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<title>Third order convergence theorem <br />for a family of Newton like methods in Banach space: Third order convergence theorem <br />for a family of Newton like methods in Banach space</title>
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<h1>Third order convergence theorem <br />for a family of Newton like methods in Banach space</h1>
<p class="authors">
<span class="author">Tugal Zhanlav\(^\ast \) Dorjgotov Khongorzul\(^\S \)</span>
</p>
<p class="date">June 15, 2014.</p>
</div>
<p>\(^\ast \)Department of Applied Mathematics, National University of Mongolia, PB 46/687 Ulaanbaatar 210646, Mongolia, e-mail: <span class="tt">tzhanlav@yahoo.com</span> </p>
<p>\(^\S \)Department of Applied Mathematics, National University of Mongolia, PB 46/687 Ulaanbaatar 210646, Mongolia, e-mail: <span class="tt">pilpalpil@gmail.com</span> </p>

<div class="abstract"><p> In this paper, we propose a family of Newton-like methods in Banach space which includes some well known third-order methods as particular cases. We establish the Newton-Kantorovich type convergence theorem for a proposed family and get an error estimate. </p>
<p><b class="bf">MSC.</b> 47H99, 65J15. </p>
<p><b class="bf">Keywords.</b> Nonlinear equations in Banach space; third order Newton like methods; recurrence relations; error bounds; convergence domain. </p>
</div>
<h1 id="a0000000002">1 Introduction</h1>
<p> Recently, many third order iterative methods free from second derivative have been derived and studied for nonlinear systems <span class="cite">
	[
	<a href="#[9]" >1</a>
	]
</span>-<span class="cite">
	[
	<a href="#[7]" >10</a>
	]
</span>. In particular, in <span class="cite">
	[
	<a href="#[9]" >1</a>
	]
</span> were suggested two Chebyshev-like (CL1, CL2) methods, while in <span class="cite">
	[
	<a href="#[7]" >10</a>
	]
</span> are considered two families of modifications of Chebyshev method (MOD1, MOD2). In <span class="cite">
	[
	<a href="#[8]" >6</a>
	]
</span> was also presented a new family of Chebyshev-type methods with a real parameter \(\theta \) \((A2_{\theta }).\) All the above mentioned methods are obtained using different approximations of second derivative in Chebyshev method. </p>
<p>In <span class="cite">
	[
	<a href="#[1]" >9</a>
	]
</span> it was proposed a family of third-order methods given by </p>
<div class="equation" id="e1">
<p>
  <div class="equation_content">
    \begin{equation} \label{e1} x_{n+1}=x_n-\tfrac {f(x_n)}{\left(1+\frac{1}{2a}\right)f’(x_n)-\frac{1}{2a}f’\big(x_n+a\frac{f(x_n)}{f'(x_n)}\big)}, \qquad a\in \mathbb {R}\setminus \{ 0\}  \end{equation}
  </div>
  <span class="equation_label">1</span>
</p>
</div>
<p> for solving nonlinear scalar equations \(f(x)=0.\) In this study, we consider a generalization of methods (<a href="#e1">1</a>) in Banach space, which is used to solve the nonlinear operator equation </p>
<div class="equation" id="e2">
<p>
  <div class="equation_content">
    \begin{equation} \label{e2} F(x)=0. \end{equation}
  </div>
  <span class="equation_label">2</span>
</p>
</div>
<p> Suppose that \(F\) is defined on an open convex domain \(\Omega \) of a Banach space \(X\) with values in a Banach space \(Y,\) \(F'(x)\) is a Frechet derivative in \(\Omega ,\) and \(F'(x)^{-1}\) exists. The generalization of methods (<a href="#e1">1</a>) is </p>
<div class="displaymath" id="a0000000003">
  \begin{align}  y_n &  = x_n - F’(x_n)^{-1}F(x_n), \nonumber \\ z_n&  =(1+a)x_{n}-ay_n, \qquad a\neq 0 \nonumber \\ x_{n+1} &  = x_n-\left[\left(1+\tfrac {1}{2a}\right)F’(x_n)-\tfrac {1}{2a}F’(z_n)\right]^{-1}F(x_n).\label{e3} \end{align}
</div>
<p> Thus we have a family of methods (<a href="#e3">3</a>) for solving nonlinear equation (<a href="#e2">2</a>). We consider some particular cases of (<a href="#e3">3</a>). Let \(a=-1\). Then (<a href="#e3">3</a>) leads to </p>
<div class="displaymath" id="e4">
