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\markboth{S.\,Herdem\,and \.{I}.\,B\"{u}y\"{u}kyaz{\i}c{\i}}{Weighted approximation by $q$-Ibragimov-Gadjiev
operators}
\title[Weighted approximation by $q$-Ibragimov-Gadjiev
operators]{Weighted approximation by $q$-Ibragimov-Gadjiev
operators}

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\author[S.\,Herdem, İ.\,Büyükyazıcı]{Serap Herdem\affil{1}\comma\corrauth and \.{I}brahim B\"{u}y\"{u}kyaz{\i}c{\i}\affil{2}}
\address{\affilnum{1} Department of Mechanical Engineering, Piri Reis University, Postane Mahallesi, Eflatun Sk. No:8, 34\,940 Tuzla, Istanbul, Turkey\\
%Department of Mechanical Engineering, Faculty of Engineering, Piri Reis University, Istanbul, Turkey
\affilnum{2} Department of Mathematics, Faculty of Science, Ankara
University, D\"{o}gol Caddesi, 06\,100, Tando$\it{\breve{q}}$an, Ankara, Turkey}
\emails{{\tt skaya@pirireis.edu.tr}\,\,(S.\,Herdem), {\tt ibuyukyazici@gmail.com} (\.{I}.\,B\"{u}y\"{u}kyaz{\i}c{\i})}


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%%%%% Begin Abstract %%%%%%%%%%%
\begin{abstract}
The present paper deals with a $q$-generalization of
Ibragimov-Gadjiev operators, which were constructed by Ibragimov and Gadjiev in
1970. We study convergence properties on the interval $\left[
0,\infty \right)  $ and obtain the rate of convergence in terms of a weighted modulus
of continuity. We also give some representation formulas of the operators
using $q$-derivatives.
\end{abstract}
%%%%% end %%%%%%%%%%%

%%%%% Keywords %%%%%%%%%%%
\keywords{$q$-Ibragimov-Gadjiev operators, weighted approximation,
$q$-integers, $q$-deri\-va\-ti\-ves}

%%%% AMS subject classifications %%%%
\ams{41A25, 41A36}

%%%% maketitle %%%%%
\maketitle

%%%% Start %%%%%%
\section{Introduction}


 In 1970, \.{I}bragimov and Gadjiev \cite{r10} introduced
the following general sequence of positive linear operators defined on the
space $C\left[  0,A\right]  $:
\begin{equation}\label{1}
L_{n}\left(  f;x\right)  =\sum \limits_{\nu=0}^{\infty}f\left(  \frac{\nu
}{n^{2}\Psi_{n}\left(  0\right)  }\right)  \left[  \dfrac{\partial^{\nu}%
}{\partial u^{\nu}}K_{n}\left(  x,t,u\right)  |_{\substack{u=\alpha_{n}%
\Psi_{n}\left(  t\right)  \\t=0}}\right]  \frac{(-\alpha_{n}\Psi_{n}\left(
0\right)  )^{\nu}}{\nu!}
\end{equation}
which contains  the well-known operators of
Bernstein, Bernstein-Chlodowsky, Sz\'{a}sz and Baskakov as a particular case. Over time, these
operators, called  Ibragimov-Gadjiev operators, have been deeply
investigated and different generalizations have been obtained in numerous
papers, see $\left[ 1-4,8,9,11 \right]$    and the literature cited therein.

In \cite{r9}, an extension in $q$-Calculus of Ibragimov-Gadjiev
operators was constructed$.$ Here we briefly describe some details. First of
all, we should mention some basic definitions of $q$-Calculus. We refer to
\cite{r13} as a good guide. Let $q>0.$ For any natural number
$n\in \mathbb{N}\cup \left \{  0\right \}  ,$ the $q$-integer $\left[  n\right]
=$ $\left[  n\right]  _{q}$ is defined by:
\[
\left[  n\right]  =\frac{1-q^{n}}{1-q},\text{ \  \  \  \  \  \  \  \  \ }\left[
0\right]  =0,
\]
and the $q$-factorial $\left[  n\right]  !=$ $\left[  n\right]  _{q}!$ by%
\[
\left[  n\right]  !=\left[  1\right]  \left[  2\right]  \cdots\left[  n\right]
,\text{ \  \  \  \  \  \ }\left[  0\right]  !=1\text{\ .\ }%
\]


For integers $0\leq k\leq n$, the $q$-binomial coefficient is defined by:
\[
\left[
\begin{array}
[c]{l}%
n\\
k
\end{array}
\right]  =\frac{\left[  n\right]  !}{\left[  n-k\right]  !\left[  k\right]
!}.
\]
Clearly, for $q=1,$%
\[
\left[  n\right]  _{1}=n,\text{ \  \ }\left[  n\right]  _{1}!=n!,\text{
\  \  \ }\left[
\begin{array}
[c]{l}%
n\\
k
\end{array}
\right]  _{1}=\left(
\begin{array}
[c]{l}%
n\\
k
\end{array}
\right)  \text{.}%
\]


The $q$-derivative of a function $f:\mathbb{R}\rightarrow \mathbb{R}$ is
defined by:
\[
D_{q}f\left(  x\right)  =\left \{
\begin{array}
[c]{ll}%
\dfrac{f\left(  x\right)  -f\left(  qx\right)  }{\left(  1-q\right)  x}, &
x\neq0\\
f^{\prime}\left(  0\right)  , & x=0
\end{array}
\right.  ,%
\]
for functions which are differentiable at $x=0$ and the higher $q$-derivatives are%
\[
D_{q}^{0}f=f,\text{ }D_{q}^{n}f=D_{q}\left(  D_{q}^{n-1}f\right)  ,\text{
\ }n=1,2,3,\ldots%
\]


The product rule is%
\begin{equation}\label{2}
D_{q}\left(  f\left(  x\right)  g\left(  x\right)  \right)  =D_{q}f\left(
x\right)  g\left(  qx\right)  +D_{q}g\left(  x\right)  f\left(  x\right).
\end{equation}


The $q$-analogue of $\left(  x-a\right)  ^{n}$ is the polynomial%
\[
\left(  x-a\right)  _{q}^{n}=\left \{
\begin{array}
[c]{ll}%
1 ,&  n=0\\
\left(  x-a\right)  \left(  x-qa\right)  \cdots\left(  x-q^{n-1}a\right)  ,&
 n\geq1
\end{array}
\right.  .
\]


