Mixed-Type Hypergeometric Bernoulli–Gegenbauer Polynomials

Abstract

In this paper, we consider a novel family of the mixed-type hypergeometric Bernoulli–Gegenbauer polynomials. This family represents a fascinating fusion between two distinct categories of special functions: hypergeometric Bernoulli polynomials and Gegenbauer polynomials. We focus our attention on some algebraic and differential properties of this class of polynomials, including its explicit expressions, derivative formulas, matrix representations, matrix-inversion formulas, and other relations connecting it with the hypergeometric Bernoulli polynomials. Furthermore, we show that unlike the hypergeometric Bernoulli polynomials and Gegenbauer polynomials, the mixed-type hypergeometric Bernoulli–Gegenbauer polynomials do not fulfill either Hanh or Appell conditions.

1. Introduction

For a fixed integer $m\in \mathbb{N}$, the mixed-type hypergeometric Bernoulli–Gegenbauer polynomials ${\mathcal{V}}_n^{[m-1,\alpha ]}\left(x\right)$ of order $\alpha \in (-1/2,\infty )$, where $n\ge 0$, are defined through generating the functions and series expansions as follows:

$$\left(\frac{z^m{e}^{xz}}{e^z-{\sum }_{l=0}^{m-1}\frac{z^l}{l!}}\right){\left(1-\frac{xz}{\pi }+\frac{z^2}{4{\pi }^2}\right)}^{-\alpha }=\sum _{n=0}^{\infty }{\mathcal{V}}_n^{[m-1,\alpha ]}\left(x\right)\frac{z^n}{n!},$$

(1)

where $\left|z\right|<2\pi$, $\left|x\right|\le 1$, and $\alpha \in (-1/2,\infty )\setminus \left\{0\right\}$.

$$\left(\frac{z^m{e}^{xz}}{e^z-{\sum }_{l=0}^{m-1}\frac{z^l}{l!}}\right)\left(\frac{2\pi -xz}{1-\frac{xz}{\pi }+\frac{z^2}{4{\pi }^2}}\right)=\sum _{n=0}^{\infty }{\mathcal{V}}_n^{[m-1,0]}\left(x\right)\frac{z^n}{n!},\left|z\right|<2\pi ,\left|x\right|\le 1.$$

(2)

The polynomials ${\left\{{\mathcal{V}}_n^{[m-1,\alpha ]}\left(x\right)\right\}}_{n\ge 0}$ represent a fascinating fusion between two classes of special functions: hypergeometric Bernoulli polynomials and Gegenbauer polynomials.

A significant amount of research has been conducted on various generalizations and analogs of the Bernoulli polynomials and the Bernoulli numbers. For a comprehensive treatment of the diverse aspects, including summation formulas and applications, interested readers can refer to recent works [1,2]. Inspired by recent articles [3,4,5,6,7] where authors explore analytic and numerical aspects of hypergeometric Bernoulli polynomials, hypergeometric Euler polynomials, generalized mixed-type Bernoulli–Gegenbauer polynomials, and Lagrange-based hypergeometric Bernoulli polynomials, this article focuses on the algebraic and differential properties of the polynomials ${\left\{{\mathcal{V}}_n^{[m-1,\alpha ]}\left(x\right)\right\}}_{n\ge 0}$. These properties include their explicit expressions, derivative formulas, matrix representations, matrix-inversion formulas, and other relationships connecting them with hypergeometric Bernoulli polynomials.

The paper is organized as follows. Section 2 provides relevant information about hypergeometric Bernoulli polynomials and Gegenbauer polynomials. Section 3 is dedicated to the study of the main algebraic and analytic properties of the HBG polynomials (1) and (2), which are summarized in Theorems 1–4, and Proposition 6.

2. Background and Previous Results

Throughout this paper, let $\mathbb{N}$, ${\mathbb{N}}_0$, $\mathbb{Z}$, $\mathbb{R}$, and $\mathbb{C}$ denote, respectively, the sets of natural numbers, non-negative integers, integers, real numbers, and complex numbers. As usual, we always use the principal branch for complex powers, in particular, $1^{\alpha }=1$ for $\alpha \in \mathbb{C}$. Furthermore, the convention $0^0=1$ is adopted.

For $\lambda \in \mathbb{C}$ and $k\in \mathbb{Z}$, we use the notations ${\lambda }^{\left(k\right)}$ and ${\left(\lambda \right)}_k$ for the rising and falling factorials, respectively, i.e.,

$${\lambda }^{\left(k\right)}=\left\{\begin{array}{cc}1,& \mathrm{if}k=0,\\ \prod _{i=1}^k(\lambda +i-1),& \mathrm{if}k\ge 1,\\ 0,& \mathrm{if}k<0,\end{array}\right.$$

and

$${\left(\lambda \right)}_k=\left\{\begin{array}{cc}1,& \mathrm{if}k=0,\hfill \\ \prod _{i=1}^k(\lambda -i+1),& \mathrm{if}k\ge 1,\hfill \\ 0,& \mathrm{if}k<0.\hfill \end{array}\right.$$

From now on, we denote by ${\mathbb{P}}_n$ the linear space of polynomials with real coefficients and a degree less than or equal to n. Moreover, to present some of our results, we require the use of the generalized multinomial theorem (cf. [8,9] and the references therein).

2.1. Hypergeometric Bernoulli Polynomials

For a fixed $m\in \mathbb{N}$, the hypergeometric Bernoulli polynomials are defined by means of the following generating function [5,10,11,12,13,14]:

$${\displaystyle \frac{z^m{e}^{xz}}{e^z-{\sum }_{l=0}^{m-1}\frac{z^l}{l!}}=\sum _{n=0}^{\infty }{B}_n^{[m-1]}\left(x\right)\frac{z^n}{n!},\left|z\right|<2\pi ,}$$

(3)

and the hypergeometric Bernoulli numbers are defined by $B_n^{[m-1]}:=B_n^{[m-1]}\left(0\right)$ for all $n\ge 0$. The hypergeometric Bernoulli polynomials also are called generalized Bernoulli polynomials of level m [5,6]. It is clear that if $m=1$ in (3), then we obtain the definition of the classical Bernoulli polynomials $B_n\left(x\right)$ and classical Bernoulli numbers, respectively, i.e., $B_n\left(x\right)=B_n^{\left[0\right]}\left(x\right)$ and $B_n=B_n^{\left[0\right]}$, respectively, for all $n\ge 0$.

The first four hypergeometric Bernoulli polynomials are as follows:

$$\begin{array}{c}{B}_0^{[m-1]}\left(x\right)=m!,\hfill \\ B_1^{[m-1]}\left(x\right)=m!\left(x-\frac{1}{m+1}\right),\hfill \\ B_2^{[m-1]}\left(x\right)=m!\left(x^2-\frac{2}{m+1}x+\frac{2}{{(m+1)}^2(m+2)}\right),\hfill \\ B_3^{[m-1]}\left(x\right)=m!\left(x^3-\frac{3}{m+1}{x}^2+\frac{6}{{(m+1)}^2(m+2)}x+\frac{6(m-1)}{{(m+1)}^3(m+2)(m+3)}\right).\hfill \end{array}$$

The following results summarize some properties of the hypergeometric Bernoulli polynomials (cf. [5,6,11,12,15]).

