A Voronovskaja-type formula for the $q$-Meyer-König and Zeller operators

Tiberiu Trif$^{\ast }$

May 17, 2011.

$^{\ast }$Babeş-Bolyai University, Faculty of Mathematics and Computer Science, Kogălniceanu no. 1, 400084 Cluj-Napoca, Romania, e-mail: ttrif@math.ubbcluj.ro

A Voronovskaja-type formula for the $q$-Meyer-König and Zeller operators is presented.

MSC. 41A36, 41A25.

Keywords. Meyer-König and Zeller operators, rate of convergence, $q$-calculus.

1 Introduction and notation

Starting from the identity

\begin{equation*} (1-x)^{n+1}\sum _{k=0}^\infty \tbinom {n+k}{k}x^k=1 \quad \mbox{for all } x\in [0,1), \end{equation*}

W. Meyer-König and K. Zeller [ 12 ] defined a sequence of linear positive operators associating with each continuous real-valued function defined on $[0,1]$ a so-called “Bernstein power series”. In the slight modification by E. W. Cheney and A. Sharma [ 4 ] , the Meyer-König and Zeller operators are defined for every $n\in \mathbb {N}$ (the set of all positive integers) and every $f\in C[0,1]$ by

\begin{eqnarray*} M_nf(x) & :=& \sum _{k=0}^\infty f\left(\tfrac {k}{n+k}\right) m_{n,k}(x) \quad \mbox{if } x\in [0,1), \\ M_nf(1) & :=& f(1), \end{eqnarray*}

where

\begin{equation*} m_{n,k}(x):=\tbinom {n+k}{k}x^k(1-x)^{n+1}. \end{equation*}

Let $q{\gt}0$. For every $n\in \{ 0,1,2,\ldots \} $ the $q$-integer $[n]_q$ is defined by

\begin{equation*} [0]_q := 0 \quad \mbox{and}\quad [n]_q := 1+q+\cdots +q^{n-1} \quad \mbox{if } n\geq 1. \end{equation*}

The $q$-factorial $[n]_q!$ is defined by

\begin{equation*} [0]_q! := 1 \quad \mbox{and}\quad [n]_q! := [1]_q[2]_q\cdots [n]_q \quad \mbox{if } n\geq 1. \end{equation*}

For all nonnegative integers $n$ and $k$ with $n\geq k$, the Gaussian binomial coefficient (or $q$-binomial coefficient ) $\genfrac {[}{]}{0pt}{1}{n}{k}_q$ is defined by

\begin{equation*} \genfrac {[}{]}{0pt}{1}{n}{k}_q:=\tfrac {[n]_q!}{[k]_q![n-k]_q!}\, . \end{equation*}

Clearly, when $q=1$ we have

\begin{equation*} [n]_1=n, \quad [n]_1!=n! \quad \mbox{and}\quad \genfrac {[}{]}{0pt}{1}{n}{k}_1=\tbinom {n}{k}. \end{equation*}

Throughout the rest of the paper $q$ denotes a positive real number such that $0{\lt}q{\lt}1$. In order to simplify the notation, whenever it is not necessary to mention explicitly $q$, we write $[n]$, $[n]!$ and $\genfrac {[}{]}{0pt}{1}{n}{k}$ instead of $[n]_q$, $[n]_q!$ and $\genfrac {[}{]}{0pt}{1}{n}{k}_q$, respectively. Likewise, in order to emphasize the analogy with the classical $M_n$-operators, we make use (for any positive integer $n$) of the following notation:

\begin{equation*} (a+b)_q^n:=(a+b)(a+qb)\cdots (a+q^{n-1}b). \end{equation*}

We set also $(a+b)_q^0:=1$.

