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<h1>On the semilocal convergence of derivative free methods for solving nonlinear equations\(^{\bullet }\)</h1>
<p class="authors">
<span class="author">I. K. Argyros\(^{\ast }\) Hongmin Ren\(^{\S }\)</span>
</p>
<p class="date">September 24, 2011</p>
</div>
<p>\(^{\ast }\)Cameron University, Dept. of Mathematics Sciences, Lawton, OK 73505, USA,<br />e-mail: <span class="ttfamily">iargyros@cameron.edu</span>. </p>
<p>\(^{\S }\)Hangzhou Polytechnic, College of Information and Engineering, Hangzhou 311402, Zhejiang, P.R.China, e-mail: <span class="ttfamily">rhm65@126.comb.xy.uv</span>. </p>
<p>\(^{\bullet }\)The research of the second author has been supported in part by National Natural Science Foundation of China (Grant No. 10871178), Natural Science Foundation of Zhejiang Province of China (Grant No. Y606154), and Scientific Research Fund of Zhejiang Provincial Education Department of China (Grant No. 20071362). </p>

<div class="abstract"><p> We introduce a <i class="it">Derivative Free Method</i> (DFM) for solving nonlinear equations in a Banach space setting. We provide a semilocal convergence analysis for DFM using recurrence relations. Numerical examples validating our theoretical results are also provided in this study to show that DFM is faster than other derivative free methods [9] using similar information. </p>
<p><b class="bf">MSC.</b> 65J15, 65G99, 47H99, 49M15. </p>
<p><b class="bf">Keywords.</b> Banach space, derivative free method, Newton’s method, divided difference, recurrence relations. </p>
</div>
<h1 id="a0000000002">1  Introduction</h1>
<p>In this study we are concerned with the problem of approximating a locally unique solution \(x^{\star }\) of an equation </p>
<div class="equation" id="1.1">
<p>
  <div class="equation_content">
    \begin{equation}  \label{1.1}F(x) =0, \end{equation}
  </div>
  <span class="equation_label">1.1</span>
</p>
</div>
<p> where \(F\) is a Fréchet–differentiable operator defined on a non–empty, open subset \(\mathcal{D}\) of a Banach space \(\mathcal{X}\) with values in a Banach space \(\mathcal{Y}\). </p>
<p>A large number of problems in applied mathematics and engineering are solved by finding the solutions of certain equations. For example, dynamic systems are mathematically modeled by difference or differential equations, and their solutions usually represent the states of the systems. For the sake of simplicity, we assume that a time–invariant system is driven by the equation \(\dot{x}=Q(x)\), for some suitable operator \(Q\), where \(x\) is the state. Then the equilibrium states are determined by solving equation (<a href="#1.1">1.1</a>). Similar equations are used in the case of discrete systems. The unknowns of engineering equations can be functions (difference, differential, and integral equations), vectors (systems of linear or nonlinear algebraic equations), or real or complex numbers (single algebraic equations with single unknowns). Except in special cases, the most commonly used solution methods are iterative. In fact, starting from one or several initial approximations a sequence is constructed that converges to a solution of the equation. Iteration methods are also applied for solving optimization problems. In such cases, the iteration sequences converge to an optimal solution of the problem at hand. Since all of these methods have the same recursive structure, they can be introduced and discussed in a general framework. </p>
<p>A classic iterative process for solving nonlinear equations is Chebyshev’s method (see [5], [9], <span class="cite">
	[
	<a href="#673-EH-C" >14</a>
	]
</span>, <span class="cite">
	[
	<a href="#GN" >17</a>
	]
</span>): </p>
<div class="displaymath" id="a0000000003">
  \[  \left\{  \begin{array}[c]{l}x_{0} \in \mathcal{D},\\ y_{k} = x_{k} - F^{\prime }(x_{k})^{-1}\, \,  F(x_{k}),\\ x_{k+1} = y_{k} - \tfrac {1}{2}\, \,  F^{\prime }(x_{k})^{-1} F^{\prime \prime }(x_{k})(y_{k}-x_{k})^{2},\quad k\geq 0. \end{array} \right.  \]
</div>
<p> This one-point iterative process depends explicitly on the two first derivatives of \(F\) (namely, \(x_{n+1}=\psi (x_{n},F(x_{n}),F^{\prime }(x_{n}),F^{\prime \prime }(x_{n}))\)). Ezquerro and Hernández introduced in&#160;<span class="cite">
	[
	<a href="#673-EH-C" >14</a>
	]
</span> some modifications of Chebyshev’s method that avoid the computation of the second derivative of \(F\) and reduce the number of evaluations of the first derivative of \(F\). Actually, these authors have obtained a modification of the Chebyshev iterative process which only need to evaluate the first derivative of \(F\), (namely, \(x_{n+1}=\overline{\psi }(x_{n},F^{\prime }(x_{n})\)), but with third-order of convergence [14]. In this paper we recall this method as the <i class="it">Chebyshev–Newton–type method</i> (CNTM) and it is written as follows: </p>
<div class="displaymath" id="a0000000004">
  \[  \left\{  \begin{array}[c]{l}x_{0} \in \mathcal{D} ,\\ y_{k}= x_{k} - F^{\prime }(x_{k})^{-1} \, \,  F(x_{k}),\\ z_{k}= x_{k} + a \, \,  (y_{k} - x_{k} )\\ x_{k+1}= x_{k} - \displaystyle \tfrac {1}{a^{2}}\, \,  F^{\prime }(x_{k})^{-1} \, \,  ((a^{2} + a -1)\, \, F(x_{k}) + F(z_{k})),\quad k\geq 0, \end{array} \right.  \]
</div>
<p>where \(F^{\prime }(x)\) \((x\in \mathcal{D})\) is the Fréchet–derivative of \(F\). </p>
<p>There is an interest in constructing families of iterative processes free of derivatives. To obtain a new family in [9] we considered an approximation of the first derivative of \(F\) from a divided difference of first order, that is, \(F^{\prime }(x_{n})\approx [x_{n-1},x_{n},F]\), where, \([x,y;F]\) is a divided difference of order one for the operator \(F\) at the points \(x\), \(y \in \mathcal{D}\). Then, we introduce the <i class="it">Chebyshev–Secant–type method</i> (CSTM) </p>
<div class="displaymath" id="a0000000005">
  \[  \left\{  \begin{array}[c]{l}x_{-1}, \, \,  x_{0} \in \mathcal{D} ,\\ y_{k}= x_{k} - B_{k}^{-1} \, \,  F(x_{k}), \quad B_{k}=[x_{k-1},x_{k};F],\\ z_{k}= x_{k} + a \, \,  (y_{k}-x_{k}),\\ x_{k+1}= x_{k} - B_{k}^{-1} \, \,  (b\, \, F(x_{k}) + c\, \,  F(z_{k})),\quad k\geq 0, \end{array} \right.  \]
</div>
<p> where \(a\), \(b\), \(c\) are non–negative parameters to be chosen so that sequence \(\{ x_{k} \} \) converges to \(x^{\star }\). Note that CSTM is reduced to the <i class="it">secant method</i> (SM) if \(a=0\), \(b=c=1/2\), and \(y_{k} = x_{k+1}\). Moreover, if \(x_{k-1} = x_{k}\), and \(F\) is differentiable on \(\mathcal{D}\), then, \(F^{\prime }(x_{k}) = [x_{k}, x_{k} ;F]\), and CSTM reduces to <i class="it">Newton’s method</i> (NM). </p>
<p>We provided a semilocal convergence analysis for CSTM using recurrence sequences, and also illustrated its effectiveness through numerical examples. Bosarge and Falb <span class="cite">
	[
	<a href="#4" >10</a>
	]
</span>, Dennis <span class="cite">
	[
	<a href="#623-6" >13</a>
	]
</span>, Potra <span class="cite">
	[
	<a href="#623-11" >23</a>
	]
</span>, Argyros <span class="cite">
	[
	<a href="#623-1" >1</a>
	]
</span>–<span class="cite">
	[
	<a href="#623-4" >5</a>
	]
</span>, Hernández <i class="it">et al.</i> <span class="cite">
	[
	<a href="#623-7" >15</a>
	]
</span> and others <span class="cite">
	[
	<a href="#8" >16</a>
	]
</span>, <span class="cite">
	[
	<a href="#10" >22</a>
	]
</span>, <span class="cite">
	[
	<a href="#12" >26</a>
	]
</span>, have provided sufficient convergence conditions for the SM based on Lipschitz-type conditions on divided difference operator (see, also relevant works in <span class="cite">
	[
	<a href="#arg-hil-sub" >8</a>
	]
</span>–<span class="cite">
	[
	<a href="#623-6" >13</a>
	]
</span>, <span class="cite">
	[
	<a href="#623-8" >18</a>
	]
</span>, <span class="cite">
	[
	<a href="#623-10" >21</a>
	]
</span>, <span class="cite">
	[
	<a href="#623-12" >24</a>
	]
</span>, <span class="cite">
	[
	<a href="#623-13" >27</a>
	]
</span>). </p>
<p>In this paper, we continue the study of inverse free iterative processes. We introduce the <i class="it">derivative free method</i> (DFM): </p>
<div class="displaymath" id="a0000000006">
  \[  \left\{  \begin{array}[c]{l}x_{-1}, \, \,  x_{0} \in \mathcal{D} ,\\ y_{k}= x_{k} - A_{k}^{-1} \, \,  F(x_{k}), \quad A_{k}=[2x_{k}-x_{k-1},x_{k-1};F],\\ z_{k}= x_{k} + a \, \,  (y_{k}-x_{k}),\\ x_{k+1}= x_{k} - A_{k}^{-1} \, \,  (b\, \, F(x_{k}) + c\, \,  F(z_{k})),\quad k\geq 0. \end{array} \right.  \]
</div>
<p> Note that DFM reduces to the <i class="it">Kurchatov-type method</i> (KTM) </p>
<div class="displaymath" id="a0000000007">
  \[  x_{k+1}= x_{k} - A_{k}^{-1} F(x_{k}),  \]
</div>
<p> if \(a=0,b=c=0.5\), and \(y_{k}=x_{k+1}\) [20], [25]. </p>
<p>In this special case the quadratic convergence of KTM was first established in [20], [25] and then in [6], [7] under different sets of sufficient conditions. We provide a semilocal convergence analysis for DFM. Then, we give numerical examples to show that DFM is faster than CSTM. In particular, two numerical examples are also provided. Firstly, we consider a scalar equation where the main study of the paper is applied. Secondly, we discretize a nonlinear integral equation and approximate a numerical solution using DFM. </p>
<h1 id="sec:2">2 Semilocal convergence analysis of DFM</h1>
<p> We shall show the semilocal convergence of DFM under the following conditions </p>
<ol class="enumerate">
  <li><p>\(F \,  : \,  \mathcal{D} \subseteq \mathcal{X} \longrightarrow \mathcal{Y}\) is a Fréchet–differentiable operator, and there exists divided difference denoted by \([x,y;F]\) satisfying </p>
<div class="displaymath" id="a0000000008">
  \[  [x,y;F] (x-y) = F(x) - F(y) \quad \mathrm{for \, \,  all} \quad x,y \in \mathcal{D} ;  \]
</div>
</li>
  <li><p>There exist \(x_{-1}\) and \(x_{0}\) in \(\mathcal{D}\) and \(\beta {\gt}0\) such that </p>
<div class="displaymath" id="a0000000009">
  \[  A_{0} ^{-1} = [2x_{0}-x_{-1}, x_{-1};F]^{-1} \in \mathcal{L} (\mathcal{Y} , \mathcal{X})  \]
</div>
<p> exists and </p>
<div class="displaymath" id="a0000000010">
  \[  0 {\lt} \parallel A_{0} ^{-1} \parallel \leq \beta ;  \]
</div>
</li>
  <li><p>There exists \(d {\gt}0\) such that </p>
<div class="displaymath" id="a0000000011">
  \[  \parallel x_{0} - x_{-1} \parallel \leq d ;  \]
</div>
</li>
  <li><p>There exists \(\eta {\gt}0\) such that </p>
<div class="displaymath" id="a0000000012">
  \[  0 {\lt} \parallel A_{0}^{-1} \,  F(x_{0} ) \parallel \leq \eta ;  \]
</div>
</li>
  <li><p>There exists constant \(M {\gt}0\), such that for all \(x\), \(y\), \(u\), \(v\) in \(\mathcal{D}\) </p>
<div class="displaymath" id="a0000000013">
  \[  \parallel [x,y;F] - [u,v;F] \parallel \leq \displaystyle \tfrac {M}{2 } \,  (\parallel x- u \parallel + \parallel y -v \parallel ) ;  \]
</div>
</li>
  <li><p>For \(a \in [0, 1]\), \(b \in [0,1]\) and \(c{\gt}0\) given in DFM, we suppose </p>
<div class="displaymath" id="a0000000014">
  \[  (1-a) \,  c = 1-b ;  \]
</div>
</li>
</ol>
<ol class="enumerate">
  <li><div class="displaymath" id="a0000000015">
  \[  \alpha =2\,  {\bigg(} 1 + d_{0} + a \,  c \,  \gamma \,  (a+ 2\, d_{0}) {\bigg)} \,  \gamma {\lt} 1 ,  \]
</div>
<p> where, </p>
<div class="displaymath" id="a0000000016">
  \[  \gamma = \displaystyle \tfrac {\beta \,  M \,  \eta }{ 2}, \quad d_{0} = \displaystyle \tfrac {d}{\eta } ;  \]
</div>
</li>
  <li><div class="displaymath" id="a0000000017">
  \[  \overline{U} (x_{0} , R=r \,  \eta ) = \{ x \in \mathcal{X} \,  : \,  \parallel x- x_{0} \parallel \leq R \}  \subseteq \mathcal{D} ,  \]
</div>
<p> for some \(r{\gt}1\) to be precised later in Theorem <a href="#T.2.4">5</a>; </p>
</li>
</ol>
<ol class="enumerate">
  <li><div class="displaymath" id="a0000000018">
  \[  x,y\in \mathcal{D}\Rightarrow 2\,  y-x\in \mathcal{D}.  \]
</div>
</li>
</ol>
<p>Delicate condition \((C_{9})\) is certainly satisfied, if \(\mathcal{D}=\mathcal{X}\). It is also satisfied, if \((C_{9})\) is replaced by </p>
<ol class="enumerate">
  <li><div class="displaymath" id="a0000000019">
  \[  \overline{U}(x_{0},3\, R)\subseteq \mathcal{D}.  \]
</div>
</li>
</ol>
<p>Indeed, if \(x,y\in \overline{U}(x_{0},R)\), then </p>
<div class="displaymath" id="a0000000020">
  \[  \| 2\, y-x-x_{0}\| \le \| y-x_{0}\| +\| y-x\| \le 2\, \| y-x_{0}\| +\| x-x_{0}\| \le 3\, R.  \]
</div>
<p> That is, \(2y-x\in \overline{U}(x_{0},3R)\). </p>
<p>We note by <b class="bfseries">(\(\mathcal{C}\))</b> the set of conditions <b class="bfseries">(\(\mathcal{C}_{1}\))</b>–<b class="bfseries">(\(\mathcal{C}_{9}\))</b>. </p>
<p><div class="definition_thmwrapper " id="D.2.1">
  <div class="definition_thmheading">
    <span class="definition_thmcaption">
    Definition
    </span>
    <span class="definition_thmlabel">1</span>
  </div>
  <div class="definition_thmcontent">
  <p> Let \(\gamma \) and \(d_{0}\) as defined in <b class="bfseries"><span class="rm">(\(\mathcal{C}_{7}\))</span></b>. It is convenient to define for \(\mu _{0}= w_{0} =1\), \(q_{-1} =d_{0}\), and \(n \geq 0\), the following sequences </p>
<div class="displaymath" id="a0000000021">
  \[  p_{n} = a \,  c \,  \gamma \,  \mu _{n} (a \,  w_{n} + 2\,  q_{n-1} ) \,  w_{n} ,  \]
</div>
<div class="displaymath" id="a0000000022">
  \[  q_{n} = p_{n} + w_{n} ,  \]
</div>
<div class="displaymath" id="a0000000023">
  \[  \mu _{n+1} = \displaystyle \tfrac {\mu _{n}}{1 - 2\, \gamma \,  \mu _{n} \,  (q_{n-1} + q_{n})} ,  \]
</div>
<div class="displaymath" id="a0000000024">
  \[  c_{n} = \displaystyle \tfrac {M}{2 } \,  ( (q_{n} + 2\,  q_{n-1}) \,  q_{n} + a \,  c \,  (a \,  w_{n} + 2\,  q_{n-1}) \,  w_{n} ) ,  \]
</div>
<p> and </p>
<div class="displaymath" id="a0000000025">
  \[  w_{n+1} = \gamma \,  \mu _{n+1} \,  ((q_{n} + 2\,  q_{n-1}) \,  q_{n} + a \,  c \,  (a \,  w_{n} + 2\,  q_{n-1} ) \,  w_{n} ) .  \]
</div>
<p>Note that </p>
<div class="displaymath" id="a0000000026">
  \[  w_{n+1} = \beta \,  \eta \,  \mu _{n+1} \,  c_{n} .  \]
</div>

