Conditional Expanding of Functions by q-Lidstone Series

Abstract

This paper characterizes those functions given by convergent q-Lidstone series expansion. We give the necessary and sufficient conditions so that the entire function $f\left(z\right)$ has such an expansion, in which case convergence is uniform on compact sets.

1. Introduction

In 1929, Lidstone [1] introduced a generalization of Taylor’s theorem that approximates an entire function f in a neighborhood of two points instead of one, that is

$$f\left(x\right)=\sum _{n=0}^{\infty }\left[f^{\left(2n\right)}\left(1\right){\mathsf{\Lambda }}_n\left(x\right)+f^{\left(2n\right)}\left(0\right){\mathsf{\Lambda }}_n(1-x)\right],$$

(1)

where ${\mathsf{\Lambda }}_n\left(x\right)$ is the unique polynomial of degree $2n+1$ determined by the conditions

$$\begin{array}{c}{\mathsf{\Lambda }}_0\left(x\right)=x;{\mathsf{\Lambda }}_n\left(0\right)={\mathsf{\Lambda }}_n\left(1\right)=0(n\ge 1);\\ {\mathsf{\Lambda }}_n^{"}\left(x\right)={\mathsf{\Lambda }}_{n-1}\left(x\right)(n\ge 1),\end{array}$$

In [2], Boas and Buck show that any entire function of order less than $\pi$ has an absolutely convergent Lidstone series representation. Buckholtz et al. [3,4] proved that the condition

$$f^{\left(n\right)}\left(0\right)=o\left({\pi }^n\right)(n\to \infty )$$

is both necessary and sufficient for (1) to hold. Several authors, including Boas [5], Poritsky [6], Schoenberg [7], Whittaker [8], and Widder [9], have presented different necessary and sufficient conditions for the representation of functions by Lidstone series (1).

Recently, this type of series was generalized by q-calculus. The quantum calculus (q-calculus or Jackson calculus [10]) is an extension of the traditional calculus. There are many new developments and applications of the q-calculus in many areas, such as ordinary fractional calculus, optimal control problems, solutions of the q-difference equations, q-integral equations, q-fractional integral inequalities, q-transform analysis, and many more (see, e.g., [11,12,13,14,15]).

Ismail and Mansour [16] introduced a q-analog of Lidstone series by the form

$$f\left(z\right)=\sum _{n=0}^{\infty }\left[D_{q^{-1}}^{2n}f\left(1\right)A_n\left(z\right)-D_{q^{-1}}^{2n}f\left(0\right)B_n\left(z\right)\right],$$

(2)

where $D_{q^{-1}}$ is the Jackson q-difference operator defined by

$$D_{q^{-1}}f\left(z\right)=\left\{\begin{array}{cc}{\displaystyle \frac{f\left(z\right)-f\left(\frac{z}{q}\right)}{z(1-\frac{1}{q})}},\hfill & z\ne 0;\hfill \\ f^{\prime }\left(0\right),\hfill & z=0,\hfill \end{array}\right.$$

provided that f is differentiable at zero (see [17]). The polynomials ${\left(A_n\right)}_n$ and ${\left(B_n\right)}_n$ are q-analogs of the Lidstone polynomials ${\left({\mathsf{\Lambda }}_n\left(z\right)\right)}_n$ and ${\left({\mathsf{\Lambda }}_n(1-z)\right)}_n$, and defined by the generating functions

$${\displaystyle \frac{E_q\left(zw\right)-E_q(-zw)}{E_q\left(w\right)-E_q(-w)}}=\sum _{n=0}^{\infty }{A}_n\left(z\right)w^{2n},$$

(3)

$${\displaystyle \frac{E_q\left(zw\right)E_q(-w)-E_q(-zw)E_q\left(w\right)}{E_q\left(w\right)-E_q(-w)}}=\sum _{n=0}^{\infty }{B}_n\left(z\right)w^n,$$

(4)

respectively, where $E_q\left(z\right)$ is Jackson’s q-exponential function defined by

$$E_q\left(z\right)=\prod _{k=0}^{\infty }(1+(1-q)q^{k}z),0<q<1,z\in \mathbb{C}.$$

(5)

Moreover,

$$A_0\left(z\right)=z,B_0\left(z\right)=z-1,$$

(6)

and for $n\in \mathbb{N}$, $A_n\left(z\right)$ and $B_n\left(z\right)$ satisfy the q-difference equation

$$D_{q^{-1}}^2{y}_n\left(z\right)=y_{n-1}\left(z\right)\mathrm{with}{y}_n\left(0\right)=y_n\left(1\right)=0.$$

(7)

It is worth mentioning that Al-Towailb [18] has constructed another q-type Lidstone theorem by expanding a class of entire functions in terms of q-derivatives of even orders at 0 and q-derivatives of odd orders at 1. Moreover, in [19], we introduced the q-Lidstone expansion in (2) for the symmetric q-difference operator ${\delta }_q$ defined by

$${\delta }_{q}f\left(z\right)=f\left(q^{\frac{1}{2}}z\right)-f\left(q^{-\frac{1}{2}}z\right)$$

(see [20]). For detailed properties of the q-analogs of Lidstone series and polynomials, readers may also refer to the literature (see [21,22,23]).

Our aim is to provide necessary and sufficient conditions for expanding an entire function, $f\left(z\right)$, in a q-Lidstone expansion of the form

$$f\left(z\right)=s_0{A}_0\left(z\right)-s_1{B}_0\left(z\right)+s_2{A}_1\left(z\right)-s_3{B}_1\left(z\right)+\dots ,$$

(8)

where ${\left\{s_n\right\}}_n$ is a complex sequence defined by

$$s_{2n}=D_{q^{-1}}^{2n}f\left(1\right),s_{2n+1}=D_{q^{-1}}^{2n}f\left(0\right)(n\in \mathbb{N}).$$

(9)

We also give a sufficient condition for expanding entire functions in a different arrangement of series (8). More precisely, we will provide a sufficient condition on f so that

$$\sum _{n=0}^{\infty }{D}_{q^{-1}}^{2n}f\left(1\right)A_n\left(z\right)-\sum _{n=0}^{\infty }{D}_{q^{-1}}^{2n}f\left(0\right)B_n\left(z\right)$$

(10)

converges to $f\left(z\right)$, uniformly on each compact subset of the plane.

This paper is organized as follows: The following section gives the essential notions and some basic results on q-analysis. In Section 3, we derive the asymptotic of q-Lidstone polynomials $A_n\left(z\right)$ and $B_n\left(z\right)$ for sufficiently large n. Section 4 provides a necessary and sufficient condition on an entire function f so that the q-Lidstone series expansion converges to $f\left(z\right)$ uniformly on compact subsets of the plane. Conclusions are drawn in Section 5.

2. Preliminaries

In this section, we recall some notions, definitions, and basic results in the area of q-calculus for $0<q<1$ which we need in our investigations (see [17,24]).

Let $n\in {\mathbb{N}}_0$ (the set of non-negative integers). The q-shifted fractional ${(a;q)}_n$ is defined by

$${(a;q)}_{\infty }=\prod _{j=0}^{\infty }(1-a{q}^j),{(a;q)}_n:=\frac{{(a;q)}_{\infty }}{{(a{q}^n;q)}_{\infty }}(a\in \mathbb{C}).$$

The q-numbers ${\left[n\right]}_q$ and q-factorial ${\left[n\right]}_q!$ are defined as:

$${\left[n\right]}_q=\frac{1-q^n}{1-q},{\left[n\right]}_q!={\left[n\right]}_q{[n-1]}_q\dots {\left[1\right]}_q.$$

Jackson’s q-trigonometric functions ${\mathrm{Sin}}_{q}z$ and ${\mathrm{Cos}}_{q}z$ are defined by

$$\begin{array}{cc}\hfill {\mathrm{Sin}}_{q}z& :=\frac{E_q\left(iz\right)-E_q(-iz)}{2i}=\sum _{n=0}^{\infty }{(-1)}^n\frac{q^{n(2n+1)}}{{(q;q)}_{2n+1}}{\left(z(1-q)\right)}^{2n+1},\hfill \\ \hfill {\mathrm{Cos}}_{q}z& :=\frac{E_q\left(iz\right)+E_q(-iz)}{2}=\sum _{n=0}^{\infty }{(-1)}^n\frac{q^{n(2n-1)}}{{(q;q)}_{2n}}{\left(z(1-q)\right)}^{2n},\hfill \end{array}$$

(11)

and q-analogs of the hyperbolic functions ${\mathrm{Sin}\mathrm{h}}_{q}z$ and ${\mathrm{Cos}\mathrm{h}}_{q}z$ are given by

$${\mathrm{Sin}\mathrm{h}}_q\left(z\right):=-i{\mathrm{Sin}}_q\left(iz\right),{\mathrm{Cos}\mathrm{h}}_q\left(z\right):={\mathrm{Cos}}_q\left(iz\right).$$

(12)

In [25], Ismail proved that the zeros of the second Jackson q-Bessel functions are real and simple. Moreover, since the q-sine and q-cosine functions in (11) has a representation in terms of the second Jackson q-Bessel function of order $1/2$ and $-1/2$, respectively, cf. [17], their zeros are real and simple. Therefore, from now on, we use ${\left\{{\xi }_k\right\}}_{k\in \mathbb{N}}$ to denote the positive zeros of ${\mathrm{Sin}}_{q}z$ arranged in increasing order of magnitude. Based on the results in [25,26,27], El-Guindy and Mansour in [28] pointed out that the real positive zeros of ${\mathrm{Sin}}_{q}z$ have the asymptotic

$${\xi }_k\sim A{q}^{-2k},A:=\frac{q^{\frac{-3}{2}}}{1-q}ask\to \infty .$$

(13)

Furthermore, they introduced the following asymptotic result.

