Generalized q-Bernoulli Polynomials Generated by Jackson q-Bessel Functions

Abstract

In this paper, we introduce the polynomials $B^{(k)}_{n,\alpha }(x;q)$ generated by a function including Jackson q-Bessel functions $J^{(k)}_{\alpha }(x;q)$ $(k=1,2,3),\,\alpha >-1$. The cases $\alpha =\pm \frac{1}{2}$ are the q-analogs of Bernoulli and Euler$^{,}$s polynomials introduced by Ismail and Mansour for $(k=1,2)$, Mansour and Al-Towalib for $(k=3)$. We study the main properties of these polynomials, their large n degree asymptotics and give their connection coefficients with the q-Laguerre polynomials and little q-Legendre polynomials.

1 Introduction and Preliminaries

The Bernoulli polynomials $\left( B_{n}(x)\right) _{n}$ are defined by the generating function

$$\begin{aligned} \frac{te^{xt}}{e^{t}-1}=\sum _{n=0}^{\infty }B_{n}(x)\frac{t^{n}}{n!},\quad \mid t\mid <2\pi . \end{aligned}$$

In a series of papers, Frappier [9,10,11] studied the generalized Bernoulli polynomials $B_{n,\alpha }(x)$, defined by the generating function

$$\begin{aligned} \frac{e^{(x-\frac{1}{2})t}}{g_{\alpha }(\frac{it}{2})}=\sum _{n=0}^{\infty }B_{n,\alpha }(x)\frac{t^{n}}{n!},\quad \mid t\mid <2 j_{1,\alpha }, \end{aligned}$$

(1.1)

where

$$\begin{aligned} g_{\alpha }(t)=2^{\alpha }\Gamma (\alpha +1)\frac{ J_{\alpha }(t)}{t^{\alpha }}, \end{aligned}$$

$J_{\alpha }(t)$ is the Bessel function of the first kind of order $\alpha ,$ and $j_{1,\alpha }$ is the smallest positive zero of $J_{\alpha }(t)$. Ismail and Mansour, see [19], introduced a pair of q-analogs of the Bernoulli polynomials by the generating functions

$$\begin{aligned} \begin{aligned} \frac{te_{q}(xt)}{e_{q}(\frac{t}{2})E_{q}(\frac{t}{2})-1}&=\sum _{n=0}^{\infty }b_{n}(x;q)\frac{t^{n}}{[n]_{q}!},\\ \frac{t E_{q}(xt)}{e_{q}(\frac{t}{2})E_{q}(\frac{t}{2})-1}&=\sum _{n=0}^{\infty }B_{n}(x;q)\frac{t^{n}}{[n]_{q}!}.\end{aligned} \end{aligned}$$

(1.2)

They also defined a pair of q-analogs of the Euler polynomials by the generating functions

$$\begin{aligned} \begin{aligned} \frac{2e_{q}(xt)}{E_{q}(\frac{t}{2})e_{q}(\frac{t}{2})+1}&=\sum _{n=0}^{\infty }e_{n}(x;q)\frac{t^{n}}{[n]_{q}!},\\ \frac{2 E_{q}(xt)}{E_{q}(\frac{t}{2})e_{q}(\frac{t}{2})+1}&=\sum _{n=0}^{\infty }E_{n}(x;q)\frac{t^{n}}{[n]_{q}!},\end{aligned} \end{aligned}$$

(1.3)

where

$$\begin{aligned}\left[ n\right] _q!=\frac{(q;q)_{n}}{(1-q)^{n}}\quad (n\in {\mathbb {N}}), \quad (a;q)_{n}=\left\{ \begin{array}{ll} 1, &{}{} {n=0;} \\ \displaystyle \prod _{k=0}^{n-1}(1-aq^{k}), &{}{} {n\in {\mathbb {N}},} \end{array} \right. \end{aligned}$$

and $a\in {\mathbb {C}}$, see [12]. The functions $E_q(x)$ and $e_q(x)$ are the q-analogs of the exponential functions defined by

$$\begin{aligned} \begin{aligned} E_q(x):= (-x(1-q);q)_{\infty }&=\sum _{n=0}^{\infty }\frac{q^{\frac{n(n-1)}{2}}(1-q)^{n}x^{n}}{(q;q)_{n}}, \,\, x\in {\mathbb {C}}, \\ e_q(x):=\frac{1}{(x(1-q);q)_{\infty }}&=\sum _{n=0}^{\infty }\frac{(1-q)^{n}x^{n}}{(q;q)_{n}}, \,\, \mid x\mid < \frac{1}{1-q}, \end{aligned} \end{aligned}$$

(1.4)

see [12].

In [22], Mansour and Al-Towalib introduced q-analogs of Bernoulli and Euler polynomials by the generating functions

$$\begin{aligned} \begin{aligned}&\frac{t \exp _{q}(xt)\exp _{q}(\frac{-t}{2})}{\exp _{q}(\frac{t}{2})-\exp _{q}(\frac{-t}{2})}=\sum _{n=0}^{\infty }{\tilde{B}}_{n}(x;q)\frac{t^{n}}{[n]_{q}!},\\&\frac{2 \exp _{q}(xt)\exp _{q}(\frac{-t}{2})}{\exp _{q}(\frac{t}{2})+\exp _{q}(\frac{-t}{2})}=\sum _{n=0}^{\infty }{\tilde{E}}_{n}(x;q)\frac{t^{n}}{[n]_{q}!}, \end{aligned} \end{aligned}$$

(1.5)

where

$$\begin{aligned} \exp _{q}(x)=\sum _{n=0}^{\infty }q^{\frac{n(n-1)}{4}}\frac{x^{n}}{[n]_{q}!},\quad x\in {\mathbb {C}}, \end{aligned}$$

is a q-analog of the exponential function. This q-exponential function has the property $\lim _{q\rightarrow 1}\exp _{q}(x) = e^{x}$ for $x \in {\mathbb {C}}$. It is an entire function of x of order zero, see [12, Eq. (1.3.27), p. 12].

In this paper, we use ${\mathbb {N}}$ to denote the set of positive integers and ${\mathbb {N}}_{0}$ to denote the set of non-negative integers. Throughout this paper, unless otherwise is stated, q is a positive number that is less than one. We follow Gasper and Rahman [12] to define the q-shifted factorial, the q-binomial coefficients, and the q-gamma function. The q-integer number $[n]_{q}$ is defined by

$$\begin{aligned}{}[n]_{q}=\frac{1-q^n}{1-q},\quad n\in {\mathbb {N}}_0. \end{aligned}$$

Jackson in [20] defined the q-difference operator by

$$\begin{aligned} D_q f(z)= \dfrac{f(qz)-f(z)}{z(q-1)}, \quad z\ne 0. \end{aligned}$$

The symmetric q-difference operator is defined by, see [8, 12],

$$\begin{aligned} \delta _{q,z} f(z)= \frac{{}f(q^{\frac{1}{2}}z)-f(q^{\frac{-1}{2}}z)}{(q^{\frac{1}{2}}-q^{\frac{-1}{2}})z}, \quad z\ne 0. \end{aligned}$$

The q-trigonometric functions $\sin _{q}z,\, \cos _{q}z,\,Sin_{q}z$ and $Cos_{q} z$ are defined by

$$\begin{aligned} \begin{aligned}&\sin _{q}z=\dfrac{e_{q}(iz)-e_{q}(-iz)}{2i},\quad \cos _{q}z=\dfrac{e_{q}(iz)+e_{q}(-iz)}{2},\,\,\mid z\mid <1,\\&Sin_{q}z=\dfrac{E_{q}(iz)-E_{q}(-iz)}{2i},\quad Cos_{q}z=\dfrac{E_{q}(iz)+E_{q}(-iz)}{2},\,\,z\in {\mathbb {C}}, \end{aligned} \end{aligned}$$

see [5, 12]. The q-sine and cosine functions $S_{q}(z),\,\, C_{q}(z)$ are defined by the q-Euler formula

$$\begin{aligned} \exp _{q}(iz):=C_{q}(z)+iS_{q}(z), \end{aligned}$$

where

$$\begin{aligned} \begin{aligned} C_{q}(z)=\sum _{n=0}^{\infty }(-1)^{n}\frac{q^{n(n-\frac{1}{2})}}{[2n]_{q}!}z^{2n},\quad S_{q}(z)=\sum _{n=0}^{\infty }(-1)^{n}\frac{q^{n(n+\frac{1}{2})}}{[2n+1]_{q}!}z^{2n+1}, \end{aligned} \end{aligned}$$

cf. [8, p. 2]. The hyperbolic functions $Sh_{q}(z)$ and $Ch_{q}(z)$ are defined for $z\in {\mathbb {C}}$ by

$$\begin{aligned} \begin{aligned}&Sh_{q}(z):=-iS_{q}(iz)=\frac{\exp _{q}(z)-\exp _{q}(-z)}{2},\\&Ch_{q}(z):= C_{q}(iz)=\frac{\exp _{q}(z)+\exp _{q}(-z)}{2}. \end{aligned} \end{aligned}$$

(1.6)

There are three known q-analogs of the Bessel function that are due to Jackson [20]. These are denoted by $J^{(k)}_{\alpha }(t;q)\, (k=1,2,3)$ and defined by

$$\begin{aligned} J_{\alpha }^{(1)}(t;q)= & {} \frac{(q^{\alpha +1};q)_{\infty }}{(q;q)_{\infty }}\sum _{n=0}^{\infty }(-1)^n\frac{(\frac{t}{2})^{2n+\alpha }}{(q;q)_{n}(q^{\alpha +1};q)_{n}} \quad (\mid t\mid <2), \\ J_{\alpha }^{(2)}(t;q)= & {} \frac{(q^{\alpha +1};q)_{\infty }}{(q;q)_{\infty }}\sum _{n=0}^{\infty }(-1)^n\frac{q^{n(\alpha +n)}(\frac{t}{2})^{2n+\alpha }}{(q;q)_{n}(q^{\alpha +1};q)_{n}}\quad (t\in {\mathbb {C}} ), \\ J_{\alpha }^{(3)}(t;q)= & {} \frac{(q^{\alpha +1};q)_{\infty }}{(q;q)_{\infty }}\sum _{n=0}^{\infty }(-1)^n\frac{q^{\frac{n(n+1)}{2}}t^{2n+\alpha }}{(q;q)_{n}(q^{\alpha +1};q)_{n}} \quad (t\in {\mathbb {C}}). \end{aligned}$$

For convenience, we set

$$\begin{aligned} \begin{aligned}&{\mathcal {J}}_{\alpha }^{(k)} (t;q):=\frac{(q;q)_{\infty }}{(q^{\alpha +1};q)_{\infty }}(\frac{t}{2})^{-\alpha } J_{\alpha }^{(k)} (t;q)\quad ( k=1,2),\\&{\mathcal {J}}_{\alpha }^{(3)} (t;q):=\frac{(q;q)_{\infty }}{(q^{\alpha +1};q)_{\infty }}t^{-\alpha } J_{\alpha }^{(3)}(t;q).\end{aligned} \end{aligned}$$

(1.7)

The functions ${\mathcal {J}}_{\alpha }^{(k)} (t;q)\, (k=1,2,3)$ are called the modified Jackson q-Bessel functions. From now on, we use $(j^{(k)}_{m,\alpha })_{m=1}^{\infty }$ to denote the positive zeros of $J^{(k)}_{\alpha }(\cdot ;q^{2})$ arranged in increasing order of magnitude. Consequently, $j^{(k)}_{1,\alpha }$ is the smallest positive zero of $J^{(k)}_{\alpha }(\cdot ;q^{2})\,( k=1,2,3)$.

This paper is organized as follows. In Sect. 2, we introduce three q-analogs of the generalized Bernoulli polynomials defined in (1.1). The generating functions of these q-analogs include the three q-analogs of Jackson q-Bessel functions mentioned above. We also include the main properties of these q-analogs. Section 3 introduces a q-Fourier expansion for the generalized Bernoulli numbers related to the first and second Jackson q-Bessel functions. Also, their large n degree asymptotic is derived. Finally, in Sect. 4 as an application, we introduce the connection coefficients between q-analogs and certain q-orthogonal polynomials.

2 Generalized q-Bernoulli Polynomials Generated by Jackson q-Bessel Functions

This section introduces three q-analogs of the generalized Bernoulli polynomials introduced by Frappier in [9,10,11].

Definition 2.1

The generalized q-Bernoulli polynomials $B^{(k)}_{n,\alpha }(x;q)\,( k=1,2,3)$ are defined by the generating functions

$$\begin{aligned}&\frac{e_{q}(xt)e_{q}(\frac{-t}{2})}{g^{(1)}_{\alpha }(it;q)}=\sum _{n=0}^{\infty }B^{(1)}_{n,\alpha }(x;q)\frac{t^{n}}{[n]_{q}!},\quad \mid t\mid < \frac{ j^{(1)}_{1,\alpha }}{1-q}, \end{aligned}$$

(2.1)

$$\begin{aligned}&\frac{E_{q}(xt)E_{q}(\frac{-t}{2})}{g^{(2)}_{\alpha }(it;q)}=\sum _{n=0}^{\infty }B^{(2)}_{n,\alpha }(x;q)\frac{t^{n}}{[n]_{q}!},\quad \mid t\mid <\frac{ j^{(2)}_{1,\alpha }}{1-q}, \end{aligned}$$

(2.2)

$$\begin{aligned}&\frac{\exp _{q}(xt)\exp _{q}(\frac{-t}{2})}{g^{(3)}_{\alpha }(it;q)}=\sum _{n=0}^{\infty }B^{(3)}_{n,\alpha }(x;q)\frac{t^{n}}{[n]_{q}!}, \quad \mid t\mid <\frac{2q^{\frac{1}{4}}j^{(3)}_{1,\alpha }}{1-q}, \end{aligned}$$

(2.3)

where $g^{(k)}_{\alpha }(t;q) \,( k=1,2,3 )$ are the functions defined for $(k=1,2)$ by

$$\begin{aligned} \begin{aligned} g^{(k)}_{\alpha }(t;q):=(1+q)^{\alpha }\Gamma _{q^{2}}(\alpha +1)\,\left( \frac{t}{2}\right) ^{-\alpha }\,J_{\alpha }^{(k)}(t(1-q);q^{2})={\mathcal {J}}_{\alpha }^{(k)} (t(1-q);q^{2}), \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} g^{(3)}_{\alpha }(t;q)\,{{:\!}=}\,&(1+q)^{\alpha }\Gamma _{q^{2}}(\alpha +1)\,\left( \frac{ q^{\frac{-1}{4}}t}{2}\right) ^{-\alpha }\,J_{\alpha }^{(3)}\left( \frac{t}{2}(1-q)q^{\frac{-1}{4}};q^{2}\right) \\ =\,&{\mathcal {J}}_{\alpha }^{(3)}\left( \frac{t}{2}(1-q)q^{\frac{-1}{4}};q^{2}\right) . \end{aligned} \end{aligned}$$

Since the generating functions in (1.2), (1.3), and (1.5) can be written as

$$\begin{aligned}&\frac{te_{q}(xt)e_{q}(\frac{-t}{2})}{2\sinh _{q}\frac{t}{2}}=\sum _{n=0}^{\infty }b_{n}(x;q)\frac{t^{n}}{[n]_{q}!},\nonumber \\&\frac{t E_{q}(xt)E_{q}(\frac{-t}{2})}{2Sinh_{q}\frac{t}{2}}=\sum _{n=0}^{\infty }B_{n}(x;q)\frac{t^{n}}{[n]_{q}!}, \end{aligned}$$

(2.4)

$$\begin{aligned}&\frac{e_{q}(xt)e_{q}(\frac{-t}{2})}{\cosh _{q}\frac{t}{2}}=\sum _{n=0}^{\infty }e_{n}(x;q)\frac{t^{n}}{[n]_{q}!},\nonumber \\&\frac{ E_{q}(xt)E_{q}(\frac{-t}{2})}{Cosh_{q}\frac{t}{2}}=\sum _{n=0}^{\infty }E_{n}(x;q)\frac{t^{n}}{[n]_{q}!}, \end{aligned}$$

(2.5)

and

$$\begin{aligned} \begin{aligned}&\frac{t \exp _{q}(xt)\exp _{q}(\frac{-t}{2})}{2Sh_{q}(\frac{t}{2})}=\sum _{n=0}^{\infty }{\tilde{B}}_{n}(x;q)\frac{t^{n}}{[n]_{q}!},\\&\frac{ \exp _{q}(xt)\exp _{q}(\frac{-t}{2})}{Ch_{q}(\frac{t}{2})}=\sum _{n=0}^{\infty }{\tilde{E}}_{n}(x;q)\frac{t^{n}}{[n]_{q}!}, \end{aligned} \end{aligned}$$

(2.6)

then, if we substitute with $\alpha =\pm \frac{1}{2}$ in (2.1), (2.2), and (2.3), we obtain the q-Bernoulli and Euler polynomials defined in (2.4), (2.5) and (2.6), respectively.

