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<title>Asymptotic formula in simultaneous approximation for certain Ismail-May-Baskakov operators: Asymptotic formula in simultaneous approximation for certain Ismail-May-Baskakov operators</title>
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<h1>Asymptotic formula in simultaneous approximation for certain Ismail-May-Baskakov operators</h1>
<p class="authors">
<span class="author">Vijay Gupta\(^\ast \) Michael Th. Rassias\(^{\ast \ast }\)</span>
</p>
<p class="date">June 30, 2021; accepted: August 29, 2021; published online: February 17, 2022.</p>
</div>
<div class="abstract"><p> The present study deals with a modification of Ismail-May operators with weights of Beta basis functions. We estimate weighted Korovkin’s theorem and difference estimates between two operators and establish an asymptotic formula for the derivative of a function. </p>
<p><b class="bf">MSC.</b> 41A25, 41A30 </p>
<p><b class="bf">Keywords.</b> Ismail-May operators, Baskakov basis, asymptotic formula, simultaneous approximation, weighted Korovkin’s theorem. </p>
</div>
<p>\(^\ast \)Department of Mathematics, Netaji Subhas University of Technology, Sector 3 Dwarka, New Delhi 110078, India, e-mail: <span class="tt">vijaygupta2001@hotmail.com</span> </p>
<p>\({\ast \ast }\)Department of Mathematics and Engineering Sciences, Hellenic Military Academy, 16673 Vari Attikis, Greece, <br />\({\ast \ast }\)Moscow Institute of Physics and Technology, 141700 Dolgoprudny, Institutskiy per, d. 9, Russia, <br />\({\ast \ast }\)Institute for Advanced Study, Program in Interdisciplinary Studies, 1 Einstein Dr, Princeton, NJ 08540, USA, e-mail: <span class="tt">mthrassias@yahoo.com</span> </p>
<h1 id="a0000000002">1 The Operator</h1>
<p>In 1978 Ismail and May <span class="cite">
	[
	<a href="#isma" >13</a>
	]
</span> (see also <span class="cite">
	[
	<a href="#IS" >12</a>
	]
</span>) studied certain operators of exponential type and introduced some new exponential operators. The following sequence of operators constitutes one of the families of linear positive operator proposed in <span class="cite">
	[
	<a href="#isma" >13</a>
	, 
	(3.14)
	]
</span>: </p>
<div class="equation" id="AE1">
<p>
  <div class="equation_content">
    \begin{equation} \label{AE1} (R_nf)(x) = \sum _{\nu =0}^\infty r_{n,\nu }(x)F_{n,\nu }, \end{equation}
  </div>
  <span class="equation_label">1</span>
</p>
</div>
<p> where \(F_{n,\nu }=f\left(\tfrac {\nu }{n}\right)\) and </p>
<div class="displaymath" id="a0000000003">
  \[ r_{n,\nu }(x)=e^{-(n+\nu )x/(1+x)}\tfrac {n(n+\nu )^{\nu -1}}{\nu !}\left(\tfrac {x}{1+x}\right)^\nu . \]
</div>
<p> These operators are exponential type operators as </p>
<div class="displaymath" id="a0000000004">
  \[ x(1+x)^2[(R_nf)(x)]^\prime =\sum _{\nu =0}^\infty (\nu -nx)r_{n,\nu }(x)f\left(\tfrac {\nu }{n}\right). \]
</div>
<p> In recent studies, Gupta and Rassias <span class="cite">
	[
	<a href="#vgmtr" >10</a>
	]
</span> presented a collection of the moments of several linear positive operators, Agratini <span class="cite">
	[
	<a href="#oag" >2</a>
	]
</span> estimated approximation properties of integral-type operators. The operators \(R_n\) are not suitable in order to approximate integral functions. Additionally, it is observed in <span class="cite">
	[
	<a href="#vgga" >6</a>
	]
</span> that obtaining the usual Durrmeyer variant of the operators (<a href="#AE1">1</a>) is not possible due to the non-convergence of the integral </p>
<div class="displaymath" id="a0000000005">
  \[ \int _0^\infty r_{n,\nu }(u)du. \]
</div>
<p>In the recent work <span class="cite">
	[
	<a href="#vgga" >6</a>
	]
</span>, the hybrid link operators were introduced and the approximation behavior was discussed. We consider here new hybrid operators with weights \(b_{n+2,\nu -1}(t)\) of slightly modified Baskakov basis functions, which for \(x\in [0,\infty )\) are defined as </p>
<div class="displaymath" id="AEB1">
  \begin{eqnarray}  (V_nf)(x)& =& \sum _{\nu =0}^\infty r_{n,\nu }(x) G_{n,\nu }(f),\label{AEB1} \end{eqnarray}
</div>
<p> where </p>
<div class="displaymath" id="a0000000006">
  \[ G_{n,\nu }(f) = \left\{  \begin{array}{lr} (n+1)\int _0^\infty b_{n+2,\nu -1}(u)f(u)du&  : 1\le \nu {\lt}\infty \\ f(0) &  : \nu =0 \end{array} \right. \]
