Generating Functions for q-Apostol Type Frobenius–Euler Numbers and Polynomials

Abstract

The aim of this paper is to construct generating functions, related to nonnegative real parameters, for q-Eulerian type polynomials and numbers (or q-Apostol type Frobenius–Euler polynomials and numbers). We derive some identities for these polynomials and numbers based on the generating functions and functional equations. We also give multiplication formula for the generalized Apostol type Frobenius–Euler polynomials.

1. Introduction

Throughout this paper we assume that $q\in \mathbb{C}$, the set of complex numbers, with $\left|q\right|<1$ and

$$\left[x\right]=[x:q]=\left\{\begin{array}{c}\frac{1-q^x}{1-q}\text{,}\text{if}q\ne 1\\ x\text{,}\text{if}q=1\end{array}\right.$$

We use the following standard notions:

$\mathbb{N}=\left\{1,2,\cdots \right\}$, ${\mathbb{N}}_0=\left\{0,1,2,\cdots \right\}=\mathbb{N}\cup \left\{0\right\}$ and also, as usual $\mathbb{R}$ denotes the set of real number, ${\mathbb{R}}^{+}$ denotes the set of positive real number and $\mathbb{C}$ denotes the set of complex numbers.

In this section, we define q-Apostol type Frobenius–Euler polynomials and numbers, related to nonnegative real parameters. numbers).

Definition 1 

Let a, $b\in {\mathbb{R}}^{+}$ ($a\ne b$) and $u\in \mathbb{C}\setminus \left\{1\right\}$. A q-Apostol type Frobenius–Euler numbers

$${\mathcal{H}}_n(u;a,b;\lambda ;q)(\lambda ,q\in \mathbb{C})$$

are defined by means of the following generating function:

$$\begin{array}{c}\hfill F_{\lambda ,q}(t;u,a,b)=\left(1-\frac{a^t}{u}\right)\sum _{n=0}^{\infty }{\left(\frac{\lambda }{u}\right)}^n{b}^{\left[n\right]t}\\ \hfill =\sum _{n=0}^{\infty }{\mathcal{H}}_n(u;a,b;\lambda ;q)\frac{t^n}{n!}\end{array}$$

(1)

where

$$\left|\frac{\lambda }{u}\right|<1$$

Definition 2 

Let a, $b\in {\mathbb{R}}^{+}$ ($a\ne b$) and $u\in \mathbb{C}\setminus \left\{1\right\}$. A q-Apostol type Frobenius–Euler polynomials

$${\mathcal{H}}_n(x;u;a,b;\lambda ;q)(\lambda \in \mathbb{C})$$

are defined by means of the following generating function:

$$F_{\lambda ,q}(x,t;u,a,b)=\left(1-\frac{a^t}{u}\right)\sum _{n=0}^{\infty }{\left(\frac{\lambda }{u}\right)}^n{b}^{[n+x]t}=\sum _{n=0}^{\infty }{\mathcal{H}}_n(x;u;a,b;\lambda ;q)\frac{t^n}{n!}$$

(2)

where

$$\left|\frac{\lambda }{u}\right|<1$$

with, of course

$${\mathcal{H}}_n(0;u;a,b;\lambda ;q)={\mathcal{H}}_n(u;a,b;\lambda ;q)$$

where ${\mathcal{H}}_n(u;a,b;\lambda ;q)$ denotes the q-Apostol type Frobenius–Euler numbers.