Lemma 2.2

For $n\in {\mathbb {N}}_{0}$ and $Re \,\alpha >-1$,

$$\begin{aligned} \frac{e_{q}(\frac{-t}{2})}{ g^{(1)}_{\alpha }(it;q)}=\frac{E_{q}(\frac{-t}{2})}{ g^{(2)}_{\alpha }(it;q)},\,\,\, \mid t \mid <\frac{1}{1-q}\min \{ j^{(1)}_{1,\alpha }, j^{(2)}_{1,\alpha },2\}. \end{aligned}$$

Proof

Hahn in [14] proved the identity

$$\begin{aligned} J^{(2)}_{\alpha }(t;q)=\left( \frac{-t^{2}}{4};q\right) _{\infty }\,J^{(1)}_{\alpha }(t;q),\quad \mid t\mid <2. \end{aligned}$$

(2.7)

Since

$$\begin{aligned} \begin{aligned}&g^{(k)}_{\alpha }(it;q)=(1+q)^{\alpha }\Gamma _{q^{2}}(\alpha +1)\,\left( \frac{it}{2}\right) ^{-\alpha }\,J_{\alpha }^{(k)}(it(1-q);q^{2})\,\, (k=1,2),\end{aligned} \end{aligned}$$

(2.8)

then, substituting from (2.8) into (2.7), we conclude that

$$\begin{aligned} g^{(2)}_{\alpha }(it;q)=\left( \frac{t^{2}}{4}(1-q)^{2};q^{2}\right) _{\infty }\,{g^{(1)}_{\alpha }(it;q)}= E_{q}\left( \frac{t}{2}\right) E_{q}\left( \frac{-t}{2}\right) \, {g^{(1)}_{\alpha }(it;q)}.\nonumber \\ \end{aligned}$$

(2.9)

Hence

$$\begin{aligned} \begin{aligned} \frac{E_{q}(\frac{-t}{2})}{ g^{(2)}_{\alpha }(it;q)}= \frac{E_{q}(\frac{-t}{2})}{ E_{q}(\frac{t}{2})E_{q}(\frac{-t}{2})\, g^{(1)}_{\alpha }(it;q)}=\frac{e_{q}(\frac{-t}{2})}{ g^{(1)}_{\alpha }(it;q)},\end{aligned} \end{aligned}$$

which completes the proof. $\square$

Definition 2.3

The generalized q-Bernoulli numbers ${\beta }_{n,\alpha }(q),$ $\beta ^{(3)}_{n,\alpha }(q)$ are defined respectively in terms of the generating functions

$$\begin{aligned} \frac{e_{q}(\frac{-t}{2})}{g^{(1)}_{\alpha }(it;q)}&=\frac{E_{q}(\frac{-t}{2})}{g^{(2)}_{\alpha }(it;q)} =\sum _{n=0}^{\infty }\beta _{n,\alpha }(q)\frac{t^{n}}{[n]_{q}!}, \end{aligned}$$

(2.10)

$$\begin{aligned} \frac{\exp _{q}(\frac{-t}{2})}{g^{(3)}_{\alpha }(it;q)}&=\sum _{n=0}^{\infty }\beta ^{(3)}_{n,\alpha }(q)\frac{t^{n}}{[n]_{q}!}. \end{aligned}$$

(2.11)

Proposition 2.4

For $n\in {\mathbb {N}}$, we have

$$\begin{aligned} B^{(k)}_{2n+1,\alpha }\left( \frac{1}{2};q\right) =0\quad (k=1,2,3). \end{aligned}$$

Proof

If we substitute with $x=\frac{1}{2}$ in Eqs. (2.1)–(2.3), we find that their left hand side are even functions. Therefore, the coefficients of the odd powers of $t^{n}$ on the right hand sides of Eqs. (2.1)–(2.3) vanish. This proves the proposition. $\square$

Proposition 2.5

For $k\in \{1,2,3\}$ and $n\in {\mathbb {N}}$, the polynomials $B_{n,\alpha }^{(k)}(x;q)$ have the representation $B_{0,\alpha }^{(k)}(x;q)=1$,

$$\begin{aligned} B^{(1)}_{n,\alpha }(x;q)= & {} \sum _{k=0}^{n}\left[ \begin{array}{c} n \\ k \\ \end{array} \right] _{q} \beta _{n-k,\alpha } (q) x^{k}, \end{aligned}$$

(2.12)

$$\begin{aligned} \quad \quad \, B^{(2)}_{n,\alpha }(x;q)= & {} \sum _{k=0}^{n}\left[ \begin{array}{c} n \\ k \\ \end{array} \right] _{q} q^{\frac{k(k-1)}{2}}\beta _{n-k,\alpha }(q) x^{k}, \end{aligned}$$

(2.13)

$$\begin{aligned} \quad \quad \, B^{(3)}_{n,\alpha }(x;q)= & {} \sum _{k=0}^{n}\left[ \begin{array}{c} n \\ k \\ \end{array} \right] _{q} q^{\frac{k(k-1)}{4}}\beta ^{(3)}_{n-k,\alpha }(q) x^{k}. \end{aligned}$$

(2.14)

Proof

We prove the case $(k=1)$. The proofs for $(k=2,3)$ are similar and are omitted. Substituting with the series representation of $e_{q}(x)$ from (1.4) into (2.1) gives

$$\begin{aligned} \begin{aligned} \sum _{n=0}^{\infty }B^{(1)}_{n,\alpha }(x;q)\frac{t^{n}}{[n]_{q}!}&=\frac{e_{q}(\frac{-t}{2})}{g^{(1)}_{\alpha }(it;q)}e_{q}(xt)\\&=\left( \sum _{n=0}^{\infty } \beta _{n,\alpha }(q)\frac{t^{n}}{[n]_{q}!}\right) \left( \sum _{n=0}^{\infty }\frac{(xt)^{n}}{[n]_{q}!}\right) . \end{aligned} \end{aligned}$$

Hence

$$\begin{aligned} \sum _{n=0}^{\infty }B^{(1)}_{n,\alpha }(x;q)\frac{t^{n}}{[n]_{q}!}= \sum _{n=0}^{\infty }\frac{t^{n}}{[n]_{q}!}\sum _{k=0}^{n} \left[ \begin{array}{c} n \\ k \\ \end{array} \right] _{q} \beta _{n-k,\alpha }\,x^{k}, \end{aligned}$$

(2.15)

where we applied the Cauchy product formula. Equating the nth power of t in (2.15), we obtain (2.12). $\square$

Proposition 2.6

For $n\in {\mathbb {N}}$ and $k\in \{1,2,3\}$, the polynomials $B_{n,\alpha }^{(k)}(x;q)$ satisfy the q-difference equations

$$\begin{aligned} D_{q,x}\,B^{(1)}_{n,\alpha }(x;q)&=[n]_{q}\,B^{(1)}_{n-1,\alpha }(x;q), \end{aligned}$$

(2.16)

$$\begin{aligned} D_{q^{-1},x}\,B^{(2)}_{n,\alpha }(x;q)&=[n]_{q}\,B^{(2)}_{n-1,\alpha }(x;q), \end{aligned}$$

(2.17)

$$\begin{aligned} \delta _{q,x}\,B^{(3)}_{n,\alpha }(x;q)&=[n]_{q}B^{(3)}_{n-1,\alpha }(x;q). \end{aligned}$$

(2.18)

Proof

We only prove the case ($k=1$) and the proofs of $(k=2,3)$ are similar. Calculating the q-derivative of both sides of (2.1) with respect to the variable x and taking into consideration that

$$\begin{aligned} D_{q,x}\,e_{q}(xt)=t\,e_{q}(xt), \end{aligned}$$

we obtain

$$\begin{aligned} \frac{t e_{q}(xt)e_{q}(\frac{-t}{2})}{g^{(1)}_{\alpha }(it;q)}=\sum _{n=1}^{\infty }D_{q,x} B^{(1)}_{n,\alpha }(x;q)\frac{t^{n}}{[n]_{q}!}. \end{aligned}$$

Therefore,

$$\begin{aligned} \sum _{n=0}^{\infty } B^{(1)}_{n,\alpha }(x;q)\frac{t^{n+1}}{[n]_{q}!}=\sum _{n=1}^{\infty }D_{q,x} B^{(1)}_{n,\alpha }(x;q)\frac{t^{n}}{[n]_{q}!}. \end{aligned}$$

(2.19)

Equating the corresponding nth power of t in (2.19), we obtain (2.16). $\square$

Corollary 2.7

Let $n\in {\mathbb {N}}$ and k be a positive integer such that $k\le n$. Then for $x\in {\mathbb {C}},$

$$\begin{aligned} \begin{aligned} D^{k}_{q,x}\,\frac{B^{(1)}_{n,\alpha }(x;q)}{[n]_{q}!}&=\frac{B^{(1)}_{n-k,\alpha }(x;q) }{[n-k]_{q}!},\\ D^{k}_{q^{-1},x}\,\frac{B^{(2)}_{n,\alpha }(x;q)}{[n]_{q}!}&=\frac{B^{(2)}_{n-k,\alpha }(x;q) }{[n-k]_{q}!},\\ \delta ^{k}_{q,x}\,\frac{B^{(3)}_{n,\alpha }(x;q)}{[n]_{q}!}&=\frac{B^{(3)}_{n-k,\alpha }(x;q)}{[n-k]_{q}!}.\end{aligned} \end{aligned}$$

Proof

The proofs follow from Proposition 2.6 and the mathematical induction. $\square$

Proposition 2.8

For $\mid t\mid <\frac{1}{1-q}\min \{ j^{(1)}_{1,\alpha }, j^{(2)}_{1,\alpha },2\}$,

$$\begin{aligned}&\sum _{n=0}^{\infty } B^{(1)}_{n,\alpha }\left( \frac{1}{2};q\right) \frac{t^{n}}{[n]_{q}!}=\frac{1}{g^{(2)}_{\alpha }(it;q)}. \end{aligned}$$

(2.20)

$$\begin{aligned}&\sum _{n=0}^{\infty } B^{(2)}_{n,\alpha }\left( \frac{1}{2};q\right) \frac{t^{n}}{[n]_{q}!}=\frac{1}{g^{(1)}_{\alpha }(it;q)}. \end{aligned}$$

(2.21)

Proof

Set $x=\frac{1}{2}$ in (2.1), we obtain

$$\begin{aligned} \frac{e_{q}(\frac{t}{2})e_{q}(\frac{-t}{2})}{g^{(1)}_{\alpha }(it;q)}=\sum _{n=0}^{\infty }B^{(1)}_{n,\alpha }\left( \frac{1}{2};q\right) \frac{t^{n}}{[n]_{q}!}. \end{aligned}$$

(2.22)

Substituting from (2.9) into (2.22), we obtain (2.20). Similarly, we can prove (2.21). $\square$

The following Lemma from [22] gives the reciprocal of $\exp _{q}(z)$ in a certain domain.

Lemma 2.9

Let $z\in \Omega , \,\displaystyle \Omega := \{ z\in {\mathbb {C}}: \, \mid 1-\exp _{q}(-z) \mid < 1\}$. Then

$$\begin{aligned} \dfrac{1}{\exp _{q}(z)}:=\sum _{n=0}^{\infty } c_n\,z^{n}, \end{aligned}$$

where

$$\begin{aligned} c_n=\, \sum _{k=1}^{n} (-1)^{k}\sum _{\begin{array}{c} {s_1+s_2+\ldots +s_k=n} \\ {s_i>0\,(i=1,\ldots ,k)} \end{array}} \dfrac{ q^{ \sum _{i=1}^k s_i(s_i-1)/4 }}{[s_1]_q! [s_2]_q!\ldots [s_{k}]_q!}. \end{aligned}$$

(2.23)

Proposition 2.10

For $Re\,\alpha >-1$ and $t\in \displaystyle \Omega = \{ t\in {\mathbb {C}}: \, \mid 1-\exp _{q}(-t)\mid < 1\}$,

$$\begin{aligned} \frac{1}{g^{(3)}_{\alpha }(it;q)}=\sum _{n=0}^{\infty }t^{n} \sum _{k=0}^{n} \frac{(-1)^{k}\,c_{k}}{2^{k}[n-k]_{q}!}\beta ^{(3)}_{n-k,\alpha }(q), \end{aligned}$$

(2.24)

where $c_{n}$ is defined in (2.23).

Proof

Substitute with $x=0$ in Eq. (2.3). This gives

$$\begin{aligned} \frac{\exp _{q}(\frac{-t}{2})}{g^{(3)}_{\alpha }(it;q)}=\sum _{n=0}^{\infty }\beta ^{(3)}_{n,\alpha }(q)\frac{t^{n}}{[n]_{q}!}. \end{aligned}$$

From Lemma 2.9,

$$\begin{aligned} \begin{aligned} \frac{1}{g^{(3)}_{\alpha }(it;q)}&=\frac{1}{\exp _{q}(\frac{-t}{2})}\sum _{n=0}^{\infty }\beta ^{(3)}_{n,\alpha }(q)\frac{t^{n}}{[n]_{q}!}\\&= \left( \sum _{n=0}^{\infty }c_{n}\frac{(-1)^{n}t^{n}}{2^{n}}\right) \left( \sum _{n=0}^{\infty }\beta ^{(3)}_{n,\alpha }(q)\frac{t^{n}}{[n]_{q}!}\right) . \end{aligned} \end{aligned}$$

Applying the Cauchy product formula, we obtain (2.24) and completes the proof. $\square$

Theorem 2.11

For $n\in {\mathbb {N}}_{0}$ and $x\in {\mathbb {C}}$,

$$\begin{aligned} \sum _{k =0}^{n}\left[ \begin{array}{c} n \\ k \\ \end{array} \right] _{q} B^{(1)}_{k,\alpha } (-x;q) B^{(2)}_{n-k,\alpha } (x;q)= \sum _{k =0}^{n}\left[ \begin{array}{c} n \\ k \\ \end{array} \right] _{q} \beta _{k,\alpha }(q) \beta _{n-k,\alpha } (q). \end{aligned}$$

Proof

If we replace x by $-x$ in (2.1), then

$$\begin{aligned} \frac{e_{q}(-xt)e_{q}(\frac{-t}{2})}{g^{(1)}_{\alpha }(it;q)}=\sum _{n=0}^{\infty }B^{(1)}_{n,\alpha }(-x;q)\frac{t^{n}}{[n]_{q}!}. \end{aligned}$$

(2.25)

Since $e_{q}(-xt)E_{q}(xt)=1,$ then multiplying (2.2) by (2.25) gives

$$\begin{aligned} \frac{E_{q}(\frac{-t}{2})e_{q}(\frac{-t}{2})}{g^{(2)}_{\alpha }(it;q)g^{(1)}_{\alpha }(it;q)}=\left( \sum _{n=0}^{\infty } B^{(1)}_{n,\alpha }(-x;q)\frac{t^{n}}{[n]_{q}!}\right) \left( \sum _{n=0}^{\infty }B^{(2)}_{n,\alpha }(x;q)\frac{t^{n}}{[n]_{q}!}\right) . \end{aligned}$$

From (2.10), we obtain

$$\begin{aligned} \left( \sum _{n=0}^{\infty } \beta _{n,\alpha }(q)\frac{t^{n}}{[n]_{q}!}\right) ^{2}=\sum _{n=0}^{\infty }\frac{t^{n}}{[n]_{q}!}\sum _{k=0}^{n} \left[ \begin{array}{c} n \\ k \\ \end{array} \right] _{q} B^{(1)}_{k,\alpha } (-x;q) B^{(2)}_{n-k,\alpha } (x;q). \end{aligned}$$

Hence

$$\begin{aligned}&\sum _{n=0}^{\infty }\frac{t^{n}}{[n]_{q}!}\sum _{k=0}^{n} \left[ \begin{array}{c} n \\ k \\ \end{array} \right] _{q} \beta _{k,\alpha }(q) \beta _{n-k,\alpha }(q) \nonumber \\&\quad = \sum _{n=0}^{\infty }\frac{t^{n}}{[n]_{q}!}\sum _{k=0}^{n} \left[ \begin{array}{c} n \\ k \\ \end{array} \right] _{q} B^{(1)}_{k,\alpha } (-x;q) B^{(2)}_{n-k,\alpha } (x;q). \end{aligned}$$

(2.26)

So, equating the nth power of t in (2.26), we obtain the required result. $\square$

Proposition 2.12

For $n\in {\mathbb {N}}_{0},$ $x\in {\mathbb {C}}$ and $q\ne 0,$

$$\begin{aligned} B^{(2)}_{n,\alpha } (x;q)=q^{\frac{n(n-1)}{2}} B^{(1)}_{n,\alpha } \left( x;\frac{1}{q}\right) . \end{aligned}$$

(2.27)

In particular,

$$\begin{aligned} \beta _{n,\alpha } (q)=q^{\frac{n(n-1)}{2}} \beta _{n,\alpha } \left( \frac{1}{q}\right) . \end{aligned}$$

(2.28)

Proof

Replacing q by $\frac{1}{q}$ on the generating function in (2.1) and using $E_{q}(x)=e_{\frac{1}{q}}(x)$, we obtain

$$\begin{aligned} \frac{E_{q}(xt)E_{q}(\frac{-t}{2})}{g^{(1)}_{\alpha }(it;\frac{1}{q})}=\sum _{n=0}^{\infty }B^{(1)}_{n,\alpha }\left( x;\frac{1}{q}\right) \frac{t^{n}}{[n]_{\frac{1}{q}}!}. \end{aligned}$$

(2.29)

Since

$$\begin{aligned} \begin{aligned} g_{\alpha }^{(1)}\left( it;\frac{1}{q}\right)&=\sum _{n=0}^{\infty }\frac{(1-q^{-1})^{2n}(\frac{t}{2})^{2n}}{(q^{-2};q^{-2})_{n}(q^{-2\alpha -2};q^{-2})_{n}}\\&=\sum _{n=0}^{\infty }\frac{(1-q)^{2n}q^{2n(n+\alpha )}(\frac{t}{2})^{2n}}{(q^{2},q^{2\alpha +2};q^{2})_{n}}=g_{\alpha }^{(2)}(it;q), \end{aligned} \end{aligned}$$

where we used the identity $(a;q^{-1})_{n}=(a^{-1};q)_{n}(-a)^{n}q^{-\frac{n(n-1)}{2}}$. Since $[n]_{1/q}!=q^{\frac{n(1-n)}{2}}[n]_{q}!$, then (2.29) takes the form

$$\begin{aligned} \frac{E_{q}(xt)E_{q}(\frac{-t}{2})}{g^{(2)}_{\alpha }(it;q)}=\sum _{n=0}^{\infty }B^{(1)}_{n,\alpha }\left( x;\frac{1}{q}\right) q^{\frac{n(n-1)}{2}}\frac{t^{n}}{[n]_{q}!}. \end{aligned}$$

Therefore,

$$\begin{aligned} \sum _{n=0}^{\infty }B^{(2)}_{n,\alpha }(x;q)\frac{t^{n}}{[n]_{q}!}=\sum _{n=0}^{\infty }B^{(1)}_{n,\alpha }\left( x;\frac{1}{q}\right) \frac{q^{\frac{n(n-1)}{2}}t^{n}}{[n]_{q}!}. \end{aligned}$$

(2.30)

Equating the coefficients of $t^{n}$ in (2.30) gives (2.27) and substituting with $x=0$ into (2.27) yields directly (2.28). $\square$

Al-Salam in [3] introduced the polynomials

$$\begin{aligned} H_{n}(x):=\sum _{k=0}^{n} \, \left[ \begin{array}{c} n \\ k \\ \end{array} \right] _{q}\,x^{k} ,\quad G_{n}(x):=\sum _{k=0}^{n} \, \left[ \begin{array}{c} n \\ k \\ \end{array} \right] _{q}\, q^{k^{2}-nk} x^{k}. \end{aligned}$$

(2.31)

He also proved that

$$\begin{aligned}&E_{q}(x) E_{q}(-x)=\sum _{n=0}^{\infty } q^{\frac{n(n-1)}{2}} G_{n}(-1)\frac{x^{n}}{[n]_{q}!} ,\quad x \in \,{\mathbb {C}}, \end{aligned}$$

(2.32)

$$\begin{aligned}&e_{q}(x) e_{q}(-x)=\sum _{n=0}^{\infty } H_{n}(-1)\frac{x^{n}}{[n]_{q}!} ,\quad \mid x \mid \,<\,\frac{1}{1-q}. \end{aligned}$$

(2.33)

The following theorem introduces connection relations between the polynomials $B^{(1)}_{n,\alpha }(x;q)$ and $B^{(2)}_{n,\alpha }(x;q)$.