  \begin{eqnarray} \label{e4} y_n &  = &  x_n-F’(x_n)^{-1}F(x_n),\nonumber \\ x_{n+1} &  = &  x_{n}-\tfrac {1}{2}\left[F’(x_n)+F’(y_n)\right]^{-1}F(x_n), \end{eqnarray}
</div>
<p> which was proposed by Q.Wu and Y.Zhao in <span class="cite">
	[
	<a href="#[4]" >7</a>
	]
</span>. They established third-order convergence of this method by using majorizing function and obtained the error estimate. It should be mentioned that the iteration (<a href="#e4">4</a>) for scalar equation was given also in <span class="cite">
	[
	<a href="#[10]" >5</a>
	]
</span>. Let \(a=-\tfrac {1}{2}\). Then (<a href="#e3">3</a>) leads to </p>
<div class="displaymath" id="a0000000004">
  \begin{eqnarray*}  y_n &  = &  x_n -\tfrac 12F’(x_n)^{-1}F(x_n),\nonumber \\ x_{n+1} &  = &  x_n-F’(y_n)^{-1}F(x_n). \end{eqnarray*}
</div>
<p> This is a generalization of the third order method proposed by the Frontini and Sormani <span class="cite">
	[
	<a href="#[5]" >2</a>
	]
</span> for scalar case. Thus, the proposed iteration (<a href="#e3">3</a>) can be considered as a generalization of well known iterations. </p>
<p>We prove Newton-Kantorovich type convergence theorem for the family of methods (<a href="#e3">3</a>) to show that it has third order convergence by using recurrent relations <span class="cite">
	[
	<a href="#[2]" >3</a>
	]
</span>-<span class="cite">
	[
	<a href="#[3]" >4</a>
	]
</span> and get the error bounds. Finally, some examples are provided to show the application of the proposed method. </p>
<h1 id="a0000000005">2 Preliminaries</h1>
<p>Let us assume that \(F'(x_0)^{-1} \in L(Y,X)\) exists for some \(x_0 \in \Omega ,\) where \(L(Y,X)\) is a set of bounded linear operators from \(Y\) into \(X.\) Moreover, we suppose that (see <span class="cite">
	[
	<a href="#[3]" >4</a>
	]
</span>) </p>
<div class="displaymath" id="a0000000006">
  \begin{align}  \| \Gamma _0\|  & = \| F’(x_0)^{-1}\|  \leq \beta , \tag {c$_1$} \\ \| y_0-x_0\|  & = \| \Gamma _0F(x_0)\| \leq \eta , \tag {c$_2$} \\ \| F”(x)\|  & \leq M, \qquad x \in \Omega , \tag {c$_3$} \\ \| F”(x)-F”(y)\|  & \leq K\| x-y\| , \quad x,y \in \Omega , \quad K{\gt}0. \tag {c$_4$} \end{align}
</div>
<p> Let \(F\) be a nonlinear twice Frechet differentiable operator in an open convex domain \(\Omega .\) We denote \(\Gamma _n= F'(x_n)^{-1},\) </p>
<div class="displaymath" id="e5">
  \begin{align} \label{e5} a_0& =M\beta \eta , \\ f(x)& =\tfrac {2-x}{2-3x}, \qquad 0{\lt}x{\lt}\tfrac 23, \nonumber \\ g(x)& =\tfrac {x^2}{(2-x)^2}d,\nonumber \end{align}
</div>
<p> where </p>
<div class="displaymath" id="e6">
  \begin{align} \label{e6} d& =\tfrac 37+\tfrac {\omega }{4}(1+|a|), \\ \omega &  = \tfrac {K}{M^2m}, \nonumber \\ m& =\min _n\| \Gamma _n\|  {\gt} 0 \nonumber \end{align}
</div>
<p> and define the sequence </p>
<div class="displaymath" id="a0000000007">
  \begin{equation*}  a_{n+1}=f(a_n)^2g(a_n)a_n. \end{equation*}
</div>
<p> We need following technical lemmas, whose proofs are trivial <span class="cite">
	[
	<a href="#[3]" >4</a>
	]
</span>. <div class="lemma_thmwrapper " id="a0000000008">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">1</span>
  </div>
  <div class="lemma_thmcontent">
  <p>Let \(f\) and \(g\) be two real functions given in <span class="rm">(<a href="#e5">8</a>)</span>. Then <br />(i) \(f(x)\) and \(g(x)\) are increasing and \(f(x){\gt}1\) for \(x\in (0,\tfrac 23)\)<br />(ii) \(f(\gamma x){\lt}f(x)\) and \(g(\gamma x) {\lt} \gamma ^2 g(x)\) for \(\gamma \in (0,1).\) </p>

  </div>
</div> <div class="lemma_thmwrapper " id="a0000000009">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">2</span>
  </div>
  <div class="lemma_thmcontent">
  <p>Let \(f^2(a_0)g(a_0){\lt}1.\) Then the sequence \(\{ a_n\} \) is decreasing. </p>

  </div>
</div> <div class="lemma_thmwrapper " id="a0000000010">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">3</span>
  </div>
  <div class="lemma_thmcontent">
  <p>If \(0{\lt}a_0{\lt}\tfrac {2}{3+\sqrt{d}}\), then \(f^2(a_0)g(a_0){\lt}1.\) </p>

  </div>
</div> <div class="lemma_thmwrapper " id="lem4">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">4</span>