By using the following well-known identity

$$[n+x]=\left[x\right]+q^x\left[n\right]$$

in Equation (2), we derive the following functional equation:

$$F_{\lambda ,q}(x,t;u,a,b)=F_{\lambda ,q}(q^{x}t;u,a,b)b^{t\left[x\right]}$$

(3)

By using Equation (3), we arrive at the following theorem:

Theorem 1 

Let $n\in {\mathbb{N}}_0$. Then we have

$${\mathcal{H}}_n(x;u;a,b;\lambda ;q)=\sum _{k=0}^n\left(\begin{array}{c}n\\ k\end{array}\right)q^{kx}{\left(\left[x\right]lnb\right)}^{n-k}{\mathcal{H}}_k(u;a,b;\lambda ;q)$$

By using Theorem 1, one can easily obtain the following result:

Corollary 1 

Let $n\in {\mathbb{N}}_0$. Then we have

$${\mathcal{H}}_n(x;u;a,b;\lambda ;q)={\left(\left[x\right]lnb+q^x\mathcal{H}(u;a,b;\lambda ;q)\right)}^n$$

(4)

with the usual convention of replacing ${\mathcal{H}}^j$ by ${\mathcal{H}}_j$.

Remark 1. 

One can easily see that

$$\begin{array}{ll}{\displaystyle \underset{q\to 1}{lim}}& F_{\lambda ,q}(x,t;u,a,b)=\frac{\left(a^t-u\right)b^{xt}}{\lambda b^t-u}\\ & ={\displaystyle \sum _{n=0}^{\infty }}{\mathcal{H}}_n(x;u;a,b;\lambda )\frac{t^n}{n!}\end{array}$$

(cf. [1]).

Remark 2. 

In their special case when ${\lambda }^r=1$ ($\lambda \ne 1$) and $a=1$ and $b=e$, the q-Apostol type Frobenius–Euler polynomials ${\mathcal{H}}_n(x;u;1,e;\lambda ;q)$ are reduced to the twisted q-Frobenius–Euler polynomials (cf. [2]).

Remark 3. 

Substituting $a=1$, $\lambda =1$ and $b=e$ into Equation (2), then we get the generating function of q-Frobenius–Euler numbers and polynomials respectively:

$$\left(1-\frac{1}{u}\right)\sum _{n=0}^{\infty }\frac{e^{\left[n\right]t}}{u^n}=\sum _{n=0}^{\infty }{H}_n(u,q)\frac{t^n}{n!}$$

(5)

and

$$\left(1-\frac{1}{u}\right)\sum _{n=0}^{\infty }\frac{e^{[n+x]t}}{u^n}=\sum _{n=0}^{\infty }{H}_n(x,u,q)\frac{t^n}{n!}$$

where $\mid u\mid >1$ (cf. [2,3]). Substituting $a=1$, $\lambda =1$ and $b=e$ into Equation (4), in this paper, we also generalize the Carlitz’s q-Frobenius–Euler polynomials $H_k(x,u,q)$ as follows:

$$H_n(u,x,q)={(q^{x}H+\left[x\right])}^n$$

(cf. [2,4,5,6,7,8,9]) and the Carlitz’s q-Frobenius–Euler numbers $H_k(u,q)$, which can be determined inductively by (cf. [4,6,7,8,9,10,11,12,13]):

$$H_0(u,q)=1$$

and for $n\ge 1$,

$${(qH+1)}^n-u{H}_n(u,q)=0$$

(cf. [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43]).

Remark 4. 

Substituting $u=-1$ in Equation (5), we have

$$2\sum _{n=0}^{\infty }{(-1)}^n{e}^{\left[n\right]t}=\sum _{n=0}^{\infty }{E}_n\left(q\right)\frac{t^n}{n!}$$

and also $H_n(x,-1,q)=E_n(x,q)$ (cf. [21]). If $q\to 1$, then Equation (5) reduces to the generating function for the classical Frobenius–Euler numbers $H_n\left(u\right)$:

$$\frac{1-u}{e^t-u}=\sum _{n=0}^{\infty }{H}_n\left(u\right)\frac{t^n}{n!}$$

(cf. [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43]).