Theorem 2.13

For $n \in {\mathbb {N}}_{0}$,

$$\begin{aligned} B^{(1)}_{n,\alpha }(x;q)&=\sum _{k=0}^{n} \, \left[ \begin{array}{c} n \\ k \\ \end{array} \right] _{q}\, x^{k}H_{k}(-1) B^{(2)}_{n-k,\alpha }(x;q), \end{aligned}$$

(2.34)

$$\begin{aligned} B^{(2)}_{n,\alpha }(x;q)&=\sum _{k=0}^{n} \, \left[ \begin{array}{c} n \\ k \\ \end{array} \right] _{q}\, q^{\frac{k(k-1)}{2}}x^{k}G_{k}(-1) B^{(1)}_{n-k,\alpha }(x;q). \end{aligned}$$

(2.35)

Proof

Since $E_{q}(xt) e_{q}(-xt)= 1,\,\mid xt \mid <\frac{1}{1-q},$ then from (2.9), the generating function of $B^{(1)}_{n,\alpha }(x;q)$ can be represented as

$$\begin{aligned} \begin{aligned} \frac{e_{q}(xt)e_{q}(\frac{-t}{2})}{g^{(1)}_{\alpha }(it;q)}=\frac{E_{q}(xt) E_{q}(\frac{-t}{2})}{g^{(2)}_{\alpha }(it;q)}e_{q}(xt)e_{q}(-xt). \end{aligned} \end{aligned}$$

From (2.1), (2.2) and (2.33), we obtain

$$\begin{aligned} \begin{aligned} \sum _{n=0}^{\infty }B^{(1)}_{n,\alpha }(x;q)\frac{t^{n}}{[n]_{q}!}&= \left( \sum _{n=0}^{\infty }B^{(2)}_{n,\alpha }(x;q)\frac{t^{n}}{[n]_{q}!}\right) \left( \sum _{n=0}^{\infty } H_{n}(-1)\frac{(xt)^{n}}{[n]_{q}!}\right) \\&=\sum _{n=0}^{\infty } \frac{t^{n}}{[n]_{q}!} \sum _{k=0}^{n} \, \left[ \begin{array}{c} n \\ k \\ \end{array} \right] _{q}\, x^{k}H_{k}(-1) B^{(2)}_{n-k,\alpha }(x;q).\end{aligned}\nonumber \\ \end{aligned}$$

(2.36)

Therefore, equating the coefficients of the nth power of t in the series of the outside parts of (2.36) gives (2.34). The proof for $B^{(2)}_{n,\alpha }(x;q)$ follows similarly from the generating function of $B^{(2)}_{n,\alpha }(x;q)$ and the identity (2.32), and is omitted. $\square$

Theorem 2.14

Let x and $\alpha$ be two complex numbers, with $Re\,\alpha >-1$, and n a positive integer. Then

$$\begin{aligned} \begin{aligned} \sum _{k = 0}^{[\frac{n}{2}]} \, \frac{ (1-q)^{2k} \,B^{(1)}_{n-2k,\alpha }(-\frac{x}{2};q)}{2^{2k}\,[n-2k]_{q}!\, (q^{2},q^{2n+\alpha };q^{2})_{k}}&=\frac{(-1/2)^{n}}{[n]_{q}!} H_{n}(x),\\ \sum _{k =0 }^{[\frac{n}{2}]} \, \frac{ (1-q)^{2k} q^{2k(k+\alpha )}\,B^{(2)}_{n-2k,\alpha }(-\frac{x}{2};q)}{2^{2k}\,[n-2k]_{q}!\, (q^{2},q^{2n+\alpha };q^{2})_{k}}&=\frac{(-1/2)^{n}}{[n]_{q}!}q^{\frac{n(n-1)}{2}} G_{n}(x),\\ \sum _{k = 0}^{[\frac{n}{2}]} \, \frac{ (1-q)^{2k} q^{k^{2}+k/2}\,B^{(3)}_{n-2k,\alpha }(-\frac{x}{2};q)}{2^{2k}\,[n-2k]_{q}!\, (q^{2},q^{2\alpha +2};q^{2})_{k}}&=\frac{(-1/2)^{n}}{[n]_{q}!}q^{\frac{n(n-1)}{4}} (-xq^{\frac{1-n}{2}};q)_{n} . \end{aligned} \end{aligned}$$

(2.37)

Proof

We can write the generating function of the polynomials $B^{(1)}_{n,\alpha }(x;q)$ as

$$\begin{aligned} \begin{aligned} e_{q}(xt)e_{q}\left( \frac{-t}{2}\right)&= g^{(1)}_{\alpha }(it;q)\sum _{n=0}^{\infty } B^{(1)}_{n,\alpha }(x;q)\frac{t^{n}}{[n]_{q}!}\\&=\left( \sum _{n=0}^{\infty } \frac{(1-q)^{2n}t^{2n}}{2^{2n}(q^{2},q^{2\alpha +2};q^{2})_{n}}\right) \,\left( \sum _{n=0}^{\infty } B^{(1)}_{n,\alpha }(x;q)\frac{t^{n}}{[n]_{q}!}\right) .\end{aligned}\nonumber \\ \end{aligned}$$

(2.38)

On one hand, applying the Cauchy product formula in (2.38), we obtain

$$\begin{aligned} e_{q}(xt)e_{q}\left( \frac{-t}{2}\right) =\sum _{n=0}^{\infty }t^{n}\sum _{k=0}^{[\frac{n}{2}]}\frac{(1-q)^{2k}B^{(1)}_{n-2k,\alpha }(x;q)}{2^{2k}[n-2k]_{q}!(q^{2},q^{2\alpha +2};q^{2})_{k}} . \end{aligned}$$

On the other hand, using the series representation of $e_{q}(x)$ in (1.4) followed by the Cauchy product formula, and using (2.31) yields

$$\begin{aligned} e_{q}(xt)e_{q}\left( \frac{-t}{2}\right) =\sum _{n=0}^{\infty }\frac{t^{n}}{[n]_{q}!}\,\left( \frac{-1}{2}\right) ^{n} H_{n}(-2x). \end{aligned}$$

(2.39)

Hence

$$\begin{aligned} \sum _{n=0}^{\infty }\frac{t^{n}}{[n]_{q}!}\,\left( \frac{-1}{2}\right) ^{n} H_{n}(-2x)=\sum _{n=0}^{\infty }t^{n}\sum _{k=0}^{[\frac{n}{2}]}\frac{(1-q)^{2k}B^{(1)}_{n-2k,\alpha }(x;q)}{2^{2k}[n-2k]_{q}!(q^{2},q^{2\alpha +2};q^{2})_{k}},\nonumber \\ \end{aligned}$$

(2.40)

equating the coefficients of $t^{n}$ in (2.40), we get

$$\begin{aligned} \frac{(\frac{-1}{2})^{n}\,H_{n}(-2x)}{[n]_{q}!}=\sum _{k=0}^{[\frac{n}{2}]}\frac{(1-q)^{2k}B^{(1)}_{n-2k,\alpha }(x;q)}{2^{2k}[n-2k]_{q}!(q^{2},q^{2\alpha +2};q^{2})_{k}}. \end{aligned}$$

(2.41)

Replacing $x \,by\,\, \frac{-x}{2}$ in (2.41) gives

$$\begin{aligned} \frac{(\frac{-1}{2})^{n}\,H_{n}(x)}{[n]_{q}!}=\sum _{k=0}^{[\frac{n}{2}]}\frac{(1-q)^{2k}B^{(1)}_{n-2k,\alpha }(\frac{-x}{2};q)}{2^{2k}[n-2k]_{q}!(q^{2},q^{2\alpha +2};q^{2})_{k}}, \end{aligned}$$

which readily completes the proof for $B^{(1)}_{n,\alpha }(x;q)$. The proofs for $B^{(2)}_{n,\alpha }(x;q)$ and $B^{(3)}_{n,\alpha }(x;q)$ are similar and are omitted. $\square$

If we set $x=0$ in (2.37), we obtain the following recurrence relations for $\beta _{n,\alpha }(q)$ and $\beta ^{(3)}_{n,\alpha }(q),$

$$\begin{aligned} \begin{aligned} \sum _{k = 0}^{[\frac{n}{2}]} \, \frac{ (1-q)^{2k} \,\beta _{n-2k,\alpha }(q)}{2^{2k}\,[n-2k]_{q}!\, (q^{2},q^{2\alpha +2};q^{2})_{k}}\,&=\frac{(\frac{-1}{2})^{n}}{[n]_{q}!}\quad (n\in {\mathbb {N}}),\\ \quad \sum _{k = 0}^{[\frac{n}{2}]} \, \frac{(1-q)^{2k} q^{k^{2}+k/2}\,\beta ^{(3)}_{n-2k,\alpha }(q)}{2^{2k}\,[n-2k]_{q}!\, (q^{2},q^{2\alpha +2};q^{2})_{k}}\,&=\frac{q^{\frac{n(n-1)}{4}}(\frac{-1}{2})^{n}}{[n]_{q}!}\quad (n\in {\mathbb {N}}). \end{aligned} \end{aligned}$$

(2.42)

As a consequence of the recursive relations in (2.42), and the fact that

$$\begin{aligned} \beta _{0,\alpha }(q)=\beta ^{(3)}_{0,\alpha }(q)=1, \end{aligned}$$

we can prove that

$$\begin{aligned} \beta _{1,\alpha }(q)&=-\frac{1}{2},\quad \beta _{2,\alpha }(q)=\frac{q(1-q^{2\alpha +1})}{4(1-q^{2\alpha +2})},\quad \beta _{3,\alpha }(q)=\frac{-q^{3}(1-q^{2\alpha -1})}{8(1-q^{2\alpha +2})},\\ \beta _{4,\alpha }(q)&=\frac{1}{16}-\frac{(q+q^{3})(1-q^{3})(1-q^{2\alpha +1})}{16(1-q^{2\alpha +2})^{2}}-\frac{(1-q)(1-q^{3})}{16(q^{2\alpha +2};q^{2})_{2}},\\ \beta _{5,\alpha }(q)&=\frac{(1+q^{2})(1-q^{5})(q^{3}-q^{2\alpha +2})}{32(1-q^{2\alpha +2})^{2}}+\frac{(1-q^{3})(1-q^{5})}{32(q^{2\alpha +2};q^{2})_{2}} -\frac{1}{32}, \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} \beta ^{(3)}_{1,\alpha }(q)&=\frac{-1}{2},\quad \quad \beta ^{(3)}_{2,\alpha }(q)= \frac{q^{1/2}(1-q^{2\alpha +2})-q^{3/2}(1-q)}{4(1-q^{2\alpha +2})},\\ \beta ^{(3)}_{3,\alpha }(q)&=\frac{-q^{3/2}(q^{3}-q^{2\alpha +2})}{8(1-q^{2\alpha +2})}, \\ \beta ^{(3)}_{4,\alpha }(q)&=\frac{q^{3}(q^{2\alpha +2};q^{2})_{2}(1-q^{2\alpha +2}) }{16(1-q^{2\alpha +2})^{2}(1-q^{2\alpha +4})}-\frac{[3]_{q}q^{5}(1-q)^{2}(1-q^{2\alpha +2})}{16(1-q^{2\alpha +2})^{2}(1-q^{2\alpha +4})}\\&\quad +\,\frac{[4]_{q}[3]_{q}q^{3/2} (1-q^{2\alpha +4})\left( q^{1/2}(1-q^{2\alpha +2})-q^{3/2}(1-q)\right) }{16(1-q^{2\alpha +2})^{2}(1-q^{2\alpha +4})}, \\ \quad \beta ^{(3)}_{5,\alpha }(q)&=\frac{[5]_{q}q^{3}(1-q)(1+q^{2})(q^{3}-q^{2\alpha +2})(1-q^{2\alpha +4})}{32(1-q^{2\alpha +2})^{2}(1-q^{2\alpha +4})} \\&\quad +\,\frac{[5]_{q}q^{5}(1-q)(1-q^{3})(1-q^{2\alpha +2})}{32(1-q^{2\alpha +2})^{2}(1-q^{2\alpha +4})}-\frac{q^{5}(1-q^{2\alpha +2})^{2}(1-q^{2\alpha +4})}{32(1-q^{2\alpha +2})^{2}(1-q^{2\alpha +4})}.\end{aligned} \end{aligned}$$

Theorem 2.15

For $n\in \,{\mathbb {N}}_{0}$ and complex numbers a and x,

$$\begin{aligned} B^{(1)}_{n,\alpha }(x;q)&=\sum _{k=0}^{n}\left[ \begin{array}{c} n \\ k \\ \end{array} \right] _{q}\,(a;q)_{k} \,x^{k} B^{(1)}_{n-k,\alpha }(ax;q), \end{aligned}$$

(2.43)

$$\begin{aligned} B^{(2)}_{n,\alpha }(x;q)&=\sum _{k=0}^{n}\left[ \begin{array}{c} n \\ k \\ \end{array} \right] _{q}\,(-a)^{k}(1/a;q)_{k} \,x^{k} B^{(2)}_{n-k,\alpha }(ax;q). \end{aligned}$$

(2.44)

Proof

The proof of (2.43) follows from the generating function (2.1) since

$$\begin{aligned} \begin{aligned} \frac{e_{q}(xt)e_{q}(\frac{-t}{2})}{g^{(1)}_{\alpha }(it;q)}=\frac{e_{q}(tax)e_{q}(\frac{-t}{2})}{g^{(1)}_{\alpha }(it;q)}\,\frac{e_{q}(tx)}{e_{q}(atx)},\quad \mid tx \mid <\frac{1}{1-q}. \end{aligned} \end{aligned}$$

From the q-binomial theorem (see [12, Eq.(1.3.2), p. 8]), we can prove that

$$\begin{aligned} \frac{e_{q}(tx)}{e_{q}(atx)}=\sum _{n=0}^{\infty }\frac{(a;q)_{n}}{(q;q)_{n}}((1-q)tx)^{n},\quad \mid tx \mid <\frac{1}{1-q}. \end{aligned}$$

Therefore,

$$\begin{aligned} \begin{aligned} \sum _{n=0}^{\infty }B^{(1)}_{n,\alpha }(x;q)\frac{t^{n}}{[n]_{q}!}&=\left( \sum _{n=0}^{\infty }B^{(1)}_{n,\alpha }(ax;q)\frac{t^{n}}{[n]_{q}!}\right) \,\, \left( \sum _{n=0}^{\infty }\frac{(a;q)_{n}}{[n]_{q}!}(tx)^{n}\right) \\&=\sum _{n=0}^{\infty }\frac{t^{n}}{[n]_{q}!}\sum _{k=0}^{n}\left[ \begin{array}{c} n \\ k \\ \end{array} \right] _{q}\, (a;q)_{k}x^{k} B^{(1)}_{n-k,\alpha }(ax;q), \end{aligned}\nonumber \\ \end{aligned}$$

(2.45)

where we used the Cauchy product formula. Equating the coefficients of $t^{n}$ in (2.45), we obtain (2.43). The proof for $B^{(2)}_{n,\alpha }(x;q)$ is similar and is omitted. $\square$

