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# Taylor theory associated with Hahn difference operator

## Abstract

In this paper, we establish Taylor theory based on Hahn’s difference operator $$D_{q,\omega}$$ which is defined by $$D_{q,\omega}f(t)=\frac{f(qt+\omega)-f(t)}{t(q-1)+\omega}$$, $$t\neq\frac {\omega}{1-q}$$, where $$q\in(0,1)$$ and ω is a positive number.

## Introduction and preliminaries

Let $$q\in(0,1)$$, $$\omega>0$$ and $${\omega_{0}:=\frac{\omega}{1-q}}$$. Let f be a function defined on an interval I of $$\mathbb {R}$$ which contains $$\omega_{0}$$. Hahn  introduced his difference operator which is defined by

$$D_{q,\omega}f(t):=\frac{f(qt+\omega)-f(t)}{t(q-1)+\omega}, \quad\text{if t\neq\omega_{0},}$$
(1.1)

and $$D_{q,\omega}f(\omega_{0}):=f'(\omega_{0})$$, provided that f is differentiable at $$\omega_{0}$$ in the usual sense. In this case we call $$D_{q,\omega}f$$ the $$q, \omega$$-derivative and that f is $$q, \omega$$-differentiable at t whenever $$D_{q,\omega}f(t)$$ exists. Finally, we say that f is $$q, \omega$$-differentiable, i.e., throughout I if $$D_{q,\omega}f(\omega_{0})$$ exists.

Hahn difference operator unifies the two most well-known quantum difference operators: the Jackson q-difference operator , which is defined by

$$D_{q}f(t)=\frac{f(qt)-f(t)}{t(q-1)},\quad \text{if t\neq0, 0< q< 1;}$$
(1.2)

and the forward difference $$\Delta_{\omega}$$, which is defined by

$$\Delta_{\omega}f(t)=\frac{f(t+\omega)-f(t)}{\omega}, \quad\text{t\in \mathbb {R}, \omega>0,}$$
(1.3)

see [4, 5, 14, 15]. Hahn operator has attracted the attention of several researchers and a variety of results can be found in papers [1, 2, 6, 1622]. In  Annaby and Mansour proved analytically the q-Taylor series associated with $$D_{q}$$, introduced by Jackson , of an analytic function in some complex domain. In the present paper, we establish an overarching $$q, \omega$$-Taylor theory associated with Hahn difference operator $$D_{q,\omega}$$. In this theory the Hahn difference operator $$D_{q,\omega}$$ replaces the differentiation operator in the usual Taylor series.

First, we introduce some preliminary results and some notations. Let f, g be $$q, \omega$$-differentiable at $$t\in I$$, then

\begin{aligned}& D_{q,\omega}(f+g) (t)=D_{q,\omega}f(t)+D_{q,\omega}g(t), \end{aligned}
(1.4)
\begin{aligned}& D_{q,\omega}(fg) (t)=D_{q,\omega}\bigl(f(t) \bigr)g(t)+f(qt+\omega)D_{q,\omega}g(t), \end{aligned}
(1.5)
\begin{aligned}& D_{q,\omega}(f/g) (t)=\frac{D_{q,\omega}(f(t))g(t)-f(t)D_{q,\omega }g(t)}{g(t)g(qt+\omega)} \end{aligned}
(1.6)

provided that in (1.6), $$g(t)g(qt+\omega)\neq0$$ [1, 2]. Also, for $$n\in\mathbb {N}$$, the following relations hold:

\begin{aligned}& D_{q,\omega}(\alpha t+\beta)^{n}=\alpha\sum _{k=0}^{n-1}\bigl(\alpha(qt+\omega )+ \beta\bigr)^{k}(\alpha t+\beta)^{n-k-1}, \end{aligned}
(1.7)
\begin{aligned}& D_{q,\omega}(\alpha t+\beta)^{-n}=-\alpha\sum _{k=0}^{n-1}\bigl(\alpha (qt+\omega)+ \beta\bigr)^{-n+k}(\alpha t+\beta)^{-k-1}, \end{aligned}
(1.8)

where $$\alpha, \beta\in\mathbb {R}$$, see [1, 2].

The q-shifted factorial $$(b;q)_{n}$$ for a complex number b and $$n\in \mathbb {N}_{0}=\mathbb {N}\cup\{0\}$$ is defined to be

$$(b;q)_{n}=\left \lbrace \textstyle\begin{array}{l@{\quad} l} \prod_{j=1}^{n}(1-bq^{j-1}), & \text{if n\in \mathbb {N},}\\ 1, & \text{if n=0.} \end{array}\displaystyle \right .$$

The limit $$\lim_{n\to\infty}(b;q)_{n}$$ is denoted by $$(b;q)_{\infty}$$. Moreover $$(b;q)_{n}$$ has the representation 

$$(b;q)_{n}=\sum_{k=0}^{n}(-1)^{k} \left ( \textstyle\begin{array}{c} n\\ k \end{array}\displaystyle \right )_{q}q^{\frac{k(k-1)}{2}}b^{k}.$$
(1.9)

The q-binomial coefficients 

$$\left ( \textstyle\begin{array}{c} n\\ k \end{array}\displaystyle \right )_{q}= \frac{(q;q)_{n}}{(q;q)_{k}(q;q)_{n-k}}$$

satisfy the following property:

$$\left ( \textstyle\begin{array}{c} n+1\\ k \end{array}\displaystyle \right )_{q}=\left ( \textstyle\begin{array}{c} n\\ k \end{array}\displaystyle \right )_{q}q^{k}+\left ( \textstyle\begin{array}{c} n\\ k-1 \end{array}\displaystyle \right )_{q}=\left ( \textstyle\begin{array}{c} n\\ k \end{array}\displaystyle \right )_{q}+\left ( \textstyle\begin{array}{c} n\\ k-1 \end{array}\displaystyle \right )_{q}q^{n+1-k}.$$
(1.10)

For $$n\in\mathbb {N}_{0}$$ and $$0< q<1$$, the q-analogues of the natural numbers of the factorial function and of the semifactorial function [7, 13] are defined by

$$[n]_{q}=\frac{1-q^{n}}{1-q},\quad n\in\mathbb {N}_{0}, 0< q< 1,$$
(1.11)

and

$$[n]_{q}!=\prod_{k=1}^{n} [k]_{q},\qquad _{q}!:=1,\quad 0< q< 1.$$
(1.12)

$$[x-a]_{n}$$ is defined by

$$[x-a]_{n}=(x-a) (x-aq) \bigl(x-aq^{2} \bigr)\cdots\bigl(x-aq^{n-1}\bigr),\quad n\ge1,\qquad [x-a]_{0}=1.$$
(1.13)

The following formula was obtained by Euler :

$$[x-a]_{n}=\sum_{k=0}^{n} \left ( \textstyle\begin{array}{c} n\\ k \end{array}\displaystyle \right )_{q}q^{\frac{k(k-1)}{2}}x^{n-k}(-a)^{k}.$$
(1.14)

The q-gamma function  is defined by

$$\varGamma_{q}(z)=\dfrac{(q;q)_{\infty}}{(q^{z};q)_{\infty}}(1-q)^{1-z},\quad 0< q< 1,$$

where $$z\in\mathbb {C}\setminus\{-n:n\in\mathbb {N}_{0}\}$$. Here, we take the principal values of $$q^{z}$$ and $$(1-q)^{1-z}$$. In particular

$$\varGamma_{q}(n+1)=\dfrac{(q;q)_{n}}{(1-q)^{n}},\quad n\in\mathbb {N}.$$

It is known that, for $$x>0$$, $$\varGamma_{q}(x)$$ is the unique logarithmically convex function that satisfies the functional equation:

$$\varGamma_{q}(x+1)=[x]_{q}\varGamma_{q}(x),\qquad \varGamma_{q}(1)=1.$$

In , Aldowah introduced the $$q,\omega$$-integral of f from a to b as follows.