</div>
<div class="displaymath" id="a0000000007">
  \[ b_{n,\nu }(u)=\textstyle {{n+\nu -1}\choose {\nu }}\tfrac {u^\nu }{(1+u)^{n+\nu }}. \]
</div>
<p> For simplicity, we use the notation </p>
<div class="displaymath" id="a0000000008">
  \[ v_{n,\nu }(u)=(n+1) b_{n+2,\nu -1}(u) \]
</div>
<p> throughout the paper. </p>
<p>The operators (<a href="#AEB1">2</a>) reproduce linear functions. The investigation of convergence of operators in ordinary and simultaneous approximation has been an active area of research over recent years. We mention here some of the related work viz. <span class="cite">
	[
	<a href="#pnkj" >3</a>
	]
</span>, <span class="cite">
	[
	<a href="#vggc" >7</a>
	]
</span>, <span class="cite">
	[
	<a href="#noor" >8</a>
	]
</span>, <span class="cite">
	[
	<a href="#vgtmr" >9</a>
	]
</span>, <span class="cite">
	[
	<a href="#vgpn" >11</a>
	]
</span>. In the present paper, we estimate direct result in weighted norm, difference of these operators with discrete operators and prove a formula of asymptotic kind for derivatives of the operators \(V_n.\) </p>
<h1 id="a0000000009">2 Some Lemmas</h1>
<p> <div class="lemma_thmwrapper " id="L1">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">1</span>
  </div>
  <div class="lemma_thmcontent">
  <p> If </p>
<div class="displaymath" id="a0000000010">
  \[  U_{n,q}(x)=\sum _{\nu =0}^\infty r_{n,\nu }(x)\left(\tfrac {\nu }{n}-x\right)^q,\quad q\in {N}\cup \{ 0\} ,  \]
</div>
<p> then the moments satisfy the relation: </p>
<div class="displaymath" id="a0000000011">
  \[ nU_{n,q+1}(x)=x(1+x)^2\left[(U_{n,q}(x))^\prime +qU_{n,q-1}(x)\right]. \]
</div>
<p> Also, one can observe that </p>
<div class="displaymath" id="a0000000012">
  \[ U_{n,q}(x)=O_x(n^{-[(q+1)/2]}), \]
</div>
<p> where \([s]\) stands for the integral part of \(s\). </p>
<p>Furthermore, for \(e_i(t)=t^i, i=0,1,2,\ldots \), we have </p>
<div class="displaymath" id="a0000000013">
  \[ n(R_ne_{q+1})(x)=x(1+x)^2[(R_ne_q)(x)]^\prime +nx(R_ne_q)(x). \]
</div>

  </div>
</div> </p>
<p><div class="lemma_thmwrapper " id="L2">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">2</span>
  </div>
  <div class="lemma_thmcontent">
  <p> If we denote \(\mu _{n,q}(x)=(V_n(e_1-xe_0)^q)(x)\) for \(q\in {N}\) and \(x\ge 0\), then </p>
<div class="displaymath" id="a0000000014">
  \begin{eqnarray*}  (n-q) \mu _{n,q+1}(x)& =& x(1+x)^2\left[[\mu _{n,q}(x)]^\prime +q\mu _{n,q-1}(x)\right]\\ & & +q\mu _{n,q}(x)+qx(1+x) \mu _{n,q-1}(x). \end{eqnarray*}
</div>
<p> In particular, we have </p>
<div class="displaymath" id="a0000000015">
  \[ \mu _{n,0}(x)=1, \mu _{n,1}(x)=0, \mu _{n,2}(x)=\tfrac {x(1+x)(2+x)}{n-1}. \]
</div>
<p> In general for all \(x\in [0,\infty )\), we have \(\mu _{n,q}(x)=O_x(n^{-[(q+1)/2]}).\) </p>

  </div>
</div> <div class="proof_wrapper" id="a0000000016">
  <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 </p>
<div class="displaymath" id="a0000000017">
  \[ x(1+x)^2 r_{n,\nu }^\prime (x)=(\nu -n x)r_{n,\nu }(x) \]
</div>
<p> and </p>
<div class="displaymath" id="a0000000018">
  \[ u(1+u)v_{n,\nu }^\prime (u)=[(\nu -1)-(n+2)u]v_{n,\nu }(u), \]
</div>
<p> we have </p>
<div class="displaymath" id="a0000000019">
  \begin{eqnarray*} & & x(1+x)^2[\mu _{n,q}(x)]^\prime =\\ & =& \sum _{\nu =1}^{\infty }x(1+x)^2r_{n,\nu }^\prime (x)\int _0^\infty v_{n,\nu }(u)(e_1-xe_0)^qdu\\ & & +n e^{-n x/(1+x)}(-x)^{q+1}+q (-x)^{q}(1+x)^2e^{-n x/(1+x)}\\ & & -qx(1+x)^2\sum _{\nu =1}^{\infty }r_{n,\nu }(x)\int _0^\infty v_{n,\nu }(u)(e_1-xe_0)^{q-1}du\\ & =&  \sum _{\nu =1}^{\infty }(\nu -n x)r_{n,\nu }(x)\int _0^\infty v_{n,\nu }(u)(e_1-xe_0)^qdu\\ & & +n e^{-n x/(1+x)}(-x)^{q+1}-qx(1+x)^2 \mu _{n,q-1}(x)\\ & =&  \sum _{\nu =1}^{\infty }r_{n,\nu }(x)\int _0^\infty u(1+u)v_{n,\nu }^\prime (u)(e_1-xe_0)^qdu+(n+2)[ \mu _{n,q+1}(x)\\ & & -(-x)^{q+1}e^{-n x/(1+x)}]+(1+2x)[\mu _{n,q}(x)-(-x)^qe^{-n x/(1+x)}]\\ & & +n e^{-n x/(1+x)}(-x)^{q+1}-qx (1+x)^2 \mu _{n,q-1}(x). \end{eqnarray*}
</div>
<p> Thus, we have </p>
<div class="displaymath" id="a0000000020">
