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<h1>On an iterative algorithm of Uľm-type for solving equations</h1>
<p class="authors">
<span class="author">Ioannis K. Argyros\(^\ast \) Sanjay K. Khattri\(^\S \)</span>
</p>
<p class="date">September 12, 2012.</p>
</div>
<p>\(^\ast \)Department of Mathematical Sciences, Cameron University, Lawton, Oklahoma 73505-6377, USA, email: <span class="tt">iargyros@cameron.edu</span>. </p>
<p>\(^\S \)Department of Engineering, Stord Haugesund University College, Norway, email: <span class="tt">sanjay.khattri@hsh.no</span> </p>

<div class="abstract"><p> We provide a semilocal convergence analysis of an iterative algorithm for solving nonlinear operator equations in a Banach space setting. This algorithm is of order \(1.839\ldots \), and has already been studied in <span class="cite">
	[
	<a href="#three" >3</a>
	, 
	<a href="#eight" >8</a>
	, 
	<a href="#eighteen" >18</a>
	, 
	<a href="#twenti" >20</a>
	]
</span>. Using our new idea of recurrent functions we show that a finer analysis is possible with sufficient convergence conditions that can be weaker than before, and under the same computational cost. Numerical examples are also provided in this study. </p>
<p><b class="bf">MSC.</b> 65G99; 65H10; 65B05; 47H17; 49M15 </p>
<p><b class="bf">Keywords.</b> Banach space, iterative algorithm, semilocal convergence, divided difference of operator, Fréchet-derivative. </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 equation: </p>
<div class="equation" id="eq:11">
<p>
  <div class="equation_content">
    \begin{equation} \label{eq:11} F(x) = 0, \end{equation}
  </div>
  <span class="equation_label">1</span>
</p>
</div>
<p> where \(F\) is a nonlinear operator defined on an open subset \(D\) of a Banach space \(X\) with values in a Banach space \(Y\). Many problems in computational mathematics can be written in the form <a href="#eq:11" class="eqref">1</a> <span class="cite">
	[
	<a href="#eight" >8</a>
	, 
	<a href="#fourteen" >14</a>
	, 
	<a href="#sixteen" >16</a>
	]
</span>. Potra in <span class="cite">
	[
	<a href="#eighteen" >18</a>
	]
</span> used the Uľm-type method <span class="cite">
	[
	<a href="#twenti" >20</a>
	]
</span> (UTM): </p>
<div class="displaymath" id="eq:utm2">
  \begin{alignat}{3} x_{n+1}& =x_n-A_{n}^{-1}F(x_n)\quad (n\ge {0})\quad (x_{-2},x_{-1},x_{0}\in {D}),\label{eq:12} \intertext {where,} A_n & =[x_n,x_{n-1};F]+[x_{n-2},x_n;F]-[x_{n-2},x_{n-1};F],\label{eq:utm2} \end{alignat}
</div>
<p> to provide a local as well as a semilocal convergence analysis under hypotheses on the first \([\cdot ,\cdot ;F]\) and second \([\cdot ,\cdot ,\cdot ;F]\) order divided differences of operator \(F\). </p>
<p>Here, an operator belonging to the space \(L(X,Y)\) (the Banach space of linear and bounded operators from \(X\) into \(Y\)) is called a divided difference of order one for the operator \(F:X\rightarrow {Y}\) on the points \(x,y\in {X}\) if the following properties hold: </p>
<div class="displaymath" id="eq:prop4">
  \begin{alignat}{4} [x,y;F](y-x) & = F(y)-F(x)\quad \text{for}\quad x\neq {y};\label{eq:prop1} \intertext {if $F$ is Fr\'{e}chet-differentiable at $x\in X$, then} [x,x;F] & = F^\prime (x).\label{eq:prop2} \intertext {An operator belonging to the space $L(X,L(X,Y))$ denoted by $[x,y,z;F]$ is called a divided difference of order two for the operator $F:X\rightarrow {Y}$ on the points $x,y,z\in {X}$ if:} [x,y,z;F](z-x) & = [y,z;F]-[x,y;F],\label{eq:prop3} \intertext {for the distinct points $x,y,z$ if $F$ is twice Fr\'{e}chet-differentiable at $x\in {X}$, then} [x,x,x;F] & =\tfrac {1}{2}F^{\prime \prime }(x). \label{eq:prop4} \end{alignat}
</div>
<p> Potra showed that the \(R-\)order of the method is given by the positive solution of the scalar equations: </p>
<div class="equation" id="a0000000003">
<p>
  <div class="equation_content">
    \begin{equation}  t^3-t^2-t-1=0, \end{equation}
  </div>
  <span class="equation_label">4</span>
</p>
</div>
<p> which is approximately \(1.839\ldots \). Other methods using divided differences of order can be found in [1-21], and references therein. </p>
<p>Here, we are motivated by optimization considerations, and we show that it is possible to provide under the same computational cost an analysis with the following advantages: </p>
<p><b class="bf">Semilocal case:</b> </p>
<ol class="enumerate">
  <li><p>finer error bounds on he distances \(\Vert {x_{n+1}-x_n}\Vert \), \(\Vert {x_{n}-x^\star }\Vert \) (\(n\ge {0}\)), </p>
</li>
  <li><p>weaker sufficient convergence conditions and, </p>
</li>
