Open Access

A certain \((p,q)\)-derivative operator and associated divided differences

  • Serkan Araci1Email author,
  • Uğur Duran2,
  • Mehmet Acikgoz2 and
  • Hari M Srivastava3, 4
Journal of Inequalities and Applications20162016:301

https://doi.org/10.1186/s13660-016-1240-8

Received: 18 August 2016

Accepted: 10 November 2016

Published: 23 November 2016

Abstract

Recently, Sofonea (Gen. Math. 16:47-54, 2008) considered some relations in the context of quantum calculus associated with the q-derivative operator \(D_{q}\) and divided difference. As applications of the post-quantum calculus known as the \((p,q)\)-calculus, we derive several relations involving the \((p,q)\)-derivative operator and divided differences.

Keywords

q-calculus \((p,q)\)-calculus divided differences \((p,q)\)-derivative operator \((p,q)\)-Leibniz rule principle of mathematical induction

MSC

11B68 11B83 81S40

1 Introduction

The quantum calculus has many applications in the fields of special functions and many other areas (see [17]). Further there is possibility of extension of the q-calculus to post-quantum calculus denoted by the \((p,q)\)-calculus. Actually such an extension of quantum calculus cannot be obtained directly by substitution of q by \(q/p\) in q-calculus. When the case \(p=1\) in \((p,q)\)-calculus, the q-calculus may be obtained (see [6, 7]). Recently, Chakrabarti and Jagannathan [8] introduced a consideration of the \(( p,q ) \)-integer in order to generalize or unify several forms of q-oscillator algebras well known in the physics literature related to the representation theory of single-paramater quantum algebras (see also [35] and [9]). They also considered the necessary elements of the \(( p,q ) \)-calculus involving \(( p,q ) \)-exponential, \(( p,q ) \)-integration and the \(( p,q ) \)-differentiation. Corcino [10] developed the theory of a \(( p,q ) \)-extension of the binomial coefficients and also established some properties parallel to those of the ordinary and q-binomial coefficients, which is comprised horizontal generating function, the triangular, vertical, and the horizontal recurrence relations and the inverse and the orthogonality relations. Sadjang [11] investigated some properties of the \(( p,q ) \)-derivatives and the \(( p,q ) \)-integrations. Sadjang [11] also provided two suitable polynomial bases for the \(( p,q ) \)-derivative and gave various properties of these bases.

The \(( p,q )\)-number is given by
$$ [ n ] _{p,q}=\frac{p^{n}-q^{n}}{p-q} \quad( p\neq q ), $$
which is a natural generalization of the q-number: that is, we have (cf. [10] and [11])
$$ \lim_{p\rightarrow1} [ n ] _{p,q}:= [ n ] _{q}. $$
It is clear that the notation \([ n ] _{p,q}\) is symmetric, that is,
$$ [ n ] _{p,q}= [ n ] _{q,p}. $$
The \((p,q)\)-Gauss binomial coefficients given by
$$ \begin{bmatrix} n\\ k \end{bmatrix} _{p,q}=\frac{ [ n ] _{p,q}!}{ [ n-k ] _{p,q}! [ k ] _{p,q}!} \quad( n\geqq k ) $$
and the \((p,q)\)-factorial given by
$$ [ n ] _{p,q}!= [ n ] _{p,q} [ n-1 ] _{p,q}\cdots [ 2 ] _{p,q} [ 1 ] _{p,q}\quad ( n\in \mathbb{N} ) $$
are also known from [10] and [11]. Further, the \((p,q)\)-analogs of Pascal’s identity are given by
$$\begin{aligned} \begin{bmatrix} n+1\\ k \end{bmatrix} _{p,q} &=p^{k} \begin{bmatrix} n\\ k \end{bmatrix} _{p,q}+q^{n-k} \begin{bmatrix} n\\ k-1 \end{bmatrix} _{p,q} \\ &=q^{k} \begin{bmatrix} n\\ k \end{bmatrix} _{p,q}+p^{n-k} \begin{bmatrix} n\\ k-1 \end{bmatrix} _{p,q}, \end{aligned}$$
where \(k\in \{ 0,1,2,\ldots,n \}\) (cf. [10] and [11]).

