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<title>An improved semilocal convergence analysis for the midpoint method: An improved semilocal convergence analysis for the midpoint method</title>
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<h1>An improved semilocal convergence analysis for the midpoint method</h1>
<p class="authors">
<span class="author">Ioannis K. Argyros\(^\ast \) Sanjay K. Khattri\(^\S \)</span>
</p>
<p class="date">February 16, 2013.</p>
</div>
<p>\(^\ast \)Department of Mathematical Sciences, Cameron University, Lawton, Oklahoma 73505-6377, USA, e-mail: <span class="tt">iargyros@cameron.edu</span>. </p>
<p>\(^\S \)Department of Engineering, Stord Haugesund University College, Norway, e-mail: <span class="tt">sanjay.Khattri@hsh.no</span>. </p>

<div class="abstract"><p> We expand the applicability of the midpoint method for approximating a locally unique solution of nonlinear equations in a Banach space setting. Our majorizing sequences are finer than the known results in scientific literature <span class="cite">
	[
	<a href="#1" >1</a>
	, 
	<a href="#3" >3</a>
	, 
	<a href="#4" >4</a>
	, 
	<a href="#5" >10</a>
	, 
	<a href="#6" >11</a>
	, 
	<a href="#7" >12</a>
	, 
	<a href="#8" >13</a>
	, 
	<a href="#9" >14</a>
	, 
	<a href="#10" >15</a>
	, 
	<a href="#11" >16</a>
	, 
	<a href="#19" >24</a>
	, 
	<a href="#20" >25</a>
	, 
	<a href="#21" >26</a>
	, 
	<a href="#23" >28</a>
	]
</span> and the convergence criteria can be weaker. Finally, numerical work is reported that compares favorably to the existing approaches in the literature [6, 8–16, 24–26,28]. </p>
<p><b class="bf">MSC.</b> 65B05, 65G99, 65J15, 65N30, 65N35, 65H10, 47H17, 49M15. </p>
<p><b class="bf">Keywords.</b> Midpoint method, semilocal convergence, majorization sequence, Banach space, Fréchet-derivative. </p>
</div>
<h1 id="intro">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} \mathcal F(x)=0, \end{equation}
  </div>
  <span class="equation_label">1.1</span>
</p>
</div>
<p> where, \(\mathcal F\) is a twice Fréchet differentiable operator defined on a convex subset \(\mathbf{D}\) of a Banach space \(\mathbf{X}\) with values in a Banach space \({\mathbf{Y}}\). Numerous problems in science and engineering can be reduced to solving the above equation <span class="cite">
	[
	<a href="#18" >23</a>
	, 
	<a href="#27" >32</a>
	]
</span>. Consequently, solving these equations is an important scientific field of research. In many situations, finding a closed form solution for the non-linear equation <a href="#eq:11" class="eqref">1.1</a> is not possible. Therefore, iterative solution techniques are employed for solving these equations. </p>
<p>The study about convergence analysis of iterative methods is usually divided into two categories: semi-local and local convergence analysis. The semilocal convergence analysis is based upon the information around an initial point to give criteria ensuring the convergence of the iterative procedure. While the local convergence analysis is based on the information around a solution to find estimates of the radii of convergence balls. In this paper, we study the semilocal convergence of the midpoint method defined as </p>
<div class="displaymath" id="eq:12">
  \begin{align} \label{eq:12} y_n & = x_n - \mathcal F^\prime (x_n)^{-1}\mathcal F(x_n),\\ x_{n+1} & = x_n - \mathcal F^\prime \Big(\tfrac {x_n+y_n}{2}\Big)^{-1} \mathcal F(x_n), \  \quad \textrm{for each}\quad n = 0,1,2,\ldots , \nonumber \end{align}
</div>
<p> where \(x_0\in \mathbf{D}\) is an initial point. Here, \(\mathcal F^\prime (x)\) denotes the first Fréchet-derivative of the operator \(\mathcal F\) <span class="cite">
	[
	<a href="#18" >23</a>
	, 
	<a href="#27" >32</a>
	]
</span>. It is well-known that the Midpoint method is cubically convergent and it has a long history <span class="cite">
	[
</span>]22,23,24,25,26,27. Let \(U(w,R)\) and \(\overline{U}(w,R)\) stand, respectively, for the open and closed balls in \(\mathbf{X}\) with center \(w\) and radius \(R{\gt}0\). Let the space of bounded linear operators from \(\mathbf{X}\) into \({\mathbf{Y}}\) be denoted by \({L}(\mathbf{X},{\mathbf{Y}})\). The following set of (<b class="bfseries">C</b>) conditions have been used </p>