### Definition 1.1

Let I be any interval of $$\mathbb {R}$$ containing $$\omega_{0}$$. Assume that $$f:I\to\mathbb {R}$$ is a function, and let $$a, b\in I$$ such that $$a< b$$. The $$q, \omega$$-integral of f from a to b is defined by

$$\int_{a}^{b}f(t)\,d_{q, \omega}t:= \int_{\omega_{0}}^{b}f(t)\,d_{q, \omega }t- \int_{\omega_{0}}^{a}f(t)\,d_{q, \omega}t,$$
(1.15)

where

$$\int_{\omega_{0}}^{x}f(t)\,d_{q, \omega}t:= \bigl(x(1-q)-\omega\bigr)\sum_{k=0}^{\infty}q^{k}f\bigl(xq^{k}+\omega[k]_{q}\bigr),\quad x \in I,$$
(1.16)

provided that the series converges at $$x=a$$ and $$x=b$$. In this case f is called $$q, \omega$$-integrable over $$[a,b]$$ for all $$a, b\in I$$.

### Lemma 1.2

([1, 2])

Let$$f, g:I\to\mathbb {R}$$be$$q, \omega$$-integrable on$$I, k\in\mathbb {R}$$and$$a, b, c\in I$$, $$a< c< b$$. Then

1. (i)

$$\int_{a}^{a}f(t)\,d_{q, \omega}t=0$$,

2. (ii)

$$\int_{a}^{b}kf(t)\,d_{q, \omega}t=k\int_{a}^{b}f(t)\,d_{q, \omega}t$$,

3. (iii)

$$\int_{a}^{b}f(t)\,d_{q, \omega}t=-\int_{b}^{a}f(t)\,d_{q, \omega}t$$,

4. (iv)

$$\int_{a}^{b}f(t)\,d_{q, \omega}t=\int_{a}^{c}f(t)\,d_{q, \omega }t+\int_{c}^{b}f(t)\,d_{q, \omega}t$$,

5. (v)

$$\int_{a}^{b}(f(t)+g(t))\,d_{q, \omega}t=\int_{a}^{b}f(t)\,d_{q, \omega}t+\int_{a}^{b}g(t)\,d_{q, \omega}t$$.

### Lemma 1.3

([1, 2])

If$$f:I\to\mathbb {R}$$is continuous at$$\omega_{0}$$, then$$\{f(sq^{k}+\omega [k]_{q})\}_{k\in\mathbb {N}}$$converges uniformly to$$f(\omega_{0})$$onI.

### Corollary 1.4

([1, 2])

If$$f:I\to\mathbb {R}$$is continuous at$$\omega_{0}$$, then$$\sum_{k=0}^{\infty}|f((sq^{k})+\omega[k]_{q})|$$converges uniformly onI, and consequentlyfis$$q, \omega$$-integrable overI.

### Lemma 1.5

([1, 2])

If$$f, g:I\to\mathbb {R}$$are continuous at$$\omega_{0}$$, then

$$\int_{a}^{b}f(t)D_{q, \omega}\bigl(g(t) \bigr)\,d_{q, \omega}t=f(t)g(t)|_{a}^{b}- \int _{a}^{b}D_{q, \omega}\bigl(f(t) \bigr)g(qt+\omega)\,d_{q, \omega}t,\quad a, b \in I.$$
(1.17)

### Theorem 1.6

([1, 2])

Assume that$$f:I\to\mathbb {R}$$is continuous at$$\omega_{0}$$. Define

$$F(x):= \int_{\omega_{0}}^{x}f(t)\,d_{q, \omega}t.$$

ThenFis continuous at$$\omega_{0}$$. Furthermore, $$D_{q, \omega}F(x)$$exists for every$$x\in I$$and$$D_{q, \omega}F(x)=f(x)$$. Conversely,

$$\int_{a}^{b}D_{q, \omega}f(t) \,d_{q, \omega}t=f(b)-f(a),\quad a, b\in I.$$

## Main results

We define the $$q, \omega$$-derivative of higher order in the usual way. That is, the nth $$q, \omega$$-derivative, $$n\in\mathbb {N}$$, of $$f:I\to \mathbb {R}$$ is the function $$D^{n}_{q, \omega}f:I\to\mathbb {R}$$ given by $$D^{n}_{q, \omega}f:=D_{q, \omega}(D^{n-1}_{q, \omega}f)$$, provided $$D^{n-1}_{q, \omega}f$$ is $$q, \omega$$-differentiable on I and $$D^{0}_{q, \omega}f=f$$. We consider the following linear spaces:

\begin{aligned}& \begin{aligned}C^{n}&=C^{n}(I,\mathbb {R})\\&:=\bigl\{ f:I\to\mathbb {R} \mid f \text{ is differentiable n-times and f^{(i)} are continuous} , i=1,2,\ldots,n \bigr\} ,\end{aligned} \\& \begin{aligned}C^{n}_{q,\omega}&=C^{n}_{q,\omega}(I, \mathbb {R})\\&:=\bigl\{ f:I\to\mathbb {R} \mid f \text{ is q, \omega-differentiable n-times and D^{n}_{q,\omega}f is continuous at } \omega_{0} \bigr\} ,\end{aligned} \end{aligned}

and

\begin{aligned} C^{\infty}_{q,\omega}&=C^{\infty}_{q,\omega}(I, \mathbb {R})\\&:=\{f:I\to \mathbb {R} \mid f \text{ is q, \omega-differentiable infinitely many times at \omega_{0}} \}. \end{aligned}

Our target is to obtain Taylor expansion of a function f defined on an interval I that contains $$\omega_{0}$$ associated with Hahn difference operator. We need the following lemmas in proving our main results.

### Lemma 2.1

Letfbe a function defined onI. Then, for$$x\neq\omega_{0}$$, thenth$$q, \omega$$derivative$$(D^{n}_{q, \omega}f)(x)$$can be expressed as

\begin{aligned} \bigl(D^{n}_{q, \omega}f\bigr) (x)&= \bigl(x(q-1)+\omega\bigr)^{-n}q^{-\frac{n(n-1)}{2}}\sum _{k=0}^{n}\left ( \textstyle\begin{array}{c} n\\ k \end{array}\displaystyle \right )_{q}(-1)^{k}q^{\frac{k(k-1)}{2}} f\bigl(xq^{n-k}+\omega[n-k]_{q}\bigr). \end{aligned}
(2.1)

### Proof

For $$n=1$$, the formula above yields (1.1). Assume that formula (2.1) is true for $$n=m$$. By relations (1.5), (1.8), and (1.10), we have

\begin{aligned} \bigl(D^{m+1}_{q, \omega}f\bigr) (x)&=D_{q, \omega} \Biggl[\bigl(x(q-1)+\omega \bigr)^{-m}q^{-\frac{m(m-1)}{2}} \sum_{k=0}^{m}\left ( \textstyle\begin{array}{c} m\\ k \end{array}\displaystyle \right )_{q}(-1)^{k}q^{\frac{k(k-1)}{2}} \\ &\quad \times f\bigl(xq^{m-k}+\omega[m-k]_{q}\bigr) \Biggr] \\ &=-(q-1)\sum_{j=0}^{m-1}\bigl((qx+ \omega) (q-1)+\omega\bigr)^{-m+j}\bigl(x(q-1)+\omega \bigr)^{-j-1} \\ &\quad\times q^{-\frac{m(m-1)}{2}}\sum_{k=0}^{m} \left ( \textstyle\begin{array}{c} m\\ k \end{array}\displaystyle \right )_{q}(-1)^{k}q^{\frac{k(k-1)}{2}}f \bigl(xq^{m-k}+\omega[m-k]_{q}\bigr) \\ &\quad+ \bigl((qx+\omega) (q-1)+\omega\bigr)^{-m} q^{-\frac{m(m-1)}{2}}\sum _{k=0}^{m}\left ( \textstyle\begin{array}{c} m\\ k \end{array}\displaystyle \right )_{q}(-1)^{k}q^{\frac{k(k-1)}{2}} \\ &\quad\times D_{q, \omega}f\bigl(xq^{m-k}+\omega[m-k]_{q} \bigr) \\ &=q^{-\frac{m(m-1)}{2}}q^{-m} \Biggl[-(q-1)\sum _{j=0}^{m-1}q^{j}\bigl(x(q-1)+\omega \bigr)^{-m-1} \\ &\quad \times\sum_{k=0}^{m}\left ( \textstyle\begin{array}{c} m\\ k \end{array}\displaystyle \right )_{q}(-1)^{k}q^{\frac{k(k-1)}{2}}f \bigl(xq^{m-k}+\omega[m-k]_{q}\bigr) \\ &\quad +\bigl(x(q-1)+\omega\bigr)^{-m-1}\sum_{k=0}^{m} \left ( \textstyle\begin{array}{c} m\\ k \end{array}\displaystyle \right )_{q}(-1)^{k}q^{\frac{k(k-1)}{2}} \\ &\quad \times \bigl(f\bigl(xq^{m-k+1}+\omega[m-k+1]_{q}\bigr)-f \bigl(xq^{m-k}+\omega[m-k]_{q}\bigr) \bigr) \Biggr]. \end{aligned}