  <li><p>an at least as precise information on the location of the solution \(x^\star \). </p>
</li>
</ol>
<p><b class="bf">Local case:</b> </p>
<ol class="enumerate">
  <li><p>finer error bounds on the distances involved, </p>
</li>
  <li><p>and at least as large radius of convergence. </p>
</li>
</ol>
<p> The semilocal convergence is provided in §2 followed by local in §3. Numerical examples are also provided in §4. </p>
<h1 id="a0000000004">2 Semilocal Convergence Analysis for (UTM)</h1>
<p> We need the following result on majorizing sequence for (UTM). <div class="lemma_thmwrapper " id="lemma1">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">1</span>
  </div>
  <div class="lemma_thmcontent">
  <p> Let \(\alpha ,\phi ,\gamma ,a,b,c,p,\) and \(q\) be given non-negative constants. Assume: </p>
<div class="displaymath" id="eq:23">
  \begin{align}  \phi +\gamma {\lt}p+q\, c,\label{eq:21} \intertext {then, the polynomial $g$ given by} g(s) = (qc+\alpha +\phi )s^2+(p+\alpha +\gamma )s+\phi +\gamma -p-qc,\label{eq:22} \intertext {has a unique positive root $\delta \in (0,1)$;} \intertext {moreover, suppose} \alpha (b+c)+\phi (a+c)+\gamma (a+b){\lt}1; \label{eq:23}\\ \delta _0\le {\delta };\\ f_2({\delta })\le {0}; \label{eq:25} \end{align}
</div>
<p> where, </p>
<div class="displaymath" id="eq:26">
  \begin{align}  \delta _{0}& =\tfrac {pc+qb(a+b)}{1-\left[\alpha (b+c)+\phi (a+c)+\gamma (a+b)\right]}, \label{eq:26} \intertext {and} f_2(s) & = psc + q(s+1)s^2 + \alpha \left[(1+s+s^2)c+(1+s)c+b\right]\nonumber \\ & {}\qquad +\phi \left[(1+s+s^2)c+a+b+c\right]+\gamma \left[ (1+s)c+a+2b+c\right]-1.\nonumber \end{align}
</div>
<p> Then, scalar sequence \(\left\{ t_n\right\} \) \((n\ge {-2})\) given by </p>
<div class="displaymath" id="a0000000005">
  \begin{gather}  t_{-2}=0,\quad t_{-1}=a,\quad t_{0}=a+b, \quad t_1=a+b+c,\nonumber \\ t_{n+2}=t_{n+1}+M_{n+1}\left(t_{n+1}-t_n\right), \nonumber \intertext {where} M_{n+1} = \tfrac {p\left(t_{n+1}-t_n\right)+q\left(t_n-t_{n-2}\right)(t_n-t_{n-1})/\mu _n}{\mu _{n+1}}, \end{gather}
</div>
<p> is non-decreasing, bounded from above by </p>
<div class="equation" id="eq:28">
<p>
  <div class="equation_content">
    \begin{equation} \label{eq:28} t^{\star \star } = \tfrac {c}{1-\delta }+a+b, \end{equation}
  </div>
  <span class="equation_label">8</span>
</p>
</div>
<p> and converges to its unique least upper bound \(t^\star \) such that \(t\in \left[0,t^{\star \star }\right]\). Moreover, the following estimates hold for all \(n\ge {0}:\) </p>
<div class="displaymath" id="eq:210">
  \begin{gather}  0 {\le } {t_{n+2}-t_{n+1}} {\le }\delta \left(t_{n+1}-t_n\right)\le \delta ^{n+1}c \quad \left(n\ge {0}\right)\label{eq:29} \intertext {and} 0{\le } {t^\star -t_n} {\le } \tfrac {\delta ^n\, c}{1-\delta }\quad (n\ge {1}), \label{eq:210} \end{gather}
</div>
<p> where, </p>
<div class="displaymath" id="a0000000006">
  \begin{align}  \mu _{n+1} =&  1-\left[\alpha \left(t_{n+1}+t_n-2a-b\right)+\phi \left(t_{n+1}+t_{n-1}-a-b \right)\right.\\ & \left. +\gamma \left(t_n+t_{n-1}-a-b-c\right)\right].\nonumber \end{align}
</div>

  </div>
</div> <div class="proof_wrapper" id="a0000000007">
  <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 shall show using induction that </p>
<div class="displaymath" id="eq:213">
  \begin{align}  M_{n+1}& \le \delta \label{eq:212} \intertext {and} \mu _{n+1}& {\lt} 1,\label{eq:213} \end{align}
</div>
<p> hold for all \(n\). It will then follow that (2.9) also holds. Estimates <a href="#eq:213" class="eqref">11</a> and <a href="#eq:213" class="eqref">11</a> hold true for \(n=0\), by (2.3) and (2.4), respectively. Let us assume <a href="#eq:210" class="eqref">9</a>, <a href="#eq:213" class="eqref">11</a> and <a href="#eq:213" class="eqref">11</a> hold for all \(k\le {n}\). It then follows from the induction hypotheses: </p>
<div class="displaymath" id="a0000000008">
  \begin{align*}  t_{k+2}& \le t_{k+1}+\delta \left(t_{k+1}-t_k\right)\le t_k+\delta \left(t_k-t_{k-1}\right)+ \delta \left(t_{k+1}-t_{k}\right) \\ & \le t_1 + \delta \left(t_1-t_0\right)+\cdots +\delta \left(t_{k+1}-t_k\right) \\ & \le a+ b + c + \delta c + \cdots + \delta ^{k+1}c \\ & = a+b+\tfrac {1-\delta ^{k+2}}{1-\delta }c {\lt} a+b+\tfrac {c}{1-\delta }= t^{\star \star }. \end{align*}
</div>
<p> Estimates <a href="#eq:213" class="eqref">11</a> and <a href="#eq:213" class="eqref">11</a> will be true if </p>
<div class="displaymath" id="a0000000009">
  \begin{align*}  p\delta ^k c & + q\delta ^{k-1}c\left(\delta ^{k-1}+\delta ^{k-2}\right)c \le \delta - \delta \left[\alpha \left(t_{k+1}+t_k-t_0-t_{-1}\right)\right.\nonumber \\ & \left.+\phi \left(t_{k+1}+t_{k-1}-t_{0}-t_{-2} \right)+\gamma \left(t_k+t_{k-1}-t_{-1}-t_{-2}\right)\right] \nonumber \end{align*}