<ol class="enumerate">
  <li><p>There exists \(x_0\in \mathbf{D}\) such that \(F^\prime (x_0)^{-1}\in {L}({\mathbf{Y}},\mathbf{X})\). </p>
</li>
  <li><p>\(\left\Vert {\mathcal F^\prime (x_0)^{-1}\mathcal F(x_0)}\right\Vert \le \eta \). </p>
</li>
  <li><p>\(\left\Vert {\mathcal F^\prime (x_0)^{-1}\mathcal F^{\prime \prime }(x)}\right\Vert \le {L}\) for all \(x\in \mathbf{D}\). </p>
</li>
  <li><p>\(\left\Vert {\mathcal F^\prime (x_0)^{-1}\left(\mathcal F^{\prime \prime }(x) - \mathcal F^{\prime \prime }(y)\right)}\right\Vert \le \mathcal{M}\left\Vert {x-y}\right\Vert \), for all \(x,y\in \mathbf{D}\). </p>
</li>
</ol>
<p> The following sufficient convergence criteria have been given in connection to the (<b class="bfseries">C</b>) conditions </p>
<div class="displaymath" id="eq:14">
  \begin{align}  \eta & \le \tfrac {4\mathcal{M}+{L}^2 - {L}\,  \sqrt{{L}^2 + 2\mathcal{M}}} {3\mathcal{M}({L}+ \sqrt{{L}^2 + 2\mathcal{M}})}\quad \textrm{\cite{1,3,4,18,19,20,21}}\label{eq:13} \intertext {or} \eta & \le \tfrac {1}{2\, \mathcal{K}}\quad \textrm{\cite{7,9}},\label{eq:14} \intertext {where} \mathcal{K}& = {L}\,  \sqrt{1+\tfrac {7\, \mathcal{M}}{6\, {L}^2}}.\nonumber \end{align}
</div>
<p> However, simple numerical examples can be used to show that criteria <a href="#eq:14" class="eqref">1.3</a> and <a href="#eq:14" class="eqref">1.3</a> are unsatisfied but the midpoint method <a href="#eq:12" class="eqref">1.2</a> still converges to the solution \(x^\star \). As an example, let \(\mathbf{X}={\mathbf{Y}}=\mathbb {R}\), \(x_0=1\) and \(\mathbf{D}= [\zeta ,2-\zeta ]\) for \(\zeta \in (0,1)\). Define function \(\mathcal F\) on \(\mathbf{D}\) by </p>
<div class="equation" id="eq:00">
<p>
  <div class="equation_content">
    \begin{equation} \label{eq:00} \mathcal F(x) = x^5 -\zeta . \end{equation}
  </div>
  <span class="equation_label">1.3</span>
</p>
</div>
<p> Then, using conditions (<b class="bfseries">C</b>), we get </p>
<div class="displaymath" id="a0000000002">
  \begin{equation*}  \eta = \tfrac {(1-\zeta )}{5},\quad {L}= 4(2-\zeta )^3,\quad \mathcal{M}= 12(2-\zeta )^2. \end{equation*}
</div>
<figure id="fig:1">
  <div class="centered"><img src="img-0001.png" alt="\includegraphics[scale=0.75]{fig_2.png}" style="width:718.5px; height:614.25px" />
 <figcaption>
  <span class="caption_title">Figure</span> 
  <span class="caption_ref">1.</span> 
  <span class="caption_text">Convergence criteria <a href="#eq:14" class="eqref">1.3</a> and <a href="#eq:14" class="eqref">1.3</a> for <a href="#eq:00" class="eqref">1.3</a>.</span> 
</figcaption>  </div>

</figure>
<p>Figure <a href="#fig:1">1.</a> plots the criteria <a href="#eq:14" class="eqref">1.3</a> and <a href="#eq:14" class="eqref">1.3</a> for the problem <a href="#eq:00" class="eqref">1.3</a>. The curve (defined by the right hand side of the inequality <a href="#eq:14" class="eqref">1.3</a>) intersect the line \(\eta \) (see Figure <a href="#fig:1">1.</a>) at \(\zeta \approx 0.73\) while the curve (defined by the right hand side of the criteria <a href="#eq:14" class="eqref">1.3</a>) intersect the \(\eta \) line at \(\zeta \approx 0.72\). We notice in the Figure <a href="#fig:1">1.</a> that for \(\zeta {\lt} 0.72\) the criteria <a href="#eq:14" class="eqref">1.3</a> and <a href="#eq:14" class="eqref">1.3</a> are not satisfied. However, one may see that the method <a href="#eq:12" class="eqref">1.2</a> is convergent. For additional examples, see the Section 4. </p>
<p>In our work we expand the applicability of the midpoint method <a href="#eq:12" class="eqref">1.2</a>, in cases where <a href="#eq:14" class="eqref">1.3</a> or <a href="#eq:14" class="eqref">1.3</a> are not satisfied, using the (<b class="bfseries">C</b>) conditions together with the following center Lipschitz condition </p>
<ol class="enumerate">
  <li><p>\(\left\Vert {\mathcal F^\prime (x_0)^{-1}\left(\mathcal F^{\prime }(x) - \mathcal F^\prime (x_0)\right)}\right\Vert \le {L}_0 \left\Vert {x-x_0}\right\Vert \) for all \(x\in \mathbf{D}\). </p>
</li>
</ol>
