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<title>A Voronovskaya type theorem for \(q\)-Szasz-Mirakyan-Kantorovich operators: A Voronovskaya type theorem for \(q\)-Szasz-Mirakyan-Kantorovich operators</title>
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<h1>A Voronovskaya type theorem for \(q\)-Szasz-Mirakyan-Kantorovich operators</h1>
<p class="authors">
<span class="author">Gülen Baþcanbaz-Tunca\(^{1}\) Ayşegül Erençin\(^{2}\)</span>
</p>
<p class="date">January 10, 2011.</p>
</div>
<p>\(^{1}\)Ankara University, Faculty of Science, Department of Mathematics, 06100, Tandoan, Ankara, Turkey, e-mail: <span class="ttfamily">tunca@science.ankara.edu.tr</span>. </p>
<p>\(^{2}\)Abant Ýzzet Baysal University, Faculty of Arts and Sciences, Department of Mathematics, 14280, Bolu, Turkey, e-mail: <span class="ttfamily">erencina@hotmail.com</span>. </p>

<div class="abstract"><p> In this work, we consider a Kantorovich type generalization of <br />\(q\)-Szasz-Mirakyan operators via Riemann type \(q\)-integral and prove a Voronovskaya type theorem by using suitable machinery of \(q\)-calculus. </p>
<p><b class="bf">MSC.</b> 41A25, 41A36 </p>
<p><b class="bf">Keywords.</b> \(q\)-Szasz-Mirakyan-Kantorovich operator, Riemann type \(q\)-integral, Voronovskaya type theorem. </p>
</div>
<h1 id="a0000000002">1 Introduction</h1>
<p> In <span class="cite">
	[
	<a href="#bu" >5</a>
	]
</span>, Butzer introduced the Kantorovich variant of the Szasz-Mirakyan operators as </p>
<div class="displaymath" id="a0000000003">
  \begin{equation*}  K_{n}(f;x)=n{\rm e}^{-nx}\sum ^{\infty }_{k=0}\tfrac {(nx)^{k}}{k!}\int ^{\tfrac {k+1}{n}}_{\tfrac {k}{n}}f(t){\rm d}t;\quad x\geq 0 \end{equation*}
</div>
<p> for each \(n\in \mathbb {N}\) and \(f\in L^{1}(0,\infty )\), the space of integrable functions on unbounded interval \([0,\infty ).\) The \(L^{p}(1{\lt}p{\lt}\infty )\) saturation and inverse theorems for the Szasz-Mirakyan-Kantorovich operators were studied by Totik in <span class="cite">
	[
	<a href="#vt" >19</a>
	]
</span>. Bezier variant of the Szasz-Mirakyan-Kantorovich operators were introduced by Gupta, Vasishtha and Gupta in <span class="cite">
	[
	<a href="#gvg" >8</a>
	]
</span>, where the rate of convergence of these operators for functions of bounded variation was measured. Moreover, approximation properties of these operators for locally bounded functions have been investigated recently by Gupta and Xiao-Ming Zeng in <span class="cite">
	[
	<a href="#gx" >9</a>
	]
</span>. In <span class="cite">
	[
	<a href="#oov" >7</a>
	]
</span>, Duman, Özarslan and Della Vecchia constructed a modified Szasz-Mirakyan-Kantorovich operators and introduced a better error estimation and a Voronovskaya type theorem. Another interesting modification of Szasz-Mirakyan-Kantorovich operators has been carried out by Nowak and Sikorska-Nowak in <span class="cite">
	[
	<a href="#now" >15</a>
	]
</span>, where the operators are defined for Denjoy-Perron integrable functions \(f\). The authors have obtained an estimate for the rate of pointwise convergence for such operators at the Lebesgue Denjoy points of \(f\). </p>
<p>In \(1996\), Phillips’ generalization of Bernstein polynomials by using \(q\)-integers <span class="cite">
	[
	<a href="#ph" >16</a>
	]
</span> has brought up a new point of view to the theory of approximation by linear positive operators. There are numerous works dealing with this approach. Some are in <span class="cite">
	[
	<a href="#ag" >2</a>
	]
</span>, <span class="cite">
	[
	<a href="#ad" >3</a>
	]
</span>, <span class="cite">
	[
	<a href="#aa" >4</a>
	]
</span>, <span class="cite">
	[
	<a href="#dd" >6</a>
	]
</span>, <span class="cite">
	[