Proposition 1

([5], Proposition 1). For a fixed $m\in \mathbb{N}$, let ${\left\{B_n^{[m-1]}\left(x\right)\right\}}_{n\ge 0}$ be the sequence of hypergeometric Bernoulli polynomials. Then the following statements hold:

(a) 

Summation formula. For every $n\ge 0$,

$$B_n^{[m-1]}\left(x\right)=\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)B_k^{[m-1]}{x}^{n-k}.$$

(4)

(b) 

Differential relations (Appell polynomial sequences). For $n,j\ge 0$ with $0\le j\le n$, we have

$${\left[B_n^{[m-1]}\left(x\right)\right]}^{\left(j\right)}=\frac{n!}{(n-j)!}{B}_{n-j}^{[m-1]}\left(x\right).$$

(5)

(c) 

Inversion formula. ([12], Equation (2.6)) For every $n\ge 0$,

$$x^n=\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)\frac{k!}{(m+k)!}{B}_{n-k}^{[m-1]}\left(x\right).$$

(6)

(d) 

Recurrence relation. ([12], Lemma 3.2) For every $n\ge 1$,

$$B_n^{[m-1]}\left(x\right)=\left(x-\frac{1}{m+1}\right)B_{n-1}^{[m-1]}\left(x\right)-\frac{1}{n(m-1)!}\sum _{k=0}^{n-2}\left(\genfrac{}{}{0pt}{}{n}{k}\right)B_{n-k}^{[m-1]}{B}_k^{[m-1]}\left(x\right).$$

(e) 

Integral formulas.

$$\begin{array}{cc}\hfill {\int }_{x_0}^{x_1}{B}_n^{[m-1]}\left(x\right)dx& =\frac{1}{n+1}\left[B_{n+1}^{[m-1]}\left(x_1\right)-B_{n+1}^{[m-1]}\left(x_0\right)\right]\hfill \\ & =\sum _{k=0}^n\frac{1}{n-k+1}\left(\genfrac{}{}{0pt}{}{n}{k}\right)B_k^{[m-1]}({\left(x_1\right)}^{n-k+1}-{\left(x_0\right)}^{n-k+1}).\hfill \\ & \\ \hfill B_n^{[m-1]}\left(x\right)& =n{\int }_0^x{B}_{n-1}^{[m-1]}\left(t\right)dt+B_n^{[m-1]}.\hfill \end{array}$$

(f) 

([12], Theorem 3.1) Differential equation. For every $n\ge 1$, the polynomial $B_n^{[m-1]}\left(x\right)$ satisfies the following differential equation

$$\begin{array}{cc}\hfill \frac{B_n^{[m-1]}}{n!}{y}^{\left(n\right)}+\frac{B_{n-1}^{[m-1]}}{(n-1)!}{y}^{(n-1)}+\cdots +\frac{B_2^{[m-1]}}{2!}{y}^{″}+(m-1)!\left(\frac{1}{m+1}-x\right)y^{\prime }+n(m-1)!y& =0.\hfill \end{array}$$

As a straightforward consequence of the inversion Formula (6), the following expected algebraic property is obtained.

Proposition 2

([5], Proposition 2). For a fixed $m\in \mathbb{N}$ and each $n\ge 0$, the set

$\left\{B_0^{[m-1]}\left(x\right),B_1^{[m-1]}\left(x\right),\dots ,B_n^{[m-1]}\left(x\right)\right\}$ is a basis for ${\mathbb{P}}_n$, i.e.,

$${\mathbb{P}}_n=\mathrm{span}\left\{B_0^{[m-1]}\left(x\right),B_1^{[m-1]}\left(x\right),\dots ,B_n^{[m-1]}\left(x\right)\right\}.$$

Let $\zeta \left(s\right)$ be the Riemann zeta function defined by

$$\zeta \left(s\right)=\sum _{n=1}^{\infty }\frac{1}{n^s},\Re \left(s\right)>1.$$

The following result provides a formula for evaluating $\zeta \left(2r\right)$ in terms of the hypergeometric Bernoulli numbers.

Proposition 3

([6], Theorem 3.3). For a fixed $m\in \mathbb{N}$ and any $r\in \mathbb{N}$, the following identity holds.

$$\zeta \left(2r\right)=\frac{{(-1)}^{r-1}{2}^{2r-1}{\pi }^{2r}{B}_{2r}^{[m-1]}}{m!\left(2r\right)!}+{\Delta }_r^{[m-1]},$$

where

$${\Delta }_r^{[m-1]}=\frac{{(-1)}^{r-1}{2}^{2r-1}{\pi }^{2r}}{m!}\left[\frac{B_{2r}^{[m-1]}\left(1\right)-B_{2r}^{[m-1]}}{2\left(2r\right)!}-\frac{B_{2r+1}^{[m-1]}\left(1\right)-B_{2r+1}^{[m-1]}}{(2r+1)!}-\sum _{j=1}^{r-1}\frac{\left(B_{2r-2j+1}^{[m-1]}\left(1\right)-B_{2r-2j+1}^{[m-1]}\right)}{(2r-2j+1)!}\frac{B_{2j}}{\left(2j\right)!}\right].$$

2.2. Gegenbauer Polynomials

For $\alpha >-\frac{1}{2}$, we denote by ${\left\{{\widehat{C}}_n^{\left(\alpha \right)}\left(x\right)\right\}}_{n\ge 0}$ the sequence of Gegenbauer polynomials, orthogonal on $[-1,1]$ with respect to the measure $d\mu \left(x\right)={(1-x^2)}^{\alpha -\frac{1}{2}}dx$ (cf. [16], Chapter IV), normalized by

$${\widehat{C}}_n^{\left(\alpha \right)}\left(1\right)=\frac{\mathsf{\Gamma }(n+2\alpha )}{n!\mathsf{\Gamma }\left(2\alpha \right)}.$$

More precisely,

$${\int }_{-1}^1{\widehat{C}}_n^{\left(\alpha \right)}\left(x\right){\widehat{C}}_m^{\left(\alpha \right)}\left(x\right)d\mu \left(x\right)={\int }_{-1}^1{\widehat{C}}_n^{\left(\alpha \right)}\left(x\right){\widehat{C}}_m^{\left(\alpha \right)}\left(x\right){(1-x^2)}^{\alpha -\frac{1}{2}}dx=M_n^{\alpha }{\delta }_{n,m},n,m\ge 0,$$

where the constant $M_n^{\alpha }$ is positive. It is clear that the normalization above does not allow $\alpha$ to be zero or a negative integer. Nevertheless, the following limits exist for every $x\in [-1,1]$ (see [16], (4.7.8))

$$\underset{\alpha \to 0}{lim}{\widehat{C}}_0^{\left(\alpha \right)}\left(x\right)=T_0\left(x\right),\underset{\alpha \to 0}{lim}\frac{{\widehat{C}}_n^{\left(\alpha \right)}\left(x\right)}{\alpha }=\frac{2}{n}{T}_n\left(x\right),$$

where $T_n\left(x\right)$ is the nth Chebyshev polynomial of the first kind. In order to avoid confusing notation, we define the sequence ${\left\{{\widehat{C}}_n^{\left(0\right)}\left(x\right)\right\}}_{n\ge 0}$ as follows:

$${\widehat{C}}_0^{\left(0\right)}\left(1\right)=1,{\widehat{C}}_n^{\left(0\right)}\left(1\right)=\frac{2}{n},{\widehat{C}}_n^{\left(0\right)}\left(x\right)=\frac{2}{n}{T}_n\left(x\right),n\ge 1.$$

We denote the nth monic Gegenbauer orthogonal polynomial by

$$C_n^{\left(\alpha \right)}\left(x\right)={\left(k_n^{\alpha }\right)}^{-1}{\widehat{C}}_n^{\left(\alpha \right)}\left(x\right),$$

where the constant $k_n^{\alpha }$ (cf. [16], Formula (4.7.31)) is given by

$$\begin{array}{cc}\hfill k_n^{\alpha }& =\frac{2^n\mathsf{\Gamma }(n+\alpha )}{n!\mathsf{\Gamma }\left(\alpha \right)},\alpha \ne 0,\hfill \\ \hfill k_n^0& =\underset{\alpha \to 0}{lim}\frac{k_n^{\alpha }}{\alpha }=\frac{2^n}{n},n\ge 1.\hfill \end{array}$$

Then for $n\ge 1$, we have

$$C_n^{\left(0\right)}\left(x\right)=\underset{\alpha \to 0}{lim}{\left(k_n^{\alpha }\right)}^{-1}{\widehat{C}}_n^{\left(\alpha \right)}\left(x\right)=\frac{1}{2^{n-1}}{T}_n\left(x\right).$$

(7)

Gegenbauer polynomials are closely connected with axially symmetric potentials in n dimensions (cf. [4] and the references cited therein), and contain the Legendre and Chebyshev polynomials as special cases. Furthermore, they inherit practically all the formulas known in the classical theory of Legendre polynomials.