  </div>
</div> </p>
<p>We need an Ostrowski–type approximations for DFM. The proof is given in [5], [9]. </p>
<p><div class="lemma_thmwrapper " id="L.2.2">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">2</span>
  </div>
  <div class="lemma_thmcontent">
  <p> Assume sequence \(\{ x_{k} \} \) generated by <span class="rm">DFM</span> is well defined, \((1-a) \, \,  c= 1-b\) holds for \(a \in [0,1]\), \(b\in [0,1]\), and \(c{\gt} 0\). </p>
<p>Then, the following items hold for all \(k\geq 0\): </p>
<div class="displaymath" id="2.1">
  \begin{align}  \label{2.1}F(z_{k}) &  =(1\! -\! a)F(x_{k})\! +\!  a \displaystyle \int _{0}^{1} (F^{\prime }(x_{k} + a(y_{k} - x_{k} ))\! -\! F^{\prime }(x_{k})) (y_{k}\! -\! x_{k})\mathrm{d}t +\\ &  \quad +a (F^{\prime }(x_{k}) - A_{k})(y_{k} - x_{k} ),\nonumber \end{align}
</div>
<div class="displaymath" id="a0000000027">
  \begin{align}  x_{k+1}- y_{k} &  =-acA_{k}^{-1}{\bigg(} \displaystyle \int _{0}^{1} (F^{\prime }(x_{k} + a(y_{k} - x_{k} )) - F^{\prime }(x_{k}))(y_{k} - x_{k})\mathrm{d}t +\\ &  \quad +(F^{\prime }(x_{k})- A_{k})(y_{k}- x_{k} ) {\bigg)},\nonumber \end{align}
</div>
<p> and </p>
<div class="displaymath" id="a0000000028">
  \begin{align}  F(x_{k+1}) &  =\displaystyle \int _{0}^{1} (F^{\prime }(x_{k} + t (x_{k+1}-x_{k} ))-F^{\prime }(x_{k}))(x_{k+1}- x_{k})\mathrm{d}t +\nonumber \\ &  \quad +(F^{\prime }(x_{k})- A_{k})(x_{k+1}-x_{k} )- ac{\bigg(} \displaystyle \int _{0}^{1} (F^{\prime }(x_{k} + at(y_{k} - x_{k} )) -\\ &  \quad -F^{\prime }(x_{k}))(y_{k} - x_{k}){\rm d}t + (F^{\prime }(x_{k}) - A_{k})(y_{k} - x_{k}) {\bigg)}.\nonumber \end{align}
</div>