Let p and q be elements of complex numbers and \(D=D_{p,q}\subset \mathbb{C}\) such that \(x\in D\) implies \(px\in D\) and \(qx\in D\). Here, in this investigation, we give the following two definitions which involve a post-quantum generalization of Sofonea’s work [1].

Definition 1

Let \(0<\vert q\vert <\vert p\vert \leqq1\). A given function \(f:D_{p,q}\rightarrow\mathbb{C}\) is called \((p,q)\)-differentiable under the restriction that, if \(0\in D_{p,q}\), then \(f^{\prime} ( 0 )\) exists.

Definition 2

Let \(0<\vert q\vert <\vert p\vert \leqq1\). A given function \(f:D_{p,q}\rightarrow\mathbb{C}\) is called \((p,q)\)-differentiable of order n, if and only if \(0\in D_{p,q}\) implies that \(f^{ ( n ) } ( 0 ) \) exists.

The \((p,q)\)-derivative operator of a function f is defined by
$$ D_{p,q}f ( x ) =\frac{f ( px ) -f ( qx ) }{ ( p-q ) x} \quad( x\neq0 ) $$
(1.1)
and
$$ ( D_{p,q}f ) ( 0 ) =f^{\prime} ( 0 ), $$
provided that the function f is differentiable at 0. We note that
$$ D_{p,q}=D_{q,p}. $$
Furthermore,
$$ ( D_{p,q}fg ) ( x ) =g ( px ) ( D_{p,q}f ) ( x ) +f ( qx ) ( D_{p,q}g ) ( x ) $$
(1.2)
and
$$ \biggl( D_{p,q}\frac{f}{g} \biggr) ( x ) =\frac{g ( px ) ( D_{p,q}f ) ( x ) -f ( px ) ( D_{p,q}g ) ( x ) }{g ( px ) g ( qx ) }\quad \bigl( g ( px ) g ( qx ) \neq0 \bigr) $$
(1.3)
hold true for the linear operator \(D_{p,q}\) (cf. [11]).
The divided differences at a system of distinct points \(x_{0},x_{1},\ldots ,x_{n}\) are denoted by \([ x_{0},x_{1},\ldots,x_{n};f ]\). In fact, we have (see [1] and [2])
$$ [ x_{0},x_{1},\ldots,x_{n};f ] =\sum _{k=0}^{n}\frac{f ( x_{k} ) }{\mathop{\prod^{n}_{(i\neq k)}}\limits_{\phantom{aa}i=0} ( x_{k}-x_{i} ) }. $$
(1.4)

In the next part of the paper, we obtain some potentially useful results and relations between the \((p,q)\)-derivative operator and divided differences. The results presented here provide a good generalization of the above-mentioned Sofonea results.

2 Main results

Let us consider the points
$$ x_{k}=p^{k}q^{n-k}x\quad ( k=0,1,\ldots,n ) $$
as follows:
$$ x_{0}=q^{n}x, \qquad x_{1}=q^{n-1}px,\qquad\ldots,\qquad x_{n-1}=qp^{n-1}x,\qquad x_{n}=p^{n}x. $$

We now state the following theorem.

Theorem 1

Let p and q be complex numbers with
$$ 0< \vert q\vert < \vert p\vert \leqq1\quad \textit{and}\quad f:D_{p,q}\rightarrow\mathbb{C}. $$
Then, by taking the knots \(x_{k}=p^{k}q^{n-k}x\),
$$\begin{aligned} & \bigl[ q^{n}x,q^{n-1}px,\ldots,qp^{n-1}x,p^{n}x;f \bigr] \\ &\quad =\frac{1}{q^{\binom{n}{2}} [ n ] _{{p,q}}!x^{n} ( p-q ) ^{n}}\sum_{k=0}^{n} ( -1 ) ^{n-k} \begin{bmatrix} n\\ k \end{bmatrix} _{p,q}p^{\frac{-k ( 2n-k-1 ) }{2}}q^{\binom{k}{2}}f \bigl( xp^{k}q^{n-k} \bigr) . \end{aligned}$$
(2.1)