  \begin{eqnarray*} & & x(1+x)^2\left[[\mu _{n,q}(x)]^\prime +q\mu _{n,q-1}(x)\right]=\\ & =&  \sum _{\nu =1}^{\infty }r_{n,\nu }(x)\int _0^\infty [(e_1\! \! -\! \! xe_0)^2\! \! +\! \! (1\! \! +\! \! 2x)(e_1\! \! -\! \! xe_0)\! \! +\! \! x(1\! +\! x)]v_{n,\nu }^\prime (u)(e_1\! \! -\! \! xe_0)^qdu\\ & & +(n+2)[ \mu _{n,q+1}(x)-(-x)^{q+1}e^{-n x/(1+x)}]\\ & & +(1+2x)[\mu _{n,q}(x)-(-x)^qe^{-n x/(1+x)}]+n e^{-n x/(1+x)}(-x)^{q+1}. \end{eqnarray*}
</div>
<p> Integrating by parts the last integral, we obtain that </p>
<div class="displaymath" id="a0000000021">
  \begin{eqnarray*} & & x(1+x)^2\left[[\mu _{n,q}(x)]^\prime +q\mu _{n,q-1}(x)\right]\\ & =&  -(q+2) \mu _{n,q+1}(x)-(q+1)(1+2x)\mu _{n,q}(x)-qx(1+x) \mu _{n,q-1}(x)\\ & & +(n+2) \mu _{n,q+1}(x)+(1+2x)\mu _{n,q}(x). \end{eqnarray*}
</div>
<p> This completes the proof of the recurrence relation. The other consequences follow from the recurrence relation. <div class="proof_wrapper" id="a0000000022">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<p><div class="lemma_thmwrapper " id="L3">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">3</span>
  </div>
  <div class="lemma_thmcontent">
  <p> If we denote \(T_{n,q}(x)=(V_ne_q)(x)\) for \(q\in {N}\) and \(x\ge 0\), then </p>
<div class="displaymath" id="a0000000023">
  \begin{eqnarray*}  (n-q) T_{n,q+1}(x)& =& x(1+x)^2[T_{n,q}(x)]^\prime +(q+nx)T_{n,q}(x). \end{eqnarray*}
</div>
<p> In general, we have </p>
<div class="displaymath" id="a0000000024">
  \begin{eqnarray*}  T_{n,q}(x)& =& x^q+\tfrac {q(q-1)}{2}x^{q-1}\tfrac {(1+x)(2+x)}{n-q+1}+O_x(n^{-2}). \end{eqnarray*}
</div>

  </div>
</div> The proof of the lemma follows along the lines of Lemma <a href="#L1">1</a>, we thus omit the details. </p>
<p><div class="lemma_thmwrapper " id="L4">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">4</span>
    <span class="lemma_thmtitle"><span class="cite">
	[
	<a href="#vgga" >6</a>
	]
</span></span>
  </div>
  <div class="lemma_thmcontent">
  <p>  There exist the polynomials \(\phi _{i,j,q}(x)\) independent of \(n\) and \( \nu \) such that </p>
<div class="displaymath" id="a0000000025">
  \[ [x^q(1+x)^{2q}]\tfrac {\partial ^q}{\partial x^q}[r_{n,\nu }(x)]=\sum _{{2i+j\le q}\atop {i,j\ge 0}}n^{i}(\nu -nx)^j\phi _{i,j,q}(x)[r_{n,\nu }(x)]. \]
</div>

  </div>
</div> </p>
<p><div class="lemma_thmwrapper " id="L5">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">5</span>
  </div>
  <div class="lemma_thmcontent">
  <p> If \(\mu _r^{J_{n,\nu }}=\sum _{i=0}^r {r\choose i} (-1)^i J_{n,\nu }(e_{r-i})[ J_{n,\nu }(e_{1})]^i\), then we have </p>
<div class="displaymath" id="a0000000026">
  \[ \mu _m^{F_{n,\nu }}=0,m=1,2,3,\ldots , \  \   \  \  \mu _2^{G_{n,\nu }}=\tfrac {\nu ^2+n\nu }{n^2(n-1)}, \  \  \mu _3^{G_{n,\nu }}=\tfrac {4 \nu ^3+6n\nu ^2+2n^2\nu }{n^3(n-1)(n-2)} \]
</div>
<p> and </p>
<div class="displaymath" id="a0000000027">
  \[  \mu _4^{G_{n,\nu }}=\tfrac {3(6+n) \nu ^4+ 6(6 + n)n \nu ^3+3(8 + n)n^2 \nu ^2+6 n^3 \nu }{n^4(n-1)(n-2)(n-3)}.  \]
</div>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000028">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> By definition, we have </p>
<div class="displaymath" id="a0000000029">
  \begin{eqnarray*}  b^{F_{n,\nu }}& =& F_{n,\nu }(e_1)=\tfrac {\nu }{n}. \end{eqnarray*}
</div>
<p> By simple computation, we get </p>
<div class="displaymath" id="a0000000030">
  \begin{eqnarray*}  \mu _m^{F_{n,\nu }}& =& 0, m=1,2,3,\ldots \end{eqnarray*}
</div>
<p> Moreover </p>
<div class="displaymath" id="a0000000031">
  \begin{eqnarray*}  G_{n,\nu }(e_q)& =& (n+1)\int _0^\infty b_{n+2,\nu -1}(u)u^q du\\ & =& (n+1) \textstyle {n+\nu \choose \nu -1}\displaystyle \int _0^\infty \tfrac {u^{\nu +q-1}}{(1+u)^{n+\nu +1}}du=\tfrac {(\nu +q-1)!(n-q)!}{(\nu -1)!n!} \end{eqnarray*}
</div>
<p>Thus </p>
<div class="displaymath" id="a0000000032">
  \begin{eqnarray*}  b^{G_{n,\nu }}& =& G_{n,\nu }(e_1)=\tfrac {\nu }{n}, \end{eqnarray*}
</div>
<p> and </p>
<div class="displaymath" id="a0000000033">
  \begin{eqnarray*}  \mu _2^{G_{n,\nu }}& =& \sum _{i=0}^2 \textstyle {2\choose i} (-1)^i G_{n,\nu }(e_{2-i})[ G_{n,\nu }(e_{1})]^i\nonumber \\ & =& G_{n,\nu }(e_{2})-[G_{n,\nu }(e_{1})]^2\\ & =& \tfrac {\nu ^2+n\nu }{n^2(n-1)}. \end{eqnarray*}
</div>
<div class="displaymath" id="a0000000034">