The $q$-exponential functions are given as follows:%
\begin{align}
e_{q}\left(  x\right) & =\sum \limits_{k=0}^{\infty}\frac{x^{k}}{\left[
k\right]  !}=\frac{1}{\left(  1-\left(  1-q\right)  x\right)  _{q}^{\infty}%
},\text{ }\left \vert x\right \vert <\frac{1}{1-q},\text{ }\left \vert
q\right \vert <1,\label{3}%
\\
E_{q}\left(  x\right)  &=\sum \limits_{k=0}^{\infty}q^{\frac{k\left(
k-1\right)  }{2}}\frac{x^{k}}{\left[  k\right]  !}=\left(  1+\left(
1-q\right)  x\right)  _{q}^{\infty},\text{ }x\in \mathbb{R},\text{ }\left \vert
q\right \vert <1.\label{4} %
\end{align}
 It is clear from  equations \eqref{3} and \eqref{4} that%
\[
e_{q}\left(  x\right)  E_{q}\left(  -x\right)  =1.
\]


The following $q$-generalization of Taylor's formula was introduced by Jackson (see \cite{r6}).



\begin{theorem} If the function\textbf{\ }$f$ can be expanded in a convergent power series and if $q$ is not a root of unity, then%

\[
f\left(  x\right)  =\sum \limits_{n=0}^{\infty}D_{q}^{n}f\left(  a\right)
\frac{\left(  x-a\right)  _{q}^{n}}{\left[  n\right]  !}.
\]
\end{theorem}

In \cite{r9}, the authors introduced a $q$-analogue of the
Ibragimov-Gadjiev operators that are called the $q$-Ibragimov-Gadjiev operators as follows:

Let $0<q<1$ and let $A>0$ be a given number and $\left \{  \psi_{n}\left(
t\right)  \right \}  $  a sequence of functions in $C\left[  0,A\right]  $
such that $\psi_{n}\left(  t\right)  >0$ for each $t\in \left[  0,A\right]  ,$
$\lim \limits_{n\rightarrow \infty}\dfrac{1}{[n]^{2}\psi_{n}\left(  0\right)
}=0.$ Also, let $\left \{  \alpha_{n}\right \}  $ be a sequence of positive real
numbers such that $\lim \limits_{n\rightarrow \infty}\dfrac{\alpha_{n}}{\left[
n\right]  }=1.$

Assume that a function $K_{n,\nu}^{q}\left(  x,t,u\right)  $ ($x,t\in \left[
0,A\right]  ,$ $-\infty<u<\infty$) depending on the three parameters $n,\nu$
and $q$ satisfies the following conditions:
\begin{itemize}
\item[({\em i})] The function $K_{n,\nu}^{q}\left(  x,t,u\right)  $ is infinitely
$q$-differentiable with respect to $u$ for fixed $x,t\in \left[  0,A\right]
,$
\item[({\em ii})] For any $x\in \left[  0,A\right]  $ and for all $n\in
\mathbb{N},$%
\[
\sum \limits_{\nu=0}^{\infty}q^{\frac{\nu \left(  \nu-1\right)  }{2}}\left[
D_{q,u}^{\nu}K_{n,\nu}^{q}\left(  x,t,u\right)  |_{\substack{u=\alpha_{n}%
\psi_{n}\left(  t\right)  \\t=0}}\right]  \frac{(-q\alpha_{n}\psi_{n}\left(
0\right)  )^{\nu}}{[\nu]!}=1,
\]
\item[({\em iii})] $\left \{  \left(  -1\right)  ^{\nu}D_{q,u}^{\nu}K_{n,\nu}%
^{q}\left(  x,t,u\right)  |_{\substack{u=\alpha_{n}\psi_{n}\left(  t\right)
\\t=0}}\right \}  \geq0$ $\left(  \nu,n=1,2,\ldots, x\in \left[  0,A\right]
\right)  ,$

(This notation means that if the $q$-derivative with respect to $u$ is taken
$\nu$ times, then one sets $u=\alpha_{n}\psi_{n}\left(  t\right)  $ and $t=0$.)
\item[({\em iv})] $D_{q,u}^{\nu}K_{n,\nu}^{q}\left(  x,t,u\right)
|_{\substack{u=\alpha_{n}\psi_{n}\left(  t\right)  \\t=0}}=-\left[  n\right]
xq^{1-\nu}\left[  D_{q,u}^{\nu-1}K_{n+m,\nu-1}^{q}\left(  x,t,u\right)
|_{\substack{u=\alpha_{n}\psi_{n}\left(  t\right)  \\t=0}}\right]  $

$\left(  \nu,n=1,2,\ldots,  x\in \left[  0,A\right]  \right)  ,$ where $m$
is a number such that $m+n$ is zero or a natural number.
\end{itemize}
According to these conditions, the $q$-Ibragimov-Gadjiev operators have the
following form:
\begin{equation}
L_{n}\left(  f;q;x\right)  =\!\sum \limits_{\nu=0}^{\infty}q^{\frac{\nu \left(
\nu-1\right)  }{2}}f\left(  \frac{[\nu]}{[n]^{2}\psi_{n}\left(  0\right)
}\right)  \!\!\left[  D_{q,u}^{\nu}K_{n,\nu}^{q}\left(  x,t,u\right)
|_{\substack{u=\alpha_{n}\psi_{n}\left(  t\right)  \\t=0}}\right]\!\!
\frac{(-q\alpha_{n}\psi_{n}\left(  0\right)  )^{\nu}}{[\nu]!}\label{5}%
\end{equation}
for $x\in \mathbb{R}_{+\text{ }}$and any function $f$ defined on $\mathbb{R}%
_{+\text{ }}.$