  </div>
  <div class="lemma_thmcontent">
  <p> Let \(0{\lt}a_0{\lt}\tfrac {2}{3+\sqrt{d}}\) and define \(\gamma =a_1/a_0.\) Then<br />(i) \(\gamma = f^2(a_0)g(a_0)\in (0,1)\); <br />\((ii_n)\) \(a_n\leq \gamma ^{3^n-1}a_{n-1}\leq \gamma ^{\tfrac {3^n-1}{2}}a_0\); <br />\((iii_n)\) \(f(a_n)g(a_n)\leq \tfrac {\gamma ^{3^{n}}}{f(a_0)}=\Delta \gamma ^{3^n}, \qquad \Delta = \tfrac {1}{f(a_0)}{\lt}1.\) </p>

  </div>
</div> </p>
<h1 id="a0000000011">3 Convergence study</h1>
<p>According to (<a href="#e3">3</a>), we have </p>
<div class="equation" id="e7">
<p>
  <div class="equation_content">
    \begin{equation} \label{e7} x_1-x_0=-A_0^{-1}F(x_0), \end{equation}
  </div>
  <span class="equation_label">10</span>
</p>
</div>
<p> where </p>
<div class="equation" id="e8">
<p>
  <div class="equation_content">
    \begin{equation} \label{e8} A_0=(1+\tfrac {1}{2a})F'(x_0)-\tfrac {1}{2a}F'(z_0). \end{equation}
  </div>
  <span class="equation_label">11</span>
</p>
</div>
<p> Using the following formula </p>
<div class="displaymath" id="a0000000012">
  \begin{equation*}  F’(z_0)=F’(x_0)+\int _{x_0}^{z_0}F”(x){\rm d}x \end{equation*}
</div>
<p> in (<a href="#e8">11</a>), we obtain </p>
<div class="displaymath" id="a0000000013">
  \begin{equation*}  A_0=F’(x_0)(I-P_0), \end{equation*}
</div>
<p> where </p>
<div class="displaymath" id="a0000000014">
  \begin{equation*}  P_0=\tfrac {1}{2a}\Gamma _0\int _{x_0}^{z_0}F”(x){\rm d}x. \end{equation*}
</div>
<p> If we notice that \(M||\Gamma _0||||\Gamma _0F(x_0)||\leq a_0{\lt}\tfrac 23,\) then follows </p>
<div class="displaymath" id="a0000000015">
  \begin{equation*}  ||P_0||\leq \tfrac {1}{2|a|}||\Gamma _0||M|a|||\Gamma _0F(x_0)||\leq \tfrac {a_0}{2}{\lt}\tfrac {1}{3}, \end{equation*}
</div>
<p> which shows the existence of \(A_0^{-1}\) </p>
<div class="displaymath" id="a0000000016">
  \begin{equation*}  A_0^{-1}=(I-P_0)^{-1}\Gamma _0, \end{equation*}
</div>
<p> where \(P_0=\tfrac 12 \Gamma _0F''(\xi _0)\Gamma _0F(x_0).\) So, from (<a href="#e7">10</a>) we get </p>
<div class="displaymath" id="a0000000017">
  \begin{equation*}  \| x_1-x_0\| \leq \tfrac {1}{1-\frac{a_0}{2}}\| \Gamma _0F(x_0)\|  \leq \tfrac {\eta }{1-\frac{a_0}{2}} {\lt} \tfrac {\eta }{(1-\frac{a_0}{2})(1-\gamma \Delta )}=R\eta , \end{equation*}
</div>
<p> where \(R=\tfrac {1}{\left(1-\frac{a_0}{2}\right)(1-\gamma \Delta )}\). This means that \(y_0, x_1 \in B(x_0,R\eta )=\{ x\in X: \| x-x_0\| {\lt}R\eta \} .\) </p>
<p>In these conditions, we prove the following statements for \(n\geq 1\): </p>
<div class="displaymath" id="a0000000018">
  \begin{align}  \| \Gamma _n\|  &  \leq f(a_{n-1})\| \Gamma _{n-1}\| , \tag {$I_n$}\\ \| \Gamma _nF(x_n)\|  &  \leq f(a_{n-1})g(a_{n-1})\| \Gamma _{n-1}F(x_{n-1})\| , \tag {$II_n$} \\ M\| \Gamma _n\| \| \Gamma _nF(x_n)\|  &  \leq a_n, \tag {$III_n$} \\ \| x_{n+1}-x_n\|  &  \leq \tfrac {1}{1-a_n/2}\| \Gamma _nF(x_n)\| , \tag {$IV_n$} \\ y_n, x_{n+1} &  \in B(x_0,R\eta ). \tag {$V_n$} \end{align}
</div>
<p> Assuming \(\tfrac {a_0}{1-a_0/2}{\lt}1\) which is valid for \(a_0{\lt}2/3\) and \(x_1\in \Omega \) we have </p>
<div class="displaymath" id="a0000000019">
  \begin{equation*}  \| I-\Gamma _0F’(x_1)\| \leq \| \Gamma _0\| \| F’(x_0)-F’(x_1)\| \leq \| \Gamma _0\| M\| x_0-x_1\| \leq \tfrac {a_0}{1-\frac{a_0}{2}}{\lt}1. \end{equation*}
</div>
<p> Then, by the Banach lemma, \(\Gamma _1\) is defined and satisfies </p>
<div class="displaymath" id="a0000000020">
  \begin{equation*}  \| \Gamma _1\| \leq \tfrac {\| \Gamma _0\| }{1-\| \Gamma _0\| \| F’(x_0)-F’(x_1)\| }\leq \tfrac {a_0}{1-\frac{a_0}{1-\frac{a_0}{2}}}\| \Gamma _0\| =f(a_0)\| \Gamma _0\| . \end{equation*}
</div>
<p> Taking into account <span class="rm">(<a href="#e3">3</a>)</span> and the Taylor formula of \(x_n, y_n \in \Omega ,\) we have </p>
<div class="equation" id="e9">
<p>
  <div class="equation_content">
    \begin{equation} \label{e9} F(x_{n+1})=F(y_n)+F'(y_n)(x_{n+1}-y_n)+\int _{y_n}^{x_{n+1}}\! \!  F''(x)(x_{n+1}-x){\rm d}x, \  n=0,1,\ldots \end{equation}