</div>
<p> or </p>
<div class="displaymath" id="eq:214">
  \begin{multline} \label{eq:214} p\delta ^{k}c+q\delta ^{k-1}\left(\delta ^{k-1}+\delta ^{k-2}\right)c^2+\delta \left[ \alpha \left(\tfrac {1-\delta ^{k+1}}{1-\delta }c+\tfrac {1-\delta ^{k}}{1-\delta } c+b\right)+\right.\\ \left.+\phi \left(\tfrac {1-\delta ^{k+1}}{1-\delta }c+\tfrac {1-\delta ^{k-1}}{1-\delta } +a+b\right)+\gamma \left(\tfrac {1-\delta }{1-\delta }c+\tfrac {1-\delta ^{k-1}}{1-\delta } c+a+2b\right)\right]-\delta \le 0. \end{multline}
</div>
<p> Estimate <a href="#eq:214" class="eqref">10</a> motivates us to introduce functions \(f_k \  (k\ge {2})\) on \([0,\infty )\) (for \(\delta =s\)) by: </p>
<div class="displaymath" id="eq:215">
  \begin{multline} \label{eq:215} f_k(s) = pcs^{k-1}+qc^2 s^{k-1}(s+1) + \alpha \left[(1+s+\cdots +s^k)c \right. \\ \left. +(1+s+\cdots +s^{k-1})c +b\right] +\phi \left[(1+s+\cdots +s^k)c\right. \\ \left. +(1+s+\cdots +s^{k-2})c+a+b\right]\\ +\gamma \left[(1+s+\cdots +s^{k-1})c+(1+s+\cdots +s^{k-2})c+a+2b\right]-1. \end{multline}
</div>
<p> We shall show instead of <a href="#eq:213" class="eqref">11</a> and <a href="#eq:213" class="eqref">11</a> that </p>
<div class="equation" id="eq:216">
<p>
  <div class="equation_content">
    \begin{equation} \label{eq:216} f_k(\delta )\le {0}. \end{equation}
  </div>
  <span class="equation_label">16</span>
</p>
</div>
<p> We need a relationship between two consecutive functions \(f_k:\) </p>
<div class="displaymath" id="a0000000010">
  \begin{align}  f_{k+1}(s)&  = pcs^k+pcs^{k-1}-pcs^{k-1}+qc^2s^k\left(s+1\right)\nonumber \\ & \quad +qc^2s^{k-1}\left(s+1\right)-qc^2s^{k-1} \left(s+1\right)\nonumber \\ & \quad +\alpha \left[(1+s+\cdots +s^k+s^{k+1})c+(1+s+\cdots +s^{k-1 } +s^k)c+b\right] \nonumber \\ & \quad +\phi \left[(1+s+\cdots +s^k+s^{k+1})c \right. \nonumber \\ & \quad \left.+(1+s+\cdots +s^{k-2}+s^{k-1})c+a+b\right] \nonumber \\ & \quad +\gamma \left[(1+s+\cdots +s^k+s^{k+1})c\nonumber \right. \\ & \quad \left. +(1+s+\cdots +s^{k-2}+s^{k-1} )c+a+2b\right]-1\nonumber \\ & = f_k(s)+g(s)s^{k-1}c,\label{eq:217} \end{align}
</div>
<p> where, function \(g\) is given by <a href="#eq:23" class="eqref">5</a>. In view of <a href="#eq:23" class="eqref">5</a> and <a href="#eq:217" class="eqref">17</a> we get </p>
<div class="equation" id="eq:218">
<p>
  <div class="equation_content">
    \begin{equation} \label{eq:218} f_k(\delta ) = f_2(\delta )\quad (k\ge {2}). \end{equation}
  </div>
  <span class="equation_label">18</span>
</p>
</div>
<p> Hence, <a href="#eq:216" class="eqref">16</a> is true if \(f_2(\delta )\le {0}\), which is true by <a href="#eq:25" class="eqref">7</a>. </p>
<p>Define function \(f_\infty \) on \([0,1)\) by </p>
<div class="equation" id="eq:219">
<p>
  <div class="equation_content">
    \begin{equation}  f_\infty (s)=\lim _{n\to \infty }f_n(s).\label{eq:219} \end{equation}
  </div>
  <span class="equation_label">19</span>
</p>
</div>
<p> Then, we have by <a href="#eq:216" class="eqref">16</a> and <a href="#eq:219" class="eqref">19</a>: </p>
<div class="equation" id="eq:220">
<p>
  <div class="equation_content">
    \begin{equation}  f_{\infty }(\infty )=\lim _{n\to \infty }f_n(\delta )\le \lim _{n\to \infty }{0}=0.\label{eq:220} \end{equation}
  </div>
  <span class="equation_label">20</span>
</p>
</div>
<p> The induction for <a href="#eq:210" class="eqref">9</a>, <a href="#eq:213" class="eqref">11</a> and <a href="#eq:213" class="eqref">11</a> is completed. Hence, we showed sequence \(\{ t_n\} \) is non-decreasing, bounded above by \(t^{\star \star }\) and as such it converges to \(t^\star \). Finally, estimate <a href="#eq:210" class="eqref">9</a> follows from </p>
<div class="displaymath" id="a0000000011">
  \begin{alignat}{3} 0& \le {t_{k+m}-t_n}=\left(t_{k+m}-t_{k+m-1}\right)\nonumber \\ &  \quad +\left(t_{k+m-1}-t_{k+m-2}\right)+\cdots +\left(t_{k+1}-t_{k}\right)\nonumber \\ & \le \left(\delta ^{k+m-1}+\delta ^{k+m-2}+\cdots +\delta ^{k}\right)c = \tfrac {1-\delta ^{k+m}}{1-\delta }\delta ^kc,\label{eq:221} \end{alignat}
</div>
<p> by letting \(m\to \infty \). That completes the proof of the lemma. <div class="proof_wrapper" id="a0000000012">
  <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 can show the main semilocal convergence result for (UTM). <div class="theorem_thmwrapper " id="a0000000013">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">2</span>
  </div>
  <div class="theorem_thmcontent">