	<a href="#hv" >11</a>
	]
</span>, <span class="cite">
	[
	<a href="#ma" >12</a>
	]
</span>, <span class="cite">
	[
	<a href="#mg" >13</a>
	]
</span>, <span class="cite">
	[
	<a href="#ti" >20</a>
	]
</span>, <span class="cite">
	[
	<a href="#he" >21</a>
	]
</span> and references therein. </p>
<p>In this work, we consider a Kantorovich type generalization of \(q\)-Szasz-Mirakyan operators, introduced by Aral in <span class="cite">
	[
	<a href="#aa" >4</a>
	]
</span> with the help of the Riemann type \(q\)-integral defined in <span class="cite">
	[
	<a href="#mrs" >14</a>
	]
</span> and prove a Voronovskaya type theorem. Observe that a different form of \(q\)-Szasz-Mirakyan-Kantorovich operators has been studied in <span class="cite">
	[
	<a href="#mg" >13</a>
	]
</span> by Mahmudov and Gupta. </p>
<h1 id="a0000000004">2 Notation and construction of the operators </h1>
<p>In this section, we first mention some well-known definitions of \(q\)-calculus. Next introduce the \(q\)-Szasz-Mirakyan-Kantorovich operators via Riemann type \(q\)-integral. </p>
<p>Let \(q{\gt}0\) be a fixed real number and \(n\) be a nonnegative integer. The respective definitions of the \(q\)-integer \([n]_{q}\) and the \(q\)-factorial \([n]_{q}!\) are (see <span class="cite">
	[
	<a href="#grr" >1</a>
	]
</span>) </p>
<div class="displaymath" id="a0000000005">
  \begin{equation*}  [n]_{q}=\begin{cases} \displaystyle \tfrac {1-q^{n}}{1-q} & \text{if $q\neq 1$}\\ n & \text{if $q=1$ }\end{cases}, \end{equation*}
</div>
<div class="displaymath" id="a0000000006">
  \begin{equation*}  [n]_{q}!=\begin{cases} \displaystyle [1]_{q}[2]_{q}\cdots [n]_{q}& \text{if $n\geq 1$} \\ 1 & \text{if $n=0$} \end{cases}. \end{equation*}
</div>
<p> The two \(q\)-analogues of the exponential function \({\rm e}^{x}\) are given by </p>
<div class="displaymath" id="a0000000007">
  \begin{equation*}  e_{q}(x)=\sum ^{\infty }_{n=0}\tfrac {1}{[n]_{q}!}x^{n}=\tfrac {1}{((1-q)x;q)_{\infty }};\quad |x|{\lt}\tfrac {1}{1-q},\quad |q|{\lt}1 \end{equation*}
</div>
<p> and </p>
<div class="displaymath" id="a0000000008">
  \begin{equation*}  E_{q}(x)=\sum ^{\infty }_{n=0}\tfrac {q^{\tfrac {n(n-1)}{2}}}{[n]_{q}!}x^{n}=(-(1-q)x;q)_{\infty };\quad x\in \mathbb {R},\quad |q|{\lt}1, \end{equation*}
</div>
<p> where \(\displaystyle (x;q)_{\infty }=\prod ^{\infty }_{k=1}(1-xq^{k-1})\) (see <span class="cite">
	[
	<a href="#kac" >10</a>
	]
</span>). We know that </p>
<div class="displaymath" id="a0000000009">
  \begin{equation*}  e_{q}(x)E_{q}(x)=1 \end{equation*}
</div>
<p> and </p>
<div class="displaymath" id="a0000000010">
  \begin{equation*}  \displaystyle \lim _{q\rightarrow 1^{-}}e_{q}(x)=\lim _{q\rightarrow 1^{-}}E_{q}(x)={\rm e}^{x}. \end{equation*}
</div>
<p>In <span class="cite">
	[
	<a href="#aa" >4</a>
	]
</span>, Aral defined the \(q\)-Szasz-Mirakyan operators as follows: </p>
<div class="equation" id="a0000000011">
<p>
  <div class="equation_content">
    \begin{equation}  S^{q}_{n}(f;x)=E_{q}\left(-[n]_{q}\tfrac {x}{b_{n}}\right) \sum ^{\infty }_{k=0} f\left(\tfrac {[k]_{q}b_{n}}{[n]_{q}}\right)\tfrac {([n]_{q}x)^{k}}{[k]_{q}!(b_{n})^{k}},\quad x\in \left[0,\tfrac {b_{n}}{(1-q)[n]_{q}}\right) \end{equation}
  </div>
  <span class="equation_label">2.5</span>
</p>
</div>
<p> where \(n\in \mathbb {N}\), \(0{\lt}q{\lt}1\), \(f\in C[0,\infty )\) and \((b_{n})\) is an increasing sequence of positive numbers such that \(\displaystyle \lim _{n\rightarrow \infty }b_{n}=\infty .\) He proved a Voronovskaya type theorem using \(q\)-derivatives and also obtained convergence properties of these operators and their derivatives. Note that the operators defined by \((2.1)\) are the \(q\)-extension of the Szasz-Chlodovsky operators introduced in <span class="cite">