Proof

For \(0\leqq l< k\), we have
$$ x_{k}-x_{l}=xp^{l}q^{n-k} ( p-q ) [ k-l ] _{p,q} $$
and, for \(k< l\leqq n\), we find that
$$ x_{k}-x_{l}=xp^{k}q^{n-l} ( q-p ) [ l-k ] _{p,q}. $$
Since
$$\begin{aligned} \mathop{\prod_{l=0}}\limits_{l\neq k}^{n} ( x_{k}-x_{l} ) & =\prod_{l=0}^{k-1} ( x_{k}-x_{l} ) \prod_{l=k+1}^{n} ( x_{k}-x_{l} ) \\ & =x^{n}p^{ ( n-k ) k} ( -1 ) ^{n-k} ( p-q ) ^{n}q^{k ( n-k ) +\binom{n-k}{2}} [ k ] _{{p,q}}!p^{k ( n-k ) +\binom {k}{2}} [ n-k ] _{{p,q}}! \\ & = ( -1 ) ^{n-k} ( p-q ) ^{n}x^{n}p^{k ( 2n-k-1 ) /2}q^{\binom{n}{2}-\binom{k}{2}} [ k ] _{{p,q}}![ n-k ] _{{p,q}}!, \end{aligned}$$
we have the following consequence from (1.4):
$$ [ x_{0},x_{1},\ldots,x_{n};f ] = \frac{q^{-\binom{n}{2}}}{ [ n] _{{p,q}}!x^{n} ( p-q ) ^{n}}\sum_{k=0}^{n} ( -1 ) ^{n-k} \begin{bmatrix} n\\ k \end{bmatrix} _{p,q} p^{-k ( 2n-k-1 ) /2}q^{\binom{k}{2}}f \bigl( xp^{k}q^{n-k} \bigr) . $$
Therefore, the proof of Theorem 1 is completed. □
By using the following expressions:
$$ D_{p,q}^{0}=I, \qquad D_{p,q}^{1}=D_{p,q}\quad \mbox{and}\quad D_{p,q}^{k}=D_{p,q}D_{p,q}^{k-1}, $$
we now give a representation of the operator \(D_{p,q}^{n}\) as in Theorem 2 below.

Theorem 2

Let the function \(f:D_{p,q}\rightarrow\mathbb{C}\) be \((p,q) \)-differentiable of order n. Then
$$ \bigl( D_{p,q}^{n}f \bigr) ( x ) =\frac{q^{-\binom{n}{2}}}{ x^{n} ( p-q ) ^{n}}\sum _{k=0}^{n} ( -1 ) ^{n-k} \begin{bmatrix} n\\ k \end{bmatrix} _{p,q} \frac{q^{\binom{k}{2}}f ( xp^{k}q^{n-k} ) }{p^{k ( 2n-k-1 ) /2}}. $$
(2.2)