<p> We shall refer to (<b class="bfseries">C</b>\(_1\))-(<b class="bfseries">C</b>\(_5\)) as the (<b class="bfseries">H</b>) conditions. </p>
<p>As can be inferred from the studies [1–28], several techniques are usually employed for analyzing the convergence of iterative methods. Among these, the most popular technique is based on the concept of majorizing sequences. In the studies that lead to the convergence conditions <a href="#eq:14" class="eqref">1.3</a> and <a href="#eq:14" class="eqref">1.3</a> the computation of the upper bounds on \(\left\Vert {\mathcal F^\prime (x_n)^{-1}\mathcal F^\prime (x_0)}\right\Vert \) was based on (<b class="bfseries">C</b>\(_3\)) and the estimate </p>
<div class="equation" id="eq:15">
<p>
  <div class="equation_content">
    \begin{equation} \label{eq:15} \left\Vert {\mathcal F^\prime (x_n)^{-1}\mathcal F^\prime (x_0)}\right\Vert \le \tfrac {1}{1-{L}\Vert {x_n-x_0}\Vert }. \end{equation}
  </div>
  <span class="equation_label">1.4</span>
</p>
</div>
<p> Instead of (<b class="bfseries">C</b>\(_3\)) we use the more precise and less expensive condition (<b class="bfseries">C</b>\(_5\)) which leads to </p>
<div class="equation" id="eq:16">
<p>
  <div class="equation_content">
    \begin{equation} \label{eq:16} \left\Vert {\mathcal F^\prime (x_n)^{-1}\mathcal F^\prime (x_0)}\right\Vert \le \tfrac {1}{1-{L}_0\Vert {x_n-x_0}\Vert }. \end{equation}
  </div>
  <span class="equation_label">1.5</span>
</p>
</div>
<p> Note that </p>
<div class="equation" id="eq:17">
<p>
  <div class="equation_content">
    \begin{equation} \label{eq:17} {L}_0 \le {L}\end{equation}
  </div>
  <span class="equation_label">1.6</span>
</p>
</div>
<p> holds in general and \({L}/{L}_0\) can be arbitrarily large <span class="cite">
	[
	<a href="#17" >22</a>
	, 
	<a href="#18" >23</a>
	]
</span>. This change in the study of the semilocal convergence of the midpoint method leads to tighter error estimates on the distances \(\Vert {y_n-x_n}\Vert \), \(\Vert {x_{n+1}-y_n}\Vert \), \(\Vert {y_n-x^\star }\Vert \), \(\Vert {x_n-x^\star }\Vert \) and weaker convergence criteria. </p>
<p>The rest of the paper is organized as follows. Section 2 develop results on majorizing sequences for the midpoint method <a href="#eq:12" class="eqref">1.2</a>, where as in the Section 3 we present the semilocal convergence of the Midpoint method. Numerical examples are given in the concluding section 4. </p>
<h1 id="a0000000003">2 Majorizing Sequences</h1>
<p> In this section, we study the convergence of scalar sequences that are majorizing for the Midpoint method <a href="#eq:12" class="eqref">1.2</a>. Let the positive constants be \({L}_0{\gt}0\), \({L}{\gt}0\), \(\mathcal{M}\ge {0}\) and \(\eta {\gt}0\). It is convenient for us to define functions \(\gamma \), \(a\), \(\alpha \), \({h}_i\), \(i = 1,2,3\) by </p>
<div class="displaymath" id="eq:21">
  \begin{gather}  \gamma (t) = \dfrac {{L}\, t }{2\left[1-\frac{{L}_0\, t}{2}\right]}, \quad \gamma = \gamma (\eta ),\label{eq:21} \\ a(t) = \tfrac {1}{24} \left(12 {L}\gamma (t)^2 + 12 {L}\gamma (t) + 7\mathcal{M}\eta \right),\quad a = a(\eta ),\label{eq:22}\\ \alpha (t) = \dfrac {a(t)t}{\left[1-\frac{{L}_0}{2}(1+\gamma (t))t\right]},\quad \alpha = \alpha (\eta ),\label{eq:23}\\ {h}_1(t) = a(t)t +\tfrac {{L}_0}{2}(1+\gamma (t))t -1 \label{eq:24}\\ {h}_2(t) = \tfrac {{L}}{2}\alpha (t)\, t + \tfrac {\gamma (t){L}_0}{2} \left[2(1+\gamma (t))+\alpha (t)\right] t -\gamma (t)\label{eq:25} \intertext {and} {h}_3(t) = a(t) t + {L}_0(1+\gamma (t)) (1+\alpha (t))t-1.\label{eq:26} \end{gather}
</div>
<p> We denote the minimal positive zeros of the functions \({h}_1\), \({h}_2\) and \({h}_3\) by \(\eta _1\), \(\eta _2\) and \(\eta _3\), respectively. Note that \(\alpha (t)\) is well defined on \((0,\eta _1)\) by the choice of \(\eta _1\). Let us set </p>
<div class="equation" id="eq:27">
<p>
  <div class="equation_content">
    \begin{equation}  \eta _0 = \min \{ \eta _1,\eta _2,\eta _3\} . \label{eq:27} \end{equation}
  </div>
  <span class="equation_label">2.12</span>
</p>
</div>
<p> Then, for all \(t\in (0,\eta _0)\) we have </p>
<div class="displaymath" id="eq:28">