For every $n\in \mathbb {N}$ one has (see, for instance, G. E. Andrews, R. Askey, R. Roy [ 2 , Corollary 10.2.2 ] )

\begin{equation} \label{basic-q-identity} (1-x)_q^{n+1}\sum _{k=0}^\infty \genfrac {[}{]}{0pt}{1}{n+k}{k}x^k=1 \quad \mbox{for all } x\in [0,1). \end{equation}

Starting from this identity, L. Lupaş [ 9 ] introduced a $q$-generalization of the $M_n$-operators. The $q$-Meyer-König and Zeller operators are defined for every $n\in \mathbb {N}$ and every $f\in C[0,1]$ by

\begin{eqnarray*} M_{n,q}f(x) & :=& \sum _{k=0}^\infty f\left(\tfrac {[k]}{[n+k]}\right) m_{{n},{k},q}(x) \quad \mbox{if } x\in [0,1), \\ M_{n,q}f(1) & :=& f(1), \end{eqnarray*}

where

\begin{equation*} m_{{n},{k},q}(x):=\genfrac {[}{]}{0pt}{1}{n+k}{k}x^k(1-x)_q^{n+1}. \end{equation*}

The $M_{n,q}$-operators and other similar ones have been intensively investigated in the last decade by many authors (see, for instance, [ 5 ] , [ 6 ] , [ 7 ] , [ 8 ] , [ 10 ] , [ 13 ] , [ 14 ] , [ 15 ] , [ 16 ] , [ 19 ] , [ 20 ] , [ 21 ] ). We merely mention here that in [ 19 , Lemma 2.1 ] it has been proved that for all $n\in \mathbb {N}$, $n\geq 3$ and all $x\in [0,1]$ one has

\begin{equation*} M_{n,q}e_2(x)=x^2+\tfrac {x(1-x)(1-q^nx)}{[n-1]}-R_{n,q}(x), \end{equation*}

where

\begin{equation*} 0\leq R_{n,q}(x)\leq \tfrac {q^{n-1}(1+q)}{[n-1][n-2]}\, x(1-x)(1-qx)(1-q^nx). \end{equation*}

Here $e_k$ $(k=0,1,2,\ldots )$ denotes, as usual, the monomial $e_k(t):=t^k$.

2 The second moment for the $q$-Meyer-König and Zeller operators

J. A. H. Alkemade [ 1 ] was the first who derived an explicit expression for $M_n e_2$ in terms of a hypergeometric series. More precisely, he proved that

\begin{equation} \label{Eq-Alkemade} M_n e_2(x)=x^2+\tfrac {x(1-x)^2}{n+1}\, _2F_1(1,2;n+2;x) \end{equation}

for all $n\in \mathbb {N}$ and all $x\in [0,1)$. Moreover, 2 holds also for $x=1$ if $n\geq 2$. In 2 the notation $_2F_1(a,b;c;x)$ is used for the sum of the hypergeometric series

\begin{equation} \label{hypergeometric} \sum _{k=0}^\infty \tfrac {(a)_k(b)_k}{(c)_k k!}\, x^k, \end{equation}

where

\begin{equation*} (y)_0 := 1 \quad \mbox{and}\quad (y)_k := y(y+1)\cdots (y+k-1) \quad \mbox{if } k\geq 1. \end{equation*}

The series 3 converges for $|x|{\lt}1$ and if $c-a-b{\gt}0$ also for $x=1$.

The $q$-analogue of the hypergeometric series 3 is the basic $q$-hypergeometric series

\begin{equation} \label{q-hypergeometric} \sum _{k=0}^\infty \tfrac {(\alpha ;q)_k(\beta ;q)_k} {(\gamma ;q)_k(q;q)_k}\, x^k, \end{equation}

where

\begin{equation*} (y;q)_k:=(1-y)(1-qy)\cdots (1-q^{k-1}y)=(1-y)_q^k. \end{equation*}

The series 4 converges for $|x|{\lt}1$. Its sum is usually denoted by

\[ _2\phi _1\left(\begin{array}{c} {\alpha },\, {\beta } \\ {\gamma } \end{array};\, q,\, {x}\right). \]

Set $[a]:=[a]_q:=\tfrac {1-q^a}{1-q}$ for every real number $a$. Set also

\begin{equation*} [a]_0 := [a]_{0,q}:= 1 \quad \mbox{and}\quad [a]_k := [a]_{k,q}:= [a][a+1]\cdots [a+k-1] \quad \mbox{if } k\geq 1, \end{equation*}

and note that

\begin{equation*} (q^a;q)_k=(1-q)^k[a]_k \end{equation*}

for every real number $a$ and every nonnegative integer $k$. Therefore one has

\begin{equation*} _2\phi _1\left(\begin{array}{c} {q^a},\, {q^b} \\ {q^c} \end{array};\, q,\, {x}\right)=\sum _{k=0}^\infty \tfrac {[a]_k[b]_k} {[c]_k[k]!}\, x^k \quad \mbox{for all } x\in (-1,1). \end{equation*}

