^k) = \mathcal{H}_{m'}(1, \zeta_n, \dots, \zeta_n^k)$, we assume also that $m = m'$. Since

$$ \{1, \zeta_n, \dots, \zeta_n^k\} = \{1, \zeta_n, \dots, \zeta_n^{n-1}, \zeta_n^n, \dots, \zeta_n^{n+k'}\} = \{1, \zeta_n, \dots, \zeta_n^{n-1}, \underbrace{\zeta_n^n, \zeta_n, \dots, \zeta_n^{k'}}_{=1} \}, $$

the identity (2.3) of Lemma (2.1) yields

$$ \mathcal{H}_m(1, \zeta_n, \dots, \zeta_n^k) = \sum_{s=0}^{m} \mathcal{H}_s(1, \zeta_n, \dots, \zeta_n^{k'}) \mathcal{H}_{m-s}(1, \zeta_n, \dots, \zeta_n^{n-1}). $$

Observe that $m - s + n - 1 \ge n$ when $s < m$. Therefore, from the first part of the proof it follows that $\mathcal{H}_{m-s}(1, \zeta_n, \dots, \zeta_n^{n-1}) = 0$ for $0 \le s < m$. Thus, the previous sum reduces to

$$ \mathcal{H}_m(1, \zeta_n, \dots, \zeta_n^k) = \mathcal{H}_m(1, \zeta_n, \dots, \zeta_n^{k'}) \mathcal{H}_0(1, \zeta_n, \dots, \zeta_n^{n-1}) = \mathcal{H}_m(1, \zeta_n, \dots, \zeta_n^{k'}), $$

as required.

□

*Proof of Corollary 1.6.* In view of (1.7) we can see that

$$ \begin{aligned} \frac{\mathcal{H}_n(1, x, \dots, x^r)}{\mathcal{H}_m(1, x, \dots, x^r)} &= \frac{\prod_{s=1}^{n} \frac{1-x^{r+s}}{1-x^s}}{\prod_{s=1}^{m} \frac{1-x^{r+s}}{1-x^s}} = \prod_{s=m+1}^{n} \frac{1-x^{r+s}}{1-x^s} = \prod_{s=1}^{n-m} \frac{1-x^{r+m+s}}{1-x^{m+s}} \\ &= \prod_{s=1}^{n-m} \frac{1-x^{r+m+s}}{1-x^s} \frac{1-x^s}{1-x^{m+s}} = \frac{\mathcal{H}_{n-m}(1, x, \dots, x^{r+m})}{\mathcal{H}_{n-m}(1, x, \dots, x^m)}, \end{aligned} $$

as claimed.

□

# 3 Appendix

In this section we have collected some further properties of the complete homogeneous symmetric polynomials. The first proposition is quoted from [7].

**Proposition 3.1** Given $x_0, x_1, \dots, x_m, x_{m+1} \in \mathbb{C}$, with $x_0 \ne x_{m+1}$, for every $n \in \mathbb{N}$ we have that

$$ \mathcal{H}_{n-1}(x_0, x_1, \dots, x_{m+1}) = \frac{\mathcal{H}_n(x_1, \dots, x_{m+1}) - \mathcal{H}_n(x_0, x_1, \dots, x_m)}{x_{m+1} - x_0}, \quad (3.1) $$

and

$$ \mathcal{H}_{n-1}(x_0, x_1, \dots, x_m, x_0) = \left[ \frac{\partial}{\partial t} \mathcal{H}_n(t, x_1, \dots, x_m) \right]_{t=x_0}. $$

*Proof* By applying (2.1) of Lemma 2.1 twice we can write

$$ \begin{aligned} \mathcal{H}_n(x_0, x_1, \dots, x_{m+1}) &= x_0 \mathcal{H}_{n-1}(x_0, x_1, \dots, x_{m+1}) + \mathcal{H}_n(x_1, \dots, x_m, x_{m+1}) \\ &= x_{m+1} \mathcal{H}_{n-1}(x_0, x_1, \dots, x_{m+1}) + \mathcal{H}_n(x_0, x_1, \dots, x_m). \end{aligned} $$

Thus, (3.1) follows immediately after subtracting the last two equations. The other formula follows from (3.1) by taking $x_{m+1} = tx_0$, with $t \ne 1$, so that $\mathcal{H}_n(x_1, \dots, x_m, tx_0) = \mathcal{H}_n(tx_0, x_1, \dots, x_m)$, and letting $t \to 1$. □
Now we prove an interesting identity for complete homogeneous symmetric polynomials that was first established by Cornelius [3]. Here we propose a different proof by using (3.1). To this end, let us first introduce the following notation:

$$ \mathbb{C}_*^{m+1} := \{(x_0, x_1, \dots, x_m) \in \mathbb{C}^{m+1} : i \neq j \Rightarrow x_i \neq x_j\} $$

for any $m \in \mathbb{N}$.

