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<title>On Newton’s method using recurrent functions under hypotheses up to the second Fréchet derivative: On Newton’s method using recurrent functions under hypotheses up to the second Fréchet derivative</title>
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<h1>On Newton’s method using recurrent functions under hypotheses up to the second Fréchet derivative</h1>
<p class="authors">
<span class="author">Ioannis K. Argyros\(^\ast \) Saïd Hilout\(^{\ast \ast }\)</span>
</p>
<p class="date">March 26, 2012</p>
</div>
<p>\(^\ast \)Cameron University, Department of Mathematics Sciences, Lawton, OK 73505, USA, e-mail: <span class="tt">iargyros@cameron.edu</span>. </p>
<p>\(^{\ast \ast }\)Poitiers University, Laboratoire de Mathématiques et Applications, Bd. Pierre et Marie Curie, Téléport 2, B.P. 30179, 86962 Futuroscope Chasseneuil Cedex, France,<br />e-mail: <span class="tt">said.hilout@math.univ-poitiers.fr</span>. </p>

<div class="abstract"><p> We provide semilocal result for the convergence of Newton method to a locally unique solution of an equation in a Banach space setting using hypotheses up to the second Fréchet–derivatives and our new idea of recurrent functions. The advantages of such conditions over earlier ones in some cases are: finer bounds on the distances involved, and a better information on the location of the solution. </p>
<p><b class="bf">MSC.</b> 65B05, 65G99, 65J15, 65N30, 65N35, 65H10, 47H17, 49M15. </p>
<p><b class="bf">Keywords.</b> Newton’s method, recurrent functions, Banach space, semilocal convergence, Fréchet–derivative, majorizing sequence, Lipschitz/center–Lipschitz conditions, radius of convergence. </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="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 twice Fréchet–differentiable operator defined on a open convex subset \(\mathcal{D}\) of a Banach space \(\mathcal{X}\) with values in a Banach space \(\mathcal{Y}\). </p>
<p>Computational sciences have received substantial and significant interest of researchers in recent years in several areas such as engineering sciences, dynamical systems, economic equilibrium theory and mathematical programming. Various problems can be solved using the computational sciences by passing first through mathematical modeling and then later looking for the solution iteratively <span class="cite">
	[
	<a href="#623-4" >4</a>
	]
</span>, <span class="cite">
	[
	<a href="#arg-hil-polimetrica2" >5</a>
	]
</span>. 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, 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–when 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. </p>
<p>The famous Newton’s method </p>
<div class="equation" id="1.2">
<p>
  <div class="equation_content">
    \begin{equation}  \label{1.2} x_{n+1}=x_n - F'(x_n)^{-1}\, F(x_n) \quad (n \geq 0), \quad (x_0 \in \mathcal{D}) \end{equation}
  </div>
  <span class="equation_label">1.2</span>
</p>
</div>
<p> has long played a central role in approximating solutions \(x^\star \) of nonlinear equations and systems. Here \(F'(x_n)\) denotes the Fréchet-derivative of operator \(F\) evaluated at \(x=x_n\) (\(n \geq 0\)) <span class="cite">
	[
	<a href="#623-4" >4</a>
	]
</span>, <span class="cite">
	[
	<a href="#623-9" >10</a>
	]
</span>. The geometric interpretation of Newton’s method is well known, if \(F\) is a real function. In such a case \(x_{n+1}\) is the point where the tangential line \(y- F(x_{n}) = F'(x_{n})\, \,  (x-x_{n})\) of function \(F(x_n)\) at the point (\(x_{n},F(x_{n})\)) intersects the \(x\)–axis. </p>
<p>Local and semilocal convergence theorems for the quadratic convergence of Newton’s method to \(x^\star \) have been given under several assumptions by various authors <span class="cite">
	[
	<a href="#633-1" >1</a>
	]
</span>–<span class="cite">
	[
	<a href="#633-8" >11</a>
	]
</span>. For example, Lipschitz conditions have been used on the Fréchet–derivative \(F'(x)\) of \(F\) (\(x \in \mathcal{D}\)) <span class="cite">
	[
	<a href="#633-1" >1</a>
	]
</span>, <span class="cite">
	[
	<a href="#633-2" >2</a>
	]
</span>, <span class="cite">
	[
	<a href="#623-8" >9</a>
	]
</span>–<span class="cite">
	[
	<a href="#633-8" >11</a>
	]
</span>, or center-Lipschitz conditions on the second-derivative \(F''(x)\) of \(F\) (\(x \in \mathcal{D}\)) <span class="cite">
	[
	<a href="#633-1" >1</a>
	]
</span>–<span class="cite">
	[
	<a href="#633-8" >11</a>
	]
</span>. Here, we use center-Lipschitz conditions on both first and second Fréchet-derivatives of \(F\) and recurrent functions. This particular combination has several advantages over the previously mentioned works. That is why we provide new semilocal convergence theorems for Newton’s method. </p>
<p>In particular, assume the Lipschitz condition: </p>
<div class="equation" id="1.3-sept">
<p>
  <div class="equation_content">
    \begin{equation}  \label{1.3-sept} \parallel F'(x_0)^{-1} \, \,  (F'(x)- F'(y) ) \parallel \leq d \, \,  \parallel x-y \parallel , \end{equation}
  </div>
  <span class="equation_label">1.3</span>
</p>
</div>
<p> for all \(x,y\) in \(\mathcal{D}\) and </p>
<div class="displaymath" id="a0000000003">
  \[  \parallel F'(x_0)^{-1} \, \,  F(x_0) \parallel \leq \eta .  \]
</div>
<p>Then, we arrive at the famous for its simplicity and clarity Kantorovich hypothesis for the semilocal convergence of Newton’s method (see <span class="cite">
	[
	<a href="#623-4" >4</a>
	]
</span>, <span class="cite">
	[
	<a href="#arg-hil-polimetrica2" >5</a>
	]
</span>, <span class="cite">
	[
	<a href="#623-9" >10</a>
	]
</span>): </p>
<div class="equation" id="1.4-sept">
<p>
  <div class="equation_content">
    \begin{equation}  \label{1.4-sept} h_K = d \, \,  \eta \leq \displaystyle \tfrac {1}{2}. \end{equation}
  </div>
  <span class="equation_label">1.4</span>
</p>
</div>
<p>In view of (<a href="#1.3-sept">1.3</a>), there exists \(b \geq 0\), such that center-Lipschitz condition </p>
<div class="displaymath" id="a0000000004">
  \[  \parallel F'(x_0)^{-1} \, \,  (F'(x)- F'(x_0) ) \parallel \leq b \, \,  \parallel x-x_0 \parallel  \]
</div>
<p> for all \(x \in \mathcal{D}\). </p>
<p>Clearly, </p>
<div class="displaymath" id="a0000000005">
  \[  b \leq d  \]
</div>
<p> holds in general and \(\displaystyle \tfrac {d}{b}\) can be arbitrarily large <span class="cite">
	[
	<a href="#623-4" >4</a>
	]
</span>, <span class="cite">
	[
	<a href="#arg-hil-complexity" >6</a>
	]
</span>. </p>
<p>Note that in practice the computation of Lipschitz constant \(d\) requires that of \(b\). Therefore, the introduction of the center-Lipschitz condition is not an additional hypothesis. Condition (<a href="#1.3-sept">1.3</a>) is exclusively used in the literature to obtain upper bounds on the norms \(\parallel F'(x_n)^{-1}\, \,  F'(x_0) \parallel \) (\(n \geq 1\)). In particular, if \(x \in \mathcal{D}_0 = U (x_0, \displaystyle \tfrac {1}{d}) \subseteq \mathcal{D}\), \(d \neq 0\), we obtain using (<a href="#1.3-sept">1.3</a>): </p>
<div class="equation" id="1.5-sept">
<p>
  <div class="equation_content">
    \begin{equation}  \label{1.5-sept} \parallel F'(x_0)^{-1} \, \,  (F'(x)- F'(x_0) ) \parallel \leq d \, \,  \parallel x- x_0 \parallel < 1. \end{equation}
  </div>
  <span class="equation_label">1.5</span>
</p>
</div>
<p>In view of (<a href="#1.5-sept">1.5</a>) and the Banach lemma on invertible operator <span class="cite">
	[
	<a href="#623-4" >4</a>
	]
</span>, we conclude that \(F'(x)^{-1}\) exists on \(\mathcal{D}_0\) and </p>
<div class="equation" id="1.6-sept">
<p>
  <div class="equation_content">
    \begin{equation}  \label{1.6-sept} \parallel F'(x)^{-1} \, \, (F'(x_0) \parallel \leq \displaystyle \tfrac {1}{1- d \, \,  \parallel x - x_0 \parallel }. \end{equation}
  </div>
  <span class="equation_label">1.6</span>
</p>
</div>
<p>The usage of estimate (<a href="#1.6-sept">1.6</a>) and several majorizing techniques all lead to condition (<a href="#1.4-sept">1.4</a>) <span class="cite">
	[
	<a href="#623-4" >4</a>
	]
</span>–<span class="cite">
	[
	<a href="#arg-hil-IJCM" >7</a>
	]
</span>. </p>
<p>It is clear that (<a href="#1.3-sept">1.3</a>) is overused, when it comes to obtaining estimate (<a href="#1.6-sept">1.6</a>). Using the needed center-Lipschitz condition, we arrive at the more precise estimate (for \(b {\lt} d\)): </p>
<div class="equation" id="1.7-sept">
<p>
  <div class="equation_content">
    \begin{equation}  \label{1.7-sept} \parallel F'(x)^{-1} \, \, F'(x_0) \parallel \leq \displaystyle \tfrac {1}{1- b \, \,  \parallel x - x_0 \parallel }. \end{equation}
  </div>
  <span class="equation_label">1.7</span>
</p>
</div>
<p>Using (<a href="#1.7-sept">1.7</a>) instead of (<a href="#1.6-sept">1.6</a>) and our new idea of recurrent functions, we showed that (<a href="#1.4-sept">1.4</a>) can always be replaced by </p>
<div class="equation" id="1.8-sept">
<p>
  <div class="equation_content">
    \begin{equation}  \label{1.8-sept} h_A= d_0 \, \,  \eta \leq \displaystyle \tfrac {1}{2}, \end{equation}