H. Wang [ 21 , Theorem 1 ] proved that for every $n\in \mathbb {N}$ and all $x\in [0,1]$ one has

\begin{equation} \label{second-moment-Mnq-Wang} M_{n,q}e_2(x)=x^2+\tfrac {x(1-x)}{[n+1]}\left(1- \tfrac {q^{n+2}[n]x}{[n+2]}\, _2\phi _1\left(\begin{array}{c} {q},\, {q^2} \\ {q^{n+3}} \end{array};\, q,\, {q^{n+1}x}\right) \right). \end{equation}

But a simple computation shows that

\begin{equation} \label{Wang-Trif} 1-\tfrac {q^{n+2}[n]x}{[n+2]}\, _2\phi _1\left(\begin{array}{c} {q},\, {q^2} \\ {q^{n+3}} \end{array};\, q,\, {q^{n+1}x}\right) = (1-q^nx)_2\phi _1\left(\begin{array}{c} {q},\, {q^2} \\ {q^{n+2}} \end{array};\, q,\, {q^n x}\right). \end{equation}

By 5 and 6 we get

\begin{equation} \label{second-moment-Mnq} M_{n,q}e_2(x)=x^2+\tfrac {x(1-x)(1-q^nx)}{[n+1]}\, _2\phi _1\left(\begin{array}{c} {q},\, {q^2} \\ {q^{n+2}} \end{array};\, q,\, {q^n x}\right), \end{equation}

a formula which is closer to 2 than 5.

By using 7, one can easily derive estimates for the second moment
$M_{n,q}e_2(x)-x^2$. Set

\begin{equation*} \Phi _m(x):=\sum _{k=m}^\infty \tfrac {[2]_k}{[n+2]_k}\, x^k \end{equation*}

for every $m\in \mathbb {N}$ and all $x\in [0,1)$.

Theorem 1

For all $n,m\in \mathbb {N}$ with $n\geq 2$ and all $x\in [0,q^{n-1}]$ one has

\begin{equation} \label{estimate-Phi-m} \Phi _m(x)\leq \tfrac {[m+1]!}{[n-1][n+2]_{m-1}}\, x^m \end{equation}

and

\begin{eqnarray} \label{estimates-phi-21} & & \sum _{k=0}^m \tfrac {[2]_k}{[n+2]_k}\, x^k \leq \, _2\phi _1\left(\begin{array}{c} {q},\, {q^2} \\ {q^{n+2}} \end{array};\, q,\, {x}\right) \leq \\ & & \qquad \leq \sum _{k=0}^{m-1} \tfrac {[2]_k}{[n+2]_k}\, x^k +\tfrac {[m+1]!}{[n-1][n+2]_{m-1}}\, x^m. \nonumber \end{eqnarray}

Proof . We have

\begin{eqnarray*} \Phi _m(x) & =& \tfrac {[m+1]!}{[n+2]_m}\, x^m \sum _{k=m}^\infty \tfrac {[m+2]_{k-m}}{[n+m+2]_{k-m}}\, x^{k-m} \\ & =& \tfrac {[m+1]!}{[n+2]_m}\, x^m \sum _{k=0}^\infty \tfrac {[m+2]_k}{[n+m+2]_k}\, x^k \\ & =& \tfrac {[m+1]!}{[n+2]_m}\, x^m\, _2\phi _1\left(\begin{array}{c} {q},\, {q^{m+2}} \\ {q^{n+m+2}} \end{array};\, q,\, {x}\right) \\ & \leq & \tfrac {[m+1]!}{[n+2]_m}\, x^m\, _2\phi _1\left(\begin{array}{c} {q},\, {q^{m+2}} \\ {q^{n+m+2}} \end{array};\, q,\, {q^{n-1}}\right). \end{eqnarray*}