  </div>
  <span class="equation_label">17</span>
</p>
</div>
<p> Also </p>
<div class="displaymath" id="a0000000021">
  \begin{align*}  F(y_n) &  = F(x_n)+F’(x_n)(y_n-x_n)+\int _{x_n}^{y_n}F”(x)(y_n-x){\rm d}x\\ &  =\int _0^1F”(x_n+t(y_n-x_n))(y_n-x_n)^2(1-t){\rm d}t \\ F’(y_n)(x_{n+1}-y_n) &  = \tfrac 12\int _0^1F”(x_n-at(y_n-x_n))(x_n-y_n)(x_{n+1}-x_n){\rm d}t. \end{align*}
</div>
<p> Substituting the last two expressions into (<a href="#e9">17</a>), we get </p>
<div class="displaymath" id="a0000000022">
  \begin{eqnarray*}  F(x_{n+1}) &  = &  \int _0^1F”(x_n+t(y_n-x_n))(y_n-x_n)^2(1-t){\rm d}t + \\ & &  +\tfrac 12\int _0^1F”(x_n-at(y_n-x_n))(x_n-y_n)(x_{n+1}-y_n){\rm d}t \\ & &  + \int _0^1F”(x_n+t(y_n-x_n))(y_n-x_n)(x_{n+1}-y_n){\rm d}t\\ & &  +\int _0^1F”(y_n+t(x_{n+1}-y_n))(x_{n+1}-y_n)^2(1-t){\rm d}t. \end{eqnarray*}
</div>
<p> From (<a href="#e3">3</a>) we also obtain </p>
<div class="displaymath" id="a0000000023">
  \begin{equation*}  x_{n+1}-y_n=\tfrac {\Gamma _n}{2a}\int _{x_n}^{z_n}F”(x)(x_{n+1}-x_n){\rm d}x \end{equation*}
</div>
<p> and </p>
<div class="displaymath" id="a0000000024">
  \begin{equation*}  x_{n+1}-x_n=(I-P_n)^{-1}(y_n-x_n)=y_n-x_n+P_n(I-P_n)(y_n-x_n), \end{equation*}
</div>
<p> where </p>
<div class="displaymath" id="a0000000025">
  \begin{equation*}  P_n=\tfrac {\Gamma _n}{2a}\int _{x_n}^{z_n}F”(x){\rm d}x. \end{equation*}
</div>
<p> Taking into account </p>
<div class="displaymath" id="a0000000026">
  \begin{align*} & \int _0^1F”(x_n-at(y_n-x_n))(x_{n+1}-y_n)(x_{n+1}-x_n){\rm d}t =\\ &  =-\int _0^1F”(x_n-at(y_n-x_n))(y_{n}-x_n)^2{\rm d}t\\ & \quad -\int _0^1F”(x_n-at(y_n-x_n))(y_{n}-x_n)P_n(I-P_n)^{-1}(y_n-x_n){\rm d}t \end{align*}
</div>
<p> we obtain </p>
<div class="displaymath" id="e10">
  \begin{eqnarray} \label{e10} F(x_{n+1}) &  = &  \int _0^1[F”(x_n+t(y_n-x_n))-F”(x_n)](y_{n}-x_n)^2(1-t){\rm d}t\nonumber \\ & & +\tfrac 12\int _0^1[F”(x_n)-F”(x_n-at(y_n-x_n))](y_{n}-x_n)^2{\rm d}t\nonumber \\ & &  -\tfrac 12\int _0^1F”(x_n-at(y_n-x_n))(y_{n}-x_n)P_n(I-P_n)^{-1}(y_{n}-x_n){\rm d}t\nonumber \\ & &  +\int _0^1F”(x_n+t(y_n-x_n))(y_{n}-x_n)(x_{n+1}-y_n){\rm d}t\nonumber \\ & &  +\int _0^1F”(y_n+t(x_{n+1}-y_n))(x_{n+1}-y_n)^2(1-t){\rm d}t. \end{eqnarray}
</div>
<p> From (<a href="#e10">18</a>) for \(n=0,\) we obtain \(||\Gamma _0F(x_n)||\); therefore </p>
<div class="displaymath" id="a0000000027">
  \begin{equation*}  ||\Gamma _1F(x_1)||\leq ||\Gamma _1F’(x_0)||||\Gamma _0F(x_1)||\leq f(a_0)g(a_0)||\Gamma _0F(x_0)||. \end{equation*}
</div>
<p> So, \((II_1)\) is true. To prove \((III_1)\) and \((IV_1)\), notice that </p>
<div class="displaymath" id="a0000000028">
  \begin{equation*}  M\| \Gamma _1\| \| \Gamma _1F(x_1)\| \leq M f^2(a_0)g(a_0)\eta \beta = f^2(a_0)g(a_0)a_0=a_1 \end{equation*}
</div>
<p> and </p>
<div class="displaymath" id="a0000000029">
  \begin{equation*}  \| x_2-x_1\| \leq \| A_1^{-1}F(x_1)\|  \end{equation*}
</div>
<p> where </p>
<div class="displaymath" id="a0000000030">
  \begin{equation*}  A_1=F’(x_1)+\tfrac {1}{2a}(F’(x_1)-F’(z_1))=F’(x_1)[I-P_1]. \end{equation*}
</div>
<p> Since </p>
<div class="displaymath" id="a0000000031">
  \begin{equation*}  \| P_1\| =\tfrac 12\| \Gamma _1F”(\eta _1)\Gamma _1F(x_1)\| \leq \tfrac {a_1}{2}{\lt}1, \end{equation*}
</div>
<p> there exists \(A_1^{-1}=(I-P_1)^{-1}\Gamma _1,\) thereby we get </p>
<div class="displaymath" id="a0000000032">
  \begin{equation*}  \| x_2-x_1\| \leq \tfrac {1}{1-a_1/2}\| \Gamma _1F(x_1)\| \leq \tfrac {f(a_0)g(a_0)}{1-a_0/2}\eta =\tfrac {\Delta \gamma }{1-a_0/2}\eta . \end{equation*}
</div>
<p> Consequently, we obtain </p>
<div class="displaymath" id="a0000000033">
  \begin{equation*}  \| x_2-x_0\| \leq \| x_2-x_1\| +\| x_1-x_0\| \leq \tfrac {\Delta \gamma }{1-a_0/2}\eta +\tfrac {\eta }{1-a_0/2}=\tfrac {1+\Delta \gamma }{1-a_0/2}\eta {\lt}R\eta . \end{equation*}
</div>
<p> Analogously, we get </p>
<div class="displaymath" id="a0000000034">