Lemma 2.16

For $n\in {\mathbb {N}}_{0}$, $Re \,\alpha >-1$, and $\mid \frac{(1-q)t}{2}\mid <1$,

$$\begin{aligned} g^{(1)}_{\alpha }(it;q)E_{q}(\frac{t}{2})= g^{(2)}_{\alpha }(it;q)e_{q}(\frac{t}{2}) ={}_{2}\phi _{1}\,(q^{\alpha +\frac{1}{2}},-q^{\alpha +\frac{1}{2}};q^{2\alpha +1};q,\frac{(1-q)t}{2}).\nonumber \\ \end{aligned}$$

(2.46)

Proof

From Lemma 2.2, we conclude that

$$\begin{aligned} \begin{aligned} g^{(2)}_{\alpha }(it;q) e_{q}\left( \frac{t}{2}\right) =g^{(1)}_{\alpha }(it;q)E_{q}\left( \frac{t}{2}\right) .\end{aligned} \end{aligned}$$

From the series representations of $E_{q}(x)$ and $g^{(1)}_{\alpha }(it;q)$ in (1.4) and (2.38), respectively we obtain

$$\begin{aligned} \begin{aligned} g^{(1)}_{\alpha }(it;q)E_{q}(\frac{t}{2})&=\left( \sum _{n=0}^{\infty }\frac{(1-q)^{2n}t^{2n}}{2^{2n}(q^{2},q^{2\alpha +2};q^{2})_{n}}\right) \left( \sum _{n=0}^{\infty }\frac{q^{\frac{n(n-1)}{2}}(1-q)^{n}t^{n}}{2^{n}(q;q)_{n}}\right) \\&=\sum _{n=0}^{\infty }\frac{(1-q)^{n}q^{\frac{n(n-1)}{2}}t^{n}}{2^{n}} \sum _{k=0}^{[\frac{n}{2}]}\frac{q^{2k^{2}-2nk +k} }{(q;q)_{n-2k}(q^{2},q^{2\alpha +2};q^{2})_{k}}\\&=\sum _{n=0}^{\infty }\frac{q^{\frac{n(n-1)}{2}}(1-q)^{n}t^{n}}{2^{n}(q;q)_{n}}\sum _{k=0}^{[\frac{n}{2}]} \frac{q^{2k}(q^{-n};q)_{2k} }{(q^{2},q^{2\alpha +2};q^{2})_{k}},\end{aligned} \end{aligned}$$

where we used the identity, see [12, Eq. (1.2.32), p. 6],

$$\begin{aligned} \begin{aligned} (a;q)_{n-k}=\frac{(a;q)_{n}}{(a^{-1}q^{1-n};q)_{k}}(-qa^{-1})^{k}q^{\frac{k(k-1)}{2} -nk}\quad (k=0,1,\ldots ,n). \end{aligned} \end{aligned}$$

(2.47)

Therefore, using the identity $(a;q)_{2n}=(a;q^{2})_{n}(aq;q^{2})_{n}$ yields

$$\begin{aligned} \begin{aligned} g^{(1)}_{\alpha }(it;q)E_{q}\left( \frac{t}{2}\right)&=\sum _{n=0}^{\infty }\frac{q^{\frac{n(n-1)}{2}}(1-q)^{n}(\frac{t}{2})^{n}}{(q;q)_{n}}\sum _{k=0}^{[\frac{n}{2}]} \frac{q^{2k}(q^{-n};q^{2})_{k}(q^{-n+1};q^{2})_{k}}{(q^{2};q^{2\alpha +2};q^{2})_{k}}\\&= \sum _{n=0}^{\infty }\dfrac{q^{\frac{n(n-1)}{2}}(1-q)^{n}(\frac{t}{2})^{n}}{(q;q)_{n}}\,\,_{2}\phi _{1}(q^{-n},q^{-n+1};q^{2\alpha +2};q^{2},q^{2}). \end{aligned} \end{aligned}$$

Since

$$\begin{aligned} _{2}\phi _{1}\,(q^{-n},q^{1-n};qb^{2};q^{2},q^{2})=\frac{(b^{2};q^{2})_{n}}{(b^{2};q)_{n}}q^{\frac{-n(n-1)}{2}}\,\,(n\in {\mathbb {N}}), \end{aligned}$$

see [12, p. 26], then

$$\begin{aligned} \begin{aligned} g^{(1)}_{\alpha }(it;q)E_{q}\left( \frac{t}{2}\right)&=\sum _{n=0}^{\infty }\dfrac{(q^{\alpha +\frac{1}{2}};q)_{n}(-q^{\alpha +\frac{1}{2}};q)_{n}(\frac{(1-q)t}{2})^{n}}{ (q;q)_{n}(q^{2\alpha +1};q)_{n}}\\&={}_{2}\phi _{1}\,\left( q^{\alpha +\frac{1}{2}},-q^{\alpha +\frac{1}{2}};q^{2\alpha +1};q,\frac{(1-q)t}{2}\right) . \end{aligned}\nonumber \\ \end{aligned}$$

(2.48)

Hence from Lemma 2.2 and (2.48), we obtain (2.46) and completes the proof. $\square$

Theorem 2.17

Let $\alpha$ be a complex number such that $Re\,\alpha >-1$. Then

$$\begin{aligned} \sum _{m=0}^{n}\left[ \begin{array}{c} n \\ m \\ \end{array} \right] _{q}\dfrac{(q^{2\alpha +1};q^{2})_{m}B^{(1)}_{n-m,\alpha }(x;q)}{2^{m} (q^{2\alpha +1};q)_{m}}&= x^{n},\\ \sum _{m=0}^{n}\left[ \begin{array}{c} n \\ m \\ \end{array} \right] _{q}\dfrac{(q^{2\alpha +1};q^{2})_{m}B^{(2)}_{n-m,\alpha }(x;q)}{2^{m} (q^{2\alpha +1};q)_{m}}&=q^{\frac{n(n-1)}{2}}x^{n},\\ \sum _{m=0}^{n}(-\frac{1}{2})^{m}\left( \sum _{k = 0}^{[\frac{m}{2}]}\, \frac{ \,q^{k^{2}+k/2}\,(1-q)^{2k}\,c_{m-2k}}{\,(q^{2},q^{2\alpha +2};q^{2})_{k}}\right) \frac{B^{(3)}_{n-m,\alpha }(x;q)}{[n-m]_{q}!}&= \frac{q^{\frac{n(n-1)}{4}}x^{n}}{[n]_{q}!}, \end{aligned}$$

where $(c_{k})_{k}$ are the coefficients defined in (2.23).

Proof

We can write Eq. (2.1) in the form

$$\begin{aligned} \begin{aligned} e_{q}(xt)&=E_{q}\left( \frac{t}{2}\right) g^{(1)}_{\alpha }(it;q)\sum _{n=0}^{\infty } B^{(1)}_{n,\alpha }(x;q)\frac{t^{n}}{[n]_{q}!}\\&= \left( \sum _{n=0}^{\infty } d_{n} t^{n}\right) \,\left( \sum _{n=0}^{\infty } B^{(1)}_{n,\alpha }(x;q)\frac{t^{n}}{[n]_{q}!}\right) .\end{aligned}\nonumber \\ \end{aligned}$$

(2.49)

From Lemma 2.16, we obtain

$$\begin{aligned} \begin{aligned} g^{(1)}_{\alpha }(it;q)E_{q}\left( \frac{t}{2}\right) =\sum _{n=0}^{\infty } d_{n}t^{n}, \end{aligned} \end{aligned}$$

(2.50)

where

$$\begin{aligned} d_{n}=\dfrac{(1-q)^{n}(q^{2\alpha +1};q^{2})_{n}}{2^{n}(q;q)_{n} (q^{2\alpha +1};q)_{n}} . \end{aligned}$$

(2.51)

Now, applying the Cauchy product formula in (2.49) gives

$$\begin{aligned} \begin{aligned} e_{q}(xt)=\sum _{n=0}^{\infty }t^{n}\,\sum _{m=0}^{n}\frac{d_{m} B^{(1)}_{n-m,\alpha }(x;q) }{ [n-m]_{q}!}= \sum _{n=0}^{\infty }\frac{(xt)^{n}}{[n]_{q}!}. \end{aligned} \end{aligned}$$

(2.52)

Equating the coefficients of the nth power of t in (2.52) gives

$$\begin{aligned} \sum _{m=0}^{n}\frac{d_{m} B^{(1)}_{n-m,\alpha }(x;q) }{ [n-m]_{q}!}=\frac{x^{n}}{[n]_{q}!}. \end{aligned}$$

(2.53)

Substituting from (2.51) into (2.53), we get the result for $B^{(1)}_{n,\alpha }(x;q)$. Similarly, we can prove the result for $B^{(k)}_{n,\alpha }(x;q)\,(k=2,3)$. $\square$

Theorem 2.18

Let n be a positive integer and x be a complex number. If $Re\,\alpha >-1$, then

$$\begin{aligned}&B^{(1)}_{n,\alpha }(x;q)-(-1)^{n} B^{(1)}_{n,\alpha }(-x;q)\nonumber \\&\quad =\sum _{k=0}^{n}\left[ \begin{array}{c} n \\ k \\ \end{array} \right] _{q}\left( \left( \frac{-1}{2}\right) ^{k}H_{k}(-2x)-\left( \frac{1}{2}\right) ^{k}H_{k}(2x)\right) B^{(2)}_{n-k,\alpha }\left( \frac{1}{2};q\right) ,\nonumber \\ \end{aligned}$$

(2.54)

$$\begin{aligned}&B^{(2)}_{n,\alpha }(x;q)-(-1)^{n} B^{(2)}_{n,\alpha }(-x;q)\nonumber \\&\quad =\sum _{k=0}^{n}\left[ \begin{array}{c} n \\ k \\ \end{array} \right] _{q}q^{\frac{k(k-1)}{2}}\left( (\frac{-1}{2})^{k}G_{k}(-2x)-(\frac{1}{2})^{k}G_{k}(2x)\right) B^{(1)}_{n-k,\alpha }(\frac{1}{2};q).\nonumber \\ \end{aligned}$$

(2.55)

Proof

We give only the proof of (2.54) since the proof of (2.55) is similar. From (2.1),

$$\begin{aligned} \frac{e_{q}(xt)e_{q}(\frac{-t}{2})}{g^{(1)}_{\alpha }(it;q)}-\frac{e_{q}(xt)e_{q}(\frac{t}{2})}{g^{(1)}_{\alpha }(-it;q)} =\sum _{n=0}^{\infty }B^{(1)}_{n,\alpha }(x;q)\frac{t^{n}}{[n]_{q}!}-\sum _{n=0}^{\infty }B^{(1)}_{n,\alpha }(-x;q)\frac{(-t)^{n}}{[n]_{q}!}.\nonumber \\ \end{aligned}$$

(2.56)

Since

$$\begin{aligned} g^{(1)}_{\alpha }(-it;q)=g^{(1)}_{\alpha }(it;q), \end{aligned}$$

then Eq. (2.56) can be written as

$$\begin{aligned} \frac{e_{q}(xt)e_{q}(\frac{-t}{2}) -e_{q}(xt)e_{q}(\frac{t}{2})}{g^{(1)}_{\alpha }(it;q)} =\sum _{n=0}^{\infty }\left[ B^{(1)}_{n,\alpha }(x;q)-(-1)^{n}B^{(1)}_{n,\alpha }(-x;q)\right] \frac{t^{n}}{[n]_{q}!}.\nonumber \\ \end{aligned}$$

(2.57)

Replacing x, t by $-x, -t$, respectively in (2.39) gives

$$\begin{aligned} e_{q}(xt)e_{q}(\frac{t}{2})= \sum _{n=0}^{\infty }\frac{t^{n}}{[n]_{q}!}(\frac{1}{2})^{n}H_{n}(2x). \end{aligned}$$

(2.58)

From (2.39) and (2.58), the left hand side of (2.57) can be written as

$$\begin{aligned} \begin{aligned} \frac{e_{q}(xt)e_{q}(\frac{-t}{2}) -e_{q}(xt)e_{q}(\frac{t}{2})}{g^{(1)}_{\alpha }(it;q)}&=\frac{1}{g^{(1)}_{\alpha }(it;q)}\sum _{n=0}^{\infty }\frac{t^{n}}{[n]_{q}!}\\&\quad \times \left( \left( \frac{-1}{2}\right) ^{n}H_{n}(-2x)-\left( \frac{1}{2}\right) ^{n}H_{n}(2x)\right) .\end{aligned} \end{aligned}$$

Therefore, by (2.21) and the Cauchy product formula, we get

$$\begin{aligned} \begin{aligned}&\frac{e_{q}(xt)e_{q}(\frac{-t}{2}) -e_{q}(xt)e_{q}(\frac{t}{2})}{g^{(1)}_{\alpha }(it;q)}\\&=\sum _{n=0}^{\infty }\frac{t^{n}}{[n]_{q}!}\sum _{k=0}^{n} \left[ \begin{array}{c} n \\ k \\ \end{array} \right] _{q}\left( \left( \frac{-1}{2}\right) ^{k}H_{k}(-2x)-\left( \frac{1}{2}\right) ^{k}H_{k}(2x)\right) B^{(2)}_{n-k,\alpha }\left( \frac{1}{2};q\right) . \end{aligned}\nonumber \\ \end{aligned}$$

(2.59)

Since the left hand side of (2.57) and (2.59) are equal, then equating the coefficients of $t^{n}$ on the right hand sides of (2.57) and (2.59) yields (2.54) and completes the proof. $\square$

Proposition 2.19

If $\alpha _{0}>-1$ satisfies the condition

$$\begin{aligned} q^{2(\alpha _{0}+1)}(1-q)^{2}<(1-q^{2})(1-q^{2\alpha _{0}+2}), \end{aligned}$$

(2.60)

then $(t/2)^{-\alpha }J_{\alpha }^{(2)}(t(1-q);q^{2})$ has no zeros in $\mid t\mid \le 1$ for all $\alpha \ge \alpha _{0}.$

Proof

Set

$$\begin{aligned} F(t):= \frac{(q;q)_{\infty }}{(q^{\alpha +1};q)_{\infty }}(t/2)^{-\alpha }J_{\alpha }^{(2)}(t(1-q);q^{2}) =\sum _{k=0}^{\infty }\frac{(-1)^{k}q^{2k(k+\alpha )}(1-q)^{2k}}{2^{2k}(q^{2},q^{2\alpha +2};q^{2})_{k}}t^{2k}, \end{aligned}$$

and

$$\begin{aligned} a_{k}:=\frac{q^{2k(k+\alpha )}(1-q)^{2k}}{2^{2k}(q^{2},q^{2\alpha +2};q^{2})_{k}}. \end{aligned}$$

Then, under hypothesis (2.60) and since $0< q < 1$,

$$\begin{aligned} q^{2(\alpha +1)}(1-q^{2})\le & {} q^{2(\alpha _{0}+1)}(1-q^{2})<(1-q^{2})(1-q^{2\alpha _{0}+2})\\ {}\le & {} (1-q^{2})(1-q^{2\alpha +2}), \end{aligned}$$

holds whenever $\alpha \ge \alpha _{0}$. Hence

$$\begin{aligned} \frac{a_{k+1}}{a_{k}}=\frac{q^{4k+2(\alpha +1)}(1-q)^{2}}{4(1-q^{2k+2})(1-q^{2k+2\alpha +2})} \le \frac{q^{2(\alpha +1)}(1-q)^{2}}{4(1-q^{2})(1-q^{2\alpha +2})}<1, \end{aligned}$$

for $t\in {\mathbb {R}},\, \mid t\mid \le 1$

$$\begin{aligned} \begin{aligned} F(t)=\sum _{k=0}^{\infty }t^{2k}(a_{2k}-a_{2k+1}t^{2})\ge (a_{0}-a_{1}t^{2})\ge (a_{0}-a_{1})>0. \end{aligned} \end{aligned}$$

This proves that F(t) has no zeros on $[-1,1]$, since F(t) has only real zeros, then F(t) has no zeros in the unit disk. i.e $\mid F(t)\mid >0,\,\,for\,\,\mid t\mid \le 1.$ $\square$

Corollary 2.20

There exists $\alpha _{0}>-1$ such that $J_{\alpha }^{(2)}(t(1-q);q^{2})$ has no zeros in the unit disk for all $\alpha \ge \alpha _{0}$.