By $\nu-$multiple application of property $\left(  iv\right)  ,$ $L_{n}\left(
f;q;x\right)  $ can be reduced to the form:%
\begin{align}
L_{n}\left(  f;q;x\right)     =&\sum \limits_{\nu=0}^{\infty}f\left(  \frac
{[\nu]}{[n]^{2}\psi_{n}\left(  0\right)  }\right)  \frac
{[n][n+m][n+2m]\cdots\left[  n+\left(  \nu-1\right)  m\right]  }{[\nu
]!}\nonumber \\
& \times(xq\alpha_{n}\psi_{n}\left(  0\right)  )^{\nu}K_{n+\nu m,0}^{q}\left(
x,0,\alpha_{n}\psi_{n}\left(  0\right)  \right)  .\label{6}%
\end{align}
Note that for $q=1$, the operators $L_{n}\left(  f;q;.\right)  $ are classical
Ibragimov-Gadjiev operators. In $\left[  9\right]  $,  the authors showed that
by taking $q_{n}\in \left(  0,1\right)  $ such that $\lim \limits_{n\rightarrow
\infty}q_{n}=1,$ $L_{n}\left(  f;q;.\right)  $ satisfies the conditions of
the Bohman-Korovkin theorem, and obtained the other convergence properties on the
space $C\left[  0,A\right]  .$ Furthermore, they showed that  in special cases, this sequence of
operators consist of $q$-positive linear operators. For
instance, by choosing $K_{n,\nu}^{q}\left(  x,t,u\right)  =\left(
1-\dfrac{q^{1-\nu}ux}{1+t}\right)  _{q}^{n}$ \ and $\alpha_{n}=\dfrac{\left[
n\right]  }{q},$ $\psi_{n}\left(  0\right)  =\dfrac{1}{\left[  n\right]  },$
the operators defined by \eqref{5} are transformed into $q$-Bernstein
polynomials which were introduced by Phillips \cite{r15}, for
$\alpha_{n}=\dfrac{\left[  n\right]  }{q},$ $\psi_{n}\left(  0\right)
=\dfrac{1}{\left[  n\right]  b_{n}}$ $(\lim \limits_{n\rightarrow \infty}%
b_{n}=\infty,\lim \limits_{n\rightarrow \infty}\dfrac{b_{n}}{\left[  n\right]
}=0)$ the operators become $q-$Chlodowsky polynomials defined by Karsl\i \ and
Gupta \cite{r14}. For $K_{n,\nu}^{q}\left(  x,t,u\right)
=E_{q}\left(  -\left[  n\right]  \left(  t+q^{1-\nu}ux\right)  \right)  ,$
$\alpha_{n}=\dfrac{\left[  n\right]  }{q},$ $\psi_{n}\left(  0\right)
=\dfrac{1}{[n]b_{n}}$ $\left(  \lim \limits_{n\rightarrow \infty}b_{n}%
=\infty,\lim \limits_{n\rightarrow \infty}\dfrac{b_{n}}{\left[  n\right]
}=0\right)  $, the $q-$analogue of the classical Sz\'{a}sz-Mirakjan operators,
which were defined by Aral \cite{r5}, can be obtained.

The aim of this paper is to establish Korovkin type theorems for continuous and
unbounded functions defined on $\left[  0,\infty \right)  $ by the
$q$-Ibragimov-Gadjiev operators. We also give some representations of these
operators using the $q$-differences and divided differences.


\section{Preliminary results}

 We define a suitable q-difference operator as follows:%
\begin{equation}
\triangle_{q}^{0}\bigskip f_{\nu}=f_{\nu},\text{ }\Delta_{q}^{r}f_{\nu}%
=\Delta_{q}^{r-1}f_{\nu+1}-q^{r-1}\Delta_{q}^{r-1}f_{\nu}\text{ , }%
r\geq1,\label{7}%
\end{equation}
where $f_{\nu}=f\left(  \frac{[\nu]}{[n]^{2}\psi_{n}\left(  0\right)
}\right)  ,$ $\nu \in%
%TCIMACRO{\U{2115} }%
%BeginExpansion
\mathbb{N}
%EndExpansion
_{0}$.

Now, suppose that in
addition to conditions $\left(  i\right)  -\left(  iv\right)  $ function $K_{n,\nu}^{q}\left(  x,t,u\right)  $   satisfies
\begin{itemize}
\item[({\em v})] $K_{n,\nu}^{q}\left(  x,t,u\right)  $ is infinitely $q-
$ differentiable with respect to $x$ for fixed $t\in \left[  0,A\right]  ,$
$u\in%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$
\[
\text{ }D_{q,x}K_{n,\nu}^{q}\left(  x,t,u\right)  |_{\substack{u=\alpha
_{n}\psi_{n}\left(  t\right)  \\t=0}}=-\left[  n\right]  \alpha_{n}\psi
_{n}\left(  0\right)  q^{1-\nu}K_{n+m,\nu-1}^{q}\left(  x,0,\alpha_{n}\psi
_{n}\left(  0\right)  \right),
\]
where $m$ is the natural number defined in $\left(  iv\right)$,
\item[({\em vi})] $K_{n,\nu}^{q}(0,0,u)=1$ \ for all $n,\nu$ and $u\in%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
.$

\end{itemize}

In what follows we take another look at our operators.

\bigskip

\begin{theorem} Let $q\in \left(  0,1\right)  $. The generalized
q-Ibragimov-Gadjiev operators may be expressed in the form:%
\begin{align*}
L_{n}\left(  f;q;x\right)     =&\sum \limits_{\nu=0}^{\infty}\triangle_{q}^{\nu
}\bigskip f_{0}\frac{[n][n+m]\cdots[n+\left(  \nu-1\right)  m]}{[\nu]!}%
(q\alpha_{n}\psi_{n}\left(  0\right)  )^{\nu}\nonumber \\
& \times K_{n+\nu m,-\nu}^{q}\left(  0,0,\alpha_{n}\psi_{n}\left(  0\right)
\right)  x^{\nu},
\end{align*}
$x\in \left[  0,A\right]  $, where $\triangle_{q}^{\nu}\bigskip f_{0}$ is defined
as in \eqref{7}.
\end{theorem}

\begin{proof} Applying a $q$-derivative operator to \eqref{6} and
taking into account product rule \eqref{2} and property
$\left(  v\right)$, we have:
\begin{align*}
&D_{q,x}L_{n}\left(  f;q;x\right) \\
&\quad =\sum \limits_{\nu=0}^{\infty}\!\!\triangle_{q}^{1}  f_{\nu}\!\frac{[n][n+m]\cdots\left[  n+\nu m\right]  }{[\nu]!}(q\alpha_{n}\psi_{n}\left(
0\right)  )^{\nu+1}K_{n+(\nu+1)m,-1}^{q}\!\!\left(  x,0,\alpha_{n}\psi_{n}\left(
0\right)  \right)  x^{\nu}
\end{align*}
for $n\in%
%TCIMACRO{\U{2115} }%
%BeginExpansion
\mathbb{N}
%EndExpansion
$ and $x\in \left[  0,A\right]$. By induction with respect to $r$ we can prove:
\begin{align*}
&D_{q,x}^{r}L_{n}\left(  f;q;x\right)  \\
&\quad =\sum \limits_{\nu=0}^{\infty}\triangle_{q}^{r} f_{\nu}%
\frac{[n][n+m]\cdots [n+\left(  \nu+r-1\right)  m]}{[\nu]!}(q\alpha_{n}\psi
_{n}\left(  0\right)  )^{\nu+r}\\
& \qquad \times K_{n+(\nu+r)m,-r}^{q}\left(  x,0,\alpha_{n}%
\psi_{n}\left(  0\right)  \right)  x^{\nu}.
\end{align*}
From the expansion of the above series by choosing $x=0,$ we observe that all
 terms except the first one are zero,
\begin{align*}
 D_{q,x}^{r}L_{n}\left(  f;q;0\right)   =&\triangle_{q}^{r} f_{0}\frac{[n][n+m]\cdots [n+\left(  r-1\right)
m]}{[0]!}(q\alpha_{n}\psi_{n}\left(  0\right)  )^{r}\\
&   \times K_{n+rm,-r}^{q}\left(
0,0,\alpha_{n}\psi_{n}\left(  0\right)  \right)  ,r=0,1,2,\ldots
\end{align*}