  </div>
</div> </p>
<p>The following relates DFM with scalar sequences introduced in Definition <a href="#D.2.1">1</a>. </p>
<p><div class="lemma_thmwrapper " id="L.2.3">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">3</span>
  </div>
  <div class="lemma_thmcontent">
  <p> Under the <b class="bfseries"><span class="rm">(\(\mathcal{C}\))</span></b> conditions, we assume: </p>
<div class="displaymath" id="a0000000029">
  \[  x_{n} \in \mathcal{D} \quad \mathrm{and} \,  \, \,  2\,  \gamma \,  \mu _{n} \,  (q_{n-1} + q_{n}) {\lt} 1 \quad (n \geq 0).  \]
</div>
<p> Then, the following items hold for all \(n \geq 0\): </p>
<ul class="itemize">
  <li><p>\(\parallel A_{n} ^{-1}\parallel \leq \mu _{n} \,  \beta ,\) </p>
</li>
  <li><p>\(\parallel y_{n} - x_{n} \parallel = \parallel A_{n} ^{-1} \,  F(x_{n}) \parallel \leq w _{n} \,  \eta ,\) </p>
</li>
  <li><p>\(\parallel x_{n+1} - y_{n} \parallel \leq p _{n} \,  \eta ,\) </p>
</li>
  <li><p>\(\parallel x_{n+1} - x_{n} \parallel \leq q _{n} \,  \eta .\) </p>
</li>
</ul>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000030">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> We use induction. </p>
<p>We have \(\parallel y_{0} - x_{0} \parallel \leq \eta \), and \(\parallel z_{0} - x_{0} \parallel \leq a \,  \eta \), so that \(x_{0}\), \(z_{0} \in \mathcal{D}\). </p>
<p>Items \(\mathbf{(I_{0})}\) and \(\mathbf{(II_{0})}\) hold by <b class="bfseries">(\(\mathcal{C} _{2}\))</b> and <b class="bfseries">(\(\mathcal{C} _{4}\))</b>, respectively. To prove \(\mathbf{(III_{0})}\), we use Lemma <a href="#L.2.2">2</a> for \(n=0\) to obtain by <b class="bfseries">(\(\mathcal{C} _{2}\))</b>–<b class="bfseries">(\(\mathcal{C} _{5}\))</b> </p>
<div class="displaymath" id="a0000000031">
  \[ \begin{array}[c]{lll}\parallel x_{1} - y_{0} \parallel &  \leq &  a \,  c \,  \parallel A_{0} ^{-1}\parallel \,  \displaystyle \tfrac {M}{2} \,  (a \,  \parallel y_{0} - x_{0} \parallel +2\,  \parallel x_{0} - x_{-1}\parallel ) \,  \parallel y_{0} - x_{0}\parallel \\ &  \leq &  \displaystyle \tfrac {a \,  c \,  \beta \,  M}{2} \,  (a \,  \eta + 2\,  d ) \,  \eta \\ &  = &  a \,  c \,  \gamma \,  (a + 2d_{0}) \,  \eta = p_{0} \,  \eta . \end{array}  \]
</div>
<p> Moreover, </p>
<div class="displaymath" id="a0000000032">
  \[  \parallel x_{1} - x_{0} \parallel \leq \parallel x_{1} - y_{0} \parallel + \parallel y_{0} - x_{0} \parallel \leq p_{0} \,  \eta + \eta = (1+p_{0}) \,  \eta = q_{0} \,  \eta ,  \]
</div>
<p> which implies \(\mathbf{(IV_{0})}\). Note also that \(z_{1} \in \mathcal{D}\). Following an inductive argument, assume \(x_{k} \in \mathcal{D}\), and \(2\,  \gamma \,  \mu _{k} \,  (q _{k-1} + q_{k}) {\lt} 1\). Then, we have </p>
<div class="displaymath" id="a0000000033">
  \begin{align*} &  \parallel A_{k} ^{-1} \parallel \,  \parallel A_{k+1} - A_{k} \parallel \leq \\ &  \leq \parallel A_{k} ^{-1} \parallel \,  \tfrac {M}{2}[\parallel 2\,  (x_{k+1}-x_{k})-(x_{k}-x_{k-1}) \parallel +\parallel x_{k}-x_{k-1} \parallel ]\\ &  \leq 2\,  \parallel A_{k} ^{-1} \parallel \,  \displaystyle \tfrac {M}{2} \,  ( \parallel x_{k} - x_{k-1} \parallel + \parallel x_{k+1} - x_{k} \parallel )\\ &  \leq \displaystyle 2\,  \tfrac { \beta \,  M}{2} \,  \mu _{k} \,  (q _{k-1} + q_{k}) \,  \eta = 2\,  \gamma \,  \mu _{k} \,  (q _{k-1} + q_{k}) {\lt} 1 . \end{align*}
</div>
<p>It follows from the Banach lemma on invertible operators [1], [5], [19] that \(A_{k+1} ^{-1} \) exists, and </p>
<div class="displaymath" id="a0000000034">
  \[ \begin{array}[c]{lll}\parallel A_{k+1} ^{-1} \parallel &  \leq &  \displaystyle \tfrac {\parallel A_{k} ^{-1} \parallel } {1 - 2\,  \parallel A_{k} ^{-1} \parallel \,  \displaystyle \tfrac {M}{2} \,  ( \parallel x_{k} - x_{k-1} \parallel + \parallel x_{k+1} - x_{k} \parallel ) }\\ &  \leq &  \displaystyle \tfrac {\beta \,  \mu _{k} } {1- 2\,  \gamma \,  \mu _{k} \,  (q _{k-1} + q_{k})} = \mu _{k+1} \,  \beta , \end{array}  \]
</div>
<p> which shows \(\mathbf{(I_{k+1})}\). Using Lemma <a href="#L.2.2">2</a>, <b class="bfseries">(\(\mathcal{C} _{5}\))</b>, and the induction hypotheses, we get </p>
<div class="equation" id="1-1nov">
<p>
  <div class="equation_content">
    \begin{equation}  \label{1-1nov}\begin{array}[c]{lll}\parallel F (x_{k+1} ) \parallel &  \leq &  \displaystyle \tfrac {M}{2} \,  \parallel x_{k+1} - x_{k} \parallel ^{2} + \displaystyle M \,  \parallel x_{k+1} - x_{k} \parallel \,  \parallel x_{k} - x_{k-1} \parallel +\\ & &  + ac{\bigg(} \displaystyle \tfrac {a \,  M}{2} \,  \parallel y _{k} - x_{k} \parallel ^{2} + \displaystyle M \,  \parallel x_{k} - x_{k-1} \parallel \,  \parallel y _{k} - x_{k} \parallel {\bigg)}\\ &  \leq &  \displaystyle \tfrac {M}{2} \,  q_{k} ^{2} \,  \eta ^{2} + \displaystyle M \,  q_{k} \,  \eta \,  q_{k-1} \,  \eta + a\,  c \,  \left( \displaystyle \tfrac {a \,  M}{2} \,  w_{k} ^{2} \,  \eta ^{2}+ \displaystyle M \,  q_{k-1} \,  \eta \,  w_{k} \,  \eta \right) \\ &  = &  c_{k} \,  \eta ^{2} . \end{array} \end{equation}
  </div>
  <span class="equation_label">2.4</span>
</p>
</div>
<p>Then, we get </p>
<div class="displaymath" id="a0000000035">
  \[  \parallel y_{k+1} - x _{k+1} \parallel \leq \parallel A_{k+1} ^{-1} \parallel \,  \parallel F(x_{k+1})\parallel \leq \mu _{k+1} \,  \beta \,  c_{k} \,  \eta ^{2} = w_{k+1} \,  \eta .  \]
</div>
<p>Moreover, by Lemma <a href="#L.2.2">2</a>, we have </p>
<div class="displaymath" id="a0000000036">
  \[ \begin{array}[c]{l}\parallel x_{k+2} - y _{k+1} \parallel \leq \\ \leq a \,  c \parallel A_{k+1} ^{-1} \parallel \,  \displaystyle \tfrac {M}{2} \,  {\bigg(} a \,  \parallel y_{k+1} - x _{k+1} \parallel +2\,  \parallel x_{k+1} - x _{k} \parallel {\bigg)} \,  \parallel y_{k+1} - x _{k+1} \parallel \\ \leq a \,  c \,  \mu _{k+1} \,  \displaystyle \tfrac {\beta \,  M}{2} \,  (a \,  w_{k+1} + 2\,  q _{k})\,  w_{k+1}\, \eta ^{2} = p_{k+1}\,  \eta , \end{array}  \]
</div>
<p> and consequently, </p>
<div class="displaymath" id="a0000000037">
  \[  \parallel x_{k+2} - x _{k+1} \parallel \leq \parallel x_{k+2} - y _{k+1} \parallel + \parallel y_{k+1} - x _{k+1} \parallel \leq (p_{k+1} + w_{k+1}) \,  \eta = q_{k+1} \,  \eta .  \]
</div>
<p>This completes the proof of Lemma <a href="#L.2.3">3</a>. <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> </p>
<p>We shall establish the convergence of sequence \(\{  x_{n} \} \) generated by DFM. This can be achieved by showing that \(\{  q_{n} \} \) is a Cauchy sequence, if the following conditions hold for \(n \geq 0\): </p>
<ol class="enumerate">
  <li><p>\(x_{n} \in \mathcal{D}\), </p>
</li>
</ol>
<p>and </p>
<ol class="enumerate">
  <li><p>\(2\,  \gamma \,  \mu _{n} \,  (q_{n-1} + q_{n} ) {\lt} 1 .\) </p>
</li>
</ol>
<p>In the next result, we show the Cauchy property for sequence \(\{  q_{n} \} \). </p>
<p><div class="lemma_thmwrapper " id="L.2.4">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">4</span>
  </div>
  <div class="lemma_thmcontent">
  <p> Assume <b class="bfseries"><span class="rm">(\(\mathcal{C} _{8}\))</span></b>. Note that \(\alpha \in [0,1)\) implies \(2\,  \gamma \,  (q_{-1} + q_{0}) {\lt} 1\). Then, scalar sequence: </p>
<ol class="enumerate">
  <li><p>\(\{ \mu _{n}\} \) is increasing. </p>
</li>
  <li><p>\(\{  q_{n} \} \) is decreasing and \(\displaystyle \lim _{n\longrightarrow \infty } q_{n} =0\). </p>
</li>
</ol>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000039">
  <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>
<ol class="enumerate">
  <li><p>We show using induction that all scalar sequences involved are positive. By Definition <a href="#D.2.1">1</a>, and <b class="bfseries">(\(\mathcal{C} _{8}\))</b>, we have for \(j=0\): \(\mu _{j} \), \(p_{j}\), \(q_{j}\), \(w_{j}\), \(c_{j}\), and \(1- 2\,  \gamma \,  \mu _{j} \,  (q_{j-1} + q_{j} )\) are positive. Assume \(\mu _{k} \), \(p_{k}\), \(q_{k}\), \(w_{k}\), \(c_{k}\), and \(1- 2\,  \gamma \,  \mu _{k} \,  (q_{k-1} + q_{k} )\) are positive for all \(k \leq n\). Since \(c_{k} {\gt}0\), it follows from the definition of the scalar sequences that \(w_{k+1}\), \(\mu _{k+1}\), \(p_{k+1}\), \(c_{k+1}\) have the same sign. Assume the common sign to be negative. Then </p>
<div class="displaymath" id="a0000000040">