  </div>
  <span class="equation_label">1.8</span>
</p>
</div>
<p> where, </p>
<div class="displaymath" id="a0000000006">
  \[  d_0= \displaystyle \tfrac {1}{8} \, \,  (d+ 4 \, \,  b + \sqrt{d^2 + 8\, \,  b \, \,  d} ).  \]
</div>
<p>Note that </p>
<div class="displaymath" id="a0000000007">
  \[  h_K \leq \displaystyle \tfrac {1}{2} \Longrightarrow h_A \leq \displaystyle \tfrac {1}{2}  \]
</div>
<p> but not necessarily vice versa, unless \(b=d\). </p>
<p>In this case, finer errors bounds on the distances \(\parallel x_{n+1} -x_n \parallel \), \(\parallel x_n - x^\star \parallel \) (\(n\geq 0\)) and an at least as precise information on the location of the solution than the ones provided by the Newton-Kantorovich theorem <span class="cite">
	[
	<a href="#623-4" >4</a>
	]
</span>, are also obtained. </p>
<p>Hence, the applicability of Newton’s method for solving nonlinear equations has been extended and under the same computational cost. </p>
<p>The idea of introducing recurrent functions is a logical consequence of proving the convergence of the majorizing sequence \(\{  \varsigma _n\} \) by showing: </p>
<div class="displaymath" id="a0000000008">
  \[  0 \leq \varsigma _{n+1} - \varsigma _n \leq \epsilon \, \,  (\varsigma _n - \varsigma _{n-1} ), \quad (n \geq 1)  \]
</div>
<p> where, \(\{ \varsigma _n \} \) is given by </p>
<div class="displaymath" id="a0000000009">
  \[  \begin{array}{l} \varsigma _0= 0 ,\qquad \varsigma _1= \eta , \quad \varsigma _{n+1} = \varsigma _{n} + \displaystyle \tfrac {d\, \,  (\varsigma _{n} - \varsigma _{n-1})^2 } {2 \, \,  (1 - b \, \,  \varsigma _{n})}, \quad (n\geq 1) \end{array}  \]
</div>
<p> and </p>
<div class="displaymath" id="a0000000010">
  \[  \epsilon = \displaystyle \tfrac {2\, \,  d} {d+ \sqrt{d^2 + 8 \, \,  b \, \,  d}} .  \]
</div>
<p>It turns out that the idea of using the center-Lipschitz condition in combination with recurrent functions can be used when (<a href="#1.7-sept">1.7</a>) also replaces the corresponding estimates by Huang <span class="cite">
	[
	<a href="#623-8" >9</a>
	]
</span> and Gutiérrez <span class="cite">
	[
	<a href="#639-10" >8</a>
	]
</span>, (see (<a href="#2.68">2.68</a>), (<a href="#2.71">2.71</a>) and Proposition <a href="#P.2.6">6</a>), which are using hypotheses on the second Fréchet–derivative \(F''(x)\) (\(x\in \mathcal{D}\)) of operator \(F\). </p>
<p>The advantages of this approach over the works by Huang <span class="cite">
	[
	<a href="#623-8" >9</a>
	]
</span> and Gutiérrez <span class="cite">
	[
	<a href="#639-10" >8</a>
	]
</span> can be (as in the case already stated above) under the same or weaker hypotheses: </p>
<ol class="enumerate">
  <li><p>Finer error bounds on the distances \(\parallel x_{n+1} - x_n \parallel \), \(\parallel x_n - x^\star \parallel \) (\(n \geq 0\)); </p>
</li>
  <li><p>Better information on the location of the solution \(x^\star \). </p>
</li>
</ol>
<h1 id="a0000000011">2 Semilocal convergence analysis of Newton’s method</h1>
<p>Let \(a{\gt}0\), \(b{\gt}0\), \(c{\gt}0\) and \(\eta {\gt}0\) be given constants. It is convenient for us to define the polynomial \(g\) on \([0, +\infty )\) by </p>
<div class="equation" id="2.1">
<p>
  <div class="equation_content">
    \begin{equation}  \label{2.1} g(s) = 2 \, \,  b \, \,  s^2 + c \,  \,  s - (c+a \, \,  n). \end{equation}
  </div>
  <span class="equation_label">2.1</span>
</p>
</div>
<p> Assume: </p>
<div class="equation" id="2.2">
<p>
  <div class="equation_content">
    \begin{equation}  \label{2.2} \eta < \delta _1 = \displaystyle \tfrac {2 \, \,  b}{a} . \end{equation}
  </div>
  <span class="equation_label">2.2</span>
</p>
</div>
<p> Then, the polynomial \(g\) has a unique root \(\delta /2\) in \((0,1)\). We have </p>
<div class="equation" id="2.3">
<p>
  <div class="equation_content">
    \begin{equation}  \label{2.3} g(0) = -(c + a \, \,  \eta ) < 0 \end{equation}
  </div>
  <span class="equation_label">2.3</span>
</p>
</div>
<p> and </p>
<div class="equation" id="2.4">
<p>
  <div class="equation_content">
    \begin{equation}  \label{2.4} g(1) = 2 \, \,  b - a \, \,  \eta > 0 \quad ({\rm {by}}\, \,  (\ref{2.2})). \end{equation}
  </div>
  <span class="equation_label">2.4</span>
</p>
</div>
<p> It follows from the Intermediate Value Theorem, (<a href="#2.3">2.3</a>) and (<a href="#2.4">2.4</a>) that there exists a root \(\displaystyle \tfrac {\delta }{2} \in (0,1)\) of the polynomial \(g\), given by </p>
<div class="equation" id="2.5">
<p>
  <div class="equation_content">
    \begin{equation}  \label{2.5} \displaystyle \delta = \displaystyle \tfrac {4(c + a \, \,  \eta )}{c + \sqrt{c^2 + 8 \, \,  b \, \, (c + a \, \,  \eta )}}. \end{equation}
  </div>
  <span class="equation_label">2.5</span>
</p>
</div>
<p> We also have </p>
<div class="equation" id="2.6">
<p>
  <div class="equation_content">
    \begin{equation}  \label{2.6} g'(s) = 4 \, \,  b \, \,  s + c > 0 \quad (s>0). \end{equation}
  </div>
  <span class="equation_label">2.6</span>
</p>
</div>
<p> That is, the polynomial \(g\) crosses the positive-axis only once. Hence, \(\delta \) is the only root of the polynomial \(g\) on \((0,1)\). Let us define the polynomial \(p\) on \([0, +\infty )\) by </p>
<div class="equation" id="2.7">
<p>
  <div class="equation_content">
    \begin{equation}  \label{2.7} p(s) = \displaystyle \tfrac {1}{3} \, \,  a \, \,  s^2 + (c + \delta \, \,  b)s - \delta . \end{equation}
  </div>
  <span class="equation_label">2.7</span>
</p>
</div>
<p> The polynomial \(p\) has also a unique positive root, given by </p>
<div class="equation" id="2.8">
<p>
  <div class="equation_content">
    \begin{equation}  \label{2.8} \delta _2 = \displaystyle \tfrac {2 \, \,  \delta }{c + \delta \, \,  b + \sqrt{(c + \delta \, \,  b)^2 + \tfrac {4}{3} \, \,  a \, \,  \delta }}. \end{equation}
  </div>
  <span class="equation_label">2.8</span>
</p>
</div>
<p> Let us also define </p>
<div class="equation" id="2.9">
<p>
  <div class="equation_content">
    \begin{equation}  \label{2.9} \delta _3 = \displaystyle \tfrac {1}{b} (1 - \displaystyle \tfrac {\delta }{2}) \end{equation}
  </div>
  <span class="equation_label">2.9</span>
</p>
</div>
<p> and set </p>
<div class="equation" id="2.10">
<p>
  <div class="equation_content">
    \begin{equation}  \label{2.10} \eta _0 = \min \{  \delta _1, \delta _2, \delta _3 \} . \end{equation}
  </div>
  <span class="equation_label">2.10</span>
</p>
</div>
<p>We can show the following result on majorizing sequences for Newton’s method (<a href="#1.2">1.2</a>). <div class="lemma_thmwrapper " id="L.2.1">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">1</span>
  </div>
  <div class="lemma_thmcontent">
  <p> Let \(a {\gt} 0\), \(b {\gt} 0\), \(c {\gt} 0\) and \(\eta {\gt} 0\) be given constants. </p>
<p>Assume: </p>
<div class="equation" id="2.11">
<p>
  <div class="equation_content">
    \begin{equation}  \label{2.11} \eta \leq \eta _0 , \end{equation}
  </div>
  <span class="equation_label">2.11</span>
</p>
</div>
<p> where \(\eta _0\) is defined in <span class="rm">(<a href="#2.10">2.10</a>)</span>. Inequality <span class="rm">(<a href="#2.11">2.11</a>)</span> is strict if \(\eta _0 = \delta _1\). </p>
<p>Then, scalar sequence \(\{  t_n \} \) (\(n \geq 0\)), generated by </p>
<div class="equation" id="2.12">
<p>
  <div class="equation_content">
    \begin{equation}  \label{2.12} \begin{array}{l} t_0= 0 ,\qquad t_1= \eta , \\ t_{n+2} = t_{n+1} + \displaystyle \tfrac {a \, \,  {\bigg(} \displaystyle \tfrac {1}{3} \, \,  (t_{n+1} - t_n) + t_n {\bigg)} +c } {2 \, \,  (1 - b \, \,  t_{n+1})} \, \,  (t_{n+1} -t_{n}) ^2 \end{array} \end{equation}
  </div>
  <span class="equation_label">2.12</span>
</p>
</div>
<p> is increasing, bounded from above by </p>
<div class="equation" id="2.13">
<p>
  <div class="equation_content">
    \begin{equation}  \label{2.13} t^{\star \star } = \displaystyle \tfrac {2\, \,  \eta }{2 - \delta } \end{equation}
  </div>
  <span class="equation_label">2.13</span>
</p>
</div>
<p> and converges to its unique least upper bound \(t^\star \), with </p>
<div class="equation" id="2.14">
<p>
  <div class="equation_content">
    \begin{equation}  \label{2.14} t^\star \in [0, t^{\star \star } ]. \end{equation}
  </div>
  <span class="equation_label">2.14</span>
</p>
</div>
<p>Moreover the following estimates hold for all \(n\geq 0\): </p>
<div class="equation" id="2.15">
<p>
  <div class="equation_content">
    \begin{equation}  \label{2.15} 0 < t_{n+2} - t_{n+1} \leq \displaystyle \tfrac {\delta }{2} \, \,  (t_{n+1} - t_{n}) \leq {\bigg(} \displaystyle \tfrac {\delta }{2} {\bigg)} ^{n+1} \, \,  \eta \end{equation}
  </div>
  <span class="equation_label">2.15</span>
</p>
</div>
<p> and </p>
<div class="equation" id="2.16">
<p>
  <div class="equation_content">
    \begin{equation}  \label{2.16} t^\star - t_n \leq \displaystyle \tfrac {2 \, \,  \eta }{2 - \delta } \, \,  {\bigg(} \displaystyle \tfrac {\delta }{2} {\bigg)}^n. \end{equation}
  </div>
  <span class="equation_label">2.16</span>
</p>
</div>

