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

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.

1 Introduction

The quantum calculus has many applications in the fields of special functions and many other areas (see [1–7]). 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 [3–5] 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.