  \begin{align*}  \mu _3^{G_{n,\nu }}& =\sum _{i=0}^3 \textstyle {3\choose i} (-1)^i G_{n,\nu }(e_{3-i})[ G_{n,\nu }(e_{1})]^i\nonumber \\ & =G_{n,\nu }(e_{3})-3G_{n,\nu }(e_{2})G_{n,\nu }(e_{1})+3[G_{n,\nu }(e_{1})]^3-[G_{n,\nu }(e_{1})]^3\\ & =\tfrac {1}{n^3(n-1)(n-2)}\left[n^2\nu (\nu \! \! +\! \! 2)(\nu \! \! +\! \! 1)\nu \! \! -\! \! 3n(n\! \! -\! \! 2)(\nu \! \! +\! \! 1)\nu ^2+2(n\! \! -\! \! 1)(n\! \! -\! \! 2)\nu ^3\right]\\ & =\tfrac {4 \nu ^3+6n\nu ^2+2n^2\nu }{n^3(n-1)(n-2)} \end{align*}
</div>
<p> and </p>
<div class="displaymath" id="a0000000035">
  \begin{align*}  \mu _4^{G_{n,\nu }}& =\sum _{i=0}^4 \textstyle {4\choose i} (-1)^i G_{n,\nu }(e_{4-i})[ G_{n,\nu }(e_{1})]^i\nonumber \\ & =G_{n,\nu }(e_{4})-4G_{n,\nu }(e_{3})G_{n,\nu }(e_{1})+6G_{n,\nu }(e_{2})[G_{n,\nu }(e_{1})]^2\\ \nonumber & -4G_{n,\nu }(e_{1})[G_{n,\nu }(e_{1})]^3+[G_{n,\nu }(e_{1})]^4\nonumber \\ & =\tfrac {1}{n^4(n-1)(n-2)(n-3)}\biggl[n^3(\nu \! +\! 3)(\nu \! +\! 2)(\nu \! +\! 1)\nu \! -\! 4n^2(n\! -\! 3)(\nu \! +\! 2)(\nu \! +\! 1)\nu ^2\\ & \quad +6n(n-2)(n-3)(\nu +1)\nu ^3-3\nu ^4(n-1)(n-2)(n-3)\biggr]\\ & =\tfrac {1}{n^4(n-1)(n-2)(n-3)}\biggl[6 n^3 \nu + 3(8 + n) n^2\nu ^2 + 6( 6 + n)n \nu ^3 + 3(6 + n) \nu ^4\biggr]. \end{align*}
</div>
<p> The other estimates follow similarly, we thus omit the details. <div class="proof_wrapper" id="a0000000036">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<h1 id="a0000000037">3 Direct Estimates and Asymptotic Formula</h1>
<p>Let </p>
<div class="displaymath" id="a0000000038">
  \[ B_{m}[0,\infty )=\{ f:|f(x)|\le c_f(1+x^m), \forall x\in [0,\infty )\} ,\quad m{\gt}3 \]
</div>
<p> where \(c_f\) is an absolute constant dependent on \(f,\) but independent of \(x.\) Let </p>
<div class="displaymath" id="a0000000039">
  \[ C_{m}[0,\infty )=C[0,\infty )\cap B_{m}[0,\infty ). \]
</div>
<p> For each \(f\in C_{m}[0,\infty )\) we consider modulus of continuity (see <span class="cite">
	[
	<a href="#ispir" >14</a>
	]
</span>) as </p>
<div class="displaymath" id="a0000000040">
  \[ \Omega (f,\delta )=\sup _{|\triangle x|{\lt}\delta ,x\in R^+}\tfrac {|f(x+\triangle x)-f(x)|}{(1+h^m)(1+x^m)}. \]
</div>
<p> Additionally, \(C^*_{m}[0,\infty )\) denotes the subspace of continuous functions \(f \in B_{m}[0,\infty )\), for which </p>
<div class="displaymath" id="a0000000041">
  \[ \lim _{x\to \infty } |f(x)|(1+x^m)^{-1}{\lt}\infty . \]
</div>
<p> We consider the norm by </p>
<div class="displaymath" id="a0000000042">
  \[ \Vert f\Vert _m=\sup _{0\le x{\lt}\infty } \tfrac {|f(x)|}{(1+x^m)}. \]
</div>
<p><div class="theorem_thmwrapper " id="T1">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">6</span>
  </div>
  <div class="theorem_thmcontent">
  <p> If \(f\in C^*_m\left[ 0,\infty \right)\), then </p>
<div class="displaymath" id="a0000000043">
  \[ \displaystyle \lim _{n\to \infty }\Vert (V_nf)-f\Vert _m=0. \]
</div>

  </div>
</div> <div class="proof_wrapper" id="a0000000044">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> By the weighted Korovkin theorem due to Gadjiev <span class="cite">
	[
	<a href="#gad" >4</a>
	]
</span>, we know that if \(f\in C^*_m\left[ 0,\infty \right)\), satisfying </p>
<div class="displaymath" id="a0000000045">
  \[ \lim _{n\to \infty }\Vert (V_ne_i)-e_i\Vert _m=0, \  \  i=0,1,2, \]
</div>
<p> then we have </p>
<div class="displaymath" id="a0000000046">
  \[ \lim _{n\to \infty }\Vert (V_nf)(x)-f(x)\Vert _m=0. \]
</div>
<p> In order to prove the result, we use <a href="#L3">lemma 3</a>. Clearly, the result holds for \(i=0,1\). Thus, in order to complete the proof we proceed for </p>
<div class="displaymath" id="a0000000047">
  \begin{eqnarray*} \lim _{n\to \infty }\Vert (V_ne_2)(x)-e_2\Vert _2& =& \lim _{n\to \infty } \tfrac {1}{(1+x^m)}\biggl[x^2+\tfrac {x(1+x)(2+x)}{n-1}-x^2\biggr]\\ & =& \lim _{n\to \infty } \tfrac {1}{(1+x^m)}\biggl[\tfrac {x(1+x)(2+x)}{n-1}\biggr]=0. \end{eqnarray*}
</div>
<p> This completes the proof of the theorem. <div class="proof_wrapper" id="a0000000048">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<p><div class="theorem_thmwrapper " id="T2">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">7</span>
  </div>
  <div class="theorem_thmcontent">
  <p> For \(f^{(s)}\in C_B[0,\infty )\) , \(0\le s\le 4\), \(s\in \mathbf{N}\), \(x\in [0,\infty )\) and \(n\in \mathbf{N}\), we have </p>
<div class="displaymath" id="a0000000049">