This implies that

\begin{aligned} \bigl(D^{m+1}_{q, \omega}f\bigr) (x)&= q^{-\frac{m(m-1)}{2}}q^{-m}\bigl(x(q-1)+\omega\bigr)^{-m-1} \Biggl[-(q-1)\sum_{j=0}^{m-1}q^{j} \\ &\quad \times\sum_{k=0}^{m}\left ( \textstyle\begin{array}{c} m\\ k \end{array}\displaystyle \right )_{q}(-1)^{k}q^{\frac{k(k-1)}{2}}f \bigl(xq^{m-k}+\omega[m-k]_{q}\bigr) \\ &\quad+ \sum_{k=0}^{m}\left ( \textstyle\begin{array}{c} m\\ k \end{array}\displaystyle \right )_{q}(-1)^{k}q^{\frac{k(k-1)}{2}} \bigl(f\bigl(xq^{m-k+1}+\omega [m-k+1]_{q}\bigr) \\ &\quad- f\bigl(xq^{m-k}+\omega[m-k]_{q}\bigr) \bigr) \Biggr] \\ &= q^{-\frac{m(m+1)}{2}}\bigl(x(q-1)+\omega\bigr)^{-m-1} \Biggl[-(q-1) \frac {q^{m}-1}{q-1}\sum_{k=0}^{m} \left ( \textstyle\begin{array}{c} m\\ k \end{array}\displaystyle \right )_{q} \\ &\quad\times (-1)^{k}q^{\frac{k(k-1)}{2}}f\bigl(xq^{m-k}+ \omega[m-k]_{q}\bigr) + \sum_{k=0}^{m} \left ( \textstyle\begin{array}{c} m\\ k \end{array}\displaystyle \right )_{q}(-1)^{k} \\ & \quad\times q^{\frac{k(k-1)}{2}} \bigl(f\bigl(xq^{m-k+1}+\omega [m-k+1]_{q}\bigr)-f\bigl(xq^{m-k}+\omega[m-k]_{q} \bigr) \bigr) \Biggr] \\ &= q^{-\frac{m(m+1)}{2}}\bigl(x(q-1)+\omega\bigr)^{-m-1} \Biggl[-q^{m}\sum_{k=0}^{m} \left ( \textstyle\begin{array}{c} m\\ k \end{array}\displaystyle \right )_{q}(-1)^{k}q^{\frac{k(k-1)}{2}} \\ & \quad\times f\bigl(xq^{m-k}+\omega[m-k]_{q}\bigr) + \sum _{k=0}^{m}\left ( \textstyle\begin{array}{c} m\\ k \end{array}\displaystyle \right )_{q}(-1)^{k}q^{\frac{k(k-1)}{2}} \\ & \quad\times f\bigl(xq^{m-k+1}+\omega[m-k+1]_{q}\bigr) \Biggr] \\ &=q^{-\frac {m(m+1)}{2}}\bigl(x(q-1)+\omega\bigr)^{-m-1} \Biggl[-q^{m}\sum_{k=1}^{m+1} \left ( \textstyle\begin{array}{c} m\\ k-1 \end{array}\displaystyle \right )_{q}(-1)^{k-1} \\ & \quad\times q^{\frac{(k-1)(k-2)}{2}}f\bigl(xq^{m-k+1}+\omega[m-k+1]_{q} \bigr) \\ & \quad+\sum_{k=0}^{m}\left ( \textstyle\begin{array}{c} m\\ k \end{array}\displaystyle \right )_{q}(-1)^{k}q^{\frac{k(k-1)}{2}}f \bigl(xq^{m-k+1}+\omega[m-k+1]_{q}\bigr) \Biggr] \\ &= q^{-\frac{m(m+1)}{2}}\bigl(x(q-1)+\omega\bigr)^{-m-1} \Biggl[\sum _{k=1}^{m+1}\left ( \textstyle\begin{array}{c} m\\ k-1 \end{array}\displaystyle \right )_{q}q^{m-k+1}(-1)^{k} \\ & \quad\times q^{\frac{k(k-1)}{2}}f\bigl(xq^{m-k+1}+\omega[m-k+1]_{q} \bigr) \\ & \quad+\sum_{k=0}^{m}\left ( \textstyle\begin{array}{c} m\\ k \end{array}\displaystyle \right )_{q}(-1)^{k}q^{\frac{k(k-1)}{2}}f \bigl(xq^{m-k+1}+\omega[m-k+1]_{q}\bigr) \Biggr] \\ &=q^{-\frac{m(m+1)}{2}} \bigl(x(q-1)+\omega\bigr)^{-m-1} \Biggl[(-1)^{m+1}q^{\frac{m(m+1)}{2}}f(x) \\ & \quad+\sum_{k=1}^{m} \left(\left ( \textstyle\begin{array}{c} m\\ k-1 \end{array}\displaystyle \right )_{q}q^{m-k+1}+ \left ( \textstyle\begin{array}{c} m\\ k \end{array}\displaystyle \right )_{q} \right) \\ & \quad\times (-1)^{k}q^{\frac{k(k-1)}{2}}f\bigl(xq^{m-k+1}+ \omega[m-k+1]_{q}\bigr) \\ & \quad+f\bigl(xq^{m+1}+\omega[m+1]_{q}\bigr) \Biggr]. \end{aligned}

That is,

\begin{aligned} \bigl(D^{m+1}_{q, \omega}f\bigr) (x)&=q^{-\frac{m(m+1)}{2}}\bigl(x(q-1)+\omega \bigr)^{-m-1} \Biggl[(-1)^{m+1}q^{\frac{m(m+1)}{2}}f(x) \\ &\quad+ \sum_{k=1}^{m}\left ( \textstyle\begin{array}{c} m+1\\ k \end{array}\displaystyle \right )_{q}(-1)^{k}q^{\frac{k(k-1)}{2}} f\bigl(xq^{m-k+1}+\omega[m-k+1]_{q}\bigr) \\ &\quad +f\bigl(xq^{m+1}+\omega[m+1]_{q}\bigr) \Biggr] \\ &= q^{-\frac{m(m+1)}{2}}\bigl(x(q-1)+\omega\bigr)^{-m-1}\sum _{k=0}^{m+1} \Biggl[\left ( \textstyle\begin{array}{c} m+1\\ k \end{array}\displaystyle \right )_{q}(-1)^{k}q^{\frac{k(k-1)}{2}} \\ & \quad\times f\bigl(xq^{m-k+1}+\omega[m-k+1]_{q}\bigr) \Biggr]. \end{aligned}

Therefore relation (2.1) is true at $$n=m+1$$ and by induction it is true for every $$n\in\mathbb {N}$$. □

In the following result, a formula of the nth derivative of a power series of center zero is given.

### Lemma 2.2

Assume that a functionfhas the power series expansion$$f(x)=\sum_{k=0}^{\infty}a_{k}x^{k}$$, $$x\in I$$. Then

\begin{aligned} \bigl(D^{n}_{q, \omega}f\bigr) (x)&=(1-q)^{-n} {\sum_{k=0}^{\infty}\frac {a_{n+k}}{(1-q)^{k}}\sum_{m=0}^{k}}(-1)^{m} \left ( \textstyle\begin{array}{c} n+k\\ n+m \end{array}\displaystyle \right ) \\ &\quad\times \bigl(x(q-1)+\omega \bigr)^{m}(\omega)^{k-m} \bigl(q^{m+1};q\bigr)_{n}, \quad x\neq\omega _{0}, n\in\mathbb {N}_{0}. \end{aligned}
(2.2)