  \begin{gather}  \alpha \in (0,1),\label{eq:28} \\ h_1(t) {\lt} 0, \label{eq:29}\\ h_2(t) \le 0 \label{eq:210} \intertext {and} h_3(t)\le {0}. \label{eq:211} \end{gather}
</div>
<p> We can show the following result on the convergence of majorizing sequences for the Midpoint method. <div class="lemma_thmwrapper " id="a0000000004">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">2.1</span>
  </div>
  <div class="lemma_thmcontent">
  <p>Let the positive constants be \({L}_0{\gt}0\), \({L}{\gt}0\), \(\mathcal{M}\ge {0}\) and \(\eta {\gt}0\). Suppose that  <div class="equation" id="eq:212">
<p>
  <div class="equation_content">
    \begin{equation} \label{eq:212} \eta \left\{  \begin{aligned} & \le \eta _0 \quad & \textrm{if}\quad & \eta _0 \neq \eta _1,\\ & < \eta _0 \quad & \textrm{if}\quad & \eta _0 = \eta _1. \end{aligned} \right. \end{equation}
  </div>
  <span class="equation_label">2.16</span>
</p>
</div> Then, scalar sequence \(\{ t_n\} \) generated by </p>
<div class="equation" id="eq:213">
<p>
  <div class="equation_content">
    \begin{equation} \label{eq:213} \begin{gathered} t_0 = 0, \quad s_0 = \eta , \quad t_{n+1} = s_n + \dfrac {{L}(s_n-t_n)^2}{2\left[ 1- \frac{{L}_0}{2}(s_n+t_n)\right] },\\ s_{n+1} = t_{n+1} + \dfrac {12{L}(t_{n+1}-s_n)^2 + \frac{6{L}^2(s_n-t_n)^3}{1-\frac{{L}_0}{2}(t_n+t_n)}(s_n-t_n)^3 + {7\mathcal{M}}(s_n-t_n)^3} {24(1-{L}_0\, t_{n+1})} \end{gathered} \end{equation}
  </div>
  <span class="equation_label">2.17</span>
</p>
</div>
<p> is increasing, bounded from above by </p>
<div class="equation" id="eq:214">
<p>
  <div class="equation_content">
    \begin{equation} \label{eq:214} t^{\star \star } = \left(\tfrac {1+\gamma }{1-\alpha }\right)\eta \end{equation}
  </div>
  <span class="equation_label">2.18</span>
</p>
</div>
<p> and converges to its unique least upper bound \(t^\star \) which satisfies </p>
<div class="equation" id="eq:215">
<p>
  <div class="equation_content">
    \begin{equation} \label{eq:215} 0\le t^\star \le t^{\star \star }. \end{equation}
  </div>
  <span class="equation_label">2.19</span>
</p>
</div>
<p> Moreover, the following estimates hold for each \(n=0,1,2,\ldots \) </p>
<div class="displaymath" id="eq:217">
  \begin{gather}  0 {\lt} t_{n+1}-s_n\le \gamma (s_n-t_n)\le \gamma \alpha ^n\eta \label{eq:216} \intertext {and} 0 {\lt} s_{n+1}-t_{n+1}\le \alpha (s_n-t_n)\le \alpha ^{n+1}\eta .\label{eq:217} \end{gather}
</div>

  </div>
</div> <div class="proof_wrapper" id="a0000000005">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> We use mathematical induction to prove <a href="#eq:217" class="eqref">2.20</a> and <a href="#eq:217" class="eqref">2.20</a>. Estimates <a href="#eq:217" class="eqref">2.20</a> and <a href="#eq:217" class="eqref">2.20</a> hold for \(n=0\) by <a href="#eq:21" class="eqref">2.7</a>-<a href="#eq:23" class="eqref">2.9</a> and <a href="#eq:213" class="eqref">2.17</a>, since </p>
<div class="displaymath" id="a0000000006">
  \begin{align}  s_1-t_1 & = \dfrac {12{L}(t_{1}-s_0)^3 + \frac{6{L}^2(s_0-t_0)^3}{1-\frac{{L}_0}{2}(t_0+s_0)} + {7\mathcal{M}}(s_0-t_0)^3} {24(1-{L}_0\, t_{1})} \nonumber \\ & \le \dfrac {12 {L}\gamma ^2+12{L}\gamma +7\mathcal{M}(s_0-t_0)}{24\left[1-\frac{{L}_0}{2}(1+\gamma )\eta \right]}(s_0-t_0)^2 \nonumber \nonumber \\ & \le \dfrac {a}{1-\frac{{L}_0}{2}(1+\gamma )}(s_0-t_0)(s_0-t_0 ) \le \alpha (s_0-t_0) = \alpha \eta . \label{eq:218} \end{align}
</div>
<p> Let us assume that <a href="#eq:217" class="eqref">2.20</a> and <a href="#eq:217" class="eqref">2.20</a> hold for all \(k\le \eta \). Then, we have </p>
<div class="displaymath" id="a0000000007">
  \begin{gather}  t_{k+1}-s_k\le \gamma (s_k-t_k)\le \gamma \alpha ^k\eta ,\nonumber \\ s_{k+1}-t_{k+1}\le \alpha (s_k-t_k)\le \alpha ^{k+1}\eta ,\nonumber \end{gather}
</div>
<div class="displaymath" id="a0000000008">