But, for $|\gamma /\alpha \beta |{\lt}1$ it holds that (see [ 2 , Corollary 10.9.2 ] )

\begin{equation*} _2\phi _1\left(\begin{array}{c} {\alpha },\, {\beta } \\ {\gamma } \end{array};\, q,\, {\gamma /\alpha \beta }\right)= \tfrac {(\gamma /\alpha ;q)_\infty (\gamma /\beta ;q)_\infty } {(\gamma ;q)_\infty (\gamma /\alpha \beta ;q)_\infty }\, , \end{equation*}

where

\begin{equation*} (y;q)_\infty :=\prod _{k=0}^\infty (1-q^ky). \end{equation*}

Taking this into account, we have

\begin{equation*} _2\phi _1\left(\begin{array}{c} {q},\, {q^{m+2}} \\ {q^{n+m+2}} \end{array};\, q,\, {q^{n-1}}\right)= \tfrac {(q^{n+m+1};q)_\infty (q^n;q)_\infty } {(q^{n+m+2};q)_\infty (q^{n-1};q)_\infty }= \tfrac {[n+m+1]}{[n-1]}\, . \end{equation*}

Consequently,

\begin{equation*} \Phi _m(x)\leq \tfrac {[m+1]!}{[n+2]_m}\cdot \tfrac {[n+m+1]}{[n-1]}\, x^m= \tfrac {[m+1]!}{[n-1][n+2]_{m-1}}\, x^m. \end{equation*}

The left inequality in 9 is obvious, while the right one follows immediately by 8. $\Box $

By means of Theorem 1 we deduce the following estimates of the second moment for the $M_{n,q}$-operators. These estimates are quite similar to those obtained by M. Becker and R. J. Nessel [ 3 ] for the classical $M_n$-operators.

Corollary 2

For every $n\in \mathbb {N}$, $n\geq 2$ and all $x\in [0,1]$ one has

\begin{align*} & \tfrac {x(1-x)(1-q^nx)}{[n+1]} \left( 1+\tfrac {(1+q)q^nx} {[n+2]}\right)\leq M_{n,q}e_2(x)-x^2 \leq \\ & \leq \tfrac {x(1-x)(1-q^nx)}{[n+1]} \left( 1+ \tfrac {(1+q)q^nx}{[n-1]}\right). \end{align*}

Proof â–¼

By Theorem 1 with $m=1$ it follows that

\begin{equation*} 1+\tfrac {1+q}{[n+2]}\, q^nx \leq \, _2\phi _1\left(\begin{array}{c} {q},\, {q^2} \\ {q^{n+2}} \end{array};\, q,\, {q^nx}\right) \leq 1+\tfrac {1+q}{[n-1]}\, q^nx. \end{equation*}

This inequality and 7 yield the conclusion.

3 A Voronovskaja-type formula for the $q$-Meyer-König and Zeller operators

The goal of this section is to establish a Voronovskaja-type formula for the $M_{n,q}$-operators. Such a formula is lacking in the literature. In order to derive it, we need the following auxiliary results whose proofs are postponed to the end of the section.

Lemma 3

For every $n\in \mathbb {N}$, $n\geq 3$ and all $x\in [0,1]$ one has

\begin{equation} \label{lemma6-eq1} M_{n,q}e_3(x)=xM_{n,q}e_2(x)+\tfrac {2qx^2(1-x)(1-q^nx)}{[n-1]} +R_{n,q}(x), \end{equation}

where

\begin{equation} \label{lemma6-eq2} |R_{n,q}(x)|\leq \tfrac {9}{[n-1][n+2]}\, . \end{equation}

Lemma 4

For every $n\in \mathbb {N}$, $n\geq 3$ and all $x\in [0,1]$ one has

\begin{equation*} M_{n,q}e_4(x)=xM_{n,q}e_3(x)+\tfrac {3q^2x^3(1-x)(1-q^nx)}{[n-1]} +\widetilde{R}_{n,q}(x), \end{equation*}

where

\begin{equation*} |\widetilde{R}_{n,q}(x)|\leq \tfrac {C}{[n-1][n+3]}\, , \end{equation*}

$C$ being an absolute constant (i.e., not depending on $n$, $q$ or $x$).