**Proposition 3.2** (Cornelius [3]) For all $(x_0, x_1, \dots, x_m) \in \mathbb{C}_*^{m+1}$ and any fixed $n \in \mathbb{N}_0$ we have that

$$ \mathcal{H}_n(x_0, x_1, \dots, x_m) = \sum_{j=0}^{m} \frac{x_j^{m+n}}{\Delta_j(x_0, x_1, \dots, x_m)}, \qquad (3.2) $$

where $\Delta_j(x_0, x_1, \dots, x_m) := \prod_{\substack{i=0 \\ i \neq j}}^{m} (x_j - x_i)$.

In this case, we say that $\mathcal{H}_n$ satisfies the Cornelius identity in $\mathbb{C}_*^{m+1}$.

*Proof* We use induction on $m$ to prove that (3.2) holds for all $m \in \mathbb{N}$, i.e., for every fixed $n \in \mathbb{N}_0$, the polynomial $\mathcal{H}_n$ satisfies the Cornelius identity in $\mathbb{C}_*^{m+1}$ for all $m \in \mathbb{N}$.

When $m=1$ and $n \in \mathbb{N}_0$, we can see at once that for every $x_0, x_1 \in \mathbb{C}$,

$$ x_1^{n+1} - x_0^{n+1} = (x_1 - x_0) \sum_{j=0}^{n} x_1^j x_0^{n-j}. $$

When $x_0 \neq x_1$, the right-hand side of (3.2) is

$$ \frac{x_0^{n+1}}{x_0 - x_1} + \frac{x_1^{n+1}}{x_1 - x_0} = \frac{x_1^{n+1} - x_0^{n+1}}{x_1 - x_0} = \sum_{j=0}^{n} x_1^j x_0^{n-j} = \mathcal{H}_n(x_0, x_1) $$

as required.

Assuming that $\mathcal{H}_n$, with $n \in \mathbb{N}_0$, satisfies the Cornelius identity in $\mathbb{C}_*^{m+1}$, we have to prove that it satisfies the identity also in $\mathbb{C}_*^{m+2}$. By applying (3.1) and the induction hypothesis, we can write

$$ \begin{aligned} \mathcal{H}_n(x_0, x_1, \dots, x_{m+1}) &= \frac{\mathcal{H}_{n+1}(x_0, x_1, \dots, x_m) - \mathcal{H}_{n+1}(x_1, \dots, x_m, x_{m+1})}{x_0 - x_{m+1}} \\ &= \frac{1}{x_0 - x_{m+1}} \left( \sum_{j=0}^{m} \frac{x_j^{m+n+1}}{\Delta_j(x_0, \dots, x_m)} - \sum_{j=1}^{m+1} \frac{x_j^{m+n+1}}{\Delta_j(x_1, \dots, x_{m+1})} \right) \\ &= \frac{1}{x_0 - x_{m+1}} \sum_{j=1}^{m} x_j^{m+n+1} \left( \frac{1}{\Delta_j(x_0, \dots, x_m)} - \frac{1}{\Delta_j(x_1, \dots, x_{m+1})} \right) \\ &\quad + \frac{1}{x_0 - x_{m+1}} \left( \frac{x_0^{m+n+1}}{\Delta_0(x_0, x_1, \dots, x_m)} - \frac{x_{m+1}^{m+n+1}}{\Delta_{m+1}(x_1, \dots, x_m, x_{m+1})} \right) \\ &:= \Sigma_1 + \Sigma_2. \end{aligned} $$

Note that for any $j \in \{1, \dots, m\}$, one has

$$ \Delta_j(x_0, \dots, x_m) = (x_j - x_0) \Delta_j(x_1, \dots, x_m) $$
$$ \Delta_j(x_1, \dots, x_{m+1}) = (x_j - x_{m+1})\Delta_j(x_1, \dots, x_m). \quad (3.3) $$

