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 which is a natural generalization of the q-number: that is, we have (cf. [10] and [11]) It is clear that the notation \([ n ] _{p,q}\) is symmetric, that is,
$$ [ n ] _{p,q}=\frac{p^{n}-q^{n}}{p-q} \quad( p\neq q ), $$
$$ \lim_{p\rightarrow1} [ n ] _{p,q}:= [ n ] _{q}. $$
$$ [ n ] _{p,q}= [ n ] _{q,p}. $$
Anzeige
The \((p,q)\)-Gauss binomial coefficients given by and the \((p,q)\)-factorial given by are also known from [10] and [11]. Further, the \((p,q)\)-analogs of Pascal’s identity are given by where \(k\in \{ 0,1,2,\ldots,n \}\) (cf. [10] and [11]).
$$ \begin{bmatrix} n\\ k \end{bmatrix} _{p,q}=\frac{ [ n ] _{p,q}!}{ [ n-k ] _{p,q}! [ k ] _{p,q}!} \quad( n\geqq k ) $$
$$ [ n ] _{p,q}!= [ n ] _{p,q} [ n-1 ] _{p,q}\cdots [ 2 ] _{p,q} [ 1 ] _{p,q}\quad ( n\in \mathbb{N} ) $$
$$\begin{aligned} \begin{bmatrix} n+1\\ k \end{bmatrix} _{p,q} &=p^{k} \begin{bmatrix} n\\ k \end{bmatrix} _{p,q}+q^{n-k} \begin{bmatrix} n\\ k-1 \end{bmatrix} _{p,q} \\ &=q^{k} \begin{bmatrix} n\\ k \end{bmatrix} _{p,q}+p^{n-k} \begin{bmatrix} n\\ k-1 \end{bmatrix} _{p,q}, \end{aligned}$$
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.
Anzeige
The \((p,q)\)-derivative operator of a function f is defined by and provided that the function f is differentiable at 0. We note that Furthermore, and hold true for the linear operator \(D_{p,q}\) (cf. [11]).
$$ D_{p,q}f ( x ) =\frac{f ( px ) -f ( qx ) }{ ( p-q ) x} \quad( x\neq0 ) $$
(1.1)
$$ ( D_{p,q}f ) ( 0 ) =f^{\prime} ( 0 ), $$
$$ D_{p,q}=D_{q,p}. $$
$$ ( D_{p,q}fg ) ( x ) =g ( px ) ( D_{p,q}f ) ( x ) +f ( qx ) ( D_{p,q}g ) ( x ) $$
(1.2)
$$ \biggl( D_{p,q}\frac{f}{g} \biggr) ( x ) =\frac{g ( px ) ( D_{p,q}f ) ( x ) -f ( px ) ( D_{p,q}g ) ( x ) }{g ( px ) g ( qx ) }\quad \bigl( g ( px ) g ( qx ) \neq0 \bigr) $$
(1.3)
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 as follows:
$$ x_{k}=p^{k}q^{n-k}x\quad ( k=0,1,\ldots,n ) $$
$$ x_{0}=q^{n}x, \qquad x_{1}=q^{n-1}px,\qquad\ldots,\qquad x_{n-1}=qp^{n-1}x,\qquad x_{n}=p^{n}x. $$
We now state the following theorem.