Lemma 1. 

Let α and β be real positive numbers such that $\alpha +\beta <2$. Let A be the constant in (13) and γ is any positive number less than α and β. Then, for $q^{-\alpha }{\xi }_n<R_n<q^{\beta }{\xi }_{n+1}$,

$$\left|\frac{1}{Sin{h}_q\left(z\right)}\right|=o\left({\left(\frac{A{q}^{\beta }}{(1+q^{\gamma })}\right)}^n{R}_n^{-n}\right),\left|z\right|=R_n,n\to \infty .$$

Furthermore, in [28], the authors proved that

$$\sum _{n=1}^{\infty }\frac{1}{{\mathrm{Sin}}_q^{\prime }\left({\xi }_n\right)}=-\frac{1}{2}.$$

(14)

The following lemma proves the absolute convergence of the series on (14).

Lemma 2. 

If ${\left({\xi }_n\right)}_n$ are the set of positive zeros of $Si{n}_q\left(z\right)$, then for each $n\in \mathbb{N}$,

$$|{\mathrm{Sin}}_q^{\prime }\left({\xi }_n\right)|\ge 2q^{-2{(n+1)}^2}{q}^{2({ϵ}_n+1)(n+1)}\frac{1}{{(q;q)}_{2n+3}}\prod _{j=1}^{\infty }{(1-q^{4j})}^3,$$

(15)

for some positive numbers ${\xi }_n$ such that ${\displaystyle \underset{n\to \infty }{lim}{ϵ}_n=0}$. Moreover, ${\displaystyle \sum _{n=1}^{\infty }\frac{1}{{\mathrm{Sin}}_q^{\prime }\left({\xi }_n\right)}}$ is an absolutely convergent series.

Proof. 

First, note that for any constant c, we have

$$\sum _{k=0}^{\infty }{(-1)}^k{q}^{k(2k+1)}\frac{c}{{[2k+1]}_q!}{\xi }_n^{2k}=c\frac{{\mathrm{Sin}}_q{\xi }_n}{{\xi }_n}=0.$$

So, we obtain

$${\mathrm{Sin}}_q^{\prime }{\xi }_n=\sum _{k=0}^{\infty }{(-1)}^k{q}^{k(2k+1)}\frac{2k+1}{{[2k+1]}_q!}{\xi }_n^{2k}=\sum _{k=0}^{\infty }{(-1)}^k{q}^{k(2k+1)}\frac{2(k-n-1)}{{[2k+1]}_q!}{\xi }_n^{2k}.$$

(16)

From (13), we may assume that ${\xi }_n=A{q}^{-2n+{ϵ}_n}<1$, with $A=\frac{q^{-3/2}}{(1-q)}$ and ${\displaystyle \underset{n\to \infty }{lim}{ϵ}_n=0}$. Therefore,

$$\begin{array}{cc}\hfill {\mathrm{Sin}}_q^{\prime }{\xi }_n& =\sum _{k=0}^{\infty }{(-1)}^k{q}^{k(2k+1)}\frac{2(k-n-1)}{{[2k+1]}_q!}{A}^{2k}{q}^{-4nk+2k{ϵ}_n}\hfill \\ & =(1-q)q^{-2{(n+1)}^2}\sum _{k=0}^{\infty }{(-1)}^k\frac{2(k-n-1)}{{(q;q)}_{2k+1}}{q}^{2{(k-n-1)}^2}{q}^{2k+2k{ϵ}_n}\hfill \\ & =t_n\sum _{m=-(n+1)}^{\infty }{(-1)}^m\frac{2m}{{(q^{2n+3};q)}_{2m+1}}{q}^{2m^2}{q}^{2({ϵ}_n+1)m}=t_n{S}_n,\hfill \end{array}$$

(17)

where

$$\begin{array}{cc}\hfill t_n:=& {(-1)}^{n+1}{q}^{-2{(n+1)}^2}{q}^{2({ϵ}_n+1)(n+1)}\frac{1}{{(q^2;q)}_{2n+2}};\hfill \\ \hfill S_n:=& \sum _{m=-(n+1)}^{\infty }{(-1)}^m\frac{2m}{{(q^{2n+3};q)}_{2m+1}}{q}^{2m(m+1)+2{ϵ}_n}.\hfill \end{array}$$

(18)

Let p be a positive integer; then,

$$S_n=\sum _{m=-(n+1)}^{-(2p+2)}{F}_{m,n}+\sum _{m=-(2p+1)}^{2p}{F}_{m,n}+\sum _{m=2p+1}^{\infty }{F}_{m,n},$$

where

$$F_{m,n}:={(-1)}^m\frac{2m}{{(q^{2n+3};q)}_{2m+1}}{q}^{2m(m+1)+2{ϵ}_n}.$$

Consequently,

$$\begin{array}{cc}\hfill |S_n|& \ge \left|\sum _{m=-(2p+1)}^{2p}{F}_{m,n}\right|-\sum _{m=-(n+1)}^{-(2p+2)}|F_{m,n}|-\sum _{m=2p+1}^{\infty }\left|F_{m,n}\right|.\hfill \end{array}$$

(19)

Since the estimate in (19) is independent of p, the last two sums on (19) tend to zero as $n\to \infty$. Therefore,

$$\underset{n\to \infty }{lim\; inf}\left|S_n\right|\ge \underset{p\to \infty }{lim}\left|2\sum _{m=-(2p+1)}^{2p}{(-1)}^{m}m{q}^{2m(m+1)}\right|.$$

However,

$$\begin{array}{cc}\hfill \sum _{m=-(2p+1)}^{2p}{(-1)}^{m}m{q}^{2m(m+1)}& =\sum _{-(2p+1)}^{-1}{(-1)}^{m}m{q}^{2m(m+1)}+\sum _{m=0}^{2p}{(-1)}^{m}m{q}^{2m(m+1)}\hfill \\ & =\sum _1^{2p+1}{(-1)}^{m-1}m{q}^{2m(m+1)}+\sum _{m=0}^{2p}{(-1)}^{m}m{q}^{2m(m+1)}\hfill \\ & =\sum _0^{2p}{(-1)}^m(2m+1)q^{2m(m+1)}.\hfill \end{array}$$

Thus,

$$\underset{n\to \infty }{lim\; inf}\left|S_n\right|\ge 2\left|\sum _0^{\infty }{(-1)}^m(2m+1)q^{2m(m+1)}\right|=\prod _{j=1}^{\infty }{(1-q^{4j})}^3,$$

(20)

where we used the identity due to Jacobi, see ([29], p. 500) ,

$$\sum _{i=0}^{\infty }{(-1)}^i(2i+1)q^{\frac{i(i+1)}{2}}=\prod _{i=1}^{\infty }{(1-q^i)}^3>0.$$

(21)

So, from (18) and (17), we deduce that $|{\mathrm{Sin}}_q^{\prime }{\xi }_n|\ge 2{\prod }_{i=1}^{\infty }{(1-q^{4i})}^3{t}_n$, i.e., the series ${\sum }_{n=0}^{\infty }\frac{1}{{\mathrm{Sin}}_q^{\prime }{\xi }_n}$ is absolutely convergent. □

Let $\mathcal{F}$ denote the class of all entire functions f that satisfy

$$D_{q^{-1}}^{n}f\left(0\right)=o\left({\xi }_1^n\right)(n\to \infty ),$$

where ${\xi }_1$ is the smallest positive zero of ${\mathrm{Sin}}_q\left(z\right)$. We define a norm of $\mathcal{F}$ by

$${\parallel f\parallel }_{\mathcal{F}}:=\underset{n\in {\mathbb{N}}_0}{sup}{\displaystyle \frac{|\left(D_{q^{-1}}^{n}f\right)\left(0\right)|}{{\xi }_1^n}}.$$

(22)

It is worth noting that if f is analytic at zero, then

$$\frac{f^{\left(n\right)}\left(0\right)}{n!}=q^{\frac{n(n-1)}{2}}\frac{D_{q^{-1}}^{n}f\left(0\right)}{{\left[n\right]}_q!}(n\in {\mathbb{N}}_0).$$

(23)

Therefore, $(\mathcal{F},\parallel f{\parallel }_{\mathcal{F}})$ is a normed space.