	[
	<a href="#zs" >18</a>
	]
</span> by Stypinski. From <span class="cite">
	[
	<a href="#aa" >4</a>
	]
</span> we have </p>
<div class="equation" id="a0000000012">
<p>
  <div class="equation_content">
    \begin{equation}  \begin{split} & S^{q}_{n}(e_{0};x)=1\\ & S^{q}_{n}(e_{1};x)=x\\ & S^{q}_{n}(e_{2};x)=qx^{2}+\tfrac {b_{n}}{[n]_{q}}x\\ & S^{q}_{n}(e_{3};x)=q^{3}x^{3}+(q^{2}+2q) \tfrac {b_{n}}{[n]_{q}}x^{2}+\left(\tfrac {b_{n}}{[n]_{q}}\right)^{2}x\\ & S^{q}_{n}(e_{4};x)\! =\! q^{6}x^{4}\! +\! (q^{5}+2q^{4}\! +\! 3q^{3}) \tfrac {b_{n}}{[n]_{q}}x^{3}\! +\! (q^{3}+3q^{2}+3q)\left(\tfrac {b_{n}}{[n]_{q}}\right)^{2}x^{2}\! +\! \left(\tfrac {b_{n}}{[n]_{q}}\right)^{3}x, \end{split} \end{equation}
  </div>
  <span class="equation_label">2.6</span>
</p>
</div>
<p> where \(e_{m}(t):=t^{m},\) \(m=0,1,2,3,4.\) </p>
<p>Now assume that \(0{\lt}a{\lt}b\), \(0{\lt}q{\lt}1\) and \(f\) is a real-valued function. The \(q\)-Jackson integral of \(f\) over the interval \([0,b]\) and over a general interval \([a,b]\) are defined by (see<span class="cite">
	[
	<a href="#kac" >10</a>
	]
</span>) </p>
<div class="displaymath" id="a0000000013">
  \begin{equation*}  \int ^{b}_{0}f(x){\rm d}_{q}x=(1-q)b\sum ^{\infty }_{j=0}f\left(bq^{j}\right)q^{j} \end{equation*}
</div>
<p> and </p>
<div class="displaymath" id="a0000000014">
  \begin{equation*}  \int ^{b}_{a}f(x){\rm d}_{q}x=\int ^{b}_{0}f(x){\rm d}_{q}x-\int ^{a}_{0}f(x){\rm d}_{q}x, \end{equation*}
</div>
<p> respectively, provided the series converge. It is easily seen that the inequality \(f(x)\geq 0\) for all \(x\in [a,b]\) does not ensure that \(\int ^{b}_{a}f(x){\rm d}_{q}x\geq 0\). This fact may not be convenient in order to guarantee the positivity of our operator. On the other hand, since the \(q\)-Jackson integral contains two infinite sums, in obtaining the \(q\)-analogues of some well-known integral inequalities which are used to estimate the order of approximation of linear positive operators involving \(q\)-Jackson integral, some problems are encountered. Because of these circumstances, we consider the Riemann type \(q\)-integral introduced in <span class="cite">
	[
	<a href="#mrs" >14</a>
	]
</span> </p>
<div class="equation" id="a0000000015">
<p>
  <div class="equation_content">
    \begin{equation}  R_{q}(f;a,b)=\int ^{b}_{a}f(x){\rm d}^{R}_{q}x=(1-q)(b-a)\sum ^{\infty }_{j=0}f\left(a+(b-a)q^{j}\right)q^{j} \end{equation}
  </div>
  <span class="equation_label">2.12</span>
</p>
</div>
<p> to construct a Kantorovich variant of the operator defined by \((2.1).\) </p>
<p>With this motivation, we now introduce the Kantorovich type generalization of the \(q\)-Szasz-Mirakyan operators as follows: </p>
<div class="equation" id="a0000000016">
<p>
  <div class="equation_content">
    \begin{equation}  K_{n,q}(f;x)\! =\! E_{q}\left(\! -[n]_{q}\tfrac {x}{b_{n}}\! \right)\!  \sum ^{\infty }_{k=0}\tfrac {([n]_{q}x)^{k}}{[k]_{q}!(b_{n})^{k}}\tfrac {[n]_{q}}{q^{k}b_{n}} \! \int ^{\tfrac {[k+1]_{q}b_{n}}{[n]_{q}}}_{\tfrac {[k]_{q}b_{n}}{[n]_{q}}}f(t){\rm d}^{R}_{q}t, \end{equation}
  </div>
  <span class="equation_label">2.13</span>
</p>
</div>
<p> where \(x\in \big[0,\tfrac {b_{n}}{(1-q)[n]_{q}}\big)\), \(n\in \mathbb {N}\), \(0{\lt}q{\lt}1\), \(f\) is Riemann type \(q\)-integrable on \(\big[\tfrac {[k]_{q}b_{n}}{[n]_{q}},\tfrac {[k+1]_{q}b_{n}}{[n]_{q}}\big]\) and \((b_{n})\) is an increasing sequence of positive numbers having the property \(\displaystyle \lim _{n\rightarrow \infty }b_{n}=\infty .\) </p>
<h1 id="a0000000017">3 Auxiliary results </h1>
<p> In this section, we give some lemmas which will be used to prove our main result. </p>
<p><div class="lem_thmwrapper " id="a0000000018">
