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# Generalized q-Taylor formulas

Advances in Difference Equations20162016:162

https://doi.org/10.1186/s13662-016-0894-3

• Received: 27 February 2016
• Accepted: 12 June 2016
• Published:

## Abstract

In this paper, new generalized q-Taylor formulas involving both Riemann-Liouville and Caputo q-difference operators are constructed. Some applications with solutions of fractional q-difference equations are also given.

## Keywords

• q-difference operator
• generalized q-Taylor formula
• Riemann-Liouville fractional q-derivative
• Caputo fractional q-derivative

• 41A58
• 39A13
• 26A33

## 1 Introduction

A q-analogue of Taylor series was introduced by Jackson :
$$f(x)=\sum_{n=0}^{\infty}\frac{(1-q)^{n}}{(q;q)_{n}}D_{q}^{n}f(a) [x-a]_{n},$$
(1.1)
where $$0< q<1$$, $$D_{q}$$ is the q-derivative, and
$$[x-a]_{n}:= (x-a) (x-qa)\cdots \bigl(x-q^{n-1}a \bigr),\quad n \geqslant1, [x-a]_{0}:=1.$$
Al-Salam and Verma  introduced the following q-interpolation series:
$$f(x)=\sum_{n=0}^{\infty}(-1)^{n}q^{-n(n-1)/2} \frac {(1-q)^{n}}{(q;q)_{n}}D_{q}^{n}f\bigl(aq^{-n}\bigr) [x-a]_{n}.$$
(1.2)
Al-Salam and Verma gave only formal proofs for (1.2); see [1, 2]. Analytic proofs of (1.1) and (1.2) were given in .

Results of generalized Taylor formulas involving the classical fractional derivative may be found in [4, 5]. In , a generalized Taylor formula involving the classical Riemann-Liouville fractional derivative of order α is deduced, whereas the generalized Taylor formula in  contains Caputo fractional derivative of order α, where $$0<\alpha\leq1$$.

In , a q-Taylor formula in terms of Riemann-Liouville fractional q-derivative $$D_{q,a}^{\alpha}$$ of order α is obtained. This result can be stated as follows.

### Theorem A

()

Let f be a function defined on $$(0,b)$$ and $$\alpha\in(0,1)$$. Then f can be expanded in the form
\begin{aligned} f(x) =&\sum_{k=0}^{n-1} \frac{(D_{q,a}^{\alpha+k}f)(c)}{ \Gamma_{q}(\alpha+k+1)}(x-c)^{(\alpha+k)} \\ &{}+\frac{1}{\Gamma_{q}(\alpha)} \int_{a}^{c} (x-c)^{(\alpha-1)} \bigl(D_{q,a}^{\alpha}f\bigr) (t) \,d_{q}t \\ &{} -K(a) (x-c)^{(\alpha-1)}+\bigl(I_{q,c}^{\alpha+n}D_{q,a}^{\alpha+n}f \bigr) (x), \end{aligned}
(1.3)
where $$0< a< c< x< b$$, and $$K(a)$$ does not depend on x.

Also, in , a generalized q-Taylor formula in fractional q-calculus is established and used in deriving certain q-generating functions for the basic hyper-geometric functions.

In this paper, we give generalized Taylor formulas involving Riemann-Liouville fractional q-derivatives of order α and Caputo fractional q-derivatives of order α; see (4.3) and (4.4). We also give sufficient conditions that guarantee that the remainders of these formulas vanish to get infinite expansions.

In the following section, we give a brief account of the q-notations and notions that will be used throughout this paper. In Section 3, we give q-analogues of mean value theorems on $$[0,a]$$. In Section 4, we give generalized q-Taylor formulas involving both Riemann-Liouville fractional q-derivative and Caputo fractional q-derivative. Then conditions for infinite expansion for some functions are given. In the last section, we apply the obtained results in solving certain q-difference equations.

## 2 Notation and preliminaries

In the following, q is a positive number, $$q<1$$. We follow  for the definition of the q-shifted factorial, Jackson q-integral, q-derivative, q-gamma function $$\Gamma_{q}(z)$$, and q-beta function $$B_{q}(\alpha,\beta)$$. Also, we follow  for the definition of the q-derivative at zero and the q-regular at zero functions.

The following q-integral is useful and will be used in the sequel:
$$\int_{0}^{x}(qt/x;q)_{\beta-1}t^{\alpha-1} \,d_{q}t=x^{\alpha }B_{q}(\alpha,\beta),\quad \alpha, \beta,x>0;$$
(2.1)
it can be proved by setting $$\xi=t/x$$.
By $$L_{q}^{1}(0,a)$$, $$a>0$$, we mean the Banach space of all functions defined on $$(0,a]$$ such that
$$\Vert f\Vert := \int_{0}^{a}\bigl\vert f(t)\bigr\vert \,d_{q}t< \infty,$$
(2.2)
where two functions in $$L_{q}^{1}(0,a)$$ are considered to be the same function if they have the same values at the sequence $$\{aq^{n}\}_{n=0}^{\infty}$$.
Let $$\mathcal{L}_{q}^{1}(0,a)$$ denote the space of all functions f defined on $$(0,a]$$ such that $$f\in L_{q}^{1}(0,x)$$ for all $$x\in(qa,a]$$. The space $$\mathcal{AC}_{q}[0,a]$$ is the space of all functions f defined on $$[0,a]$$ such that f is q-regular at zero and
$$\sum_{j=0}^{\infty}\bigl\vert f \bigl(tq^{j}\bigr)-f\bigl(tq^{j+1}\bigr)\bigr\vert < \infty, \quad t\in (qa,a].$$
(2.3)
A characterization of the space $$\mathcal{AC}_{q}[0,a]$$ is given as follows (see ).