  <p>Let \(F\) be a nonlinear operator defined on an open subset \(D\) of a Banach space \(X\) with values in a Banach space \(Y\). Let \(\left[\cdot ,\cdot ;F\right]\), \(\left[\cdot ,\cdot ,\cdot ;F\right]\) be divided differences of first and second order of \(F\) on \(D\), respectively. Let \(x_{-2}, x_{-1}, x_{0}\in {D}\) be three given points from \(D\), and assume \(A_0\) is invertible. Let \(a, b, c, p, q, \alpha , \phi , \gamma \) be non-negative numbers such that for all \(x, y, z, u, v\in {D}:\) </p>
<div class="displaymath" id="eq:222">
  \begin{align}  \left\Vert {x_{-1}-x_0}\right\Vert \le {a}, \quad \Vert {x_{-2}-x_{-1}}\Vert &  \le {b},\quad \left\Vert {A_0^{-1}F(x_0)}\right\Vert \le {c},\label{eq:222}\\ \left\Vert {A_{0}^{-1}\left(\left[x,y;F\right]-\left[u,v;F\right]\right) }\right\Vert & \le {p}\left(\left\Vert {x-u}\right\Vert +\left\Vert {y-v}\right\Vert \right),\label{eq:223} \\ \left\Vert {A_{0}^{-1}\left(\left[x,y,z;F\right]-\left[u,v,z;F\right]\right) }\right\Vert & \le {q}\left\Vert {x-u}\right\Vert ,\label{eq:224}\\ \left\Vert {A_{0}^{-1}\left(\left[x,y;F\right]-\left[x_{0},x_{-1},;F\right]\right) }\right\Vert & \le {\alpha } \left(\left\Vert {x-x_{0}}\right\Vert +\left\Vert {y-x_{-1}}\right\Vert \right), \label{eq:225} \\ \left\Vert {A_{0}^{-1}\left(\left[x,y;F\right]-\left[x_{-2},x_{0};F\right]\right) }\right\Vert & \le {\phi }\left(\left\Vert {x-x_{-2}}\right\Vert +\left\Vert {y-x_{0}}\right\Vert \right), \label{eq:226} \\ \left\Vert {A_{0}^{-1}\left(\left[x,y;F\right]-\left[x_{-2},x_{-1};F\right]\right) }\right\Vert & \le {\gamma }\left(\left\Vert {x-x_{-2}}\right\Vert +\left\Vert {y-x_{-1}}\right\Vert \right), \label{eq:227} \end{align}
</div>
<p> hypotheses of Lemma <span class="rm">1</span> hold and </p>
<div class="equation" id="eq:228">
<p>
  <div class="equation_content">
    \begin{equation} \label{eq:228} \overline{U}(x_0,t^\star ) {:=} \left\{ x\in {X}:\left\Vert {x-x^\star }\right\Vert \subseteq {t^\star }\right\} \subseteq {D}. \end{equation}
  </div>
  <span class="equation_label">28</span>
</p>
</div>
<p> Then, sequence \(\{ x_n\} \) \((n\ge {-2})\) generated by (UTM) is well defined, remains in \(\overline{U}(x_0,t^\star )\) for all \(n\ge {0}\) and converges to a solution \(x^\star \in {\overline{U}(x_0,t^\star )}\) of equation \(F(x)=0\). Moreover, the following estimates hold for all \(n\ge {0}:\) </p>
<div class="displaymath" id="eq:229">
  \begin{alignat}{3} \left\Vert {x_{n+2}-x_{n+1}}\right\Vert &  \le L_{n+1}{\left\Vert {x_{n+1}-x_{n}}\right\Vert }\le t_{n+2}-t_{n+1},\label{eq:229}\\ \left\Vert {x_{n}-x^\star }\right\Vert & \le t^\star -t_n,\label{eq:230} \end{alignat}
</div>
<p> where </p>
<div class="displaymath" id="a0000000014">
  \begin{alignat}{3} L_{n+1} & = \frac{p{\left\Vert {x_{n+1}-x_{n}}\right\Vert }+q{\left\Vert {x_{n}-x_{n-2}}\right\Vert }\, \left\Vert { x_{n}-x_{n-1}}\right\Vert }{d_{n+1}},\nonumber \\ d_{n+1} & = 1\! -\! \left[\alpha \left(\left\Vert {x_{n+1}-x_{0}}\right\Vert + \left\Vert {x_{n}-x_{-1}}\right\Vert \right) \! +\! \phi \left(\left\Vert {x_{n-1}-x_{-2}} \right\Vert \! +\!  \left\Vert {x_{n+1}-x_{0}}\right\Vert \right)\right.\nonumber \\ & \left. \quad + \gamma \left(\left\Vert {x_{n-1}-x_{-2}}\right\Vert + \left\Vert {x_{n}-x_{-1}}\right\Vert \right)\right]. \end{alignat}
</div>
<p> Furthermore, if there exists \(r\ge {t^\star }\) such that </p>
<div class="displaymath" id="eq:233">
  \begin{alignat}{3} U(x_0,r) & \subseteq {D}\label{eq:232} \intertext {and} \phi (t^\star +r)+(\phi +\gamma )(a+b) & \le 1,\label{eq:233} \end{alignat}
</div>
<p> then, the solution \(x^\star \) is unique in \(U(x_0,r)\). </p>

  </div>
</div> <div class="proof_wrapper" id="a0000000015">
  <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 shall show using induction on \(k\ge {0}:\) </p>
<div class="equation" id="eq:234">
<p>
  <div class="equation_content">
    \begin{equation} \label{eq:234} \left\Vert {t_{k+1}-t_k}\right\Vert \le {t_{k+1} - t_k}. \end{equation}
  </div>
  <span class="equation_label">33</span>
</p>
</div>
<p> Estimate <a href="#eq:234" class="eqref">33</a> holds for \(k=-2,-1,0\) by <a href="#eq:26" class="eqref">8</a>, and <a href="#eq:222" class="eqref">22</a>. We also have have \(x_{-2},x_{-1},x_{0}\in \overline{U}(x_0,t^\star )\). Let us assume <a href="#eq:234" class="eqref">33</a>, and \(x_k\in \overline{U}(x_0,t^\star )\) hold for all \(n\le {k+1}\). We have using <a href="#eq:225" class="eqref">25</a>-<a href="#eq:227" class="eqref">27</a>: </p>
<div class="displaymath" id="a0000000016">