  </div>
</div> <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 shall show using induction on \(m\): </p>
<div class="equation" id="2.17">
<p>
  <div class="equation_content">
    \begin{equation}  \label{2.17} \begin{array}{lll} 0 & <&  t_{m+2} - t_{m+1} \\ &  = &  \displaystyle \tfrac {a \, \,  {\bigg(} \displaystyle \tfrac {1}{3} \, \,  (t_{m+1} - t_m) ^2+ t_m \, \,  (t_{m+1} - t_m) {\bigg)} +c \, \,  (t_{m+1} - t_m)} {2 \, \,  (1 - b \, \,  t_{m+1})} \, \,  (t_{m+1} -t_{m}) \\ & \leq &  \displaystyle \tfrac {\delta }{2} \, \,  (t_{m+1} -t_{m}) \end{array} \end{equation}
  </div>
  <span class="equation_label">2.17</span>
</p>
</div>
<p> and </p>
<div class="equation" id="2.18">
<p>
  <div class="equation_content">
    \begin{equation}  \label{2.18} b \, \,  t_{m+1} < 1. \end{equation}
  </div>
  <span class="equation_label">2.18</span>
</p>
</div>
<p>Estimates (<a href="#2.17">2.17</a>) and (<a href="#2.18">2.18</a>) hold for \(m=0\), since </p>
<div class="equation" id="2.19">
<p>
  <div class="equation_content">
    \begin{equation}  \label{2.19} \displaystyle \tfrac {a \, \,  {\bigg(} \displaystyle \tfrac {1}{3} \, \,  (t_{1} - t_0) ^2 + t_0 \, \,  (t_{1} - t_0) {\bigg)} +c \, \,  (t_{1} - t_0)} {1 - b \, \,  t_{1}} = \displaystyle \tfrac { \displaystyle \tfrac {1}{3} \, \,  a \, \,  \eta ^2+ c \, \,  \eta } {1 - b \, \,  \eta } = \delta _0 \leq \delta \end{equation}
  </div>
  <span class="equation_label">2.19</span>
</p>
</div>
<p> and </p>
<div class="equation" id="2.20">
<p>
  <div class="equation_content">
    \begin{equation}  \label{2.20} b \, \,  t_1 = b \, \,  \eta < 1 , \end{equation}
  </div>
  <span class="equation_label">2.20</span>
</p>
</div>
<p> by the choice of \(\delta \) and (<a href="#2.11">2.11</a>). </p>
<p>Let us assume (<a href="#2.17">2.17</a>)–(<a href="#2.18">2.18</a>) for all \(n \leq m+1\). Then, we get from (<a href="#2.15">2.15</a>): </p>
<div class="equation" id="2.21">
<p>
  <div class="equation_content">
    \begin{equation}  \label{2.21} t_{m+2}\leq \displaystyle \tfrac {1- {\bigg(}\displaystyle \tfrac {\delta }{2} {\bigg)}^{m+2} }{1- \displaystyle \tfrac {\delta }{2} } \, \,  \eta < \displaystyle \tfrac {2 \, \,  \eta }{2 - \delta } = t^{\star \star } . \end{equation}
  </div>
  <span class="equation_label">2.21</span>
</p>
</div>
<p>We shall show (<a href="#2.19">2.19</a>) and (<a href="#2.20">2.20</a>), if </p>
<div class="displaymath" id="2.22">
  \begin{align} \label{2.22} &  a \, \,  {\bigg\{ } \displaystyle \tfrac {1}{3} \, \,  {\bigg(} {\bigg(} \displaystyle \tfrac {\delta }{2} {\bigg)}^{m} \, \,  \eta {\bigg)}^2+\displaystyle \tfrac {1- {\bigg(} \displaystyle \tfrac {\delta }{2} {\bigg)} ^{m} } {1- \displaystyle \tfrac {\delta }{2} } \, \,  {\bigg(} \displaystyle \tfrac {\delta }{2} {\bigg)} ^m \, \,  \eta ^2 {\bigg\} }\\ & \quad + c \, \,  {\bigg(} \displaystyle \tfrac {\delta }{2} {\bigg)} ^m \, \,  \eta + b \, \,  \delta \, \,  \displaystyle \tfrac {1- {\bigg(} \displaystyle \tfrac {\delta }{2} {\bigg)} ^{m+1} } {1- \displaystyle \tfrac {\delta }{2} } \, \,  \eta - \delta \leq 0 .\nonumber \end{align}
</div>
<p>Estimates (<a href="#2.22">2.22</a>) motivates us to define polynomials \(f_m\) (\(m \geq 1\)), for \(s= \displaystyle \tfrac {\delta }{2} \) and show instead: </p>
<div class="equation" id="2.23">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.23} \begin{array}{lll} f_m(s)& =&  \displaystyle \tfrac {a \, \,  \eta ^2}{3} \, \,  s^{2 \,  m -1 } + a\, \,  (1+ s + \cdots + s^{m-1})\, \,  s^{m-1} \, \,  \eta ^2 \\ & & +\  c \, \,  s^{m-1} \, \,  \eta + 2 \, \,  b \, \,  (1+ s + \cdots +s^{m} ) \, \,  \eta - 2 \leq 0. \end{array} \end{equation}
  </div>
  <span class="equation_label">2.23</span>
</p>
</div>
<p>We need a relationship between two consecutive functions \(f_m\): </p>
<div class="equation" id="2.24">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.24} \begin{array}{lll} f_{m+1}(s) & =&  \displaystyle \tfrac {a \, \,  \eta ^2}{3} \, \,  s^{2 \,  m +1 } + a\, \,  (1+ s + \cdots + s^{m-1} + s^m)\, \,  s^{m} \, \,  \eta ^2 \\ & &  +\  c \, \,  s^{m} \, \,  \eta + 2 \, \,  b \, \,  (1+ s + \cdots +s^{m} + s^{m+1} ) \, \,  \eta - 2 \\ & =&  \displaystyle \tfrac {a \, \,  \eta ^2}{3} \, \,  s^{2 \,  m +1 } + \displaystyle \tfrac {a \, \,  \eta ^2}{3} \, \,  s^{2 \,  m -1 } - \displaystyle \tfrac {a \, \,  \eta ^2}{3} \, \,  s^{2 \,  m -1 } \\ & & +\  a\, \,  (1+ s + \cdots + s^{m-1} + s^{m-1})\, \,  s^{m-1} \, \,  \eta ^2\\ & &  -\  a\, \,  (1+ s + \cdots + s^{m-1} + s^{m-1})\, \,  s^{m-1} \, \,  \eta ^2 \\ & &  +\  a\, \,  (1+ s + \cdots + s^{m-1} + s^m)\, \,  s^{m} \, \,  \eta ^2 + c \, \,  s^{m-1} \, \,  \eta - c \, \,  s^{m-1} \, \,  \eta \\ & &  +\  c \, \,  s^{m} \, \,  \eta + 2\, \,  b\, \,  (1+ s + \cdots + s^m) \, \,  \eta + 2 \, \,  b \, \,  s^{m+1} \, \,  \eta - 2\\ & =&  f_m(s) +g_m(s) + g(s) \, \,  s^{m-1} \, \,  \eta , \end{array} \end{equation}
  </div>
  <span class="equation_label">2.24</span>
</p>
</div>
<p> where, the function \(g\) is given by (<a href="#2.11">2.11</a>) and </p>
<div class="equation" id="2.25">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.25} g_m(s)= {\bigg(} \displaystyle \tfrac {1}{3} \, \,  s^2 + s + \displaystyle \tfrac {2}{3} {\bigg)} \, \,  s^{2 \,  m -1} \, \,  \eta ^2 \, \,  a \geq 0, \quad (s\geq 0) . \end{equation}
  </div>
  <span class="equation_label">2.25</span>
</p>
</div>
<p>In view of (<a href="#2.1">2.1</a>), (<a href="#2.24">2.24</a>) and (<a href="#2.25">2.25</a>) we have </p>
<div class="equation" id="2.26">
<p>
  <div class="equation_content">
    \begin{equation}  \label{2.26} f_{m+1} {\bigg(}\displaystyle \tfrac {\delta }{2} {\bigg)} \geq f_m {\bigg(}\displaystyle \tfrac {\delta }{2} {\bigg)}. \end{equation}
  </div>
  <span class="equation_label">2.26</span>
</p>
</div>
<p> We shall show, instead of (<a href="#2.22">2.22</a>) </p>
<div class="equation" id="etoile">
<p>
  <div class="equation_content">
    \begin{equation} \label{etoile} f_m {\bigg(} \displaystyle \tfrac {\delta }{2} {\bigg)} \leq 0 \qquad (m \geq 1). \end{equation}
  </div>
  <span class="equation_label">2.27</span>
</p>
</div>
<p> Define the function \(f_\infty \) on \([0, 1)\) by </p>
<div class="equation" id="2.28">
<p>
  <div class="equation_content">
    \begin{equation}  \label{2.28} f_\infty (s) = \displaystyle \lim _{n \to \infty } f_n (s). \end{equation}
  </div>
  <span class="equation_label">2.28</span>
</p>
</div>
<p> Then, using (<a href="#2.23">2.23</a>), we have: </p>
<div class="equation" id="2.29">
<p>
  <div class="equation_content">
    \begin{equation}  \label{2.29} f_\infty (s) = 2 {\bigg[} \displaystyle \tfrac {b \, \,  \eta }{1-s} - 1 {\bigg]}. \end{equation}
  </div>
  <span class="equation_label">2.29</span>
</p>
</div>
<p> It also follows from (<a href="#2.26">2.26</a>) that </p>
<div class="equation" id="2.30">
<p>
  <div class="equation_content">
    \begin{equation}  \label{2.30} f_\infty {\bigg(} \displaystyle \tfrac {\delta }{2} {\bigg)} \geq f_m {\bigg(} \displaystyle \tfrac {\delta }{2} {\bigg)}. \end{equation}
  </div>
  <span class="equation_label">2.30</span>
</p>
</div>
<p> In view of (<a href="#etoile">2.27</a>) and (<a href="#2.30">2.30</a>), it is enough to show </p>
<div class="equation" id="2.31">
<p>
  <div class="equation_content">
    \begin{equation}  \label{2.31} f_\infty {\bigg(} \displaystyle \tfrac {\delta }{2} {\bigg)} \leq 0, \end{equation}
  </div>
  <span class="equation_label">2.31</span>
</p>
</div>
<p> which is true, since \(\eta \leq \delta _3\). This completes the induction. </p>
<p>Therefore, the sequence \(\{  t_n \} \) is non–decreasing, bounded above by \(t^{\star \star } \), given by (<a href="#2.13">2.13</a>) and converges to its unique least upper bound \(t^\star \) satisfying (<a href="#2.14">2.14</a>). Finally, estimate (<a href="#2.16">2.16</a>) follows from (<a href="#2.15">2.15</a>) by using standard majorization techniques <span class="cite">
	[
	<a href="#623-4" >4</a>
	]
</span>, <span class="cite">
	[
	<a href="#623-9" >10</a>
	]
</span>. </p>
<p>That completes the proof of Lemma <a href="#L.2.1">1</a>. </p>
<p>Let us define functions \(\bar{f}_m\) by: </p>
<div class="equation" id="2.32">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.32} \begin{array}{lll} \overline{f}_m(s)& =&  a\, \,  {\bigg(} \displaystyle \tfrac {1}{3} \, \,  s^{2\, m-1}+ (1+ s + \cdots + s^{m-1} )\, \,  s^{m-1} {\bigg)} \, \,  \eta ^2 \\ & &  +\  c \, \,  s^{m-1} \, \,  \eta + 2 \, \,  c \, \,  (1+ s + \cdots +s^{m} ) \, \,  \eta \\ & &  +\  a\, \,  (1+ s + \cdots + s^{m} )^2 \, \,  \eta ^2 - 2 . \end{array} \end{equation}
  </div>
  <span class="equation_label">2.32</span>
</p>
</div>
<p>Then, we have as in Lemma <a href="#L.2.1">1</a>: </p>
<div class="equation" id="2.33">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.33} \overline{f}_{m+1}(s)= \overline{f}_m(s) + \overline{g}_m (s)+ \overline{g} (s) \, \,  s^{m-1} \, \,  \eta , \end{equation}
  </div>
  <span class="equation_label">2.33</span>
</p>
</div>
<p> where, </p>
<div class="displaymath" id="a0000000013">
  \begin{eqnarray*} \begin{array}{lll} \overline{g} _m(s)& =&  {\bigg(} \displaystyle \tfrac {a}{3} \, \,  s^{m+2}+ a \, \,  \eta \, \,  s^{m+3} + 2 \, \,  a \, \,  \eta \, \,  (1+ s + \cdots + s^{m} )\, \,  s^{2} \\ & & \quad + \displaystyle \tfrac {2}{3} \, \, \eta \, \,  s^{m}+ a \, \,  \eta ^{m+1} {\bigg)} \, \,  s^{m-1} \, \,  \eta {\gt}0 \qquad (s{\gt}0) \end{array}\end{eqnarray*}
</div>
<p> and </p>
<div class="equation" id="2.34">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.34} \overline{g} (s) = 2 \, \,  c \, \,  s^2 + c \, \,  s - (c + a \, \,  \eta ). \end{equation}
  </div>
  <span class="equation_label">2.34</span>
</p>
</div>
<p>We also have </p>
<div class="equation" id="2.35">
<p>
  <div class="equation_content">
    \begin{equation}  \label{2.35} \overline{p} (s) = \displaystyle \tfrac {a}{6} \, \,  (2 + 3 \, \,  \delta ) \, \,  s^2 + c \, \,  (1 + \delta ) \, \,  s - \delta \end{equation}
  </div>
  <span class="equation_label">2.35</span>
</p>
</div>
<p> and </p>
<div class="equation" id="2.36">
<p>
  <div class="equation_content">
    \begin{equation}  \label{2.36} \overline{f}_\infty (s) = \displaystyle \tfrac {2 \, \,  c \, \,  \eta }{1 - s} + a {\bigg( } \displaystyle \tfrac {1}{1 - s} {\bigg) }^2 \, \,  \eta ^2 - 2 . \end{equation}
  </div>
  <span class="equation_label">2.36</span>
</p>
</div>
<p>Moreover, we obtain: </p>
<div class="displaymath" id="a0000000014">
  \begin{eqnarray*} \begin{array}{lll} \overline{\delta }_1 & =&  \displaystyle \tfrac {2 \, \,  c}{a}, \\ \overline{\delta }_2 &  = &  \displaystyle \tfrac {2 \, \,  \delta \, \,  (2 + 3 \, \,  \delta )}{c \, \,  (1 + \delta ) + \sqrt{(c \, \,  (1 + \delta ))^2 + \displaystyle \tfrac {2}{3} \, \,  a \, \,  \delta \, \,  (2 + 3 \delta )}}, \\ \overline{\delta }_3 &  = &  \displaystyle \tfrac {2 - \delta }{c + \sqrt{c^2 + 2 \, \,  a} } \quad {\rm {and}} \\ \displaystyle \tfrac {\overline{\delta }}{2} &  = &  \displaystyle \tfrac {2 \, \,  (c + a \, \,  \eta )}{c + \sqrt{c^2 + 8 \, \,  c \, \,  (c + a \, \,  \eta )} } . \end{array}\end{eqnarray*}
</div>
<p>Set </p>
<div class="equation" id="2.37">
<p>
  <div class="equation_content">
    \begin{equation}  \label{2.37} \overline{\eta }_0 = \min \{  \overline{\delta }_1, \overline{\delta }_2, \overline{\delta }_3 \} . \end{equation}
  </div>
  <span class="equation_label">2.37</span>
</p>
</div>
<p>Then, with the above changes and simply following the proof of Lemma <a href="#L.2.1">1</a>, we can provide another majorizing sequence result for Newton’s method (<a href="#1.2">1.2</a>): <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> Let \(a {\gt} 0\), \(b {\gt} 0\), \(c {\gt} 0\) and \(\eta {\gt} 0\) be given constants. </p>
<p>Assume: </p>
<div class="equation" id="2.38">
<p>
  <div class="equation_content">
    \begin{equation}  \label{2.38} \eta \leq \overline{\eta }_0 . \end{equation}
  </div>
  <span class="equation_label">2.38</span>
</p>
</div>
<p> Inequality <span class="rm">(<a href="#2.38">2.38</a>)</span> is strict if \(\overline{\eta }_0 = \overline{\delta }_1\). </p>
<p>Then, scalar sequence \(\{  v_n \} \) (\(n \geq 0\)), given by </p>
<div class="equation" id="2.39">
<p>
  <div class="equation_content">
    \begin{equation}  \label{2.39} \begin{array}{l} v_0= 0 ,\qquad v_1= \eta , \\ v_{n+2} = v_{n+1} + \displaystyle \tfrac {a \, \,  {\bigg(} \displaystyle \tfrac {1}{3} \, \,  (v_{n+1} - v_n) + v_n {\bigg)} +c } {2 \, \,  (1 - c \, \,  v_{n+1} - \displaystyle \tfrac {a}{2}\, \,  v_{n+1} ^2)} \, \,  (v_{n+1} -v_{n}) ^2 \end{array} \end{equation}
  </div>
  <span class="equation_label">2.39</span>
</p>
</div>
<p> is increasing, bounded from above by \( t^{\star \star } = \displaystyle \tfrac {2 \, \,  \eta }{2 - \bar{\delta }}\) and converges to its unique least upper bound \(v^\star \) satisfying \(v^{\star } \in [0, t^{\star \star } ]\). </p>
<p>Moreover the following estimates hold for all \(n\geq 0\): </p>
<div class="equation" id="2.40">
<p>
  <div class="equation_content">
    \begin{equation}  \label{2.40} 0 < v_{n+2} - v_{n+1} \leq \displaystyle \tfrac { \overline{\delta } }{2} \, \,  (v_{n+1} - v_{n}) \leq {\bigg(} \displaystyle \tfrac { \overline{\delta } }{2} {\bigg)} ^{n+1} \, \,  \eta \end{equation}
  </div>
  <span class="equation_label">2.40</span>
</p>
</div>
<p> and </p>
<div class="equation" id="2.41">
<p>
  <div class="equation_content">
    \begin{equation}  \label{2.41} v^{\star } - v_n \leq \displaystyle \tfrac {2 \, \,  \eta }{2 - \overline{\delta }} \, \,  {\bigg(} \displaystyle \tfrac {\overline{\delta }}{2} {\bigg)} \, \,  \eta . \end{equation}
  </div>
  <span class="equation_label">2.41</span>
</p>
</div>