  \begin{eqnarray*} & & |((V_nf)-(R_nf))(x)|\le \\ & \le &  \tfrac {\| f^{iv}\| }{24(n-1)(n-2)(n-3)}\biggl[6 x + 3(8 + n)\left( x^2+\tfrac {x(1+x)^2}{n}\right)\\ & & + 6( 6 + n)\left(x^3+\tfrac {3x^2(1+x)^2}{n}+\tfrac {x(1+x)^3(1+3x)}{n^2}\right)\\ & &  + 3(6 + n) \left(x^4+\tfrac {6x^3(1+x)^2}{n}+\tfrac {x^2(1+x)^3(7+15x)}{n^2}+\tfrac {x(1+x)^4(1+10x+15x^2)}{n^3}\right)\biggr]\\ & & +\tfrac {\|  f^{\prime \prime \prime }\| }{3(n-1)(n-2)}\left[2x^3+\tfrac {6x^2(1+x)^2}{n}+\tfrac {2x(1+x)^3(1+3x)}{n^2}+3x^2+\tfrac {3x(1+x)^2}{n}+x\right]\\ & & + \tfrac {\| f^{\prime \prime }\| }{2(n-1)}\left[x^2+\tfrac {x(1+x)^2}{n}+x\right]. \end{eqnarray*}
</div>
<p> <div class="proof_wrapper" id="a0000000050">
  <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 <a href="#L5">lemma 5</a> and <a href="#L1">lemma 1</a>, we have </p>
<div class="displaymath" id="a0000000051">
  \begin{eqnarray*}  \alpha (x)& =& \sum _{\nu =0}^{\infty }r_{n,\nu }(x) (\mu _{4}^{F_{n,\nu }}+\mu _4^{G_{n,\nu }})\\ & =& \tfrac {1}{24n^4(n-1)(n-2)(n-3)}\sum _{\nu =0}^{\infty }r_{n,\nu }(x)\biggl[6 n^3 \nu + 3(8 + n) n^2\nu ^2 \\ & & + 6( 6 + n)n \nu ^3 + 3(6 + n) \nu ^4\biggr]\\ & =& \tfrac {1}{24(n-1)(n-2)(n-3)}\biggl[6 (R_ne_1)(x) + 3(8 + n)(R_ne_2)(x)\\ & & + 6( 6 + n)(R_ne_3)(x)+ 3(6 + n) (R_ne_4)(x)\biggr]\\ & =& \tfrac {1}{24(n-1)(n-2)(n-3)}\biggl[6 x + 3(8 + n)\left( x^2+\tfrac {x(1+x)^2}{n}\right)\\ & & + 6( 6 + n)\left(x^3+\tfrac {3x^2(1+x)^2}{n}+\tfrac {x(1+x)^3(1+3x)}{n^2}\right)\\ & &  + 3(6\! \!  +\! \!  n) \left(x^4\! +\! \tfrac {6x^3(1+x)^2}{n}\! +\! \tfrac {x^2(1+x)^3(7+15x)}{n^2}\! +\! \tfrac {x(1+x)^4(1+10x+15x^2)}{n^3}\right)\biggr]. \end{eqnarray*}
</div>
<p> Additionally, </p>
<div class="displaymath" id="a0000000052">
  \begin{eqnarray*}  \beta (x)& =& \tfrac {1}{3!}\sum _{\nu =0}^{\infty }r_{n,\nu }(x)\left| \mu _{3}^{F_{n,\nu }}-\mu _{3}^{G_{n,\nu }}\right|\\ & =& \tfrac {1}{3!}\sum _{\nu =0}^{\infty }r_{n,\nu }(x)\tfrac {4 \nu ^3+6n\nu ^2+2n^2\nu }{n^3(n-1)(n-2)}\\ & =& \tfrac {1}{3(n-1)(n-2)}\left[2(R_ne_3)(x)+3(R_ne_2)(x)+(R_ne_1)(x)\right]\\ & =& \tfrac {1}{3(n-1)(n-2)}\left[2x^3+\tfrac {6x^2(1+x)^2}{n}+\tfrac {2x(1+x)^3(1+3x)}{n^2}+3x^2+\tfrac {3x(1+x)^2}{n}+x\right]. \end{eqnarray*}
</div>
<p> Furthermore, we have </p>
<div class="displaymath" id="a0000000053">
  \begin{eqnarray*}  \gamma (x)& =& \tfrac {1}{2!}\sum _{\nu =0}^{\infty }r_{n,\nu }(x)\left| \mu _{2}^{F_{n,\nu }}-\mu _2^{G_{n,\nu }} \right|\\ & =& \tfrac {1}{2!}\sum _{\nu =0}^{\infty }r_{n,\nu }(x)\tfrac {\nu ^2+n\nu }{n^2(n-1)}\\ & =& \tfrac {1}{2(n-1)}[(R_ne_2)(x)+(R_ne_1)(x)]=\tfrac {1}{2(n-1)}\left[x^2+\tfrac {x(1+x)^2}{n}+x\right]. \end{eqnarray*}
</div>
<p> Finally, we get </p>
<div class="displaymath" id="a0000000054">
  \begin{eqnarray*}  \delta ^2(x)& =& \sum _{\nu =0}^{\infty }r_{n,\nu }(x)(b^{F_{n,\nu }}-b^{G_{n,\nu }})^2=0. \end{eqnarray*}
</div>
<p> Following <span class="cite">
	[
	<a href="#vgwil" >5</a>
	, 
	Theorem 3
	]
</span> and <span class="cite">
	[
	<a href="#acu" >1</a>
	, 
	Theorem 4
	]
</span>, we derive that </p>
<div class="displaymath" id="a0000000055">
  \begin{eqnarray*}  |((V_nf)-(R_nf))(x)| & \le & \| f^{iv}\|  \alpha (x)+\|  f^{\prime \prime \prime }\|  \beta (x)+\| f^{\prime \prime }\|  \gamma {(x)}\\ & &  +2\omega (f,\delta (x)). \end{eqnarray*}
</div>
<p> Substituting the values of the above estimates, we get the desired result. <div class="proof_wrapper" id="a0000000056">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>

  </div>
</div> </p>
<p><div class="theorem_thmwrapper " id="T3">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">8</span>
  </div>
  <div class="theorem_thmcontent">
  <p> Let \(f\in C[0,\infty )\) with \(|f(t)|\leq C (1+t)^{\gamma }\) for some \(\gamma {\gt}0, t\ge 0.\) If \(f^{(q+2)}\) exists at a point \(x\in (0,\infty ),\) then we have </p>
<div class="displaymath" id="a0000000057">
  \begin{eqnarray*}  \lim _{n\rightarrow \infty }n(V_n^{(q)}f)(x) -f^{(q)}(x))& =& \tfrac {q(q-1)(q-2)}{2}f^{(q-1)}(x)+\tfrac {3q(q-1)(x+1)}{2}f^{(q)}(x)\\ & & +\tfrac {q[(q^2+2)x^2+3(q+1)x+2]}{2}f^{(q+1)}(x)\\ & & +\tfrac {x(1+x)(2+x)}{2}f^{(q+2)}(x). \end{eqnarray*}
</div>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000058">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> Applying Taylor’s expansion, we have From Taylor’s theorem, we have </p>