Proof

Theorem 2 is proved by making use of the following results:
$$ ( D_{p,q}f ) ( x ) =\frac{f ( qx ) -f ( px ) }{ ( q-p ) x}=\frac{f ( qx ) }{qx-px}+ \frac{f ( px ) }{px-qx}= [ 1 ] _{p,q}! [ qx,px;f ] $$
and
$$\begin{aligned} & \bigl( D_{p,q}^{2}f \bigr) ( x ) \\ &\quad =\frac{ ( D_{p,q}f ) ( qx ) - ( D_{p,q}f ) ( px ) }{ ( q-p ) x} \\ &\quad =\frac{\frac{f ( q^{2}x ) -f ( pqx ) }{ ( q-p ) qx}-\frac{f ( pqx ) -f ( p^{2}x ) }{ ( q-p ) px}}{ ( p-q ) x} \\ &\quad = ( p+q ) \biggl[ \frac{f ( q^{2}x ) }{ ( q^{2}-p^{2} ) ( q-p ) x^{2}q}-\frac{f ( pqx ) }{( q-p ) ^{2}x^{2}pq}+\frac{f ( p^{2}x ) }{ ( q^{2}-p^{2} ) ( q-p ) x^{2}p} \biggr] \\ & \quad= [ 2 ] _{p,q}! \bigl[ q^{2}x,pqx,p^{2}x;f \bigr] . \end{aligned}$$
Continuing this process, we deduce
$$ \bigl( D_{p,q}^{n}f \bigr) ( x ) = [ n ] _{p,q}! \bigl[ q^{n}x,q^{n-1}px,\ldots,qp^{n-1}x,p^{n}x;f \bigr] $$
(2.3)
by using the following formula:
$$ [ x_{0},x_{1},\ldots,x_{n};\cdot] = \frac{ [ x_{1},x_{2},\ldots,x_{n};\cdot ] - [ x_{0},x_{1},\ldots ,x_{n-1};\cdot ] }{x_{n}-x_{0}}. $$
It follows from Theorem 1 that
$$ \bigl( D_{p,q}^{n}f \bigr) ( x ) =q^{-\binom{n}{2}}x^{-n} ( p-q ) ^{-n}\sum_{k=0}^{n} ( -1 ) ^{n-k} \begin{bmatrix} n\\ k \end{bmatrix} _{p,q} p^{-k ( 2n-k-1 ) /2}q^{\binom{k}{2}}f \bigl( xp^{k}q^{n-k} \bigr) , $$
which completes the proof of Theorem 2. □
In the case when
$$ f ( x ) =x^{n} $$
in Theorem 2, we get the following corollary.

Corollary 1

The following result holds true:
$$ ( p-q ) ^{n}=\sum_{k=0}^{n} \begin{bmatrix} n\\ k \end{bmatrix} _{p,q}p^{\binom{k+1}{2}}q^{\binom{n-k+1}{2}}\frac{ ( -1 ) ^{n-k}}{ [ n ] _{p,q}!}. $$

We now consider the \(( p,q ) \)-analog of the Leibniz rule to represent it by means of the divided differences. First of all, we need to get the \(( p,q ) \)-analog of the Leibniz rule by the following lemma.

Lemma

Let the functions \(f:D_{p,q}\rightarrow \mathbb{C}\) and \(g:D_{p,q}\rightarrow\mathbb{C}\) be \((p,q)\)-differentiable of order n. Then
$$ D_{p,q}^{n} ( fg ) ( x ) =\sum_{k=0}^{n} \begin{bmatrix} n\\ k \end{bmatrix} _{p,q} D_{p,q}^{k} ( f ) \bigl( xp^{n-k} \bigr) D_{p,q}^{n-k} ( g ) \bigl( xq^{k} \bigr). $$

Proof

The lemma can easily be proved by applying the principle of mathematical induction. We, therefore, omit the proof of the lemma. □

We now state the \(( p,q )\)-Leibniz rule by using divided differences as follows.

Theorem 3

Let the functions \(f:D_{p,q}\rightarrow\mathbb{C}\) and \(g:D_{p,q}\rightarrow\mathbb{C}\) be \((p,q)\)-differentiable of order n. Then \((fg ) (x)\) is also \((p,q)\)-differentiable of order n and
$$\begin{aligned} D_{p,q}^{n} ( fg ) ( x ) ={}& [ n ] _{p,q}!\sum _{k=0}^{n} \bigl[ q^{n}x,q^{n-1}px, \ldots,q^{n-k+1}p^{k-1}x,q^{n-k}p^{k}x;f \bigr]\\ &{} \cdot\bigl[ q^{n-k}p^{k}x,q^{n-k-1}p^{k+1}x, \ldots,qp^{n-1}x,p^{n}x;g \bigr] . \end{aligned}$$

Proof

Our assertion in Theorem 3 follows from equation (2.3) and the above lemma. The details involved are being omitted here. □

Now also we give a function at a point \(p^{n}x\) by binomial expression and \((p,q)\)-derivative of order k.