  </div>
</div> </p>
<p>Below is the main semilocal convergence theorem’s method involving twice Fréchet differentiable operator and using center-Lipschitz conditions. <div class="theorem_thmwrapper " id="T.2.3">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">3</span>
  </div>
  <div class="theorem_thmcontent">
  <p> Let \(F \,  : \,  \mathcal{D}\subseteq \mathcal{X}\longrightarrow \mathcal{Y}\) be a twice Fréchet differentiable operator. </p>
<p>Assume there exist a point \(x_0\in \mathcal{D}\); constants \(\eta {\gt} 0\), \(a{\gt}0\), \(b{\gt}0\), \(c{\gt}0\), such that for all \(x \in \mathcal{D}\): </p>
<div class="displaymath" id="2.42">
  \begin{eqnarray} & &  F’ (x_0)^{-1} \in \mathcal{L}(\mathcal{Y},\mathcal{X}) , \label{2.42} \\ & & \parallel F’ (x_0)^{-1}\, \,  F(x_0) \parallel \leq \eta , \label{2.43}\\ & &  \parallel F’(x_0)^{-1} [F’(x)- F’(x_0) ] \parallel \leq b \, \,  \parallel x- x_0 \parallel , \label{2.44} \\ & & \parallel F’ (x_0)^{-1}\, \,  F”(x_0) \parallel \leq c , \label{2.45}\\ & &  \parallel F’(x_0)^{-1} [F”(x)- F”(x_0) ] \parallel \leq a \, \,  \parallel x- x_0 \parallel , \label{2.46}\\ & & \overline{U} (x_0 ,t^\star ) = \{  x \in \mathcal{X}, \, \,  \parallel x- x_0 \parallel \leq t^\star \} \subseteq \mathcal{D}\label{2.47} \end{eqnarray}
</div>
<p> and hypotheses of Lemma <span class="rm"><a href="#L.2.1">1</a></span> hold. </p>
<p>Then the sequence \(\{  x_n\} \) defined by Newton’s method <span class="rm">(<a href="#1.2">1.2</a>)</span> is well defined, remains in \( \overline{U} (x_0 ,t^\star ) \) for all \(n\geq 0\) and converges to a unique solution \(x^\star \in \overline{U} (x_0 ,t^\star ) \) of equation \(F(x) =0\). </p>
<p>Moreover the following estimates hold for all \(n \geq 0\): </p>
<div class="displaymath" id="a0000000015">
  \begin{eqnarray}  \parallel x_{n +2}\! -\!  x_{n +1} \parallel &  \leq &  \tfrac { a \, \,  {\bigg(} \displaystyle \tfrac {1}{6} \, \,  \parallel x_{n+1} \! -\! x_n \parallel \! +\!  \displaystyle \tfrac {1}{2} \, \,  \parallel x_{n} \! -\! x_0 \parallel {\bigg)} \! +\!  \displaystyle \tfrac {c}{2} } {1 - b \, \,  \parallel x_{n+1} - x_0 \parallel }\, \,  \parallel x_{n+1} -x_n \parallel ^2 \nonumber \\ & \leq &  \tfrac { a \, \,  {\bigg(} \displaystyle \tfrac {1}{6} \, \,  ( t_{n+1} -t_n ) + \displaystyle \tfrac {1}{2} \, \,  ( t_{n} -t_0 ) {\bigg)} + \displaystyle \tfrac {c}{2} } {1 - b \, \, ( t_{n+1} - t_0 ) }\, \,  ( t_{n+1} -t_n ) ^2 \label{2.48} \\ &  = &  t_{n+2} - t_{n+1} \label{2.49} \end{eqnarray}
</div>
<p> and </p>
<div class="equation" id="2.50">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.50} \parallel x_{n }- x^\star \parallel \leq t^\star - t_n , \end{equation}
  </div>
  <span class="equation_label">2.50</span>
</p>
</div>
<p> where, sequence \(\{ t_n \} \) (\(n\geq 0\)) is given by <span class="rm">(<a href="#2.12">2.12</a>)</span>. </p>
<p>Furthermore, if there exists \(R \geq t^\star \), such that </p>
<div class="equation" id="2.51">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.51} U(x_0, R) \subseteq \mathcal{D}\end{equation}
  </div>
  <span class="equation_label">2.51</span>
</p>
</div>
<p> and </p>
<div class="equation" id="2.52">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.52} b \, \,  (t^\star + R) \leq 2 . \end{equation}
  </div>
  <span class="equation_label">2.52</span>
</p>
</div>
<p>The solution \(x^\star \) is unique in \(U(x_0, R) \). </p>