### Proof

It is clear that Eq. (2.2) is true for $$n=0$$. From Eq. (2.1) and relation (1.9), we have, for $$n\in\mathbb {N}$$,

\begin{aligned} \bigl(D^{n}_{q, \omega}f\bigr) (x)&=\bigl(x(q-1)+\omega \bigr)^{-n}q^{-\frac{n(n-1)}{2}}\sum_{k=0}^{n} \left ( \textstyle\begin{array}{c} n\\ k \end{array}\displaystyle \right )_{q}(-1)^{k}q^{\frac{k(k-1)}{2}} \\ & \quad\times\sum_{j=0}^{\infty}a_{j} \bigl(xq^{n-k}+\omega[n-k]_{q}\bigr)^{j} \\ &=\bigl(x(q-1)+\omega\bigr)^{-n}q^{-\frac{n(n-1)}{2}}\sum _{j=0}^{\infty}\frac {a_{j}}{(1-q)^{j}}\sum _{r=0}^{j}(-1)^{r}\left ( \textstyle\begin{array}{c} j\\ r \end{array}\displaystyle \right )q^{nr} \\ &\quad\times \bigl(x(q-1)+\omega \bigr)^{r}(\omega)^{j-r}\sum _{k=0}^{n}\left ( \textstyle\begin{array}{c} n\\ k \end{array}\displaystyle \right )_{q}(-1)^{k}q^{\frac{k(k-1)}{2}}q^{-kr} \\ &= \bigl(x(q-1)+\omega\bigr)^{-n}q^{-\frac{n(n-1)}{2}}\sum _{j=0}^{\infty}\frac {a_{j}}{(1-q)^{j}}\sum _{r=0}^{j}(-1)^{r}\left ( \textstyle\begin{array}{c} j\\ r \end{array}\displaystyle \right )q^{nr} \\ & \quad\times\bigl(x(q-1)+\omega \bigr)^{r}(\omega)^{j-r} \bigl(q^{-r};q\bigr)_{n}. \end{aligned}

Then

\begin{aligned} \bigl(D^{n}_{q,\omega}f\bigr) (x)&=(-1)^{n} \bigl(x(q-1)+\omega\bigr)^{-n}q^{-\frac {n(n-1)}{2}}\sum _{j=n}^{\infty}\frac{a_{j}}{(1-q)^{j}}\sum _{r=n}^{j}(-1)^{r}q^{nr} \left ( \textstyle\begin{array}{c} j\\ r \end{array}\displaystyle \right ) \\ & \quad\times\bigl(x(q-1)+\omega \bigr)^{r}(\omega)^{j-r}q^{-rn+\frac {n(n-1)}{2}} \bigl(q^{r-n+1};q\bigr)_{n} \\ &= (-1)^{n}\bigl(x(q-1)+\omega\bigr)^{-n}\sum _{j=n}^{\infty}\frac{a_{j}}{(1-q)^{j}}\sum _{r=n}^{j}(-1)^{r}\left ( \textstyle\begin{array}{c} j\\ r \end{array}\displaystyle \right ) \\ & \quad\times\bigl(x(q-1)+\omega \bigr)^{r}(\omega )^{j-r} \bigl(q^{r-n+1};q\bigr)_{n} \\ &= (-1)^{n}\bigl(x(q-1)+\omega\bigr)^{-n}\sum _{k=0}^{\infty}\frac {a_{n+k}}{(1-q)^{n+k}}\sum _{r=n}^{n+k}(-1)^{r}\left ( \textstyle\begin{array}{c} n+k\\ r \end{array}\displaystyle \right ) \\ &\quad\times \bigl(x(q-1)+\omega \bigr)^{r}(\omega )^{n+k-r} \bigl(q^{r-n+1};q\bigr)_{n} \\ &=(-1)^{n}\bigl(x(q-1)+\omega\bigr)^{-n}\sum _{k=0}^{\infty}\frac{a_{n+k}}{(1-q)^{n+k}}\sum _{m=0}^{k}(-1)^{n+m}\left ( \textstyle\begin{array}{c} n+k\\ n+m \end{array}\displaystyle \right ) \\ &\quad\times \bigl(x(q-1)+\omega \bigr)^{n+m}(\omega )^{k-m} \bigl(q^{m+1};q\bigr)_{n} \\ &= (1-q)^{-n}\sum_{k=0}^{\infty}\frac{a_{n+k}}{(1-q)^{k}}\sum_{m=0}^{k}(-1)^{m} \left ( \textstyle\begin{array}{c} n+k\\ n+m \end{array}\displaystyle \right ) \bigl(x(q-1)+\omega \bigr)^{m} \\ &\quad\times (\omega)^{k-m}\bigl(q^{m+1};q\bigr)_{n} \\ &= (1-q)^{-n}\sum_{k=0}^{\infty} \frac{a_{n+k}}{(1-q)^{k}} \Biggl[(-1)^{k}\bigl(x(q-1)+\omega \bigr)^{k}\bigl(q^{k+1};q\bigr)_{n} \\ & \quad+\sum_{m=0}^{k-1}(-1)^{m} \left ( \textstyle\begin{array}{c} n+k\\ n+m \end{array}\displaystyle \right ) \bigl(x(q-1)+\omega \bigr)^{m}(\omega)^{k-m}\bigl(q^{m+1};q \bigr)_{n} \Biggr]. \end{aligned}

□

The following result includes a useful formula for the nth derivative of a power series of center $$\omega_{0}$$.

### Lemma 2.3

Assume that a functionfhas the power series expansion$$f(x)=\sum_{k=0}^{\infty}a_{k}(x-\omega_{0})^{k}$$, $$x\in I$$. Then

$$D^{n}_{q, \omega}f(x)=\bigl(x(1-q)-\omega \bigr)^{-n} {\sum_{k=0}^{\infty}a_{n+k}(x-\omega_{0})^{n+k} \bigl(q^{k+1};q\bigr)_{n}},\quad x\neq\omega_{0}.$$
(2.3)

### Proof

It is clear that Eq. (2.3) is true for $$n=0$$. From Eq. (2.1) and relation (1.9), we have, for $$n\in\mathbb {N}$$,

\begin{aligned} \bigl(D^{n}_{q, \omega}f\bigr) (x)&=\bigl(x(q-1)+\omega \bigr)^{-n}q^{-\frac{n(n-1)}{2}}\sum_{k=0}^{n} \left [\left ( \textstyle\begin{array}{c} n\\ k \end{array}\displaystyle \right )_{q}(-1)^{k}q^{\frac{k(k-1)}{2}} \right . \\ &\quad\times \left .\sum_{j=0}^{\infty}a_{j}\bigl(xq^{n-k}+\omega[n-k]_{q}- \omega_{0}\bigr)^{j}\right ]. \end{aligned}

From this it follows that

\begin{aligned} \bigl(D^{n}_{q, \omega}f\bigr) (x)&=\bigl(x(q-1)+\omega \bigr)^{-n}q^{-\frac{n(n-1)}{2}}\sum_{k=0}^{n} \left [\left ( \textstyle\begin{array}{c} n\\ k \end{array}\displaystyle \right )_{q}(-1)^{k}q^{\frac{k(k-1)}{2}} \right . \\ &\quad\times \left .\sum_{j=0}^{\infty}a_{j}q^{nj-kj}(x-\omega_{0})^{j} \right ] \\ &=\bigl(x(q-1)+\omega \bigr)^{-n}q^{-\frac{n(n-1)}{2}}\sum _{j=0}^{\infty}\Biggl[ a_{j}q^{nj}(x- \omega _{0})^{j} \\ & \quad\times\sum_{k=0}^{n}\left ( \textstyle\begin{array}{c} n\\ k \end{array}\displaystyle \right )_{q}(-1)^{k}q^{\frac{k(k-1)}{2}}q^{-kj} \Biggr] \\ &=\bigl(x(q-1)+\omega \bigr)^{-n}q^{-\frac{n(n-1)}{2}}\sum _{j=0}^{\infty}a_{j}q^{nj}(x- \omega _{0})^{j}\bigl(q^{-j};q \bigr)_{n} \\ &=\bigl(x(q-1)+\omega\bigr)^{-n}q^{-\frac{n(n-1)}{2}}\sum _{j=n}^{\infty}\bigl[ a_{j}q^{nj}(x- \omega_{0})^{j}(-1)^{n}q^{-nj+\frac {n(n-1)}{2}} \\ & \quad\times \bigl(q^{j-n+1};q\bigr)_{n} \bigr] \\ &=\bigl(x(1-q)-\omega\bigr)^{-n}\sum_{k=0}^{\infty}a_{n+k}(x-\omega_{0})^{n+k} \bigl(q^{k+1};q\bigr)_{n}. \end{aligned}

□

One of the important questions: Is there a relation between the nth $$q, \omega$$ derivative and the usual nth derivative? The answer is in the following lemma.