Theorem 1
Let
p
and
q
be complex numbers with
Then, by taking the knots
\(x_{k}=p^{k}q^{n-k}x\),
$$ 0< \vert q\vert < \vert p\vert \leqq1\quad \textit{and}\quad f:D_{p,q}\rightarrow\mathbb{C}. $$
$$\begin{aligned} & \bigl[ q^{n}x,q^{n-1}px,\ldots,qp^{n-1}x,p^{n}x;f \bigr] \\ &\quad =\frac{1}{q^{\binom{n}{2}} [ n ] _{{p,q}}!x^{n} ( p-q ) ^{n}}\sum_{k=0}^{n} ( -1 ) ^{n-k} \begin{bmatrix} n\\ k \end{bmatrix} _{p,q}p^{\frac{-k ( 2n-k-1 ) }{2}}q^{\binom{k}{2}}f \bigl( xp^{k}q^{n-k} \bigr) . \end{aligned}$$
(2.1)
Proof
For \(0\leqq l< k\), we have and, for \(k< l\leqq n\), we find that
$$ x_{k}-x_{l}=xp^{l}q^{n-k} ( p-q ) [ k-l ] _{p,q} $$
$$ x_{k}-x_{l}=xp^{k}q^{n-l} ( q-p ) [ l-k ] _{p,q}. $$
Since we have the following consequence from (1.4): Therefore, the proof of Theorem 1 is completed. □
$$\begin{aligned} \mathop{\prod_{l=0}}\limits_{l\neq k}^{n} ( x_{k}-x_{l} ) & =\prod_{l=0}^{k-1} ( x_{k}-x_{l} ) \prod_{l=k+1}^{n} ( x_{k}-x_{l} ) \\ & =x^{n}p^{ ( n-k ) k} ( -1 ) ^{n-k} ( p-q ) ^{n}q^{k ( n-k ) +\binom{n-k}{2}} [ k ] _{{p,q}}!p^{k ( n-k ) +\binom {k}{2}} [ n-k ] _{{p,q}}! \\ & = ( -1 ) ^{n-k} ( p-q ) ^{n}x^{n}p^{k ( 2n-k-1 ) /2}q^{\binom{n}{2}-\binom{k}{2}} [ k ] _{{p,q}}![ n-k ] _{{p,q}}!, \end{aligned}$$
$$ [ x_{0},x_{1},\ldots,x_{n};f ] = \frac{q^{-\binom{n}{2}}}{ [ n] _{{p,q}}!x^{n} ( p-q ) ^{n}}\sum_{k=0}^{n} ( -1 ) ^{n-k} \begin{bmatrix} n\\ k \end{bmatrix} _{p,q} p^{-k ( 2n-k-1 ) /2}q^{\binom{k}{2}}f \bigl( xp^{k}q^{n-k} \bigr) . $$
By using the following expressions: we now give a representation of the operator \(D_{p,q}^{n}\) as in Theorem 2 below.
$$ D_{p,q}^{0}=I, \qquad D_{p,q}^{1}=D_{p,q}\quad \mbox{and}\quad D_{p,q}^{k}=D_{p,q}D_{p,q}^{k-1}, $$
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: and Continuing this process, we deduce by using the following formula: It follows from Theorem 1 that which completes the proof of Theorem 2. □
$$ ( D_{p,q}f ) ( x ) =\frac{f ( qx ) -f ( px ) }{ ( q-p ) x}=\frac{f ( qx ) }{qx-px}+ \frac{f ( px ) }{px-qx}= [ 1 ] _{p,q}! [ qx,px;f ] $$
$$\begin{aligned} & \bigl( D_{p,q}^{2}f \bigr) ( x ) \\ &\quad =\frac{ ( D_{p,q}f ) ( qx ) - ( D_{p,q}f ) ( px ) }{ ( q-p ) x} \\ &\quad =\frac{\frac{f ( q^{2}x ) -f ( pqx ) }{ ( q-p ) qx}-\frac{f ( pqx ) -f ( p^{2}x ) }{ ( q-p ) px}}{ ( p-q ) x} \\ &\quad = ( p+q ) \biggl[ \frac{f ( q^{2}x ) }{ ( q^{2}-p^{2} ) ( q-p ) x^{2}q}-\frac{f ( pqx ) }{( q-p ) ^{2}x^{2}pq}+\frac{f ( p^{2}x ) }{ ( q^{2}-p^{2} ) ( q-p ) x^{2}p} \biggr] \\ & \quad= [ 2 ] _{p,q}! \bigl[ q^{2}x,pqx,p^{2}x;f \bigr] . \end{aligned}$$
$$ \bigl( D_{p,q}^{n}f \bigr) ( x ) = [ n ] _{p,q}! \bigl[ q^{n}x,q^{n-1}px,\ldots,qp^{n-1}x,p^{n}x;f \bigr] $$
(2.3)
$$ [ x_{0},x_{1},\ldots,x_{n};\cdot] = \frac{ [ x_{1},x_{2},\ldots,x_{n};\cdot ] - [ x_{0},x_{1},\ldots ,x_{n-1};\cdot ] }{x_{n}-x_{0}}. $$
$$ \bigl( D_{p,q}^{n}f \bigr) ( x ) =q^{-\binom{n}{2}}x^{-n} ( p-q ) ^{-n}\sum_{k=0}^{n} ( -1 ) ^{n-k} \begin{bmatrix} n\\ k \end{bmatrix} _{p,q} p^{-k ( 2n-k-1 ) /2}q^{\binom{k}{2}}f \bigl( xp^{k}q^{n-k} \bigr) , $$
In the case when in Theorem 2, we get the following corollary.