  \begin{align} & \left\Vert {A_0}^{-1}\left(A_{k+1}-A_0\right)\right\Vert =\nonumber \\ & = \left\Vert {A_0^{-1}}\left(\left[x_{k+1},x_k;F\right]-\left[x_{0},x_{-1};F\right]\right)\right. \nonumber \\ & {}\quad \left.+{A_0^{-1}}\left(\left[x_{k-1},x_{k+1};F\right]-\left[x_{-2},x_{0};F\right]\right) +{A_0^{-1}}\left(\left[x_{k-1},x_k;F\right]-\left[x_{-2},x_{-1};F\right] \right)\right\Vert \nonumber \\ & \le \alpha \left(\left\Vert {x_{k+1}-x_0}\right\Vert +\left\Vert {x_{k}-x_{-1}}\right\Vert \right) + \phi \left(\left\Vert {x_{k-1}-x_{-2}}\right\Vert +\left\Vert {x_{k+1}-x_{0}}\right\Vert \right) \nonumber \\ & {} \quad + \gamma \left(\left\Vert {x_{k-1}-x_{-2}}\right\Vert +\left\Vert {x_{k}-x_{-1}}\right\Vert \right) \nonumber \\ & \le \alpha \left(t_{k+1}+t_k-2a-b\right) +\phi \left(t_{k+1}+t_{k-1}-a-b\right)+\nonumber \\ & {} \qquad +\gamma \left(t_{k+1}+t_{k-1}-a-b-c\right) {\lt} 1 \qquad (\textrm{by}\eqref{eq:214}).\label{eq:235} \end{align}
</div>
<p> It follows from <a href="#eq:235" class="eqref">34</a> and the Banach lemma on invertible operators <span class="cite">
	[
	<a href="#eight" >8</a>
	, 
	<a href="#fourteen" >14</a>
	]
</span> that \(A_{k+1}^{-1}\) exists and </p>
<div class="displaymath" id="a0000000017">
  \begin{align}  \left\Vert {A_{k+1}^{-1}A_{0}}\right\Vert & \le d_{k+1}^{-1} \nonumber \\ & \le \big\{  1-\left[\alpha \left(t_{k+1}+t_k-2a-b\right)+\phi \left(t_{k+1}+t_{k-1}-a-b\right)\right.\nonumber \\ & \left.\quad +\gamma \left(t_{k+1}+t_{k-1} -a-b-c\right)\right]\big\} ^{-1} \label{eq:236} \end{align}
</div>
<p> In view of <a href="#eq:utm2" class="eqref">2</a>, <a href="#eq:236" class="eqref">35</a>, <a href="#eq:237" class="eqref">36</a>, we have: </p>
<div class="displaymath" id="a0000000018">
  \begin{align}  \left\Vert {x_{k+2}-x_{k+1}}\right\Vert & = \left\Vert \left(A_{k+1}^{-1}A_{0}\right)\left(A_{0}^{-1}F(x_{k+1})\right) \right\Vert \nonumber \\ & \le \left\Vert {A_{k+1}^{-1}A_{0}} \right\Vert \left\Vert {A_{0}^{-1}F(x_{k+1})}\right\Vert \nonumber \\ & \le L_{k+1}\left\Vert {x_{k+1}-x_k}\right\Vert \le t_{k+2}-t_{k+1},\nonumber \end{align}
</div>
<p> which shows <a href="#eq:229" class="eqref">29</a> and <a href="#eq:234" class="eqref">33</a> for all \(n\). By Lemma 2.1, sequence \(\{ x_n\} \) is Cauchy in a Banach space \(X\) and as such it converges to some \(x^\star \in \overline{U}(x_0,t^\star )\) (since \(\overline{U}(x_0,t^\star )\) is a closed set). Estimate <a href="#eq:230" class="eqref">30</a> follows from <a href="#eq:229" class="eqref">29</a> by using standard majorization techniques <span class="cite">
	[
	<a href="#eight" >8</a>
	, 
	<a href="#fourteen" >14</a>
	, 
	<a href="#sixteen" >16</a>
	]
</span>. </p>
<p>Using <a href="#eq:223" class="eqref">23</a>, <a href="#eq:224" class="eqref">24</a> and the induction hypotheses, we obtain in turn </p>
<div class="displaymath" id="a0000000019">
  \begin{align} & \left\Vert {A_0^{-1}\left(\left[x_k,x_{k+1};F\right]-A_k\right)}\right\Vert = \nonumber \\ & = \left\Vert {A_0^{-1}} \left( \left[x_k,x_{k+1};F\right] - \left[x_k,x_{k};F\right] + \left[x_k,x_{k};F\right] -A_k \right) \right\Vert \nonumber \\ & = \Big\| {A_0^{-1}} \left\{  \left[x_k,x_{k+1};F\right] - \left[x_k,x_{k};F\right] - \left(\left[x_k,x_k,x_{k-1};F\right] \right. \right. \nonumber \\ & {}\quad \left. \left. - \left[x_{k-2},x_k,x_{k-1};F\right]\right) \left(x_k-x_{k-1}\right) \right\}  \Big\| \nonumber \\ & \le p \left\Vert {x_{k+1}-x_k} \right\Vert + q \left\Vert {x_{k}-x_{k-2}} \right\Vert \left\Vert {x_{k}-x_{k-1}} \right\Vert \nonumber \\ & \le p \left(t_{k+1}-t_k\right) + q \left(t_{k}-t_{k-2}\right) \left(t_{k}-t_{k-1}\right). \label{eq:237} \end{align}
</div>
<p> We also need the estimate: </p>
<div class="displaymath" id="a0000000020">
  \begin{align}  \left\Vert {x_{k+2}-x_{k+1}}\right\Vert & = \left\Vert {\left(A_{k+1}^{-1}A_0\right)\left(A_{0}^{-1}F\left(x_{k+1}\right)\right)}\right\Vert \nonumber \\ & = \left\Vert {\left(A_{k+1}^{-1}A_0\right)A_0^{-1}\left(F\left(x_{k+1}\right)- F\left(x_{k}\right) - A_k\left(x_{k+1}-x_k\right)\right)}\right\Vert \nonumber \\ & \le \left\Vert {A_{k+1}^{-1}A_0}\right\Vert \left\Vert {A_0^{-1} \left(\left[x_k,x_{k+1};F\right]-A_k\right)}\right\Vert \left\Vert {x_{k+1}-x_k}\right\Vert .\label{eq:238} \end{align}
</div>
<p> The fact that \(x^\star \) is a solution of equation \(F(x)=0\) follows by letting \(k\to \infty \) in the estimate: </p>
<div class="displaymath" id="a0000000021">
  \begin{align}  \left\Vert {A_{0}^{-1}F\left(x_{k+1}\right)}\right\Vert & = \left\Vert {A_{0}^{-1}\left(\left[x_k,_{k+1};F\right]-A_k\right)\left(x_{k+1}-x_k\right) } \right\Vert \nonumber \\ & \le p \left\Vert {x_{k+1}-x_k}\right\Vert ^2 + q \left\Vert {x_{k}-x_{k-2}}\right\Vert \, \left\Vert {x_{k}-x_{k-1}}\right\Vert \, \left\Vert {x_{k+1}-x_k}\right\Vert .\label{eq:239} \end{align}