  </div>
</div> <div class="proof_wrapper" id="a0000000016">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>Let us prove that: </p>
<div class="equation" id="2.53">
<p>
  <div class="equation_content">
    \begin{equation}  \label{2.53} \parallel x_{k+1} - x _k \parallel \leq t_{k+1} - t_{k} \end{equation}
  </div>
  <span class="equation_label">2.53</span>
</p>
</div>
<p> and </p>
<div class="equation" id="2.54">
<p>
  <div class="equation_content">
    \begin{equation}  \label{2.54} \overline{U} (x_{k+1}, t^\star - t_{k+1} ) \subseteq \overline{U} (x_{k},t^\star - t_{k} ) \end{equation}
  </div>
  <span class="equation_label">2.54</span>
</p>
</div>
<p> hold for all \(k \geq 0\). </p>
<p>For every \(z \in \overline{U} (x_{1}, t^\star - t_{1} ) \), </p>
<div class="displaymath" id="a0000000017">
  \[  \begin{array}{lll} \parallel z- x_0\parallel & \leq &  \parallel z- x_1 \parallel + \parallel x_1 - x_0\parallel \\ & \leq &  (t^\star - t_1) + (t_1 - t_0) = t^\star - t_0, \end{array}  \]
</div>
<p> implies \(z \in \overline{U} (x_{0}, t^\star - t_0 ) \). Since, also </p>
<div class="displaymath" id="a0000000018">
  \[ \parallel x_1 - x_0\parallel = \parallel F'(x_0)^{-1} \,  F(x_0) \parallel \leq \eta = t_1 -t_0,  \]
</div>
<p> estimates (<a href="#2.53">2.53</a>) and (<a href="#2.54">2.54</a>) hold for \(k=0\). </p>
<p>Given they hold for \(n= 0,1,\cdots , k\), then we have: </p>
<div class="equation" id="2.55">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.55} \parallel x_{k+1} - x_0\parallel \leq \displaystyle \sum _{i=1}^{k+1} \parallel x_i - x_{i-1} \parallel \leq \displaystyle \sum _{i=1}^{k+1} (t_{i} - t_{i-1}) = t_{k+1} -t_0 = t_{k+1} \end{equation}
  </div>
  <span class="equation_label">2.55</span>
</p>
</div>
<p> and </p>
<div class="equation" id="2.56">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.56} \begin{array}{lll} \parallel x_k + \theta \,  (x_{k+1} - x_k ) - x_0\parallel & \leq &  t_k + \theta \,  (t_{k+1} - t_{k}) \\ & \leq &  t^\star , \\ \end{array} \end{equation}
  </div>
  <span class="equation_label">2.56</span>
</p>
</div>
<p> for all \(\theta \in [0,1]\). </p>
<p>Using (<a href="#1.2">1.2</a>), we obtain the approximation </p>
<div class="equation" id="2.57">
<p>
  <div class="equation_content">
    \begin{equation}  \label{2.57} \begin{array}{lll} F(x_{k+1} ) & =&  F(x_{k+1}) - F(x_{k}) -F’(x_{k}) \,  ( x_{k+1} -x_k ) \\ & =&  \displaystyle \int _0 ^1 [F’(x_k + \theta \, (x_{k+1} -x_k )) -F’(x_k)]\,  \, \,  ( x_{k+1} -x_k ) \, \,  {\rm d} \theta \\ & =&  \displaystyle \int _0 ^1 F”(x_k + \theta \, (x_{k+1} -x_k )) \, \,  (1- \theta ) \, \,  ( x_{k+1} -x_k ) ^2 \, \,  {\rm d} \theta . \\ \end{array} \end{equation}
  </div>
  <span class="equation_label">2.57</span>
</p>
</div>
<p>Then, we get by (<a href="#2.45">2.45</a>), (<a href="#2.46">2.46</a>) and (<a href="#2.47">2.47</a>): </p>
<div class="equation" id="2.58">
<p>
  <div class="equation_content">
    \begin{equation}  \label{2.58} \begin{array}{l} \parallel F’(x_0)^{-1} \,  F(x_{k+1}) \parallel \\ \leq \displaystyle \int _0 ^1 {\bigg(} \parallel F’(x_0)^{-1} \,  \,  [F”(x_k + \theta \, (x_{k+1} -x_k )) -F”(x_0)]\parallel \\ \quad + \parallel F’(x_0)^{-1} \, \,  F”(x_0) \parallel {\bigg)} \, \,  \parallel x_{k+1} - x_k \parallel ^2 \, \,  ( 1- \theta ) \, \,  {\rm d} \theta \\ \leq {\bigg\{ } a\, \, {\bigg(} \displaystyle \int _0 ^1 \parallel x_k \! -\!  x_0\parallel + \theta \, \,  \parallel x_{k+1} \! -\!  x_k \parallel \, \,  ( 1- \theta ) \, \,  {\rm d} \theta {\bigg)} + \displaystyle \tfrac {c}{2} {\bigg\} } \, \,  \parallel x_{k+1} - x_k \parallel ^2 \\ \leq \displaystyle \tfrac {a}{6} \, \,  \parallel x_{k+1} - x_k \parallel ^3+ \displaystyle \tfrac {a}{2} \, \, \parallel x_{k} - x_0 \parallel \, \,  \parallel x_{k+1} - x_k \parallel ^2 + \displaystyle \tfrac {c}{2} \, \,  \parallel x_{k+1} - x_k \parallel ^2 \\ \leq {\bigg\{ } a\, \,  {\bigg(} \displaystyle \tfrac {1}{6} \, \,  ( t_{k+1} - t_k )+ \displaystyle \tfrac {1}{2} \, \,  ( t_{k} - t_0 ) {\bigg)} + \displaystyle \tfrac {c}{2} {\bigg\}  } \, \,  ( t_{k+1} - t_k )^2. \end{array} \end{equation}
  </div>
  <span class="equation_label">2.58</span>
</p>
</div>
<p>Using (<a href="#2.44">2.44</a>), we obtain: </p>
<div class="equation" id="2.59">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.59} \begin{array}{lll} \parallel F’(x_0) ^{-1} \,  \,  (F’(x_{k+1}) -F’(x_0))\parallel & \leq &  b \,  \,  \parallel x _{k+1}-x_0 \parallel \\ & \leq &  b \, \,  t_{k+1} \leq b \, \,  t^\star < 1 . \end{array} \end{equation}
  </div>
  <span class="equation_label">2.59</span>
</p>
</div>
<p>It follows from the Banach lemma on invertible operators <span class="cite">
	[
	<a href="#623-4" >4</a>
	]
</span>, <span class="cite">
	[
	<a href="#623-9" >10</a>
	]
</span> and (<a href="#2.59">2.59</a>) that \(F'(x_{k+1}) ^{-1} \) exists and </p>
<div class="equation" id="2.60">
<p>
  <div class="equation_content">
    \begin{equation}  \label{2.60} \begin{array}{lll} \parallel F’(x_{k+1}) ^{-1} \,  \,  F’(x_0) \parallel & \leq &  (1- b \,  \,  \parallel x _{k+1}-x_0 \parallel ) ^{-1} \\ & \leq &  (1- b \,  t_{k+1} )^{-1}. \end{array} \end{equation}
  </div>
  <span class="equation_label">2.60</span>
</p>
</div>
<p>Therefore, by (<a href="#1.2">1.2</a>), (<a href="#2.58">2.58</a>) and (<a href="#2.60">2.60</a>), we obtain in turn: </p>
<div class="equation" id="2.61">
<p>
  <div class="equation_content">
    \begin{equation}  \label{2.61} \begin{array}{lll} \parallel x_{k+2} - x_{k+1} \parallel & =&  \parallel F’(x_{k+1})^{-1} \,  F (x_{k+1}) \parallel \\ & \leq &  \parallel F’(x_{k+1})^{-1} \,  F’ (x_{0}) \parallel \,  \,  \parallel F’(x_{0})^{-1} \,  F(x_{k+1}) \parallel \\ & \leq &  t_{k+2} -t_{k+1}. \end{array} \end{equation}
  </div>
  <span class="equation_label">2.61</span>
</p>
</div>
<p>Thus for every \(z \in \overline{U} (x_{k+2}, t^\star - t_{k+2} )\), we have: </p>
<div class="equation" id="2.62">
<p>
  <div class="equation_content">
    \begin{equation}  \label{2.62} \begin{array}{lll} \parallel z - x _{k+1} \parallel & \leq &  \parallel z - x_{k+2} \parallel + \parallel x_{k+2} - x_{k+1} \parallel \\ & \leq &  t^\star - t_{k+2} + t_{k+2} - t_{k+1} =t^\star - t_{k+1 } . \end{array} \end{equation}
  </div>
  <span class="equation_label">2.62</span>
</p>
</div>
<p>That is, </p>
<div class="equation" id="2.63">
<p>
  <div class="equation_content">
    \begin{equation}  \label{2.63} z \in \overline{U} (x_{k+1}, t^\star -t_{k+1} ). \end{equation}
  </div>
  <span class="equation_label">2.63</span>
</p>
</div>
<p>Estimates (<a href="#2.60">2.60</a>) and (<a href="#2.63">2.63</a>) imply that (<a href="#2.53">2.53</a>) and (<a href="#2.54">2.54</a>) hold for \(n=k+1\). The proof of (<a href="#2.53">2.53</a>) and (<a href="#2.54">2.54</a>) is now complete by induction. </p>
<p>Lemma <a href="#L.2.1">1</a> implies that sequence \(\{ t _n\} \) is a Cauchy sequence. From (<a href="#2.53">2.53</a>) and (<a href="#2.54">2.54</a>) \(\{ x_n\} \) (\(n \geq 0 \)) become a Cauchy sequence too 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="#2.50">2.50</a>) follows from (<a href="#2.49">2.49</a>) by using standard majorization techniques <span class="cite">
	[
	<a href="#623-4" >4</a>
	]
</span>, <span class="cite">
	[
	<a href="#623-9" >10</a>
	]
</span>. </p>
<p>Moreover, by letting \(k \to \infty \) in (<a href="#2.58">2.58</a>), we obtain \(f(x^\star ) = 0\). Finally, to show uniqueness: let \(y ^\star \) be a solution of equation \(F(x) =0\) in \(U(x_0, R)\). It follows from (<a href="#2.44">2.44</a>) for \(x=y^\star + \theta \,  (x^\star - y^\star )\), \(\theta \in [0,1]\), the estimate: </p>
<div class="displaymath" id="a0000000019">
  \begin{align*} & \parallel \displaystyle F’(x_0) ^{-1} \,  \displaystyle \int _0 ^1 (F’(y^\star + \theta \, (x^\star - y^\star ) )- F’(x_0)) \,  \parallel \, \,  {\rm d} \theta \\ &  \leq b \, \,  \displaystyle \int _0 ^1 \parallel y^\star + \theta \, (x^\star - y^\star ) - x_0 \parallel \,  {\rm d} \theta \\ & \leq b \,  \,  \displaystyle \int _0 ^1 (\theta \parallel x^\star - x_0\parallel + (1- \theta )\,  \parallel y^\star - x_0 \parallel ) \,  {\rm d} \theta \\ & {\lt} \displaystyle \tfrac {b}{2} \,  ( t^\star + R) \leq 1, \quad ({\rm by} \, \,  (\ref{2.52})) \end{align*}
</div>
<p> and the Banach lemma on invertible operators implies that the linear operator \(\mathcal{M} = \int _0 ^1 F'(y ^\star + \theta \, \,  (x^\star - y ^\star ))\,  {\rm d}\theta \) is invertible. Using the identity \( 0 = F(x^\star ) - F(y^\star ) = \mathcal{M} \,  (x^\star - y^\star ), \) we deduce \(x^\star = y^\star \). </p>
<p>Similarly, we show the uniqueness in \(\overline{U} (x_0, t^\star ) \) using (<a href="#2.52">2.52</a>). </p>
<p>That completes the proof of Theorem <a href="#T.2.3">3</a>. </p>
<p><div class="remark_thmwrapper " id="R.2.4">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">4</span>
  </div>
  <div class="remark_thmcontent">
  <p> The conclusions of Theorem <a href="#T.2.3">3</a> hold if (<a href="#2.44">2.44</a>) is dropped from the hypotheses and Lemma <a href="#L.2.1">1</a>, \(\{ t_n \} \), \(t^\star \) are replaced by Lemma <a href="#L.2.2">2</a>, \(\{ v_n \} \), \(v^{\star }\), respectively. Indeed, we have: <span class="qed">â–¡</span></p>