### Lemma 2.4

If$$f\in C^{n+1}$$, then

1. (i)

$$D^{m}_{q, \omega}f$$exists onIand is continuous at$$\omega _{0}$$for all$$m=1,2,\ldots,n+1$$;

2. (ii)

for$$1\le m\le n+1$$,

$$D^{m}_{q, \omega}f(\omega_{0})= \frac{[m]_{q}!}{m!}f^{(m)}(\omega_{0}),$$
(2.4)

where$$f^{(m)}$$is the usualmth derivative off.

### Proof

The proof is by induction. The $$q, \omega$$ derivative $$D_{q, \omega}f$$ exists and $$D_{q, \omega}f(\omega_{0})=f'(\omega_{0})$$. Also $$D_{q, \omega}f$$ is continuous at $$\omega_{0}$$. Indeed,

$$\lim_{x\to\omega_{0}}D_{q, \omega}f(x)=\lim_{t\to\omega_{0}} \frac {f(qx+\omega)-f(x)}{x(q-1)+\omega}=f'(\omega_{0})=D_{q, \omega}f( \omega_{0}).$$

Now, we assume that (i) and (ii) hold for all $$m=1,2,\ldots,l$$, where $$l\leq n$$ and we want to prove that they are true at $$m=l+1$$. By Lemma 2.1, we conclude that

\begin{aligned} \lim_{x\to\omega_{0}}D^{l+1}_{q, \omega}f(x)&=\lim _{x\to\omega_{0}}\frac {1}{(x(q-1)+\omega)^{l+1}q^{\frac{l(l+1)}{2}}} \Biggl[\sum _{k=0}^{l+1}\left ( \textstyle\begin{array}{c} l+1\\ k \end{array}\displaystyle \right )_{q}(-1)^{k}q^{\frac{k(k-1)}{2}} \\ & \quad\times f\bigl(xq^{l-k+1}+\omega [l-k+1]_{q}\bigr) \Biggr] \\ &= \lim_{x\to\omega_{0}}\sum_{k=0}^{l+1} \biggl[\frac{ \tbinom {l+1}{k} _{q}(-1)^{k}q^{\frac{k(k-1)}{2}}q^{(l+1)(l-k+1)}}{(q-1)^{l+1}(xq^{l-k+1}+ \omega[l-k+1]_{q}-\omega_{0})^{l+1} {q^{\frac{l(l+1)}{2}}}} \\ & \quad\times f\bigl(xq^{l-k+1}+\omega[l-k+1]_{q}\bigr) \biggr]. \end{aligned}

Applying L’Hopital rule $$l+1$$ times and using relations (1.12), (1.13), and (1.14), we get

\begin{aligned} \lim_{x\to\omega_{0}}D^{l+1}_{q, \omega}f(x)&=\lim _{x\to\omega_{0}}\frac {1}{(q-1)^{l+1}(l+1)! {q^{\frac{l(l+1)}{2}}}}\sum_{k=0}^{l+1} \Biggl[\left ( \textstyle\begin{array}{c} l+1\\ k \end{array}\displaystyle \right )_{q}(-1)^{k}q^{\frac{k(k-1)}{2}} \\ & \quad\times q^{(l+1)(l-k+1)}f^{(l+1)}\bigl(xq^{l-k+1}+ \omega[l-k+1]_{q}\bigr) \Biggr] \\ &= \frac{\sum_{k=0}^{l+1} \tbinom {l+1}{k} _{q}(-1)^{k}q^{\frac{k(k-1)}{2}}(q^{l+1})^{l-k+1}f^{(l+1)}(\omega_{0})}{ (q-1)^{l+1}(l+1)! {q^{\frac{l(l+1)}{2}}}} \\ &= {\frac{[q^{l+1}-1]_{l+1}f^{(l+1)}(\omega_{0})}{(q-1)^{l+1}(l+1)! {q^{\frac{l(l+1)}{2}}}}} \\ &= {\frac{(q^{l+1}-1)(q^{l+1}-q)(q^{l+1}-q^{2})\cdots (q^{l+1}-q^{l})f^{(l+1)}(\omega_{0})}{(q-1)^{l+1}(l+1)! {q^{0+1+2+\cdots+(l-1)+l}}}} \\ &= {\frac{(q^{l+1}-1)(q^{l}-1)(q^{l-1}-1)\cdots(q-1)f^{(l+1)}(\omega _{0})}{(q-1)^{l+1}(l+1)!}} \\ &= {\frac{_{q}_{q}\cdots[l]_{q}[l+1]_{q}f^{(l+1)}(\omega_{0})}{(l+1)!}} \\ &= {\frac{[l+1]_{q}!}{(l+1)!}f^{(l+1)}(\omega_{0})}. \end{aligned}

On the other hand, we conclude that

\begin{aligned} D^{l+1}_{q, \omega}f(\omega_{0})&=\lim _{x\to\omega_{0}}\frac{D^{l}_{q, \omega}f(x)-D^{l}_{q, \omega}f(\omega_{0})}{x-\omega_{0}} \\ &=\lim_{x\to\omega_{0}}\frac{d}{dx} \biggl[ \frac{\sum_{k=0}^{l}\tbinom {l}{k}_{q}(-1)^{k}q^{\frac{k(k-1)}{2}}f(xq^{l-k}+\omega [l-k]_{q})}{(x(q-1)+\omega)^{l}q^{\frac{l(l-1)}{2}}} \biggr] \\ &= \lim_{x\to\omega_{0}}\frac{1}{(x(q-1)+\omega)^{l+1} {q^{\frac {l(l-1)}{2}}}}\sum _{k=0}^{l}\left ( \textstyle\begin{array}{c} l\\ k \end{array}\displaystyle \right ) _{q}(-1)^{k}q^{\frac{k(k-1)}{2}} \\ &\quad\times \bigl[\bigl(x(q-1)+\omega \bigr)q^{l-k}f' \bigl(xq^{l-k}+\omega[l-k]_{q}\bigr)-l(q-1)f \bigl(xq^{l-k}+\omega[l-k]_{q}\bigr) \bigr]. \end{aligned}

Again, applying L’Hopital rule $$l+1$$ times and using relations (1.12), (1.13), and (1.14), we get

\begin{aligned} D^{l+1}_{q, \omega}f(\omega_{0})&=\lim _{x\to\omega_{0}}\frac {1}{(q-1)^{l+1}(l+1)! {q^{\frac{l(l-1)}{2}}}}\sum_{k=0}^{l} \Biggl[\left ( \textstyle\begin{array}{c} l\\ k \end{array}\displaystyle \right )_{q}(-1)^{k}q^{\frac{k(k-1)}{2}}q^{(l+1)(l-k)} \\ & \quad\times (q-1)f^{(l+1)}\bigl(xq^{l-k}+\omega[l-k]_{q} \bigr) \Biggr] \\ &= {\frac{[q^{l+1}-1]_{l}(q-1)f^{(l+1)}(\omega_{0})}{ (q-1)^{l+1}(l+1)! {q^{\frac{l(l-1)}{2}}}}} \\ &= {\frac {[l+1]_{q}!}{(l+1)!}f^{(l+1)}(\omega_{0})}. \end{aligned}

Therefore,

$$\lim_{x\to\omega_{0}}D^{l+1}_{q, \omega}f(x)=D^{l+1}_{q, \omega}f( \omega_{0})= {\frac{[l+1]_{q}!}{(l+1)!}f^{(l+1)}( \omega_{0})}.$$

□

### Corollary 2.5

Assume that f has the power series expansion

$$f(x)=\sum_{n=0}^{\infty}a_{n}(x-\omega_{0})^{n},\quad x\in I.$$

Then

$$a_{n}= \frac{D^{n}_{q, \omega}f(\omega_{0})}{[n]_{q}!},\quad n\in\mathbb {N}.$$
(2.5)

### Proof

By Lemma 2.4, we have

$$a_{n}=\frac{f^{(n)}(\omega_{0})}{n!}=\frac{D^{n}_{q, \omega}f(\omega_{0})}{[n]_{q}!}.$$

□

Now we define the two variable polynomials $$H_{n}(x,t)$$, $$x, t\in I$$, to be

$$H_{0}(x,t):=1,\qquad H_{n}(x,t):=\prod _{j=0}^{n-1}\bigl(x-h^{j}(t) \bigr),$$
(2.6)

where $$h^{j}(t)=tq^{j}+\omega[j]_{q}, t\in I$$ is the jth order iteration of $$h(t)=qt+\omega$$, which uniformly converges to $$\omega_{0}$$ on I.