### Theorem B

Let f be a function defined on $$[0,a]$$. Then $$f\in\mathcal{AC}_{q}[0,a]$$ if and only if there exist a constant c and a function ϕ in $$\mathcal{L}_{q}^{1}[0,a]$$ such that
$$f\in \mathcal{AC}_{q}[0,a]\quad \Longleftrightarrow\quad f(x)=c+ \int_{0}^{x}\phi(u) \,d_{q}u,\quad x\in[0,a].$$
(2.4)
Moreover, c and ϕ are uniquely determined by $$c=f(0)$$ and $$\phi(x)=D_{q}f(x)$$ for all $$x\in(0,a]$$.
The Riemann-Liouville fractional q-integral operator is introduced in  by Al-Salam through
$$I_{q}^{\alpha}f(x):=\frac{x^{\alpha-1}}{\Gamma_{q}(\alpha)} \int _{0}^{x} (qt/x;q)_{\alpha-1}f(t) \,d_{q}t,\quad \alpha\notin \{-1,-2,\ldots\}.$$
(2.5)
In , the generalized Riemann-Liouville fractional q-integral operator for $$\alpha\in\mathbb{R^{+}}$$ is given as
$$I_{q,a}^{\alpha}f(x):=\frac{x^{\alpha-1}}{\Gamma_{q}(\alpha)} \int _{a}^{x} (qt/x;q)_{\alpha-1}f(t) \,d_{q}t.$$
(2.6)
Using the definition of the q-integral, (2.5) reduces to
$$I_{q}^{\alpha}f(x)=x^{\alpha}(1-q)^{\alpha} \sum_{n=0}^{\infty }q^{n} \frac{(q^{\alpha};q)_{n}}{(q;q)_{n}}f\bigl(xq^{n}\bigr),$$
(2.7)
which is valid for all α. For example,
$$I_{q}^{\alpha}x^{\beta-1}= \frac{\Gamma_{q}(\beta)}{\Gamma_{q}(\beta +\alpha)}x^{\alpha+\beta-1}.$$
(2.8)
This basic Riemann-Liouville fractional q-integral was also given later by Agarwal . In the same paper, he introduced the following semigroup property:
$$I_{q}^{\alpha}I_{q}^{\beta}f(x)=I_{q}^{\beta}I_{q}^{\alpha}f(x)=I_{q}^{\alpha +\beta}f(x),\quad \alpha, \beta\geq0.$$
(2.9)
The generalized Riemann-Liouville fractional q-derivative is given in  by
$$D_{q,a}^{\alpha}f(x)=D_{q}I_{q,a}^{1-\alpha}f(x), \quad a\ge0,$$
(2.10)
and $$D_{q,0}^{\alpha}f(x)=D_{q}^{\alpha}f(x)$$. The Caputo fractional q-derivative of order α, $$0<\alpha\leq1$$, is (see)
$${}^{c}D_{q}^{\alpha}f(x):=I_{q}^{1-\alpha}D_{q}f(x).$$
(2.11)

Let $$\mathcal{AC}_{q}^{(k)}[0,a]$$, $$k\in N$$, be the space of all functions f defined on $$[0,a]$$ such that $$f, D_{q} f,\ldots, D^{k-1}_{q} f$$ are q-regular at zero and $$D^{k-1}_{q} f\in \mathcal{AC}_{q}[0,a]$$.

For $$\alpha> 0$$, let $$k = \ulcorner\alpha\urcorner$$, where is the ceiling function. Then the Riemann-Liouville fractional derivative $${D}_{q}^{\alpha}f(x)$$ exists if (see )
$$f\in\mathcal{L}_{q}^{1}[0;a],\quad I_{q}^{k-\alpha}D_{q}^{k}f \in\mathcal{AC}_{q}^{(k)}[0,a],$$
and $${}^{c}D_{q}^{\alpha}f(x)$$ exists if $$f\in\mathcal{AC}_{q}^{(k)}[0,a]$$.

The following results are proved in  for any $$\alpha>0$$; the result for the case $$0<\alpha<1$$ is introduced in the following theorems without proof.