Proof

Since for a fixed $q\in (0,1)$,

$$\begin{aligned} \displaystyle \lim _{\alpha \rightarrow \infty }q^{2\alpha +2}=0,\,\,\displaystyle \lim _{\alpha \rightarrow \infty }(1-q^{2})(1-q^{2\alpha +2})=(1-q^{2}), \end{aligned}$$

then there exists $\alpha _{0}>-1$ such that the condition (2.60) holds for all $\alpha \ge \alpha _{0}$. Consequently from Proposition 2.19, $J_{\alpha }^{(2)}(t(1-q);q^{2})$ has no zeros in the unit disk for all $\alpha \ge \alpha _{0}$. $\square$

Theorem 2.21

For $n\in {\mathbb {N}}$,

$$\begin{aligned}&\lim _{\alpha \rightarrow \infty }\,B^{(2)}_{n,\alpha }(x;q)=\left( -\frac{1}{2}\right) ^{n}q^{\frac{n(n-1)}{2}}G_{n}(-2x), \end{aligned}$$

(2.61)

$$\begin{aligned}&\lim _{\alpha \rightarrow \infty }\,B^{(1)}_{n,\alpha }(x;q)=x^{n}\left( \frac{1}{2x};q\right) _{n}. \end{aligned}$$

(2.62)

Proof

Taking the limit on both sides of Eq. (2.2) as $\alpha \rightarrow \infty$ we get

$$\begin{aligned} \lim _{\alpha \rightarrow \infty } \frac{E_{q}(xt)E_{q}(\frac{-t}{2})}{ g^{(2)}_{\alpha }(it;q)}= \lim _{\alpha \rightarrow \infty }\,\sum _{n=0}^{\infty } B^{(2)}_{n,\alpha }(x;q)\frac{t^{n}}{[n]_{q}!}. \end{aligned}$$

(2.63)

From Corollary 2.20, there exists $\alpha _{0}>-1$ such that $g^{(2)}_{\alpha }(it;q)$ has no zeros in $\mid t\mid \le 1$ for all $\alpha \ge \alpha _{0}$. This means that $\dfrac{E_{q}(xt)E_{q}(\frac{-t}{2})}{ g^{(2)}_{\alpha }(it;q)}$ is analytic in $\mid t\mid \le 1$ for all $\alpha \ge \alpha _{0}$. Therefore, we can interchange the limit with the summation in (2.63) when $\mid t\mid \le 1$ to obtain

$$\begin{aligned} \frac{E_{q}(xt)E_{q}(\frac{-t}{2})}{\displaystyle \lim _{\alpha \rightarrow \infty } g^{(2)}_{\alpha }(it;q)}= \sum _{n=0}^{\infty } \lim _{\alpha \rightarrow \infty }\, B^{(2)}_{n,\alpha }(x;q)\frac{t^{n}}{[n]_{q}!}. \end{aligned}$$

Since

$$\begin{aligned} \lim _{\alpha \rightarrow \infty } g^{(2)}_{\alpha }(it;q)=1,\quad E_{q}(xt)E_{q}(yt)=\sum _{n=0}^{\infty }\frac{q^{n(n-1)/2}(ty)^{n}}{[n]_{q}!}G_{n}(\frac{x}{y}), \end{aligned}$$

then from (2.32)

$$\begin{aligned} \begin{aligned} \sum _{n=0}^{\infty }\lim _{\alpha \rightarrow \infty } B^{(2)}_{n,\alpha }(x;q)\frac{t^{n}}{[n]_{q}!} =\sum _{n=0}^{\infty }\frac{q^{n(n-1)/2}t^{n}}{[n]_{q}!}\left( \frac{-1}{2}\right) ^{n}G_{n}(-2x).\end{aligned} \end{aligned}$$

(2.64)

Equating the coefficients of $t^{n}$ in (2.64) gives (2.61). The proof of (2.62) follows directly from the relation (2.9) since

$$\begin{aligned} 1=\displaystyle \lim _{\alpha \rightarrow \infty }g^{(2)}_{\alpha }(it;q)= E_{q}\left( \frac{t}{2}\right) E_{q}\left( \frac{-t}{2}\right) \displaystyle \lim _{\alpha \rightarrow \infty }g^{(1)}_{\alpha }(it;q). \end{aligned}$$

Hence

$$\begin{aligned} \displaystyle \lim _{\alpha \rightarrow \infty }g^{(1)}_{\alpha }(it;q)=e_{q}\left( \frac{t}{2}\right) e_{q}\left( \frac{-t}{2}\right) ,\,\,\mid t(1-q)\mid <2. \end{aligned}$$

Therefore, computing the limit in both sides of (2.1) gives

$$\begin{aligned} \frac{e_{q}(xt)}{e_{q}(\frac{t}{2})}= \sum _{n=0}^{\infty } \lim _{\alpha \rightarrow \infty }\, B^{(1)}_{n,\alpha }(x;q)\frac{t^{n}}{[n]_{q}!}. \end{aligned}$$

From the q-binomial theorem (see [12, Eq.(1.3.2), p. 8]), we have

$$\begin{aligned} \frac{e_{q}(xt)}{e_{q}(\frac{t}{2})}= \frac{(\frac{t}{2}(1-q);q)_{\infty }}{(xt(1-q);q)_{\infty }}=\sum _{n=0}^{\infty } \frac{(\frac{1}{2x};q)_{n}}{(q;q)_{n}}(xt(1-q))^{n},\,\,\mid xt(1-q)\mid <1. \end{aligned}$$

Hence

$$\begin{aligned} \sum _{n=0}^{\infty } \lim _{\alpha \rightarrow \infty }\, B^{(1)}_{n,\alpha }(x;q)\frac{t^{n}}{[n]_{q}!}= \sum _{n=0}^{\infty }\frac{(xt)^{n}}{[n]_{q}!}(\frac{1}{2x};q)_{n}, \end{aligned}$$

(2.65)

equating the coefficients of $t^{n}$ in (2.65) yields the required result. $\square$

Corollary 2.22

For $n\in {\mathbb {N}}$,

$$\begin{aligned} \lim _{\alpha \rightarrow \infty }\,\beta _{n,\alpha }(q)=(-1)^{n}2^{-n}q^{\frac{n(n-1)}{2}}. \end{aligned}$$

(2.66)

Proof

Since

$$\begin{aligned} \lim _{x\rightarrow 0}x^{n}\left( \frac{1}{2x};q\right) _{n}= & {} \lim _{x\rightarrow 0}x^{n}\prod _{k=0}^{n-1}\left( 1-\frac{q^{k}}{2x}\right) =\lim _{x\rightarrow 0}\prod _{k=0}^{n-1}\left( x-\frac{q^{k}}{2}\right) \\ {}= & {} (-1)^{n}2^{-n}q^{\frac{n(n-1)}{2}}, \end{aligned}$$

then substituting with $x=0$ into (2.62) yields (2.66). $\square$

Lemma 2.23

Let $\alpha _{0}>-1$. If $q^{3/2}(1-q)^{2}<(1-q^{2})(1-q^{2\alpha _{0}+2})$, then $(q^{\frac{1}{4}}t/2)^{-\alpha }J_{\alpha }^{(3)}(\frac{t}{2}(1-q)q^{\frac{-1}{4}};q^{2})$ has no zeros in $\mid t\mid \le 1$ for all $\alpha \ge \alpha _{0}.$

Proof

The proof is similar to the proof of Proposition 2.19 and is omitted. $\square$

Theorem 2.24

For $n\in {\mathbb {N}}$,

$$\begin{aligned} \lim _{\alpha \rightarrow \infty }\!\!B^{(3)}_{n,\alpha }\!(x;q)\!= & {} \!{} q^{\frac{n(n-1)}{4}}\!\left( \frac{-1}{2}\right) ^{n}\!\sum _{k=0}^{[\frac{n}{2}]}\frac{(-1)^{k}q^{k(n-k+3)} (q^{-n};q)_{2k}}{(q^{2};q^{2})_{k}}\! (2xq^{\frac{1-n}{2}};q)_{n-2k},\nonumber \\ \end{aligned}$$

(2.67)

$$\begin{aligned} \lim _{\alpha \rightarrow \infty }\,\beta ^{(3)}_{n,\alpha }(q)= & {} q^{\frac{n(n-1)}{4}}\left( \frac{-1}{2}\right) ^{n}\sum _{k=0}^{[\frac{n}{2}]}\frac{(-1)^{k} q^{k(n-k+3)}(q^{-n};q)_{2k}}{(q^{2};q^{2})_{k}}. \nonumber \\ \end{aligned}$$

(2.68)

Proof

Taking the limit as $\alpha \rightarrow \infty$ on both sides of (2.3), we obtain

$$\begin{aligned} \displaystyle \lim _{\alpha \rightarrow \infty } \frac{\exp _{q}(xt)\exp _{q}(\frac{-t}{2})}{ g^{(3)}_{\alpha }(it;q)}= \lim _{\alpha \rightarrow \infty }\,\sum _{n=0}^{\infty } B^{(3)}_{n,\alpha }(x;q)\frac{t^{n}}{[n]_{q}!}. \end{aligned}$$

(2.69)

We can choose $\alpha _{0}>-1$ such that

$$\begin{aligned} q^{3/2}\le \frac{(1-q^{2})}{1-q}\frac{(1-q^{2\alpha _{0}+2})}{1-q}\le \frac{(1-q^{2})}{1-q}\frac{(1-q^{2\alpha +2})}{1-q}, \end{aligned}$$

for all $\alpha \ge \alpha _{0}$. Hence from Lemma 2.23, the function $g^{(3)}_{\alpha }(it;q)$ does not vanish on the unit disk, and the left hand side of (2.69) is analytic for $\mid t\mid \le 1$. Therefore, we can interchange the limit as $\alpha \rightarrow \infty$ with the summation in (2.69) to obtain

$$\begin{aligned} \frac{\exp _{q}(xt)\exp _{q}(\frac{-t}{2})}{\displaystyle \lim _{\alpha \rightarrow \infty } g^{(3)}_{\alpha }(it;q)}= \sum _{n=0}^{\infty } \lim _{\alpha \rightarrow \infty }\, B^{(3)}_{n,\alpha }(x;q)\frac{t^{n}}{[n]_{q}!}. \end{aligned}$$

Since

$$\begin{aligned} \begin{aligned} \lim _{\alpha \rightarrow \infty } g^{(3)}_{\alpha }(it;q)&= \sum _{n=0}^{\infty }\lim _{\alpha \rightarrow \infty }\frac{q^{n^{2}+\frac{n}{2}}(1-q)^{2n}}{(q^{2},q^{2\alpha +2};q^{2})_{n}}(\frac{t}{2})^{2n} \\&=\sum _{n=0}^{\infty }\frac{q^{n^{2}+\frac{n}{2}}(1-q)^{2n}(\frac{t}{2})^{2n}}{(q^{2};q^{2})_{n}} =\left( -\frac{q^{\frac{3}{2}}(1-q)^{2}t^{2}}{4};q^{2}\right) _{\infty }.\end{aligned} \end{aligned}$$

Hence

$$\begin{aligned} \frac{ \exp _{q}(xt)\exp _{q}(\frac{-t}{2})}{\left( -\frac{q^{\frac{3}{2}}(1-q)^{2}t^{2}}{4};q^{2}\right) _{\infty }}= \sum _{n=0}^{\infty } \lim _{\alpha \rightarrow \infty }\, B^{(3)}_{n,\alpha }(x;q)\frac{t^{n}}{[n]_{q}!}. \end{aligned}$$

(2.70)

But

$$\begin{aligned} \begin{aligned} \exp _{q}(xt)\exp _{q}\left( \frac{-t}{2}\right)&=\sum _{n=0}^{\infty }\frac{(\frac{-1}{2})^{n}q^{\frac{n(n-1)}{4}}t^{n}}{[n]_{q}!} (2xq^{\frac{1-n}{2}};q)_{n}. \end{aligned} \end{aligned}$$

Therefore,

$$\begin{aligned} \begin{aligned}&\left( \frac{1}{(- q^{\frac{3}{2}}(1-q)^{2}t^{2}/4;q^{2})_{\infty }}\right) \left( \exp _{q}(xt)\exp _{q}(\frac{-t}{2})\right) \\&=\left( \sum _{n=0}^{\infty }\frac{(-1)^{n}q^{\frac{3}{2}n} (\frac{(1-q)t}{2})^{2n}}{(q^{2};q^{2})_{n}}\right) \left( \sum _{n=0}^{\infty }\frac{q^{\frac{n(n-1)}{4}}(\frac{-t}{2})^{n}}{[n]_{q}!} (2xq^{\frac{1-n}{2}};q)_{n}\right) \\&=\sum _{n=0}^{\infty }q^{\frac{n(n-1)}{4}}\left( \frac{-t(1-q)}{2}\right) ^{n}\sum _{k=0}^{[\frac{n}{2}]}\frac{(-1)^{k}q^{\frac{3}{2}k}q^{k^{2}-nk+k/2}}{(q^{2};q^{2})_{k}(q;q)_{n-2k}} (2xq^{\frac{1-n}{2}};q)_{n-2k}\\&=\sum _{n=0}^{\infty }\frac{q^{\frac{n(n-1)}{4}}(\frac{-t}{2})^{n}}{[n]_{q}!}\sum _{k=0}^{[\frac{n}{2}]}\frac{(-1)^{k} q^{k(n-k+3)}(q^{-n};q)_{2k}}{(q^{2};q^{2})_{k}} (2xq^{\frac{1-n}{2}};q)_{n-2k}. \end{aligned}\nonumber \\ \end{aligned}$$

(2.71)

Substituting from (2.71) into (2.70) and equating the coefficients of $t^{n}$ yields (2.67). The proof of (2.68) follows directly by setting $x=0$ in (2.67). $\square$

Theorem 2.25

Let $\alpha$ be a complex number such that $Re\,\alpha >-1$. Then for $n\in {\mathbb {N}},\,n\,\ge \,2$,

$$\begin{aligned} \begin{aligned} \beta _{n,\alpha }(q)&=-\frac{[n]_{q}!(1-q)^{2}}{4}\sum _{k=0}^{[\frac{n}{2}]-1}\frac{\left( (1-q)/2\right) ^{2k}\beta _{n-2k-2,\alpha }(q)}{[n-2k-2]_{q}!(q^{2},q^{2\alpha +2};q^{2})_{k+1}} +\frac{(-1)^{n}}{2^{n}},\\ \beta _{n,\alpha }^{(3)}(q)&=-\frac{[n]_{q}!q^{3/2}(1-q)^{2}}{4}\sum _{k=0}^{[\frac{n}{2}]-1}\frac{q^{k^{2}+5k/2}\left( (1-q)/2\right) ^{2k}\beta ^{(3)}_{n-2k-2,\alpha }(q)}{[n-2k-2]_{q}!(q^{2},q^{2\alpha +2};q^{2})_{k+1}}\\ {}&\quad +\frac{(-1)^{n}q^{\frac{n(n-1)}{4}}}{2^{n}}.\end{aligned} \end{aligned}$$

(2.72)

Proof

We give in detail the proof of $\beta _{n,\alpha }(q)$ in (2.72). The proof for $\beta _{n,\alpha }^{(3)}(q)$ is similar. Since

$$\begin{aligned} \frac{e_{q}(\frac{-t}{2})}{g^{(1)}_{\alpha }(it;q)}=\sum _{n=0}^{\infty }\beta _{n,\alpha }(q)\frac{t^{n}}{[n]_{q}!}, \end{aligned}$$

(2.73)

then

$$\begin{aligned} \frac{e_{q}(\frac{-t}{2})}{g^{(1)}_{\alpha }(it;q)}-e_{q} \left( \frac{-t}{2}\right) =\sum _{n=0}^{\infty }\beta _{n,\alpha }(q)\frac{t^{n}}{[n]_{q}!}- e_{q} \left( \frac{-t}{2}\right) . \end{aligned}$$

Consequently, from the series representation of $e_{q}(t)$ in (1.4), we get

$$\begin{aligned} \frac{e_{q}(\frac{-t}{2})}{g^{(1)}_{\alpha }(it;q)}\left( 1-g^{(1)}_{\alpha }(it;q)\right) =\sum _{n=0}^{\infty }\left( \beta _{n,\alpha }(q) -\frac{(-1)^{n}}{2^{n}}\right) \frac{t^{n}}{[n]_{q}!}. \end{aligned}$$

(2.74)

Since

$$\begin{aligned} \begin{aligned} \left( g^{(1)}_{\alpha }(it;q)-1\right) =t^{2} \sum _{m=0}^{\infty }\frac{(1-q)^{2m+2}t^{2m}}{2^{2m+2}(q^{2},q^{2\alpha +2};q^{2})_{m+1}},\end{aligned} \end{aligned}$$

(2.75)

then substituting from (2.75) into (2.74) and using (2.73), we obtain

$$\begin{aligned} \begin{aligned}&\left( \sum _{n=0}^{\infty }\beta _{n,\alpha }(q)\frac{t^{n}}{[n]_{q}!}\right) \left( -t^{2} \sum _{n=0}^{\infty }\frac{(1-q)^{2n+2}t^{2n}}{2^{2n+2}(q^{2},q^{2\alpha +2};q^{2})_{n+1}}\right) \\&\quad =\sum _{n=0}^{\infty }\left( \beta _{n,\alpha }(q)- \frac{(-1)^{n}}{2^{n}}\right) \frac{t^{n}}{[n]_{q}!}.\end{aligned} \end{aligned}$$

Therefore, by the Cauchy product formula

$$\begin{aligned} \begin{aligned}&-\frac{(1-q)^{2}}{4}\sum _{n=2}^{\infty }t^{n}\sum _{k=0}^{[\frac{n}{2}]-1}\frac{(1-q)^{2k}\beta _{n-2k-2,\alpha }(q)}{2^{2k}[n-2k-2]_{q}!(q^{2},q^{2\alpha +2};q^{2})_{k+1}} \\&\quad =\sum _{n=0}^{\infty }\left( \beta _{n,\alpha }(q)-\frac{(-1)^{n}}{2^{n}}\right) \frac{t^{n}}{[n]_{q}!}.\end{aligned}\nonumber \\ \end{aligned}$$

(2.76)

Equating the coefficient of $t^{n}$ in (2.76), we get the required result and the theorem follows. $\square$

The following theorem gives a recursive relations between the polynomials $B^{(k)}_{n,\alpha }(x;q)$ and $B^{(k)}_{n,\alpha +1}(x;q)\,(k=2,3)$.