Proposition 4

([17], cf. Proposition 2.1). Let ${\left\{C_n^{\left(\alpha \right)}\right\}}_{n\ge 0}$ be the sequence of monic Gegenbauer orthogonal polynomials. Then the following statements hold.

(a) 

Three-term recurrence relation.

$$x{C}_n^{\left(\alpha \right)}\left(x\right)=C_{n+1}^{\left(\alpha \right)}\left(x\right)+{\gamma }_n^{\left(\alpha \right)}{C}_{n-1}^{\left(\alpha \right)}\left(x\right),\alpha >-\frac{1}{2},\alpha \ne 0,$$

(8)

with initial conditions $C_{-1}^{\left(\alpha \right)}\left(x\right)=0$, $C_0^{\left(\alpha \right)}\left(x\right)=1$ and recurrence coefficients ${\gamma }_0^{\left(\alpha \right)}\in \mathbb{R}$,${\gamma }_n^{\left(\alpha \right)}=\frac{n(n+2\alpha -1)}{4(n+\alpha )(n+\alpha -1)}$, $n\in \mathbb{N}$.

(b) 

For every $n\in \mathbb{N}$ (see [16], (4.7.15))

$$h_n^{\alpha }:={\parallel C_n^{\left(\alpha \right)}\parallel }_{\mu }^2={\int }_{-1}^1{\left[C_n^{\left(\alpha \right)}\left(x\right)\right]}^{2}d\mu \left(x\right)=\pi 2^{1-2\alpha -2n}\frac{n!\mathsf{\Gamma }(n+2\alpha )}{\mathsf{\Gamma }(n+\alpha +1)\mathsf{\Gamma }(n+\alpha )}.$$

(9)

(c) 

Rodrigues formula.

$${(1-x^2)}^{\alpha -\frac{1}{2}}{C}_n^{\left(\alpha \right)}\left(x\right)=\frac{{(-1)}^n\mathsf{\Gamma }(n+2\alpha )}{\mathsf{\Gamma }(2n+2\alpha )}\frac{d^n}{d{x}^n}\left[{(1-x^2)}^{n+\alpha -\frac{1}{2}}\right],x\in (-1,1).$$

(d) 

Structure relation (see [16], (4.7.29)). For every $n\ge 2$

$$C_n^{(\alpha -1)}\left(x\right)=C_n^{\left(\alpha \right)}\left(x\right)+{\xi }_{n-2}^{\left(\alpha \right)}{C}_{n-2}^{\left(\alpha \right)}\left(x\right),$$

where

$${\xi }_n^{\left(\alpha \right)}=\frac{(n+2)(n+1)}{4(n+\alpha +1)(n+\alpha )},n\ge 0.$$

(e) 

For every $n\in \mathbb{N}$ (see [16], Formula (4.7.14))

$$\frac{d}{dx}{C}_n^{\left(\alpha \right)}\left(x\right)=n{C}_{n-1}^{(\alpha +1)}\left(x\right).$$

(f) 

For every $n\in \mathbb{N}$ (see [18], Proposition 2.1)

$$\frac{d}{dx}{C}_n^{\left(0\right)}\left(x\right)=\frac{n}{2}{C}_{n-1}^{\left(1\right)}\left(x\right).$$

As is well known, the monic Gegenbauer orthogonal polynomials admit other different definitions [16,19,20,21]. In order to deal with the definitions (1) and (2) of the HBG polynomials, we also are interested in the definition of the monic Gegenbauer orthogonal polynomials by means of the following generating functions:

$${\left(1-\frac{xz}{\pi }+\frac{z^2}{4{\pi }^2}\right)}^{-\alpha }=\sum _{n=0}^{\infty }\frac{\mathsf{\Gamma }(n+\alpha )}{{\pi }^n\mathsf{\Gamma }\left(\alpha \right)}{C}_n^{\left(\alpha \right)}\left(x\right)\frac{z^n}{n!},\left|z\right|<2\pi ,\left|x\right|\le 1,\alpha \in (-1/2,\infty )\setminus \left\{0\right\},$$

(10)

and

$$\frac{2\pi -xz}{1-\frac{xz}{\pi }+\frac{z^2}{4{\pi }^2}}=\sum _{n=0}^{\infty }\frac{1}{{\pi }^{n-1}}{C}_n^{\left(0\right)}\left(x\right)z^n=\sum _{n=0}^{\infty }\frac{\mathsf{\Gamma }(n+1)}{{\pi }^{n-1}}{C}_n^{\left(0\right)}\left(x\right)\frac{z^n}{n!},\left|z\right|<2\pi ,\left|x\right|\le 1.$$

(11)

Remark 1.

Note that (10) and (11) are suitable modifications of the generating functions for the Gegenbauer polynomials ${\widehat{C}}_n^{\left(\alpha \right)}\left(x\right)$:

$$\begin{array}{cc}\hfill {\left(1-2xz+z^2\right)}^{-\alpha }& =\sum _{n=0}^{\infty }{\widehat{C}}_n^{\left(\alpha \right)}\left(x\right)z^n,\left|z\right|<1,\left|x\right|\le 1,\alpha \in (-1/2,\infty )\setminus \left\{0\right\},\hfill \\ \hfill \frac{1-xz}{1-xz+z^2}& =1+\sum _{n=1}^{\infty }\frac{n}{2}{\widehat{C}}_n^{\left(0\right)}\left(x\right)z^n,\left|z\right|<1,\left|x\right|\le 1.\hfill \end{array}$$

3. The Polynomials ${\mathcal{V}}_{\mathit{n}}^{[\mathit{m}-\mathit{1},\mathsf{\alpha }]}\left(\mathit{x}\right)$ and Their Properties

Now, we can proceed to investigate some relevant properties of the HBG polynomials.

Proposition 5.

For $\alpha \in (-1/2,\infty )$, let ${\left\{{\mathcal{V}}_n^{[m-1,\alpha ]}\left(x\right)\right\}}_{n\ge 0}$ be the sequence of HBG polynomials of order α. Then the following explicit formulas hold.

$${\mathcal{V}}_n^{[m-1,\alpha ]}\left(x\right)=\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)\frac{\mathsf{\Gamma }(k+\alpha )}{{\pi }^k\mathsf{\Gamma }\left(\alpha \right)}{C}_k^{\left(\alpha \right)}\left(x\right)B_{n-k}^{[m-1]}\left(x\right),n\ge 0,\alpha \ne 0,$$

(12)

$${\mathcal{V}}_n^{[m-1,0]}\left(x\right)=\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)\frac{k!}{{\pi }^{k-1}}{C}_k^{\left(0\right)}\left(x\right)B_{n-k}^{[m-1]}\left(x\right),n\ge 0.$$

(13)

Proof. 

On account of the generating functions (1) and (10), it suffices to make a suitable use of Cauchy product of series in order to deduce the expression (12).