Lemma 3. 

If $f\in \mathcal{F}$, then

$$\left|f\left(z\right)\right|\le {\parallel f\parallel }_{\mathcal{F}}{E}_q({\xi }_1\left|z\right|)(z\in \mathbb{C}),$$

where $E_q\left(z\right)$ is Jackson’s q-exponential function defined in (5).

Proof. 

From the Maclaurin series expansion and Equation (23), we have

$$f\left(z\right)=\sum _{m=0}^{\infty }\frac{f^{\left(m\right)}\left(0\right)}{m!}{z}^m=\sum _{m=0}^{\infty }{q}^{\frac{m(m-1)}{2}}\frac{D_{q^{-1}}^{m}f\left(0\right)}{{\left[m\right]}_q!}{z}^m.$$

(24)

Then, from (22), we obtain

$$\begin{array}{cc}\hfill \left|f\right(z\left)\right|& =\left|\sum _{m=0}^{\infty }\frac{D_{q^{-1}}^{m}f\left(0\right)}{{\xi }_1^m}{q}^{\frac{m(m-1)}{2}}\frac{{\left({\xi }_{1}z\right)}^m}{{\left[m\right]}_q!}\right|\hfill \\ \hfill & \le {\parallel f\parallel }_{\mathcal{F}}\sum _{m=0}^{\infty }{q}^{\frac{m(m-1)}{2}}\frac{({\xi }_1{\left|z\right|)}^m}{{\left[m\right]}_q!}={\parallel f\parallel }_{\mathcal{F}}{E}_q\left({\xi }_1\right|z\left|\right).\hfill \end{array}$$

This completes the proof. □

In the following, we introduce some essential properties for the q-Lidstone polynomials $A_n\left(z\right)$ and $B_n\left(z\right)$ defined in (3) and (4), respectively.

Let ${\left\{p_n(z;q)\right\}}_n$ be the sequence defined by

$$p_{2n}(z;q)=A_n\left(z\right),p_{2n+1}(z;q)=-B_n\left(z\right),$$

(25)

and ${\left\{L_n\right\}}_n$ be the sequence of linear functionals defined by

$$L_{2n}\left(f\right):=D_{q^{-1}}^{2n}f\left(1\right),L_{2n+1}\left(f\right):=D_{q^{-1}}^{2n}f\left(0\right).$$

(26)

Then, the expansion (2) can be written in the form

$$f\left(z\right)=\sum _{n=0}^{\infty }{L}_n\left(f\right)p_n(z;q).$$

(27)

By using (24), we can write $L_n$ as

$$L_0\left(f\right)=f\left(1\right),L_n\left(f\right)=\sum _{m=n-1}^{\infty }{D}_{q^{-1}}^{m}f\left(0\right)L_{nm}(n\in \mathbb{N}),$$

(28)

where $L_{nm}={\delta }_{m,n-1}$ for odd n, where ${\delta }_{i,j}$ is the Kronecker’s delta, and

$$L_{nm}=\left\{\begin{array}{cc}{\displaystyle \frac{q^{\frac{(m-n)(m-n-1)}{2}}}{{[m-n]}_q!}},\hfill & m\ge n;\hfill \\ 0,\hfill & m<n,\hfill \end{array}\right.$$

for even n. Consequently,

$$0\le L_{nm}<\frac{1}{{[m-n]}_q!},for0\le n\le m.$$

(29)

From (28), we have

$$L_m\left(z^j\right)=\left\{\begin{array}{cc}{q}^{-\frac{j(j-1)}{2}}{L}_{jm},& m\le j+1,\\ 0,& m>j+1.\end{array}\right.$$

Therefore, by using (27), we obtain

$$\frac{q^{\frac{j(j-1)}{2}}{z}^j}{{\left[j\right]}_q!}=\sum _{m=0}^{j+1}{L}_{jm}{p}_m(z;q)(j\in {\mathbb{N}}_0).$$

(30)

Remark 1. 

From (6), (7) and (25), we have

$$p_0\left(0\right)=0,p_1\left(0\right)=1and{p}_n\left(0\right)=0foralln>1.$$

Moreover, from (7), we conclude that $D_{q^{-1}}^2{p}_n(z;q)=p_{n-2}(z;q)$. Therefore, by mathematical induction on k, we can verify that

$$D_{q^{-1}}^{2k}{p}_n(z;q)=p_{n-2k}(z;q)(n\ge 2k).$$

(31)

This implies $L_n\left(p_k\right)={\delta }_{n,k}$, and

$$D_{q^{-1}}^{2k}{p}_n\left(0\right)=p_{n-2k}\left(0\right),D_{q^{-1}}^{2k+1}{p}_n\left(0\right)=D_{q^{-1}}{p}_{n-2k}\left(0\right).$$

(32)

3. Asymptotic Expansions for the q-Lidstone Polynomials

In this section, we use the contour integration and Darboux method (see [30]) for determining the asymptotic behavior of the q-Lidstone polynomials $A_n\left(z\right)$ and $B_n\left(z\right)$ defined in (3) and (4) for sufficiently large n.

Consider the generating function

$$G_q(z,w):=\frac{{\mathrm{Sin}\mathrm{h}}_q\left(zw\right)}{w^{m+1}{\mathrm{Sin}\mathrm{h}}_q\left(w\right)}.$$

(33)

We choose a contour ${\Gamma }_n$, as shown in Figure 1, such that $G_q(z,w)$ is analytic in an open set containing ${\Gamma }_n$ and its interior, except for the poles inside ${\Gamma }_n$.

Proposition 1. 

Let ${\left\{{\xi }_k\right\}}_{k\in \mathbb{N}}$ be the sequence of the positive zeros of ${\mathrm{Sin}}_q\left(z\right)$. Then

$${\oint }_{{\Gamma }_n}{G}_q(z,w)dw=\left(2\pi i\right)\left[{\tilde{A}}_m\left(z\right)+2cos\left(\frac{\pi }{2}m\right)\sum _{k=1}^n\frac{{\mathrm{Sin}}_q\left({\xi }_{k}z\right)}{{\left({\xi }_k\right)}^{m+1}{\mathrm{Sin}}_q^{\prime }\left({\xi }_k\right)}\right],$$

(34)

where $G_q(z,w)$ is defined in (33), and ${\left\{{\tilde{A}}_m\right\}}_m$ is the sequence of function defined by

$${\tilde{A}}_{2m}\left(z\right)=A_m\left(z\right),{\tilde{A}}_{2m+1}\left(z\right)=0.$$

Proof. 