  <div class="lem_thmheading">
    <span class="lem_thmcaption">
    Lemma
    </span>
    <span class="lem_thmlabel">1</span>
  </div>
  <div class="lem_thmcontent">
  <p>Let \(m\) be a nonnegative integer. Then we have </p>
<div class="displaymath" id="a0000000019">
  \begin{equation*}  I_{n,k}(e_{m}):=\int ^{\tfrac {[k+1]_{q}b_{n}}{[n]_{q}}}_{\tfrac {[k]_{q}b_{n}}{[n]_{q}}}t^{m}{\rm d}^{R}_{q}t =\tfrac {q^{k}b_{n}}{[n]_{q}}\sum ^{m}_{l=0}\tbinom {m}{l}\left(\tfrac {[k]_{q}b_{n}}{[n]_{q}}\right)^{m-l}C_{m,l}(q,b_{n}) \end{equation*}
</div>
<p> where \(e_{m}(t):=t^{m}\) and </p>
<div class="equation" id="a0000000020">
<p>
  <div class="equation_content">
    \begin{equation}  C_{m,l}(q,b_{n})=\left(\tfrac {b_{n}}{[n]_{q}}\right)^{l}\sum ^{m-l}_{s=0}\tbinom {m-l}{s}\tfrac {(q-1)^{s}}{[l+s+1]_{q}}. \end{equation}
  </div>
  <span class="equation_label">3.1</span>
</p>
</div>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000021">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>From \((2.3)\) and binomial formula it follows that </p>
<div class="displaymath" id="a0000000022">
  \begin{equation*} \begin{split}  I_{n,k}(e_{m})& =(1-q)\tfrac {q^{k}b_{n}}{[n]_{q}}\sum ^{\infty }_{j=0}\left(\tfrac {[k]_{q}b_{n}}{[n]_{q}} +\left(\tfrac {b_{n}}{[n]_{q}}+(q-1)\tfrac {[k]_{q}b_{n}}{[n]_{q}}\right)q^{j}\right)^{m}q^{j}\\ & =(1-q)\tfrac {q^{k}b_{n}}{[n]_{q}}\sum ^{m}_{i=0}\sum ^{\infty }_{j=0}(q^{j})^{i+1}\tbinom {m}{i} \left(\tfrac {b_{n}}{[n]_{q}}+(q-1)\tfrac {[k]_{q}b_{n}}{[n]_{q}}\right)^{i}\left(\tfrac {[k]_{q}b_{n}}{[n]_{q}}\right)^{m-i}\\ & =\tfrac {q^{k}b_{n}}{[n]_{q}}\sum ^{m}_{i=0}\tbinom {m}{i}\tfrac {1}{[i+1]_{q}}\left(\tfrac {b_{n}}{[n]_{q}}+(q-1 )\tfrac {[k]_{q}b_{n}}{[n]_{q}}\right)^{i} \left(\tfrac {[k]_{q}b_{n}}{[n]_{q}}\right)^{m-i}\\ & =\tfrac {q^{k}b_{n}}{[n]_{q}}\sum ^{m}_{i=0}\tbinom {m}{i}\tfrac {1}{[i+1]_{q}}\left(\tfrac {[k]_{q}b_{n}}{[n]_{q}}\right)^{m-i} \sum ^{i}_{l=0}\tbinom {i}{l}\left(\tfrac {b_{n}}{[n]_{q}}\right)^{l}\left((q-1)\tfrac {[k]_{q}b_{n}}{[n]_{q}}\right)^{i-l}\\ & =\tfrac {q^{k}b_{n}}{[n]_{q}}\sum ^{m}_{i=0}\tbinom {m}{i}\tfrac {1}{[i+1]_{q}}\sum ^{i}_{l=0}\tbinom {i}{l} \left(\tfrac {[k]_{q}b_{n}}{[n]_{q}}\right)^{m-l}\left(\tfrac {b_{n}}{[n]_{q}}\right)^{l}(q-1)^{i-l}. \end{split}\end{equation*}
</div>
<p> Interchanging the order of summations over \(i\) and \(l\) and then taking \(s=i-l\) in the obtained result, one may write </p>
<div class="displaymath" id="a0000000023">
  \begin{equation*} \begin{split}  I_{n,k}(e_{m})& =\tfrac {q^{k}b_{n}}{[n]_{q}}\sum ^{m}_{l=0}\sum ^{m}_{i=l}\tbinom {m}{i}\tbinom {i}{l}\left(\tfrac {[k]_{q}b_{n}}{[n]_{q}}\right)^{m-l} \left(\tfrac {b_{n}}{[n]_{q}}\right)^{l} \tfrac {(q-1)^{i-l}}{[i+1]_{q}}\\ & =\tfrac {q^{k}b_{n}}{[n]_{q}}\sum ^{m}_{l=0}\sum ^{m}_{i=l}\tfrac {m!}{(m-i)!}\tfrac {1}{l!(i-l)!}\left(\tfrac {[k]_{q}b_{n}}{[n]_{q}}\right)^{m-l} \left(\tfrac {b_{n}}{[n]_{q}}\right)^{l} \tfrac {(q-1)^{i-l}}{[i+1]_{q}}\\ & =\tfrac {q^{k}b_{n}}{[n]_{q}}\sum ^{m}_{l=0}\sum ^{m-l}_{s=0}\tfrac {m!}{l!(m-l)!}\tfrac {(m-l)!}{s!(m-l-s)!}\left(\tfrac {[k]_{q}b_{n}}{[n]_{q}}\right)^{m-l} \left(\tfrac {b_{n}}{[n]_{q}}\right)^{l}\tfrac {(q-1)^{s}}{[l+s+1]_{q}}=\end{split}\end{equation*}
</div>
<div class="displaymath" id="a0000000024">
  \begin{equation*} \begin{split} & =\tfrac {q^{k}b_{n}}{[n]_{q}}\sum ^{m}_{l=0}\tbinom {m}{l}\left(\tfrac {[k]_{q}b_{n}}{[n]_{q}}\right)^{m-l}\left(\tfrac {b_{n}}{[n]_{q}}\right)^{l} \sum ^{m-l}_{s=0}\tbinom {m-l}{s}\tfrac {(q-1)^{s}}{[l+s+1]_{q}}\\ & =\tfrac {q^{k}b_{n}}{[n]_{q}}\sum ^{m}_{l=0}\tbinom {m}{l}\left(\tfrac {[k]_{q}b_{n}}{[n]_{q}}\right)^{m-l} C_{m,l}(q,b_{n}), \end{split}\end{equation*}
</div>