Similarly, taking into account the generating functions (2) and (11), we can use an analogous reasoning to the previous one to obtain expression (13). □

Thus, the suitable use of (8) and (12) allow us to check that for $\alpha \in (-1/2,\infty )\setminus \left\{0\right\}$, the first five HBG polynomials are:

$$\begin{array}{cc}\hfill {\mathcal{V}}_0^{[m-1,\alpha ]}\left(x\right)=& m!v_0\left(\alpha \right),\hfill \\ & \\ \hfill {\mathcal{V}}_1^{[m-1,\alpha ]}\left(x\right)=& m!\left[v_1\left(\alpha \right)x-\frac{1}{m+1}\right],\hfill \\ & \\ \hfill {\mathcal{V}}_2^{[m-1,\alpha ]}\left(x\right)=& m!\left[v_2\left(\alpha \right)x^2-\frac{2(\pi +\alpha )}{\pi (m+1)}x+\frac{4{\pi }^2(\alpha +1)+\alpha {(m+1)}^2(m+2)}{2{\pi }^2{(m+1)}^2(m+2)(1+\alpha )}\right],\hfill \\ & \\ \hfill {\mathcal{V}}_3^{[m-1,\alpha ]}\left(x\right)=& m!\left[v_3\left(\alpha \right)x^3-\frac{3}{m+1}{v}_2\left(\alpha \right)x^2+3\left(\frac{2}{{(m+1)}^2(m+2)}\left(1+\frac{\alpha }{\pi }\right)-\frac{\alpha }{2{\pi }^2}\left(1+\frac{(1+\alpha )}{\pi }\right)\right)x\right.\hfill \\ & \left.+3\left(\frac{2(m-1)}{{(m+1)}^3(m+2)(m+3)}-\frac{\alpha }{2{\pi }^2(m+1)}\right)\right],\hfill \end{array}$$

$$\begin{array}{cc}\hfill {\mathcal{V}}_4^{[m-1,\alpha ]}\left(x\right)=& m!\left[v_4\left(\alpha \right)x^4-\frac{4}{m+1}{v}_3\left(\alpha \right)x^3+3\left(\frac{m-2}{(m+1)(m+2)}+\frac{8\alpha }{\pi {(m+1)}^2(m+2)}-\frac{\alpha }{{\pi }^2}-\frac{2(1+\alpha )\alpha }{{\pi }^3}\right.\right.\hfill \\ & \left.-\frac{(2+\alpha )(1+\alpha )\alpha }{{\pi }^4}\right)x^2+6\left(\frac{5-m}{{(m+1)}^2(m+2)(m+3)}+\frac{4(m-1)\alpha }{\pi {(m+1)}^3(m+2)(m+3)}+\frac{\alpha }{{\pi }^2(m+1)}\right.\hfill \\ & \left.\left.+\frac{(1+\alpha )\alpha }{{\pi }^3(m+1)}\right)x^2-\frac{6(m^3-3m^2-6m+36)}{{(m+1)}^2{(m+2)}^2(m+3)(m+4)}+\frac{6(1+2\alpha )\alpha }{{\pi }^2{(m+1)}^2(m+2)}+\frac{3(1+\alpha )\alpha }{4{\pi }^4}\right],\hfill \end{array}$$

where ${\displaystyle v_n\left(\alpha \right)=\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)\frac{{\alpha }^{\left(k\right)}}{{\pi }^k}}$, $0\le n\le 4$.

In contrast to the hypergeometric Bernoulli polynomials and Gegenbauer polynomials, the HBG polynomials neither satisfy a Hanh condition nor an Appell condition. More precisely, we have the following result.

Theorem 1.

For $\alpha \in (-1/2,\infty )$, let ${\left\{{\mathcal{V}}_n^{[m-1,\alpha ]}\left(x\right)\right\}}_{n\ge 0}$ be the sequence of HBG polynomials of order α. Then we have

$$\frac{d}{dx}{\mathcal{V}}_{n+1}^{[m-1,\alpha ]}\left(x\right)=(n+1)\left[\frac{\alpha }{\pi }{\mathcal{V}}_n^{[m-1,\alpha +1]}\left(x\right)+{\mathcal{V}}_n^{[m-1,\alpha ]}\left(x\right)\right],\alpha \ne 0,$$

(14)

$$\frac{d}{dx}{\mathcal{V}}_{n+1}^{[m-1,0]}\left(x\right)=(n+1)\left[{\mathcal{V}}_n^{[m-1,0]}\left(x\right)+\frac{1}{2}\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)\frac{(k+1)!}{{\pi }^k}{C}_k^{\left(1\right)}\left(x\right)B_{n-k}^{[m-1]}\left(x\right)\right],\alpha =0.$$

(15)

Proof. 

From (12), we have

$${\mathcal{V}}_{n+1}^{[m-1,\alpha ]}\left(x\right)=\sum _{k=0}^{n+1}\left(\genfrac{}{}{0pt}{}{n+1}{k}\right)\frac{\mathsf{\Gamma }(k+\alpha )}{{\pi }^k\mathsf{\Gamma }\left(\alpha \right)}{C}_k^{\left(\alpha \right)}\left(x\right)B_{n+1-k}^{[m-1]}\left(x\right),$$

differentiating this last equation, and using part (e) of Proposition 4, (14) follows. □

Furthermore, it is possible to establish an integral formula connecting the HBG polynomials with the monic Gegenbauer polynomials. This integral formula allows us to deduce a concise expression for the Fourier coefficients of the HBG polynomials in terms of the basis of monic Gegenbauer polynomials.

Lemma 1.

For $\alpha \in (-1/2,\infty )$, let ${\left\{{\mathcal{V}}_n^{[m-1,\alpha ]}\left(x\right)\right\}}_{n\ge 0}$ be the sequence of HBG polynomials of order α. Then, the following formula holds.

$${\int }_{-1}^1{\mathcal{V}}_n^{[m-1,\alpha ]}\left(x\right)C_n^{\left(\alpha \right)}\left(x\right)d\mu \left(x\right)=\left\{\begin{array}{c}\frac{m!n!\mathsf{\Gamma }(n+2\alpha )}{{\pi }^{2\alpha +2n}\mathsf{\Gamma }(n+\alpha +1)\mathsf{\Gamma }(n+\alpha )}\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)\frac{\mathsf{\Gamma }(k+\alpha )}{{\pi }^{k-1}\mathsf{\Gamma }\left(\alpha \right)},\alpha \ne 0,\hfill \\ \\ \frac{m!\pi }{2^n}\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)\frac{k!}{{\pi }^{k-1}},\alpha =0,\hfill \end{array}\right.$$

(16)

whenever $n\ge 0$.

Proof. 

In order to obtain (16), it suffices to use the orthogonality properties of the monic Gegenbauer polynomials (4), (7), (9), (12) and (13). □

Regarding the zero distribution of these polynomials, the numerical evidence indicates that this distribution does not align with the behavior of either Bernoulli hypergeometric polynomials or Gegenbauer polynomials. For instance, in Figure 1, the plots for the zeros of ${\mathcal{V}}_{28}^{[m-1,\alpha ]}\left(x\right)$ and ${\mathcal{V}}_{30}^{[m-1,\alpha ]}\left(x\right)$ are shown for $m=2$ and $\alpha =-\frac{1}{4}$.

As expected, the symmetry property of Gegenbauer polynomials is not inherited by the HBG polynomials. For instance, Figure 2 displays the induced mesh of ${\mathcal{V}}_j^{[m-1,\alpha ]}\left(x\right)$ for $m=2$, $\alpha =1$, and $j=1,\dots ,21$. Each point on this mesh takes the form $(x_j^{[m-1,\alpha ]},j)$, $j=1,\dots ,21$. In contrast, Figure 3 displays the induced mesh of $C_j^{\left(\alpha \right)}\left(x\right)$ for $\alpha =1$, and $j=1,\dots ,19$.