<div class="displaymath" id="a0000000059">
  \begin{eqnarray}  (V_n^{(q)}f)(x)& =& \sum _{v=0}^{q+2}\tfrac {f^{(v)}(x)}{v!}\bigg(\tfrac {\partial ^q}{\partial \kappa ^{q}}(V_n(e_1-xe_0)^{v})(\kappa )\bigg)_{\kappa =x} \nonumber \\ & & + \bigg(\tfrac {\partial ^q}{\partial \kappa ^{q}}(V_n\psi (t,x)(e_1-xe_0)^{q+2})(\kappa )\bigg)_{\kappa =x}\nonumber \\ & =& \sum _{v=0}^{q+2}\tfrac {f^{(v)}(x)}{v!}\sum _{j=0}^{v}\textstyle {v\choose j}(-x)^{v-j}\bigg(\tfrac {\partial ^q}{\partial \kappa ^{q}}T_{n,j}(\kappa )\bigg)_{\kappa =x}\nonumber \\ & & + \bigg(\tfrac {\partial ^q}{\partial \kappa ^{q}}(V_n\psi (u,x)(e_1-xe_0)^{q+2})(\kappa )\bigg)_{\kappa =x}\nonumber \\ & :=& I_1+I_2\nonumber \end{eqnarray}
</div>
<p> where the function \(\psi (u,x)\rightarrow 0\) as \(u\rightarrow x.\) Obviously \(I_1\) is estimated as </p>
<div class="displaymath" id="a0000000060">
  \begin{eqnarray*}  I_1& =& \tfrac {f^{(q-1)}(x)}{(q-1)!}\bigg(\tfrac {\partial ^q}{\partial \kappa ^{q}}T_{n,q-1}(\kappa )\bigg)_{\kappa =x}\nonumber \\ & & +\tfrac {f^{(q)}(x)}{q!}\left[q(-x)\bigg(\tfrac {\partial ^q}{\partial \kappa ^{q}}T_{n,q-1}(\kappa )\bigg)_{\kappa =x} +\bigg(\tfrac {\partial ^q}{\partial \kappa ^{q}}T_{n,q}(\kappa )\bigg)_{\kappa =x}\right]\nonumber \\ & & +\tfrac {f^{(q+1)}(x)}{(q+1)!}\bigg[\tfrac {(q+1)q}{2}x^2\bigg(\tfrac {\partial ^q}{\partial \kappa ^{q}}T_{n,q-1}(\kappa )\bigg)_{\kappa =x} \! +\! (q+1)(\! -\! x)\bigg(\tfrac {\partial ^q}{\partial \kappa ^{q}}T_{n,q}(\kappa )\bigg)_{\kappa =x}\nonumber \\ & & +\bigg(\tfrac {\partial ^q}{\partial \kappa ^{q}}T_{n,q+1}(\kappa )\bigg)_{\kappa =x}\biggr]\nonumber \\ & & +\tfrac {f^{(q+2)}(x)}{(q+2)!}\biggl[\tfrac {(q+2)(q+1)q}{3!}(-x)^3 \bigg(\tfrac {\partial ^q}{\partial \kappa ^{q}}T_{n,q-1}(\kappa )\bigg)_{\kappa =x} \\ & & +\tfrac {(q+2)(q+1)}{2}x^2 \bigg(\tfrac {\partial ^q}{\partial \kappa ^{q}}T_{n,q}(\kappa )\bigg)_{\kappa =x}\nonumber \\ & & +(q+2)(-x)\bigg(\tfrac {\partial ^q}{\partial \kappa ^{q}}T_{n,q+1}(\kappa )\bigg)_{\kappa =x}+ \bigg(\tfrac {\partial ^q}{\partial \kappa ^{q}}T_{n,q+2}(\kappa )\bigg)_{\kappa =x}\biggr].\nonumber \end{eqnarray*}
</div>
<p> Applying <a href="#L3">lemma 3</a>, we have </p>
<div class="displaymath" id="a0000000061">
  \begin{eqnarray*}  I_1& =& \tfrac {f^{(q-1)}(x)}{(q-1)!}\left[\tfrac {q!(q-1)(q-2)}{2(n-q+2)}\right]+f^{(q)}(x)\left[1+\tfrac {3nq(q-1)(x+1)}{2(n-q+2)(n-q+1)}+O(n^{-2})\right]\\ & & +f^{(q+1)}(x)\left[\tfrac {nq[(q^2+2)x^2+3(q+1)x+2]}{2(n-q)(n-q+2)}+O(n^{-2})\right]\nonumber \\ & & +\tfrac {f^{(q+2)}(x)}{(q+2)!}\biggl[\tfrac {x^3}{2(n-q+1)}+\tfrac {3x^2}{2(n-q-1)}+\tfrac {nx}{(n-q)(n-q-1)}\biggr].\nonumber \end{eqnarray*}
</div>
<p> Thus </p>
<div class="displaymath" id="a0000000062">
  \begin{eqnarray}  \lim _{n\rightarrow \infty }n((V_n^{(q)}f)(x)-f^{(q)}(x))& =& \tfrac {q(q-1)(q-2)}{2}f^{(q-1)}(x)+\tfrac {3q(q-1)(x+1)}{2}f^{(q)}(x)\nonumber \\ & & +\tfrac {q[(q^2+2)x^2+3(q+1)x+2]}{2}f^{(q+1)}(x)\nonumber \\ & & +\tfrac {x(1+x)(2+x)}{2}f^{(q+2)}(x)+\lim _{n\rightarrow \infty }n I_2.\nonumber \end{eqnarray}
</div>
<p>Next, in view of <a href="#L4">lemma 4</a>, we have </p>
<div class="displaymath" id="a0000000063">
  \begin{align*} & |I_2|\leq \\ & \leq \left(\sum _{\nu =1}^{\infty }\sum _{{2i+j\leq q}\atop { i,j\geq 0}}n^{i}|\nu \! \! -\! \! n \kappa |^{j}\tfrac {|\phi _{i,j,q}(\kappa )|}{\kappa ^q(1+\kappa )^{2q}} r_{n,\nu }(\kappa )\int _0^{\infty }v_{n,\nu }(u)|\psi (u,x)|.|u\! \! -\! \! x|^{q+2} du\right)_{\kappa =x}\nonumber \\ & \quad +|\psi (0,x)(-x)^{q+2}| \bigg(\tfrac {\partial ^q}{\partial \kappa ^{q}}r_{n,0}(\kappa )\bigg)_{\kappa =x}\nonumber \\ & := E_1+E_2.\nonumber \end{align*}
</div>
<p>Because of the fact \(\psi (u,x)\rightarrow 0\) as \(u\rightarrow x,\) for a given \(\epsilon {\gt}0\) there will exist a \(\delta {\gt}0\) such that \(|\psi (u,x)|{\lt}\epsilon \) whenever \(|u-x|{\lt}\delta .\) </p>
<p>Furthermore, if \(m \ge \{ \gamma ,q+2\} \), and \(m\) a positive integer </p>
<div class="displaymath" id="a0000000064">
  \[ |(u-x)^{q+2}\psi (u,x)|\leq M |u-x|^m \]
</div>
<p> for \(|u-x|\geq \delta .\) Thus </p>
<div class="displaymath" id="a0000000065">