  \[ \begin{array}[c]{l}q_{k-1}+ q_{k} +q_{k+1} {\lt} q_{k-1} + q_{k}\\ \Longrightarrow 1- 2\,  \gamma \,  \mu _{k} \,  (q_{k-1}+ q_{k} +q_{k+1}) {\gt} 1 -2\,  \gamma \,  \mu _{k} \,  (q_{k-1}+ q_{k} )\\ \Longrightarrow \displaystyle \tfrac {1- 2\,  \gamma \,  \mu _{k} \,  (q_{k-1}+ q_{k} +q_{k+1})}{ 1 - 2\,  \gamma \,  \mu _{k} \,  (q_{k-1}+ q_{k} )} {\gt} 1 . \end{array}  \]
</div>
<p> But it follows from the definition of sequence \(\{ \mu _{k} \} \) that </p>
<div class="displaymath" id="a0000000041">
  \[ \begin{array}[c]{l}1- 2\,  \gamma \,  \mu _{k+1} \,  (q_{k} + q_{k+1} ) = \displaystyle \tfrac {1-2\,  \gamma \,  \mu _{k} \,  (q_{k-1}+ 2\,  q_{k} +q_{k+1})}{ 1 - 2\,  \gamma \,  \mu _{k} \,  (q_{k-1}+ q_{k} )}\\ \begin{array}[c]{lll}\Longrightarrow 1-2\,  \gamma \,  \mu _{k+1} \,  q_{k+1} &  = &  \displaystyle \tfrac {1-2\, \gamma \,  \mu _{k} \,  (q_{k-1}+ 2\,  q_{k} +q_{k+1})}{ 1 -2\,  \gamma \,  \mu _{k} \,  (q_{k-1}+ q_{k} )} + 2\,  \gamma \,  \mu _{k+1} \,  q_{k}\\ &  = &  \displaystyle \tfrac {1- 2\,  \gamma \,  \mu _{k} \,  (q_{k-1}+ q_{k} +q_{k+1})}{ 1 -2\,  \gamma \,  \mu _{k} \,  (q_{k-1}+ q_{k} )} {\gt} 1, \end{array}\end{array}  \]
</div>
<p> which is a contradiction, since we get \(2\,  \gamma \,  \mu _{k+1} \,  q_{k+1} {\lt} 0\), but \(\mu _{k+1}, \,  q_{k+1}\) have the same sign, and \(\gamma {\gt}0\). The induction is then completed. </p>
<p>By the definition of sequence \(\{ \mu _{n}\} \) and \(\mu _{0} =1\), we have </p>
<div class="displaymath" id="a0000000042">
  \[ \begin{array}[c]{l}1- 2\,  \gamma \,  \mu _{k} \,  ( q_{k-1} + q_{k}) = \displaystyle \tfrac {\mu _{k}}{\mu _{k+1}}\\ \Longrightarrow q_{k-1} + q_{k} = \displaystyle \tfrac {1}{2\,  \gamma } \,  (\displaystyle \tfrac {1}{ \mu _{k}} - \displaystyle \tfrac {1}{\mu _{k+1} })\\ \Longrightarrow \displaystyle \sum _{i=0} ^{k-1} ( q_{i-1} + q_{i} ) = \displaystyle \tfrac {1}{2\,  \gamma } \,  (\displaystyle \tfrac {1}{ \mu _{0}} - \displaystyle \tfrac {1}{\mu _{k} }) = \displaystyle \tfrac {1}{2\,  \gamma } \,  (1 - \displaystyle \tfrac {1}{\mu _{k} })\\ \Longrightarrow \mu _{k} = \displaystyle \tfrac {1}{1 -2\,  \gamma \,  \displaystyle \sum _{i=0} ^{k-1} ( q_{i-1} + q_{i} ) } . \end{array}  \]
</div>
<p>But \(1- 2\,  \gamma \,  \displaystyle \sum _{i=0} ^{k-1} ( q_{i-1} + q_{i} )\) decreases. Therefore, sequence \(\{ \mu _{k} \} \) increases, and consequently \(\mu _{k} \geq \mu _{0} =1\). </p>
</li>
  <li><p>We have that sequence \(\mu _{k} {\gt}1\) is increasing, so that \(0 \leq \displaystyle \tfrac {1}{\mu _{k}} \leq 1\). Since \(\{ \displaystyle \tfrac {1}{\mu _{k}} \} \) is monotonic on a compact set, it converges to \(\displaystyle \tfrac {1}{\mu _{\infty }}\). Then, we have </p>
<div class="displaymath" id="a0000000043">
  \[  \displaystyle \lim _{k \longrightarrow \infty } (q_{k-1} + q_{k}) = \displaystyle \tfrac {1}{2\,  \gamma } \,  \displaystyle \lim _{k \longrightarrow \infty } (\displaystyle \tfrac {1}{\mu _{k}} - \displaystyle \tfrac {1}{\mu _{k+1}}) = \displaystyle \tfrac {1}{2\,  \gamma } \,  (\displaystyle \tfrac {1}{\mu _{\infty }} - \displaystyle \tfrac {1}{\mu _{\infty }})=0 .  \]
</div>
</li>
</ol>
<p>This completes the proof of Lemma <a href="#L.2.4">4</a>. </p>
<p>We can show the main semilocal convergence theorem for DFM. </p>
<p><div class="theorem_thmwrapper " id="T.2.4">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">5</span>
  </div>
  <div class="theorem_thmcontent">
  <p> Let \(F \,  : \,  \mathcal{D} \subseteq \mathcal{X} \longrightarrow \mathcal{Y}\) be a Fréchet–differentiable operator defined on a non–empty open, convex domain \(\mathcal{D}\) of a Banach space \(\mathcal{X}\), with values in a Banach space \(\mathcal{Y}\). Assume that the <span class="rm">(\({\mathcal{C}}\))</span> conditions hold. Then, sequence \(\{  x_{n} \} \) \((n\geq -1)\), generated by <span class="rm">DFM</span>, is well defined, remains in \(\overline{U} (x_{0} , r \,  \eta )\) for all \(n\geq 0\), and converges to a solution \(x^{\star }\in \overline{U} (x_{0} , r \,  \eta )\) of equation \(F(x) =0\), where, </p>
<div class="equation" id="def-r">
<p>
  <div class="equation_content">
    \begin{equation}  \label{def-r}r= \displaystyle \sum _{n=0}^{\infty } q_{n} . \end{equation}
  </div>
  <span class="equation_label">2.5</span>
</p>
</div>
<p>Moreover, the following estimate holds </p>
<div class="displaymath" id="a0000000044">
  \[  \parallel x_{n} - x^{\star }\parallel \leq \displaystyle \sum _{k=n+1}^{\infty } q_{k} \,  \eta {\lt} r \,  \eta .  \]
</div>
<p>Furthermore, \(x^{\star }\) is the unique solution of \(F(x)=0\) in \(U(x_{0} , r_{0}) \cap \mathcal{D} \), provided that \(r_{0} \geq r \,  \eta \), where, </p>
<div class="equation" id="def-r0">
<p>
  <div class="equation_content">
    \begin{equation}  \label{def-r0}r_{0} = \displaystyle \tfrac {2}{ \beta \,  M} - 2\,  d - r \,  \eta . \end{equation}
  </div>
  <span class="equation_label">2.6</span>
</p>
</div>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000045">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>According to Lemmas <a href="#L.2.3">3</a>, and <a href="#L.2.4">4</a>, sequence \(\{ x_{n} \} \) is of Cauchy (\(\{  q_{n} \} \) is of Cauchy) in a Banach space \(\mathcal{X}\), and it converges to some \(x^{\star }\in \overline{U} (x_{0} , r \,  \eta )\) (since, \(\overline{U} (x_{0} , r \,  \eta )\) is a closed set). The sequence \(\{ \mu _{n} \} \) is bounded above. Indeed, we have </p>
<div class="displaymath" id="a0000000046">
  \[  \mu _{n} = \displaystyle \tfrac {1}{ 1 - 2\,  \gamma \,  \displaystyle \sum _{i=0}^{n-1} (q_{i-1}+q_{i}) } \leq \displaystyle \tfrac {1}{ 1 - 2\,  \gamma \,  \displaystyle \sum _{i=0}^{\infty } (q_{i-1}+q_{i}) } ,  \]
</div>
<p> and \(\displaystyle \lim _{n \longrightarrow \infty } q_{n} =0\), which imply \(\displaystyle \lim _{n \longrightarrow \infty } c_{n} =0\). By letting \(n \longrightarrow \infty \) in (<a href="#1-1nov">2.4</a>), we get \(F(x^{\star }) =0\). </p>
<p>We also have </p>
<div class="equation" id="2-1nov">
<p>
  <div class="equation_content">
    \begin{equation}  \label{2-1nov}\parallel x_{n+1} - x_{0} \parallel \leq \displaystyle \sum _{i=0}^{n} \parallel x_{i+1} - x_{i} \parallel \leq \displaystyle \sum _{i=0}^{n} q_{i} \,  \eta < r \,  \eta , \end{equation}
  </div>
  <span class="equation_label">2.7</span>
</p>
</div>
<p> which imply \(x_{n} \in \overline{U} (x_{0} , r \,  \eta ) \). Consequently, we obtain \(x^{\star }\in \overline{U} (x_{0} , r \,  \eta )\).<br /></p>
<p>Finally, we shall show the uniqueness of the solution \(x^{\star }\) in \(U(x_{0} ,r_{0})\). Let \(y^\star \) be a solution of equation \(F(x)=0\) in \(U(x_{0} , r_{0})\). Define linear operator </p>
<div class="displaymath" id="a0000000047">
  \[  \mathcal{L} = \displaystyle \int _{0} ^{1} F^{\prime }(x^{\star }_{t}) \mathrm{d}t , \quad \mathrm{where} \quad x^{\star }_{t} =x^{\star }+ t \,  (y^\star - x^{\star }) .  \]
</div>
<p>We shall show \({\mathcal{L} } ^{-1}\) exists. Using (\({\mathcal{C} _{2}}\)) and (\({\mathcal{C} _{7}}\)), we get </p>
<div class="displaymath" id="a0000000048">
  \begin{align}  \parallel A _{0}^{-1} \parallel \,  \parallel A_{0} - \mathcal{L} \parallel &  \leq \displaystyle \tfrac {\beta \,  M }{2 } \displaystyle \int _{0}^{1} ( \parallel 2\,  x_{0}-x_{-1} - x^{\star }_{t}\parallel + \parallel x_{-1} - x^{\star }_{t} \parallel \mathrm{d}t\nonumber \\ &  \leq \displaystyle 2\,  \tfrac {\beta \,  M }{2 } \,  \displaystyle \int _{0}^{1} ( \parallel x_{0} - x_{-1} \parallel + 2 \,  \parallel x_{0} - x^{\star }_{t} \parallel )\mathrm{d}t\label{2.18}\\ &  \leq \displaystyle \tfrac {\beta \,  M }{2 } \,  (2\,  d + \parallel x_{0} - x^{\star }\parallel +\parallel y^\star - x_{0}\parallel )\nonumber \\ &  {\lt} \displaystyle \tfrac {\beta \,  M }{2 } \,  (2\,  d + r \,  \eta + r_{0}) =1.\nonumber \end{align}
</div>
<p>It follows from (<a href="#2.18">2.8</a>), and the Banach lemma on invertible operators, that \(\mathcal{L} \) is invertible. </p>
<p>Finally, in view of the equality </p>
<div class="displaymath" id="a0000000049">
  \[  0 = F(y^\star ) - F(x^{\star }) = \mathcal{L} \,  (y^\star - x^{\star }),  \]
</div>
<p> we obtain </p>
<div class="displaymath" id="a0000000050">
  \[  x^{\star }= y^\star .  \]
</div>
<p>This completes the proof of Theorem <a href="#T.2.4">5</a>. </p>
<p><div class="remark_thmwrapper " id="R.2.6">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">6</span>
  </div>
  <div class="remark_thmcontent">
  