<p> where each \(C_{m,l}(q,b_{n})\) is given by \((3.1).\) Thus the proof is completed. </p>
<p>Now, by means of the Lemma 1 we can evaluate \(K_{n,q}(e_{m};x).\) <div class="lem_thmwrapper " id="a0000000025">
  <div class="lem_thmheading">
    <span class="lem_thmcaption">
    Lemma
    </span>
    <span class="lem_thmlabel">2</span>
  </div>
  <div class="lem_thmcontent">
  <p>Let \(m\) be a nonnegative integer. Then for the operator \(K_{n,q}(f;x)\) defined by \((2.4)\), we have </p>
<div class="displaymath" id="a0000000026">
  \begin{equation*}  K_{n,q}(e_{m};x)=\sum ^{m}_{l=0}\tbinom {m}{l}C_{m,l}(q,b_{n})S^{q}_{n}(e_{{m-l}};x), \end{equation*}
</div>
<p> where \(C_{m,l}(q,b_{n})\) and \(S^{q}_{n}\) are given by \((3.1)\) and \((2.1),\) respectively. </p>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000027">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>Taking into account the definitions of \(K_{n,q}\) and \(S^{q}_{n}\) we immediately obtain that </p>
<div class="displaymath" id="a0000000028">
  \begin{equation*} \begin{split}  K_{n,q}(e_{m};x)& =E_{q}\left(-[n]_{q}\tfrac {x}{b_{n}}\right) \sum ^{\infty }_{k=0}\tfrac {([n]_{q}x)^{k}}{[k]_{q}!(b_{n})^{k}}\tfrac {[n]_{q}}{q^{k}b_{n}}I_{n,k}(e_{m})\\ & =\sum ^{m}_{l=0}\tbinom {m}{l}C_{m,l}(q,b_{n})\left\{ E_{q}\left(-[n]_{q}\tfrac {x}{b_{n}}\right)\sum ^{\infty }_{k=0} \tfrac {([n]_{q}x)^{k}}{[k]_{q}!(b_{n})^{k}}\left(\tfrac {[k]_{q}b_{n}}{[n]_{q}}\right)^{m-l}\right\} \\ & =\sum ^{m}_{l=0}\tbinom {m}{l}C_{m,l}(q,b_{n})S^{q}_{n}(e_{{m-l}};x). \end{split}\end{equation*}
</div>

<p><div class="cor_thmwrapper " id="a0000000029">
  <div class="cor_thmheading">
    <span class="cor_thmcaption">
    Corollary
    </span>
    <span class="cor_thmlabel">1</span>
  </div>
  <div class="cor_thmcontent">
  <p>For the operator \(K_{n,q}(f;x)\) defined by \((2.4)\) we have </p>
<div class="equation" id="a0000000030">
<p>
  <div class="equation_content">
    \begin{equation}  K_{n,q}(e_{0};x)=C_{0,0}(q,b_{n}) \end{equation}
  </div>
  <span class="equation_label">3.15</span>
</p>
</div>
<div class="equation" id="a0000000031">
<p>
  <div class="equation_content">
    \begin{equation}  K_{n,q}(e_{1};x)=C_{1,0}(q,b_{n})x+C_{1,1}(q,b_{n}), \end{equation}
  </div>
  <span class="equation_label">3.16</span>
</p>
</div>
<div class="equation" id="a0000000032">
<p>
  <div class="equation_content">
    \begin{equation}  K_{n,q}(e_{2};x)=C_{2,0}(q,b_{n})qx^{2}+\left\{ \tfrac {b_{n}}{[n]_{q}}C_{2,0}(q,b_{n})+2C_{2,1}(q,b_{n})\right\} x+C_{2,2}(q,b_{n}) \end{equation}
  </div>
  <span class="equation_label">3.17</span>
</p>
</div>
<div class="equation" id="a0000000033">
<p>
  <div class="equation_content">
    \begin{equation}  \begin{split}  K_{n,q}(e_{3};x)=& C_{3,0}(q,b_{n})q^{3}x^{3}+\left\{ (q^{2}+2q)\tfrac {b_{n}}{[n]_{q}}C_{3,0}(q,b_{n})+3qC_{3,1}(q,b_{n})\right\} x^{2}\\ & +\left\{ \left(\tfrac {b_{n}}{[n]_{q}}\right)^{2}C_{3,0}(q,b_{n})+3\tfrac {b_{n}}{[n]_{q}}C_{3,1}(q,b_{n})+3C_{3,2}(q,b_{n})\right\} x\\ & +C_{3,3}(q,b_{n}) \end{split} \end{equation}
  </div>
  <span class="equation_label">3.18</span>
</p>
</div>
<p> and </p>
<div class="equation" id="a0000000034">
<p>
  <div class="equation_content">
    \begin{equation}  \begin{split}  K_{n,q}(e_{4};x)=& C_{4,0}(q,b_{n})q^{6}x^{4}+ \left\{ (q^{5}+2q^{4}+3q^{3})\tfrac {b_{n}}{[n]_{q}}C_{4,0}(q,b_{n})\right.\\ & +4q^{3}C_{4,1}(q,b_{n})\Big\} x^{3}+\left\{ (q^{3}+3q^{2}+3q)\left(\tfrac {b_{n}}{[n]_{q}}\right)^{2}C_{4,0}(q,b_{n})\right.\\ & \left.+4(q^{2}+2q)\tfrac {b_{n}}{[n]_{q}}C_{4,1}(q,b_{n})+6C_{4,2}(q,b_{n})\right\} x^{2}\\ & +\left\{ \left(\tfrac {b_{n}}{[n]_{q}}\right)^{3}C_{4,0}(q,b_{n})+4\left(\tfrac {b_{n}}{[n]_{q}}\right)^{2}C_{4,1}(q,b_{n})\right.\\ & \left.+6\tfrac {b_{n}}{[n]_{q}}C_{4,2}(q,b_{n})+4C_{4,3}(q,b_{n})\right\} x+C_{4,4}(q,b_{n}).\\ \end{split} \end{equation}