  </div>
</div> <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 twice Fréchet differentiable operator. </p>
<p>Assume hypotheses of Lemma <span class="rm"><a href="#L.2.2">2</a></span> hold and there exist a point \(x_0\in \mathcal{D}\), a constants \(\eta {\gt} 0\), \(a{\gt}0\) and \(c{\gt}0\), such that for all \(x \in \mathcal{D}\): </p>
<div class="displaymath" id="a0000000020">
  \begin{eqnarray} & &  F’ (x_0)^{-1} \in \mathcal{L}(\mathcal{Y},\mathcal{X}) , \nonumber \\ & & \parallel F’ (x_0)^{-1}\, \,  F(x_0) \parallel \leq \eta , \nonumber \\ & & \parallel F’ (x_0)^{-1}\, \,  F”(x_0) \parallel \leq c , \nonumber \\ & &  \parallel F’(x_0)^{-1} [F”(x)- F”(x_0) ] \parallel \leq a \, \,  \parallel x- x_0 \parallel , \nonumber \\ & & \overline{U} (x_0 , v^{\star } ) \subseteq \mathcal{D}, \nonumber \end{eqnarray}
</div>
<p> where, \(v^{\star }\) is given in Lemma <span class="rm"><a href="#L.2.2">2</a></span>. </p>
<p>Then sequence \(\{  x_n\} \) defined by Newton’s method <span class="rm">(<a href="#1.2">1.2</a>)</span> is well defined, remains in \( \overline{U} (x_0 , v^{\star } ) \) for all \(n\geq 0\) and converges to a unique solution \(x^\star \in \overline{U} (x_0 , v^{\star } ) \) of equation \(F(x) =0\). </p>
<p>Moreover the following estimates hold for all \(n \geq 0\): </p>
<div class="equation" id="2.64">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.64} \parallel x_{n +2}- x_{n +1} \parallel \leq v_{n+1} - v_{n} \end{equation}
  </div>
  <span class="equation_label">2.64</span>
</p>
</div>
<p> and </p>
<div class="equation" id="2.65">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.65} \parallel x_{n }- x^\star \parallel \leq v_{n+2} - v_{n+1} , \end{equation}
  </div>
  <span class="equation_label">2.65</span>
</p>
</div>
<p> where, sequence \(\{ v_n \} \) (\(n\geq 0\)) is given by <span class="rm">(<a href="#2.39">2.39</a>)</span>. </p>