For any $\alpha \in (-1/2,\infty )$, it is possible to deduce interesting relations connecting the HBG polynomials ${\mathcal{V}}_n^{[m-1,\alpha ]}\left(x\right)$ and the hypergeometric Bernoulli polynomials $B_n^{[m-1]}\left(x\right)$. The following two results concern these relations.

Proposition 6.

For a fixed $m\in \mathbb{N}$, let ${\mathcal{V}}_n^{[m-1,\alpha ]}\left(x\right)$ be the nth HBG polynomial of order $\alpha \in (-1/2,\infty )\setminus \left\{0\right\}$. Then, the following relation is satisfied:

$${\displaystyle \sum _{n=0}^{\infty }{B}_n^{[m-1]}\left(x\right)\frac{z^n}{n!}=\sum _{n=0}^{\infty }\sum _{0\le j,k\le \left|\alpha \right|}\frac{{(-1)}^j}{2^{2k}{\pi }^{2k+j}}\left(\genfrac{}{}{0pt}{}{\alpha }{j,k}\right)x^j{\mathcal{V}}_n^{[m-1,\alpha ]}\left(x\right)\frac{z^{n+2k+j}}{n!}}.$$

(17)

Proof. 

On the account of generalized multinomial theorem, we deduce that

$${\left(1-\frac{xz}{\pi }+\frac{z^2}{4{\pi }^2}\right)}^{\alpha }=\sum _{0\le j,k\le \left|\alpha \right|}\frac{{(-1)}^j}{2^{2k}{\pi }^{2k+j}}\left(\genfrac{}{}{0pt}{}{\alpha }{j,k}\right)x^j{z}^{2k+j}.$$

(18)

Next, (1), (3) and (18) imply that

$$\begin{array}{ccc}\hfill \sum _{n=0}^{\infty }{B}_n^{[m-1]}\left(x\right)\frac{z^n}{n!}& =& {\left(1-\frac{xz}{\pi }+\frac{z^2}{4{\pi }^2}\right)}^{\alpha }\sum _{n=0}^{\infty }{\mathcal{V}}_n^{[m-1,\alpha ]}\frac{z^n}{n!}\hfill \\ \hfill & =& \left(\sum _{0\le j,k\le \left|\alpha \right|}\frac{{(-1)}^j}{2^{2k}{\pi }^{2k+j}}\left(\genfrac{}{}{0pt}{}{\alpha }{j,k}\right)x^j{z}^{2k+j}\right)\sum _{n=0}^{\infty }{\mathcal{V}}_n^{[m-1,\alpha ]}\frac{z^n}{n!}.\hfill \end{array}$$

(19)

Since the sum on the right-hand side of (18) is finite, (17) follows directly from (19). □

Theorem 2.

For a fixed $m\in \mathbb{N}$, the HBG polynomials ${\mathcal{V}}_n^{[m-1,0]}\left(x\right)$ are related with the hypergeometric Bernoulli polynomials $B_n^{[m-1]}\left(x\right)$ by means of the following identities.

$$\begin{array}{c}2\pi B_0^{[m-1]}\left(x\right)={\mathcal{V}}_0^{[m-1,0]}\left(x\right),\hfill \\ 2\pi B_1^{[m-1]}\left(x\right)-x{B}_0^{[m-1]}\left(x\right)={\mathcal{V}}_1^{[m-1,0]}\left(x\right)-\frac{x}{\pi }{\mathcal{V}}_0^{[m-1,0]}\left(x\right),\hfill \\ 2\pi B_n^{[m-1]}\left(x\right)-nx{B}_{n-1}^{[m-1]}\left(x\right)={\mathcal{V}}_n^{[m-1,0]}\left(x\right)-\frac{nx}{\pi }{\mathcal{V}}_{n-1}^{[m-1,0]}\left(x\right)+\frac{n(n-1)}{4{\pi }^2}{\mathcal{V}}_{n-2}^{[m-1,0]}\left(x\right),n\ge 2.\hfill \end{array}$$

(20)

Proof. 

From the identities (2) and (3), we have

$$\left(2\pi -xz\right)\sum _{n=0}^{\infty }{B}_n^{[m-1]}\left(x\right)\frac{z^n}{n!}=\left(1-\frac{xz}{\pi }+\frac{z^2}{4{\pi }^2}\right)\sum _{n=0}^{\infty }{\mathcal{V}}_n^{[m-1,0]}\left(x\right)\frac{z^n}{n!}.$$

Multiplying, respectively, the left-hand side of the above expression by $\left(2\pi -xz\right)$ and the right-hand side by $\left(1-\frac{xz}{\pi }+\frac{z^2}{4{\pi }^2}\right)$, we obtain the following equivalent expression:

$$\begin{array}{c}2\pi B_0^{[m-1]}\left(x\right)+2\pi B_1^{[m-1]}\left(x\right)z-x{B}_0^{[m-1]}\left(x\right)z+\sum _{n=2}^{\infty }\left(2\pi B_n^{[m-1]}\left(x\right)-nx{B}_{n-1}^{[m-1]}\left(x\right)\right)\frac{z^n}{n!}\hfill \\ ={\mathcal{V}}_0^{[m-1,0]}\left(x\right)+{\mathcal{V}}_1^{[m-1,0]}\left(x\right)z-\frac{x}{\pi }{\mathcal{V}}_0^{[m-1,0]}\left(x\right)z\\ \hfill +\sum _{n=2}^{\infty }\left({\mathcal{V}}_n^{[m-1,0]}\left(x\right)-\frac{nx}{\pi }{\mathcal{V}}_{n-1}^{[m-1,0]}\left(x\right)+\frac{n(n-1)}{4{\pi }^2}{\mathcal{V}}_{n-2}^{[m-1,0]}\left(x\right)\right)\frac{z^n}{n!}.\end{array}$$

(21)

Therefore, by comparing the coefficients on both sides of (21), we obtain the identities (20). □

Remark 2.

When $\alpha =r\in \mathbb{N}$, Equation (18) becomes

$${\left(1-\frac{xz}{\pi }+\frac{z^2}{4{\pi }^2}\right)}^r=\sum _{j+k=r}\frac{{(-1)}^j}{2^{2k}{\pi }^{2k+j}}\left(\genfrac{}{}{0pt}{}{r}{j,k}\right)x^j{z}^{2k+j}.$$

Thus, for $r=1$ we can combine the above identity with (17), and obtain the following connecting relations:

$$\begin{array}{cc}\hfill & B_0^{[m-1]}\left(x\right)={\mathcal{V}}_0^{[m-1,1]}\left(x\right)\hfill \\ \hfill & B_1^{[m-1]}\left(x\right)={\mathcal{V}}_1^{[m-1,1]}\left(x\right)-\frac{x}{\pi }{\mathcal{V}}_0^{[m-1,1]}\left(x\right)\hfill \\ \hfill & B_n^{[m-1]}\left(x\right)={\mathcal{V}}_n^{[m-1,1]}\left(x\right)-\frac{nx}{\pi }{\mathcal{V}}_{n-1}^{[m-1,1]}\left(x\right)+\frac{n(n-1)}{4{\pi }^2}{\mathcal{V}}_{n-2}^{[m-1,1]}\left(x\right),n\ge 2,\hfill \end{array}$$

(22)