### Lemma 2.6

For$$n\in\mathbb {N}$$and$$x, t\in I$$, we have

\begin{aligned}& {}_{t}D_{q, \omega}H_{n}(x,t)=-[n]_{q}H_{n-1} \bigl(x,h(t)\bigr), \end{aligned}
(2.7)
\begin{aligned}& {}_{x}D_{q, \omega}H_{n}(x,t)=[n]_{q}H_{n-1}(x,t), \end{aligned}
(2.8)

where$${}_{t}D_{q, \omega}$$is the$$q, \omega$$-derivative with respect tot,

$$I^{n}_{q, \omega}(1)=\frac{H_{n}(x,a)}{\varGamma_{q}(n+1)},$$

where$$I^{n}_{q, \omega}$$is the$$q, \omega$$-integral

$$I^{n}_{q, \omega}f(x):= \int_{a}^{x} \int_{a}^{x_{n-1}} \int_{a}^{x_{n-2}}\cdots \int _{a}^{x_{1}}f(s)\,d_{q, \omega}s \,d_{q, \omega}x_{1}\cdots d_{q, \omega }x_{n-2} \,d_{q, \omega}x_{n-1}.$$

Now, we establish Taylor’s theorem based on Hahn difference operator.

### Theorem 2.7

Letfbe a function defined onI. If$$f\in C^{n}_{q,\omega}$$for some$$n\in\mathbb {N}$$, then for$$x, a\in I$$,

$$f(x)=\sum_{k=0}^{n-1} \frac{D^{k}_{q, \omega}f(a)}{[k]_{q}!}H_{k}(x,a)+R_{n}(x,a),$$
(2.9)

where

$$R_{n}(x,a)= \int_{a}^{x}\frac{D^{n}_{q, \omega}f(t)}{[n-1]_{q}!} H_{n-1}\bigl(x,h(t)\bigr)\,d_{q, \omega}t.$$
(2.10)

### Proof

We prove relation (2.9) by induction. The right-hand side (R.H.S) of (2.9) at $$n=1$$ is

\begin{aligned} R.H.S&=f(a)H_{0}(x,a)+R_{1}(x,a) \\ &=f(a)+ \int_{a}^{x}D_{q, \omega}f(t) \,d_{q, \omega}t=f(x). \end{aligned}

Assume that relation (2.9) is true for $$n=m$$, that is,

$$f(x)=\sum_{k=0}^{m-1} \frac{D^{k}_{q, \omega}f(a)}{[k]_{q}!}H_{k}(x,a)+R_{m}(x,a),$$

where $$R_{m}(x,a)= \int_{a}^{x}\frac{D^{m}_{q, \omega}f(t)}{[m-1]_{q}!} H_{m-1}(x,h(t))\,d_{q, \omega}t$$. We integrate by parts in the remainder term $$R_{m}(x,a)$$. We obtain

\begin{aligned} R_{m}(x,a)&= \int_{a}^{x}\frac{D^{m}_{q, \omega}f(t)}{[m-1]_{q}!} H_{m-1}\bigl(x,h(t)\bigr)\,d_{q, \omega}t \\ &=- \int_{a}^{x}\frac{D^{m}_{q, \omega}f(t)}{[m-1]_{q}!} \frac{_{t}D_{q, \omega}H_{m}(x,t)}{[m]_{q}}\,d_{q, \omega}t \\ &=-\frac {D^{m}_{q, \omega}f(t)}{[m]_{q}!}H_{m}(x,t)|_{a}^{x}+ \int_{a}^{x}\frac{D^{m+1}_{q, \omega}f(t)}{[m]_{q}!} H_{m}\bigl(x,h(t)\bigr)\,d_{q, \omega}t \\ &=D^{m}_{q, \omega}f(a)\frac {H_{m}(x,a)}{[m]_{q}!}+R_{m+1}(x,a). \end{aligned}

Then

$$f(x)=\sum_{k=0}^{m} \frac{D^{k}_{q, \omega}f(a)}{[k]_{q}!}H_{k}(x,a)+R_{m+1}(x,a).$$

Therefore, relation (2.9) is true for $$n=m+1$$, then it is true for every $$n\in\mathbb {N}$$. □

As a direct consequence of the previous theorem, we deduce the following theorem.

### Theorem 2.8

Let$$f\in C^{\infty}_{q,\omega}$$. If for$$x, a\in I$$, $$\lim_{n\to\infty } R_{n}(x,a)=0$$, then$$f(x)$$has the following expansion:

$$f(x)=\sum_{k=0}^{\infty} \frac{D^{k}_{q, \omega}f(a)}{[k]_{q}!}H_{k}(x,a).$$
(2.11)

Furthermore, if$$\lim_{n\to\infty} R_{n}(x,a)=0$$uniformly with respect toxin some subinterval ofI, then the series given by (2.11) is uniformly convergent in this subinterval.

### Corollary 2.9

Let$$f\in C^{\infty}_{q, \omega}$$. If for$$x\in I$$, $$\lim_{n\to\infty} R_{n}(x,\omega_{0})=0$$, then$$f(x)$$has the following expansion:

$$f(x)=\sum_{k=0}^{\infty} \frac{D^{k}_{q, \omega}f(\omega _{0})}{[k]_{q}!}(x-\omega_{0})^{k}.$$

### Theorem 2.10

Let$$f\in C^{\infty}_{q, \omega}$$. Assume that there is a nonnegative sequence$$\{M_{n}\}$$such that

1. (i)

$$|D^{n}_{q, \omega}f(h^{m}(y))|\le C M_{n}$$, $$n, m\in\mathbb {N}_{0}$$, $$y\in I$$, for some$$C>0$$;

2. (ii)

$${\lim_{n\to\infty} \frac{M_{n+1}}{M_{n}}=M}$$exists.

Thenfhas the$$q, \omega$$-Taylor expansion

$$f(x)=\sum_{k=0}^{\infty} \frac{D^{k}_{q, \omega}f(a)}{[k]_{q}!}H_{k}(x,a)$$
(2.12)

for every$${x\in(\omega_{0}-\frac{1}{M(1-q)},\omega_{0}+\frac{1}{M(1-q)})}$$when$$M>0$$ (respectively$${x\in I}$$when$$M=0$$).

### Proof

We can write $$R_{n}(x,a)$$ as follows:

$$R_{n}(x,a)= R_{1,n}(x,\omega_{0})-R_{2,n}(x;a, \omega_{0}),$$

where

$$R_{1,n}(x,\omega_{0}):=\frac{1}{\varGamma_{q}(n)} \int_{\omega_{0}}^{x} H_{n-1}\bigl(x,h(t) \bigr)D^{n}_{q, \omega}f(t)\,d_{q, \omega}t$$

and

$$R_{2,n}(x;a,\omega_{0}):=\frac{1}{\varGamma_{q}(n)} \int_{\omega_{0}}^{a} H_{n-1}\bigl(x,h(t) \bigr)D^{n}_{q, \omega}f(t)\,d_{q, \omega}t.$$

From (1.16), we have

\begin{aligned} R_{1,n}(x,\omega_{0})&=\bigl(x(1-q)-\omega\bigr) {\sum _{m=0}^{\infty}q^{m} \frac{1}{\varGamma_{q}(n)}H_{n-1}\bigl(x,h^{m+1}(x) \bigr)D^{n}_{q, \omega}f\bigl(h^{m}(x)\bigr)} \\ &= \frac{1}{\varGamma_{q}(n)}\bigl(x(1-q)-\omega\bigr) {\sum _{m=0}^{\infty}} \Biggl[ q^{m}\prod _{r=0}^{n-2}\bigl(x-\bigl[xq^{m+1+r}+[m+1+r]_{q} \omega\bigr]\bigr) \\ &\quad\times D^{n}_{q, \omega }f\bigl(h^{m}(x)\bigr) \Biggr] \\ &= \frac{(1-q)(x-\omega_{0})}{[n-1]_{q}!} {\sum_{m=0}^{\infty}q^{m}(x-\omega _{0})^{n-1}\prod _{r=0}^{n-2}\bigl(1-q^{m+r+1} \bigr)D^{n}_{q, \omega}f\bigl(h^{m}(x)\bigr)} \\ &= \frac{(1-q)(x-\omega_{0})^{n}}{[n-1]_{q}!} {\sum_{m=0}^{\infty}q^{m}\bigl(q^{m+1};q\bigr)_{n-1}D^{n}_{q, \omega}f \bigl(h^{m}(x)\bigr)} \\ &= \frac{(1-q)^{n}(x-\omega_{0})^{n} }{(q;q)_{n-1}} {\sum_{m=0}^{\infty}q^{m}\bigl(q^{m+1};q\bigr)_{n-1}D^{n}_{q, \omega}f \bigl(h^{m}(x)\bigr).} \end{aligned}