  </div>
  <span class="equation_label">3.22</span>
</p>
</div>
<p> In the equalities \((3.2)-(3.6)\), the coefficients \(C_{m,l}(q,b_{n})\) for \(m=0,1,2,3,4\), \(l=0,1,2,3,4\) are listed below: </p>
<div class="displaymath" id="a0000000035">
  \begin{equation*} \begin{split} & C_{0,0}(q,b_{n})=1,\\ & C_{1,0}(q,b_{n})=1+\tfrac {q-1}{[2]_{q}},\\ & C_{1,1}(q,b_{n})=\tfrac {1}{[2]_{q}}\tfrac {b_{n}}{[n]_{q}},\\ & C_{2,0}(q,b_{n})=1+\tfrac {2(q-1)}{[2]_{q}}+\tfrac {(q-1)^{2}}{[3]_{q}},\\ & C_{2,1}(q,b_{n})=\left(\tfrac {1}{[2]_{q}}+\tfrac {q-1}{[3]_{q}}\right)\tfrac {b_{n}}{[n]_{q}},\\ & C_{2,2}(q,b_{n})=\tfrac {1}{[3]_{q}}\left(\tfrac {b_{n}}{[n]_{q}}\right)^{2},\\ & C_{3,0}(q,b_{n})=1+\tfrac {3(q-1)}{[2]_{q}}+\tfrac {3(q-1)^{2}}{[3]_{q}}+\tfrac {(q-1)^{3}}{[4]_{q}},\\ & C_{3,1}(q,b_{n})=\left(\tfrac {1}{[2]_{q}}+\tfrac {2(q-1)}{[3]_{q}}+\tfrac {(q-1)^{2}}{[4]_{q}}\right)\tfrac {b_{n}}{[n]_{q}},\\ & C_{3,2}(q,b_{n})=\left(\tfrac {1}{[3]_{q}}+\tfrac {q-1}{[4]_{q}}\right)\left(\tfrac {b_{n}}{[n]_{q}}\right)^{2},\\ & C_{3,3}(q,b_{n})=\tfrac {1}{[4]_{q}}\left(\tfrac {b_{n}}{[n]_{q}}\right)^{3},\\ & C_{4,0}(q,b_{n})=1+\tfrac {4(q-1)}{[2]_{q}}+\tfrac {6(q-1)^{2}}{[3]_{q}}+\tfrac {4(q-1)^{3}}{[4]_{q}}+\tfrac {(q-1)^{4}}{[5]_{q}}, \end{split}\end{equation*}
</div>
<div class="displaymath" id="a0000000036">
  \begin{equation*} \begin{split} & C_{4,1}(q,b_{n})=\left(\tfrac {1}{[2]_{q}}+\tfrac {3(q-1)}{[3]_{q}}+\tfrac {3(q-1)^{2}}{[4]_{q}}+\tfrac {(q-1)^{3}}{[5]_{q}}\right)\tfrac {b_{n}}{[n]_{q}},\\ & C_{4,2}(q,b_{n})=\left(\tfrac {1}{[3]_{q}}+\tfrac {2(q-1)}{[4]_{q}}+\tfrac {(q-1)^{2}}{[5]_{q}}\right)\left(\tfrac {b_{n}}{[n]_{q}}\right)^{2},\\ & C_{4,3}(q,b_{n})=\left(\tfrac {1}{[4]_{q}}+\tfrac {q-1}{[5]_{q}}\right)\left(\tfrac {b_{n}}{[n]_{q}}\right)^{3},\\ & C_{4,4}(q,b_{n})=\tfrac {1}{[5]_{q}}\left(\tfrac {b_{n}}{[n]_{q}}\right)^{4}. \end{split}\end{equation*}
</div>

  </div>
</div> Using Lemma 2 and \((2.2)\), it can be proved by direct calculation. So we omit it. </p>
<p>In the light of Corollary 1 and the linearity of \(K_{n,q}\), we now introduce the following Lemma without proof. </p>
<p><div class="lem_thmwrapper " id="a0000000037">
  <div class="lem_thmheading">
    <span class="lem_thmcaption">
    Lemma
    </span>
    <span class="lem_thmlabel">3</span>
  </div>
  <div class="lem_thmcontent">
  <p>For the operator \(K_{n,q}(f;x)\) defined by \((2.4)\), we have </p>
<div class="equation" id="a0000000038">
<p>
  <div class="equation_content">
    \begin{equation}  K_{n,q}(\varphi _{x};x)=\left\{ C_{1,0}(q,b_{n})-1\right\} x+C_{1,1}(q,b_{n}) \end{equation}
  </div>
  <span class="equation_label">3.44</span>
</p>
</div>
<div class="equation" id="a0000000039">
<p>
  <div class="equation_content">
    \begin{equation}  \begin{split}  K_{n,q}(\varphi ^{2}_{x};x)=& \left\{ qC_{2,0}(q,b_{n})-2C_{1,0}(q,b_{n})+1\right\} x^{2}\\ & +\left\{ \tfrac {b_{n}}{[n]_{q}}C_{2,0}(q,b_{n})+2C_{2,1}(q,b_{n})-2C_{1,1}(q,b_{n})\right\} x+C_{2,2}(q,b_{n}) \end{split} \end{equation}
  </div>
  <span class="equation_label">3.45</span>
</p>
</div>
<p> and </p>
<div class="equation" id="a0000000040">
<p>
  <div class="equation_content">