Hence, as a straightforward consequence of (17) and (20), the HBG polynomials ${\mathcal{V}}_n^{[m-1,1]}\left(x\right)$ and ${\mathcal{V}}_n^{[m-1,0]}\left(x\right)$ are related by means of the following identities:

$$\begin{array}{c}2\pi {\mathcal{V}}_0^{[m-1,1]}\left(x\right)={\mathcal{V}}_0^{[m-1,0]}\left(x\right)\\ 2\pi {\mathcal{V}}_1^{[m-1,1]}\left(x\right)-3x{\mathcal{V}}_0^{[m-1,1]}\left(x\right)={\mathcal{V}}_1^{[m-1,0]}\left(x\right)-\frac{x}{\pi }{\mathcal{V}}_0^{[m-1,0]}\left(x\right)\\ 2\pi {\mathcal{V}}_n^{[m-1,1]}\left(x\right)-3nx{\mathcal{V}}_{n-1}^{[m-1,1]}\left(x\right)+\left(\frac{n(n-1)}{2\pi }+\frac{n(n-1)x^2}{\pi }\right){\mathcal{V}}_{n-2}^{[m-1,1]}\left(x\right)-\frac{n(n-1)(n-2)x}{4{\pi }^2}{\mathcal{V}}_{n-3}^{[m-1,1]}\left(x\right)\\ ={\mathcal{V}}_n^{[m-1,0]}\left(x\right)-\frac{nx}{\pi }{\mathcal{V}}_{n-1}^{[m-1,0]}\left(x\right)+\frac{n(n-1)}{4{\pi }^2}{\mathcal{V}}_{n-2}^{[m-1,0]}\left(x\right),n\ge 2.\end{array}$$

(23)

Using (12), (13), and employing a matrix approach, we can obtain a matrix representation for ${\mathcal{V}}_n^{[m-1,\alpha ]}\left(x\right)$, $n\ge 0$. In order to implement that, we follow some ideas from [4,5].

First of all, we must point out that for $r=0,1,\dots ,n$, Equations (12) and (13) allow us to deduce the following matrix form of ${\mathcal{V}}_r^{[m-1,\alpha ]}\left(x\right)$:

$${\mathcal{V}}_r^{[m-1,\alpha ]}\left(x\right)={\mathbf{C}}_r^{\left(\alpha \right)}\left(x\right){\mathbf{B}}^{[m-1]}\left(x\right),r=0,1,\dots ,n,$$

(24)

where

$${\displaystyle {\mathbf{C}}_r^{\left(\alpha \right)}\left(x\right)=\left\{\begin{array}{cc}\left[\begin{array}{ccccccc}\left(\genfrac{}{}{0pt}{}{r}{r}\right)\frac{\mathsf{\Gamma }(r+\alpha )}{{\pi }^r\mathsf{\Gamma }\left(\alpha \right)}{C}_r^{\left(\alpha \right)}\left(x\right)& \left(\genfrac{}{}{0pt}{}{r}{r-1}\right)\frac{\mathsf{\Gamma }(r-1+\alpha )}{{\pi }^{r-1}\mathsf{\Gamma }\left(\alpha \right)}{C}_{r-1}^{\left(\alpha \right)}\left(x\right)& \cdots & C_0^{\left(\alpha \right)}\left(x\right)& 0& \cdots & 0\end{array}\right],\hfill & \mathrm{if}\alpha \ne 0,\hfill \\ & \\ \left[\begin{array}{ccccccc}\left(\genfrac{}{}{0pt}{}{r}{r}\right)\frac{r!}{{\pi }^{r-1}}{C}_r^{\left(0\right)}\left(x\right)& \left(\genfrac{}{}{0pt}{}{r}{r-q}\right)\frac{(r-1)!}{{\pi }^{r-2}}{C}_{r-1}^{\left(0\right)}\left(x\right)& \cdots & C_0^{\left(0\right)}\left(x\right)& 0& \cdots & 0\end{array}\right],\hfill & \mathrm{if}\alpha =0,\hfill \end{array}\right.}$$

the null entries of the matrix ${\mathbf{C}}_r^{\left(\alpha \right)}\left(x\right)$ appear $(n-r)$-times, and the matrix ${\mathbf{B}}^{[m-1]}\left(x\right)$ is given by ${\mathbf{B}}^{[m-1]}\left(x\right)={\left[\begin{array}{cccc}{B}_0^{[m-1]}\left(x\right)& B_1^{[m-1]}\left(x\right)& \cdots & B_n^{[m-1]}\left(x\right)\end{array}\right]}^T$.

Now, for $\alpha \in (-1/2,\infty )$, let ${\mathbf{C}}^{\left(\alpha \right)}\left(x\right)$ be the $(n+1)\times (n+1)$ whose rows are precisely the matrices ${\mathbf{C}}_r^{\left(\alpha \right)}\left(x\right)$ for $r=0,1,\dots ,n$. That is,

$${\mathbf{C}}^{\left(\alpha \right)}\left(x\right)=\left[\begin{array}{cccc}{C}_0^{\left(\alpha \right)}\left(x\right)& 0& \cdots & 0\\ \\ \left(\genfrac{}{}{0pt}{}{1}{1}\right)\frac{\mathsf{\Gamma }(1+\alpha )}{\pi \mathsf{\Gamma }\left(\alpha \right)}{C}_1^{\left(\alpha \right)}\left(x\right)& C_0^{\left(\alpha \right)}\left(x\right)& \cdots & 0\\ \\ \left(\genfrac{}{}{0pt}{}{2}{2}\right)\frac{\mathsf{\Gamma }(2+\alpha )}{{\pi }^2\mathsf{\Gamma }\left(\alpha \right)}{C}_2^{\left(\alpha \right)}\left(x\right)& \left(\genfrac{}{}{0pt}{}{2}{1}\right)\frac{\mathsf{\Gamma }(1+\alpha )}{\pi \mathsf{\Gamma }\left(\alpha \right)}{C}_1^{\left(\alpha \right)}\left(x\right)& \cdots & 0\\ \\ ⋮& ⋮& \ddots & ⋮\\ \\ \left(\genfrac{}{}{0pt}{}{n}{n}\right)\frac{\mathsf{\Gamma }(n+\alpha )}{{\pi }^n\mathsf{\Gamma }\left(\alpha \right)}{C}_n^{\left(\alpha \right)}\left(x\right)& \left(\genfrac{}{}{0pt}{}{n}{n-1}\right)\frac{\mathsf{\Gamma }(n-1+\alpha )}{{\pi }^{n-1}\mathsf{\Gamma }\left(\alpha \right)}{C}_{n-1}^{\left(\alpha \right)}\left(x\right)& \cdots & C_0^{\left(\alpha \right)}\left(x\right)\end{array}\right],\alpha >-\frac{1}{2},\alpha \ne 0,$$

and from (7):

$${\mathbf{C}}^{\left(0\right)}\left(x\right)=\left[\begin{array}{cccc}1& 0& \cdots & 0\\ \\ \left(\genfrac{}{}{0pt}{}{1}{1}\right)\pi T_1\left(x\right)& 1& \cdots & 0\\ \\ \left(\genfrac{}{}{0pt}{}{2}{2}\right)\frac{1}{\pi }{T}_2\left(x\right)& \left(\genfrac{}{}{0pt}{}{2}{1}\right)T_1\left(x\right)& \cdots & 0\\ \\ ⋮& ⋮& \ddots & ⋮\\ \\ \left(\genfrac{}{}{0pt}{}{n}{n}\right)\frac{n!}{{\left(2\pi \right)}^{n-1}}{T}_n\left(x\right)& \left(\genfrac{}{}{0pt}{}{n}{n-1}\right)\frac{(n-1)!}{{\left(2\pi \right)}^{n-2}}{T}_{n-1}\left(x\right)& \cdots & 1\end{array}\right].$$

It is clear that the matrix ${\mathbf{C}}^{\left(\alpha \right)}\left(x\right)$ is a lower triangular matrix for each $x\in \mathbb{R}$, so that $det\left({\mathbf{C}}^{\left(\alpha \right)}\left(x\right)\right)=1.$ Therefore, ${\mathbf{C}}^{\left(\alpha \right)}\left(x\right)$ is a nonsingular matrix for each $x\in \mathbb{R}$ and $\alpha \in (-1/2,\infty )$.

Theorem 3.