Hence, choosing $a=0$ in Theorem 1, we obtain
\begin{align}
L_{n}\left(  f;q;x\right)    =&\sum \limits_{\nu=0}^{\infty}\triangle_{q}^{\nu
}\bigskip f_{0}\frac{[n][n+m]\cdots [n+\left(  \nu-1\right)  m]}{[\nu]!}%
(q\alpha_{n}\psi_{n}\left(  0\right)  )^{\nu}\nonumber \\
& \times K_{n+(\nu)m,-\nu}^{q}\left(  0,0,\alpha_{n}\psi_{n}\left(  0\right)
\right)  x^{\nu},\label{9}%
\end{align}
which completes the proof.
\end{proof}

Now, we will give another representation of operators \eqref{5} in terms of divided differences, using the following theorem given in
\cite[p. 44]{r16}. Here, $\left[  x_{0,}x_{1,}%
\ldots, x_{k;};f\right]  $ denotes the divided difference of the function $f$ with
respect to distinct points in the domain of $f$,
\begin{align*}
\left[  x_{0};f\right]  =&f\left(  x_{0}\right)  ,\left[  x_{0},x_{1};f\right]
=\frac{f\left(  x_{1}\right)  -f\left(  x_{0}\right)  }{x_{1}-x_{0}},\ldots ,
\\
\left[  x_{0},x_{1},\ldots ,x_{k};f\right]  =&\frac{\left[  x_{1},\ldots,x_{k}%
;f\right]  -\left[  x_{0},\ldots,x_{k-1};f\right]  }{x_{k}-x_{0}}.
\end{align*}

\begin{theorem} For all $j,k\geq0,$%
\[
f\left[  x_{j},x_{j+1,}\ldots ,x_{j+k}\right]  =\frac{\Delta_{q}^{k}f_{j}%
}{q^{k\left(  2j+k-1\right)  /2}\left[  k\right]  !},%
\]
where $x_{j}=\left[  j\right]  $ and $\left[  k\right]  !=\left[  k\right]
\left[  k-1\right]  \cdots\left[  1\right]  .$
\end{theorem}

\bigskip

\begin{corollary} The $q$-Ibragimov-Gadjiev operators can be represented as:%
\begin{align}
L_{n}\left(  f;q;x\right)     =&\sum \limits_{\nu=0}^{\infty}\frac{(q\alpha
_{n}x)^{\nu}}{\left[  n\right]  ^{2\nu}}\left[  0,\frac{[1]}{[n]^{2}\psi
_{n}\left(  0\right)  },\frac{[2]}{[n]^{2}\psi_{n}\left(  0\right)
},\ldots ,\frac{[\nu]}{[n]^{2}\psi_{n}\left(  0\right)  };f\right] \nonumber \\
& \times q^{\frac{\left(  \nu-1\right)  \nu}{2}}[n][n+m]\cdots[n+\left(
\nu-1\right)  m]K_{n+\nu m,-\nu}^{q}\left(  0,0,\alpha_{n}\psi_{n}\left(
0\right)  \right)  .\  \nonumber
\end{align}
\end{corollary}

\begin{proof} The proof is obvious from  equality \eqref{9} and Theorem 3.
\end{proof}

\begin{remark} It will be noted that for $q=1,$ this is the
result obtained by Ibragimov and Gadjiev \cite{r10} for the
operators defined by \eqref{1}.
\end{remark}


\begin{lemma} For the operators defined by \eqref{5}, the
following equalities hold:
\begin{align}
L_{n}\left(  1;q;x\right)  =&1,\label{10}
\\
L_{n}\left(  t;q;x\right)  =&\frac{q\alpha_{n}}{\left[  n\right]  }x,\label{11}\\
L_{n}\left(  t^{2};q;x\right)  =&\left(  \frac{q\alpha_{n}}{\left[  n\right]
}x\right)  ^{2}\frac{q\left[  n+m\right]  }{[n]}+\left(  \frac{q\alpha_{n}}{\left[  n\right]  }x\right)  \frac{1}{[n]^{2}\psi_{n}\left(  0\right)
},\label{12}%
\\
L_{n}\left(  t^{3};q;x\right)     =&\frac{q^{6}\alpha_{n}^{3}\left[
n+m\right]  \left[  n+2m\right]  x^{3}}{[n]^{5}}+\frac{\left(  q^{4}%
+2q^{3}\right)  \alpha_{n}^{2}\left[  n+m\right]  x^{2}}{[n]^{5}\psi
_{n}\left(  0\right)  }  +\frac{q\alpha_{n}x}{[n]^{5}\psi_{n}^{2}\left(  0\right)  },\label{13}\\
L_{n}\left(  t^{4};q;x\right)     =&\frac{q^{10}\alpha_{n}^{4}\left[
n+m\right]  \left[  n+2m\right]  \left[  n+3m\right]  x^{4}}{[n]^{7}%
}\nonumber \\
&+\frac{\left(  q^{8}+2q^{7}+3q^{6}\right)  \alpha_{n}^{3}\left[  n+m\right]
\left[  n+2m\right]  x^{3}}{[n]^{7}\psi_{n}\left(  0\right)  }\nonumber \\
& +\frac{\left(  q^{5}+3q^{4}+3q^{3}\right)  \alpha_{n}^{2}\left[  n+m\right]
x^{2}}{[n]^{7}\psi_{n}^{2}\left(  0\right)  }+\frac{q\alpha_{n}x}{[n]^{7}%
\psi_{n}^{3}\left(  0\right)  }.\label{14}%
\end{align}