  </div>
</div> <div class="proof_wrapper" id="a0000000021">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>We proceed as in the proof of Theorem <a href="#T.2.3">3</a> until (<a href="#2.58">2.58</a>). Then use (<a href="#2.45">2.45</a>) and (<a href="#2.46">2.46</a>) (instead of (<a href="#2.44">2.44</a>)) to obtain in turn: </p>
<div class="displaymath" id="2.66">
  \begin{align} \label{2.66} & \parallel F’(x_0) ^{-1} \,  \,  (F’(x_{k+1}) -F’(x_0))\parallel \\ &  =\parallel \displaystyle \int _0 ^1 F’(x_0) ^{-1} \, \,  (F”(x_0 + t \, (x_{k+1} - x_0 ) )- F”(x_0)) \,  \, \,  {\rm d} t \, \,  (x_{k+1} - x_0)\nonumber \\ & \quad +F’(x_0) ^{-1} \,  \,  F”(x_{0}) \, \,  (x_{k+1} - x_0)\parallel \nonumber \\ & \leq \displaystyle \int _0 ^1 a \, \,  t \, \,  {\rm d}t \, \,  \parallel x_{k+1} - x_0 \parallel ^2 + c \, \,  \parallel x_{k+1} - x_0 \parallel \nonumber \\ & \leq \displaystyle \tfrac {a}{2} \, \,  v_{k+1} ^2 + c \, \,  v_{k+1} {\lt} 1.\nonumber \end{align}
</div>
<p>It follows from (<a href="#2.66">2.66</a>) and the Banach lemma on invertible operators, that \(F'(x_{k+1})^{-1}\) exists and </p>
<div class="equation" id="2.67">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.67} \parallel F'(x_{k+1}) ^{-1} \,  \,  F'(x_0) \parallel \leq (1- c \, \,  v_{k+1} - \displaystyle \tfrac {a}{2} \, \,  v_{k+1} ^2) ^{-1} . \end{equation}
  </div>
  <span class="equation_label">2.67</span>
</p>
</div>
<p>The rest of the proof follows as in the proof of Theorem <a href="#T.2.3">3</a>, with (<a href="#2.67">2.67</a>) replacing (<a href="#2.60">2.60</a>) until the uniqueness part. </p>
<p>Let \(y^\star \) be a solution of equation \(F(x)=0\) in \(\overline{U} (x_0, t^\star )\). Then, since </p>
<div class="equation" id="2.68">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.68} \begin{array}{lll} y^\star - x_{k+1} & =&  y^\star - x_k + F’(x_k)^{-1} \, \,  F(x_k) \\ & =&  - F’(x_k) ^{-1} \,  \,  ( F(y^\star ) -F(x_{k}) -F’(x_k) \, \,  (y^\star - x_k)) \\ & =&  (- F’(x_0) ^{-1} \,  \,  F’(x_k))^{-1} \\ & &  \displaystyle \int _0 ^1 F’(x_0) ^{-1} \, \,  F”(x_k + t \, \,  (y^\star - x_k) ) \, \,  (1-t)\, \,  {\rm d}t \, \,  (y^\star - x_k) ^2 \end{array} \end{equation}
  </div>
  <span class="equation_label">2.68</span>
</p>
</div>
<p> and </p>
<div class="displaymath" id="a0000000022">
  \[  \parallel y^\star - x_0 \parallel \leq v^{\star } - v_0 ,  \]
</div>
<p> we obtain: </p>
<div class="equation" id="2.69">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.69} \parallel x_k - y^\star \parallel \leq v^{\star } - v_k , \end{equation}
  </div>
  <span class="equation_label">2.69</span>
</p>
</div>
<p> which leads to \(\displaystyle \lim _{k \longrightarrow \infty } x_k = y^\star \). But, we showed \(\displaystyle \lim _{k \longrightarrow \infty } x_k = x^\star \). Hence, we deduce \(x^\star = y^\star \). </p>
<p>That completes the proof of Theorem <a href="#T.2.4">5</a>. </p>
<p>We can now compare majorizing sequences \(\{ t_n\} \) and \(\{ v_n \} \) (\(n \geq 0\)): <div class="proposition_thmwrapper " id="P.2.6">
  <div class="proposition_thmheading">
    <span class="proposition_thmcaption">
    Proposition
    </span>
    <span class="proposition_thmlabel">6</span>
  </div>
  <div class="proposition_thmcontent">
  <p> Assume: </p>
<div class="equation" id="2.70">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.70} b< \displaystyle \tfrac {a}{2 }\, \,  \eta +c , \end{equation}
  </div>
  <span class="equation_label">2.70</span>
</p>
</div>
<p> hypotheses of Theorems <span class="rm"><a href="#T.2.3">3</a></span> and <span class="rm"><a href="#T.2.4">5</a></span> hold. </p>
<p>Then, the following estimates hold for all \(n \geq 0\): </p>
<div class="equation" id="2.71">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.71} \parallel x_{n+2} - x_{n+1} \parallel \leq t_{n+2} - t_{n+1} < v_{n+2} - v_{n+1} \end{equation}
  </div>
  <span class="equation_label">2.71</span>
</p>
</div>
<p> and </p>
<div class="equation" id="2.72">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.72} \parallel x_n - x^\star \parallel \leq t^\star - t_n \leq v^{\star } - v_n . \end{equation}
  </div>
  <span class="equation_label">2.72</span>
</p>
</div>

  </div>
</div> <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>We only need to show using induction on the integer \(k\): </p>
<div class="equation" id="2.73">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.73} t_{k+2} - t_{k+1} < v_{k+2} - v_{k+1} . \end{equation}
  </div>
  <span class="equation_label">2.73</span>
</p>
</div>
<p>In view of (<a href="#2.23">2.23</a>), (<a href="#2.52">2.52</a>) and (<a href="#2.70">2.70</a>), we obtain for \(n=0\): </p>
<div class="displaymath" id="a0000000024">
  \[  t_2 {\lt} v_2 \quad {\rm and} \quad t_2 - t_1 {\lt} v_2 - v_1 .  \]
</div>
<p>Assuming: </p>
<div class="equation" id="2.74">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.74} t_{k+1} < v_{k+1} \end{equation}
  </div>
  <span class="equation_label">2.74</span>
</p>
</div>
<p> and </p>
<div class="equation" id="2.75">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.75} t_{k+1} - t_{k} < v_{k+1} - v_{k} , \end{equation}
  </div>
  <span class="equation_label">2.75</span>
</p>
</div>
<p> for \(k \leq n+1\), we obtain from (<a href="#2.17">2.17</a>), (<a href="#2.39">2.39</a>), (<a href="#2.74">2.74</a>) and (<a href="#2.75">2.75</a>), that (<a href="#2.73">2.73</a>) holds for all \(k \geq 0\). </p>
<p>Estimate (<a href="#2.72">2.72</a>) follows from (<a href="#2.71">2.71</a>) by using standard majorization techniques. </p>
<p>That completes the proof of Proposition <a href="#P.2.6">6</a>. </p>
<p><div class="remark_thmwrapper " id="R.2.7">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">7</span>
  </div>
  <div class="remark_thmcontent">
  <p> If equality holds in (<a href="#2.70">2.70</a>), then \(v_n = t_n\) (\(n \geq 0\)), whereas if </p>
<div class="equation" id="2.76">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.76} b > \displaystyle \tfrac {a}{2 }\, \,  \eta +c , \end{equation}
  </div>
  <span class="equation_label">2.76</span>
</p>
</div>
<p> then, the conclusions of Proposition <a href="#P.2.6">6</a> hold with sequence \(\{ t _n\} \), \(t^\star \) switching places with \(\{ v _n\} \), \(v^{\star }\), respectively in (<a href="#2.71">2.71</a>) and (<a href="#2.72">2.72</a>).<span class="qed">â–¡</span></p>