Consequently,

\begin{aligned} \bigl\vert R_{1,n}(x,\omega_{0}) \bigr\vert &\le \frac{C}{(q;q)_{\infty}} M_{n} \bigl[(1-q) \vert x-\omega _{0} \vert \bigr]^{n} {\sum _{m=0}^{\infty}q^{m}} \\ &\le \frac{CM_{n}[(1-q) \vert x-\omega_{0} \vert ]^{n}}{(q;q)_{\infty}(1-q)}. \end{aligned}

Then $$\lim_{n\to\infty}R_{1,n}(x,\omega_{0})=0$$, $$x\in(\omega_{0}-\frac {1}{M(1-q)},\omega_{0}+\frac{1}{M(1-q)})$$, when $$M>0$$ (respectively $$x\in I$$, when $$M=0$$). On the other hand, for $$a\in I$$, we have

\begin{aligned} R_{2,n}(x;a,\omega_{0})&= \frac{(a(1-q)-\omega)}{\varGamma_{q}(n)} {\sum _{m=0}^{\infty}q^{m}H_{n-1} \bigl(x,h^{m+1}(a)\bigr)D^{n}_{q, \omega}f \bigl(h^{m}(a)\bigr)}. \end{aligned}

Simple calculations show that

\begin{aligned} \bigl\vert H_{n-1}\bigl(x,h^{m+1}(a)\bigr) \bigr\vert &= \Biggl\vert \prod_{r=0}^{n-2} \bigl(x-h^{m+r+1}(a)\bigr) \Biggr\vert \\ &\le\prod_{r=0}^{n-2}\bigl[ \vert x- \omega_{0} \vert +q^{m+r+1} \vert a- \omega_{0} \vert \bigr] \\ &\le \vert x-\omega_{0} \vert ^{n-1} {e^{\sum_{r=0}^{\infty}q^{m+r+1}\frac{ \vert a-\omega _{0} \vert }{ \vert x-\omega_{0} \vert }}} \\ &\le \vert x-\omega_{0} \vert ^{n-1} {e^{\frac{ \vert a-\omega_{0} \vert }{(1-q) \vert x-\omega_{0} \vert }}}. \end{aligned}

Consequently,

\begin{aligned} \bigl\vert R_{2,n}(x,a,\omega_{0}) \bigr\vert &\le \frac{ \vert x-\omega_{0} \vert ^{n-1}(1-q) \vert a-\omega _{0} \vert }{[n-1]_{q}!}CM_{n} {e^{\frac{ \vert a-\omega_{0} \vert }{(1-q) \vert x-\omega_{0} \vert }}} {\sum _{m=0}^{\infty}q^{m}} \\ &\le \frac{C \vert a-\omega_{0} \vert M_{n}[(1-q) \vert x-\omega_{0} \vert ]^{n-1}}{(q,q)_{\infty}} {e^{\frac{ \vert a-\omega_{0} \vert }{(1-q) \vert x-\omega_{0} \vert }}}. \end{aligned}

This implies that $$\lim_{n\to\infty}R_{2,n}(x;a,\omega_{0})=0$$, $$x\in (\omega_{0}-\frac{1}{M(1-q)},\omega_{0}+\frac{1}{M(1-q)})$$, when $$M>0$$ (respectively $$x\in I$$, when $$M=0$$). Therefore

$${\lim_{n\to\infty}R_{n}(x,a)=\lim _{n\to\infty}\bigl[R_{1,n}(x,\omega _{0})-R_{2,n}(x;a, \omega_{0})\bigr]=0,}$$

$${x\in(\omega_{0}-\frac{1}{M(1-q)},\omega_{0}+\frac{1}{M(1-q)})}$$, when $$M>0$$ (respectively $$x\in I$$, when $$M=0$$). □

### Theorem 2.11

Assume thatfhas the power series expansion$$f(x)=\sum_{n=0}^{\infty}a_{n}(x-\omega_{0})^{n}$$with interval of convergence$$I_{r}=(\omega_{0}-r,\omega_{0}+r)$$, $$r>0$$. Then, for any$$a\in I_{r}$$, fhas the$$q, \omega$$-Taylor expansion

$$f(x)=\sum_{k=0}^{\infty} \frac{D^{k}_{q, \omega}f(a)}{[k]_{q}!}H_{k}(x,a),$$
(2.13)

in any closed subinterval$${\overline{I_{\alpha}}, \alpha< r}$$, where the series is absolutely and uniformly convergent on$${\overline{I_{\alpha}}, \alpha< r}$$.

### Proof

For $$n,m\in\mathbb {N}$$ and by Lemma 2.3, we get

\begin{aligned} D^{n}_{q, \omega}f\bigl(h^{m}(y)\bigr)&= \bigl(h^{m}(y) (1-q)-\omega\bigr)^{-n} {\sum _{k=0}^{\infty}a_{n+k} \bigl(h^{m}(y)-\omega_{0}\bigr)^{n+k} \bigl(q^{k+1};q\bigr)_{n}} \\ &= q^{-mn}\bigl(y(1-q)-\omega\bigr)^{-n} {\sum _{k=0}^{\infty}a_{n+k}q^{mn+mk}(y- \omega_{0})^{n+k}\bigl(q^{k+1};q \bigr)_{n}} \\ &= \frac{1}{(1-q)^{n}} {\sum_{k=0}^{\infty}a_{k}q^{mk}(y-\omega _{0})^{k} \bigl(q^{k+1};q\bigr)_{n}.} \end{aligned}

Consequently, for $$\alpha< r$$,

\begin{aligned} \bigl\vert D^{n}_{q, \omega}f\bigl(h^{m}(y) \bigr) \bigr\vert &\le \frac{1}{(1-q)^{n}} {\sum_{k=0}^{\infty}\bigl\vert a_{k}(y-\omega_{0})^{k} \bigr\vert q^{mk}} \\ &\le \frac{1}{(1-q)^{n}} {\sum_{k=0}^{\infty}\bigl\vert a_{k}\alpha^{k} \bigr\vert q^{mk}} \\ &\leq\frac{1}{(1-q)^{n}} C, y\in\overline{I_{\alpha}}, \end{aligned}

where $${ C=\sum_{k=0}^{\infty}|a_{k}\alpha^{k}| }$$. Then, by Theorem 2.10, f has the $$q, \omega$$-Taylor expansion (2.13). □

Now, we establish some properties of the $$q, \omega$$-exponential functions $$e_{q, \omega}(t)$$ and $$E_{q, \omega}(t)$$ for $$t\in\mathbb {R}$$, $$|t-\omega_{0}|<\frac{1}{1-q}$$, where

\begin{aligned} e_{q, \omega}(t)&=\frac{1}{\prod_{k=0}^{\infty}(1-q^{k}(t(1-q)-\omega ))} \\ &=\frac{1}{((t(1-q)-\omega);q)_{\infty}} \end{aligned}
(2.14)

and

\begin{aligned} E_{q, \omega}(t)&=\prod_{k=0}^{\infty}\bigl(1+q^{k}\bigl(t(1-q)-\omega\bigr)\bigr) \\ &=\bigl(-\bigl(t(1-q)-\omega\bigr);q\bigr)_{\infty}. \end{aligned}
(2.15)

Simple calculations show that the following inequalities are true:

$$\frac{e^{-\frac{q}{1-q}}}{(1-(t(1-q)-\omega))}< e_{q, \omega}(t)< \frac{e^{A}}{1-(t(1-q)-\omega)},\quad \vert t-\omega_{0} \vert < \frac{1}{1-q}$$
(2.16)

and

$$\bigl(1+\bigl(t(1-q)-\omega\bigr)\bigr)e^{-A}< E_{q, \omega}(t)< \bigl(1+\bigl(t(1-q)-\omega\bigr)\bigr)e^{\frac {q}{1-q}},\quad \vert t- \omega_{0} \vert < \frac{1}{1-q},$$
(2.17)

where $${ A=\sum_{k=1}^{\infty} \frac{q^{k}}{1-q^{k}}}$$.