  \begin{equation*}  \| y_1-x_0\| \leq \| y_1-x_1\| +\| x_1-x_0\| \leq f(a_0)g(a_0)\eta +\tfrac {1}{1-a_0/2}\eta {\lt}R\eta \end{equation*}
</div>
<p> i.e. \((IV_1)\) and \((V_1)\) are proved. Now, following an inductive procedure and assuming </p>
<div class="equation" id="e11">
<p>
  <div class="equation_content">
    \begin{equation} \label{e11} y_n,x_{n+1}\in \Omega \quad \mbox{ and }\quad \tfrac {a_n}{1-a_n/2}<1 \quad \forall n\in N, \end{equation}
  </div>
  <span class="equation_label">19</span>
</p>
</div>
<p> the items \((I_n)-(V_n)\) can be proved. Notice that \(\Gamma _n {\gt}0\) for all \(n=0,1,\ldots .\) Indeed if \(\Gamma _k=0\) for some \(k,\) then due to statement \((I_n),\) we have \(\| \Gamma _n\| =0\) for all \(n\geq k.\) As a consequence, the iteration (<a href="#e3">3</a>) terminated after \(k\)-th step, i.e. the convergence of iteration does not hold. To establish the convergence of \(\{ x_n\} \) we only have to prove that it is a Cauchy sequence and that the above assumptions (<a href="#e11">19</a>) are true. Note that </p>
<div class="displaymath" id="a0000000035">
  \begin{align*}  \tfrac {1}{1-a_n/2}\| \Gamma _nF(x_n)\|  &  \leq \tfrac {1}{1-a_0/2}f(a_{n-1})g(a_{n-1})\| \Gamma _{n-1}F(x_{n-1})\| \\ &  \leq \ldots \leq \tfrac {1}{1-a_0/2}\| \Gamma _0F(x_0)\| \prod _{k=0}^{n-1}f(a_k)g(a_k). \end{align*}
</div>
<p> As a consequence of Lemma <a href="#lem4">4</a>, it follows that </p>
<div class="displaymath" id="a0000000036">
  \begin{equation*}  \prod _{k=0}^{n-1}f(a_k)g(a_k)\leq \prod _{k=0}^{n-1}\Delta \gamma ^{3^k}=\Delta ^n\gamma ^{\tfrac {3^n-1}{2}}. \end{equation*}
</div>
<p> So, from \(\Delta {\lt}1\) and \(\gamma {\lt}1,\) we deduce that \(\prod _{k=0}^{n-1}f(a_k)g(a_k)\) converges to zero by letting \(n\to \infty .\) We are now ready to state the main result on convergence for (<a href="#e3">3</a>). </p>
<p><div class="theorem_thmwrapper " id="a0000000037">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">5</span>
  </div>
  <div class="theorem_thmcontent">
  <p>Let us assume that \(\Gamma _0=F'(x_0)^{-1}\in L(Y,X)\) exists at some \(x_0\in \Omega \) and \((c_1)-(c_4)\) are satisfied. Suppose that </p>
<div class="equation" id="e12">
<p>
  <div class="equation_content">
    \begin{equation} \label{e12} 0<a_0<\tfrac {2}{3+\sqrt{d}}\quad \mbox{ with } d \mbox{ given by {\rm (\ref{e6})}} \end{equation}
  </div>
  <span class="equation_label">20</span>
</p>
</div>
<p> Then, if \(\overline{B(x_0,R\eta )}=\{ x\in X: \| x-x_0\| \leq R\eta \} \subseteq \Omega ,\) the sequence \(\{ x_n\} \) defined in <span class="rm">(<a href="#e3">3</a>)</span> and starting at \(x_0\) has, at least, R-order three and converges to a solution \(x^*\) of the equation <span class="rm">(<a href="#e2">2</a>)</span>. In that case, the solution \(x^*\) and the iterates \(y_n,x_n\) belong to \(\overline{B(x_0,R\eta )}\) and \(x^{*}\) is the only solution of <span class="rm">(<a href="#e2">2</a>)</span> in \(B(x_0,2/M\beta -R\eta )\cap \Omega .\) Furthermore, we have the following error estimates: </p>
<div class="equation" id="e13">
<p>
  <div class="equation_content">
    \begin{equation} \label{e13} \| x_n-x^{*}\| \leq \tfrac {1}{1-\frac{a_0}{2}\gamma ^{\frac{3^n-1}{2}}}\gamma ^{\frac{3^n-1}{2}}\tfrac {\Delta ^n}{1-\Delta \gamma ^{3^n}}\eta . \end{equation}
  </div>
  <span class="equation_label">21</span>
</p>
</div>

  </div>
</div> <div class="proof_wrapper" id="a0000000038">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> Let us now prove (<a href="#e12">20</a>). From \(a_0\in (0;\tfrac {2}{3+\sqrt{d}})\) follows </p>
<div class="displaymath" id="a0000000039">
  \begin{equation*}  \tfrac {a_n}{1-\tfrac {a_n}{2}}{\lt}\tfrac {a_0}{1-\tfrac {a_0}{2}}{\lt}1. \end{equation*}
</div>
<p> In addition, as \(y_n,x_n\in B(x_0,R\eta )\) for all \(n\in N,\) then \(y_n,x_n\in \Omega ,\) \(\forall n\in N.\) Hence (<a href="#e12">20</a>) follows. Now we prove that \(\{ x_n\} \) is a Cauchy sequence. To do this, we consider \(n,m\geq 1:\) </p>
<div class="displaymath" id="e14">