The function $G_q(z,w)$ has a pole of order $m+1$ at 0 and simple poles at $\pm i{\xi }_k$ for $k=1,2,\dots ,n$. So, by using the Cauchy residue theorem, we obtain

$$\frac{1}{2\pi i}{\oint }_{{\Gamma }_n}{G}_q(z,w)dw=\frac{1}{m!}\frac{d^m}{d{z}^m}\left[{\displaystyle \frac{{\mathrm{Sin}\mathrm{h}}_q\left(zw\right)}{{\mathrm{Sin}\mathrm{h}}_q\left(w\right)}}\right]\left(0\right)+\sum _{k=1}^n\mathrm{Res}(G_q,\pm i{\xi }_k).$$

(35)

From Equation (3), we obtain

$$\frac{1}{m!}\frac{d^m}{d{z}^m}\left[{\displaystyle \frac{{\mathrm{Sin}\mathrm{h}}_q\left(zw\right)}{{\mathrm{Sin}\mathrm{h}}_q\left(w\right)}}\right]\left(0\right)=\left\{\begin{array}{cc}{A}_r\left(z\right),\hfill & \mathrm{m}=2\mathrm{r};\hfill \\ 0,\hfill & \mathrm{m}=2\mathrm{r}+1\hfill \end{array}\right..$$

(36)

For a fixed $k\in \{1,2,\dots ,n\}$,

$$\begin{array}{ccc}\hfill \mathrm{Res}(G_q,i{\xi }_k)& =& \underset{w\to i{\xi }_k}{lim}\frac{(w-i{\xi }_k)}{w^{m+1}{\mathrm{Sin}\mathrm{h}}_q\left(w\right)}{\mathrm{Sin}\mathrm{h}}_q\left(zw\right)\hfill \\ & =& \frac{{\mathrm{Sin}\mathrm{h}}_q\left(i{\xi }_{k}z\right)}{{\left(i{\xi }_k\right)}^{m+1}}\underset{w\to i{\xi }_k}{lim}\frac{(w-i{\xi }_k)}{{\mathrm{Sin}}_q\left(w\right)}\hfill \\ & =& \frac{i{\mathrm{Sin}}_q\left({\xi }_{k}z\right)}{{\left(i{\xi }_k\right)}^{m+1}{\mathrm{Sin}}_q^{\prime }\left({\xi }_k\right)}.\hfill \end{array}$$

Similarly, $\mathrm{Res}(G_q,-i{\xi }_k)={\displaystyle \frac{(-i){\mathrm{Sin}}_q\left({\xi }_{k}z\right)}{{(-i{\xi }_k)}^{m+1}{\mathrm{Sin}}_q^{\prime }\left({\xi }_k\right)}}$. Therefore,

$$\sum _{k=1}^n\mathrm{Res}(G_q,\pm i{\xi }_k)=2cos\left(\frac{\pi }{2}m\right)\sum _{k=1}^n\frac{{\mathrm{Sin}}_q\left({\xi }_{k}z\right)}{{\left({\xi }_k\right)}^{m+1}{\mathrm{Sin}}_q^{\prime }\left({\xi }_k\right)},$$

(37)

and the result follows by substituting (36) and (37) into (35). □

The following proposition proves that the contour integration on ${\Gamma }_n$ in (34) vanishes as $n\to \infty$.

Proposition 2. 

Let ${\left\{{\xi }_k\right\}}_{k\in \mathbb{N}}$ be the sequence of the positive zeros of $Si{n}_q\left(z\right)$, and let α and β be real positive numbers such that $\alpha +\beta <2$. Then, there exists $m_0\in \mathbb{N}$ such that for $q^{-\alpha }{\xi }_n<R_n<q^{\beta }{\xi }_{n+1}$

$$\underset{n\to \infty }{lim}{\oint }_{{\Gamma }_n:\left|w\right|=R_n}\frac{Sin{h}_q\left(zw\right)}{w^{m+1}Sin{h}_q\left(w\right)}dw=0\left(\right|z|\le 1,m\ge m_0).$$

Proof. 

It suffices to prove that

$$\underset{n\to \infty }{lim}{\oint }_{{\Gamma }_n}\frac{E_q\left(zw\right)}{w^{m+1}{\mathrm{Sin}\mathrm{h}}_q\left(w\right)}dw=0\left(\right|z|\le 1).$$

From (5), if $\left|z\right|\le 1$ and $\left|w\right|=R_n$, then

$$\begin{array}{c}\hfill |E_q\left(zw\right)|=\prod _{k=0}^{\infty }|1+q^k(1-q)zw|\le \prod _{k=0}^{2n+1}(1+q^k{R}_n)\prod _{k=2n+2}^{\infty }(1+q^k{R}_n).\end{array}$$

(38)

From the asymptotic (13), ${\displaystyle \underset{k\to \infty }{lim}\frac{{\xi }_k}{A{q}^{-2k}}=1}$, where $A=\frac{q^{-3/2}}{1-q}$. Consequently, for any $ϵ>0$ there exists $k_0\in \mathbb{N}$ such that

$$(1-ϵ)A{q}^{-2k}<{\xi }_k<(1+ϵ)A{q}^{-2k}forallk\ge k_0.$$

(39)

Now, set $\delta =\frac{q^{-\beta }}{A(1+ϵ)}$. Since $R_n<q^{\beta }{\xi }_{n+1}$, then from (39), we obtain

$$\begin{array}{cc}\hfill \prod _{k=0}^{2n+1}(1+q^k{R}_n)& \le \prod _{k=0}^{2n+1}(1+q^{\beta +k}{\xi }_{n+1})\le \prod _{k=0}^{2n+1}(1+\frac{q^{k-2n-2}}{\delta })\hfill \\ \hfill & \le \frac{q^{-(n+1)(2n+1)}}{{\delta }^{2n+2}}{(-q\delta ;q)}_{2n+1}.\hfill \end{array}$$

(40)

Furthermore, we can verify that

$$\begin{array}{c}\hfill \prod _{k=2n+2}^{\infty }(1+q^k{R}_n)\le {(-\frac{1}{\delta };q)}_{\infty }.\end{array}$$

(41)

Combining the inequalities (38), (40), (41) and using (39), we conclude that

$$\begin{array}{cc}\hfill |E_q\left(zw\right)|& \le \frac{q^{-(n+1)(2n+1)}}{{\delta }^{2n+2}}{(-q\delta ;q)}_{\infty }{(-\frac{1}{\delta };q)}_{\infty }\hfill \\ \hfill & \le \frac{q^{-1}}{{\delta }^{2n+1}}{\left(\frac{q^{\alpha }{R}_n}{A(1-ϵ)}\right)}^{n+3/2}{(-q\delta ;q)}_{\infty }{(-\frac{1}{\delta };q)}_{\infty }.\hfill \end{array}$$

Therefore, from Lemma 1, if $c=\frac{A{q}^{\beta }}{(1+q^{\gamma })}$, $\gamma <min\{\alpha ,\beta \}$, and

$${\alpha }_n:=\frac{c^{-n}{R}_n^n}{|{\mathrm{Sin}\mathrm{h}}_q\left(w\right)|},\left|w\right|=R_n,$$

then ${lim}_{n\to \infty }{\alpha }_n=0$. Therefore,

$$\left|\frac{E_q\left(zw\right)}{w^{m+1}{\mathrm{Sin}\mathrm{h}}_q\left(w\right)}\right|\le {\alpha }_n{R}_n^{3/2}{\left(\frac{q^{\alpha +3\beta }{A}^2{(1+ϵ)}^2}{(1-ϵ)(1+q^{\gamma })}\right)}^n{\displaystyle \frac{{(-q\delta ;q)}_{\infty }{(-\frac{1}{\delta };q)}_{\infty }}{q\delta }},$$

and then

$$\left|{\oint }_{{\Gamma }_n}\frac{E_q\left(zw\right)}{w^{m+1}{\mathrm{Sin}\mathrm{h}}_q\left(w\right)}dw\right|\le 2\pi {\alpha }_n{R}_n^{-m+3/2}{\left(\frac{q^{\alpha +3\beta }{A}^2{(1+ϵ)}^2}{(1-ϵ)(1+q^{\gamma })}\right)}^n{\displaystyle \frac{{(-q\delta ;q)}_{\infty }{(-\frac{1}{\delta };q)}_{\infty }}{q\delta }}.$$

(42)

Hence, there exists $m_0\in \mathbb{N}$ such that the limit of the right-hand side of (42) as $n\to \infty$ is zero for all $m\ge m_0$. This yields the required result. □

Propositions 1 and 2 lead to the following expansion of the polynomials ${\left(A_m\left(z\right)\right)}_m$.

Theorem 1. 

Let ${\left\{{\xi }_k\right\}}_{k\in \mathbb{N}}$ be a sequence of the positive zeros of ${\mathrm{Sin}}_q\left(z\right)$. Then, there exists $m_0\in \mathbb{N}$ such that

$$A_m\left(z\right)=2{(-1)}^{m+1}\sum _{k=1}^{\infty }\frac{Si{n}_q\left({\xi }_{k}z\right)}{{\xi }_k^{2m+1}{\mathrm{Sin}}_q^{\prime }\left({\xi }_k\right)},for\left|z\right|\le 1,$$

(43)

for all $m\ge m_0$, where ${\left(A_m\right)}_m$ is the sequence of the q-Lidstone polynomials which is defined in (3). In particular,

$$A_m\left(z\right)=2{(-1)}^{m+1}\frac{{\mathrm{Sin}}_q\left({\xi }_{1}z\right)}{{\xi }_1^{2m+1}{\mathrm{Sin}}_q^{\prime }\left({\xi }_1\right)}+o\left(\frac{1}{{\xi }_2^{2m+1}}\right)(m\to \infty ).$$

(44)

Proof. 