$$ f ( x ) =x^{n} $$
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
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: Upon setting in equation (2.4) and \(z=p^{n}x\), if we use equation (2.1), we find that as asserted by Theorem 4. □
$$\begin{aligned} f ( z ) ={}&\sum_{k=0}^{n-1} ( z-x_{0} ) ( z-x_{1} ) \cdots( z-x_{k-1} ) [ x_{0},x_{1},\ldots,x_{k};f ] \\ &{} + ( z-x_{0} ) ( z-x_{1} ) \cdots( z-x_{n-1} ) [ x_{0},x_{1},\ldots,x_{n-1},z;f ] . \end{aligned}$$
(2.4)
$$ x_{k}=p^{k}q^{n-k}x\quad (k=0,1,\ldots,n-1) $$
$$\begin{aligned} f \bigl( p^{n}x \bigr) ={}&\sum_{k=0}^{n-1} \bigl( p^{n}x-q^{n}x \bigr) \bigl( p^{n}x-q^{n-1}px \bigr) \cdots\bigl( p^{n}x-q^{n-k+1}p^{k-1}x \bigr)\\ &{}\cdot \bigl[ q^{n}x,q^{n-1}px,\ldots,q^{n-k}p^{k}x;f \bigr] \\ &{} + \bigl( p^{n}x-q^{n}x \bigr) \bigl( p^{n}x-q^{n-1}px \bigr) \cdots\bigl( p^{n}x-qp^{n-1}x \bigr) \\ &{}\cdot\bigl[ q^{n}x,q^{n-1}px,\ldots,qp^{n-1}x,p^{n}x;f \bigr] \\ ={}&\sum_{k=0}^{n-1} \bigl( p^{n}x-q^{n}x \bigr) \bigl( p^{n}x-q^{n-1}px \bigr) \cdots\bigl( p^{n}x-q^{n-k+1}p^{k-1}x \bigr) \frac{( D_{p,q}^{k}f ) ( x ) }{ [ k ] _{p,q}!} \\ &{} + \bigl( p^{n}x-q^{n}x \bigr) \bigl( p^{n}x-q^{n-1}px \bigr) \cdots\bigl( p^{n}x-qp^{n-1}x \bigr) \frac{ ( D_{p,q}^{n}f ) ( x ) }{ [ n ] _{p,q}!} \\ ={}&\sum_{k=0}^{n} \bigl( p^{n}x-q^{n}x \bigr) \bigl( p^{n}x-q^{n-1}px \bigr) \cdots\bigl( p^{n}x-q^{n-k+1}p^{k-1}x \bigr) \frac{ ( D_{p,q}^{k}f ) ( x ) }{ [ k ] _{p,q}!} \\ ={}&\sum_{k=0}^{n}x^{k}p^{\binom{k}{2}} \frac{\frac{ ( p^{n}-q^{n} ) ( p^{n-1}-q^{n-1} ) \cdots ( p-q ) }{( p-q ) ^{n}}}{\frac{ ( p^{n-k}-q^{n-k} ) ( p^{n-k-1}-q^{n-k-1} ) \cdots ( p-q ) }{ ( p-q ) ^{n-k} ( p-q ) ^{k}}}\frac{ ( D_{p,q}^{k}f ) ( x ) }{ [ k ] _{p,q}!} \\ ={}&\sum_{k=0}^{n}x^{k}p^{\binom{k}{2}} ( p-q ) ^{k}\frac{ [ n] _{p,q}!}{ [ n-k ] _{p,q}! [ k ] _{p,q}!} \bigl( D_{p,q}^{k}f \bigr) ( x ) , \end{aligned}$$
Finally, we are in a position to give the following result.