<ol class="enumerate">
  <li><p>It follows from the proof of Lemma <span class="rmfamily"><a href="#L.2.4">4</a></span> that </p>
<div class="displaymath" id="a0000000051">
  \[  \mu _{k} = \displaystyle \tfrac {1}{1 -2\,  \gamma \,  \displaystyle \sum _{i=0} ^{k-1} ( q_{i-1} + q_{i} ) },  \]
</div>
<p> so that </p>
<div class="equation" id="add-6nov">
<p>
  <div class="equation_content">
    \begin{equation}  \label{add-6nov}\sum _{i=0} ^{k-1} ( q_{i-1} + q_{i} ) = \displaystyle \tfrac {1}{2\,  \gamma } \,  (1- \displaystyle \tfrac {1}{\mu _{k}} ) . \end{equation}
  </div>
  <span class="equation_label">2.9</span>
</p>
</div>
<p>By (<a href="#add-6nov">2.9</a>), the following relation between \(\mu _{\infty }\) and \(r\) holds: </p>
<div class="displaymath" id="a0000000052">
  \[  r = .5 \,  \,  {\bigg(}-q_{-1} + \displaystyle \tfrac {1}{2\,  \gamma } \,  (1- \displaystyle \tfrac {1}{\mu _{\infty }}) {\bigg)}.  \]
</div>
<p>Set </p>
<div class="displaymath" id="a0000000053">
  \[  \overline{r} _{n} = .5 \,  {\bigg(}-q_{-1} + \displaystyle \tfrac {1}{2\,  \gamma } \,  (1- \displaystyle \tfrac {1}{\mu _{n}}) {\bigg)}, \quad \overline{r}= .5 \,  (-q_{-1} + \displaystyle \tfrac {1}{2\, \gamma } ) \quad \mathrm{and} \quad \overline{r} _{0} = \displaystyle \tfrac {2}{\beta \,  M} -2\,  d - \overline{r} \,  \eta .  \]
</div>
<p>Then, we have </p>
<div class="displaymath" id="a0000000054">
  \[  \overline{r} {\gt} r \quad \mathrm{and} \quad \overline{r} _{0} {\lt} r_{0} .  \]
</div>
<p>In view of the proof of Theorem <a href="#T.2.4">5</a>, \(\overline{r}\) can replace \(r\). However, this approach is less accurate but it avoids the computation of \(\mu _{\infty }\). </p>
</li>
  <li><p>Condition (\({\mathcal{C}} _{5}\)) implies that for \(x=y\) and \(u=v\) </p>
<div class="displaymath" id="a0000000055">
  \[  \parallel F^{\prime }(x_{0})^{-1} \, \,  (F^{\prime }(x) - F^{\prime }(u)) \parallel \leq M\, \,  \parallel x- u \parallel \quad \mathrm{for \, \,  all} \quad x,u \in \mathcal{D} .  \]
</div>
<p>Then the conclusions of <span class="cite">
	[
	<a href="#673-EH-C" >14</a>
	, 
	Theorem 4.4
	]
</span> can be obtained from Theorem <a href="#T.2.4">5</a> for </p>
<div class="displaymath" id="a0000000056">
  \[  b= \displaystyle \tfrac {a^{2} + a -1}{a^{2}}, \qquad c= \displaystyle \tfrac {1}{a^{2}} .\hfil \qed  \]
</div>
</li>
</ol>