  </div>
</div> <div class="remark_thmwrapper " id="R.2.8">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">8</span>
  </div>
  <div class="remark_thmcontent">
  <p> We can now compare our results with the ones obtained by Huang <span class="cite">
	[
	<a href="#623-8" >9</a>
	]
</span> and Gutiérrez <span class="cite">
	[
	<a href="#639-10" >8</a>
	]
</span>. </p>
<p>Huang <span class="cite">
	[
	<a href="#623-8" >9</a>
	]
</span> used (<a href="#2.42">2.42</a>), (<a href="#2.45">2.45</a>), </p>
<div class="equation" id="2.77">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.77} \parallel F'(x_0)^{-1} \, \,  (F''(x) - F''(y))\parallel \leq \alpha \, \,  \parallel x- y \parallel \end{equation}
  </div>
  <span class="equation_label">2.77</span>
</p>
</div>
<p> for all \(x,y \in \mathcal{D}\) and </p>
<div class="equation" id="2.78">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.78} 3 \, \,  \alpha ^2 \, \,  \eta + 3 \, \,  \alpha \, \,  c + c ^3 \leq (c^2 + 2 \, \,  \alpha )^{3/2} \end{equation}
  </div>
  <span class="equation_label">2.78</span>
</p>
</div>
<p> and majorizing sequence \(\{ v_n \} \) to arrive at conclusions (<a href="#2.64">2.64</a>) and (<a href="#2.65">2.65</a>). </p>
<p>Gutiérrez <span class="cite">
	[
	<a href="#639-10" >8</a>
	]
</span> weakened Huang’s conditions by using (<a href="#2.42">2.42</a>), (<a href="#2.45">2.45</a>), (<a href="#2.46">2.46</a>) (which is weaker than (<a href="#2.77">2.77</a>), since \(a \leq \alpha \)) and the condition: </p>
<div class="equation" id="2.79">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.79} 3 \, \,  a ^2 \, \,  \eta + 3 \, \,  a \, \,  c + c ^3 \leq (c^2 + 2 \, \,  a )^{3/2} \end{equation}
  </div>
  <span class="equation_label">2.79</span>
</p>
</div>
<p> and majorizing sequence \(\{ v_n \} \) to also arrive at conclusions (<a href="#2.64">2.64</a>) and (<a href="#2.65">2.65</a>). </p>
<p>Hypotheses of Lemma <a href="#L.2.1">1</a> use information on \(a\), \(b\), \(c\) and \(\eta \), Huang <span class="cite">
	[
	<a href="#623-8" >9</a>
	]
</span> uses information on \(\alpha \), \(c\) and \(\eta \), whereas, Gutiérrez <span class="cite">
	[
	<a href="#639-10" >8</a>
	]
</span> uses \(a\), \(c\) and \(\eta \). </p>
<p>Therefore, a direct comparison between the sufficient convergence conditions is not possible. However, under (<a href="#2.70">2.70</a>), our majorizing sequence \(\{ t_n\} \) is finer that \(\{ v_n\} \) (See Example 2.9). </p>
<p>Note that in this study, we have simplified the sufficient convergence conditions provided by us in <span class="cite">
	[
	<a href="#633-1" >1</a>
	]
</span>. A favorable comparison of our approach with the corresponding one given by the Newton-Kantorovich theorem for solving nonlinear equations was also given in <span class="cite">
	[
	<a href="#633-1" >1</a>
	]
</span>. The same favorable comparison extends in this study. </p>
<p>The results obtained here can be extended for \(m\) (\(m \geq 2\)) Fréchet differentiable operators <span class="cite">
	[
	<a href="#633-1" >1</a>
	]
</span>, <span class="cite">
	[
	<a href="#633-2" >2</a>
	]
</span>.<span class="qed">â–¡</span></p>

  </div>
</div> </p>
<div class="centered"> Comparison Table 1<br /><table class="tabular">
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \(n\) </p>

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

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

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p>0 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> 0.21 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> 0.21 </p>

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

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> 0.0775551724 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> 0.0807721314 </p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p>2 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> 0.0167871035 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> 0.0213332937 </p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p>3 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> 0.0006952583 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> 0.0018344379 </p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p>4 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> 0.0000025816 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> 0.0000139339 </p>

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

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> 0 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> 0.0000000008 </p>

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

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> 0 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> 0 </p>

    </td>
  </tr>
</table> </div>
<p>Comparison table 1 justifies the theoretical results of Proposition <a href="#P.2.6">6</a>, since the majorizing sequence \(\{  t_n \} \) is tighter than \(\{ v_n \} \). </p>
<p><div class="example_thmwrapper " id="E.39">
  <div class="example_thmheading">
    <span class="example_thmcaption">
    Example
    </span>
    <span class="example_thmlabel">9</span>
  </div>
  <div class="example_thmcontent">
  <p> Let \(a = b = 2\), \(c = 1.9\) and \(\eta = .21\). Then, condition (<a href="#2.79">2.79</a>) (and (<a href="#2.78">2.78</a>)) is satisfied, since </p>
<div class="displaymath" id="a0000000025">
  \[  3 \, \,  a^2 \, \,  \eta + 3 \, \,  a \, \,  c + c^3 = 20.779 {\lt} 20.99311985 = (c^2 + 2 \, \,  a)^{3/2}  \]
</div>
<p> However, using (<a href="#2.1">2.1</a>)–(<a href="#2.10">2.10</a>), we get \(\delta _1 = 2\), \(\delta _2 = .259636075\), \(\delta _3 = .220086356\), \(\delta = 1.119654576\) and \(\eta _0 = \delta _3 {\gt} \eta = .21\). Note also that \((2.70)\) holds, since </p>
<div class="equation" id="a0000000026">
<p>
  <div class="equation_content">
    \begin{equation}  b = 2 < 2.11 = \tfrac {a}{2} \eta + c. \nonumber \end{equation}
  </div>
  <span class="equation_label">2.80</span>
</p>
</div>
<p> We can compare the error estimates using (<a href="#2.12">2.12</a>) and (<a href="#2.39">2.39</a>).<span class="qed">â–¡</span></p>

  </div>
</div> </p>
<p><small class="footnotesize">  </small></p>
<div class="bibliography">
<h1>Bibliography</h1>
<dl class="bibliography">
  <dt><a name="633-1">1</a></dt>
  <dd><p><i class="sc">I.K. Argyros</i>, <i class="it">A Newton–Kantorovich theorem for equations involving \(m\)–Fréchet–differentiable operators and applications in radiative transfer</i>, J. Comp. Appl. Math., <b class="bf">131</b> (2001) nos. 1–2, pp.&#160;149–159. </p>
</dd>
  <dt><a name="633-2">2</a></dt>
  <dd><p><i class="sc">I.K. Argyros</i>, <i class="it">Concerning the convergence and application of Newton’s method under hypotheses on the first and second Fréchet derivative</i>, Comm. Appl. Nonlinear Anal., <b class="bf">11</b> (2004), pp.&#160;103–119. </p>
</dd>
  <dt><a name="623-3">3</a></dt>
  <dd><p><i class="sc">I.K. Argyros</i>, <i class="it">A unifying local–semilocal convergence analysis and applications for two–point Newton–like methods in Banach space</i>, J. Math. Anal. Appl., <b class="bf">298</b> (2004), pp.&#160;374–397. </p>
</dd>
  <dt><a name="623-4">4</a></dt>
  <dd><p><i class="sc">I.K. Argyros</i>, <i class="it">Convergence and applications of Newton–type iterations</i>, Springer–Verlag Publ., New York, 2008. </p>
</dd>
  <dt><a name="arg-hil-polimetrica2">5</a></dt>
  <dd><p><i class="sc">I.K. Argyros</i> and <i class="sc">S. Hilout</i>, <i class="it">Aspects of the computational theory for certain iterative methods</i>, Polimetrica Publisher, 2009. </p>
</dd>
  <dt><a name="arg-hil-complexity">6</a></dt>
  <dd><p><i class="sc">I.K. Argyros</i> and <i class="sc">S. Hilout</i>, <i class="it">Enclosing roots of polynomial equations and their applications to iterative processes</i>, Surveys Math. Appl., <b class="bf">4</b> (2009), pp.&#160;119–132. </p>
</dd>
  <dt><a name="arg-hil-IJCM">7</a></dt>
  <dd><p><i class="sc">I.K. Argyros</i> and <i class="sc">S. Hilout</i>, <i class="it">On the semi-local convergence of inexact Newton methods in Banach spaces</i>, J. Comput. Appl. Math., <b class="bf">228</b> (2009) no. 1, pp.&#160;434–443. </p>
</dd>
  <dt><a name="639-10">8</a></dt>
  <dd><p><i class="sc">J.M. Gutiérrez</i>, <i class="it">A new semilocal convergence theorem for Newton’s method</i>, J. Comp. Appl. Math., <b class="bf">79</b> (1997), pp.&#160;131–145. </p>
</dd>
  <dt><a name="623-8">9</a></dt>
  <dd><p><i class="sc">Z. Huang</i>, <i class="it">A note of Kantorovich theorem for Newton iteration</i>, J. Comput. Appl. Math., <b class="bf">47</b> (1993), pp.&#160;211–217. </p>
</dd>
  <dt><a name="623-9">10</a></dt>
  <dd><p><i class="sc">L.V. Kantorovich</i> and <i class="sc">G.P. Akilov</i>, <i class="it">Functional Analysis</i>, Pergamon Press, Oxford, 1982. </p>
</dd>
  <dt><a name="633-8">11</a></dt>
  <dd><p><i class="sc">F.A. Potra</i>, <i class="it">Sharp error bounds for a class of Newton–like methods</i>, Libertas Mathematica, <b class="bf">5</b> (1985), pp.&#160;71–84. </p>
</dd>
</dl>


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