    \begin{equation}  \begin{split} & K_{n,q}(\varphi ^{4}_{x};x)= \\ & =\Big\{ q^{6}C_{4,0}(q,b_{n})\! -\! 4q^{3}C_{3,0}(q,b_{n})\! +\! 6qC_{2,0}(q,b_{n}) \! - \! 4C_{1,0}(q,b_{n})\! +\! 1\Big\} x^{4}\\ & \quad +\bigg\{ (q^{5}\! +\! 2q^{4}\! +\! 3q^{3})\tfrac {b_{n}}{[n]_{q}}C_{4,0}(q,b_{n})+4q^{3}C_{4,1}(q,b_{n}) -4(q^{2}+2q)\tfrac {b_{n}}{[n]_{q}}C_{3,0}(q,b_{n})\\ & \quad \quad \  \   -\! 12qC_{3,1}(q,b_{n})\! +\! 6\tfrac {b_{n}}{[n]_{q}}C_{2,0}(q,b_{n})\! \! +\! \! 12C_{2,1}(q,b_{n})\! -\! 4C_{1,1}(q,b_{n})\bigg\} \! x^{3}\\ & \quad +\Bigg\{ (q^{3}\! +3q^{2}\! +\! 3q)\left(\tfrac {b_{n}}{[n]_{q}}\right)^{2}C_{4,0}(q,b_{n})\! +\! 4(q^{2}+2q)\tfrac {b_{n}}{[n]_{q}}C_{4,1}(q,b_{n})\\ & \quad \quad \  \  \left.+6qC_{4,2}(q,b_{n})-4\left(\tfrac {b_{n}}{[n]_{q}}\right)^{2}C_{3,0}(q,b_{n})\right.\\ & \quad \quad \  \   -12\tfrac {b_{n}}{[n]_{q}}C_{3,1}(q,b_{n})-12C_{3,2}(q,b_{n})+6C_{2,2}(q,b_{n})\Bigg\} x^{2}\\ & \quad +\Bigg\{  \left(\tfrac {b_{n}}{[n]_{q}}\right)^{3}C_{4,0}(q,b_{n})+4\left(\tfrac {b_{n}}{[n]_{q}}\right)^{2}C_{4,1}(q,b_{n}) +6\tfrac {b_{n}}{[n]_{q}}C_{4,2}(q,b_{n}) \\ & \quad \quad \  \  +4C_{4,3}(q,b_{n})-4C_{3,3}(q,b_{n})\Bigg\} x+C_{4,4}(q,b_{n}), \end{split} \end{equation}
  </div>
  <span class="equation_label">3.48</span>
</p>
</div>
<p> where \(\varphi _{x}:=e_{1}-x\) and the coefficients \(C_{m,l}(q,b_{n})\) for \(m=0,1,2,3,4\), \(l=0,1,2,3,4\) are given as in Corollary <span class="rm">1</span>. </p>

  </div>
</div> </p>
<p>Now, with the help of the Lemma 3, we can evaluate the following limits which will be essential in the proof of the main theorem. </p>
<p><div class="lem_thmwrapper " id="a0000000041">
  <div class="lem_thmheading">
    <span class="lem_thmcaption">
    Lemma
    </span>
    <span class="lem_thmlabel">4</span>
  </div>
  <div class="lem_thmcontent">
  <p>Let \((q_{n})\) be a sequence in \((0,1)\) such that \(\displaystyle \lim _{n\rightarrow \infty }q_{n}=1\) and \(\displaystyle \lim _{n\rightarrow \infty }\tfrac {b_{n}}{[n]_{q_{n}}}=0\). Assume also that there is some \(a{\gt}0\) and some \(n_{0}\in \mathbb {N}\) such that </p>
<div class="displaymath" id="a0000000042">
  \[ a\leq \tfrac {b_{n}}{(1-q_{n})[n]_{q_{n}}}\quad \text{for all} \quad n\geq n_{0}. \]
</div>
<p> Then for every \( x\in [0,a)\) we have </p>
<div class="equation" id="a0000000043">
<p>
  <div class="equation_content">
    \begin{equation}  \lim _{n\rightarrow \infty }\tfrac {[n]_{q_{n}}}{b_{n}}K_{n,q_{n}}(\varphi _{x};x)=\tfrac {1}{2} \end{equation}
  </div>
  <span class="equation_label">3.58</span>
</p>
</div>
<div class="equation" id="a0000000044">
<p>
  <div class="equation_content">
    \begin{equation}  \lim _{n\rightarrow \infty }\tfrac {[n]_{q_{n}}}{b_{n}}K_{n,q_{n}}(\varphi ^{2}_{x};x)=x \end{equation}
  </div>
  <span class="equation_label">3.59</span>
</p>
</div>
<p> and </p>
<div class="equation" id="a0000000045">
<p>
  <div class="equation_content">
    \begin{equation}  \lim _{n\rightarrow \infty }\tfrac {[n]_{q_{n}}}{b_{n}}K_{n,q_{n}}(\varphi ^{4}_{x};x)=0. \end{equation}
  </div>
  <span class="equation_label">3.60</span>
</p>
</div>

  </div>
</div> <div class="proof_wrapper" id="a0000000046">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>Using the fact \(\displaystyle \lim _{n\rightarrow \infty }b_{n}=\infty \) from \((3.7)\) it readily follows that </p>
<div class="displaymath" id="a0000000047">
  \begin{equation*} \begin{split}  \lim _{n\rightarrow \infty }\tfrac {[n]_{q_{n}}}{b_{n}}K_{n,q_{n}}(\varphi _{x};x)& =\lim _{n\rightarrow \infty }\tfrac {[n]_{q_{n}}}{b_{n}}\left\{ \left(C_{1,0}(q_{n},b_{n})-1\right)x+C_{1,1}(q_{n},b_{n})\right\} \\ & =\lim _{n\rightarrow \infty }\left(\tfrac {(q_{n}-1)[n]_{q_{n}}}{[2]_{q_{n}}b_{n}}x+\tfrac {1}{[2]_{q_{n}}}\right)\\ & =\lim _{n\rightarrow \infty }\left(\tfrac {q^{n}_{n}-1}{[2]_{q_{n}}b_{n}}x+\tfrac {1}{[2]_{q_{n}}}\right)\\ & =\tfrac {1}{2}. \end{split}\end{equation*}
</div>
<p>Now from \((3.8),\) we can write </p>
<div class="displaymath" id="a0000000048">