Theorem 2.26

If $Re\,\alpha > -1$ , $x\in {\mathbb {C}},$ and $k \in \,{\mathbb {N}}$, then

$$\begin{aligned}&\frac{B^{(2)}_{n,\alpha }(x;q)}{[n]_{q}!}=2(1-q^{2\alpha +2})\sum _{k=0}^{[\frac{n}{2}]}(-1)^{k}\frac{(1-q)^{2k}\,h^{(2)}_{k+1}(q^{2})\,}{[n-2k]_{q}!}B^{(2)}_{n-2k,\alpha +1}(x;q),\nonumber \\ \end{aligned}$$

(2.77)

$$\begin{aligned}&\frac{B^{(3)}_{n,\alpha }(x;q)}{[n]_{q}!}=(1-q^{2\alpha +2})\sum _{k=0}^{[\frac{n}{2}]}(-1)^{k}\frac{(1-q)^{2k}q^{\frac{-k}{2}}\,h^{(3)}_{k+1}(q^{2})\,}{2^{2k}[n-2k]_{q}!}B^{(3)}_{n-2k,\alpha +1}(x;q),\nonumber \\ \end{aligned}$$

(2.78)

where

$$\begin{aligned} h^{(r)}_{k}(q^{2})=\sum _{m=1}^{\infty }\frac{-2J^{(r)}_{\alpha +1}(j^{(r)}_{m,\alpha };q^{2})}{ \frac{d}{dz}J^{(r)}_{\alpha }(z;q^{2})\mid _{z=j^{(r)}_{m,\alpha }}}\left( \frac{1}{j^{(r)}_{m,\alpha }}\right) ^{2k}, \end{aligned}$$

and $(j^{(r)}_{m,\alpha })_{m=1}^{\infty }\,\,(r=2,3)$ are the positive zero of $J^{(r)}_{\alpha }( \cdot ;q^{2}).$

Proof

We start with the proof of (2.77). From [6, 13], we have the identity

$$\begin{aligned} \frac{J^{(2)}_{\alpha +1}(t;q)}{J^{(2)}_{\alpha }(t;q)}=\sum _{n=1}^{\infty }h^{(2)}_{n}(q)t^{2n-1}, \end{aligned}$$

(2.79)

where

$$\begin{aligned} h^{(2)}_{n}(q)=\sum _{m=1}^{\infty }\frac{-2J^{(2)}_{\alpha +1}(j^{(2)}_{m,\alpha };q^{2})}{ \frac{d}{dz}J^{(2)}_{\alpha }(z;q^{2})\mid _{z=j^{(2)}_{m,\alpha }} }\left( \frac{1}{j^{(2)}_{m,\alpha }}\right) ^{2n}. \end{aligned}$$

Replacing t by $it(1-q)$ and q by $q^{2}$ in (2.79), we obtain

$$\begin{aligned} \frac{1}{J^{(2)}_{\alpha }(it(1-q);q^{2})}= \frac{1}{J^{(2)}_{\alpha +1}(it(1-q);q^{2})}\sum _{n=1}^{\infty }h^{(2)}_{n}(it(1-q))^{2n-1}. \end{aligned}$$

(2.80)

Multiplying (2.80) by $E_{q}(xt) E_{q}(\frac{-t}{2})$ to obtain

$$\begin{aligned} \frac{E_{q}(xt) E_{q}(\frac{-t}{2})}{J^{(2)}_{\alpha }(it(1-q);q^{2})}= \frac{E_{q}(xt) E_{q}(\frac{-t}{2})}{J^{(2)}_{\alpha +1}(it(1-q);q^{2})}\sum _{n=1}^{\infty }h^{(2)}_{n}(q^{2})(it(1-q))^{2n-1}. \end{aligned}$$

(2.81)

Substituting from (2.8) into (2.81), we get

$$\begin{aligned} \frac{E_{q}(xt) E_{q}(\frac{-t}{2})}{g^{(2)}_{\alpha }(it;q)}=\frac{(1+q)[\alpha +1]_{q^{2}}}{(\frac{it}{2})}\frac{E_{q}(xt) E_{q}(\frac{-t}{2})}{ g^{(2)}_{\alpha +1}(it;q)}\sum _{n=1}^{\infty }h^{(2)}_{n}(q^{2})(it(1-q))^{2n-1}.\nonumber \\ \end{aligned}$$

(2.82)

Consequently,

$$\begin{aligned} \begin{aligned} \sum _{n=0}^{\infty }&B^{(2)}_{n,\alpha }(x;q)\frac{t^{n}}{[n]_{q}!}\\ {}&=\frac{2(1+q)}{it}[\alpha +1]_{q^{2}}\left( \sum _{n=0}^{\infty }B^{(2)}_{n,\alpha +1}(x;q) \frac{t^{n}}{[n]_{q}!} \right) \left( \sum _{n=1}^{\infty }h^{(2)}_{n}(q^{2})(it(1-q))^{2n-1}\right) \\ {}&=2(1-q^{2\alpha +2}) \left( \sum _{n=0}^{\infty }B^{(2)}_{n,\alpha +1}(x;q)\frac{t^{n}}{[n]_{q}!}\right) \left( \sum _{n=0}^{\infty }h^{(2)}_{n+1}(q^{2})t^{2n}(i(1-q))^{2n} \right) \\ {}&=2(1-q^{2\alpha +2})\sum _{n=0}^{\infty }t^{n}\sum _{k=0}^{[\frac{n}{2}]}(-1)^{k} \frac{(1-q)^{2k} \,h^{(2)}_{k+1}(q^{2})}{[n-2k]_{q}!}B^{(2)}_{n-2k,\alpha +1}(x;q). \end{aligned} \end{aligned}$$

Hence

$$\begin{aligned} \begin{aligned}&\sum _{n=0}^{\infty }B^{(2)}_{n,\alpha }(x;q)\frac{t^{n}}{[n]_{q}!}\\&\quad ={2(1-q^{2\alpha +2})}\sum _{n=0}^{\infty }t^{n}\sum _{k=0}^{[\frac{n}{2}]}\frac{(-1)^{k}(1-q)^{2k} \,h^{(2)}_{k+1}(q^{2})\,}{[n-2k]_{q}!}B^{(2)}_{n-2k,\alpha +1}(x;q).\end{aligned}\nonumber \\ \end{aligned}$$

(2.83)

Equating the coefficients of $t^{n}$ in (2.83), we get (2.77). The proof of (2.78) follows from the identity (see [1, Eq. (4.3), p. 1201]),

$$\begin{aligned} \begin{aligned} \frac{J^{(3)}_{\alpha +1}(t;q)}{J^{(3)}_{\alpha }(t;q)}=\sum _{n=1}^{\infty }h^{(3)}_{n}(q)t^{2n-1}, \end{aligned} \end{aligned}$$

where

$$\begin{aligned} h^{(3)}_{n}(q)=\sum _{m=1}^{\infty }\frac{-2J^{(3)}_{\alpha +1}(j^{(3)}_{m,\alpha };q^{2})}{ \frac{d}{dz}J^{(3)}_{\alpha }(z;q^{2})\mid _{z=j^{(3)}_{m,\alpha }} }\left( \frac{1}{j^{(3)}_{m,\alpha }}\right) ^{2n}, \end{aligned}$$

and by using the same technique. $\square$

3 Asymptotic Relations for the Generalized q-Bernoulli Numbers

In this section, we derive asymptotic relations for the generalized q-Bernoulli numbers defined in (2.10).

Theorem 3.1

Let n be a non negative integer and $\alpha$ be a complex number such that $Re\,\alpha >-1$. Then for $n\in {\mathbb {N}},$

$$\begin{aligned} \begin{aligned} \beta _{2n,\alpha }(q)&=2(-1)^{n+1}(q;q)_{2n} \sum _{k=1}^{\infty }\frac{Cos_{q}(\frac{j^{(2)}_{k,\alpha }}{2(1-q)})}{(j^{(2)}_{k,\alpha })^{2n+1}\, \frac{d}{dz}{\mathcal {J}}_{\alpha }^{(2)}(z;q^{2})\mid _{z=j^{(2)}_{k,\alpha }}},\\ \beta _{2n+1,\alpha }(q)&=2(-1)^{n}(q;q)_{2n+1} \sum _{k=1}^{\infty }\frac{Sin_{q}(\frac{j^{(2)}_{k,\alpha }}{2(1-q)})}{(j^{(2)}_{k,\alpha })^{2n+2}\,\frac{d}{dz}{\mathcal {J}}_{\alpha }^{(2)}(z;q^{2})\mid _{z=j^{(2)}_{k,\alpha }}}, \end{aligned} \end{aligned}$$

(3.1)

where ${\mathcal {J}}_{\alpha }^{(2)} (z;q)$ is defined in (1.7).

Proof

Since

$$\begin{aligned} G(z):=\frac{E_{q}(\frac{-z}{2})}{g^{(2)}_{\alpha }(iz;q)}=\sum _{n=0}^{\infty }\beta _{n,\alpha }(q)\frac{z^{n}}{[n]_{q}!},\quad \mid z\mid <\frac{j^{(2)}_{1,\alpha }}{1-q}, \end{aligned}$$

then

$$\begin{aligned} \frac{\beta _{n,\alpha }(q)}{[n]_{q}!}=\frac{G^{(n)}(0)}{n!},\quad n\in {\mathbb {N}}_{0}. \end{aligned}$$

Now, we integrate $f(z):=\dfrac{G(z)}{z^{n+1}},\,G(z)=\dfrac{E_{q}(\frac{-z}{2})}{g^{(2)}_{\alpha }(iz;q)}$ on the contour $\Gamma _{m},$ where $\Gamma _{m}$ is a circle of radius $R_{m},$ $\mid z_{m}\mid<\,R_{m}<\mid z_{m+1}\mid$. From the Cauchy Residue Theorem, see [2],

$$\begin{aligned} \int _{\Gamma _{m}} f(z) \,dz=2\pi i\sum \,Res(f,z_{k} ), \end{aligned}$$

where $\{z_{k}\}$ are the poles of f that lie inside $\Gamma _{m}$. The function f(z) has a pole at $z=0$ of order $n+1$ and simple poles at $\pm z_{k}$ where $z_{k}= \dfrac{ij^{(2)}_{k,\alpha }}{1-q},\,k\in {\mathbb {N}}$. Consequently,

$$\begin{aligned} I_{m}=\frac{1}{2\pi i}\int _{\Gamma _{m}} f(z) \,dz=Res(f(z),0)+\sum _{k=1}^{m}Res(f(z),\pm z_{k} ). \end{aligned}$$

(3.2)

Since

$$\begin{aligned} Res(f,0)= & {} \frac{f^{n}(0)}{n!}=\frac{\beta _{n,\alpha }(q)}{[n]_{q}!}, \\ \quad Res(f,z_{k} )= & {} \frac{E_{q}(\frac{-z_{k}}{2})}{\frac{d}{dz}g^{(2)}_{\alpha }(iz;q)\mid _{z=z_{k}}}\frac{1}{(z_{k})^{n+1}}\\ {}= & {} \frac{E_{q}(\frac{-ij^{(2)}_{k,\alpha }}{2(1-q)})}{\frac{d}{dz}{\mathcal {J}}^{(2)}_{\alpha }(z;q^{2})\mid _{z=j^{(2)}_{k,\alpha }}}\frac{(i)^{-n}(1-q)^{n}}{(j^{(2)}_{k,\alpha })^{n+1}}, \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} Res(f,-z_{k} )&=\frac{E_{q}(\frac{z_{k}}{2})}{\frac{d}{dz}g^{(2)}_{\alpha }(iz;q)\mid _{z=-z_{k}}}\frac{1}{(-z_{k})^{n+1}}\\&=\frac{E_{q}(\frac{ij^{(2)}_{k,\alpha }}{2(1-q)})}{\frac{d}{dz}{\mathcal {J}}^{(2)}_{\alpha }(z;q^{2})\mid _{z=j^{(2)}_{k,\alpha }}}\frac{(-i)^{-n}(1-q)^{n}}{(j^{(2)}_{k,\alpha })^{n+1}}. \end{aligned} \end{aligned}$$

Then Eq. (3.2) can be written as

$$\begin{aligned} \begin{aligned} I_{m}&= \frac{\beta _{n,\alpha }(q)}{[n]_{q}!}\\ {}&\quad +\sum _{k=1}^{m}2Re\,\left( (-i)^{-n}E_{q}\left( \frac{ij^{(2)}_{k,\alpha }}{2(1-q)}\right) \right) \frac{(1-q)^{n}}{(j^{(2)}_{k,\alpha })^{n+1}\frac{d}{dz}{\mathcal {J}}^{(2)}_{\alpha }(z;q^{2})\mid _{z=j^{(2)}_{k,\alpha }}}, \end{aligned}\nonumber \\ \end{aligned}$$

(3.3)

substituting into (3.3) with $-i=e^{-\frac{i\pi }{2}}$ gives

$$\begin{aligned} \begin{aligned} I_{m}&= \frac{\beta _{n,\alpha }(q)}{[n]_{q}!}+2(1-q)^{n}\cos \frac{n\pi }{2} \sum _{k=1}^{m}\frac{Cos_{q}\left( \frac{j^{(2)}_{k,\alpha }}{2(1-q)}\right) }{\frac{d}{dz}{\mathcal {J}}^{(2)}_{\alpha }(z;q^{2})\mid _{z=j^{(2)}_{k,\alpha }}}\frac{1}{(j^{(2)}_{k,\alpha })^{n+1}} \\&\quad -2(1-q)^{n}\sin \frac{n\pi }{2}\sum _{k=1}^{m}\frac{Sin_{q}\left( \frac{j^{(2)}_{k,\alpha }}{2(1-q)}\right) }{\frac{d}{dz}{\mathcal {J}}^{(2)}_{\alpha }(z;q^{2})\mid _{z=j^{(2)}_{k,\alpha }}} \frac{1}{(j^{(2)}_{k,\alpha })^{n+1}}. \end{aligned} \end{aligned}$$

Now, we show that the integral $I_{m}\rightarrow 0$ as $m\rightarrow \infty$. Bergweiller and Hayman [7] introduced the asymptotic relation for $E_{q}(z)$,

$$\begin{aligned} \mid M(r;E_{q})\mid :=\sup \{\mid E_{q}(z)\mid :\mid z\mid =r\}\sim \,e^{\frac{-(\log r)^{2}}{2\log q}},\quad \text { when} \quad r=\mid z\mid \rightarrow \infty . \end{aligned}$$

In [4], Annaby and Mansour proved that for $r=\mid z\mid \,\rightarrow \,\infty$

$$\begin{aligned} z^{-\nu }J_{\nu }^{(2)}(z;q) \sim \,\exp \left( -\frac{(\log \,r)^{2}}{2\log \,q}-\frac{\log \,2}{\log \,q}\log \,r\right) . \end{aligned}$$

Hayman in [15] introduced the higher order asymptotics of $J_{\nu }^{(2)}(z;q)$. Then, Annaby and Mansour, see [4], pointed out that the first order asymptotics of the zeros of $J_{\nu }^{(2)}(z;q^{2})$ is given by

$$\begin{aligned} j_{m,\nu }^{(2)}=2q^{-2m}q^{-\nu +1}(1+O(q^{2m})), \quad (m\rightarrow \infty ). \end{aligned}$$

Hence if $(z_{m})_{m}$ are the positive zeros of $g^{(2)}_{\alpha }(iz;q)$, then

$$\begin{aligned} \lim _{m\rightarrow \infty }\frac{z_{m}}{z_{m+1}}=\lim _{m\rightarrow \infty }\frac{j^{(2)}_{m,\nu }}{j^{(2)}_{m+1,\nu }}=q^{2}, \quad \lim _{m\rightarrow \infty }z_{m}=\infty . \end{aligned}$$

(3.4)

Let $0<\epsilon <(q^{-1}-1)$. There exists $M_{0}\in {\mathbb {N}}$ such that if $m\in {\mathbb {N}},\,m\ge M_{0}$, then

$$\begin{aligned} q^{2}(1-\epsilon )<\frac{z_{m}}{z_{m+1}}<q^{2}(1+\epsilon ). \end{aligned}$$

Hence $z_{m}<qz_{m+1}$ for all $m\ge M_{0}$. We can choose $R_{m},\,\delta :=q^{-1}\displaystyle \sup _{m\ge M_{0}}\frac{z_{m}}{z_{m+1}}$ such that $(z_{m}<\delta R_{m}<q z_{m+1} <R_{m})$. Indeed,

$$\begin{aligned} \delta =q^{-1}\displaystyle \sup _{m\ge M_{0}}\frac{z_{m}}{z_{m+1}}\ge q^{-1}\frac{z_{m}}{z_{m+1}},\,\, m\ge M_{0}. \end{aligned}$$

But $q z_{m+1} <R_{m}$ leads to $\delta >\frac{z_{m}}{R_{m}}$ and so $z_{m} <\delta R_{m}$. Now,

$$\begin{aligned} \delta =q^{-1}\displaystyle \sup _{ m\ge M_{0}}\frac{z_{m}}{z_{m+1}}\ge q^{-1}\displaystyle \lim _{m\rightarrow \infty }\frac{z_{m}}{z_{m+1}}=q^{-1}q^{2}=q. \end{aligned}$$

Also $\delta =q^{-1}\displaystyle \sup _{ m\ge M_{0}}\frac{z_{m}}{z_{m+1}}\le q(1+\epsilon )<1$. Hence $1>\delta \ge q$ and so by

$$\begin{aligned} z_{m}\le R_{m}\le \frac{q}{\delta } z_{m+1}\le z_{m+1}, \end{aligned}$$

(3.5)

the annulus $\delta R_{m}<|z| <R_{m}$ has no zeros of the function $g^{(2)}_{\alpha }(iz;q)$. Hence, from the minimum modulus principle we have

$$\begin{aligned} \begin{aligned} \mid g^{(2)}_{\alpha }(iz;q)\mid&\ge c_{1}\,e^{-\frac{(\log \delta R_{m})^{2}}{2\log q}-\frac{\log 2}{\log q}\log \delta R_{m}},\quad c_{1}>0.\\ \mid E_{q}(\frac{-z}{2}) \mid&\le \,c_{2}\,e^{\frac{-(\log \frac{R_{m}}{2})^{2}}{2\log q}},\quad c_{2}>0.\end{aligned} \end{aligned}$$