### Theorem C

Assume that $$f\in\mathcal{L}_{q}^{1}[0;a]$$ and $$I_{q}^{1-\alpha}f\in\mathcal{AC}_{q}[0,a]$$, where $$0<\alpha<1$$. Then the Riemann-Liouville fractional derivative of order α, $$0<\alpha< 1$$, exists, and
$$I_{q}^{\alpha}{D}_{q}^{\alpha }f(x)=f(x)- \frac {I_{q}^{1-\alpha}f(0)}{\Gamma_{q}(\alpha)}x^{\alpha-1.}$$
(2.12)

### Theorem D

If $$f\in\mathcal{AC}_{q}[0,a]$$, then
$$I_{q}^{\alpha}{}^{c}D_{q}^{\alpha }f(x)=I_{q}D_{q}f(x)=f(x)-f(0)$$
(2.13)
for $$0<\alpha<1$$.
It is worth mentioning that the key point in the proofs of Theorems C and D is the q-integration by parts formula:
$$\int_{0}^{b}f(t)D_{q}g(t) \,d_{q}t= (fg) (b)-\lim_{n\to\infty }(fg) \bigl(bq^{n}\bigr)- \int_{0}^{b}D_{q}f(t) g(qt) \,d_{q}t.$$
Hence, if fg is q-regular at zero, then the limit on the right-hand side is nothing but $$(fg)(0)$$.

## 3 Generalized q-mean value theorems on $$[0,a]$$

In this section, we introduce two q-analogues of the mean value theorems. The first one is for q-integrals on an interval of the form $$[0,a]$$, and the second is a mean value theorem with both of Riemann-Liouville fractional q-derivative and Caputo fractional q-derivative on $$[0,a]$$. The first one can be stated as follows.

### Theorem 3.1

(Mean value theorem for q-integrals)

Let g be a continuous function defined on $$[0,a]$$, and h be a nonnegative function defined on $$[0,a]$$ and q-regular at zero. Then
$$\int_{0}^{a}g(t)h(t) \,d_{q}t=g(\xi) \int_{0}^{a}h(t) \,d_{q}t$$
(3.1)
for some $$\xi\in[0,a]$$.

### Proof

The proof is similar to the classical case (see , p.139) and is omitted. □

The derivations of the main results of this paper mainly depend on Theorem 3.1.

### Remark 3.2

1. (1)
We cannot replace the lower end point of the q-integrals in (3.1) by arbitrary nonzero number because the inequality
$$\biggl\vert \int_{c}^{a}f(t) \,d_{q}t \biggr\vert \le \int_{c}^{a} \bigl\vert f(t) \bigr\vert \,d_{q}t,$$
holds only for $$c\in\{0,aq^{n}, n\in\mathbb{N}_{0}\}$$. In this case, (3.1) is also true.

2. (2)

There are q-analogues of mean value theorems on $$[a,b]$$ in , but all these analogues are valid only for certain values of q. For example, one of the mean value theorems for q-integrals in  is the following:

Let f, g be continuous functions on $$[a,b]$$ . Then there exists $$\widehat{q}\in(0,1)$$ such that
$$\bigl(\forall q\in(\widehat{q},1)\bigr)\ \bigl(\exists\xi\in[a,b]\bigr):\quad \int_{a}^{b}g(t)f(t) \,d_{q}t=g(\xi) \int_{a}^{b}f(t) \,d_{q}t.$$

The second theorem is a q-analogue of the mean value theorem for derivative on $$[0,a]$$. Throughout the rest of this article, we assume that $$0<\alpha<1$$.

### Theorem 3.3

1. (1)
If $$f\in \mathcal{L}_{q}^{1}[0;a]$$, $$I_{q}^{1-\alpha}f\in\mathcal{AC}_{q}[0,a]$$, and $$x^{1-\alpha}D^{\alpha}_{q}f\in C[0,a]$$, then
$$f(x)=\frac{I_{q}^{1-\alpha}f(0)}{\Gamma_{q}(\alpha)} x^{\alpha-1} +\frac{{\Gamma_{q}(\alpha)} \xi^{1-\alpha} D^{\alpha}_{q}f(\xi)}{\Gamma_{q}(2\alpha)} x^{2\alpha-1}.$$
(3.2)

2. (2)
If $$f\in\mathcal{AC}_{q}[0,a]$$ and $${}^{c}D_{q}^{\alpha}f\in C[0,a]$$, then
$$f(x)=f(0)+\frac{{}^{c}D_{q}^{\alpha}f(\xi)}{\Gamma _{q}(\alpha)} x^{\alpha}$$
(3.3)

for some ξ lying in the interval $$[0,x]$$ and all $$x\in(0,a]$$.

### Proof

We first prove (3.2). Since (see , p.494)
$$B_{q}(\alpha,\beta)=\frac{\Gamma_{q}(\alpha)\Gamma_{q}(\beta)}{\Gamma _{q}(\alpha+\beta)},$$
from (2.5), Theorem 3.1, and (2.1) we get
\begin{aligned} I_{q}^{\alpha}D^{\alpha}_{q}f(x) =&\frac{x^{\alpha-1}}{\Gamma_{q}(\alpha)} \int _{0}^{x}(qt/x;q)_{\alpha-1} t^{\alpha-1} t^{1-\alpha}D^{\alpha}_{q}f(t) \,d_{q}t \\ =&\frac{x^{\alpha-1}}{\Gamma_{q}(\alpha)} \xi^{1-\alpha}D^{\alpha}_{q}(\xi) \int_{0}^{x}(qt/x;q)_{\alpha-1} t^{\alpha-1} \,d_{q}t \\ =&\frac{{\Gamma_{q}(\alpha)} \xi^{1-\alpha} D^{\alpha}_{q}f(\xi)}{\Gamma_{q}(2\alpha)} x^{2\alpha-1} \end{aligned}
for $$0\leq\xi\leq x$$. Hence, (3.2) follows from (2.12). Similarly, using (2.13), we can prove (3.3). □