</div>
<p> Finally, to show the uniqueness part, let \(y^\star \in {U(x_0,r)}\) be a solution of equation \(F(x)=0\). We can write for \(L=\left[y^\star ,x^\star ;F\right]:\) </p>
<div class="equation" id="eq:240">
<p>
  <div class="equation_content">
    \begin{equation} \label{eq:240} F(y^\star )-F(x^\star ) = L\left(y^\star -x^\star \right). \end{equation}
  </div>
  <span class="equation_label">39</span>
</p>
</div>
<p> We shall show linear operator \(L\) is invertible. Using <a href="#eq:225" class="eqref">25</a>-<a href="#eq:227" class="eqref">27</a>, <a href="#eq:233" class="eqref">32</a> and <a href="#eq:233" class="eqref">32</a>, we have: </p>
<div class="displaymath" id="a0000000022">
  \begin{align} & \left\Vert {A_{0}^{-1}}\left(\left[x_{0},x_{-2};F\right] + \left[x_{-1},x_{0};F\right] - \left[y^\star ,x^\star ;F\right] - \left[x_{-2},x_{-1};F\right]\right)\right\Vert \le \nonumber \\ & {}\le \left\Vert {A_{0}^{-1}} \left( \left[x_{0},x_{-2};F\right] - \left[y^\star ,x^\star ;F\right] \right)\right\Vert \nonumber \\ & \quad + \left\Vert {A_{0}^{-1}} \left( \left[x_{-1},x_{0};F\right] - \left[x_{-2},x_{-1};F\right] \right)\right\Vert \nonumber \\ & {} \le \phi \left(\left\Vert {x_{0}-y^\star }\right\Vert + \left\Vert {x_{-2}-x^\star }\right\Vert \right) + \gamma \left(\left\Vert {x_{-1}-x_{-2}}\right\Vert + \left\Vert {x_{0}-x_{-1}}\right\Vert \right) \nonumber \\ & {} {\lt} \phi \left(r+t^\star +b+a\right) + \gamma \left(b+a\right)\le {1}.\label{eq:241} \end{align}
</div>
<p> In view of <a href="#eq:241" class="eqref">40</a> and the Banach lemma on invertible operators, \(L^{-1}\) exists. We deduce from <a href="#eq:240" class="eqref">39</a> that \(x^\star =y^\star .\) That completes the proof of the Theorem. <div class="proof_wrapper" id="a0000000023">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<p><div class="remark_thmwrapper " id="a0000000024">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">3</span>
  </div>
  <div class="remark_thmcontent">
  <p>\((a)\) A similar existence Theorem (without a uniqueness result) was provided in <span class="cite">
	[
	<a href="#eighteen" >18</a>
	, 
	p.91
	]
</span> using conditions <a href="#eq:222" class="eqref">22</a>-<a href="#eq:223" class="eqref">23</a>, a decreasing majorizing sequence, and some different sufficient convergence conditions. Therefore a direct comparison is not possible. However, in §4, we show that the results obtained in Theorem 2.2 can be weaker than the corresponding ones of Theorem 5.1 in <span class="cite">
	[
	<a href="#eighteen" >18</a>
	, 
	p.91
	]
</span>. </p>
<p>\((b)\) Note that \(t^{\star \star }\) given by <a href="#eq:28" class="eqref">8</a> can replace \(t^\star \) in hypotheses <a href="#eq:228" class="eqref">28</a> and <a href="#eq:233" class="eqref">32</a> of Theorem 2.2.<span class="qed">â–¡</span></p>

  </div>
</div> </p>
<h1 id="a0000000025">3 Local Convergence of (UTM)</h1>
<p> We can show the local convergence result for (UTM). <div class="theorem_thmwrapper " id="a0000000026">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">4</span>
  </div>
  <div class="theorem_thmcontent">
  <p>Let \(F\colon {D\subseteq {X\rightarrow {Y}}}\) and let \(x^\star \in {D}\) be such that \({{F^\prime (x^\star )^{-1}}}\) exists. Assume that for all \(x,y,u,v,z\in {D}:\) </p>
<div class="displaymath" id="eq:31">
  \begin{align}  \left\Vert {{F^\prime (x^\star )^{-1}}} \left(\left[x^\star ,x^\star ;F\right]-\left[x,x^\star ; F\right]\right)\right\Vert & \le p_0\left\Vert {x^\star -x}\right\Vert , \label{eq:31} \\ \left\Vert {{F^\prime (x^\star )^{-1}}} \left(\left[z,x^\star ;F\right]-\left[z,x; F\right]\right)\right\Vert & \le p_1\left\Vert {x^\star -x}\right\Vert ,\label{eq:32} \end{align}
</div>
<div class="displaymath" id="eq:33">
  \begin{align}  \left\Vert {{F^\prime (x^\star )^{-1}}} \left(\left[x,x^\star ,y;F\right]-\left[z,x^\star ,y; F\right]\right)\right\Vert & \le q_0\left\Vert {x-z}\right\Vert ,\label{eq:33} \\ \left\Vert {{F^\prime (x^\star )^{-1}}} \left(\left[u,x,y;F\right]-\left[v,x,y; F\right]\right)\right\Vert & \le q_\star \left\Vert {u-v}\right\Vert \label{eq:34} \end{align}
</div>
<p> and </p>
<div class="displaymath" id="eq:35">
  \begin{align}  {{U\left(x^\star ,r^\star \right)}} & \subseteq {D},\label{eq:35} \end{align}
</div>
<p> where </p>
<div class="displaymath" id="eq:36">
  \begin{align} & r^\star = \tfrac {2}{p_0+2p_1+\sqrt{\left(p_0+2p_1\right)^2+16\left(q_0+q_\star \right)}}.\label{eq:36} \end{align}
</div>