  </div>
</div> </p>
<h1 id="a0000000057">3 Numerical examples</h1>
<p>To illustrate the theoretical results introduced previously, we present some numerical examples. In these examples we show some situations where the results provided in the paper can be applied. </p>
<p><div class="example_thmwrapper " id="E.2.6">
  <div class="example_thmheading">
    <span class="example_thmcaption">
    Example
    </span>
    <span class="example_thmlabel">7</span>
  </div>
  <div class="example_thmcontent">
  <p> Let \(\mathcal{X} =\mathcal{Y} =\mathbb {R}^{2}\) be equipped with the max–norm. Choose: </p>
<div class="displaymath" id="a0000000058">
  \[  x_{-1}=(.999,.999)^{T}, \quad x_{0}=(1,1)^{T}, \quad \mathcal{D} = U(x_{0} , 1- \kappa ), \quad \kappa \in [0, 1) .  \]
</div>
<p>Define function \(F\) on \(D\) by </p>
<div class="equation" id="3-3.12">
<p>
  <div class="equation_content">
    \begin{equation}  \label{3-3.12}F(x) =(\theta _{1} ^{3} -\kappa , \,  \theta _{2}^{3}- \kappa )^{T} , \qquad x=(\theta _{1},\theta _{2})^{T}. \end{equation}
  </div>
  <span class="equation_label">3.1</span>
</p>
</div>
<p>The Fréchet–derivative of operator \(F\) is given by </p>
<div class="equation" id="3-amc-2">
<p>
  <div class="equation_content">
    \begin{equation}  \label{3-amc-2}F^{\prime }(x)=\left[ \begin{array}[c]{cc}3 \, \,  \theta _{1}^{2} &  0\\ 0 &  3 \, \,  \theta _{2} ^{2}\end{array} \right] , \end{equation}
  </div>
  <span class="equation_label">3.2</span>
</p>
</div>
<p> and the divided difference of \(F\) is defined by </p>
<div class="displaymath" id="a0000000059">
  \[  [y,x;F]= \displaystyle \int _{0}^{1} F^{\prime }(x+ t(y-x))\mathrm{d}t .  \]
</div>
<p> By the <b class="bfseries">(\(\mathcal{C}\))</b> conditions, Definition <a href="#D.2.1">1</a>, and Remark <a href="#R.2.6">6</a> (a), we have: </p>
<div class="displaymath" id="a0000000060">
  \[  M= 6\, \, (2- \kappa ), \quad \eta = (1- \kappa ) \, \,  \beta .  \]
</div>
<p>Let \(\kappa =.75\). Then, using Maple 13, we get for \(a=b=.5\), and \(c=1\): </p>
<div class="displaymath" id="a0000000061">
  \[  \beta =.333333222, \qquad M=7.5 ,  \]
</div>
<div class="displaymath" id="a0000000062">
  \[  q_{-1} =d=.001, \qquad \eta = .083333306,  \]
</div>
<div class="displaymath" id="a0000000063">
  \[  \gamma =.104166597 , \quad d_{0}=.012000004 , \quad \alpha = .216518950 ,  \]
</div>
<table class="tabular">
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p>\(p_{0}=.027291649 \) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \(q_{0}= p_{0} + w_{0} = 1.027291649 \) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">&nbsp;</td>
  </tr>
  <tr>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>\(\mu _{1} = 1.276355057\) </p>

    </td>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(\overline{r}_{1}=.513645824 \) </p>

    </td>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(\overline{r}_{1} \,  \eta = .042803804 \)</p>

    </td>
  </tr>
</table>
<table class="tabular">
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p>\(w_{1}=.178421441 \) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \(p_{1}=.02542729 \) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \(q_{1}=p_{1}+w_{1}= .203848731 \)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>\(\mu _{2}=1.897556100 \) </p>

    </td>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(\overline{r}_{2}=1.129216014 \) </p>

    </td>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(\overline{r}_{2} \,  \eta =.094101303 \)</p>

    </td>
  </tr>
</table>
<table class="tabular">
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p>\(w_{2}=.128802051 \) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \(p_{2}=.006009641 \) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \(q_{2}=p_{2}+w_{2}= .134811692\)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>\(\mu _{3}=2.190871176 \) </p>

    </td>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(\overline{r}_{3}= 1.298546226 \) </p>

    </td>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(\overline{r}_{3} \,  \eta = .108212149 \)</p>

    </td>
  </tr>
</table>
<table class="tabular">
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p>\(w_{3}=.023629489 \) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \(p_{3}=.000758844 \) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \(q_{3}=p_{3}+w_{3}= .024388333 \)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>\(\mu _{4}=2.362542637 \) </p>

    </td>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(\overline{r}_{4}= 1.378146238 \) </p>

    </td>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(\overline{r}_{4} \,  \eta = .114845482 \)</p>

    </td>
  </tr>
</table>
<table class="tabular">
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p>\(w_{4}=.00258294 \) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \(p_{4}=1.59131E-05 \) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \(q_{4}=p_{4}+w_{4}=.002598853 \)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>\(\mu _{5}=2.394346713 \) </p>

    </td>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(\overline{r}_{5}= 1.391639832 \) </p>

    </td>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(\overline{r}_{5} \,  \eta = .115969947 \)</p>

    </td>
  </tr>
</table>
<table class="tabular">
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p>\(w_{5}=4.9428E\! -\! 05 \) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \(p_{5}=3.21907E\! -\! 08 \) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \(q_{5}=p_{5}\! +\! w_{5}\! =\! 4.94602E\! -\! 05 \)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>\(\mu _{6}=2.397513917 \) </p>

    </td>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(\overline{r}_{6}=1.392963989 \) </p>

    </td>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(\overline{r}_{6} \,  \eta =.116080294 \)</p>

    </td>
  </tr>
</table>
<table class="tabular">
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p>\(w_{6}=9.70475E\! -\! 08 \) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \(p_{6}=1.19934E\! -\! 12 \) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \(q_{6}=p_{6}\! +\! w_{6}=9.70487E\! -\! 08 \)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>\(\mu _{7}=2.397573264 \) </p>

    </td>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(\overline{r}_{7}= 1.392988767 \) </p>

    </td>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(\overline{r}_{7} \,  \eta = .116082359 \)</p>

    </td>
  </tr>
</table>
<table class="tabular">
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p>\(w_{7}=3.59932E\! -\! 12 \) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \(p_{7}=8.72398E\! -\! 20 \) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \(q_{7}=p_{7}\! +\! w_{7}=3.59932E\! -\! 12 \)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>\(\mu _{8}=2.397573381 \) </p>

    </td>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(\overline{r}_{8}= 1.392988816 \) </p>

    </td>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(\overline{r}_{8} \,  \eta = .116082363 \)</p>