For a fixed $m\in \mathbb{N}$ and any $\alpha \in (-1/2,\infty )$, let ${\left\{{\mathcal{V}}_n^{[m-1,\alpha ]}\left(x\right)\right\}}_{n\ge 0}$ be the sequence of HBG polynomials. Then, the following matrix representation holds.

$$\begin{array}{ccc}\hfill {\mathbf{V}}^{[m-1,\alpha ]}\left(x\right)& =& {\mathbf{C}}^{\left(\alpha \right)}\left(x\right){\mathbf{B}}^{[m-1]}\left(x\right),\hfill \end{array}$$

(25)

where ${\mathbf{V}}^{[m-1,\alpha ]}\left(x\right)={\left[\begin{array}{cccc}{\mathcal{V}}_0^{[m-1,\alpha ]}\left(x\right)& {\mathcal{V}}_1^{[m-1,\alpha ]}\left(x\right)& \cdots & {\mathcal{V}}_n^{[m-1,\alpha ]}\left(x\right)\end{array}\right]}^T$.

Proof. 

For each $r=0,1,\dots ,n$, consider the matrix form (24) of ${\mathcal{V}}_r^{[m-1,\alpha ]}\left(x\right)$. Then, it is not difficult to see that the matrix ${\mathbf{V}}^{[m-1,\alpha ]}\left(x\right)$ becomes

$${\mathbf{V}}^{[m-1,\alpha ]}\left(x\right)={\left[\begin{array}{cccc}{\mathcal{V}}_0^{[m-1,\alpha ]}\left(x\right)& {\mathcal{V}}_1^{[m-1,\alpha ]}\left(x\right)& \cdots & {\mathcal{V}}_n^{[m-1,\alpha ]}\left(x\right)\end{array}\right]}^T={\mathbf{C}}^{\left(\alpha \right)}\left(x\right){\mathbf{B}}^{[m-1]}\left(x\right),$$

and (25) follows. □

The following examples show how Theorem 3 can be used.

Example 1.

Let us consider $m=1,$$n=3$, and $\alpha =1,$ then,

$$\mathbf{B}\left(x\right)={\left({\mathbf{C}}^{\left(1\right)}\left(x\right)\right)}^{-1}{\mathbf{V}}^{[0,1]}\left(x\right)={\left[\begin{array}{cccc}1& 0& 0& 0\\ \\ \frac{x}{\pi }& 1& 0& 0\\ \\ \frac{4x^2-1}{2{\pi }^2}& \frac{2x}{\pi }& 1& 0\\ \\ \frac{6x^3-3x}{{\pi }^3}& \frac{3\left(4x^2-1\right)}{2{\pi }^2}& \frac{3x}{\pi }& 1\end{array}\right]}^{-1}{\mathbf{V}}^{[0,1]}\left(x\right),$$

(26)

where

$${\mathbf{V}}^{[0,1]}\left(x\right)=\left[\begin{array}{c}1\\ \\ \left(1+\frac{1}{\pi }\right)x-\frac{1}{2}\\ \\ \left(1+\frac{2}{\pi }+\frac{1}{{\pi }^2}\right)x^2-\left(1+\frac{1}{\pi }\right)x+\frac{1}{6}-\frac{1}{2{\pi }^2}\\ \\ \left(1+\frac{3}{\pi }+\frac{6}{{\pi }^2}+\frac{6}{{\pi }^3}\right)x^3-\frac{3}{2}\left(1+\frac{2}{\pi }+\frac{2}{{\pi }^2}\right)x^2+\frac{1}{2}\left(1+\frac{1}{\pi }-\frac{3}{{\pi }^2}-\frac{6}{{\pi }^3}\right)x+\frac{3}{4{\pi }^2}\end{array}\right].$$

Since

$${\left({\mathbf{C}}^{\left(1\right)}\left(x\right)\right)}^{-1}={\left[\begin{array}{cccc}1& 0& 0& 0\\ \\ \frac{x}{\pi }& 1& 0& 0\\ \\ \frac{4x^2-1}{2{\pi }^2}& \frac{2x}{\pi }& 1& 0\\ \\ \frac{6x^3-3x}{{\pi }^3}& \frac{3\left(4x^2-1\right)}{2{\pi }^2}& \frac{3x}{\pi }& 1\end{array}\right]}^{-1}=\left[\begin{array}{cccc}1& 0& 0& 0\\ \\ -\frac{x}{\pi }& 1& 0& 0\\ \\ \frac{1}{2{\pi }^2}& -\frac{2x}{\pi }& 1& 0\\ \\ 0& \frac{3}{2{\pi }^2}& -\frac{3x}{\pi }& 1\end{array}\right],$$

then (26) becomes

$$\mathbf{B}\left(x\right)=\left[\begin{array}{c}1\\ \\ x-\frac{1}{2}\\ \\ x^2-x+\frac{1}{6}\\ \\ x^3-\frac{3}{2}{x}^2+\frac{1}{2}x\end{array}\right].$$

That is, the entries of the matrix $\mathbf{B}\left(x\right)$ are the first four classical Bernoulli polynomials.

It is worth noting that for $\alpha =m=1$, the HBG polynomials ${\mathcal{V}}_n^{[0,1]}\left(x\right)$ coincide with the GBG polynomials ${\mathcal{V}}_n^{\left(1\right)}\left(x\right)$, for all $n\ge 0$ (cf. [4]).

Example 2.

Let $m=n=3$ and $\alpha =-\frac{1}{4}$. From (25), we obtain

$$\begin{array}{cc}\hfill {\mathbf{C}}^{\left(-\frac{1}{4}\right)}\left(x\right){\mathbf{B}}^{\left[2\right]}\left(x\right)& =\left[\begin{array}{cccc}1& 0& 0& 0\\ \\ -\frac{x}{4\pi }& 1& 0& 0\\ \\ -\frac{3}{16{\pi }^2}\left(x^2-\frac{2}{3}\right)& -\frac{x}{2\pi }& 1& 0\\ \\ -\frac{21}{64{\pi }^3}\left(x^3-\frac{6x}{7}\right)& -\frac{9}{16{\pi }^2}\left(x^2-\frac{2}{3}\right)& -\frac{3x}{4\pi }& 1\end{array}\right]\left[\begin{array}{c}6\\ \\ 6x-\frac{3}{2}\\ \\ 6x^2-3x+\frac{3}{20}\\ \\ 6x^3-\frac{9x^2}{2}+\frac{9x}{20}+\frac{3}{80}\end{array}\right]\hfill \\ & =\left[\begin{array}{c}6\\ \\ -\frac{3x}{2\pi }+6x-\frac{3}{2}\\ \\ \frac{-45x^2+6{\pi }^2\left(40x^2-20x+1\right)+30\pi (1-4x)x+30}{40{\pi }^2}\\ \\ \frac{3\left(-6{\pi }^{2}x\left(40x^2-20x+1\right)+15x\left(6-7x^2\right)-15\pi \left(12x^3-3x^2-8x+2\right)+{\pi }^3\left(320x^3-240x^2+24x+2\right)\right)}{160{\pi }^3}\end{array}\right].\hfill \end{array}$$

Straightforward calculations show that this last matrix coincides with

$${\mathbf{V}}^{\left[2,-\frac{1}{4}\right]}\left(x\right)=\left[\begin{array}{c}6\\ \\ 6\left(1-\frac{1}{4\pi }\right)x-\frac{3}{2}\\ \\ 6\left(1-\frac{1}{2\pi }-\frac{3}{16{\pi }^2}\right)x^2-3\left(1-\frac{1}{4\pi }\right)x+\frac{3}{20}+\frac{3}{4{\pi }^2}\\ \\ 6\left(1-\frac{3}{4\pi }-\frac{9}{16{\pi }^2}-\frac{21}{64{\pi }^2}\right)x^3-\frac{9}{2}\left(1-\frac{1}{2\pi }-\frac{3}{16{\pi }^2}\right)x^2+\frac{9}{4}\left(\frac{1}{5}-\frac{1}{20\pi }+\frac{1}{{\pi }^2}+\frac{3}{4{\pi }^3}\right)x+\frac{3}{80}-\frac{9}{16{\pi }^2}\end{array}\right].$$

Hence, ${\mathbf{C}}^{\left(-\frac{1}{4}\right)}\left(x\right){\mathbf{B}}^{\left[2\right]}\left(x\right)={\mathbf{V}}^{\left[2,-\frac{1}{4}\right]}\left(x\right)$.