<p> Then, sequence \(\left\{ x_n\right\} \) generated by (UTM) is well defined, remains in \(U\left(x^\star ,r^\star \right)\) for all \(n\ge {0}\) and converges to \(x^\star \), provided that \(x_0\in U(x^\star ,r^\star )\). Moreover, the following estimates hold: </p>
<div class="equation" id="eq:47">
<p>
  <div class="equation_content">
    \begin{equation}  \left\Vert {x_{n+1}-x^\star }\right\Vert \le \tfrac {e_n}{h_n}\left\Vert {x_n-x^\star }\right\Vert ,\label{eq:47} \end{equation}
  </div>
  <span class="equation_label">47</span>
</p>
</div>
<p> where </p>
<div class="equation" id="eq:48">
<p>
  <div class="equation_content">
    \begin{equation}  e_n = p_1\left\Vert {x_n-x^\star }\right\Vert + q_\star \left\Vert {x_n-x_{n-2}}\right\Vert \left\Vert {x_n-x_{n-1}}\right\Vert \label{eq:48}\qquad \end{equation}
  </div>
  <span class="equation_label">48</span>
</p>
</div>
<p> and </p>
<div class="equation" id="eq:49">
<p>
  <div class="equation_content">
    \begin{equation}  h_n = 1- \left(p_0+p_1\right) \left\Vert {x_n-x^\star }\right\Vert - q_0 \left\Vert {x_n-x_{n-2}}\right\Vert \left\Vert {x_{n-1}-x^\star }\right\Vert .\label{eq:49} \end{equation}
  </div>
  <span class="equation_label">49</span>
</p>
</div>

  </div>
</div> <div class="proof_wrapper" id="a0000000027">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>It follows as the proof of Theorem 4.1 in <span class="cite">
	[
	<a href="#eight" >8</a>
	, 
	p.87
	]
</span> but uses the needed conditions <a href="#eq:31" class="eqref">41</a>-<a href="#eq:34" class="eqref">44</a> instead of: </p>
<div class="displaymath" id="eq:411">
  \begin{align}  \left\Vert {F^\prime (x^\star )^{-1}}\left(\left[x,y;F\right]-\left[u,v;F\right]\right)\right\Vert & \le p_\star \left(\left\Vert {x-u}\right\Vert +\left\Vert {y-v}\right\Vert \right)\label{eq:310} \intertext {and} \left\Vert {F^\prime (x^\star )^{-1}}\left(\left[u,x,y;F\right]-\left[v,x,y;F\right]\right)\right\Vert & \le q_\star \left\Vert {u-v}\right\Vert .\label{eq:411} \end{align}
</div>
<p> <div class="remark_thmwrapper " id="a0000000028">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">5</span>
  </div>
  <div class="remark_thmcontent">
  <p>\((a)\) Clearly </p>
<div class="displaymath" id="eq:312">
  \begin{align}  p_0 & \le p_\star , \label{eq:312}\\ p_1 & \le p_\star , \label{eq:313}\\ q_0 & \le q_\star , \label{eq:314} \end{align}
</div>
<p> hold in general, and \(p_\star /p_0\), \(p_\star /p_1\) and \(q_\star /q_0\) can be arbitrarily large <span class="cite">
	[
	<a href="#seven" >7</a>
	, 
	<a href="#eight" >8</a>
	]
</span>. If equality holds in <a href="#eq:312" class="eqref">51</a>-<a href="#eq:314" class="eqref">53</a>, then our results reduce to the ones in <span class="cite">
	[
	<a href="#eighteen" >18</a>
	]
</span>. Otherwise they constitute an improvement with advantages as noted in the Introduction of this study. \((b)\) The radius of convergence \(r^\star \) obtained in Theorem 3.1 is smaller in general than the corresponding one of Newton’s method. Indeed from the hypotheses <a href="#eq:411" class="eqref">50</a> it follows that \(F\) is Fréchet-differentiable on \(D\) and its Fréchet derivative satisfies </p>
<div class="displaymath" id="eq:316">
  \begin{align}  \left\Vert {F^\prime (x^\star )^{-1}} \left(F^\prime (x)-F^\prime (y)\right) \right\Vert & \le 2p_\star \left\Vert {x-y}\right\Vert \label{eq:315} \intertext {and} \left\Vert {F^\prime (x^\star )^{-1}} \left(F^\prime (x)-F^\prime (x^\star )\right) \right\Vert & \le 2p_2 \left\Vert {x-x^\star }\right\Vert . \label{eq:316} \end{align}
</div>
<p> The radius of convergence \(r^\star \) is then given for \(q_0=q_\star =0:\) </p>
<div class="displaymath" id="eq:319">
  \begin{align}  r_A^\star & = \tfrac {1}{2p_2+p_\star } \label{eq:317} \intertext {for} p_2 & \le {p_\star },\label{eq:318} \intertext {whereas the one obtained by Theorem 4.1 in \cite{eighteen} is given by} r_R^\star & = \tfrac {1}{3p_\star },\label{eq:319} \end{align}
</div>
<p> found by Rheinboldt in <span class="cite">
	[
	<a href="#sixteen" >16</a>
	]
</span>. Note that </p>
<div class="equation" id="eq:320">
<p>
  <div class="equation_content">
    \begin{equation}  r_R^\star \le r_A^\star . \label{eq:320} \end{equation}
  </div>
  <span class="equation_label">56</span>
</p>
</div>
<p> If strict inequality holds in <a href="#eq:319" class="eqref">55</a>, then so does in <a href="#eq:320" class="eqref">56</a>.<span class="qed">â–¡</span></p>

  </div>
</div> </p>
<h1 id="a0000000029">4 A Numerical Example</h1>