In [1], we give the following generating function, which we need in the next section:

$$\frac{t}{\lambda a^t-1}{a}^{xt}=\sum _{n=0}^{\infty }{Y}_n(x;\lambda ;a)\frac{t^n}{n!}$$

(6)

We also note that

$$Y_n(0;\lambda ;a)=Y_n(\lambda ;a)$$

If we substitute $x=0$ and $a=1$ into Equation (6), we see that

$$Y_n(\lambda ;1)=\frac{1}{\lambda -1}$$

(cf. [1])

2. Identities

In this section, we derive some identities related to the q-Apostol type Frobenius–Euler numbers and polynomials, using generating functions.

We are now ready to give explicit formulas for computing the q-Apostol type Frobenius–Euler numbers and polynomials.

Theorem 2 

Let $n\in {\mathbb{N}}_0$. Then we have

$$\begin{gathered} {\mathcal{H}}_n(u;a,b;\lambda ;q)=u{\left(\frac{lnb}{1-q}\right)}^n\sum _{k=0}^n{(-1)}^k\left(\begin{array}{c}n\\ k\end{array}\right)\frac{1}{u-\lambda q^k} \\ +\sum _{k=0}^n{(-1)}^{k+1}\left(\begin{array}{c}n\\ k\end{array}\right)\frac{{\left(lna+\frac{lnb}{1-q}\right)}^{n-k}{\left(\frac{lnb}{1-q}\right)}^k}{u-\lambda q^k} \end{gathered}$$

Proof 1 

By using Equation (1), we get

$$\sum _{n=0}^{\infty }{\mathcal{H}}_n(u;a,b;\lambda ;q)\frac{t^n}{n!}=(1-\frac{a^t}{u})b^{\frac{t}{1-q}}\sum _{n=0}^{\infty }{\left(\frac{\lambda }{u}\right)}^n{b}^{-\frac{q^n}{1-q}t}$$

From the above equation, we obtain

$$\begin{gathered} \sum _{n=0}^{\infty }{\mathcal{H}}_n(u;a,b;\lambda ;q)\frac{t^n}{n!} \\ =\sum _{n=0}^{\infty }\left(u{\left(\frac{lnb}{1-q}\right)}^n\sum _{k=0}^n{(-1)}^k\left(\begin{array}{c}n\\ k\end{array}\right)\frac{1}{u-\lambda q^k}\right)\frac{t^n}{n!} \\ +\sum _{n=0}^{\infty }\left(\sum _{k=0}^n{(-1)}^{k+1}\left(\begin{array}{c}n\\ k\end{array}\right)\frac{{\left(lna+\frac{lnb}{1-q}\right)}^{n-k}{\left(\frac{lnb}{1-q}\right)}^k}{u-\lambda q^k}\right)\frac{t^n}{n!} \end{gathered}$$

Therefore, equating the coefficients of ${\displaystyle \frac{t^n}{n!}}$ on both sides of the above equation, we obtain the desired result.

Substituting $n=0$ into Theorem 2, we have

$${\mathcal{H}}_0(u;a,b;\lambda ;q)=\frac{u-1}{u-\lambda }$$

By using Theorem 2, all the q-Apostol type Frobenius–Euler numbers are easily computed.

Remark 5. 

If we put $a=1$, $u=-1$, $\lambda =1$ and $b=e$, then Theorem 2 reduces to Theorem 1 in [21].

If we substitute $a=1$, $\lambda =1$ and $b=e$ into Theorem 2, then we obtain an explicit formula, for the q-Frobenius–Euler numbers $H_n(u;q)={\mathcal{H}}_n(u;1,e;1;q)$, which is given by the following corollary:

Corollary 2 

Let $n\in {\mathbb{N}}_0$. Then we have

$${\mathcal{H}}_n(u;1,e;1;q)=(u-1){\left(\frac{1}{1-q}\right)}^n\sum _{k=0}^n{(-1)}^k\left(\begin{array}{c}n\\ k\end{array}\right)\frac{1}{u-q^k}$$

We give an explicit formula for the q-Apostol type Frobenius–Euler polynomials as follows:

Theorem 3 

Let $n\in {\mathbb{N}}_0$. Then we have

$$\begin{gathered} {\mathcal{H}}_n(x;u;a,b;\lambda ;q)=u\sum _{k=0}^n{(-1)}^k\left(\begin{array}{c}n\\ k\end{array}\right)\frac{q^{xk}{\left(\frac{lnb}{1-q}\right)}^n}{u-\lambda q^k} \\ +\sum _{k=0}^n{(-1)}^{k+1}\left(\begin{array}{c}n\\ k\end{array}\right)\frac{q^{xk}{\left(lna+\frac{lnb}{1-q}\right)}^{n-k}{\left(\frac{lnb}{1-q}\right)}^k}{u-\lambda q^k} \end{gathered}$$

Proof 2 

Proof of this theorem is the same as that of Theorem 2, so we omit it.

Remark 6. 

If we put $a=1$, $u=-1$, $\lambda =1$ and $b=e$, then Theorem 3 reduces to Theorem 2 in [21].

By using Equation (4) and Theorem 3, we arrive at the following result:

Corollary 3 

Let $n\in {\mathbb{N}}_0$. Then we have

$$\begin{gathered} {\left(\left[x\right]lnb+q^x\mathcal{H}(u;a,b;\lambda ;q)\right)}^n \\ =\sum _{k=0}^n{(-1)}^k\frac{\left(\begin{array}{c}n\\ k\end{array}\right)u{q}^{xk}{\left(\frac{lnb}{1-q}\right)}^n}{u-\lambda q^k}+\sum _{k=0}^n{(-1)}^{k+1}\frac{\left(\begin{array}{c}n\\ k\end{array}\right)q^{xk}{\left(lna+\frac{lnb}{1-q}\right)}^{n-k}{\left(\frac{lnb}{1-q}\right)}^k}{u-\lambda q^k} \end{gathered}$$

2.1. Multiplication Formula

Here we prove multiplication formula for the q-Apostol type Frobenius–Euler numbers and polynomials. This formula is very important to investigate fundamental properties of these polynomials.

Theorem 4 

Let $n\in \mathbb{N}$. Then we have

$$\begin{gathered} n{\left[f\right]}^{1-n}{\mathcal{H}}_{n-1}(fx;u;a,b;\lambda ;q) \\ =-u\sum _{j=0}^n\left(\begin{array}{c}n\\ j\end{array}\right)Y_{n-j}\left(\frac{1}{u^{2f}};a\right)\sum _{l=1}^f\frac{{\lambda }^l}{u^{l+f+1}}{\mathcal{H}}_j\left(x+\frac{l}{f};u^f;a,b;{\lambda }^f;q^f\right) \\ +\sum _{j=0}^n\left(\begin{array}{c}n\\ j\end{array}\right)Y_{n-j}\left(\frac{1}{\left[f\right]};\frac{1}{u^{2f}};a\right)\sum _{l=1}^f\frac{{\lambda }^l}{u^{l+f+1}}{\mathcal{H}}_j\left(x+\frac{l}{f};u^f;a,b;{\lambda }^f;q^f\right) \end{gathered}$$

Proof 3 

Substituting $n=l+mf$ with $m=0,1,\dots ,\infty$, $l=1,2,\dots ,f$ into Equation (2), we get

$$\sum _{n=0}^{\infty }{\mathcal{H}}_n(x;u;a,b;\lambda ;q)\frac{t^n}{n!}=(1-\frac{a^t}{u})\sum _{m=0}^{\infty }\sum _{l=1}^f{\left(\frac{\lambda }{u}\right)}^{l+mf}{b}^{[l+mf+x]t}$$