  \begin{equation*} \begin{split}  K_{n,q_{n}}(\varphi ^{2}_{x};x)=& \left\{ qC_{2,0}(q_{n},b_{n})-2C_{1,0}(q_{n},b_{n})+1\right\} x^{2}\\ & \! +\! \left\{ \tfrac {b_{n}}{[n]_{q}}\! C_{2,0}(q_{n},b_{n})\! +\! 2C_{2,1}(q_{n},b_{n})\! \! -\! \! 2C_{1,1}(q_{n},b_{n})\right\} x\! +\! C_{2,2}(q_{n},b_{n})\\ =& \left\{ (q_{n}-1)\! +\! \tfrac {2(q_{n}-1)^{2}}{[2]_{q_{n}}}\! +\! \tfrac {q_{n}(q_{n}-1)^{2}}{[3]_{q_{n}}}\right\} x^{2}\! +\!  \left(1\! +\! \tfrac {2(q_{n}-1)}{[2]_{q_{n}}}\! +\! \tfrac {q^{2}_{n}-1}{[3]_{q_{n}}}\right)\tfrac {b_{n}}{[n]_{q_{n}}}x\\ & +\tfrac {1}{[3]_{q_{n}}}\left(\tfrac {b_{n}}{[n]_{q_{n}}}\right)^{2} \end{split}\end{equation*}
</div>
<p> and </p>
<div class="displaymath" id="a0000000049">
  \begin{equation*} \begin{split}  \tfrac {[n]_{q_{n}}}{b_{n}}\! K_{n,q_{n}}(\varphi ^{2}_{x};x) \! =& \left\{ (q_{n}-1)\! +\! \tfrac {2(q_{n}\! -\! 1)^{2}}{[2]_{q_{n}}}\! +\! \tfrac {q_{n}(q_{n}\! -\! 1)^{2}}{[3]_{q_{n}}}\right\} \!  \tfrac {[n]_{q_{n}}}{b_{n}}x^{2}\\ & +\left(1+\tfrac {2(q_{n}-1)}{[2]_{q_{n}}}\! +\! \tfrac {q^{2}_{n}\! -\! 1}{[3]_{q_{n}}}\right)x+\tfrac {1}{[3]_{q_{n}}}\! \tfrac {b_{n}}{[n]_{q_{n}}}\\ =& \left\{ (1+\tfrac {2(q_{n}-1)}{[2]_{q_{n}}}+\tfrac {q_{n}(q_{n}-1)}{[3]_{q_{n}}}\right\} \tfrac {q^{n}_{n}-1}{b_{n}}x^{2}+\left(1+\tfrac {2(q_{n}-1)}{[2]_{q_{n}}}+\tfrac {q^{2}_{n}-1}{[3]_{q_{n}}}\right)x\\ & +\tfrac {1}{[3]_{q_{n}}}\tfrac {b_{n}}{[n]_{q_{n}}}. \end{split}\end{equation*}
</div>
<p> Since \(\displaystyle \lim _{n\rightarrow \infty }q_{n}=1\), \(\displaystyle \lim _{n\rightarrow \infty }b_{n}=\infty \) and \(\displaystyle \lim _{n\rightarrow \infty }\tfrac {b_{n}}{[n]_{q_{n}}}=0\) this gives </p>
<div class="displaymath" id="a0000000050">
  \begin{equation*}  \lim _{n\rightarrow \infty }\tfrac {[n]_{q_{n}}}{b_{n}}K_{n,q_{n}}(\varphi ^{2}_{x};x)=x. \end{equation*}
</div>
<p>Similarly it can be shown that \((3.12)\) holds. </p>
<h1 id="a0000000051">4 Main result</h1>
<p> Now, we can establish a Voronovskaya type result for the operator \(K_{n,q}(f;x)\) defined by \((2.4)\). </p>
<p><div class="thm_thmwrapper " id="a0000000052">
  <div class="thm_thmheading">
    <span class="thm_thmcaption">
    Theorem
    </span>
    <span class="thm_thmlabel">1</span>
  </div>
  <div class="thm_thmcontent">
  <p>Let \((q_{n})\) be a sequence in \((0,1)\) such that \(\displaystyle \lim _{n\rightarrow \infty }q_{n}=1\) and \(\displaystyle \lim _{n\rightarrow \infty }\tfrac {b_{n}}{[n]_{q_{n}}}=0\). Assume also that there is some \(a{\gt}0\) and some \(n_{0}\in \mathbb {N}\) such that </p>
<div class="displaymath" id="a0000000053">
  \[ a\leq \tfrac {b_{n}}{(1-q_{n})[n]_{q_{n}}}\quad \text{for all} \quad n\geq n_{0}. \]
</div>
<p> Then for every \( x\in [0,a)\) and every function \(f\in C_{B}[0,\infty ),\) the space of continuous and bounded functions on \([0,\infty )\), which is twice differentiable at \(x\) one has </p>
<div class="displaymath" id="a0000000054">
  \begin{equation*}  \lim _{n\rightarrow \infty }\tfrac {[n]_{q_{n}}}{b_{n}}\left(K_{n,q_{n}}(f;x)-f(x)\right)=\tfrac {f’(x)+xf”(x)}{2}. \end{equation*}
</div>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000055">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>Follows by Lemma 4 and Theorem 1 in <span class="cite">
	[
	<a href="#sk" >17</a>
	]
</span>. </p>
<p><small class="footnotesize">  </small></p>
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</dd>
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  <dd><p><i class="sc">Mahmudov, N.I.</i> and <i class="sc">Gupta, V.</i>, <i class="it">On certain \(q\)-analogue of Szász Kantorovich operators</i>, Journal of Applied Mathematics and Computing, DOI: <span class="ttfamily">1</span>0.1007/s12190-010-0441-4. </p>
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</dd>
</dl>


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