  \begin{align} \label{e14} & \| x_{n+m}-x_n\|  \leq \\ & \leq \| x_{n+m}-x_{n+m-1}\| \! +\! \| x_{n+m-1}-x_{n+m-2}\| \! +\! \ldots +\| x_{n+1}-x_{n}\|  \nonumber \\ &  \leq \tfrac {1}{1-\tfrac {a_n}{2}}\eta \left(\prod _{k=0}^{n+m-2}\! \! f(a_k)g(a_k)\! +\! \! \prod _{k=0}^{n+m-3}\! \! f(a_k)g(a_k)+\ldots +\! \prod _{k=0}^{n-1}f(a_k)g(a_k)\right)\nonumber \\ &  \leq \tfrac {\eta }{1-\tfrac {a_n}{2}}\left(\Delta ^{n+m-1}\gamma ^{\tfrac {3^{n+m-1}-1}{2}}+\Delta ^{n+m-2}\gamma ^{\tfrac {3^{n+m-2}-1}{2}}+ \ldots +\Delta ^{n}\gamma ^{\tfrac {3^{n}-1}{2}}\right)\nonumber \\ &  {\lt} \tfrac {\eta }{1-\tfrac {a_0}{2}\gamma ^{\tfrac {3^n-1}{2}}}\gamma ^{\tfrac {3^n-1}{2}}\Delta ^{n}\tfrac {1-\gamma ^{3^nm}\Delta ^m}{1-\gamma ^{3^n}\Delta },\nonumber \end{align}
</div>
<p> Then \(\{ x_n\} \) is a Cauchy sequence. By letting \(m\to \infty \) in (<a href="#e14">22</a>), we obtain (<a href="#e13">21</a>). To prove that \(F(x^{*})=0,\) notice that \(\| \Gamma _nF(x_n)\| \to 0\) by letting \(n\to \infty .\) As \(\| F(x_n)\| \leq \| F'(x_n)\| \| \Gamma _nF(x_n)\| \) and \(\{ \| F'(x_n)\| \} \) is a bounded sequence, we deduce \(\| F(x_n)\| \to 0,\) this means \(F(x^*)=0\) by the continuity of \(F.\) </p>
<p>Now to show the uniqueness, suppose that \(y^*\in B(x_0,\tfrac {2}{M\beta }-R\eta )\cap \Omega \) is another solution of (<a href="#e2">2</a>). Then </p>
<div class="displaymath" id="a0000000040">
  \begin{equation*}  0=F(y^*)-F(x^*)=\int _0^1F’(x^*+t(y^*-x^*)){\rm d}t(y^*-x^*). \end{equation*}
</div>
<p> Using the estimate </p>
<div class="displaymath" id="a0000000041">
  \begin{align*} & \| \Gamma _0\| \int _0^{1}\| F’(x^*+t(y^*-x^*))-F’(x_0)\| {\rm d}t\leq \\ & \leq M\beta \int _0^1\| x^*+t(y^*-x^*)-x_0\| {\rm d}t\\ & \leq M\beta \int _0^1((1-t)\| x^*-x_0\| +t\| y^*-x_0\| ){\rm d}t \\ & {\lt}\tfrac {M\beta }{2}(R\eta +\tfrac {2}{M\beta }-R\eta )=1, \end{align*}
</div>
<p> we have that the operator \(\int _0^1F'(x^*+t(y^*-x^*)){\rm d}t\) has an inverse and consequently, \(y^*=x^*.\) <div class="proof_wrapper" id="a0000000042">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> It should be mentioned that in <span class="cite">
	[
	<a href="#[4]" >7</a>
	]
</span> the convergence of iteration (<a href="#e4">4</a>) was proved under conditions \((c_1)-(c_4)\) and </p>
<div class="displaymath" id="a0000000043">
  \begin{equation*}  a_0\leq \tfrac {1}{2(1+\frac{5K}{3M^2\beta })}, \end{equation*}
</div>
<p> whereas convergence of iteration (<a href="#e3">3</a>) holds under condition (<a href="#e12">20</a>).In <span class="cite">
	[
	<a href="#[7]" >10</a>
	]
</span> was found the convergence domain </p>
<div class="equation" id="e15">
<p>
  <div class="equation_content">
    \begin{equation} \label{e15} 0 < a_0 < \tfrac {1}{2d}, \end{equation}
  </div>
  <span class="equation_label">23</span>
</p>
</div>
<p> where </p>
<div class="equation" id="a0000000044">
<p>
  <div class="equation_content">
    \begin{equation}  d = \left\{  \begin{array}{l} 1+2w, \qquad \mbox{ for Chebyshev method (CM) }\\ 1+5w, \qquad \mbox{ for the first modification of CM }\\ 1+4w, \qquad \mbox{ for the second modification of CM. } \end{array}\right.\nonumber \end{equation}
  </div>
  <span class="equation_label">24</span>
</p>
</div>
<p>The comparison of (<a href="#e12">20</a>) and (<a href="#e15">23</a>) shows that the convergence domain of (<a href="#e3">3</a>) is larger than that of CM and its modifications, when \(|a|{\lt}7.\) </p>
<h1 id="a0000000045">4 Numerical results</h1>
<p>Now we present some numerical test results for the various third order, free from second derivative methods. Tests were done with a double arithmetic precision and the numbers of iterations such that \(\| x_n-x_{n-1}\| \leq 1.0e-15\) are shown below. Compared were </p>
<div class="displaymath" id="a0000000046">