We can now proceed as outlined in [5]. From the summation Formula (4) it follows

$$B_r^{[m-1]}\left(x\right)={\mathbf{M}}_r^{[m-1]}\mathbf{T}\left(x\right),r=0,1,\dots ,n,$$

where

$${\mathbf{M}}_r^{[m-1]}=\left[\begin{array}{ccccccc}\left(\genfrac{}{}{0pt}{}{r}{r}\right)B_r^{[m-1]}& \left(\genfrac{}{}{0pt}{}{r}{r-1}\right)B_{r-1}^{[m-1]}& \cdots & \left(\genfrac{}{}{0pt}{}{r}{0}\right)B_0^{[m-1]}& 0& \cdots & 0\end{array}\right],$$

(27)

the null entries of the matrix ${\mathbf{M}}_r^{[m-1]}$ appear $(n-r)$-times, and $\mathbf{T}\left(x\right)={\left[\begin{array}{cccc}1& x& \cdots & x^n\end{array}\right]}^T.$

Analogously, by (27) the matrix ${\mathbf{B}}^{[m-1]}\left(x\right)$, can be expressed as follows:

$$\begin{array}{ccc}\hfill {\mathbf{B}}^{[m-1]}\left(x\right)& =& {\mathbf{M}}^{[m-1]}\mathbf{T}\left(x\right)\hfill \\ \hfill & =& \left[\begin{array}{cccc}{B}_0^{[m-1]}& 0& \cdots & 0\\ \\ \left(\genfrac{}{}{0pt}{}{1}{1}\right)B_1^{[m-1]}& \left(\genfrac{}{}{0pt}{}{1}{0}\right)B_0^{[m-1]}& \cdots & 0\\ \\ \left(\genfrac{}{}{0pt}{}{2}{2}\right)B_2^{[m-1]}& \left(\genfrac{}{}{0pt}{}{2}{1}\right)B_1^{[m-1]}& \cdots & 0\\ \\ ⋮& ⋮& \ddots & ⋮\\ \\ \left(\genfrac{}{}{0pt}{}{n}{n}\right)B_n^{[m-1]}& \left(\genfrac{}{}{0pt}{}{n}{n-1}\right)B_{n-1}^{[m-1]}& \cdots & \left(\genfrac{}{}{0pt}{}{n}{0}\right)B_0^{[m-1]}\end{array}\right]\mathbf{T}\left(x\right).\hfill \end{array}$$

(28)

Notice that according to (27) the rows of the matrix ${\mathbf{M}}^{[m-1]}$ are precisely the matrices ${\mathbf{M}}_r^{[m-1]}$ for $r=0,\dots ,n$. Furthermore, the matrix ${\mathbf{M}}^{[m-1]}$ is a lower triangular matrix, so that $det\left({\mathbf{M}}^{[m-1]}\right)={(m!)}^{n+1}$. Therefore, ${\mathbf{M}}^{[m-1]}$ is a nonsingular matrix.

Another interesting algebraic property of the HBG polynomials is related with the following matrix-inversion formula.

Theorem 4.

For a fixed $m\in \mathbb{N}$ and any $\alpha \in (-1/2,\infty )$, let ${\left\{{\mathcal{V}}_n^{[m-1,\alpha ]}\left(x\right)\right\}}_{n\ge 0}$ be the sequence of HBG polynomials. Then, the following formula holds.

$$\mathbf{T}\left(x\right)={\left({\mathbf{Q}}^{[m-1,\alpha ]}\left(x\right)\right)}^{-1}{\mathbf{V}}^{[m-1,\alpha ]}\left(x\right),$$

(29)

where ${\mathbf{Q}}^{[m-1,\alpha ]}\left(x\right)={\mathbf{C}}^{\left(\alpha \right)}\left(x\right){\mathbf{M}}^{[m-1]}$.

Proof. 

Using the inversion Formulas (6), (25) and (28), and the nonsingularity of the matrices ${\mathbf{C}}^{\left(\alpha \right)}\left(x\right)$ and ${\mathbf{M}}^{[m-1]}$, it is possible to deduce that

$$\begin{array}{cc}\hfill \mathbf{T}\left(x\right)& ={\left({\mathbf{M}}^{[m-1]}\right)}^{-1}{\left({\mathbf{C}}^{\left(\alpha \right)}\left(x\right)\right)}^{-1}{\mathbf{V}}^{[m-1,\alpha ]}\left(x\right),\hfill \end{array}$$

and (29) follows. □

A simple and important consequence of Theorem 4 is:

Corollary 1.

For a fixed $m\in \mathbb{N}$ and any $\alpha \in (-1/2,\infty )$ the set $\left\{{\mathcal{V}}_0^{[m-1,\alpha ]}\left(x\right),\dots ,{\mathcal{V}}_n^{[m-1,\alpha ]}\left(x\right)\right\}$ is a basis for ${\mathbb{P}}_n$, $n\ge 0,$ i.e.,

$${\mathbb{P}}_n=span\left\{{\mathcal{V}}_0^{[m-1,\alpha ]}\left(x\right),{\mathcal{V}}_1^{[m-1,\alpha ]}\left(x\right),\dots ,{\mathcal{V}}_n^{[m-1,\alpha ]}\left(x\right)\right\}.$$

4. Conclusions

In the present paper, we introduced the mixed-type hypergeometric Bernoulli–Gegenbauer polynomials and analyzed some algebraic and differential properties of these polynomials, including their explicit expressions, derivative formulas, matrix representations, matrix-inversion formulas, and other relations connecting them with the hypergeometric Bernoulli polynomials. Furthermore, we demonstrated that unlike the hypergeometric Bernoulli polynomials and Gegenbauer polynomials, the HBG polynomials do not fulfill either Hanh or Appell conditions.

It is worth noting that the utilization of a matrix approach, specifically employing the operational matrix method based on hypergeometric Bernoulli polynomials, underpins several of our formulations. The matrix approaches using operational matrix methods associated with special polynomials and their practical applications constitute a relatively recent area of interest, as evidenced by the substantial body of literature (see, for instance, refs. [22,23,24,25,26,27,28] and the references therein). However, within the context of mixed special polynomials, to the best of our knowledge, there are no other published works that have adopted a similar approach, with the possible exception of a recent investigation [4].

Furthermore, we provided some examples to illustrate that the class of HBG polynomials does not generalize to the classical Bernoulli polynomials, although the latter can be recovered using Theorem 3. Unfortunately, the numerical evidence suggests that the zero distribution of the HBG polynomials does not align with the behavior of either Bernoulli hypergeometric polynomials or Gegenbauer polynomials.

Finally, by employing the determinantal approach introduced by Costabile and Longo [29], which implies that hypergeometric Bernoulli polynomials have a corresponding determinant form, and considering Theorem 3, it becomes feasible to investigate the determinantal forms associated with the HBG polynomials. Furthermore, Theorem 4 and the differential equation presented in part (f) of Proposition 1 (cf. [12], Theorem 3.1) suggest that the HBG polynomials satisfy a differential equation of order n. These two properties, along with their implications and potential applications, will be the focus of our future work.