From Proposition 1, we obtain

$$\frac{1}{2\pi i}{\oint }_{{\Gamma }_n}{G}_q(z,w)dw=A_m\left(z\right)+2{(-1)}^{m+1}\sum _{k=1}^n\frac{{\mathrm{Sin}}_q\left({\xi }_{k}z\right)}{{\xi }_k^{m+1}{\mathrm{Sin}}_q^{\prime }\left({\xi }_k\right)}.$$

Letting $n\to \infty$ and using Proposition 2, we obtain (43). The proof of (44) follows directly from (43). □

The Darboux method (see [30], pp. 309–310) states that if $f\left(z\right)=\sum a_n{z}^n$ is a meromorphic function in $\left|z\right|>0$, r is the distance from the origin to the nearest pole, and if there exists a function g such that

1.

g is analytic in $0<\left|z\right|<r$;

2.

$f-g$ is continuous on $0<\left|z\right|\le r$;

3.

the coefficients $b_n$ in the Laurent expansion of the function g have known asymptotic behavior as $n\to \infty$;

then,

$$a_n=b_n+o\left(r^{-n}\right)\mathrm{as}n\to \infty .$$

(45)

Moreover, if $f-g$ is m times continuously differentiable on the circle $\left|z\right|=r$, then

$$a_n=b_n+o\left(r^{-n}{n}^{-m}\right)\mathrm{as}n\to \infty .$$

(46)

Ismail and Mansour [16] proved that the q-Lidstone polynomial $B_n\left(z\right)$ can be written as

$$B_n\left(z\right)=\frac{2^{2n+1}}{{[2n+1]}_q!}{B}_{2n+1}(z/2;q),$$

(47)

where ${\left(B_n(z;q)\right)}_n$ are the q-Bernoulli polynomials defined by

$${\displaystyle \frac{w{E}_q\left(zw\right)E_q(-w/2)}{E_q(w/2)-E_q(-w/2)}}=\sum _{n=0}^{\infty }{B}_n(z;q)\frac{w^n}{{\left[n\right]}_q!}.$$

(48)

In [31], the authors used the Darboux method with (45) to provide the following asymptotic behavior of the q-Bernoulli polynomials $B_n(z;q)$ defined in (48) for sufficiently large n by

$$\begin{array}{cc}\hfill B_{2n}(z;q)=& \frac{{(-1)}^{n+1}{\left[2n\right]}_q!{\mathrm{Cos}}_q\left(2{\xi }_{1}z\right){\mathrm{Cos}}_q\left({\xi }_1\right)}{(1-q){\left(2{\xi }_1\right)}^{2n}{\mathrm{Sin}}_q^{\prime }\left({\xi }_1\right)}+o\left({\xi }_1^{-2n}\right),\hfill \\ \hfill B_{2n+1}(z;q)=& \frac{{(-1)}^{n+1}{[2n+1]}_q!{\mathrm{Sin}}_q\left(2{\xi }_{1}z\right){\mathrm{Cos}}_q\left({\xi }_1\right)}{(1-q){\left(2{\xi }_1\right)}^{2n+1}{\mathrm{Sin}}_q^{\prime }\left({\xi }_1\right)}+o\left({\xi }_1^{-(2n+1)}\right).\hfill \end{array}$$

(49)

We can improve the o term in (49) by using (46) to obtain

$$\begin{array}{cc}\hfill B_{2n}(z;q)=& \frac{{(-1)}^{n+1}{\left[2n\right]}_q!{\mathrm{Cos}}_q\left(2{\xi }_{1}z\right){\mathrm{Cos}}_q\left({\xi }_1\right)}{(1-q){\left(2{\xi }_1\right)}^{2n}{\mathrm{Sin}}_q^{\prime }\left({\xi }_1\right)}+o\left({\xi }_1^{-2n}{\left(2n\right)}^{-m}\right),\hfill \\ \hfill B_{2n+1}(z;q)=& \frac{{(-1)}^{n+1}{[2n+1]}_q!{\mathrm{Sin}}_q\left(2{\xi }_{1}z\right){\mathrm{Cos}}_q\left({\xi }_1\right)}{(1-q){\left(2{\xi }_1\right)}^{2n+1}{\mathrm{Sin}}_q^{\prime }\left({\xi }_1\right)}+o\left({\xi }_1^{-(2n+1)}{(2n+1)}^{-m}\right)\hfill \end{array}$$

(50)

as $n\to \infty$, where $m\in {\mathbb{N}}_0$ is fixed. Consequently, from Equation (47), we obtain the n large asymptotic of the q-Lidstone polynomial $B_n\left(z\right)$ as in the following result.

Proposition 3. 

The large n asymptotic of the q-Lidstone polynomials $B_n\left(z\right)$ defined in (4) is

$$\begin{array}{c}\hfill B_n\left(z\right)=\frac{{(-1)}^{n+1}{\mathrm{Sin}}_q\left({\xi }_{1}z\right){\mathrm{Cos}}_q\left({\xi }_1\right)}{(1-q){\left({\xi }_1\right)}^{2n+1}{\mathrm{Sin}}_q^{\prime }\left({\xi }_1\right)}+o\left({\xi }_1^{-2n}{\left(2n\right)}^{-m}\right),\end{array}$$

(51)

where ${\xi }_1$ is the smallest positive zero of ${\mathrm{Sin}}_q\left(z\right)$, and m is any non-negative integer.

Lemma 4. 

If ${\left(p_n\right)}_n$ is the sequence of polynomials defined in (25), then there exists a positive constant M such that

$$\left|D_{q^{-1}}^k{p}_n\left(0\right)\right|\le \frac{M}{{\left({\xi }_1\right)}^{n-k}},$$

(52)

where ${\xi }_1$ is the smallest positive zero of ${\mathrm{Sin}}_q(·).$

Proof. 

According to (25) and Theorem 1, we obtain

$$D_{q^{-1}}^{2r}{p}_{2n}(z;q)=D_{q^{-1}}^{2r}{A}_n\left(z\right)=2{(-1)}^{n+1}\sum _{k=1}^{\infty }{(-1)}^r\frac{{\xi }_k^{2r}{\mathrm{Sin}}_q\left({\xi }_{k}z\right)}{{\left({\xi }_k\right)}^{2n}{\mathrm{Sin}}_q^{\prime }\left({\xi }_k\right)},$$

$$D_{q^{-1}}^{2r+1}{p}_{2n}(z;q)=D_{q^{-1}}^{2r+1}{A}_n\left(z\right)=2{(-1)}^{n+1}\sum _{k=1}^{\infty }{(-1)}^r\frac{{\xi }_k^{2r+1}{\mathrm{Cos}}_q\left({\xi }_{k}z\right)}{{\left({\xi }_k\right)}^{2n}{\mathrm{Sin}}_q^{\prime }\left({\xi }_k\right)}.$$

Therefore, $D_{q^{-1}}^{2r}{p}_{2n}{(z;q)|}_{z=0}=0$, and

$$\begin{array}{cc}\hfill \left|D_{q^{-1}}^{2r+1}{p}_{2n}{(z;q)|}_{z=0}\right|& =\left|2{(-1)}^{n+1}\sum _{k=1}^{\infty }{(-1)}^r\frac{{\xi }_k^{2r+1}}{{\left({\xi }_k\right)}^{2n}{\mathrm{Sin}}_q^{\prime }\left({\xi }_k\right)}\right|\hfill \\ & \le \frac{2}{{\xi }_1^{2n-2r-1}}\sum _{k=1}^{\infty }\left|\frac{1}{{\mathrm{Sin}}_q^{\prime }\left({\xi }_k\right)}\right|.\hfill \end{array}$$

From Lemma 2, the series ${\sum }_{k=1}^{\infty }\left|\frac{1}{{\mathrm{Sin}}_q^{\prime }\left({\xi }_k\right)}\right|$ is convergent. Hence, there exists a constant $M_1>0$ such that

$$\left|D_{q^{-1}}^r{p}_{2n}(0;q)\right|=\left|D_{q^{-1}}^r{A}_n\left(0\right)\right|\le \frac{M_1}{{\xi }_1^{2n-r}}.$$

(53)

In [16], Ismail and Mansour introduced an expansion of $E_q\left(zw\right)$ as

$$E_q\left(zw\right)=\sum _{m=0}^{\infty }{B}_m\left(z\right)w^{2m}+E_q\left(w\right)\sum _{m=0}^{\infty }{A}_m\left(z\right)w^{2m}.$$