  \begin{align}  t_{n+1}& \le s_k + \gamma \alpha ^k\eta \le t_k + \alpha ^k\eta + \gamma \alpha ^k\eta \nonumber \\ & \le t_{k-1} +\alpha ^{k-1}\eta + \alpha ^k \eta + \gamma \alpha ^{k-1}\eta +\gamma \alpha ^k\eta \nonumber \\ & \le \cdots \le t_2 + (\alpha ^2 \eta + \alpha ^3\eta + \cdots +\alpha ^k \eta )+ (\gamma \alpha ^2 \eta + \cdots +\gamma \alpha ^k \eta )\nonumber \\ & \le s_1 + \gamma \alpha \eta + (\alpha ^2 \eta + \alpha ^3\eta + \cdots +\alpha ^k \eta )+ (\gamma \alpha ^2 \eta + \cdots +\gamma \alpha ^k \eta )\nonumber \\ & \le t_1 +\alpha \eta + \gamma \alpha \eta + (\alpha ^2 \eta + \alpha ^3\eta + \cdots +\alpha ^k \eta )+ (\gamma \alpha ^2\eta + \cdots +\gamma \alpha ^k \eta )\nonumber \\ & \le \eta + \gamma \eta + \alpha \eta + \gamma \alpha \eta + (\alpha ^2 \eta + \cdots +\alpha ^k \eta )+ (\gamma \alpha ^2\eta + \cdots +\gamma \alpha ^k \eta )\nonumber \\ & \le \tfrac {1-\alpha ^{k+1}}{1-\alpha } ( 1+\gamma )\eta {\lt} \tfrac {1+\gamma }{1-\alpha }\eta =t^{\star \star }, \label{eq:219} \end{align}
</div>
<p> and </p>
<div class="displaymath" id="a0000000009">
  \begin{align}  s_{k+1} \le t_{k+1} + \alpha ^{k+1}\eta & \le \left[\tfrac {1-\alpha ^{k+1}}{1-\alpha } ( 1+\gamma ) + \alpha ^{k+1} \right]\eta \nonumber \\ & {\lt} \left(\tfrac {1+\gamma }{1-\gamma }+\alpha ^{k+1} \right)\eta \le t^{\star \star }.\nonumber \end{align}
</div>
<p> Evidently, estimates <a href="#eq:217" class="eqref">2.20</a> and <a href="#eq:217" class="eqref">2.20</a> are true provided that </p>
<div class="displaymath" id="eq:221">
  \begin{align}  \dfrac {{L}(s_k-t_k)}{2\left(1-\frac{{L}_0}{2}(s_k+t_k)\right)}\le \gamma \label{eq:220} \intertext {and} \dfrac {a(s_k-t_k)}{2(1-{L}_0t_{k+1})} \le \alpha .\label{eq:221} \end{align}
</div>
<p> Inequality <a href="#eq:221" class="eqref">2.22</a> can be written as </p>
<div class="equation" id="eq:222">
<p>
  <div class="equation_content">
    \begin{equation}  \tfrac {{L}\alpha ^k\eta }{2} + \tfrac {\gamma {L}_0}{2} \left(2 \tfrac {1-\alpha ^k}{1-\alpha }(1+\gamma ) + \alpha ^k \right)\eta -\gamma \le 0.\label{eq:222} \end{equation}
  </div>
  <span class="equation_label">2.23</span>
</p>
</div>
<p> Estimate <a href="#eq:222" class="eqref">2.23</a> motivates us to define recurrent functions \(f_k\) on \([0,1)\) for each \(k=1,2,\ldots \) by </p>
<div class="equation" id="eq:223">
<p>
  <div class="equation_content">
    \begin{equation} \label{eq:223} f_k(t) = \tfrac {{L}\, t^k\eta }{2} + \tfrac {\gamma {L}_0}{2} \left(2\tfrac {1-t^k}{1-t} (1+\gamma )+t^k\right)\eta -\gamma . \end{equation}
  </div>
  <span class="equation_label">2.24</span>
</p>
</div>
<p> We need a relationship between two consecutive functions \(f_k\). We have by <a href="#eq:223" class="eqref">2.24</a> </p>
<div class="displaymath" id="a0000000010">
  \begin{align}  f_{k+1}(t) & = f_k(t) + \tfrac {{L}\, t^{k+1}\eta }{2} - \tfrac {{L}\, t^k\eta }{2} + \tfrac {\gamma {L}_0}{2}\left(2(1+\gamma )(t^k-t^{k-1})+t^{k+1}-t^k\right)\eta \nonumber \\ & = f_k(t) + (t-1)\left[\tfrac {{L}}{2}t +\gamma \, {L}_0(1+\gamma ) +\tfrac {\gamma \alpha {L}_0}{2} \right]t^{k-1}\eta . \label{eq:224} \end{align}
</div>
<p> It follows from <a href="#eq:224" class="eqref">2.25</a> that </p>
<div class="equation" id="eq:225">
<p>
  <div class="equation_content">
    \begin{equation}  f_{k+1}(t)\le f_{k}(t) \le \cdots \le f_{1}(t). \label{eq:225} \end{equation}
  </div>
  <span class="equation_label">2.26</span>
</p>
</div>
<p> In view of <a href="#eq:225" class="eqref">2.26</a>, estimate <a href="#eq:222" class="eqref">2.23</a> holds if </p>
<div class="equation" id="eq:226">
<p>
  <div class="equation_content">
    \begin{equation} \label{eq:226} f_1(\alpha )\le {0} \end{equation}
  </div>
  <span class="equation_label">2.27</span>
</p>
</div>
<p> which is true by the choice of \(\eta _2\). Similarly, estimate <a href="#eq:221" class="eqref">2.22</a> can be written as </p>
<div class="equation" id="eq:227">
<p>
  <div class="equation_content">
    \begin{equation} \label{eq:227} a\alpha ^{k}\eta + \alpha {L}_0(1+\gamma ) \left(\tfrac {1-\alpha ^{k+1}}{1-\alpha }\right)\eta - \alpha \le {0}. \end{equation}
  </div>
  <span class="equation_label">2.28</span>
</p>
</div>
<p> Define recurrent functions \(g_k\) on \([0,1)\) for each \(k = 1,2,\ldots \) by </p>
<div class="equation" id="eq:228">
<p>
  <div class="equation_content">
    \begin{equation} \label{eq:228} g_k(t) = {a}t^{k-1}\eta + {L}_0(1+\gamma ) \left(\tfrac {1-t^{k+1}}{1-t}\right)\eta -1. \end{equation}