Theorem 4

Let the function \(f:D_{p,q}\rightarrow\mathbb{C}\) be \((p,q) \)-differentiable of order n. Then
$$ f \bigl( p^{n}x \bigr) =\sum_{k=0}^{n} \begin{bmatrix} n\\ k \end{bmatrix} _{p,q}x^{k}p^{\binom{k}{2}} ( p-q ) ^{k}D_{p,q}^{k} \bigl( f ( x ) \bigr) . $$

Proof

We consider Newton’s formula as follows:
$$\begin{aligned} f ( z ) ={}&\sum_{k=0}^{n-1} ( z-x_{0} ) ( z-x_{1} ) \cdots( z-x_{k-1} ) [ x_{0},x_{1},\ldots,x_{k};f ] \\ &{} + ( z-x_{0} ) ( z-x_{1} ) \cdots( z-x_{n-1} ) [ x_{0},x_{1},\ldots,x_{n-1},z;f ] . \end{aligned}$$
(2.4)
Upon setting
$$ x_{k}=p^{k}q^{n-k}x\quad (k=0,1,\ldots,n-1) $$
in equation (2.4) and \(z=p^{n}x\), if we use equation (2.1), we find that
$$\begin{aligned} f \bigl( p^{n}x \bigr) ={}&\sum_{k=0}^{n-1} \bigl( p^{n}x-q^{n}x \bigr) \bigl( p^{n}x-q^{n-1}px \bigr) \cdots\bigl( p^{n}x-q^{n-k+1}p^{k-1}x \bigr)\\ &{}\cdot \bigl[ q^{n}x,q^{n-1}px,\ldots,q^{n-k}p^{k}x;f \bigr] \\ &{} + \bigl( p^{n}x-q^{n}x \bigr) \bigl( p^{n}x-q^{n-1}px \bigr) \cdots\bigl( p^{n}x-qp^{n-1}x \bigr) \\ &{}\cdot\bigl[ q^{n}x,q^{n-1}px,\ldots,qp^{n-1}x,p^{n}x;f \bigr] \\ ={}&\sum_{k=0}^{n-1} \bigl( p^{n}x-q^{n}x \bigr) \bigl( p^{n}x-q^{n-1}px \bigr) \cdots\bigl( p^{n}x-q^{n-k+1}p^{k-1}x \bigr) \frac{( D_{p,q}^{k}f ) ( x ) }{ [ k ] _{p,q}!} \\ &{} + \bigl( p^{n}x-q^{n}x \bigr) \bigl( p^{n}x-q^{n-1}px \bigr) \cdots\bigl( p^{n}x-qp^{n-1}x \bigr) \frac{ ( D_{p,q}^{n}f ) ( x ) }{ [ n ] _{p,q}!} \\ ={}&\sum_{k=0}^{n} \bigl( p^{n}x-q^{n}x \bigr) \bigl( p^{n}x-q^{n-1}px \bigr) \cdots\bigl( p^{n}x-q^{n-k+1}p^{k-1}x \bigr) \frac{ ( D_{p,q}^{k}f ) ( x ) }{ [ k ] _{p,q}!} \\ ={}&\sum_{k=0}^{n}x^{k}p^{\binom{k}{2}} \frac{\frac{ ( p^{n}-q^{n} ) ( p^{n-1}-q^{n-1} ) \cdots ( p-q ) }{( p-q ) ^{n}}}{\frac{ ( p^{n-k}-q^{n-k} ) ( p^{n-k-1}-q^{n-k-1} ) \cdots ( p-q ) }{ ( p-q ) ^{n-k} ( p-q ) ^{k}}}\frac{ ( D_{p,q}^{k}f ) ( x ) }{ [ k ] _{p,q}!} \\ ={}&\sum_{k=0}^{n}x^{k}p^{\binom{k}{2}} ( p-q ) ^{k}\frac{ [ n] _{p,q}!}{ [ n-k ] _{p,q}! [ k ] _{p,q}!} \bigl( D_{p,q}^{k}f \bigr) ( x ) , \end{aligned}$$
as asserted by Theorem 4. □

Finally, we are in a position to give the following result.