Corollary 2
Let
p
and
q
be complex numbers such that
Also let the function
\(f:D_{p,q}\rightarrow\mathbb{C}\)
be
\((p,q)\)-differentiable of order
n. Then
$$ 0< \vert q\vert < \vert p\vert \leqq1. $$
$$ f ( x ) =\sum_{k=0}^{n} \begin{bmatrix} n\\ k \end{bmatrix} _{p,q}q^{k ( k-n ) }p^{\binom{k+1}{2}} ( qx-px ) ^{k} \bigl( D_{p,q}^{k}f \bigr) \biggl( \frac{xp^{n-k}}{q^{k}} \biggr) . $$
Proof
Since, for \(k\in \{ 0,1,\ldots,n \} \), we have and Continuing the process, we readily observe that From Theorem 4, we thus conclude that which evidently proves Corollary 2. □
$$ \begin{bmatrix} n\\ k \end{bmatrix} _{\frac{1}{p},\frac{1}{q}}=\frac{ [ n ] _{\frac{1}{p},\frac{1}{q}}!}{ [ n-k ] _{\frac{1}{p},\frac{1}{q}}![ k ] _{\frac{1}{p},\frac{1}{q}}!}=\frac{ ( pq ) ^{-\binom{n}{2}}}{ ( pq ) ^{-\binom{n-k}{2}} ( pq ) ^{-\binom{k}{2}}} \begin{bmatrix} n\\ k \end{bmatrix} _{p,q}, $$
$$ ( D_{\frac{1}{p},\frac{1}{q}}f ) ( x ) =\frac{f ( \frac{x}{q} ) -f ( \frac{x}{p} ) }{ ( p-q ) x} ( pq ) =pq ( D_{p,q}f ) \biggl( \frac{x}{pq} \biggr) $$
$$\begin{aligned} \bigl( D_{\frac{1}{p},\frac{1}{q}}^{2}f \bigr) ( x ) & =\frac{pq ( D_{p,q}f ) ( \frac{x}{pq} ) -pq ( D_{p,q}f ) ( \frac{x}{pq} ) }{ ( \frac{1}{p}-\frac{1}{q}) x} \\ & =\frac{ ( pq ) ^{2} [ ( D_{p,q}f ) ( \frac{x}{pq^{2}} ) - ( D_{p,q}f ) ( \frac{x}{pq} ) ] }{( p-q ) x} \\ & =p^{2}q^{2} \bigl( D_{p,q}^{2}f \bigr) \biggl( \frac{x}{p^{2}q^{2}} \biggr) . \end{aligned}$$
$$ \bigl( D_{\frac{1}{p},\frac{1}{q}}^{n}f \bigr) ( x ) =p^{n}q^{n} \bigl( D_{p,q}^{n}f \bigr) \biggl( \frac{x}{p^{n}q^{n}} \biggr) . $$
(2.5)
$$ f ( x ) =\sum_{k=0}^{n} \begin{bmatrix} n\\ k \end{bmatrix} _{p,q}q^{k ( k-n ) }p^{\binom{k+1}{2}} ( qx-px ) ^{k} \bigl( D_{p,q}^{k}f \bigr) \biggl( \frac{xp^{n-k}}{q^{k}} \biggr) , $$
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.
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Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors contributed equally to this work. All authors read and approved the final manuscript.