We notice that, for a fixed $q\in (0,1)$, the sequence $(M_{n,q}f)_{n\geq 1}$ does not converge to $f$ for every $f\in C[0,1]$. For instance, by Corollary 2 it follows that

\begin{equation*} M_{n,q}e_2(x) \to x^2+(1-q)x(1-x) \quad \mbox{as } n\to \infty . \end{equation*}

In order to obtain a convergent sequence of $q$-Meyer-König and Zeller operators we must replace $q$ by a sequence $(q_n)_{n\geq 1}$ of numbers in $(0,1)$. If $(q_n)_{n\geq 1}$ satisfies

\begin{equation} \label{condition1-qn} q_n \to 1 \quad \mbox{and}\quad [n]_{q_n}=1+q_n+\cdots +q_n^{n-1} \to \infty \quad \mbox{as } n\to \infty , \end{equation}

then the sequence $(M_{n,q_n}f)_{n\geq 1}$ converges uniformly to $f$ on $[0,1]$ for all $f\in C[0,1]$ (see [ 9 , Theorem 2 ] or [ 19 , Theorem 2.2 ] ).

In order to obtain a Voronovskaja-type formula for the $M_{n,q_n}$-operators, $(q_n)_{n\geq 1}$ must satisfy an additional condition, namely

\begin{equation} \label{condition2-qn} \mbox{there exists } \lim _{n\to \infty }q_n^n=:\alpha \in [0,1]. \end{equation}

It is not difficult to construct a sequence $(q_n)_{n\geq 1}$, satisfying both 12 and 13. Indeed, it suffices to take $q_n$ such that

\begin{equation*} 1-\tfrac {1}{n}\leq q_n \leq 1-\tfrac {1}{n-1} \quad \mbox{for all } n\geq 3. \end{equation*}

Then clearly $q_n\to 1$ and $q_n^n\to e^{-1}$ as $n\to \infty $. Moreover, since

\begin{equation*} 1-\tfrac {r}{n}\leq q_n^r \quad \mbox{for } 1\leq r\leq n-1, \end{equation*}

we have

\begin{equation*} [n]_{q_n}\geq n-\tfrac {n(n-1)}{2n}=\tfrac {n+1}{2} \quad \mbox{for all } n\in \mathbb {N}. \end{equation*}

We notice also that, if $(q_n)_{n\geq 1}$ is a sequence in $(0,1)$ satisfying 13, then

\begin{equation} \label{condition3-qn} \lim _{n\to \infty }\tfrac {[n]_{q_n}}{[n-1]_{q_n}}= \lim _{n\to \infty }\tfrac {[n]_{q_n}}{[n+1]_{q_n}}=1. \end{equation}

Indeed, if $\alpha \in [0,1)$, then

\begin{equation*} \tfrac {[n]_{q_n}}{[n-1]_{q_n}}=\tfrac {1-q_n^n}{1-q_n^{n-1}} \to 1 \quad \mbox{as } n\to \infty \end{equation*}

and, analogously, $[n]_{q_n}/[n+1]_{q_n}\to 1$ as $n\to \infty $. On the other hand, if $\alpha =1$, then it is easily seen that

\begin{equation*} \lim _{n\to \infty }\tfrac {[n]_{q_n}}{n}= \lim _{n\to \infty }\tfrac {[n-1]_{q_n}}{n}= \lim _{n\to \infty }\tfrac {[n+1]_{q_n}}{n}=1, \end{equation*}

whence 14 holds also in this case.

Theorem 5

Let $(q_n)_{n\geq 1}$ be a sequence in $(0,1)$ satisfying 12 and 13. Then for every $x\in [0,1]$ and every function $f\in C[0,1]$ which is twice differentiable at $x$ one has

\begin{equation*} \lim _{n\to \infty }[n]_{q_n}\Big(M_{n,q_n}f(x)-f(x)\Big)= \tfrac {x(1-x)(1-\alpha x)}{2}\, f”(x). \end{equation*}

Proof â–¼

For all $n\in \mathbb {N}$ we have

\begin{equation*} M_{n,q_n}e_k(x)=e_k(x), \quad k=0,1. \end{equation*}

By Corollary 2, 12, 13 and 14 it follows that

\begin{equation*} \lim _{n\to \infty }[n]_{q_n}\Big(M_{n,q_n}e_2(x)-x^2\Big)= x(1-x)(1-\alpha x). \end{equation*}