  </div>
  <span class="equation_label">2.29</span>
</p>
</div>
<p> Then using <a href="#eq:228" class="eqref">2.29</a> we get </p>
<div class="equation" id="eq:229">
<p>
  <div class="equation_content">
    \begin{equation} \label{eq:229} g_{k+1}(t) = g_k(t) + (t-1)\left[{a} + {L}_0(1+t)t\right]t^{k-1}\eta . \end{equation}
  </div>
  <span class="equation_label">2.30</span>
</p>
</div>
<p> Hence, it follows from <a href="#eq:229" class="eqref">2.30</a> that </p>
<div class="equation" id="eq:230">
<p>
  <div class="equation_content">
    \begin{equation} \label{eq:230} g_{k+1}(t)\le g_{k}(t) \le \cdots \le g_{1}(t). \end{equation}
  </div>
  <span class="equation_label">2.31</span>
</p>
</div>
<p> In view of <a href="#eq:230" class="eqref">2.31</a>, instead of <a href="#eq:227" class="eqref">2.28</a>, we can show that </p>
<div class="equation" id="eq:231">
<p>
  <div class="equation_content">
    \begin{equation} \label{eq:231} g_1(\alpha ) \le {0}, \end{equation}
  </div>
  <span class="equation_label">2.32</span>
</p>
</div>
<p> which is true by the choice of \(\eta _3\). The induction for <a href="#eq:217" class="eqref">2.20</a> and <a href="#eq:217" class="eqref">2.20</a> is complete. Hence, sequence \(\{ t_n\} \) is an increasing, bounded from above by \(t^{\star \star }\) and as such it converges to its unique least upper bound \(t^\star \). The proof of the Lemma is complete. <div class="proof_wrapper" id="a0000000011">
  <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 have the following useful and obvious extension of Lemma 2.1 <div class="lemma_thmwrapper " id="a0000000012">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">2.2</span>
  </div>
  <div class="lemma_thmcontent">
  <p>Suppose there exists \(N\ge {0}\) such that </p>
<div class="displaymath" id="eq:233">
  \begin{gather}  t_0 {\lt} s_0 {\lt} t_1 {\lt} \cdots {\lt} t_N {\lt} s_N {\lt} t_{N+1} {\lt} \tfrac {1}{{L}_0} \label{eq:232}. \intertext {and} s_N-t_N \left\{  \begin{aligned} \label{eq:233} & \le \eta _0\quad \textrm{if}\quad \eta _0 \neq \eta _1 \\ & {\lt}\eta _0 \quad \textrm{if}\quad \eta _0 = \eta _1. \end{aligned} \right. \end{gather}
</div>
<p> Then, the conclusions of the Lemma 2.1 hold for sequence \(\{ t_n\} \). Moreover, the following estimates hold for each \(n=0,1,2,3,\ldots \) </p>
<div class="displaymath" id="eq:235">
  \begin{gather}  0 {\lt} t_{N+1+n}-s_{N+n} \le \gamma _{N}(s_{N+n}-t_{N+n})\label{eq:234} \intertext {and} 0 {\lt} s_{N+1+n}-t_{N+1+n} \le \alpha _{N}(s_{N+n}-t_{N+n})\label{eq:235} \end{gather}
</div>
<p> where \(\gamma _N = \gamma (s_N-t_N)\), \(\alpha _N = \alpha (s_N-t_N)\) and \(t_N^{\star \star } = \tfrac {1+\gamma _N}{1-\alpha _N}(s_N-t_N)\). </p>

  </div>
</div> <div class="remark_thmwrapper " id="a0000000013">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">2.3</span>
  </div>
  <div class="remark_thmcontent">
  <ol class="enumerate">
  <li></li>
  <li><p>Note that for \(N=0\), the Lemma 2.2 reduces to Lemma 2.1 with \(\alpha _0=\alpha \) and \(\gamma _0=\gamma \). </p>
</li>
  <li><p>Let us define sequences \(\{ r_n\} \) and \(\{ v_n\} \) by </p>
<div class="equation" id="eq:236">
<p>
  <div class="equation_content">
    \begin{equation} \label{eq:236} \begin{gathered} r_0 = 0, \quad q_0 = \eta , \quad r_1 = q_0 + \dfrac {\mathcal{K}_0(q_0-r_0)^2}{2\left(1-{{L}_3\eta }/{2}\right)},\\ q_1 = r_1 + \dfrac {\Big(12{L}_1(r_1-q_0)^2 + \frac{6{L}_2{L}_2^\prime (q_0-r_0)^3}{1-{{L}_0\, r_0}/{2}} + {7\mathcal{M}_0}(q_0-r_0)^3\Big)}{24(1-{L}_3r_1)},\\ r_{n+1} = q_n + \dfrac {{L}(q_n-r_n)^2}{2\left[1-\tfrac {{L}_0}{2} (q_n+r_n)\right]},\\ q_{n+1} = r_{n+1} + \dfrac {\Big(12{L}(r_{n+1}-q_n)^2+ \frac{6{L}^2(q_n-r_n)^3}{1-{{L}_0(r_n+q_n)}/{2}} + {7\mathcal{M}}(r_n-q_n)^3\Big)}{24(1-{L}_0r_{n+1})}\quad (n\ge 1) \end{gathered} \end{equation}
  </div>
  <span class="equation_label">2.35</span>
</p>
</div>
<p> for some \({L}_0\), \({L}_1\), \({L}_2\), \({L}_3\), \(\mathcal{K}_0\), \(\mathcal{M}_0\) such that </p>
<div class="equation" id="eq:237">
<p>
  <div class="equation_content">