From the above equation, we obtain

$$\sum _{n=0}^{\infty }{\mathcal{H}}_n(x;u;a,b;\lambda ;q)\frac{t^n}{n!}=\frac{1-\frac{a^t}{u}}{1-\frac{a^{\left[f\right]t}}{u^f}}\sum _{l=1}^f{\left(\frac{\lambda }{u}\right)}^l\sum _{m=0}^{\infty }{\mathcal{H}}_n\left(\frac{x+l}{f};u^f;a,b;{\lambda }^f;q^f\right)\frac{{\left[f\right]}^n{t}^n}{n!}$$

By using Equation (6) in the right side of the above equation, we have

$$\begin{gathered} \left[f\right]\sum _{n=0}^{\infty }n{\mathcal{H}}_{n-1}(x;u;a,b;\lambda ;q)\frac{t^n}{n!} \\ =-\frac{u}{u^{f+1}}\sum _{m=0}^{\infty }{Y}_n\left(\frac{1}{u^{2f}};a\right)\frac{{\left[f\right]}^n{t}^n}{n!}\sum _{m=0}^{\infty }\sum _{l=1}^f{\left(\frac{\lambda }{u}\right)}^l{\mathcal{H}}_n\left(\frac{x+l}{f};u^f;a,b;{\lambda }^f;q^f\right)\frac{{\left[f\right]}^n{t}^n}{n!} \\ +\frac{1}{u^{f+1}}\sum _{m=0}^{\infty }{Y}_n\left(\frac{1}{\left[f\right]};\frac{1}{u^{2f}};a\right)\frac{{\left[f\right]}^n{t}^n}{n!}\sum _{m=0}^{\infty }\sum _{l=1}^f{\left(\frac{\lambda }{u}\right)}^l{\mathcal{H}}_n\left(\frac{x+l}{f};u^f;a,b;{\lambda }^f;q^f\right)\frac{{\left[f\right]}^n{t}^n}{n!} \end{gathered}$$

Thus, by using the Cauchy product in the above equation and then equating the coefficients of ${\displaystyle \frac{t^n}{n!}}$ on both sides of the resulting equation, we get the desired result.

If we put $a=1$ in Equation (2), we simplify Theorem 4 as follows:

$$\sum _{n=0}^{\infty }{\mathcal{H}}_n(x;u;1,b;\lambda ;q)\frac{t^n}{n!}=\frac{u^f-u^{f-1}}{u^f-1}\sum _{l=1}^f{\left(\frac{\lambda }{u}\right)}^l\sum _{m=0}^{\infty }{\mathcal{H}}_n\left(\frac{x+l}{f};u^f;a,b;{\lambda }^f;q^f\right)\frac{{\left[f\right]}^n{t}^n}{n!}$$

Equating the coefficients of ${\displaystyle \frac{t^n}{n!}}$ on both sides of the resulting equation, we obtain:

$${\mathcal{H}}_n(x;u;1,b;\lambda ;q)={\left[f\right]}^n\frac{u^f-u^{f-1}}{u^f-1}\sum _{l=1}^f{\left(\frac{\lambda }{u}\right)}^l{\mathcal{H}}_n\left(\frac{x+l}{f};u^f;1,b;{\lambda }^f;q^f\right)$$

Replacing x by $fx$ in the above equation, we get the following result:

Corollary 4 

Let $n\in {\mathbb{N}}_0$. Then we have

$${\mathcal{H}}_n(fx;u;1,b;\lambda ;q)={\left[f\right]}^n\frac{u^f-u^{f-1}}{u^f-1}\sum _{l=1}^f{\left(\frac{\lambda }{u}\right)}^l{\mathcal{H}}_n\left(x+\frac{l}{f};u^f;1,b;{\lambda }^f;q^f\right)$$

(7)

Remark 7. 

If we put $u=-1$, $\lambda =1$ and $b=e$, and $q\to 1$ in Equation (7), Equation (4,13) in [1].

Remark 8. 

If we put $u=-1$, $\lambda =1$ and $b=e$, for odd f, in Equation (7), we get Theorem 4 in [21].