Further, let $\psi _x$ be the function defined by $\psi _x(t):=t-x$. We have

\begin{equation*} \Big(M_{n,q_n}\psi _x^4\Big)(x)=M_{n,q_n}e_4(x)-4xM_{n,q_n}e_3(x)+6x^2 M_{n,q_n}e_2(x)-3x^4. \end{equation*}

By Lemma 3, Lemma 4, 12, 13 and 14 it follows that

\begin{equation*} \lim _{n\to \infty }[n]_{q_n}\Big(M_{n,q_n}\psi _x^4\Big)(x)=0. \end{equation*}

Now the conclusion of the theorem is an immediate consequence of a standard result (for instance [ 11 , Theorem 3 ] or [ 17 , Theorem 1 ] ).

Proof of Lemma 3 . Clearly, the assertion holds for $x=1$. For $x\in [0,1)$ we have

\begin{eqnarray*} M_{n,q}e_3(x) & =& (1-x)_q^{n+1}\sum _{k=1}^\infty \genfrac {[}{]}{0pt}{1}{n+k}{k} x^k \tfrac {[k]^3}{[n+k]^3} \\ & =& x(1-x)_q^{n+1}\sum _{k=0}^\infty \genfrac {[}{]}{0pt}{1}{n+k}{k} x^k \tfrac {[k+1]^2}{[n+k+1]^2}\, . \end{eqnarray*}

Following P. C. Sikkema [ 18 ] in his proof of Theorem 3, we notice that

\begin{equation*} \tfrac {[k+1]^2}{[n+k+1]^2}=\tfrac {[k]^2}{[n+k]^2}+ \tfrac {2q^k[n][k]}{[n+k]^2[n+k+1]}+\tfrac {q^{2k}[n]^2} {[n+k]^2[n+k+1]^2}\, . \end{equation*}

Taking this into account, we deduce that

\begin{equation} \label{lemma6-eq3} M_{n,q}e_3(x)=xM_{n,q}e_2(x)+S_{n,q}(x)+T_{n,q}(x), \end{equation}

where

\begin{eqnarray*} S_{n,q}(x) & :=& 2[n]x(1-x)_q^{n+1} \sum _{k=1}^\infty \genfrac {[}{]}{0pt}{1}{n+k}{k} (q x)^k \tfrac {[k]}{[n+k]^2[n+k+1]}\, , \\ T_{n,q}(x) & :=& x(1-x)_q^{n+1} \sum _{k=0}^\infty \genfrac {[}{]}{0pt}{1}{n+k}{k} (q^2 x)^k \tfrac {[n]^2}{[n+k]^2[n+k+1]^2}\, . \end{eqnarray*}

Since

\begin{equation*} S_{n,q}(x)=2[n]qx^2(1-x)_q^{n+1}\sum _{k=0}^\infty \genfrac {[}{]}{0pt}{1}{n+k}{k} (q x)^k \tfrac {1}{[n+k+1][n+k+2]} \end{equation*}

and

\begin{equation*} \tfrac {1}{[n+k+1]}=\tfrac {1}{[n+k]}-\tfrac {q^{n+k}} {[n+k][n+k+1]}\, , \end{equation*}

it follows that

\begin{equation} \label{lemma6-eq4} S_{n,q}(x)=U_{n,q}(x)-V_{n,q}(x), \end{equation}

where

\begin{eqnarray*} U_{n,q}(x) & :=& 2qx^2(1-x)_q^{n+1} \sum _{k=0}^\infty \genfrac {[}{]}{0pt}{1}{n+k}{k} \tfrac {(q x)^k\, [n]}{[n+k][n+k+2]}\, , \\ V_{n,q}(x) & :=& 2q^{n+1}x^2(1-x)_q^{n+1} \sum _{k=0}^\infty \genfrac {[}{]}{0pt}{1}{n+k}{k} \tfrac {(q^2 x)^k\, [n]}{[n+k][n+k+1][n+k+2]}\, . \end{eqnarray*}