## 4 Generalized q-Taylor formula

In this section, we introduce generalized q-Taylor formulas for functions in terms of the sequential Riemann-Liouville q-derivative and the sequential Caputo fractional q-derivatives, where the sequential Riemann-Liouville q-derivative $${\mathcal{D}}_{q}^{n\alpha}$$ and Caputo fractional q-derivative $${}^{c}\mathcal{D}_{q}^{n\alpha}$$, $$n\in\mathbb{N}$$, are
$${\mathcal{D}}_{q}^{n\alpha}={D}_{q}^{\alpha} \cdots{D}_{q}^{\alpha } \quad \mbox{and}\quad {}^{c} \mathcal{D}_{q}^{n\alpha}={}^{c}D_{q}^{\alpha} \cdots {}^{c}D_{q}^{\alpha} \quad \mbox{(n times)},$$
respectively. The following lemma is important to get these formulas.

### Lemma 4.1

1. (1)
If $${\mathcal{D}}_{q}^{k\alpha}f\in\mathcal{L}_{q}^{1}[0,a]$$ and $$I_{q}^{1-\alpha}{\mathcal{D}}_{q}^{k\alpha}f\in\mathcal{AC}_{q}[0,a]$$, $$k=0,1,\ldots,n$$, then
$$I_{q}^{n\alpha}{\mathcal{D}}_{q}^{n\alpha}f(x)-I_{q}^{(n+1)\alpha }{ \mathcal{D}}_{q}^{(n+1)\alpha}f(x)=\frac{I_{q}^{1-\alpha}{\mathcal {D}}_{q}^{n\alpha}f(0)}{ \Gamma_{q}((n+1)\alpha)}x^{(n+1)\alpha-1}.$$
(4.1)

2. (2)
If $${}^{c}\mathcal{D}_{q}^{k\alpha}f \in\mathcal{AC}_{q}[0,a]$$, $$k=0,1,\ldots,n$$, then
$$I_{q}^{n\alpha}{}^{c} \mathcal{D}_{q}^{n\alpha}f(x)-I_{q}^{(n+1)\alpha }{}^{c} \mathcal{D}_{q}^{(n+1)\alpha }f(x)=\frac{{}^{c}\mathcal{D}_{q}^{n\alpha}f(0)}{ \Gamma_{q}(n\alpha+1)}x^{n\alpha}.$$
(4.2)

### Proof

We give a proof of (4.1), and the proof of (4.2) can be obtained similarly. Applying (2.12) and (2.8), we obtain
\begin{aligned} I_{q}^{n\alpha}{\mathcal{D}}_{q}^{n\alpha}f(x)-I_{q}^{(n+1)\alpha }{ \mathcal{D}}_{q}^{(n+1)\alpha}f(x) =&I_{q}^{n\alpha} \bigl({\mathcal{D}}_{q}^{n\alpha}f(x)-I_{q}^{\alpha }{D}_{q}^{\alpha} \bigl({\mathcal{D}}_{q}^{n\alpha}f(x)\bigr)\bigr) \\ =&I_{q}^{n\alpha} \biggl(\frac{I_{q}^{1-\alpha}{\mathcal{D}}_{q}^{n\alpha }f(0)}{\Gamma_{q}(\alpha)} x^{\alpha-1} \biggr) \\ =&\frac{I_{q}^{1-\alpha}{\mathcal{D}}_{q}^{n\alpha}f(0)}{\Gamma _{q}(\alpha)} I_{q}^{n\alpha}\bigl(x^{\alpha-1}\bigr) \\ =&\frac{I_{q}^{1-\alpha}{\mathcal{D}}_{q}^{n\alpha}f(0)}{\Gamma _{q}((n+1)\alpha)} x^{(n+1)\alpha-1}, \end{aligned}
and the lemma follows. □

### Theorem 4.2

(Generalized q-Taylor formulas)

1. (1)
Suppose that $${\mathcal{D}}_{q}^{k\alpha}f\in \mathcal{L}_{q}^{1}[0,a]$$, $$I_{q}^{1-\alpha}{\mathcal{D}}_{q}^{k\alpha}f\in \mathcal{AC}_{q}[0,a]$$, $$k=0,1,\ldots,n-1$$, and $$x^{1-\alpha}{\mathcal{D}}_{q}^{n\alpha}f\in C[0,a]$$. Then
$$f(x)=\sum_{k=1}^{n-1} \frac{I_{q}^{1-\alpha}{\mathcal{D}}_{q}^{k\alpha}f(0)}{ \Gamma_{q}((k+1)\alpha)}x^{(k+1)\alpha-1} +\frac{\Gamma_{q}(\alpha) \xi^{1-\alpha} {\mathcal {D}}_{q}^{n\alpha}f(\xi)}{\Gamma_{q}((n+1)\alpha)} x^{(n+1)\alpha-1}.$$
(4.3)