    \begin{equation} \label{eq:237} {L}_1 \le {L}, \quad {L}_2 \le {L}, \quad {L}_2^\prime \le {L}, \quad \mathcal{K}_0 \le {L}, \quad {L}_3 \le {L}_0\quad \textrm{and}\quad \mathcal{M}_0 \le \mathcal{M}\end{equation}
  </div>
  <span class="equation_label">2.36</span>
</p>
</div>
<p> and </p>
<div class="equation" id="eq:238">
<p>
  <div class="equation_content">
    \begin{equation}  \label{eq:238} \begin{gathered} v_0 = 0,\quad u_0 = \eta , \quad v_{n+1} = u_n + \dfrac {{L}(u_n-v_n)^2}{2\left[1-\frac{{L}_0}{2} (v_n+u_n)\right]},\\ u_{n+1} = v_{n+1} + \dfrac {12{L}(v_{n+1}-u_n)^2+ \frac{6{L}^2(u_n-v_n)^3}{1-\frac{{L}_0}{2} (v_n+u_n)} + {7\mathcal{M}}(u_n-v_n)^3} {24(1-{L}v_{n+1} )}. \end{gathered} \end{equation}
  </div>
  <span class="equation_label">2.37</span>
</p>
</div>
<p> In view of <a href="#eq:17" class="eqref">1.6</a>, <a href="#eq:213" class="eqref">2.17</a>, <a href="#eq:236" class="eqref">2.35</a>, <a href="#eq:237" class="eqref">2.36</a>–<a href="#eq:238" class="eqref">2.37</a> a simple inductive argument shows that </p>
<div class="displaymath" id="eq:239">
  \begin{gather}  r_n \le t_n \le v_n \label{eq:239}\\ q_n \le s_n \le u_n, \label{eq:240} \\ r_{n+1}-q_n \le s_{n+1}-t_n \le u_{n+1}-v_n, \label{eq:241}\\ q_{n+1}-r_{n+1} \le s_{n+1}-t_{n+1} \le u_{n+1}-v_{n+1} \label{eq:242} \intertext {and} r^\star = \lim _{n\to \infty }r_n \le t^\star \le v^\star = \lim _{n\to \infty }v_n.\label{eq:243} \end{gather}
</div>
<p> Moreover, <a href="#eq:239" class="eqref">2.38</a>-<a href="#eq:243" class="eqref">2.41</a> hold as strict inequalities for \(n\ge {1}\) if <a href="#eq:17" class="eqref">1.6</a> and <a href="#eq:237" class="eqref">2.36</a> hold as strict inequalities. Sequence \(\{ v_n\} \) was shown to be majorizing for the Midpoint method <a href="#eq:12" class="eqref">1.2</a> provided that <a href="#eq:14" class="eqref">1.3</a> or <a href="#eq:14" class="eqref">1.3</a> hold <span class="cite">
	[
	<a href="#1" >1</a>
	, 
	<a href="#3" >3</a>
	, 
	<a href="#4" >4</a>
	, 
	<a href="#5" >10</a>
	, 
	<a href="#6" >11</a>
	, 
	<a href="#7" >12</a>
	, 
	<a href="#8" >13</a>
	, 
	<a href="#9" >14</a>
	, 
	<a href="#10" >15</a>
	, 
	<a href="#11" >16</a>
	, 
	<a href="#19" >24</a>
	, 
	<a href="#20" >25</a>
	, 
	<a href="#21" >26</a>
	, 
	<a href="#23" >28</a>
	]
</span>. We shall prove in the next section that tighter sequences \(\{ r_n\} \) and \(\{ v_n\} \) are also majorizing for the Midpoint method <a href="#eq:12" class="eqref">1.2</a>. Then, certainly these majorizing sequences also converge under <a href="#eq:14" class="eqref">1.3</a> or <a href="#eq:14" class="eqref">1.3</a>. However, these sequences converge under the new convergence criteria given in the Lemma 2.1 which can be weaker that <a href="#eq:14" class="eqref">1.3</a> or <a href="#eq:14" class="eqref">1.3</a> (see Section 4). In the next Section, we shall provide the connection of \({L}_0\), \({L}_1\), \({L}_2\), \({L}_2^\prime \), \({L}_3\), \(\mathcal{K}_0\), \(\mathcal{M}_0\) to the equation <a href="#eq:11" class="eqref">1.1</a> and the Midpoint method <a href="#eq:12" class="eqref">1.2</a> so that estimates <a href="#eq:237" class="eqref">2.36</a> are satisfied. </p>
</li>
</ol>

  </div>
</div> </p>
<h1 id="a0000000014">3 Semi-local convergence of the Midpoint method</h1>
<p> We need the following Ostrowski-type representation for the Midpoint method <a href="#eq:12" class="eqref">1.2</a>. <div class="lemma_thmwrapper " id="a0000000015">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">3.1</span>
  </div>
  <div class="lemma_thmcontent">
  <p>Suppose that the Midpoint method <a href="#eq:12" class="eqref">1.2</a> is well defined for each \(n=0,1,2,\ldots \). Then, the following identities are true for each \(n=0,1,2,\ldots \) </p>
<div class="displaymath" id="eq:31">