  \begin{eqnarray*}  \mbox{ MOD1 \cite{[7]}: }\qquad y_n &  = &  x_n-\Gamma _nF(x_n)\\ z_n &  = &  (1-\theta )x_n+\theta y_n,\qquad \theta \in (0,1]\\ x_{n+1} &  = &  y_n-\tfrac {1}{2\theta }\Gamma _n(F’(z_n)-F’(x_n))(y_n-x_n)\\ \mbox{ MOD2 \cite{[7]}: }\qquad y_n &  = &  x_n-\Gamma _nF(x_n)\\ z_n &  = &  x_n-\Gamma _nF(x_n)\\ x_{n+1} &  = &  y_n-\Gamma _n\left((1+\tfrac {b}{2})F(y_n)+F(x_n)-\tfrac {b}{2}F(z_n)\right) \qquad b\in [-2,0]\\ \mbox{ CL1 \cite{[9]}: }\qquad y_n &  = &  x_n-\Gamma _nF(x_n)\\ x_{n+1} &  = &  y_n-\tfrac 12\Gamma _n(F’(y_n)-F’(x_n))(y_n-x_n)\\ \mbox{ CL2 \cite{[9]}: }\qquad y_n &  = &  x_n-\Gamma _nF(x_n)\\ x_{n+1} &  = &  y_n-\Gamma _nF(y_n) \\ \mbox{ $A2_{\theta }$ \cite{[8]}: }\qquad y_n &  = &  x_n-\Gamma _nF(x_n)\\ x_n^p &  = &  x_n-\theta \Gamma _nF(x_n)\\ y_n^p &  = &  -\tfrac 12\Gamma _n(F’(x_n^p)-F’(x_n))(x_n^p-x_n)\\ x_n^c &  = &  x_n^p+y_n^p\\ x_{n+1} &  = &  y_n-\tfrac 12\Gamma _n(F’(x_n^c)-F’(x_n))(x_n^c-x_n) \end{eqnarray*}
</div>
<p> and the proposed iteration (<a href="#e3">3</a>). </p>
<p>As a test we take the following systems of equations: </p>
<div class="displaymath" id="a0000000047">
  \begin{eqnarray*}  I. \qquad & &  x_1^2-x_2+1 = 0\\ & &  x_1+\cos (\tfrac {\pi }{2}x_2) = 0\qquad x^0=(0; 0.1) \\[.2cm] II.\qquad & &  x_1x_3+x_2x_4+x_3x_5+x_4x_6=0\\ & &  x_1x_5+x_2x_6 =0\\ & &  x_1+x_3+x_5=1\\ & &  -x_1+x_2-x_3+x_4-x_5+x_6=0 \\ & &  -3x_1-2x_2-x_3+x_5+2x_6=0\\ & &  3x_1-2x_2+x_3-x_5+2x_6=0\qquad x^0=(0; 0; 0; 1; 1; 0) \\[.2cm] III.\qquad & &  x_1^2+x_2^2 = 1\\ & &  x_1^2-x_2^2 = -0.5\qquad x^0=(0.3; 0.7) \\[.2cm] IV. \qquad & &  x_1^2-x_1-x_2^2 = 1\\ & &  sin(x_1)-x_2 = 0\qquad x^0=(0.1; 0) \\[.2cm] V. \qquad & &  x_1^2+x_2^2 = 4\\ & &  e^{x_{1}}+x_2 = 1\qquad x^0=(0.5; -1) \end{eqnarray*}
</div>
<p><small class="footnotesize"><div class="table"  id="a0000000048">
   <table class="tabular">
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black; border-left:1px solid black" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(MOD 1\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(MOD 2\)</p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(CL 1\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(CL 2\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black; border-left:1px solid black" 
        rowspan=""
        colspan="2">
      <p> \(A2_\theta \) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black; border-left:1px solid black" 
        rowspan=""
        colspan="4">
      <p> (<a href="#e3">3</a>)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center; border-right:1px solid black; border-left:1px solid black" 
        rowspan=""
        colspan="">
      <p>  Ex. </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(\theta \! =\! 0.5\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(b\! =\! -1\) </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(\theta \! =\! -1\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(\theta \! =\! 1\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(\! a=\! -1\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(a\! =\! -0.5\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(a\! =\! 0.5\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(a\! =\! 1\) </p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black; border-left:1px solid black" 
        rowspan=""
        colspan="">
      <p>I </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 7 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 7 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 8 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 7 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 6 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> - </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 6 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 6 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 6 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 5</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center; border-right:1px solid black; border-left:1px solid black" 
        rowspan=""
        colspan="">