\end{lemma}

\begin{proof} For $s=0,1,2$, the equalities of $L_{n}\left(  t^{s}%
;q;.\right)  $ were obtained in view of the definition of the operators
defined by \eqref{5} in \cite{r9}. So this part of proof
is omitted. Hence for $s=3,4$, we will prove this lemma in terms of  $q$-differences. Using  equality \eqref{7} for $s=3$ one has%
\begin{align*}
\triangle_{q}^{0}\bigskip f_{0}=&f_{0}=0
\\
\triangle_{q}^{1}\bigskip f_{0}=&f_{1}-f_{0}=\left(  \frac{1}{[n]^{2}\psi
_{n}\left(  0\right)  }\right)  ^{3}%
\\
\triangle_{q}^{2}\bigskip f_{0}=&f_{2}-\left(  1+q\right)  f_{1}+q\bigskip
f_{0}=\left(  \frac{[2]}{[n]^{2}\psi_{n}\left(  0\right)  }\right)
^{3}-\left(  1+q\right)  \left(  \frac{1}{[n]^{2}\psi_{n}\left(  0\right)
}\right)  ^{3}%
\\
\triangle_{q}^{3}\bigskip f_{0}    =&f_{3}-\left(  1+q+q^{2}\right)
f_{2}+\left(  q+q^{2}+q^{3}\right)  f_{1}-q^{3}f_{0}\\
 =&\left(  \frac{[3]}{[n]^{2}\psi_{n}\left(  0\right)  }\right)  ^{3}-\left(
1+q+q^{2}\right)  \left(  \frac{[2]}{[n]^{2}\psi_{n}\left(  0\right)
}\right)  ^{3}\\
& \mathbf{+}\left(  q+q^{2}+q^{3}\right)  \left(  \frac{1}{[n]^{2}\psi
_{n}\left(  0\right)  }\right)  ^{3}.%
\end{align*}
By using Theorem 2, we can write:%
\begin{align*}
L_{n}\left(  t^{3};q;x\right)     =&\triangle_{q}^{0}\bigskip f_{0}%
+\triangle_{q}^{1}\bigskip f_{0}\frac{[n]}{[1]!}q\alpha_{n}\psi_{n}\left(
0\right)  K_{n+m,-1}^{q}\left(  0,0,\alpha_{n}\psi_{n}\left(  0\right)
\right)  x+\\
& +\triangle_{q}^{2}\bigskip f_{0}\frac{[n][n+m]}{[2]!}(q\alpha_{n}\psi
_{n}\left(  0\right)  )^{2}K_{n+2m,-2}^{q}\left(  0,0,\alpha_{n}\psi
_{n}\left(  0\right)  \right)  x^{2}+\\
& +\triangle_{q}^{3}\bigskip f_{0}\frac{[n][n+m][n+2m]}{[3]!}(q\alpha_{n}%
\psi_{n}\left(  0\right)  )^{3}K_{n+3m,-3}^{q}\left(  0,0,\alpha_{n}\psi
_{n}\left(  0\right)  \right)  x^{3}.
\end{align*}
Taking into account  condition $\left(  vi\right)  $, by a straightforward
calculation, we obtain  equality \eqref{13}.

We now prove \eqref{14}. For $s=4$, we have:%
\begin{align*}
\triangle_{q}^{0}\bigskip f_{0}=&f_{0}=0
\\
\triangle_{q}^{1}\bigskip f_{0}=&f_{1}-f_{0}=\left(  \frac{1}{[n]^{2}\psi
_{n}\left(  0\right)  }\right)  ^{4}%
\\
\triangle_{q}^{2}\bigskip f_{0}=&f_{2}-\left(  1+q\right)  f_{1}+q
f_{0}=\left(  \frac{[2]}{[n]^{2}\psi_{n}\left(  0\right)  }\right)
^{4}-\left(  1+q\right)  \left(  \frac{1}{[n]^{2}\psi_{n}\left(  0\right)
}\right)  ^{4}%
\\
\triangle_{q}^{3}\bigskip f_{0}    =&f_{3}-\left(  1+q+q^{2}\right)
f_{2}+\left(  q+q^{2}+q^{3}\right)  f_{1}-q^{3}f_{0}\\
 =&\left(  \frac{[3]}{[n]^{2}\psi_{n}\left(  0\right)  }\right)  ^{4}-\left(
1+q+q^{2}\right)  \left(  \frac{[2]}{[n]^{2}\psi_{n}\left(  0\right)
}\right)  ^{4}+\\
& \mathbf{+}\left(  q+q^{2}+q^{3}\right)  \left(  \frac{1}{[n]^{2}\psi
_{n}\left(  0\right)  }\right)  ^{4}%
\\
\Delta_{q}^{4}f_{0}    =&f_{4}-\left(  1+q+q^{2}+q^{3}\right)  f_{3}+\left(
q+q^{2}+2q^{3}+q^{4}+q^{5}\right)  f_{2}\\
& -\left(  q^{3}+q^{4}+q^{5}+q^{6}\right)  f_{1}+q^{6}f_{0}\text{ }\\
  =&\left(  \frac{[4]}{[n]^{2}\psi_{n}\left(  0\right)  }\right)  ^{4}-\left(
1+q+q^{2}+q^{3}\right)  \left(  \frac{[3]}{[n]^{2}\psi_{n}\left(  0\right)
}\right)  ^{4} \\
& +\left(  q+q^{2}+2q^{3}+q^{4}+q^{5}\right) \!\! \left(  \frac{[2]}{[n]^{2}%
\psi_{n}\left(  0\right)  }\right)  ^{4}\!\!\!\!-\left(  q^{3}+q^{4}+q^{5}%
+q^{6}\right)  \!\!\left(  \frac{1}{[n]^{2}\psi_{n}\left(  0\right)  }\right)
^{4}.%
\end{align*}
Thus applying again Theorem 2 and  using   condition $\left(  vi\right)
,$ proceeding similarly, we get \eqref{14}.
\end{proof}

\section{Weighted approximation}

This section is devoted to obtaining weighted approximation properties of the
$q$-Ibragimov-Gadjiev operators. Note that we will use a weighted Korovkin  type
theorem proved by Gadjiev \cite{r7}. When considering that theorem we
should mention some notations: Let $\rho \left(  x\right)  =1+x^{2}$ be a
weight function and $B_{\rho}\left[  0,\infty \right)  $  the set of all
functions $f$ defined on the semi-axis $\left[  0,\infty \right)  $ satisfying
the condition $\left \vert f\left(  x\right)  \right \vert \leq M_{f}\rho \left(
x\right)  $, where $M_{f}$ is a constant depending only on $f.$ $C_{\rho
}\left[  0,\infty \right)  $ denotes the subspace of all continuous functions
belonging to $B_{\rho}\left[  0,\infty \right)  $ and $C_{\rho}^{k}\left[
0,\infty \right)  $ denotes the subspace of all functions $f\in$ $C_{\rho
}\left[  0,\infty \right)  $ with $\lim \limits_{\left \vert x\right \vert
\rightarrow \infty}\frac{f\left(  x\right)  }{\rho \left(  x\right)  }%
=k_{f}<\infty.$ Obviously $C_{\rho}^{k}\left[  0,\infty \right)  $ is a linear
normed space with the $\rho-$norm:%
\[
\left \Vert f\right \Vert _{\rho}=\sup_{x\in \left[  0,b_{n}\right)  }%
\frac{\left \vert f\left(  x\right)  \right \vert }{\rho \left(  x\right)  },%
\]
where $\left \{  b_{n}\right \}  $ is a sequence of positive numbers, which
has a finite or an infinite limit.