Finally, we can prove the following power series expansions for $$e_{q,\omega}$$ and $$E_{q,\omega}$$.

### Example 2.12

The exponential functions $$e_{q,\omega}$$ and $$E_{q,\omega}$$ defined in (2.14) and (2.15) have the following power series expansions of center $$a\in I$$:

$$e_{q,\omega}(x)=\sum_{k=0}^{\infty} \frac{e_{q, \omega }(a)}{[k]_{q}!}H_{k}(x,a), \quad \vert x-\omega_{0} \vert < \frac{1}{1-q}$$
(2.18)

and

$$E_{q,\omega}(x)=\sum_{k=0}^{\infty} \frac{q^{\frac{k(k-1)}{2}}E_{q, \omega}(h^{k}(a))}{[k]_{q}!}H_{k}(x,a),\quad x\in I,$$
(2.19)

and have the following power series expansions of center $$\omega_{0}$$:

$$e_{q, \omega}(t)=\sum_{k=0}^{\infty} \frac{1}{[k]_{q}!}(t-\omega_{0})^{k}$$
(2.20)

and

$$E_{q, \omega}(t)= \sum_{k=0}^{\infty} \frac{q^{\frac{k(k-1)}{2}}}{[k]_{q}!}(t-\omega_{0})^{k} .$$
(2.21)

Furthermore, both $$e_{q,\omega}$$ and $$E_{q,\omega}$$ are continuous.

### Proof

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

$$D^{n}_{q,\omega}e_{q, \omega}(t)=e_{q, \omega}(t).$$

Inequality (2.16) shows that $$e_{q, \omega}(t)$$ is positive and bounded on every compact subinterval of $${(\omega_{0}-\frac{1}{1-q}, \omega_{0}+\frac{1}{1-q})}$$. For fixed $$t\in{(\omega_{0}-\frac{1}{1-q}, \omega_{0}+\frac{1}{1-q})}$$, there exists $$0<\alpha\le1$$ such that $${|t(1-q)-\omega|<\alpha}$$, which implies that

\begin{aligned} \bigl\vert D^{n}_{q,\omega}e_{q, \omega}(t) \bigr\vert &\le\frac{e^{A}}{1-\alpha},\quad n\in\mathbb {N}_{0}. \end{aligned}

By Theorem 2.10, the $$q,\omega$$-Taylor expansion of $$e_{q, \omega}(t)$$ at a is given by

$$e_{q, \omega}(t)=\sum_{k=0}^{\infty} \frac{e_{q, \omega}(a)}{[k]_{q}!}H_{k}(t,a).$$
(2.22)

Since $$D^{n}_{q,\omega}e_{q, \omega}(\omega_{0})= 1$$, the $$q,\omega$$-Taylor expansion of $$e_{q, \omega}(t)$$ at $$\omega_{0}$$ is given by

$$e_{q, \omega}(t)=\sum_{k=0}^{\infty} \frac{1}{[k]_{q}!}(t-\omega_{0})^{k}.$$
(2.23)

The series in (2.23) is uniformly convergent on every compact subinterval of $$(\omega_{0}-\frac{1}{1-q}, \omega_{0}+\frac{1}{1-q})$$ by Weierstrass M-test, and consequently $$e_{q, \omega}(t)$$ is continuous.

Let $$t\in\mathbb {R}$$, $$|t-\omega_{0}|<\frac{1}{1-q}$$. First, we show that

$$D^{n}_{q,\omega}E_{q,\omega}(t)=q^{\frac {n(n-1)}{2}}E_{q,\omega} \bigl(h^{n}(t)\bigr),\quad n\in\mathbb {N}_{0}$$
(2.24)

by induction. For $$n=1$$, we have

\begin{aligned} D_{q,\omega}E_{q,\omega}(t)&=\frac{1}{t(q-1)+\omega}\Biggl[\prod _{k=0}^{\infty}\bigl(1+q^{k}(qt+ \omega) (1-q)-\omega\bigr)) \\ &\quad -\prod_{k=0}^{\infty}\bigl(1+q^{k}\bigl(t(1-q)-\omega\bigr)\bigr)\Biggr] \\ &=\frac{\prod_{k=0}^{\infty}(1+q^{k+1}(t(1-q)-\omega))}{ t(q-1)+\omega} \bigl[1- \bigl(1+t(1-q)-\omega\bigr) \bigr] \\ &=E_{q,\omega}\bigl(h(t)\bigr). \end{aligned}

Assume that formula (2.24) is true for $$n=m$$. We have

\begin{aligned} D^{m+1}_{q,\omega}E_{q,\omega}(t)&=D_{q,\omega} \bigl( D^{m}_{q,\omega }E_{q,\omega}(t)\bigr) \\ &= q^{\frac{m(m-1)}{2}}D_{q,\omega}E_{q,\omega }\bigl(h^{m}(t) \bigr) \\ &= q^{\frac{m(m-1)}{2}}\frac{1}{t(q-1)+\omega}\Biggl[\prod _{k=0}^{\infty}\bigl(1+q^{k+m+1} \bigl(t(1-q)-\omega\bigr)\bigr) \\ & \quad-\prod_{k=0}^{\infty}\bigl(1+q^{k+m}\bigl(t(1-q)-\omega\bigr)\bigr)\Biggr] \\ &= q^{\frac{m(m-1)}{2}}\frac{\prod_{k=0}^{\infty}(1+q^{k+m+1} (t(1-q)-\omega))}{t(q-1)+\omega} \\ &\quad \times\bigl[1- \bigl(1+q^{m}\bigl(t(1-q)-\omega\bigr)\bigr) \bigr] \\ &=q^{\frac{m(m+1)}{2}}\prod_{k=0}^{\infty}\bigl(1+q^{k+m+1}\bigl(t(1-q)-\omega\bigr)\bigr) \\ &= q^{\frac{m(m+1)}{2}}E_{q,\omega}\bigl(h^{m+1}(t)\bigr). \end{aligned}

Inequality (2.17) shows that $$E_{q, \omega}(t)$$ is positive and is bounded on every compact subinterval of $${(\omega_{0}-\frac{1}{1-q}, \omega_{0}+\frac{1}{1-q})}$$. Also we can see that

\begin{aligned} \bigl\vert E_{q,\omega}\bigl(h^{n}(t)\bigr) \bigr\vert & \le\prod_{k=0}^{\infty}\bigl\vert 1+q^{k+n}\bigl(t(1-q)-\omega \bigr) \bigr\vert \\ &\le\prod_{k=0}^{\infty}\bigl[1+q^{k+n}(1-q) \vert t-\omega_{0} \vert \bigr] \\ &\le\prod_{k=0}^{\infty}\bigl[1+q^{k+n}\bigr] \\ &\leq e^{\frac{1}{1-q}}. \end{aligned}

Therefore,

\begin{aligned} \bigl\vert D^{n}_{q,\omega}E_{q,\omega}(t) \bigr\vert &\le q^{\frac{n(n-1)}{2}} \bigl\vert E_{q,\omega} \bigl(h^{n}(t)\bigr) \bigr\vert \\ &\le q^{\frac{n(n-1)}{2}} e^{\frac{1}{1-q}}. \end{aligned}

By Theorem 2.10, the $$q,\omega$$-Taylor expansion of $$E_{q, \omega}(t)$$ at a is given by

$$E_{q, \omega}(t)=\sum_{k=0}^{\infty} \frac{q^{\frac {k(k-1)}{2}}E_{q, \omega}(h^{k}(a))}{[k]_{q}!}H_{k}(t,a).$$

Since $$D^{n}_{q,\omega}f(\omega_{0})= q^{\frac{n(n-1)}{2}}$$, the $$q,\omega$$-Taylor expansion of $$E_{q,\omega}(t)$$ at $$\omega_{0}$$ is given by

$$E_{q,\omega}(t)= \sum_{k=0}^{\infty} \frac{q^{\frac{k(k-1)}{2}}}{[k]_{q}!}(t-\omega_{0})^{k}.$$
(2.25)

The series in (2.25) is uniformly convergent on every compact subinterval of $$(\omega_{0}-\frac{1}{1-q}, \omega_{0}+\frac{1}{1-q})$$ and consequently $$E_{q, \omega}(t)$$ is continuous. □

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