      <p>II </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 6 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 6 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 6 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 6 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 5 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 18 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 6 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 6 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 6 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 6 </p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center; border-right:1px solid black; border-left:1px solid black" 
        rowspan=""
        colspan="">
      <p>III </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 6 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 6 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 6 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 6 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 5 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 7 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 5 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 5 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 5 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 5 </p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center; border-right:1px solid black; border-left:1px solid black" 
        rowspan=""
        colspan="">
      <p>IV </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>6 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 6 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 6 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 6 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 6 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 8 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 6 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 6 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 7 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 8</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center; border-right:1px solid black; border-left:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>V </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 7 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 6 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 7 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 7 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 6 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 15 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 6 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 6 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 6 </p>

    </td>
    <td  style="text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 6</p>

    </td>
  </tr>
</table> <figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">1</span> 
  <span class="caption_text">Numerical results.</span> 
</figcaption> 
</div></small> </p>
<h1 id="a0000000049">5 Conclusion</h1>
<p> In this work we proposed a family of Newton type methods which is free from second derivative and includes some known third order methods as particular case. Also, we proved Newton-Kantorovich type convergence theorem using recurrent relations to show that it has a R-order three convergence and obtained an error estimate. The proposed method was compared to previously known third order methods to show that it has an equivalent performance. <div class="acknowledgement_thmwrapper " id="a0000000050">
  <div class="acknowledgement_thmheading">
    <span class="acknowledgement_thmcaption">
    Acknowledgement
    </span>
  </div>
  <div class="acknowledgement_thmcontent">
  <p>This work was sponsored particularly by Foundation of Science and Technology of Ministry of Education and Science of Mongolia. </p>

  </div>
</div> </p>
<p><small class="footnotesize">  </small></p>
<div class="bibliography">
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</dd>
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</dd>
</dl>


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