<p> We provide a numerical example to show that Theorem 2.2 can be used to solve equation <a href="#eq:11" class="eqref">1</a> but not corresponding Theorem 5.1 in <span class="cite">
	[
	<a href="#eighteen" >18</a>
	]
</span>. Let \(X=Y=R^2\) be equipped with the max-norm, \(x_0=(1,1)^T,\) \(D=U(x_0,1-\lambda )\), \(\lambda \in [0,1/2)\) and define function \(F\) on \(D\) for \(x=\left(\mu _1,\mu _2\right)\) by </p>
<div class="equation" id="eq:41">
<p>
  <div class="equation_content">
    \begin{equation} \label{eq:41} F(x) = \left(\mu _1^3-\lambda ,\mu _2^3-\lambda \right)^T. \end{equation}
  </div>
  <span class="equation_label">57</span>
</p>
</div>
<p> Using <a href="#eq:41" class="eqref">57</a> we obtain the Fréchet-derivative </p>
<div class="equation" id="eq:42">
<p>
  <div class="equation_content">
    \begin{equation} \label{eq:42} F^\prime (x)=\left[\begin{matrix}  3\mu _1^2 

&  0 

\\ 0 

&  3\mu _2^2 

\end{matrix}\right]. \end{equation}
  </div>
  <span class="equation_label">58</span>
</p>
</div>
<p> Define </p>
<div class="equation" id="eq:43">
<p>
  <div class="equation_content">
    \begin{equation} \label{eq:43} [x,y;F] = \int _{0}^1F^\prime (y+t(x-y)){\rm dt}. \end{equation}
  </div>
  <span class="equation_label">59</span>
</p>
</div>
<p> Let \(x_{-2} = \left(1.02,1.02\right)^T\), \(x_{-1}=\left(1.01,1.01\right)\), \(\lambda = 0.49\). Using <a href="#eq:222" class="eqref">22</a>-<a href="#eq:227" class="eqref">27</a> and <a href="#eq:43" class="eqref">59</a>, we get </p>
<div class="displaymath" id="a0000000030">
  \begin{gather*}  a=b=0.1,\quad c = 0.170011334, \quad \left\Vert {A_0}^{-1}\right\Vert = 0.33335557, \nonumber \\ p = 3 \left\Vert {A_{0}^{-1}}\right\Vert (2-\lambda ), \quad q = 3\left\Vert {A_{0}^{-1}}\right\Vert , \nonumber \\ \alpha = \tfrac {1}{2} \left(1-\lambda +a + \left\Vert {x_0+x_{-1} }\right\Vert \right) \left\Vert {A_{0}^{-1}}\right\Vert , \\ \phi = \tfrac {1}{2} \left(1-\lambda + 2\left\Vert {x_0}\right\Vert + \left\Vert {x_{-2}-x_{0} }\right\Vert \right) \left\Vert {A_{0}^{-1}}\right\Vert , \quad \nonumber \\ \gamma = \tfrac {1}{2} \left(1-\lambda +b + \left\Vert {x_0+x_{-1} }\right\Vert \right) \left\Vert {A_{0}^{-1}}\right\Vert ,\nonumber \intertext {so} p = 1.510100673, \quad q = 1.000066671, \quad \alpha = \phi = \gamma = 0.42169478. \end{gather*}
</div>
<p> Moreover, using <a href="#eq:23" class="eqref">5</a>-<a href="#eq:25" class="eqref">7</a> and <a href="#eq:28" class="eqref">8</a>, we obtain <a href="#eq:23" class="eqref">5</a>, <a href="#eq:23" class="eqref">5</a> become </p>
<div class="displaymath" id="a0000000031">
  \begin{align}  0.84338956 & {\lt} 1.680123342, \nonumber \\ 0.160253576 & {\lt} 1,\nonumber \end{align}
</div>
<p> respectively, </p>
<div class="displaymath" id="a0000000032">
  \begin{gather*}  \delta _0 = 0.305966463, \quad \delta = 0.310470973{\gt} \delta _0, \quad f_2(\delta ) = -0.310669379 {\lt} 0 \end{gather*}
</div>
<p> and </p>
<div class="displaymath" id="a0000000033">
  \begin{gather*}  t^{\star \star } = 0.267566675.\nonumber \end{gather*}
</div>
<p> That is, the hypotheses of Theorem 2.2 are satisfied. Moreover, using Remark 2.3(b) and <a href="#eq:233" class="eqref">32</a>, we see that we can set \(r = 1-\lambda = 0.51.\) Hence, there exists a unique solution: </p>
<div class="equation" id="a0000000034">
<p>
  <div class="equation_content">
    \begin{equation} \nonumber x^\star = \left(\sqrt[3]{0.49},\sqrt[3]{0.49}\right)^T = \left(0.788373516,0.78373516\right)^T \end{equation}
  </div>
  <span class="equation_label">55</span>
</p>
</div>
<p> of equation \(F(x)=0\) in \(\overline{U}(x_0,r)\), which can be obtained as the limit of (UTM). However, hypotheses of Theorem 5.1 in <span class="cite">
	[
	<a href="#eighteen" >18</a>
	]
</span> do not hold. The sufficient convergence condition corresponding to <a href="#eq:23" class="eqref">5</a> is given by </p>
<div class="displaymath" id="eq:44">
  \begin{gather}  c \le \lambda = \tfrac {1}{3}\tfrac {p+qa+2\lambda _0}{\left(p+qa+\lambda _0\right)^2}\left[1-qa(a+b)\right]^2,\label{eq:44} \intertext {where} \lambda _0 = \left\{ \left(p+qa\right)^2+3q\left(1-qa(a+b)\right)\right\} ^{1/2}.\nonumber \end{gather}
</div>
<p> We have </p>
<div class="displaymath" id="a0000000035">
  \begin{gather}  \lambda _0 = 1.520496025,\quad \lambda = 0.08533979\nonumber \\ \intertext {{{and, so \eqref{eq:44} is violated, since}}} c = 0.170011334{\gt}\lambda = 0.08533979.\nonumber \end{gather}
</div>
<p> Hence, there is no guarantee that (UTM) starting at \(x_0\) converges to \(x^\star \). </p>
<p><small class="footnotesize">  </small></p>
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