(3.6)

Therefore, from (3.6), we conclude that

$$\begin{aligned} \begin{aligned} \mid \frac{E_{q}(\frac{-z}{2})}{g^{(2)}_{\alpha }(iz;q)} \mid&\le \,\frac{c_{2}}{c_{1}}\,\,\frac{e^{\frac{-(\log \frac{ R_{m}}{2})^{2}}{2\log q}}}{e^{-\frac{( \log \delta R_{m})^{2}}{2\log q}-\frac{\log 2}{\log q}\log \delta R_{m}}}\\&\le \,\frac{ c_{2}}{c_{1}}\,e^{\frac{1}{2\log q}\left( ((\log \delta R_{m})^{2}-(\log \frac{R_{m}}{2})^{2})\right) +\frac{\log 2}{\log q}\log \delta R_{m}} \\&\le \,\frac{ c_{2}}{c_{1}}\,e^{K}\,e^{\frac{2\log 2 \log R_{m}}{\log q}+\frac{\log \delta \log R_{m}}{\log q}}, \end{aligned} \end{aligned}$$

where

$$\begin{aligned} K=\frac{1}{2\log q}\left( (\log \delta )^{2} -(\log 2)^{2}+2\log 2\log \delta \right) . \end{aligned}$$

Now, using the ML-inequality (see [2]) to obtain

$$\begin{aligned} \begin{aligned} \mid I_{m}\mid&=\mid \int _{\Gamma _{m}}f(z)dz\mid \le (2\pi R_{m})\mid M(r;f(z))\mid \\&\le \,\frac{2\pi R_{m}c_{2}}{c_{1}}\,e^{K}\,e^{\frac{2\log 2 \log R_{m}}{\log q}+\frac{\log \delta \log R_{m}}{\log q}}\frac{1}{R_{m}^{n+1}}\\&\le \,\frac{ 2\pi c_{2}}{c_{1}}\,e^{K}\,R_{m}^{\frac{2 \log 2}{\log q}}R_{m}^{\frac{\log \delta }{\log q}-n}. \end{aligned} \end{aligned}$$

(3.7)

From (3.4) and (3.5), we have $\displaystyle \lim _{m\rightarrow \infty }R_{m}=\infty$. Also, since $0<q<1$ and $1>\delta \ge q$ then

$$\begin{aligned} R_{m}^{\frac{2 \log 2}{\log q}}\rightarrow 0\quad and \quad R_{m}^{\frac{\log \delta }{\log q}-n}\rightarrow 0 \quad as\,\, m\,\,\rightarrow \,\infty . \end{aligned}$$

Hence $\displaystyle \lim _{m\rightarrow \infty }I_{m}=0.$ Consequently,

$$\begin{aligned} \begin{aligned} \frac{\beta _{n,\alpha }(q)}{[n]_{q}!}&=-2(1-q)^{n}\cos \frac{n\pi }{2} \sum _{k=1}^{\infty }\frac{Cos_{q}\left( \frac{j^{(2)}_{k,\alpha }}{2(1-q)}\right) }{\frac{d}{dz}{\mathcal {J}}^{(2)}_{\alpha }(z;q^{2})\mid _{z=j^{(2)}_{k,\alpha }}}\frac{1}{(j^{(2)}_{k,\alpha })^{n+1}} \\&\quad +2(1-q)^{n}\sin \frac{n\pi }{2}\sum _{k=1}^{\infty }\frac{Sin_{q}\left( \frac{j^{(2)}_{k,\alpha }}{2(1-q)}\right) }{\frac{d}{dz}{\mathcal {J}}^{(2)}_{\alpha }(z;q)\mid _{z=j^{(2)}_{k,\alpha }}} \frac{1}{(j^{(2)}_{k,\alpha })^{n+1}}. \end{aligned} \end{aligned}$$

Therefore,

$$\begin{aligned} \begin{aligned} \beta _{2n,\alpha }(q)&=2(-1)^{n+1}(q;q)_{2n} \sum _{k=1}^{\infty }\frac{Cos_{q}\left( \frac{j^{(2)}_{k,\alpha }}{2(1-q)}\right) }{(j^{(2)}_{k,\alpha })^{2n+1}\, \frac{d}{dz}{\mathcal {J}}_{\alpha }^{(2)}(z;q^{2})\mid _{z=j^{(2)}_{k,\alpha }}},\\ \beta _{2n+1,\alpha }(q)&=2(-1)^{n}(q;q)_{2n+1} \sum _{k=1}^{\infty }\frac{Sin_{q}\left( \frac{j^{(2)}_{k,\alpha }}{2(1-q)}\right) }{(j^{(2)}_{k,\alpha })^{2n+2}\,\frac{d}{dz}{\mathcal {J}}_{\alpha }^{(2)}(z;q^{2})\mid _{z=j^{(2)}_{k,\alpha }}}, \end{aligned} \end{aligned}$$

which completes the proof of the theorem. $\square$

Remark 3.2

If we substitute with $\alpha =\frac{1}{2}$ in the second equation in (3.1), then $(z_{k})_{k}$ will be the positive zeros of $Sin_{q}(z)$ and consequently, the series in the left hand side vanishes which coincide with the known result that the odd Bernoulli numbers vanish $(\beta _{2n+1}(q)=0,\,n\ge \, 1)$ (see [19]). Similarly, if we set $\alpha =-\frac{1}{2}$ in the first equation in (3.1), the series in the left hand side vanishes and this coincide with the fact that the even Euler$^{,}$s numbers are zero $(E_{2n}(q)=0,\,n\ge \,1)$ (see [19]).

Corollary 3.3

The asymptotic relations of the generalized q-Bernoulli numbers $(\beta _{n,\alpha }(q))_{n},$

$$\begin{aligned} \begin{aligned} \beta _{2n,\alpha }(q)&=2(-1)^{n+1}(q;q)_{2n} \frac{Cos_{q}\left( \frac{j^{(2)}_{1,\alpha }}{2(1-q)}\right) }{(j^{(2)}_{1,\alpha })^{2n+1}\, \frac{d}{dz}{\mathcal {J}}_{\alpha }^{(2)}(z;q^{2})\mid _{z=j^{(2)}_{1,\alpha }}}\left( 1+o(1)\right) ,\\ \beta _{2n+1,\alpha }(q)&=2(-1)^{n}(q;q)_{2n+1} \frac{Sin_{q}\left( \frac{j^{(2)}_{1,\alpha }}{2(1-q)}\right) }{(j^{(2)}_{1,\alpha })^{2n+2}\,\frac{d}{dz}{\mathcal {J}}_{\alpha }^{(2)}(z;q^{2})\mid _{z=j^{(2)}_{1,\alpha }}}\left( 1+o(1)\right) , \end{aligned} \end{aligned}$$

where ${\mathcal {J}}_{\alpha }^{(2)} (z;q)$ is defined in (1.7).

Proof

The proof follows directly from Theorem 3.1. $\square$

4 Applications of the Generalized q-Bernoulli Polynomials

In this section, we introduce connection relations between the generalized q-Bernoulli polynomials $B^{(k)}_{n,\alpha }(x;q)\,(k=1,2,3)$ and the q-Laguerre and the little q-Legendre polynomials.

The q-Laguerre polynomials $L_{n}^{\alpha }(x;q)$ of degree n are defined by

$$\begin{aligned} \begin{aligned} L_{n}^{\alpha }(x;q):&=\frac{1}{(q;q)_{n}}\,_{2}\phi _{1} \left( \begin{array}{c} q^{-n},-x \\ 0 \\ \end{array} q;q^{n+\alpha +1} \right) \\&=\frac{(q^{\alpha +1};q)_{n}}{(q;q)_{n}}\sum _{k=0}^{n} \frac{(q^{-n};q)_{k}}{(q^{\alpha +1};q)_{k}}(-1)^{k}(q^{n+\alpha +1})^{k}x^{k}. \end{aligned} \end{aligned}$$

(4.1)

The Rodrigues formula is given by

$$\begin{aligned} \begin{aligned} L_{n}^{\alpha }(x;q)=\frac{(1-q)^{n}}{(q;q)_{n}}(-x;q)_{\infty }\,x^{-\alpha }\,D_{q}^{n}\left( \frac{x^{\alpha +n}}{(-x;q)_{\infty }}\right) , \end{aligned} \end{aligned}$$

(4.2)

and the orthogonality relation is

$$\begin{aligned} \begin{aligned}&\int _{0}^{\infty }\frac{x^{\alpha }}{(-x;q)_{\infty }}L_{m}^{\alpha }(x;q)L_{n}^{\alpha }(x;q)dx \\&\quad =\frac{(q^{-\alpha };q)_{\infty }}{(q;q)_{\infty }}\frac{(q^{\alpha +1};q)_{n}}{(q;q)_{n}q^{n}}\Gamma _{q}(-\alpha )\Gamma _{q}(\alpha +1)\delta _{mn},\end{aligned}\nonumber \\ \end{aligned}$$

(4.3)

$\alpha >-1$, where $\delta _{mn}$ is the Kronecker delta function, see [17, 21]. The q-Laguerre polynomials $L_{n}^{\alpha }(x;q)$ satisfy three term recurrence relation

$$\begin{aligned} \begin{aligned} -xa_{n}L_{n}^{\alpha }(x;q)=L_{n+1}^{\alpha }(x;q)-b_{n}L_{n}^{\alpha }(x;q)+d_{n}L_{n-1}^{\alpha }(x;q), \end{aligned} \end{aligned}$$

where

$$\begin{aligned} a_{n}=\frac{q^{2n+\alpha +1}}{1-q^{n+1}},\quad b_{n}=1+\frac{q(1-q^{n+\alpha }}{1-q^{n+1}},\quad d_{n}=\frac{q(1-q^{n+\alpha })}{1-q^{n+1}}. \end{aligned}$$

In the following, let $\alpha >-1$ and ${\mathbb {P}}_{n}=\{p(x):\text{ deg }\, p(x)\le n\}$ with the inner product

$$\begin{aligned} \langle p(x),g(x)\rangle = \int _{0}^{\infty }\frac{x^{\alpha }}{(-x;q)_{\infty }}p(x)g(x)dx, \end{aligned}$$

where $p(x),\,g(x)\in {\mathbb {P}}_{n}$.

Theorem 4.1

Let p(x) $\in {\mathbb {P}}_{n}.$ Then p(x) can be expanded as

$$\begin{aligned} p(x)=\sum _{m=0}^{n}C_{m}L_{m}^{\alpha }(x;q), \end{aligned}$$

where

$$\begin{aligned} C_{m}=\frac{q^{m}(1-q)^{m-1}(q^{\alpha +m+1};q)_{\infty }}{(q;q)_{\infty }}\int _{0}^{\infty } \,D_{q}^{m}\left( \frac{x^{\alpha +m}}{(-x;q)_{\infty } }\right) p(x) dx. \end{aligned}$$

Proof

Since

$$\begin{aligned} p(x)=\sum _{m=0}^{n}C_{m}L_{m}^{\alpha }(x;q), \end{aligned}$$

in order to calculate the constant $C_{m}$, we use (4.3) to obtain

$$\begin{aligned} \left\langle p(x),L_{k}^{\alpha }(x;q)\right\rangle= & {} \left\langle \sum _{m=0}^{n}C_{m}L_{m}^{\alpha }(x;q),L_{k}^{\alpha }(x;q)\right\rangle \\ {}= & {} \sum _{m=0}^{n}C_{m}\langle L_{m}^{\alpha }(x;q),L_{k}^{\alpha }(x;q)\rangle . \end{aligned}$$

Then

$$\begin{aligned} \begin{aligned} \langle p(x),L_{m}^{\alpha }(x;q)\rangle&=C_{m}\langle L_{m}^{\alpha }(x;q) ,L_{m}^{\alpha }(x;q)\rangle \\&=C_{m}\frac{(q^{\alpha +1};q)_{m}}{q^{m}(q;q)_{m}}(1-q)^{1+\alpha }\Gamma _{q}(\alpha +1). \end{aligned} \end{aligned}$$

Therefore,

$$\begin{aligned} \begin{aligned} C_{m}= \frac{q^{m}(q;q)_{m}}{(q^{\alpha +1};q)_{m}(1-q)^{1+\alpha }\Gamma _{q}(\alpha +1)}\int _{0}^{\infty }\frac{x^{\alpha }}{(-x;q)_{\infty }}L_{m}^{\alpha }(x;q) p(x) dx. \end{aligned} \end{aligned}$$

(4.4)

Using (4.2) with n replaced by m, we obtain

$$\begin{aligned} \begin{aligned} C_{m}=\frac{q^{m}(1-q)^{m-1}(q^{\alpha +m+1};q)_{\infty }}{(q;q)_{\infty }}\int _{0}^{\infty } \,D_{q}^{m}\left( \frac{x^{\alpha +m}}{(-x;q)_{\infty } }\right) p(x) dx, \end{aligned} \end{aligned}$$

and the theorem follows. $\square$

The following Lemma, see [18], is essential in the proof of Theorem 4.3.

Lemma 4.2

Let the functions f and g be defined and continuous on $[0,\infty ]$. Assume that the improper Riemann integrals of the functions f(x)g(x) and f(x/q)g(x) exist on $[0,\infty ]$. Then

$$\begin{aligned} \begin{aligned} \int _{0}^{\infty }f(x)D_{q}g(x)dx&=\frac{f(0)g(0)}{1-q}\ln q-\frac{1}{q}\int _{0}^{\infty }g(x)D_{q^{-1}}f(x)dx\\&=\frac{f(0)g(0)}{1-q}\ln q-\int _{0}^{\infty }g(qx)D_{q}f(x)dx. \end{aligned} \end{aligned}$$

Theorem 4.3

If $n\in {\mathbb {N}}$ and $x\in {\mathbb {C}},$ then

$$\begin{aligned} B^{(1)}_{n,\alpha }(x;q)&=\sum _{m=0}^{n}A_{m}\left( \sum _{k=m}^{n}q^{\frac{k(2n-k+1)}{2}}\frac{ (q^{-n},q^{-\alpha -k};q)_{k} (q^{-k};q)_{m}}{(q;q)_{k}} \beta _{n-k,\alpha } (q)\right) L_{m} ^{\alpha }(x;q),\\ B^{(2)}_{n,\alpha }(x;q)&=\sum _{m=0}^{n}A_{m}\left( \sum _{k=m}^{n}\frac{q^{nk}(q^{-n},q^{-\alpha -k};q)_{k} (q^{-k};q)_{m}}{(q;q)_{k}} \beta _{n-k,\alpha } (q)\right) L_{m} ^{\alpha }(x;q),\\ B^{(3)}_{n,\alpha }(x;q)&=\sum _{m=0}^{n}A_{m}\left( \sum _{k=m}^{n}q^{\frac{k(4n-k+1)}{4}}\frac{ (q^{-n},q^{-\alpha -k};q)_{k} (q^{-k};q)_{m}}{(q;q)_{k}} \beta ^{(3)}_{n-k,\alpha } (q)\right) L_{m} ^{\alpha }(x;q), \end{aligned}$$

where

$$\begin{aligned} A_{m}=\frac{-q^{m}(q^{\alpha +m+1},q^{-\alpha };q)_{\infty }}{(1-q)(q,q;q)_{\infty }}\frac{\pi }{sin(\alpha \pi )}. \end{aligned}$$

Proof

We prove the identity for $B^{(1)}_{n,\alpha }(x;q)$ and the proofs for $B^{(k)}_{n,\alpha }(x;q)\,(k=2,3)$ are similar. Substitute with $p(x)=B^{(1)}_{n,\alpha }(x;q)$ in (4.4). This gives

$$\begin{aligned} \begin{aligned} C_{m}= \frac{q^{m}(q;q)_{m}}{(q^{\alpha +1};q)_{m}(1-q)^{1+\alpha }\Gamma _{q}(\alpha +1)}\int _{0}^{\infty }\frac{x^{\alpha }}{(-x;q)_{\infty }}L_{m}^{\alpha }(x;q) B^{(1)}_{n,\alpha }(x;q) dx. \end{aligned}\nonumber \\ \end{aligned}$$

(4.5)

Since $\{L_{m}^{\alpha }(x;q)\}_{n\in {\mathbb {N}}}$ is an orthogonal polynomials sequence then $C_{m}=0$ for $m>n$, and

$$\begin{aligned} B^{(1)}_{n,\alpha }(x;q) =\sum _{m=0}^{n}C_{m}L_{m}^{\alpha }(x;q). \end{aligned}$$

Now, we calculate $C_{m}$. Using (2.12) in (4.5) gives

$$\begin{aligned} \begin{aligned} C_{m}&=\frac{q^{m}(q;q)_{m}}{(q^{\alpha +1};q)_{m}(1-q)^{1+\alpha }\Gamma _{q}(\alpha +1)}\sum _{k=0}^{n}\left[ \begin{array}{c} n \\ k \\ \end{array} \right] _{q} \beta _{n-k,\alpha } (q) \int _{0}^{\infty } \,\frac{x^{\alpha +k}}{(-x;q)_{\infty }}L_{m}^{\alpha }(x;q)dx.\end{aligned} \end{aligned}$$