\begin{theorem}[{see \cite{r7}}] Let $\left \{  T_{n}\right \}  $ be a sequence of linear
positive operators which are mappings from $C_{\rho}$ into $B_{\rho} $
\ satisfying the conditions:%
\[
\lim_{n\rightarrow \infty}\left \Vert T_{n}(t^{k};x)-x^{k}\right \Vert _{\rho
}=0,\ k=0,1,2.
\]
Then, for any function $f\in C_{\rho}^{k},$%
\[
\lim_{n\rightarrow \infty}\left \Vert T_{n}f-f\right \Vert _{\rho}=0.
\]
\end{theorem}


In order to study convergence properties of  operators \eqref{5}
into the weighted approximation fields we consider the following modified form
of the $q$-Ibragimov-Gadjiev operators.

Let $\left\{  b_{n}\right\}  $ be a sequence of positive numbers, which has
a finite or an infinite limit and $\left \{  \psi_{n}\left(  t\right)  \right \}  $
be a sequence of functions in $C\left[  0,b_{n}\right]  $ such that
$\psi_{n}\left(  t\right)  >0$ for each $t\in \left[  0,b_{n}\right]  ,$
satisfying the condition:%
\begin{equation}
\lim \limits_{n\rightarrow \infty}\dfrac{1}{[n]^{2}\psi_{n}\left(  0\right)
}=0.\label{15}%
\end{equation}
Also, let $\left \{  \alpha_{n}\right \}  $ be a sequence of positive real
numbers such that
\begin{equation}
\dfrac{\alpha_{n}}{\left[  n\right]  }=1+O\left(  \dfrac{1}{\left[  n\right]
^{2}\psi_{n}\left(  0\right)  }\right)  .\label{16}%
\end{equation}
Now, assume that a function $K_{n,\nu}^{q}\left(  x,t,u\right)  $
($x,t\in \left[  0,b_{n}\right]  ,$ $-\infty<u<\infty$) depending on the three
parameters $n,\nu$ and $q$ satisfies the following conditions:
\begin{itemize}
\item[(${i}^{o}$)] The function $K_{n,\nu}^{q}\left(  x,t,u\right)  $
is infinitely $q$-differentiable with respect to $u$ for fixed $x,t\in \left[
0,b_{n}\right]$,
\item[($\mathit{ii}^{o}$)] For any $x\in \left[  0,b_{n}\right]  $ and for all
$n\in \mathbb{N},$%
\[
\sum \limits_{\nu=0}^{\infty}q^{\frac{\nu \left(  \nu-1\right)  }{2}}\left[
D_{q,u}^{\nu}K_{n,\nu}^{q}\left(  x,t,u\right)  |_{\substack{u=\alpha_{n}%
\psi_{n}\left(  t\right)  \\t=0}}\right]  \frac{(-q\alpha_{n}\psi_{n}\left(
0\right)  )^{\nu}}{[\nu]!}=1,
\]
\item[($\mathit{iii}^{o}$)] $\left \{  \left(  -1\right)  ^{\nu}D_{q,u}^{\nu
}K_{n,\nu}^{q}\left(  x,t,u\right)  |_{\substack{u=\alpha_{n}\psi_{n}\left(
t\right)  \\t=0}}\right \}  \geq0$ $\left(  \nu,n=1,2,\ldots,\text{ }x\in \left[
0,b_{n}\right]  \right)  ,$
\item[($\mathit{iv}^{o}$)] $D_{q,u}^{\nu}K_{n,\nu}^{q}\left(  x,t,u\right)
|_{\substack{u=\alpha_{n}\psi_{n}\left(  t\right)  \\t=0}}=-\left[  n\right]
xq^{1-\nu}\left[  D_{q,u}^{\nu-1}K_{n+m,\nu-1}^{q}\left(  x,t,u\right)
|_{\substack{u=\alpha_{n}\psi_{n}\left(  t\right)  \\t=0}}\right]  $

$\left(  \nu,n=1,2,\ldots, x\in \left[  0,b_{n}\right]  \right)  ,$ where
$m $ is a number such that $m+n$ is zero or a natural number.
\end{itemize}
Under above conditions we call operators \eqref{5}
$L_{n}^{\ast}\left(  f;q;.\right)  .$ It is clear that $L_{n}^{\ast}\left(
f;q;.\right)  $ holds for the equalities in Lemma 1. In the case
$\lim \limits_{n\rightarrow \infty}b_{n}=A$ and $\lim \limits_{n\rightarrow
\infty}\dfrac{\alpha_{n}}{\left[  n\right]  }$ =1 we obtain
$q$-\.{I}bragimov-Gadjiev operators defined by \eqref{5}.


\begin{theorem} Let $q=q_{n}$, $0<q_{n}<1$, satisfy the condition
($q_{n})\rightarrow1$ as $n\rightarrow \infty.$ Then for each function $f\in
C_{\rho}^{k}\left[  0,\infty \right)  ,$%
\[
\lim_{n\rightarrow \infty}\left \Vert L_{n}^{\ast}\left(  f;q;.\right)
-f\right \Vert _{\rho,\left[  0,b_{n}\right]  }=0.
\]
\end{theorem}

\begin{proof} Making use of the Korovkin type theorem on weighted
approximation, it is sufficient to verify the following three conditions:%
\begin{equation}
\lim_{n\rightarrow \infty}\left \Vert L_{n}^{\ast}(t^{k};q;x)-x^{k}\right \Vert
_{\rho}=0,\ k=0,1,2.\label{17}%
\end{equation}

From  equality \eqref{10} clearly $\left \Vert L_{n}\left(
1,q_{n},x\right)  -1\right \Vert _{\rho}\rightarrow0$ as $n\rightarrow \infty$
on $\left[  0,b_{n}\right]$. From  equalities \eqref{11}
and \eqref{15} we have:%
\begin{align*}
\sup_{x\in \left[  0,b_{n}\right)  }\frac{\left \vert L_{n}^{\ast}\left(
t;q;x\right)  -x\right \vert }{1+x^{2}}  & \leq \left \vert \frac{q\alpha_{n}%
}{\left[  n\right]  }-1\right \vert \sup_{x\in \left[  0,b_{n}\right)  }\frac
{x}{1+x^{2}}\\
& \leq \left \vert \frac{q\alpha_{n}}{\left[  n\right]  }-1\right \vert,
\end{align*}
hence the condition in \eqref{17} holds for $k=1$.