2. (2)
Suppose that $${}^{c}{\mathcal{D}}_{q}^{k\alpha}f \in\mathcal{AC}_{q}[0,a]$$, $$k=0,1,\ldots,n-1$$, and $${}^{c}D^{n\alpha}_{q}f\in C[0,a]$$. Thus,
$$f(x)=\sum_{k=0}^{n-1} \frac{{}^{c}\mathcal{D}_{q}^{k\alpha }f(0)}{\Gamma_{q}(k\alpha +1)}x^{k\alpha} +\frac{{}^{c}\mathcal{D}_{q}^{n\alpha}f(\xi)}{\Gamma_{q}(n\alpha +1)}x^{n\alpha},$$
(4.4)

where $$0\leq\xi\leq x$$.

### Proof

For (4.3), applying (4.1), we obtain
\begin{aligned}& \sum_{k=0}^{n-1} \bigl[I_{q}^{k\alpha}{ \mathcal{D}}_{q}^{k\alpha }f(x)-I_{q}^{(k+1)\alpha}{ \mathcal{D}}_{q}^{(k+1)\alpha}f(x) \bigr] \\& \quad =\sum _{k=0}^{n-1}\frac{I_{q}^{1-\alpha}{\mathcal{D}}_{q}^{k\alpha}f(0)}{ \Gamma_{q}((k+1)\alpha)}x^{(k+1)\alpha-1}, \end{aligned}
(4.5)
that is,
$$f(x)=\sum_{k=0}^{n-1} \frac{I_{q}^{1-\alpha}{\mathcal{D}}_{q}^{k\alpha}f(0)}{ \Gamma_{q}((k+1)\alpha)}x^{(k+1)\alpha-1}+I_{q}^{n\alpha}{\mathcal {D}}_{q}^{n\alpha}f(x).$$
(4.6)
Applying the q-integral mean value theorem and (2.8) yield
\begin{aligned} I_{q}^{n\alpha}{\mathcal{D}}_{q}^{n\alpha}f(x) =& \frac{x^{n\alpha -1}}{\Gamma_{q}(n\alpha)} \int_{0}^{x}(qt/x;q)_{n\alpha-1} t^{\alpha-1} t^{1-\alpha }{\mathcal{D}}_{q}^{n\alpha}f(t) \,d_{q}t \\ =&\frac{x^{n\alpha-1}}{\Gamma_{q}(n\alpha)} \xi^{1-\alpha }{\mathcal{D}}_{q}^{n\alpha}f( \xi) \int_{0}^{x}(qt/x;q)_{n\alpha-1} t^{\alpha-1} \,d_{q}t \\ =&\frac{\Gamma_{q}(\alpha) \xi^{1-\alpha} {\mathcal {D}}_{q}^{n\alpha}f(\xi)}{\Gamma_{q}((n+1)\alpha)} x^{(n+1)\alpha-1} \end{aligned}
(4.7)
for some $$\xi\in[0,x]$$. Combining (4.6) and (4.7) yields (4.3).

By using (4.2), (4.4) can be treated similarly. □

A natural question arises: can we expand a function f in terms of q-fractional derivatives? That is,
$$f(x)=x^{\alpha-1}\sum_{k=0}^{\infty}c_{k}x^{k\alpha} \quad \mbox{or}\quad f(x)=\sum_{k=0}^{\infty}c_{k}x^{k\alpha}?$$
The following theorem gives the answer for such expansions with sufficient conditions for the uniform convergence.

### Theorem 4.3

Assume that $$f\in\mathcal{L}_{q}^{1}[0,a]$$ and $$x^{1-\alpha}{\mathcal{D}}_{q}^{n\alpha}f\in C[0,a]$$ for all $$n\in\mathbb{N}$$. If
$$\bigl\vert x^{1-\alpha}{\mathcal{D}}_{q}^{n\alpha}f(x) \bigr\vert \le c A^{n\alpha}, \quad \forall x\in[0,a], n\in\mathbb{N},$$
where c is a positive constant, and A is a positive number satisfying $$A<\frac{1}{a(1-q)}$$, then f has the expansion
$$f(x)=\sum_{k=0}^{\infty } \frac{I_{q}^{1-\alpha}{\mathcal{D}}_{q}^{k\alpha}f(0)}{ \Gamma_{q}((k+1)\alpha)}x^{(k+1)\alpha-1}.$$
(4.8)
Moreover, the series $$\sum_{k=0}^{\infty}\frac{I_{q}^{1-\alpha}{\mathcal{D}}_{q}^{k\alpha}f(0)}{ \Gamma_{q}((k+1)\alpha)}x^{k\alpha}$$ converges uniformly to $$x^{1-\alpha}f(x)$$ on $$[0,a]$$.