Note also that

$$ \begin{aligned} (x_0 - x_{m+1})\Delta_0(x_0, x_1, \dots, x_m) &= \Delta_0(x_0, x_1, \dots, x_{m+1}), \\ (x_{m+1} - x_0)\Delta_{m+1}(x_1, \dots, x_m, x_{m+1}) &= \Delta_{m+1}(x_0, x_1, \dots, x_{m+1}). \end{aligned} \quad (3.4) $$

Thus, we can write

$$ \begin{aligned} \Sigma_1 &= \frac{1}{x_0 - x_{m+1}} \sum_{j=1}^{m} \frac{x_j^{m+n+1}}{\Delta_j(x_1, \dots, x_m)} \left( \frac{1}{x_j - x_0} - \frac{1}{x_j - x_{m+1}} \right) = \sum_{j=1}^{m} \frac{x_j^{m+n+1}}{\Delta_j(x_0, \dots, x_{m+1})}, \\ \Sigma_2 &= \frac{x_0^{m+n+1}}{\Delta_0(x_0, \dots, x_{m+1})} + \frac{x_{m+1}^{m+n+1}}{\Delta_{m+1}(x_0, \dots, x_{m+1})}. \end{aligned} $$

Hence, we conclude that

$$ \mathcal{H}_n(x_0, x_1, \dots, x_m, x_{m+1}) = \Sigma_1 + \Sigma_2 = \sum_{j=0}^{m+1} \frac{x_j^{m+n+1}}{\Delta_j(x_0, \dots, x_{m+1})}, $$

as required.

□

Note that the particular case of (3.2) for $n = 0$ yields

$$ \mathcal{H}_0(x_0, x_1, \dots, x_m) = \sum_{j=0}^{m} \frac{x_j^m}{\Delta_j(x_0, x_1, \dots, x_m)} = 1. \quad (3.5) $$

This is a special case of the following property that might be interesting in its own right.

**Proposition 3.3** For every $(x_0, \dots, x_m) \in \mathbb{C}_*^{m+1}$, with $m \in \mathbb{N}$, and for every $0 \le k \le m$, one has

$$ S(m, k) := \sum_{j=0}^{m} \frac{x_j^k}{\Delta_j(x_0, x_1, \dots, x_m)} = \delta_{m,k}, \quad (3.6) $$

$$ \text{where } \delta_{m,k} = \begin{cases} 1 & \text{if } k = m, \\ 0 & \text{if } 0 \le k < m. \end{cases} $$

*Proof* We argue by induction on $m$. When $m = 1$ we can see at once that

$$ S(1, 0) = \frac{1}{\Delta_0(x_0, x_1)} + \frac{1}{\Delta_1(x_0, x_1)} = \frac{1}{x_0 - x_1} + \frac{1}{x_1 - x_0} = 0. $$

When $k = m = 1$, we readily obtain $S(1, 1) = \frac{x_0}{x_0 - x_1} + \frac{x_1}{x_1 - x_0} = 1$. Assuming that $S(m, k) = \delta_{m,k}$ when $m \ge 1$ and $0 \le k \le m$, we prove that $S(m+1, k) = \delta_{m+1,k}$ whenever $0 \le k \le m+1$. Since $S(m+1, m+1) = 1$ because of (3.5), we have to show that $S(m+1, k) = 0$ for $0 \le k \le m$. To this end, by using (3.4) we can write

$$ S(m+1, k) = \frac{x_0^k}{\Delta_0(x_0, \dots, x_{m+1})} + \frac{x_{m+1}^k}{\Delta_{m+1}(x_0, \dots, x_{m+1})} + \sum_{j=1}^{m} \frac{x_j^k}{\Delta_j(x_0, \dots, x_{m+1})} $$
$$ = \frac{1}{x_0 - x_{m+1}} \left( \frac{x_0^k}{\Delta_0(x_0, \dots, x_m)} - \frac{x_{m+1}^k}{\Delta_{m+1}(x_1, \dots, x_{m+1})} \right) + \Sigma_1, \quad (3.7) $$

where $\Sigma_1 := \sum_{j=1}^{m} \frac{x_j^k}{\Delta_j(x_0, \dots, x_{m+1})}$.