Using the series representation of $E_q\left(zw\right)$, we can verify that

$$B_m\left(z\right)=-q^{\left(\genfrac{}{}{0pt}{}{2m}{2}\right)}\frac{z^{2m}}{{\left[2m\right]}_q!}+\sum _{k=0}^m{\displaystyle \frac{q^{\left(\genfrac{}{}{0pt}{}{2m-2k}{2}\right)}}{{[2m-2k]}_q!}}{A}_k\left(z\right),$$

$$D_{q^{-1}}^r{B}_m\left(0\right)=-{\delta }_{r,2m}+\sum _{k=0}^m{\displaystyle \frac{q^{\left(\genfrac{}{}{0pt}{}{2m-2k}{2}\right)}}{{[2m-2k]}_q!}}{D}_{q^{-1}}^r{A}_k\left(0\right).$$

Therefore, $D_{q^{-1}}^{2m}{B}_m\left(0\right)=1$, and from (53), we obtain

$$\begin{array}{cc}\hfill |D_{q^{-1}}^r{B}_m\left(0\right)|& \le M\sum _{k=0}^m{\displaystyle \frac{q^{\left(\genfrac{}{}{0pt}{}{2m-2k}{2}\right)}}{{[2m-2k]}_q!}}\frac{1}{{\xi }_1^{2k-r}}\hfill \\ & \le \frac{M}{{\xi }_1^{2m-r}}\sum _{j=0}^m{\displaystyle \frac{q^{\left(\genfrac{}{}{0pt}{}{2j}{2}\right)}}{{\left[2j\right]}_q!}}\frac{1}{{\xi }_1^{2j}}\hfill \\ & \le \frac{M}{{\xi }_1^{2m-r}}{E}_q\left(\frac{1}{{\xi }_1}\right)(r<m).\hfill \end{array}$$

Thus, there exists a constant $M_1>0$ such that

$$\left|D_{q^{-1}}^r{p}_{2m+1}(0;q)\right|=\left|D_{q^{-1}}^r{B}_m\left(0\right)\right|\le \frac{M_1}{{\xi }_1^{2m-r}}.$$

(54)

Hence, the lemma follows from (53) and (54). □

4. The Main Results

In this section, we provide the necessary and sufficient condition in order that the entire function $f\left(z\right)$ has the expansion (8).

First, we define the sequence ${\left\{{\sigma }_n\right\}}_n$ by

$${\sigma }_{2n}=2\frac{{(-1)}^n}{{\xi }_1^{2n+1}{\mathrm{Sin}}_q^{\prime }\left({\xi }_1\right)},{\sigma }_{2n+1}=\frac{{(-1)}^{n+1}{\mathrm{Cos}}_q\left({\xi }_1\right)}{(1-q){\xi }_1^{2n+1}{\mathrm{Sin}}_q^{\prime }\left({\xi }_1\right)}.$$

(55)

From Theorem 1, Proposition 3, and the definition of $p_n(z;q)$ in (25), we obtain

$$\frac{p_n(z;q)}{{\sigma }_n}=\left\{\begin{array}{cc}{\mathrm{Sin}}_q\left({\xi }_{1}z\right)+o\left({\left(\frac{{\xi }_1}{{\xi }_2}\right)}^n\right),& n\mathrm{is}\mathrm{even};\\ \\ {\mathrm{Sin}}_q\left({\xi }_{1}z\right)+o\left(\frac{1}{n^m}\right),& n\mathrm{is}\mathrm{odd},\end{array}\right.$$

(56)

where m is a fixed positive integer. In the following, we choose $m\ge 2$.

Theorem 2. 

Let ${\left(s_n\right)}_n$ be a sequence of complex numbers. Suppose that the series

$$\sum _{n=0}^{\infty }{s}_n{p}_n(z;q)$$

(57)

converges to $f\left(z\right)$ for a single value $z_0\in (0,1)$. Then, it converges to $f\left(z\right)$ uniformly on compact subsets of $\mathbb{C}$. Furthermore,

$$s_n=L_n\left(f\right)(n\in \mathbb{N}),$$

(58)

where ${\left(L_n\left(f\right)\right)}_n$ are defined as in (28), and the function f satisfies the condition

$$D_{q^{-1}}^{n}f\left(0\right)=o\left({\xi }_1^n\right)asn\to \infty .$$

(59)

Proof. 

Suppose that the series (57) converges at $z_0$. Then, from (56),

$$0=\underset{n\to \infty }{lim}{s}_n{p}_n(z_0;q)=\underset{n\to \infty }{lim}\frac{p_n(z_0;q)}{{\sigma }_n}{\sigma }_n{s}_n={\mathrm{Sin}}_q\left({\xi }_1{z}_0\right)\underset{n\to \infty }{lim}{\sigma }_n{s}_n.$$

Since $0<z_0<1$, ${\mathrm{Sin}}_q\left({\xi }_1{z}_0\right)\ne 0$, and then ${\displaystyle \underset{n\to \infty }{lim}{\sigma }_n{s}_n=0}$. By using (56) and Weierstrass M-test, we conclude that the series

$$s\left(z\right)=\sum _{n=0}^{\infty }{\sigma }_n{s}_n\left[\frac{p_n(z;q)}{{\sigma }_n}-{\mathrm{Sin}}_q\left({\xi }_{1}z\right)\right]$$

(60)

converges uniformly on compact sets. Since the series $s\left(z\right)$ converges at $z_0$, we have

$$f\left(z_0\right)-s\left(z_0\right)={\mathrm{Sin}}_q\left({\xi }_1{z}_0\right)\sum _{n=0}^{\infty }{\sigma }_n{s}_n.$$

Therefore, ${\displaystyle \sum _{n=0}^{\infty }{\sigma }_n{s}_n}$ is convergent. This implies the uniform convergence of ${\displaystyle \sum _{n=0}^{\infty }{\sigma }_n{s}_n{\mathrm{Sin}}_q\left({\xi }_{1}z\right)}$ on compact subsets of the complex plane. In view of (60), we obtain the series (57) converges uniformly on each compact subsets of $\mathbb{C}$. Combining (9) with (26) we have already verified Equation (58).

Now, from (31) we have

$$\begin{array}{cc}\hfill \left(D_{q^{-1}}^{2k}f\right)\left(z\right)=& \sum _{n=0}^{\infty }{s}_n\left(D_{q^{-1}}^{2k}{p}_n\right)(z;q)=\sum _{n=2k}^{\infty }{s}_n{p}_{n-2k}(z;q)\hfill \\ \hfill =& \sum _{n=2k}^{\infty }\left({\sigma }_{n-2k}{s}_n\right)\frac{p_{n-2k}(z;q)}{{\sigma }_{n-2k}},\hfill \end{array}$$

(61)

and from (55), we obtain

$$\frac{{\sigma }_{n-2k}}{{\sigma }_n}={(-1)}^k\frac{{\xi }_1^{2k}}{{\mathrm{Sin}}_q^{\prime }\left({\xi }_1\right)}.$$

(62)

By substituting (62) into (61), we obtain

$$\left(D_{q^{-1}}^{2k}f\right)\left(z\right)={(-1)}^k\frac{{\xi }_1^{2k}}{{\mathrm{Sin}}_q^{\prime }\left({\xi }_1\right)}\sum _{n=2k}^{\infty }\left({\sigma }_n{s}_n\right)\frac{p_{n-2k}(z;q)}{{\sigma }_{n-2k}}.$$

So, if ${\displaystyle h_n(z;q):=\frac{p_n(z;q)}{{\sigma }_n}-{\mathrm{Sin}}_q\left({\xi }_{1}z\right)}$, then

$${(-1)}^k\frac{{\mathrm{Sin}}_q^{\prime }\left({\xi }_1\right)}{{\xi }_1^{2k}}{D}_{q^{-1}}^{2k}f\left(z\right)=\sum _{n=2k}^{\infty }\left({\sigma }_n{s}_n\right)h_{n-2k}(z;q)+{\mathrm{Sin}}_q\left({\xi }_{1}z\right)\sum _{n=2k}^{\infty }\left({\sigma }_n{s}_n\right).$$

(63)

From the asymptotic (56), we conclude that there exists a constant $K>0$ such that

$$|h_n(z;q)|<K{\left(\frac{{\xi }_1}{{\xi }_2}\right)}^n,\mathrm{for}\left|z\right|<1\mathrm{and}n\in {\mathbb{N}}_0.$$

(64)

Therefore, from (63) and (64) we have

$$\left|\frac{{\mathrm{Sin}}_q^{\prime }\left({\xi }_1\right)}{{\xi }_1^{2k}}{D}_{q^{-1}}^{2k}f\left(z\right)\right|\le \frac{K{\xi }_2}{{\xi }_2-{\xi }_1}(\underset{n\ge 2k}{sup}|{\sigma }_n{s}_n\left|\right)+\left|{\mathrm{Sin}}_q\left({\xi }_{1}z\right)\right|\left|\sum _{n=2k}^{\infty }\left({\sigma }_n{s}_n\right)\right|.$$

Consequently, the sequence ${\left(F_k(z;q)\right)}_k={\left({\xi }_1^{-2k}{D}_{q^{-1}}^{2k}f\left(z\right)\right)}_k$ is uniformly convergent on $\left|z\right|<1$. Therefore, the sequence of functions ${\left(D_{q^{-1}}{F}_k(z;q)\right)}_k$ also convergent uniformly on $\left|z\right|<1$. Furthermore,

$$\underset{n\to \infty }{lim}{F}_n(0;q)=0=\underset{n\to \infty }{lim}\left(D_{q^{-1}}{F}_n\right)(0;q),$$

which implies the function f satisfies the condition (59) and the proof is complete. □

Theorem 3. 