Similarly, using   equality \eqref{12} for $k=2$ we can write: %
\begin{align*}
\sup_{x\in \left[  0,b_{n}\right)  }\frac{\left \vert L_{n}^{\ast}\left(
t^{2};q;x\right)  -x^{2}\right \vert }{1+x^{2}}   \leq &\left \vert \left(
\frac{q\alpha_{n}}{\left[  n\right]  }\right)  ^{2}\frac{q\left[  n+m\right]
}{\left[  n\right]  }-1\right \vert \sup_{x\in \left[  0,b_{n}\right)  }%
\frac{x^{2}}{1+x^{2}}\\
& +\frac{q\alpha_{n}}{\left[  n\right]  }\frac{1}{[n]^{2}\psi_{n}\left(
0\right)  }\sup_{x\in \left[  0,b_{n}\right)  }\frac{x}{1+x^{2}}\\
  \leq& \left \vert \left(  \frac{q\alpha_{n}}{\left[  n\right]  }\right)
^{2}\frac{q\left[  n+m\right]  }{\left[  n\right]  }-1\right \vert
+\frac{q\alpha_{n}}{\left[  n\right]  }\frac{1}{[n]^{2}\psi_{n}\left(
0\right)  },%
\end{align*}
which implies that \eqref{17}  also holds for $k=2$. Therefore, the
desired result follows from Theorem 4.
\end{proof}

Now, we will give an estimate concerning the rate of convergence for the
$q$-Ibragimov-Gadjiev operators. As  known, the first modulus of continuity
$\omega \left(  f;\delta \right)  $ does not tend to zero as $\delta
\rightarrow0$ on the infinite interval. For this reason, we consider%
\[
\Omega \left(  f;\delta \right)  =\sup_{\left \vert h\right \vert \leq \delta
,x\geq0}\frac{\left \vert f\left(  x+h\right)  -f\left(  x\right)  \right \vert
}{\left(  1+h^{2}\right)  \left(  1+x^{2}\right)  }%
\]
a weighted modulus of smoothness associated to the space $C_{\rho}^{k}\left[
0,\infty \right)  .$

$\Omega \left(  f;\delta \right)  $ possesses the following properties (see \cite{r12}):
\begin{itemize}
\item[-] $\Omega \left(  f;\delta \right)  $ is a monotonically increasing function of
$\delta,$ $\delta \geq0$,
\item[-]
for every $f\in$ $C_{\rho}^{k}\left[  0,\infty \right)  ,$ $\lim \limits_{\delta
\rightarrow0}\Omega \left(  f;\delta \right)  =0$,
\item[-] for each positive value of $\lambda$%
\begin{equation}
\Omega \left(  f;\lambda \delta \right)  \leq2\left(  1+\lambda \right)  \left(
1+\delta^{2}\right)  \Omega \left(  f;\delta \right)  .\label{18}%
\end{equation}
\end{itemize}
From  inequality \eqref{18} and the definition of
$\Omega \left(  f;\delta \right)  $ we get%
\begin{equation}
\left \vert f\left(  t\right)  -f\left(  x\right)  \right \vert \leq2\left(
1+\frac{\left \vert t-x\right \vert }{\delta}\right)  \left(  1+\delta
^{2}\right)  \Omega \left(  f;\delta \right)  \left(  1+x^{2}\right)  \left(
1+\left(  t-x\right)  ^{2}\right) ,\label{19}%
\end{equation}
for every $f\in C_{\rho}^{k}\left[  0,\infty \right)  $ and $x,t\in \left[
0,\infty \right)  .$

\bigskip

\begin{theorem} Let $q=q_{n}$, $0<q_{n}<1$ satisfies the condition
 $q_{n} \rightarrow1$ as $n\rightarrow \infty.$ If $f\in$ $C_{\rho}^{k}\left[
0,\infty \right)  ,$ then the inequality%

\[
\sup_{x\geq0}\frac{\left \vert L_{n}^{\ast}\left(  f;q;x\right)  -f\left(
x\right)  \right \vert }{\left(  1+x^{2}\right)  ^{3}}\leq M\Omega \left(
f;\left(  \left[  n\right]  ^{2}\psi_{n}\left(  0\right)  \right)
^{-1/2}\right)
\]
is satisfied for a sufficiently large $n$, where $M$ is a constant independent of
$\alpha_{n},$ $\psi_{n}\left(  0\right)  .$
\end{theorem}