Since $k \le m$, by assumption we have that

$$ \frac{x_0^k}{\Delta_0(x_0, \dots, x_m)} + \sum_{j=1}^{m} \frac{x_j^k}{\Delta_j(x_0, \dots, x_m)} = \delta_{m,k}. $$

In view of (3.3),

$$ \frac{x_0^k}{\Delta_0(x_0, \dots, x_m)} = \delta_{m,k} - \sum_{j=1}^{m} \frac{x_j^k}{\Delta_j(x_0, \dots, x_m)} = \delta_{m,k} - \sum_{j=1}^{m} \frac{x_j^k(x_j - x_{m+1})}{\Delta_j(x_0, \dots, x_{m+1})}. $$

Similarly,

$$ \frac{x_{m+1}^k}{\Delta_{m+1}(x_1, \dots, x_{m+1})} = \delta_{m,k} - \sum_{j=1}^{m} \frac{x_j^k(x_j - x_0)}{\Delta_j(x_0, x_1, \dots, x_{m+1})}. $$

Therefore,

$$ \begin{aligned} & \frac{x_0^k}{\Delta_0(x_0, \dots, x_m)} - \frac{x_{m+1}^k}{\Delta_{m+1}(x_1, \dots, x_{m+1})} \\ &= \sum_{j=1}^{m} \left( \frac{x_j^k(x_j - x_0)}{\Delta_j(x_0, x_1, \dots, x_{m+1})} - \frac{x_j^k(x_j - x_{m+1})}{\Delta_j(x_0, x_1, \dots, x_{m+1})} \right) \\ &= \sum_{j=1}^{m} \frac{x_j^k(x_{m+1} - x_0)}{\Delta_j(x_0, x_1, \dots, x_{m+1})} = (x_{m+1} - x_0) \Sigma_1, \end{aligned} $$

and it is plain that the sum in (3.7) equals zero. $\square$

## Declarations

* Funding: No funding was provided for the completion of this study

* Conflict of interest/Competing interests: The authors declare no conflict of interest in this paper

* Ethics approval and consent to participate: Not applicable

* Consent for publication: Not applicable

* Data availability: Since no data sets were created or examined for this research data sharing is not applicable

* Materials availability: Not applicable

* Code availability: Not applicable

* Author contribution: The authors contributed equally to this work

## References

[1] Aguilar K, Chávez A, Garcia S, Volcic J (2022) Norms on complex matrices induced by complete homogeneous symmetric polynomials. Bull. Lond. Math. Soc. 54(6):2078–2100.
[2] Barbeau EJ (1989) Polynomials. Problem Books in Mathematics, Springer-Verlag, New York, NY.

[3] Cornelius EF (2011) Identities for complete homogeneous symmetric polynomials. Journal of Algebra, Number Theory and Applications 21(1):109–116.

[4] Egge SE (2019) An Introduction to Symmetric Functions and Their Combinatorics. American Mathematical Society, Student Mathematical Library, vol 91.

[5] Freed-Brown J, Holden M, Orrison M, Vrable M (2012) Cyclotomic polynomials, symmetric polynomials, and a generalization of Euler's totient function. Math. Mag. 85(1):44–50.

[6] Girard A (1884) Invention nouvelle en l'algèbre. Bierens de Haan, Leiden.

[7] Kiryatskii EG (2009) On a family of polynomials univalent in an angular domain. Sib. Math. J. 50(3):456–466.

[8] Macdonald IG (2008) Symmetric Functions and Hall Polynomials. Oxford Univ. Press, 2nd edition.

[9] Monov V (2008) A family of symmetric polynomials of the eigenvalues of a matrix. Linear Algebra Appl. 429(8-9):2199–2208.

[10] Nagell T (1951) Introduction to Number Theory. New York: Wiley.

[11] Sagan BE (2001) The Symmetric Group. Representations, Combinatorial Algorithms, and Symmetric Functions. Graduate Texts in Mathematics, vol 203, Springer-Verlag.

[12] Stanley RP (1999) Enumerative Combinatorics, vol 2. Cambridge University Press.

[13] Verde-Star L (2012) Transformation formulas for complete symmetric polynomials and identities for generalized binomial coefficients and q-Stirling numbers. Discrete Math. 312(15):2197–2202.

[14] Viète F (1646) Opera mathematica. F. van Schouten (ed.), Leiden.