If $f\in \mathcal{F}$, then the series

$$\sum _{n=0}^{\infty }{L}_n\left(f\right)p_n(z;q)$$

converges to f uniformly on compact subsets of $\mathbb{C}$.

Proof. 

Since the convergence with respect to the uniform norm is uniform convergence, it is enough to prove that

$$\underset{k\to \infty }{lim}\parallel f-\sum _{n=0}^k{L}_n\left(f\right)p_n(z;q)\parallel =0,$$

where $∥·∥$ is defined in (22). From (28) and (30), we obtain

$$\begin{array}{cc}\hfill \sum _{n=0}^k{L}_n\left(f\right)p_n(z;q)=& \sum _{n=0}^k\left(\sum _{j=n-1}^{\infty }{D}_{q^{-1}}^{j}f\left(0\right)L_{nj}\right)p_n(z;q)\hfill \\ \hfill =& \sum _{j=0}^{\infty }{D}_{q^{-1}}^{j}f\left(0\right)\sum _{n=0}^{min(k,j+1)}{L}_{nj}{p}_n(z;q)\hfill \\ \hfill =& \sum _{j=0}^{k-1}{D}_{q^{-1}}^{j}f\left(0\right)\sum _{n=0}^{j+1}{L}_{nj}{p}_n(z;q)+\sum _{j=k}^{\infty }{D}_{q^{-1}}^{j}f\left(0\right)\sum _{n=0}^k{L}_{nj}{p}_n(z;q)\hfill \\ \hfill =& \sum _{j=0}^{k-1}{q}^{\frac{j(j-1)}{2}}{D}_{q^{-1}}^{j}f\left(0\right)\frac{z^j}{{\left[j\right]}_q!}+\sum _{j=k}^{\infty }{D}_{q^{-1}}^{j}f\left(0\right)\sum _{n=0}^k{L}_{nj}{p}_n(z;q).\hfill \end{array}$$

(65)

Using the Maclaurin’s expansion (24) and Equation (65), we obtain

$$\parallel f-\sum _{n=0}^k{L}_n\left(f\right)p_n(z;q)\parallel =\parallel \sum _{j=k}^{\infty }{q}^{\frac{j(j-1)}{2}}{D}_{q^{-1}}^{j}f\left(0\right)\frac{z^j}{{\left[j\right]}_q!}-\sum _{j=k}^{\infty }{D}_{q^{-1}}^{j}f\left(0\right)\sum _{n=0}^k{L}_{nj}{p}_n(z;q)\parallel .$$

(66)

Now, set

$$g_k\left(z\right):=\sum _{j=k}^{\infty }{q}^{\frac{j(j-1)}{2}}{D}_{q^{-1}}^{j}f\left(0\right),r_k\left(z\right):=\sum _{j=k}^{\infty }{D}_{q^{-1}}^{j}f\left(0\right)\sum _{n=0}^k{L}_{nj}{p}_n(z;q).$$

Then, from (22), we obtain

$$\begin{array}{cc}\hfill ∥g_k\left(z\right)∥& =\underset{j\in {\mathbb{N}}_0}{sup}\left|\frac{D_{q^{-1}}^j{g}_k\left(0\right)}{{\xi }_1^j}\right|=\underset{j\ge k}{sup}\left|\frac{1}{{{\xi }_1}^j}{D}_{q^{-1}}^{j}f\left(0\right)\right|,\hfill \\ \hfill ∥r_k\left(z\right)∥& =\underset{j\in {\mathbb{N}}_0}{sup}\left|\frac{D_{q^{-1}}^j{r}_k\left(0\right)}{{\xi }_1^j}\right|=\underset{m\in {\mathbb{N}}_0}{sup}\left|\frac{1}{{{\xi }_1}^m}\sum _{j=k}^{\infty }{D}_{q^{-1}}^{j}f\left(0\right)\sum _{n=0}^k{L}_{nj}{D}_{q^{-1}}^m{p}_n\left(0\right)\right|.\hfill \end{array}$$

Now, using the inequalities (29) and (52), we conclude that there exists a constant $M>0$ such that

$$\begin{array}{cc}\hfill |\frac{1}{{{\xi }_1}^m}\sum _{j=k}^{\infty }{D}_{q^{-1}}^{j}f\left(0\right)& \sum _{n=0}^k{L}_{nj}{D}_{q^{-1}}^m{p}_n\left(0\right)|\le M|\sum _{j=k}^{\infty }\frac{1}{{\xi }_1^j}{D}_{q^{-1}}^{j}f\left(0\right)|\sum _{n=0}^k{q}^{\frac{(j-n)(j-n-1)}{2}}\frac{{\xi }_1^{j-n}}{{[j-n]}_q!}\hfill \\ \hfill <& M\underset{j\ge k}{sup}\left|\frac{1}{{\xi }_1^j}{D}_{q^{-1}}^{j}f\left(0\right)\right|\sum _{j=k}^{\infty }\sum _{n=-\infty }^k{q}^{\frac{(j-n)(j-n-1)}{2}}\frac{{\xi }_1^{j-n}}{{[j-n]}_q!}.\hfill \end{array}$$

(67)

Since

$$\sum _{j=k}^{\infty }\sum _{n=-\infty }^k{q}^{\frac{(j-n)(j-n-1)}{2}}\frac{{\xi }_1^{j-n}}{{[j-n]}_q!}=\sum _{m=0}^{\infty }\sum _{r=m}^{\infty }{q}^{\frac{r(r-1)}{2}}\frac{{\xi }_1^r}{{\left[r\right]}_q!},$$

(68)

changing the order of summations yields

$$\begin{array}{cc}\hfill \sum _{j=k}^{\infty }\sum _{n=-\infty }^k{q}^{\frac{(j-n)(j-n-1)}{2}}\frac{{\xi }_1^{j-n}}{{[j-n]}_q!}& =\sum _{r=0}^{\infty }{q}^{\frac{r(r-1)}{2}}\frac{{\xi }_1^r}{{\left[r\right]}_q!}\sum _{m=0}^{r}1\hfill \\ & =\sum _{r=0}^{\infty }(r+1)q^{\frac{r(r-1)}{2}}\frac{{\xi }_1^r}{{\left[r\right]}_q!}\hfill \\ \hfill & =E_q\left({\xi }_1\right)+{\varphi }_1\left({\xi }_1\right),\hfill \end{array}$$

(69)

where ${\varphi }_1\left(t\right)={\sum }_{r=0}^{\infty }{q}^{\frac{r(r-1)}{2}}\frac{r{t}^r}{{\left[r\right]}_q!}.$ By combining Equation (66) with the above estimates (67) and (69), we obtain

$$\parallel f-\sum _{n=0}^k{L}_n\left(f\right)p_n(z;q)\parallel \le \left[1+M(E_q\left({\xi }_1\right)+{\varphi }_1\left({\xi }_1\right))\right]\underset{j\ge k}{sup}\left|\frac{1}{{\xi }_1^j}{D}_{q^{-1}}^{j}f\left(0\right)\right|,$$

and this complete the proof. □

Theorem 4. 

Let f be an entire function. Then, the q-Lidstone series expansion

$$f\left(1\right)A_0\left(z\right)-f\left(0\right)B_0\left(z\right)+\left(D_{q^{-1}}^{2}f\right)\left(1\right)A_1\left(z\right)-\left(D_{q^{-1}}^{2}f\right)\left(0\right)B_1\left(z\right)+\dots$$

(70)

converges to f if and only if $f\in \mathcal{F}$.