Since

\begin{equation*} U_{n,q}(x)=2qx^2(1-x)_q^{n+1} \sum _{k=0}^\infty \genfrac {[}{]}{0pt}{1}{n-1+k}{k} (q x)^k \tfrac {1}{[n+k+2]} \end{equation*}

and

\begin{equation*} \tfrac {1}{[n+k+2]}=\tfrac {1}{[n-1+k]}-\tfrac {[3]q^{n-1+k}} {[n-1+k][n+k+2]}\, , \end{equation*}

we deduce that

\begin{eqnarray} \label{lemma6-eq5} U_{n,q}(x) & =& \tfrac {2qx^2(1-x)(1-q^nx)(1-q x)_q^{n-1}}{[n-1]} \sum _{k=0}^\infty \genfrac {[}{]}{0pt}{1}{n-2+k}{k} (q x)^k \\ & & \qquad -W_{n,q}(x) \nonumber \\ & =& \tfrac {2qx^2(1-x)(1-q^nx)}{[n-1]}-W_{n,q}(x), \nonumber \end{eqnarray}

where

\begin{equation*} W_{n,q}(x):=2[3]q^{n}x^2(1-x)_q^{n+1}\sum _{k=0}^\infty \genfrac {[}{]}{0pt}{1}{n-1+k}{k} \tfrac {(q^2 x)^k}{[n-1+k][n+k+2]}\, . \end{equation*}

By 15, 16 and 17 we get

\begin{eqnarray} \label{lemma6-eq6} M_{n,q}e_3(x) & =& xM_{n,q}e_2(x)+ \tfrac {2qx^2(1-x)(1-q^nx)}{[n-1]} \\ & & \qquad -W_{n,q}(x)-V_{n,q}(x)+T_{n,q}(x).\nonumber \end{eqnarray}

Now we have

\begin{eqnarray} \label{lemma6-eq7} 0 \leq W_{n,q}(x) & \leq & \tfrac {6}{[n-1][n+2]}\, (1-x)_q^{n+1} \sum _{k=0}^\infty \genfrac {[}{]}{0pt}{1}{n-1+k}{k} (q^2 x)^k \\ & \leq & \tfrac {6}{[n-1][n+2]}\, (1-x)_q^n \sum _{k=0}^\infty \genfrac {[}{]}{0pt}{1}{n-1+k}{k} x^k \nonumber \\ & =& \tfrac {6}{[n-1][n+2]}\, , \nonumber \end{eqnarray}

\begin{eqnarray} \label{lemma6-eq8} 0 \leq V_{n,q}(x) & \leq & \tfrac {2}{[n+1][n+2]}\, (1-x)_q^{n+1} \sum _{k=0}^\infty \genfrac {[}{]}{0pt}{1}{n-1+k}{k} (q^2 x)^k \\ & \leq & \tfrac {2}{[n+1][n+2]}\, (1-x)_q^n \sum _{k=0}^\infty \genfrac {[}{]}{0pt}{1}{n-1+k}{k} x^k \nonumber \\ & =& \tfrac {2}{[n+1][n+2]}\, , \nonumber \end{eqnarray}

and

\begin{eqnarray} \label{lemma6-eq9} 0 \leq T_{n,q}(x) & \leq & \tfrac {1}{[n+1]^2}\, (1-x)_q^{n+1} \sum _{k=0}^\infty \genfrac {[}{]}{0pt}{1}{n+k}{k} (q^2 x)^k \\ & \leq & \tfrac {1}{[n+1]^2}\, (1-x)_q^{n+1} \sum _{k=0}^\infty \genfrac {[}{]}{0pt}{1}{n+k}{k} x^k \nonumber \\ & =& \tfrac {1}{[n+1]^2}\, . \nonumber \end{eqnarray}

By 18, 19, 20 and 21 we conclude that 10 and 11 hold. $\Box $

Proof of Lemma 4 . Since this proof is similar to that of Lemma 3 and follows the same lines, we omit it. $\Box $

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A Voronovskaja-type formula for the \(q\)-Meyer-König and Zeller operators

1 Introduction and notation

2 The second moment for the \(q\)-Meyer-König and Zeller operators

3 A Voronovskaja-type formula for the \(q\)-Meyer-König and Zeller operators

Bibliography