\begin{proof} From \eqref{19} we can write:%
\begin{align*}
\left \vert L_{n}^{\ast}\left(  f;q;x\right)  -f\left(  x\right)  \right \vert
 \leq&2\left(  1+\delta_{n}^{2}\right)  \Omega \left(  f;\delta_{n}\right)
\left(  1+x^{2}\right)  \sum \limits_{\nu=0}^{\infty}P_{\nu,n}^{q}\left(
x\right) \\
& \times \left(  1+\frac{\left \vert \frac{[\nu]}{[n]^{2}\psi_{n}\left(
0\right)  }-x\right \vert }{\delta_{n}}\right)  \left(  1+\left(  \frac{[\nu
]}{[n]^{2}\psi_{n}\left(  0\right)  }-x\right)  ^{2}\right) \\
  \leq&4\left(  1+x^{2}\right)  \Omega \left(  f;\delta_{n}\right)  \left \{
1+\frac{1}{\delta_{n}}\sum \limits_{\nu=0}^{\infty}\left \vert \frac{[\nu
]}{[n]^{2}\psi_{n}\left(  0\right)  }-x\right \vert P_{\nu,n}^{q}\left(
x\right)  \right. \\
& +\sum \limits_{\nu=0}^{\infty}\left(  \frac{[\nu]}{[n]^{2}\psi_{n}\left(
0\right)  }-x\right)  ^{2}P_{\nu,n}^{q}\left(  x\right) \\
& \left.  +\frac{1}{\delta_{n}}\sum \limits_{\nu=0}^{\infty}\left \vert
\frac{[\nu]}{[n]^{2}\psi_{n}\left(  0\right)  }-x\right \vert \left(
\frac{[\nu]}{[n]^{2}\psi_{n}\left(  0\right)  }-x\right)  ^{2}P_{\nu,n}%
^{q}\left(  x\right)  \right \},
\end{align*}
where $P_{\nu,n}^{q}\left(  x\right)  =q^{\frac{\nu \left(  \nu-1\right)  }{2}%
}\left[  D_{q,u}^{\nu}K_{n,\nu}^{q}\left(  x,t,u\right)  |_{\substack{u=\alpha
_{n}\psi_{n}\left(  t\right)  \\t=0}}\right]  \frac{(-q\alpha_{n}\psi
_{n}\left(  0\right)  )^{\nu}}{[\nu]!}$ and $\delta_{n}>0.$ Applying
the Cauchy-Schwarz inequality we obtain%
\begin{equation}
\left \vert L_{n}^{\ast}\left(  f;q;x\right)  -f\left(  x\right)  \right \vert
\leq4\left(  1+x^{2}\right)  \Omega \left(  f;\delta_{n}\right)  \left(
1+\frac{2}{\delta_{n}}\sqrt{K_{1}}+K_{1}+\frac{1}{\delta_{n}}K_{2}\right),
\label{20}%
\end{equation}
where\bigskip%
\begin{align*}
K_{1}=&\sum \limits_{\nu=0}^{\infty}\left(  \frac{[\nu]}{[n]^{2}\psi_{n}\left(
0\right)  }-x\right)  ^{2}P_{\nu,n}^{q}\left(  x\right)
\\
K_{2}=&\sum \limits_{\nu=0}^{\infty}\left(  \frac{[\nu]}{[n]^{2}\psi_{n}\left(
0\right)  }-x\right)  ^{4}P_{\nu,n}^{q}\left(  x\right).
\end{align*}
Using Lemma 1, we have%
\begin{align*}
K_{1}  & =L_{n}^{\ast}\left(  \left(  t-x\right)  ^{2};q;x\right) \\
& =\left(  1-2q\left(  \frac{\alpha_{n}}{\left[  n\right]  }\right)  +\left(
\frac{\alpha_{n}}{\left[  n\right]  }\right)  ^{2}\frac{q^{3}\left[
n+m\right]  }{[n]}\right)  x^{2}+q\left(  \frac{\alpha_{n}}{\left[  n\right]
}\right)  \frac{1}{[n]^{2}\psi_{n}\left(  0\right)  }x
\end{align*}
and
\begin{align*}
K_{2}=&L_{n}^{\ast}\left(  \left(  t-x\right)  ^{4};q;x\right)  \\
=&\left(  \left(  \frac{\alpha_{n}}{\left[  n\right]  }\right)  ^{4}%
\frac{q^{10}\left[  n+m\right]  \left[  n+2m\right]  \left[  n+3m\right]
}{[n]^{3}}-4\left(  \frac{\alpha_{n}}{\left[  n\right]  }\right)  ^{3}%
\frac{q^{6}\left[  n+m\right]  \left[  n+2m\right]  }{[n]^{2}}\right.\\
&\left.+6\left(
\frac{\alpha_{n}}{\left[  n\right]  }\right)  ^{2}\frac{q^{3}\left[
n+m\right]  }{[n]}-4q\left(  \frac{\alpha_{n}}{\left[  n\right]  }\right)
+1\right)  x^{4}\\
&+\left(  \left(  \frac{\alpha_{n}}{\left[  n\right]  }\right)  ^{3}%
\frac{\left(  q^{8}+2q^{7}+3q^{6}\right)  \left[  n+m\right]  \left[
n+2m\right]  }{[n]^{2}}\frac{1}{[n]^{2}\psi_{n}\left(  0\right)  }\right.\\
&\left. -4\left(
\frac{\alpha_{n}}{\left[  n\right]  }\right)  ^{2}\frac{\left(  q^{4}%
+2q^{3}\right)  \left[  n+m\right]  }{[n]}\frac{1}{[n]^{2}\psi_{n}\left(
0\right)  }+6q\left(  \frac{\alpha_{n}}{\left[  n\right]  }\right)  \frac
{1}{[n]^{2}\psi_{n}\left(  0\right)  }\right)  x^{3}\\
&+\left( \! \left( \! \frac{\alpha_{n}}{\left[  n\right]  }\!\right)  ^{2}%\!\!
\frac{\left(  q^{5}+3q^{4}+3q^{3}\right)  \left[  n+m\right]  }{[n]}\frac
{1}{\left(  [n]^{2}\psi_{n}\left(  0\right)  \right)  ^{2}}-4q\left(
\frac{\alpha_{n}}{\left[  n\right]  }\right)  \frac{1}{\left(  [n]^{2}\psi
_{n}\left(  0\right)  \right)  ^{2}}\right)\!  x^{2}\\
&+q\left(  \frac{\alpha_{n}%
}{\left[  n\right]  }\right)  \frac{1}{\left(  [n]^{2}\psi_{n}\left(
0\right)  \right)  ^{3}}x.
\end{align*}
Using condition \eqref{16} we can write%
\[
K_{1}=O\left(  \frac{1}{[n]^{2}\psi_{n}\left(  0\right)  }\right)  \left(
x^{2}+x\right)
\]
and%
\[
K_{2}=O\left(  \frac{1}{[n]^{2}\psi_{n}\left(  0\right)  }\right)  \left(
x^{4}+x^{3}+x^{2}+x\right),
\]
thus by substituting these equalities \eqref{20} becomes%
\begin{align*}
\left \vert L_{n}^{\ast}\left(  f;q;x\right)  -f\left(  x\right)  \right \vert
  \leq&4\left(  1+x^{2}\right)  \Omega \left(  f;\delta_{n}\right)  \left \{
1+\frac{2}{\delta_{n}}\sqrt{O\left(  \frac{1}{[n]^{2}\psi_{n}\left(  0\right)
}\right)  \left(  x^{2}+x\right)  }\ \  +\right. \\
& \left.  +O\!\left(\!   \frac{1}{[n]^{2}\psi_{n}\left(  0\!\!\right)  }\right)\!
\left(  x^{2}\!+\!x\right) \!+\!\frac{1}{\delta_{n}}O\left( \! \frac{1}{[n]^{2}\psi
_{n}\left(  0\right)  }\!\right)  \left(  x^{4}\!+\!x^{3}\!+\!x^{2}\!+\!x\right)  \!\right \}.
\end{align*}
By choosing $\delta_{n}=\left(  \left[  n\right]  ^{2}\psi_{n}\left(
0\right)  \right)  ^{-1/2},$ for sufficiently large $n$'s, we obtain the desired result.
\end{proof}

\begin{remark} It will be noted that, for $q=1,$ this is the result
obtained by Gadjiev and Ispir \cite{r8} for Ibragimov-Gadjiev
operators defined by \eqref{1}.
\end{remark}

%%%% Acknowledgment %%%%%%%%
\section*{Acknowledgement}
The authors would like to thank the referees for their helpful
suggestions.

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%\end{linenumbers}
\end{document}