### Proof

Using (4.3), we obtain
\begin{aligned}& \Biggl\vert x^{1-\alpha}f(x)-\sum_{k=0}^{n-1} \frac {I_{q}^{1-\alpha}{\mathcal{D}}_{q}^{k\alpha}f(0)}{ \Gamma_{q}((k+1)\alpha)}x^{k\alpha}\Biggr\vert \\& \quad \le c \Gamma_{q}( \alpha)\frac{ (a A)^{n\alpha}}{\Gamma _{q}((n+1)\alpha)} \\& \quad =\frac{c \Gamma_{q}(\alpha)(q^{(n+1)\alpha };q)_{\infty}}{(q;q)_{\infty}}\frac{ (a A)^{n\alpha }}{(1-q)^{1-(n+1)\alpha}} \\& \quad =\frac{c \Gamma_{q}(\alpha)(q^{(n+1)\alpha };q)_{\infty}}{(q;q)_{\infty}(1-q)^{1-\alpha}} \bigl(a A(1-q) \bigr)^{n\alpha} \longrightarrow0 \quad \mbox{as } n\rightarrow\infty. \end{aligned}
Thus, the result follows. □

### Theorem 4.4

Assume that $${}^{c}\mathcal{D}_{q}^{n\alpha}f\in C[0,a]$$ for $$n\in \mathbb{N}$$. If
$$\bigl\vert {}^{c}\mathcal{D}_{q}^{n\alpha}f(x)\bigr\vert \le c A^{n\alpha},\quad \forall x\in[0,a], n\in\mathbb{N},$$
where c is a positive constant, and A is a positive number satisfying $$A<\frac{1}{a(1-q)}$$, then f has the expansion
$$f(x)=\sum_{k=0}^{\infty} \frac{{}^{c}\mathcal{D}_{q}^{k\alpha }f(0)}{\Gamma _{q}(k\alpha+1)}x^{k\alpha},$$
(4.9)
and the series on the right-hand side of (4.9) converges uniformly to $$f(x)$$ on $$[0,a]$$.

### Proof

The proof is similar to the proof of Theorem 4.3 and is omitted. □

### Remark 4.5

1. (1)
If a function f has the expansion
$$f(x)=\sum_{k=0}^{\infty}a_{k} x^{(k+1)\alpha-1},$$
then we can deduce that
$$a_{k}=\frac{I_{q}^{1-\alpha}{\mathcal{D}}_{q}^{k\alpha}f(0)}{ \Gamma_{q}((k+1)\alpha)}.$$
Also, if a function f has the expansion
$$f(x)=\sum_{k=0}^{\infty}b_{k} x^{k\alpha},$$
then we can deduce that
$$b_{k}=\frac{{}^{c}\mathcal{D}_{q}^{k\alpha}f(0)}{\Gamma_{q}(k\alpha+1)}.$$

2. (2)
The results of this paper are valid if f is a function defined on intervals of the form $$[-a,a]$$ or $$[-a,0]$$, where $$a>0$$. In these two cases, $$\mathcal{L}_{q}^{1}[-a,b]$$, $$b=0$$ or a, is the space of all functions defined on $$[-a,b]$$ such that
$$\sum_{k=0}^{\infty}q^{k}(1-q)\bigl\vert f\bigl(xq^{k}\bigr)\bigr\vert < \infty \quad \mbox{for all } x\in [-a,b].$$
The space $$\mathcal{AC}_{q}[-a,b]$$ is the space of all q-regular at zero functions that satisfy condition (2.3) for all $$t\in[-a,b]$$.

## 5 Examples

In this section, we apply the generalized q-Taylor formula to solve fractional q-difference equations with constant coefficients. A solution to this type of equations is introduced in  by using q-Laplace transforms. In the following examples, λ is a real number. We assume that the conditions of Theorems 4.3 and 4.4 are satisfied.

### Example 5.1

Consider the q-initial value problem
$${}^{c}D_{q}^{\alpha}y(x)=\lambda y(x),\quad y(0)=y_{0}, x>0.$$
(5.1)
We assume that $$y\in C[0,a]$$ for some $$a>0$$ to be determined later. By (5.1), $${}^{c}\mathcal{D}_{q}^{n\alpha }y(x)=\lambda^{n} y(x)$$. Consequently,
$$\bigl\vert {}^{c}\mathcal{D}_{q}^{n\alpha}y(x)\bigr\vert \leq c \vert \lambda \vert ^{n},\quad c:=\max_{x\in [0,a]} \bigl\vert y(x)\bigr\vert .$$
Hence, if we assume that $$\vert \lambda a^{\alpha}(1-q)^{\alpha} \vert <1$$, then $$y(x)$$ can be written as
$$y(x)=\sum_{n=0}^{\infty}{}^{c} \mathcal{D}_{q}^{n\alpha}y(0)\frac {x^{n\alpha }}{\Gamma_{q}(n\alpha+1)} =y_{0} e_{\alpha,1}\bigl(\lambda x^{\alpha};q\bigr), \quad x\in[0,a],$$
(5.2)
where $$e_{\nu,\mu}(z;q)$$ is one of the q-Mittag-Leffler function defined by
$$e_{\nu,\mu}(z;q)=\sum_{k=0}^{\infty} \frac{z^{k}}{\Gamma_{q}(\nu k+\mu)},\quad \vert z\vert < (1-q)^{\nu}.$$