  \begin{align}  |E_1|& \leq \sum _{\nu =1}^{\infty }\sum _{{2i+j\leq q}\atop {i,j\geq 0}}n^{i}|\nu -n x|^{j}\tfrac {|\phi _{i,j,q}(x)|}{x^q(1+x)^{2q}} {r_{n,\nu }(x)}\bigg(\epsilon \int _{|u-x|{\lt}\delta } v_{n,\nu }(u)|u-x|^{q+2}du\nonumber \\ & \quad +M\int _{|u-x|\geq \delta } v_{n,\nu }(u) |u-x|^m du\bigg)\nonumber \\ & :=J_1+J_2.\nonumber \end{align}
</div>
<p> Let us consider </p>
<div class="displaymath" id="a0000000066">
  \[ K=\displaystyle \sup _{{2i+j\leq q}\atop { i,j\geq 0}}\tfrac {|\phi _{i,j,q}(x)|}{x^q(1+x)^{2q}}. \]
</div>
<p>Using Schwarz’s inequality, <a href="#L1">lemma 1</a>, <a href="#L2">lemma 2</a> and <a href="#L3">lemma 3</a>, we deduce that </p>
<div class="displaymath" id="a0000000067">
  \begin{eqnarray*}  J_1& =& \epsilon \, \  K\sum _{\nu =1}^{\infty }\sum _{ {2i+j\leq q}\atop {i,j\geq 0}}n^{i}|\nu -n x|^{j}r_{n,\nu }(x) \bigg(\int _{0}^{\infty }v_{n,\nu }(u)du\bigg)^{1/2}\\ & & \bigg(\int _{0}^{\infty }v_{n,\nu }(u)(e_1-xe_0)^{2q+4}du\bigg)^{1/2}\nonumber \\ & \leq & \epsilon \, \  K\sum _{ {2i+j\leq q}\atop { i,j\geq 0}} n^{i+j}\bigg(\sum _{\nu =0}^{\infty }r_{n,\nu } (x)\bigg(\tfrac {\nu }{n}-x\bigg)^{2j}-x^{2j}r_{n,0}(x)\bigg)^{1/2}\nonumber \\ & & \bigg(V_n((e_1-xe_0)^{2q+4},x)-x^{2q+4}r_{n,0}(x)\bigg)^{1/2}\nonumber \\ & =& \epsilon \sum _{{2i+j\leq q}\atop {i,j\geq 0}}n^{i+j}\bigg\{ O_x\bigg(\tfrac {1}{n^j}\bigg)+O_x\bigg(\tfrac {1}{n^s}\bigg)\bigg\} ^{1/2} \bigg\{ O_x\bigg(\tfrac {1}{n^{q+2}}\bigg)+O_x\bigg(\tfrac {1}{n^p}\bigg)\bigg\} ^{1/2}, \end{eqnarray*}
</div>
<p> for any \(s,p{\gt}0\), choosing \(s\) and \(p\) such that \(s{\gt}j\) and \(p{\gt}q+2\), we obtain </p>
<div class="displaymath" id="a0000000068">
  \begin{eqnarray}  J_1& \leq &  \epsilon \sum _{{2i+j\leq q}\atop { i,j\geq 0}} n^{i+j}O_x\bigg(\tfrac {1}{n^{j/2}}\bigg)O_x\bigg(\tfrac {1}{n^{q/2+1}}\bigg)=\epsilon . O_x(n^{-1}).\nonumber \end{eqnarray}
</div>
<p>Since \(\epsilon {\gt}0\) is arbitrary, \(n J_1\rightarrow 0\) as \(n\rightarrow \infty .\) </p>
<p>Applying again Schwarz’s inequality, <a href="#L1">lemma 1</a> and <a href="#L3">lemma 3</a>, we obtain </p>
<div class="displaymath" id="a0000000069">
  \begin{eqnarray*}  J_2& \leq &  M_1\sum _{{2i+j\leq q}\atop { i,j\geq 0}}n^{i+j} \bigg(\sum _{\nu =0}^{\infty }\bigg(\tfrac {\nu }{n}-x\bigg)^{2j}r_{n,\nu }(x)-x^{2j}r_{n,0}(x)\bigg)^{1/2}\\ & & \bigg(\sum _{\nu =1}^{\infty }r_{n,\nu } (x)\int _{|u-x|\geq \delta }v_{n,\nu }(u)(e_1-xe_0)^{2m}du\bigg)^{1/2}\\ & \leq &  M_1\sum _{{2i+j\leq q}\atop {i,j\geq 0}}n^{i+j}\bigg\{ O_x\bigg(\tfrac {1}{n^j}\bigg)+O_x\bigg(\tfrac {1}{n^s}\bigg)\bigg\} ^{1/2} \bigg\{ O_x\bigg(\tfrac {1}{n^m}\bigg)+O_x\bigg(\tfrac {1}{n^l}\bigg)\bigg\} ^{1/2}, \end{eqnarray*}
</div>
<p> for any \(s,l{\gt}0\). We now choose \(s,l\) such that \( s{\gt} j, l{\gt}m\). Then, we derive that </p>
<div class="displaymath" id="a0000000070">
  \begin{eqnarray*}  J_2& \leq &  M_1 \sum _{{2i+j\leq q}\atop { i,j\geq 0}}n^{i+j}O_x\bigg(\tfrac {1}{n^{j/2}}\bigg)O_x\bigg(\tfrac {1}{n^{m/2}}\bigg)= O_x\bigg(\tfrac {1}{n^{(m-q)/2}}\bigg), \end{eqnarray*}
</div>
<p> which implies that \(n J_2\rightarrow 0\), as \(n\rightarrow \infty \), on choosing \(m {\gt}q+2.\) </p>
<p>From the above estimates of \(J_1\) and \(J_2,\) \(n E_1\rightarrow 0\), as \(n\rightarrow \infty .\) Finally, we have </p>
<div class="displaymath" id="a0000000071">
  \[  |E_2|=|\psi (0,x)(-x)^q| \bigg(\tfrac {\partial ^q}{\partial \kappa ^{q}}r_{n,0}(\kappa )\bigg)_{\kappa =x}. \]
</div>
<p> Since \(|\psi (0,x)(-x)^q|{\lt} N_1\) for some \(N_1{\gt}0,\) also </p>
<div class="displaymath" id="a0000000072">
  \[ \bigg(\tfrac {\partial ^q}{\partial \kappa ^{q}}r_{n,0}(\kappa )\bigg)_{\kappa =x}=\bigg[\tfrac {\partial ^q}{\partial \kappa ^{q}} \bigg(e^{-n \kappa /(1+\kappa )}\bigg)\bigg]_{\kappa =x} \]
</div>
<p> which yields that \(nE_2\rightarrow 0\) as \(n\rightarrow \infty .\) By collecting the estimates of \(E_1\) and \(E_2,\) we obtain \(n I_2\rightarrow 0\) as \(n\rightarrow \infty .\) Therefore, combining \(I_1\) and \(I_2,\) the required result follows. <div class="proof_wrapper" id="a0000000073">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<p><div class="corollary_thmwrapper " id="C1">
  <div class="corollary_thmheading">
    <span class="corollary_thmcaption">
    Corollary
    </span>
    <span class="corollary_thmlabel">9</span>
  </div>