Since

$$\begin{aligned} \int _{0}^{\infty }\frac{x^{\alpha }}{(-x;q)_{\infty }} L_{m}^{\alpha }(x;q)x^{k} dx=0, \quad \text {for}\,\, k<m, \end{aligned}$$

then

$$\begin{aligned} \begin{aligned} C_{m}&=\frac{q^{m}(q;q)_{m}}{(q^{\alpha +1};q)_{m}(1-q)^{1+\alpha }\Gamma _{q}(\alpha +1)}\sum _{k=m}^{n}\left[ \begin{array}{c} n \\ k \\ \end{array} \right] _{q} \beta _{n-k,\alpha } (q) \int _{0}^{\infty } \,\frac{x^{\alpha +k}}{(-x;q)_{\infty }}L_{m}^{\alpha }(x;q) dx. \end{aligned} \end{aligned}$$

From (4.2), we get

$$\begin{aligned} \begin{aligned} C_{m}&=\frac{q^{m}(1-q)^{m-1}(q^{\alpha +m+1};q)_{\infty }}{(q;q)_{\infty }}\sum _{k=m}^{n}\left[ \begin{array}{c} n \\ k \\ \end{array} \right] _{q} \beta _{n-k,\alpha } (q) \int _{0}^{\infty } \,D_{q}^{m}\left( \frac{x^{\alpha +m}}{(-x;q)_{\infty }}\right) x^{k} dx, \end{aligned} \end{aligned}$$

then applying the q-integration by part introduced in Lemma 4.2m times, we obtain

$$\begin{aligned} \begin{aligned} C_{m}=\,&\frac{(-1)^{m}q^{m}(1-q)^{m-1}(q^{\alpha +m+1};q)_{\infty }}{(q;q)_{\infty }}\\&\times \sum _{k=m}^{n}\left[ \begin{array}{c} n \\ k \\ \end{array} \right] _{q} \left( \prod _{i=0}^{m-1}q^{i-k}\right) \frac{[k]_{q}!}{[k-m]_{q}!} \beta _{n-k,\alpha } (q) \int _{0}^{\infty } \frac{x^{\alpha +k}}{(-x;q)_{\infty }}dx. \end{aligned} \end{aligned}$$

From [16, Eq. (5.4), p. 465],

$$\begin{aligned} \frac{1}{ \Gamma _{q}(z)}=\frac{sin\pi z}{\pi }\int _{0}^{\infty } \frac{ t^{-z}}{(-t(1-q);q)_{\infty }}dt, \,\,\,Re\,z>0. \end{aligned}$$

Then

$$\begin{aligned} \int _{0}^{\infty } \frac{x^{\alpha +k}}{(-x;q)_{\infty }}dx= \frac{\pi }{sin(-\alpha -k)\pi }\frac{1}{ \Gamma _{q}(-\alpha -k)}(1-q)^{\alpha +k+1}. \end{aligned}$$

Therefore,

$$\begin{aligned} \begin{aligned} C_{m}&=\frac{(-1)^{m}q^{m}(q^{\alpha +m+1};q)_{\infty }}{(1-q)(q,q;q)_{\infty }}\\&\quad \times \sum _{k=m}^{n} \left( \prod _{i=0}^{m-1}q^{i-k}\right) \frac{(q;q)_{n}(q^{-\alpha -k};q)_{\infty }}{(q;q)_{n-k}(q;q)_{k-m}}\frac{\pi }{sin(-\alpha -k)\pi }\beta _{n-k,\alpha }(q). \end{aligned}\nonumber \\ \end{aligned}$$

(4.6)

Since

$$\begin{aligned} \frac{\pi }{sin(-\alpha -k)\pi } =(-1)^{k-1}\frac{\pi }{sin(\alpha \pi )},\quad \prod _{i=0}^{m-1}q^{i-k}=q^{\frac{m(m-1)}{2}}q^{-km}, \end{aligned}$$

(4.7)

then substituting from (4.7) into (4.6), we get

$$\begin{aligned} \begin{aligned} C_{m}=\,&\frac{(-1)^{m}q^{m}q^{m(m-1)/2}(q^{\alpha +m+1};q)_{\infty }(q^{-\alpha };q)_{\infty }}{(1-q)(q,q;q)_{\infty }}\frac{\pi }{sin(\alpha \pi )}\\&\times \sum _{k=m}^{n} (-1)^{k-1}q^{-km} \frac{(q;q)_{n}(q^{-\alpha -k};q)_{k}}{(q;q)_{n-k}(q;q)_{k-m}}\beta _{n-k,\alpha }(q). \end{aligned} \end{aligned}$$

Using the relation (2.47), we obtain

$$\begin{aligned} \begin{aligned} C_{m}=\,&\frac{-q^{m}(q^{\alpha +m+1};q^{-\alpha };q)_{\infty }}{(1-q)(q,q;q)_{\infty }}\frac{\pi }{sin(\alpha \pi )}\\&\times \sum _{k=m}^{n}q^{\frac{k(2n-k+1)}{2}}\frac{ (q^{-n};q)_{k} (q^{-k};q)_{m}(q^{-\alpha -k};q)_{k}}{(q;q)_{k}} \beta _{n-k,\alpha }(q), \end{aligned} \end{aligned}$$

and this completes the proof of the theorem. $\square$

The little q-Legendre polynomials $( P_{n}(x\mid q))_{n}$ are defined by

$$\begin{aligned} \begin{aligned} P_{n}(x\mid q)&=\,_{2}\phi _{1} \left( \begin{array}{c} q^{-n},q^{n+1} \\ q \\ \end{array} q;qx \right) \\&=\sum _{k=0}^{n}\frac{(q^{-n};q)_{k}(q^{n+1};q)_{k}}{(q;q)_{k}}\frac{q^{k}x^{k}}{(q;q)_{k}}. \end{aligned} \end{aligned}$$

They satisfy the Rodrigues formula

$$\begin{aligned} P_{n}(x\mid q)=\frac{q^{n(n-1)/2}(1-q)^{n}}{(q;q)_{n}}\,D_{q^{-1}}^{n}(x^{n}(qx;q)_{n}), \quad \text{ for }\,\, n\ge 0, \end{aligned}$$

(4.8)

and the orthogonality relation

$$\begin{aligned} \int _{0}^{1}P_{m}(x\mid q)P_{n}(x \mid q)d_{q}x =\frac{(1-q)}{(1-q^{2n+1})}\delta _{mn},\quad \text{ for }\,\, m,\,n\ge 0, \end{aligned}$$

(4.9)

see [21]. Let ${\mathbb {P}}_{n}=\{g(x):\text{ deg }\, g(x)\le n\}$ with the inner product

$$\begin{aligned} \langle g(x),p(x)\rangle = \int _{0}^{1}g(x)p(x)d_{q}x, \end{aligned}$$

where $p(x),\,g(x)\in {\mathbb {P}}_{n}$.

Theorem 4.4

Let g(x) $\in {\mathbb {P}}_{n}$. Then g(x) can be represented by

$$\begin{aligned} g(x)=\sum _{k=0}^{n}C_{k}P_{k}(x\mid q), \end{aligned}$$

where

$$\begin{aligned} \begin{aligned} C_{k}=\frac{q^{k(k-1)/2}(1-q)^{k-1}(1-q^{2k+1})}{(q;q)_{k}}\int _{0}^{1}\,D_{q^{-1}}^{k}(x^{k}(qx;q)_{k})g(x)d_{q}x. \end{aligned} \end{aligned}$$

Proof

Since

$$\begin{aligned} g(x)=\sum _{k=0}^{n}C_{k}P_{k}(x\mid q), \end{aligned}$$

then by the orthogonality relation (4.9), we obtain

$$\begin{aligned} C_{k}=\frac{(1-q^{2k+1})}{(1-q)} \langle g(x),P_{k}(x \mid q)\rangle =\frac{(1-q^{2k+1})}{(1-q)}\int _{0}^{1}P_{k}(x\mid q)g(x)d_{q}x.\nonumber \\ \end{aligned}$$

(4.10)

By using (4.8), we get

$$\begin{aligned} \begin{aligned} C_{k}=\frac{q^{k(k-1)/2}(1-q)^{k-1}(1-q^{2k+1})}{(q;q)_{k}}\int _{0}^{1}\,D_{q^{-1}}^{k}(x^{k}(qx;q)_{k})g(x)d_{q}x, \end{aligned} \end{aligned}$$

which readily gives the result. $\square$

Theorem 4.5

For $n\in {\mathbb {N}}$ and $x\in {\mathbb {C}}$,

$$\begin{aligned} \begin{aligned} B^{(1)}_{n,\alpha }(x;q)&=\sum _{k=0}^{n}\lambda _{k}\left( \sum _{m=k}^{n}(-1)^{m}q^{\frac{m(2n-m+1)}{2}+1}\,\frac{(q^{-n};q)_{m}(q^{-m};q)_{k}}{(q;q)_{m+k+1}}\beta _{n-m,\alpha }(q)\right) P_{k}(x \mid q),\\ B^{(2)}_{n,\alpha }(x;q)&=\sum _{k=0}^{n}\lambda _{k}\left( \sum _{m=k}^{n}(-1)^{m}q^{nm+1}\, \frac{(q^{-n};q)_{m}(q^{-m};q)_{k}}{(q;q)_{m+k+1}}\beta _{n-m,\alpha }(q)\right) P_{k}(x \mid q),\\ B^{(3)}_{n,\alpha }(x;q)&=\sum _{k=0}^{n}\lambda _{k}\left( \sum _{m=k}^{n}(-1)^{m}q^{\frac{m(4n-m+1)}{4}+1}\, \frac{(q^{-n};q)_{m}(q^{-m};q)_{k}}{(q;q)_{m+k+1}}\beta ^{(3)}_{n-m,\alpha }(q)\right) P_{k}(x\mid q),\end{aligned} \end{aligned}$$

where

$$\begin{aligned} \lambda _{k}=q^{\frac{-k(k-3)}{2}}(1-q^{2k+1}). \end{aligned}$$

Proof

Substitute with $g(x)=B^{(1)}_{n,\alpha }(x;q)$ in (4.10), we obtain

$$\begin{aligned} \begin{aligned} C_{k} = \dfrac{(1-q^{2k+1})}{(1-q)}\int _{0}^{1}P_{k}(x \mid q) B^{(1)}_{n,\alpha }(x;q)d_{q}x. \end{aligned} \end{aligned}$$

(4.11)

Since the polynomials $\{P_{k}(x\mid q)\}$ are orthogonal, then $C_{k}=0$ for $k>n$, and

$$\begin{aligned} B^{(1)}_{n,\alpha }(x;q) =\sum _{k=0}^{n}C_{k}P_{k}(x\mid q). \end{aligned}$$

(4.12)

Set

$$\begin{aligned} B^{(1)}_{n,\alpha }(x;q)=\sum _{m=0}^{n}\left[ \begin{array}{c} n \\ m \\ \end{array} \right] _{q} \beta _{n-m,\alpha } (q) x^{m}. \end{aligned}$$

From (4.11),

$$\begin{aligned} \begin{aligned} C_{k}&=\dfrac{(1-q^{2k+1})}{(1-q)} \sum _{m=0}^{n}\left[ \begin{array}{c} n \\ m \\ \end{array} \right] _{q} \beta _{n-m,\alpha } (q)\int _{0}^{1}P_{k}(x\mid q) x^{m}d_{q}x\\&=\frac{(1-q^{2k+1})}{(1-q)} \sum _{m=k}^{n}\left[ \begin{array}{c} n \\ m \\ \end{array} \right] _{q} \beta _{n-m,\alpha } (q)\int _{0}^{1}P_{k}(x\mid q) x^{m}d_{q}x, \end{aligned} \end{aligned}$$

since

$$\begin{aligned} \int _{0}^{1} P_{k}(x\mid q) x^{m}d_{q}x=0\quad for\,\,m<k. \end{aligned}$$

Hence, by the Rodrigues formula in (4.8), we obtain

$$\begin{aligned} \begin{aligned} C_{k}&= \dfrac{(1-q^{2k+1})q^{k(k-1)/2}(1-q)^{k-1}}{(q;q)_{k}} \sum _{m=k}^{n}\left[ \begin{array}{c} n \\ m \\ \end{array} \right] _{q} \beta _{n-m,\alpha } (q)\\&\quad \times \int _{0}^{1}\,D_{q^{-1}}^{k}(x^{k}(qx;q)_{k}) x^{m}d_{q}x. \end{aligned} \end{aligned}$$

(4.13)

Using the $q^{-1}$-integration by parts

$$\begin{aligned} \int _{0}^{a}f\left( \frac{t}{q}\right) D_{q^{-1}}g(t)d_{q}t =q\left( (fg)\left( \frac{a}{q}\right) -(fg)(0) \right) -\int _{0}^{a}g(t)D_{q^{-1}}f(t)d_{q}t,\nonumber \\ \end{aligned}$$

(4.14)

where f and g are continuous functions at zero, see [5]. This gives

$$\begin{aligned} \begin{aligned} \int _{0}^{1}\,D_{q^{-1}}^{k}(x^{k}(qx;q)_{k}) x^{m}d_{q}x=&q\left[ x^{m}D_{q^{-1}}^{k-1}(x^{k}(qx;q)_{k})\right] _{0}^{\frac{1}{q}} \\&-[m]_{q}q^{1-m}\int _{0}^{1}x^{m-1}D_{q^{-1}}^{k-1}(x^{k}(qx;q)_{k})d_{q}x. \end{aligned} \end{aligned}$$

(4.15)

The first term on the right hand side of (4.15) vanishes because

$$\begin{aligned} D_{q^{-1}}(x^{k}(qx;q)_{k})=[k]_{q^{-1}}x^{k-1}(x;q)_{k}+x^{k}D_{q^{-1}}(qx;q)_{k}, \end{aligned}$$

and

$$\begin{aligned} D_{q^{-1}}^{j}(qx;q)_{k}\mid _{x=\frac{1}{q}}=a^{k} \frac{[k]_{q}!}{[k-j]_{q}!}(1;q)_{k-j}=0,\quad \text{ for }\,\,j=0,1,\ldots ,k-1. \end{aligned}$$

Therefore,

$$\begin{aligned} \int _{0}^{1}\,D_{q^{-1}}^{k}(x^{k}(qx;q)_{k}) x^{m}d_{q}x=-[m]_{q}q^{1-m}\int _{0}^{1}x^{m-1}D_{q^{-1}}^{k-1}(x^{k}(qx;q)_{k})d_{q}x.\nonumber \\ \end{aligned}$$

(4.16)

Now, applying (4.14) $k-1$ times on the right hand side of (4.16), and using that $D_{q^{-1}}^{m}(x^{k}(qx;q)_{k}= 0$ at $x=0,\, x=\frac{1}{q}$ $(m=0,1,\ldots ,k-1)$ yields

$$\begin{aligned} \begin{aligned} \int _{0}^{1}\,D_{q^{-1}}^{k}(x^{k}(qx;q)_{k}) x^{m}d_{q}x&=(-1)^{k} \left( \prod _{j=0}^{k-1}q^{1-m-j}\right) \frac{[m]_{q}!}{[m-k]_{q}!} \int _{0}^{1} x^{m}(qx;q)_{k}\,d_{q}x. \end{aligned} \end{aligned}$$

Since

$$\begin{aligned} \begin{aligned} B_{q}(x,y)&=\int _{0}^{1} t^{x-1}(qt;q)_{y-1}d_{q}t\\ {}&=\int _{0}^{1}t^{x-1}\frac{(tq;q)_{\infty }}{(tq^{y};q)_{\infty }}d_{q}t,\,\,\,Re\,(x)>0,\,Re\,(y)>0,\end{aligned} \end{aligned}$$

see [5, Eq. (1.58), p. 22], then

$$\begin{aligned} \begin{aligned} \int _{0}^{1}D_{q^{-1}}^{k}(x^{k}(qx;q)_{k}) x^{m}d_{q}x&=(-1)^{k}q^{\frac{-k(k-1)}{2}+k}q^{-mk} \frac{[m]_{q}!}{[m-k]_{q}!}B_{q}(m{+}1,k{+}1)\\ {}&= (-1)^{k}q^{\frac{-k(k-3)}{2}}q^{-mk} \frac{[m]_{q}!\Gamma _{q}(m+1)\Gamma _{q}(k+1)}{[m-k]_{q}!\Gamma _{q}(m+k+2)}\\ {}&= (-1)^{k}q^{\frac{-k(k-3)}{2}}q^{-mk} \frac{([m]_{q}!)^{2}[k]_{q}!}{[m-k]_{q}![m+k+1]_{q}!}. \end{aligned}\nonumber \\ \end{aligned}$$

(4.17)

Substituting from (4.17) into (4.13) yields

$$\begin{aligned} \begin{aligned} C_{k}&=(-1)^{k}q^{k}(1-q^{2k+1}) \sum _{m=k}^{n}q^{-mk}\frac{(q;q)_{n}(q;q)_{m}}{(q;q)_{n-m}(q;q)_{m-k}(q;q)_{m+k+1}}\beta _{n-m,\alpha }(q)\\ {}&= q^{\frac{-k(k-3)}{2}}(1-q^{2k+1}) \sum _{m=k}^{n}(-1)^{m} \frac{q^{\frac{m(2n-m+1)}{2}+1}(q^{-n};q)_{m}(q^{-m};q)_{k}}{(q;q)_{m+k+1}}\beta _{n-m,\alpha }(q), \end{aligned}\nonumber \\ \end{aligned}$$

(4.18)

where we used the identity in (2.47). Therefore, from (4.18) and (4.12), we get the required result for $B^{(1)}_{n,\alpha }(x;q)$. Similarly, we can prove the result for $B^{(k)}_{n,\alpha }(x;q)\,(k=2,3)$. $\square$