### Example 5.2

Consider the q-initial value problem
$${}^{c}\mathcal{D}_{q}^{2\alpha} y(x)=-y(x),\quad y(0)=0, \qquad {}^{c}D_{q}^{\alpha}y(0)=1.$$
(5.3)
We assume that $$y,{}^{c}D_{q}^{\alpha}y\in C[0,a]$$ for some $$a>0$$ to be determined later. From (5.3), we conclude that
$${}^{c}\mathcal{D}_{q}^{(2n+1)\alpha}y(x)=(-1)^{n} {}^{c}D_{q}^{\alpha }y(x),\qquad {}^{c} \mathcal{D}_{q} ^{(2n)\alpha}y(x)=(-1)^{n} y(x),\quad n\in \mathbb{N}.$$
Hence, if $$c=\max{ \{\max_{x\in[0,a]}\vert y(x)\vert , \max_{x\in[0,a]}\vert {}^{c}D_{q}^{\alpha}y(x)\vert \}}$$, then
$$\bigl\vert {}^{c}\mathcal{D}_{q}^{n\alpha}y(x)\bigr\vert \leq c,\quad \forall n\in\mathbb{N}.$$
Therefore, by Theorem 4.3, if a is chosen such that $$a<\frac{1}{(1-q)}$$, then
\begin{aligned} y(x) =&\sum_{n=0}^{\infty}{}^{c} \mathcal{D}_{q}^{n\alpha}y(0) \frac {x^{n\alpha }}{\Gamma_{q}(n\alpha+1)} \\ =&\sum_{n=0}^{\infty}(-1)^{n} \frac{x^{(2n+1)\alpha}}{\Gamma _{q}((2n+1)\alpha+1)}=x^{\alpha} e_{2\alpha,\alpha+1}\bigl(-x^{2\alpha};q \bigr). \end{aligned}
(5.4)
It is worth mentioning that if we set $$\alpha=1$$ in (5.4), then we get the Jackson q-sine function introduced in . Thus, we may consider the function in (5.4) as a fractional analogue of the Jackson q-sine function.

### Example 5.3

Consider the q-initial value problem
$${D}_{q}^{\alpha} y(x)=\lambda y(x),\qquad \bigl[x^{1-\alpha}y \bigr]\bigl(0^{+}\bigr)=\frac{y_{0}}{\Gamma_{q}(\alpha)}.$$
(5.5)
Hence, $${\mathcal{D}}_{q}^{n\alpha}y(x)=\lambda^{n} y(x)$$. We seek a solution y such that $$x^{1-\alpha} y(x)\in C[0,a]$$ for some a. Then
$$\bigl\vert x^{1-\alpha}{\mathcal{D}}_{q}^{n\alpha}y(x) \bigr\vert \leq c \vert \lambda \vert ^{n}, \quad c:=\max _{x\in[0,a]}\bigl\vert x^{1-\alpha}y(x)\bigr\vert .$$
We can show that
$$I_{q}^{1-\alpha}D_{q}^{\alpha}y(0)= \Gamma_{q}(\alpha)\bigl[x^{1-\alpha}y(x)\bigr]\bigl(0^{+}\bigr).$$
(5.6)
Consequently, $$I_{q}^{1-\alpha}{\mathcal{D}}_{q}^{k\alpha}y(0)=\lambda^{n}y_{0}$$. Therefore,
\begin{aligned} y(x) =&\sum_{k=0}^{\infty} \frac{I_{q}^{1-\alpha}{\mathcal {D}}_{q}^{k\alpha}y(0)}{ \Gamma_{q}((k+1)\alpha)}x^{(k+1)\alpha-1} \\ =&y_{0} x^{\alpha-1}\sum_{k=0}^{\infty} \frac{(\lambda x^{\alpha})^{k}}{\Gamma_{q}((k+1)\alpha)}=y_{0} x^{\alpha-1} e_{\alpha,\alpha}\bigl(\lambda x^{\alpha};q\bigr), \end{aligned}
where $$\vert \lambda a^{\alpha}(1-q)^{\alpha} \vert <1$$.