Proof. 

The proof follows immediately from Theorems 2 and 3. □

Remark 2. 

Theorem 4 is not applicable with the following arrangement of q-Lidstone series:

$$\sum _{n=0}^{\infty }{D}_{q^{-1}}^{2n}f\left(1\right)A_n\left(z\right)-\sum _{n=0}^{\infty }{D}_{q^{-1}}^{2n}f\left(0\right)B_n\left(z\right).$$

(71)

Notice, in the proof of Theorem 3, we used Inequality (52) to estimate the norm of the function

$$\parallel f-\sum _{n=0}^k{L}_n\left(f\right)p_n(z;q)\parallel$$

to the quantity ${\displaystyle \left[1+M(E_q\left({\xi }_1\right)+\varphi \left({\xi }_1\right))\right]\underset{j\ge k}{sup}\left|\frac{1}{{\xi }_1^j}\left(D_{q^{-1}}^{j}f\right)\left(0\right)\right|}$. This estimation is inaccurate if the q-Lidstone series has the form (2). Here, each of the two series is required to be convergent. Therefore, we can not apply the result in this case.

We end the paper by providing a sufficient condition on f such that (71) converges to $f\left(z\right)$ uniformly on each compact subset of the plane.

Theorem 5. 

Let ${\xi }_1$ be the smallest positive zero of ${\mathrm{Sin}}_q\left(z\right)$. Suppose the series

$$\sum _{n=0}^{\infty }\frac{D_{q^{-1}}^{2n}f\left(0\right)}{{{\xi }_1}^{2n}},\sum _{n=0}^{\infty }\frac{D_{q^{-1}}^{2n}f\left(1\right)}{{{\xi }_1}^{2n}}$$

(72)

are absolutely convergent. Then

$$\sum _{n=0}^{\infty }{D}_{q^{-1}}^{2n}f\left(1\right)A_n\left(z\right)-\sum _{n=0}^{\infty }{D}_{q^{-1}}^{2n}f\left(0\right)B_n\left(z\right)$$

converges uniformly on each compact subset of $\mathbb{C}$ to an entire function $f\left(z\right)$.

Proof. 

Let $z\in \mathbb{C}$. Then, from (44), there exists a constant $M>0$ such that

$$|A_n\left(z\right)-\frac{2{(-1)}^n{\mathrm{Sin}}_q\left({\xi }_{1}z\right)}{{\xi }_1^{2n+1}{\mathrm{Sin}}_q^{\prime }\left({\xi }_1\right)}|\le \frac{M}{{\xi }_2^{2n}}(n\to \infty ).$$

Therefore,

$$\sum _{n=0}^{\infty }\left|D_{q^{-1}}^{2n}f\left(1\right)A_n\left(z\right)\right|\le \frac{2\left|{\mathrm{Sin}}_q\left({\xi }_{1}z\right)\right|}{{\xi }_1{\mathrm{Sin}}_q^{\prime }\left({\xi }_1\right)}\sum _{n=0}^{\infty }\left|\frac{D_{q^{-1}}^{2n}f\left(1\right)}{{{\xi }_1}^{2n}}\right|+M\sum _{n=0}^{\infty }\left|\frac{D_{q^{-1}}^{2n}f\left(1\right)}{{\xi }_1^{2n}}\right|.$$

(73)

Since the series in (72) are absolutely convergent, then the series ${\displaystyle \sum _{n=0}^{\infty }{D}_{q^{-1}}^{2n}f\left(1\right)A_n\left(z\right)}$ converges uniformly on compact subsets of $\mathbb{C}$. Similarly, by using (51) we can verify that the series ${\displaystyle \sum _{n=0}^{\infty }{D}_{q^{-1}}^{2n}f\left(0\right)B_n\left(z\right)}$ converges uniformly on each compact subsets of $\mathbb{C}$. Now, by using Equations (6), (7) and (24), we obtain

$$\begin{array}{cc}\hfill & \sum _{n=0}^{\infty }{D}_{q^{-1}}^{2n}f\left(1\right)A_n\left(z\right)-\sum _{n=0}^{\infty }{D}_{q^{-1}}^{2n}f\left(0\right)B_n\left(z\right)\hfill \\ \hfill =& \sum _{n=0}^{\infty }\sum _{k=0}^{\infty }{q}^{\frac{k(k-1)}{2}}\frac{D_{q^{-1}}^{2n+k}f\left(0\right)}{{\left[k\right]}_q!}{A}_n\left(z\right)-\sum _{n=0}^{\infty }{D}_{q^{-1}}^{2n}f\left(0\right)B_n\left(z\right)\hfill \end{array}$$

(74)

Now the series $S_1:={\sum }_{n=0}^{\infty }{\sum }_{k=0}^{\infty }{q}^{\frac{k(k-1)}{2}}\frac{D_{q^{-1}}^{2n+k}f\left(0\right)}{{\left[k\right]}_q!}{A}_n\left(z\right)$ can be rearranged as

$$S_1=\sum _{m=0}^{\infty }{D}_{q^{-1}}^{2m}f\left(0\right)\sum _{n=0}^m{\displaystyle \frac{q^{\left(\genfrac{}{}{0pt}{}{2m-2n}{2}\right)}}{{[2m-2n]}_q!}}{A}_n\left(z\right)+\sum _{m=0}^{\infty }{D}_{q^{-1}}^{2m+1}f\left(0\right)\sum _{n=0}^m{\displaystyle \frac{q^{\left(\genfrac{}{}{0pt}{}{2m+1-2n}{2}\right)}}{{[2m+1-2n]}_q!}}{A}_n\left(z\right).$$

(75)

Combining (74) and (75) and using (30), we obtain

$$\begin{array}{c}{\displaystyle \sum _{n=0}^{\infty }}{D}_{q^{-1}}^{2n}f\left(1\right)A_n\left(z\right)-{\displaystyle \sum _{n=0}^{\infty }}{D}_{q^{-1}}^{2n}f\left(0\right)B_n\left(z\right)=\\ {\displaystyle \sum _{m=0}^{\infty }}{D}_{q^{-1}}^{2m}f\left(0\right)\left({\displaystyle \sum _{n=0}^m}{\displaystyle \frac{q^{\left(\genfrac{}{}{0pt}{}{2m-2n}{2}\right)}}{{[2m-2n]}_q!}}{A}_n\left(z\right)-B_m\left(z\right)\right)+{\displaystyle \sum _{m=0}^{\infty }}{D}_{q^{-1}}^{2m+1}f\left(0\right){\displaystyle \sum _{n=0}^m}{\displaystyle \frac{q^{\left(\genfrac{}{}{0pt}{}{2m+1-2n}{2}\right)}}{{[2m+1-2n]}_q!}}{A}_n\left(z\right)\\ ={\displaystyle \sum _{m=0}^{\infty }}{D}_{q^{-1}}^{m}f\left(0\right){\displaystyle \frac{q^{\left(\genfrac{}{}{0pt}{}{m}{2}\right)}}{{\left[m\right]}_q!}}{z}^m=f\left(z\right).\end{array}$$

This proves the series (71) has the limit $f\left(z\right)$ on the plane, and we obtain the required result. □

5. Conclusions and Future Work

We proved that the q-Lidstone series expansion

$$f\left(1\right)A_0\left(z\right)-f\left(0\right)B_0\left(z\right)+D_{q^{-1}}^{2}f\left(1\right)A_1\left(z\right)-D_{q^{-1}}^{2}f\left(0\right)B_1\left(z\right)+\dots ,$$

converges to the function $f\left(z\right)$ for each complex z if and only if $f\in \mathcal{F}$, i.e.,

$$D_{q^{-1}}^{n}f\left(0\right)=o\left({\xi }_1^n\right)asn\to \infty ,$$

where ${\xi }_1$ is the smallest positive zero of the function ${\mathrm{Sin}}_q\left(z\right)$. We also provided a sufficient condition on f so that

$$\sum _{n=0}^{\infty }{D}_{q^{-1}}^{2n}f\left(1\right)A_n\left(z\right)-\sum _{n=0}^{\infty }{D}_{q^{-1}}^{2n}f\left(0\right)B_n\left(z\right)$$

converges to $f\left(z\right)$, uniformly on each compact subset of the plane.

Another study to give a characterization of those functions on the plane given by absolutely convergent of q-Lidstone series expansion is in progress.