    </td>
  </tr>
</table>
<table class="tabular">
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p>\(w_{8}=2.61721E\! -\! 19 \) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \(p_{8}=2.35266E\! -\! 31 \) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \(q_{8}=p_{8}\! +\! w_{8}=2.61721E\! -\! 19 \)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>\(\mu _{9}=2.397573381 \) </p>

    </td>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(\overline{r}_{9}= 1.392988816 \) </p>

    </td>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(\overline{r}_{9} \,  \eta = .116082363 \)</p>

    </td>
  </tr>
</table>
<p>We can stop the process, since \(\overline{r}_{9} =\overline{r}_{8}\). Then, we set \(r\simeq \overline{r}_{9} = 1.392988816 \). Consequently </p>
<div class="displaymath" id="a0000000064">
  \[  \overline{r}_{0} = .681917904  \]
</div>
<p> and </p>
<div class="displaymath" id="a0000000065">
  \[  \mathcal{D} _{0} =U(x_{0}, .681917904 ) \cap \mathcal{D} = \mathcal{D} .  \]
</div>
<p>The hypotheses of Theorem <a href="#T.2.4">5</a> are satisfied. Hence, equation \(F(x)=0\) has a solution </p>
<div class="displaymath" id="a0000000066">
  \[  x^{\star }= ( \sqrt[3]{.75} , \sqrt[3]{.75} )^{T} = (.908560296 , .908560296)^{T} ,  \]
</div>
<p> which is unique in \(\mathcal{D} _{0}\) and can be obtained as the limit of \(\{  x_{k} \} \) starting at \(x_{0}\). </p>
<p>We can make a comparison between CSTM and DFM. Table 1 shows the comparison results for CSTM and DFM for this example. From Table 1, we can conclude that DFM is faster than CSTM. </p>
<div class="table"  id="a0000000067">
   <figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">1</span> 
  <span class="caption_text">The comparison results for CSTM and DFM</span> 
</figcaption> <div class="centered"><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" 
        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="2">
      <p> <small class="footnotesize">CSTM</small> </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="2">
      <p> <small class="footnotesize">DFM</small></p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>\({\footnotesize n}\) </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> \(\Vert y_{n}\! -\! x_{n}\Vert \) </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> \(\Vert x_{n+1}\! -\! x_{n}\Vert \) </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> \(\Vert y_{n}\! -\! x_{n}\Vert \) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p> \(\Vert x_{n+1}\! -\! x_{n}\Vert \)</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" 
        rowspan=""
        colspan="">
      <p><small class="footnotesize">0</small> </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> <small class="footnotesize">0.170170113</small> </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> <small class="footnotesize">0.001</small> </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> <small class="footnotesize">0.169999943</small> </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p> <small class="footnotesize">0.001</small></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" 
        rowspan=""
        colspan="">
      <p><small class="footnotesize">1</small> </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> <small class="footnotesize">0.026874237</small> </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> <small class="footnotesize">0.177126154</small> </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> <small class="footnotesize">0.032667629</small> </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p> <small class="footnotesize">0.177020256</small></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" 
        rowspan=""
        colspan="">
      <p><small class="footnotesize">2</small> </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> <small class="footnotesize">0.004721205</small> </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> <small class="footnotesize">0.029570382</small> </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> <small class="footnotesize">0.001370181</small> </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p> <small class="footnotesize">0.033232850</small></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" 
        rowspan=""
        colspan="">
      <p><small class="footnotesize">3</small> </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> <small class="footnotesize">0.000115413</small> </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> <small class="footnotesize">0.004813812</small> </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> <small class="footnotesize">2.19776E-06</small> </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p> <small class="footnotesize">0.001371179</small></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" 
        rowspan=""
        colspan="">
      <p><small class="footnotesize">4</small> </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> <small class="footnotesize">3.68019E-07</small> </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> <small class="footnotesize">0.000115768</small> </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> <small class="footnotesize">5.70322E-12</small> </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p> <small class="footnotesize">2.19776E-06</small></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-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p><small class="footnotesize">5</small> </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-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> <small class="footnotesize">2.71547E-11</small> </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-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> <small class="footnotesize">3.68046E-07</small> </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-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> <small class="footnotesize">1.11022E-16</small> </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> <small class="footnotesize">5.7031E-12</small></p>

    </td>
  </tr>
</table> </div> 
</div>
<p><span class="qed">â–¡</span></p>

  </div>
</div> </p>
<p><div class="example_thmwrapper " id="E.2.7">
  <div class="example_thmheading">
    <span class="example_thmcaption">
    Example
    </span>
    <span class="example_thmlabel">8</span>
  </div>
  <div class="example_thmcontent">
  
<p>In this example we present an application of the previous analysis to the Chandrasekhar equation [1], [5], [12], [19]: </p>
<div class="equation" id="chan">
<p>
  <div class="equation_content">
    \begin{equation}  \label{chan}x(s) = 1 + \tfrac {s}{4}\,  x(s) \int _{0}^{1} \tfrac {x(t)}{s+t}\mathrm{d}t, \quad s\in [0,1]. \end{equation}
  </div>
  <span class="equation_label">3.3</span>
</p>
</div>
<p> We determine where a solution is located, along with its region of uniqueness. Later, the solution is approximated by an iterative method of DFM. </p>
<p>Note that solving (<a href="#chan">3.3</a>) is equivalent to solve \(F(x)=0\), where \(F:C[0,1]\rightarrow C[0,1]\) and </p>
<div class="equation" id="Fchan">
<p>
  <div class="equation_content">
    \begin{equation}  \lbrack F(x)](s)=x(s)-1-\tfrac {s}{4}\, x(s)\int _{0}^{1}\tfrac {x(t)}{s+t}\, \mathrm{d}t,\quad s\in \lbrack 0,1].\label{Fchan}\end{equation}
  </div>
  <span class="equation_label">3.4</span>
</p>
</div>
<p> To obtain the existence of a unique solution of \(F(x)=0\), where \(F\) is given in (<a href="#Fchan">3.4</a>), we need evaluate \(d\), \(\beta \), \(\eta \), \(M\) from operator (<a href="#Fchan">3.4</a>) and the starting points \(x_{-1}\) and \(x_{0}\). In addition, from (<a href="#Fchan">3.4</a>), we have </p>
<div class="displaymath" id="a0000000068">
  \[  \lbrack F^{\prime }(x)y](s)=y(s)-\tfrac {s}{4}\, x(s)\int _{0}^{1}\tfrac {y(t)}{s+t}\, \mathrm{d}t-\tfrac {s}{4}\, y(s)\int _{0}^{1}\tfrac {x(t)}{s+t}\, \mathrm{d}t,\quad s\in \lbrack 0,1],  \]
</div>
<div class="displaymath" id="a0000000069">
  \begin{align*}  \lbrack 2\, y-x,x;F]z(s)& =\int _{0}^{1}F^{\prime }(x+2\, \tau (y-x))z(s)\, \mathrm{d}\tau \\ & =z(s)-\tfrac {1}{4}\int _{0}^{1}\tfrac {s}{s+t}(y(s)z(t)+z(s)y(t))\, \mathrm{d}t. \end{align*}
</div>
<p>On the other hand, from (<a href="#chan">3.3</a>), we infer that \(x(0)=1\), so that reasonable choices of initial approximations seem to be \(x_{-1}(s)=.99\) and \(x_{0}(s)=1\), for all \(s\in \lbrack 0,1]\), and \(d=\Vert x_{0}-x_{-1}\Vert =.01\). In consequence, </p>
<div class="displaymath" id="a0000000070">
  \[  \Vert I-A_{0}\Vert =\tfrac {1}{2}\max _{s\in \lbrack 0,1]}\left\vert \int _{0}^{1}\tfrac {s}{s+t}\mathrm{d}t\right\vert =\tfrac {1}{2}\max _{s\in \lbrack 0,1]}s\ln (1+\tfrac {1}{s})=\tfrac {\ln 2}{2}{\lt}1.  \]
</div>
<p> Hence, by the Banach lemma, there exists \(A_{0}^{-1}\) and </p>
<div class="displaymath" id="a0000000071">
  \[  \Vert A_{0}^{-1}\Vert \leq \tfrac {1}{1-\Vert I-A_{0}\Vert }\leq \tfrac {2}{2-\ln 2}=1.17718382=\beta .  \]
</div>
<p> Moreover, </p>
<div class="displaymath" id="a0000000072">
  \[  \Vert A_{0}^{-1}F(x_{0})\Vert \leq \beta \ast \tfrac {1}{4}\max _{s\in \lbrack 0,1]}s\ln (1+\tfrac {1}{s})=\beta \ast \tfrac {\ln 2}{4}=0.08859191=\eta .  \]
</div>
<p> Furthermore, </p>
<div class="displaymath" id="a0000000073">
  \[  \Vert \lbrack x,y;F]-[u,v;F]\Vert \leq \tfrac {\ln 2}{4}\left( \Vert x-u\Vert +\Vert y-v\Vert \right)\  \text{and}\  M=\tfrac {\ln 2}{2}=0.150514998.  \]
</div>
<p>If we now choose \(a=b=1/2\), \(c=1\), and using Maple 13, then </p>
<div class="displaymath" id="a0000000074">
  \[  \gamma = .007848527 , \quad q_{-1}=d=.01, \quad d_{0}=.112877124 , \quad \alpha =.017513597 {\lt} 1 ,  \]
</div>
<table class="tabular">
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p>\(p_{0}=.002848051 \) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \(q_{0}=1.002848051 \) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \(\mu _{1} =1.017825791 \)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>\(\overline{r}_{1}=.501424025 \) </p>