  <div class="corollary_thmcontent">
  <p> If \(f\in C[0,\infty )\) and \(|f(t)|\leq C (1+t)^{\gamma }\) for some \(\gamma {\gt}0, t\ge 0\) with the existence of \(f^{\prime \prime }\) in \(x\in (0,\infty ),\) then we have </p>
<div class="displaymath" id="a0000000074">
  \begin{eqnarray*}  \lim _{n\rightarrow \infty }n((V_nf)(x) -f(x))& =& \tfrac {x(x^2+3x+2)}{2}f^{\prime \prime }(x). \end{eqnarray*}
</div>

  </div>
</div> </p>
<p><small class="footnotesize">  </small></p>
<div class="bibliography">
<h1>Bibliography</h1>
<dl class="bibliography">
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</dd>
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</dd>
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  <dd><p><a href ="https://doi.org.10.1016/j.amc.2015.09.037"> <i class="sc">V. Gupta</i> and <i class="sc">G. C. Greubel</i>, <i class="it">Moment estimations of new Szász–Mirakyan–Durrmeyer operators</i>, Appl. Math. Comput., <b class="bf">271</b> (2015), pp. 540–547, <a href="https://doi.org.10.1016/j.amc.2015.09.037">https://doi.org.10.1016/j.amc.2015.09.037</a>. <img src="img-0001.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
  <dt><a name="noor">8</a></dt>
  <dd><p><a href ="https://doi.org.10.1016/j.jmaa.2005.07.036"> <i class="sc">V. Gupta</i> and <i class="sc">M. A. Noor</i>, <i class="it">Convergence of derivatives for certain mixed Szász–Beta operators</i>, J. Math. Anal. Appl., <b class="bf">321</b> (2006) no. 1, pp. 1–9, <a href="https://doi.org.10.1016/j.jmaa.2005.07.036">https://doi.org.10.1016/j.jmaa.2005.07.036</a>. <img src="img-0001.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
  <dt><a name="vgtmr">9</a></dt>
  <dd><p><a href ="https://doi.org.10.1016/j.amc.2014.11.078"> <i class="sc">V. Gupta</i> and <i class="sc">Th. M. Rassias</i>, <i class="it">Direct estimates for certain Szász type operators</i>, Appl. Math. Comput., <b class="bf">251</b> (2015), pp. 469–474, <a href="https://doi.org.10.1016/j.amc.2014.11.078">https://doi.org.10.1016/j.amc.2014.11.078</a>. <img src="img-0001.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
  <dt><a name="vgmtr">10</a></dt>
  <dd><p><a href ="https://doi.org.10.1007/978-3-030-19455-0"> <i class="sc">V. Gupta</i> and <i class="sc">M. Th. Rassias</i>, <i class="it">Moments of Linear Positive Operators and Approximation</i>, Series: SpringerBriefs in Mathematics, Springer Nature Switzerland AG, 2019, <a href="https://doi.org.10.1007/978-3-030-19455-0">https://doi.org.10.1007/978-3-030-19455-0</a>. <img src="img-0001.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
  <dt><a name="vgpn">11</a></dt>
  <dd><p><a href ="https://doi.org.10.1016/j.joems.2012.07.001"> <i class="sc">V. Gupta, D. K. Verma</i> and <i class="sc">P. N. Agrawal</i>, <i class="it">Simultaneous approximation by certain Baskakov–Durrmeyer–Stancu operators</i>, J. Egyptian Math. Soc., <b class="bf">20</b> (2012) no. 3, pp. 183–187, <a href="https://doi.org.10.1016/j.joems.2012.07.001">https://doi.org.10.1016/j.joems.2012.07.001</a>. <img src="img-0001.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
  <dt><a name="IS">12</a></dt>
  <dd><p><a href ="https:doi.org.10.1016/0021-9045(78)90105-3"> <i class="sc">M. Ismail</i>, <i class="it">Polynomials of binomial type and approximation theory</i>, J. Approx. Theory, <b class="bf">23</b> (1978), pp. 177–186, <a href="https:doi.org.10.1016/0021-9045(78)90105-3">https:doi.org.10.1016/0021-9045(78)90105-3</a>. <img src="img-0001.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
  <dt><a name="isma">13</a></dt>
  <dd><p><a href ="https://doi.org.10.1016/0022-247X(78)90090-2"> <i class="sc">M. Ismail</i> and <i class="sc">C. P. May</i>, <i class="it">On a family of approximation operators</i>, J. Math. Anal. Appl., <b class="bf">63</b> (1978), pp. 446–462, <a href="https://doi.org.10.1016/0022-247X(78)90090-2">https://doi.org.10.1016/0022-247X(78)90090-2</a>. <img src="img-0001.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
  <dt><a name="ispir">14</a></dt>
  <dd><p><i class="sc">N. Ispir</i>, <i class="it">On modified Baskakov operators on weighted spaces</i>, Turk. J. Math., <b class="bf">26</b> (2001) no. 3, pp. 355–365. </p>
</dd>
</dl>


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