Corollary 2

Let p and q be complex numbers such that
$$ 0< \vert q\vert < \vert p\vert \leqq1. $$
Also let the function \(f:D_{p,q}\rightarrow\mathbb{C}\) be \((p,q)\)-differentiable of order n. Then
$$ f ( x ) =\sum_{k=0}^{n} \begin{bmatrix} n\\ k \end{bmatrix} _{p,q}q^{k ( k-n ) }p^{\binom{k+1}{2}} ( qx-px ) ^{k} \bigl( D_{p,q}^{k}f \bigr) \biggl( \frac{xp^{n-k}}{q^{k}} \biggr) . $$

Proof

Since, for \(k\in \{ 0,1,\ldots,n \} \),
$$ \begin{bmatrix} n\\ k \end{bmatrix} _{\frac{1}{p},\frac{1}{q}}=\frac{ [ n ] _{\frac{1}{p},\frac{1}{q}}!}{ [ n-k ] _{\frac{1}{p},\frac{1}{q}}![ k ] _{\frac{1}{p},\frac{1}{q}}!}=\frac{ ( pq ) ^{-\binom{n}{2}}}{ ( pq ) ^{-\binom{n-k}{2}} ( pq ) ^{-\binom{k}{2}}} \begin{bmatrix} n\\ k \end{bmatrix} _{p,q}, $$
we have
$$ ( D_{\frac{1}{p},\frac{1}{q}}f ) ( x ) =\frac{f ( \frac{x}{q} ) -f ( \frac{x}{p} ) }{ ( p-q ) x} ( pq ) =pq ( D_{p,q}f ) \biggl( \frac{x}{pq} \biggr) $$
and
$$\begin{aligned} \bigl( D_{\frac{1}{p},\frac{1}{q}}^{2}f \bigr) ( x ) & =\frac{pq ( D_{p,q}f ) ( \frac{x}{pq} ) -pq ( D_{p,q}f ) ( \frac{x}{pq} ) }{ ( \frac{1}{p}-\frac{1}{q}) x} \\ & =\frac{ ( pq ) ^{2} [ ( D_{p,q}f ) ( \frac{x}{pq^{2}} ) - ( D_{p,q}f ) ( \frac{x}{pq} ) ] }{( p-q ) x} \\ & =p^{2}q^{2} \bigl( D_{p,q}^{2}f \bigr) \biggl( \frac{x}{p^{2}q^{2}} \biggr) . \end{aligned}$$
Continuing the process, we readily observe that
$$ \bigl( D_{\frac{1}{p},\frac{1}{q}}^{n}f \bigr) ( x ) =p^{n}q^{n} \bigl( D_{p,q}^{n}f \bigr) \biggl( \frac{x}{p^{n}q^{n}} \biggr) . $$
(2.5)
From Theorem 4, we thus conclude that
$$ f ( x ) =\sum_{k=0}^{n} \begin{bmatrix} n\\ k \end{bmatrix} _{p,q}q^{k ( k-n ) }p^{\binom{k+1}{2}} ( qx-px ) ^{k} \bigl( D_{p,q}^{k}f \bigr) \biggl( \frac{xp^{n-k}}{q^{k}} \biggr) , $$
which evidently proves Corollary 2. □

3 Conclusion

We have considered \((p,q)\)-analogs of several results investigated recently by Sofonea [1]. We have also given the \((p,q)\)-Leibniz rule and stated the \((p,q)\)-Leibniz rule by means of divided differences. Moreover, we have shown that a function f at a point \(q^{n}x\) can be generated by a linear combination of the \((p,q)\)-derivatives of order k. In the case when \(p=1\), the results derived in this paper would correspond to those based upon the relatively more familiar q-numbers.

Declarations

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 Economics, Faculty of Economics, Administrative and Social Science, Hasan Kalyoncu University
(2)
Department of Mathematics, Faculty of Arts and Science, University of Gaziantep
(3)
Department of Mathematics and Statistics, University of Victoria
(4)
China Medical University

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© Araci et al. 2016