    </td>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(\overline{r}_{1} \,  \eta =.044422112 \) </p>

    </td>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">&nbsp;</td>
  </tr>
</table>
<table class="tabular">
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p>\(w_{1}=.012741379 \) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \(p_{1}= .000102398\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \(q_{1}= .012843777\)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>\(\mu _{2}=1.034615081 \) </p>

    </td>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(\overline{r}_{2}=1.009269939 \) </p>

    </td>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(\overline{r}_{2} \,  \eta =.089413152 \)</p>

    </td>
  </tr>
</table>
<table class="tabular">
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p>\(w_{2}=.000314609 \) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \(p_{2}=3.30127E-08\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \(q_{2}=.000314642 \)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>\(\mu _{3}=1.034836224\) </p>

    </td>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(\overline{r}_{3}=1.015849148 \) </p>

    </td>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(\overline{r}_{3} \,  \eta =.089996016 \)</p>

    </td>
  </tr>
</table>
<table class="tabular">
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p>\(w_{3}=9.94684E-08 \) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \(p_{3}= 2.54212E-13 \) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \(q_{3}=9.94687E-08 \)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>\(\mu _{4}=1.034841515 \) </p>

    </td>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(\overline{r}_{4}=1.016006519 \) </p>

    </td>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(\overline{r}_{4} \,  \eta =0.090009958 \)</p>

    </td>
  </tr>
</table>
<table class="tabular">
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p>\(w_{4}=7.6268E-13 \) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \(p_{4}=6.16157E-22 \) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \(q_{4}=7.6268E-13\)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>\(\mu _{5}=1.034841516 \) </p>

    </td>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(\overline{r}_{5}=1.016006568 \) </p>

    </td>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(\overline{r}_{5} \,  \eta =0.090009963 \)</p>

    </td>
  </tr>
</table>
<table class="tabular">
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p>\(w_{5}=1.84847E-21\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \(p_{5}=1.14503E-35 \) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \(q_{5}=1.84847E-21\)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>\(\mu _{6}=1.034841516 \) </p>

    </td>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(\overline{r}_{6}=1.016006568 \) </p>

    </td>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(\overline{r}_{6} \,  \eta =0.090009963 \)</p>

    </td>
  </tr>
</table>
<p>We stop the process, since \(\overline{r}_{6} =\overline{r}_{5}\). Then, we set \(r\simeq \overline{r}_{6} =1.016006568 \). Consequently </p>
<div class="displaymath" id="a0000000075">
  \[  \overline{r}_{0} = 11.17770242 .  \]
</div>
<p> The conditions of Theorem <a href="#T.2.4">5</a> are satisfied. In consequence, equation (<a href="#chan">3.3</a>) has a solution \(x^{\star }\) in \(\{ \varphi \in C[0,1]; \Vert \varphi -1\Vert \le .090009963 \} \). </p>
<p>To obtain a numerical solution of (<a href="#chan">3.3</a>), we first discretize the problem and approach the integral by a Gauss-Legendre numerical quadrature with eight nodes, </p>
<div class="displaymath" id="a0000000076">
  \[  \int _{0}^{1}f(t)\, \mathrm{d}t\approx \sum _{j=1}^{8}w_{j}f(t_{j}),  \]
</div>
<p> where<br /></p>
<table class="tabular">
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p>\({\footnotesize t}_{1}{\footnotesize =0.019855072}\)<small class="tiny">&#8195;</small>\({\footnotesize t}_{2}{\footnotesize =0.101666761}\)<small class="tiny">&#8195;</small>\({\footnotesize t}_{3}{\footnotesize =0.237233795}\)<small class="tiny">&#8195;</small>\({\footnotesize t}_{4}{\footnotesize =0.408282679}\)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p>\({\footnotesize t}_{5}{\footnotesize =0.591717321}\)<small class="tiny">&#8195;</small>\({\footnotesize t}_{6}{\footnotesize =0.762766205}\)<small class="tiny">&#8195;</small>\({\footnotesize t}_{7}{\footnotesize =0.898333239}\)<small class="tiny">&#8195;</small>\({\footnotesize t}_{8}{\footnotesize =0.980144928}\)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p>\({\footnotesize w}_{1}{\footnotesize =0.050614268}\  {\footnotesize w}_{2}{\footnotesize =0.111190517}\)<small class="tiny"> </small>\({\footnotesize w}_{3}{\footnotesize =0.156853323}\)<small class="footnotesize"> </small>\({\footnotesize w}_{4}{\footnotesize =0.181341892}\)</p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>\({\footnotesize w}_{5}{\footnotesize =0.181341892}\)<small class="tiny"> </small>\(\  {\footnotesize w}_{6}{\footnotesize =0.156853323}\)<small class="tiny"> </small>\({\footnotesize w}_{7}{\footnotesize =0.111190517}\)<small class="tiny"> </small>\({\footnotesize w}_{8}{\footnotesize =0.050614268}\)</p>

    </td>
  </tr>
</table>
<p>If we denote \(x_{i}=x(t_{i})\), \(i=1,2,\dots , 8\), equation (<a href="#chan">3.3</a>) is transformed into the following nonlinear system: </p>
<div class="displaymath" id="a0000000077">
  \[  x_{i}=1+\tfrac {x_{i}}{4} \sum _{j=1}^{8} a_{ij}x_{j},\quad i=1,2,\dots ,8,  \]
</div>
<p> where \(\displaystyle {a_{ij}={\tfrac {{t_{i}w_{j}}}{{t_{i}+t_{j}}}}}\). </p>
<p>Denote now \(\overline{x}=(x_{1},x_{2},\ldots ,x_{8})^{T}\), \(\overline{1}=(1,1,\ldots ,1)^{T}\), \(A=(a_{ij})\) and write the last nonlinear system in the matrix form: </p>
<div class="equation" id="nonlin">
<p>
  <div class="equation_content">
    \begin{equation}  \label{nonlin}\overline{x} = \overline{1} + \tfrac {1}{4}\,  \overline{x} \odot (A\overline{x}), \end{equation}
  </div>
  <span class="equation_label">3.5</span>
</p>
</div>
<p> where \(\odot \) represents the inner product. If we choose \(\overline{x}_{0}=(1,1,\ldots ,1)^{T}\) and \(\overline{x}_{-1}=(.99,.99,\ldots ,.99)^{T}\), after eight iterations by applying method DFM with \(a=b=1/2\) and \(c=1\), and using the stopping criterion \(\| \overline{x}_{n+1}-\overline{x}_{n}\| {\lt}10^{-20}\), we obtain the numerical solution \(\overline{x}^{\star }=(x_{1}^{\star },x_{2}^{},\ldots ,x_{8}^{\star })^{T}\) given in Table&#160;<a href="#numsol">2</a>. </p>
<div class="table"  id="numsol">
   <div class="centered"><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="">
      <p>\(j\) </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> \(x^{\star }_{j}\) </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> \(j\) </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> \(x^{\star }_{j}\) </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> \(j\) </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> \(x^{\star }_{j}\) </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> \(j\) </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> \(x^{\star }_{j}\)</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>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> 1.0220…</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> 3 </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> 1.1291…</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>
    <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> 1.2102…</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> 1.2510…</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>2 </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> 1.0747…</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> 4 </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> 1.1751…</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> 1.2350…</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> 8 </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> 1.2595…</p>

    </td>
  </tr>
</table> </div> <figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">2</span> 
  <span class="caption_text">Numerical solution \(\overline{x}^{\star }=(x_{1}^{\star },x_{2}^{\star },\ldots ,x_{8}^{\star })^{T}\) of system (<a href="#nonlin">3.5</a>)</span> 
</figcaption>
</div>
<p><span class="qed">â–¡</span></p>

  </div>
</div> </p>
<h1 id="a0000000078">Conclusion</h1>
<p>We provided a semilocal convergence analysis of DFM for approximating a locally unique solution of an equation in a Banach space, which shows that DFM is faster than CSTM given in [9]. These advantages are obtained under the same computational cost as in [9]. DFM is also a useful derivative free alternative to the usage of <i class="it">Newton’s method</i> (NW). The latter method requires Fréchet-derivative operator at each step but its computation may be too expensive or impossible, especially if the analytic representation of Fréchet-derivative operator involved is unavailable [5]–[7], [14]–[17], [19], [20], [22], [25]. </p>
<p><small class="footnotesize">  </small></p>
<div class="bibliography">
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</dl>


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