### Example 5.4

Consider the q-initial value problem
$$\mathcal{D}^{2\alpha}_{q} y(x)=-\lambda y(x),\qquad \bigl[x^{1-\alpha}y\bigr]\bigl(0^{+}\bigr)=\frac{y_{1}}{\Gamma_{q}(\alpha)},\qquad \bigl[x^{1-\alpha}{D}_{q}^{\alpha}y\bigr]\bigl(0^{+}\bigr)= \frac{y_{2}}{\Gamma_{q}(\alpha)}.$$
(5.7)
Thus,
$${\mathcal{D}}_{q}^{2n\alpha}y(x)=(-\lambda)^{n} y(x),\qquad { \mathcal {D}}_{q}^{(2n+1)\alpha}y(x)=(-\lambda)^{n} D^{\alpha}_{q}y(x).$$
For a solution y such that $$x^{1-\alpha} y(x), x^{1-\alpha} D^{\alpha}_{q}y(x)\in C[0,a]$$ for some a, we have
$$\bigl\vert x^{1-\alpha}{\mathcal{D}}^{n\alpha}_{q}y(x) \bigr\vert \leq c \vert \lambda \vert ^{n},\quad c:=\max\Bigl\{ \max _{x\in[0,a]}\bigl\vert x^{1-\alpha}y(x)\bigr\vert , \max _{x\in [0,a]}\bigl\vert x^{1-\alpha} D^{\alpha}_{q}y(x) \bigr\vert \Bigr\} .$$
Also,
$$I_{q}^{1-\alpha}{\mathcal{D}}_{q}^{2n\alpha}y(0)=(- \lambda)^{n} y_{1},\qquad I_{q}^{1-\alpha}{ \mathcal{D}}_{q}^{(2n+1)\alpha}y(0)=(-\lambda)^{n}y_{2}.$$
Consequently,
\begin{aligned} y(x) =&x^{\alpha-1}\sum _{k=0}^{\infty}\frac{I_{q}^{1-\alpha}{\mathcal {D}}_{q}^{2k\alpha}y(0)}{ \Gamma_{q}((2k+)\alpha)}x^{2k\alpha}+x^{\alpha-1} \sum_{k=0}^{\infty }\frac{I_{q}^{1-\alpha}{\mathcal{D}}_{q}^{(2k+1)\alpha}y(0)}{ \Gamma_{q}((2k+2)\alpha)}x^{(2k+1)\alpha} \\ =&y_{1} x^{\alpha-1}\sum_{k=0}^{\infty} \frac{(-\lambda)^{k} x^{2k\alpha}}{\Gamma_{q}((2k+1)\alpha)} + y_{2} x^{\alpha-1}\sum _{k=0}^{\infty}\frac{(-\lambda)^{k} x^{(2k+1)\alpha}}{\Gamma_{q}((2k+2)\alpha)} \\ =&y_{1} x^{\alpha-1}e_{2\alpha,\alpha}\bigl(-\lambda x^{2\alpha};q\bigr)+ y_{2} x^{2\alpha-1} e_{2\alpha,2\alpha} \bigl(-\lambda x^{2\alpha};q\bigr), \end{aligned}
where $$\vert \lambda a^{\alpha}(1-q)^{\alpha} \vert <1$$.

### Example 5.5

Consider the initial value problem
$$D_{q}^{\alpha}y(x)=\lambda q^{\alpha(1-\alpha)} y \bigl(q^{\alpha}x\bigr),\qquad \bigl[x^{1-\alpha}y\bigr]\bigl(0^{+}\bigr)= \frac{1}{\Gamma_{q}(\alpha)}.$$
(5.8)
Applying
$$D_{q,x}^{\alpha}y(x\beta)=\beta\bigl(D_{q}^{\alpha}y \bigr) (x\beta),$$
(5.9)
on (5.8) $$n-1$$ times, we obtain
$${\mathcal{D}}_{q}^{n\alpha}y(x)= \bigl(\lambda q^{\alpha(1-\alpha)} \bigr)^{n} q^{\frac{n(n-1)\alpha}{2}}y\bigl(xq^{n\alpha}\bigr).$$
(5.10)
For a solution y such that $$x^{1-\alpha} y(x)\in C[0,a]$$, we have
$$\bigl\vert x^{1-\alpha}{\mathcal{D}}^{n\alpha}_{q}y(x) \bigr\vert \leq c \vert \lambda \vert ^{n}q^{\frac{n(n-1)\alpha}{2}},\quad c:=\max _{x\in[0,a]}\bigl\vert x^{1-\alpha}y(x)\bigr\vert ,$$
and
$$I_{q}^{1-\alpha}{\mathcal{D}}_{q}^{n\alpha}y(0)= \lambda^{n}q^{\frac {n(n-1)\alpha}{2}}.$$
Therefore,
\begin{aligned} y(x) =&\sum_{k=0}^{\infty} \frac{I_{q}^{1-\alpha}{\mathcal {D}}_{q}^{k\alpha}y(0)}{ \Gamma_{q}((k+1)\alpha)}x^{(k+1)\alpha-1} \\ =&x^{\alpha-1}\sum_{k=0}^{\infty}q^{\frac{n(n-1)\alpha}{2}} \frac{ (\lambda x^{\alpha})^{k}}{\Gamma_{q}((k+1)\alpha)} \\ =&x^{\alpha-1}E_{\alpha,\alpha}\bigl(\lambda x^{\alpha};q\bigr), \end{aligned}
where, in general, $$E_{\alpha,\beta}(z;q)$$ is a second q-analogue of Mittag-Leffler function defined by
$$E_{\alpha,\beta}(z)=\sum_{n=0}^{\infty}q^{\frac{n(n-1)\alpha }{2}} \frac{z^{n}}{\Gamma_{q}(n\alpha+\beta)},\quad z\in\mathbb{R}.$$
Hence, a can be taken to be any positive value in this example. For some derived properties for these q-analogues of Mittag-Leffler functions, see  and the references therein.

## Declarations

### Acknowledgements

The author is grateful to the referees for their valuable comments and suggestions, which have improved the manuscript in its present form.

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

## Authors’ Affiliations

(1)
Department of Mathematics, Faculty of Science, Cairo University, Giza, Egypt
(2)
Department of Mathematics, Faculty of Basic Education, PAAET, Ardiyah, Kuwait

## References

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