  \begin{align} \label{eq:31} &  \mathcal F(x_{n+1}) = \\ &  =\int _{0}^1\mathcal F^{\prime \prime }(y_n+\theta (x_{n+1}-y_n)) (1-\theta )\mathop{d\theta }(x_{n+1}-y_n)^2 \nonumber \\ & \quad +\tfrac {1}{4}\int _0^1 \mathcal F^{\prime \prime }( \tfrac {(x_n+y_n)}{2} +\tfrac {\theta }{2}(y_{n}-x_n)) (y_n-x_n) \mathop{d\theta } \mathcal F^\prime (\tfrac {x_n+y_n}{2})^{-1}\times \nonumber \\ & \qquad \times \int _0^1 \mathcal F^{\prime \prime }(x_n + \tfrac {\theta }{2}(y_n-x_n))\mathop{d\theta } (y_n-x_n)^2 \nonumber \\ & \quad + \int _0^1 \left[ \mathcal F^{\prime \prime } (x_n + \theta (y_n-x_n)) (1-\theta ) - \tfrac {1}{2}\mathcal F^{\prime \prime }(x_n+\tfrac {\theta }{2}(y_n-x_n) )\right]\mathop{d\theta } (y_n-x_n)^2.\nonumber \end{align}
</div>
<p> and </p>
<div class="displaymath" id="a0000000016">
  \begin{multline*}  x_{n+1}-y_n = -\tfrac {1}{2} \mathcal F^\prime (\tfrac {x_n+y_n}{2})^{-1} \int _0^1 \mathcal F^{\prime \prime }(x_n + \tfrac {\theta }{2}(y_n-x_n))\mathop{d\theta } (y_n-x_n)^2. \end{multline*}
</div>

  </div>
</div> <div class="proof_wrapper" id="a0000000017">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> The proof of <a href="#eq:31" class="eqref">3.42</a> can be found in [1–4]. Using <a href="#eq:12" class="eqref">1.2</a>, we get in turn that </p>
<div class="displaymath" id="a0000000018">
  \begin{align*} & x_{n+1}-y_n = \\ & = \mathcal F^\prime (x_n)^{-1}\mathcal F(x_n) - \mathcal F^\prime (\tfrac {x_n+y_n}{2})^{-1}\mathcal F(x_n) \\ & = \left( \mathcal F^\prime (x_n)^{-1} - \mathcal F^\prime (\tfrac {x_n+y_n}{2})^{-1} \right) \mathcal F(x_n) \\ & =\mathcal F^\prime (\tfrac {x_n+y_n}{2})^{-1}\left[\mathcal F^\prime (\tfrac {x_n+y_n}{2}) \mathcal F^\prime (x_n)^{-1} - \mathcal{I}\right] \\ & = \mathcal F^\prime (\tfrac {x_n+y_n}{2})^{-1}\left[ \mathcal F^\prime (\tfrac {x_n+y_n}{2}) - \mathcal F^\prime (x_n) \right]\mathcal F^\prime (x_n)^{-1}\mathcal F(x_n) \\ & = \mathcal F^\prime (\tfrac {x_n+y_n}{2})^{-1} \int _0^1 \mathcal F^{\prime \prime }(x_n+\theta (\tfrac {x_n+y_n}{2}-x_n)) (\tfrac {x_n+y_n}{2}-x_n)[-(y_n-x_n)]\mathop{d\theta }\\ & = -\tfrac {1}{2} \mathcal F^\prime (\tfrac {x_n+y_n}{2})^{-1} \int _0^1 \mathcal F^{\prime \prime }(x_n+\tfrac {\theta }{2}{y_n-x_n})\mathop{d\theta } (y_n-x_n)^2. \end{align*}
</div>
<p> The proof of the Lemma is complete. <div class="proof_wrapper" id="a0000000019">
  <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 semi-local convergence result for the Midpoint method <a href="#eq:12" class="eqref">1.2</a> under the (<b class="bfseries">H</b>) conditions. <div class="theorem_thmwrapper " id="a0000000020">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">3.2</span>
  </div>
  <div class="theorem_thmcontent">
  <p>Suppose that the (<b class="bfseries">H</b>) conditions and those of the Lemma 2.1 hold. Moreover, suppose that </p>
<div class="equation" id="eq:32">
<p>
  <div class="equation_content">
    \begin{equation} \label{eq:32} \overline{U}(x_0,t^\star )\subseteq \mathbf{D}. \end{equation}
  </div>
  <span class="equation_label">3.43</span>
</p>
</div>
<p> Then, sequence \(\{ x_n\} \) generated by the Midpoint method <a href="#eq:12" class="eqref">1.2</a> 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 \(\mathcal F(x)=0\). Moreover, the following estimates hold </p>
<div class="displaymath" id="eq:33">
  \begin{gather}  \Vert {y_n-x_n}\Vert \le s_n - t_n, \label{eq:33}\\ \Vert {x_{n+1}-y_n}\Vert \le t_{n+1} - s_n, \label{eq:34}\\ \Vert {x_n-x^\star }\Vert \le t^\star - t_n \label{eq:35} \intertext {and} \Vert {y_n-x^\star }\Vert \le t^\star - s_n. \label{eq:36} \end{gather}
</div>
<p> Furthermore, if there exists \(R\ge {t^\star }\) such that </p>
<div class="equation" id="eq:37">
<p>
  <div class="equation_content">
    \begin{equation} \label{eq:37} {\overline{U}(x_0,R) \subseteq \mathbf{D}} \end{equation}
  </div>
  <span class="equation_label">3.47</span>
</p>
</div>
<p> and </p>
<div class="equation" id="eq:38">
<p>
  <div class="equation_content">
    \begin{equation} \label{eq:38} \dfrac {{L}_0}{2}(t^\star +R)<1, \end{equation}
  </div>
  <span class="equation_label">3.48</span>
</p>
</div>
<p> then, the solution \(x^\star \) is unique in \(\overline{U}(x_0,R)\) </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 shall prove that <a href="#eq:37" class="eqref">3.47</a> and <a href="#eq:38" class="eqref">3.48</a> hold using mathematical induction. Using <a href="#eq:12" class="eqref">1.2</a>, (<b class="bfseries">C</b>\(_2\)) and <a href="#eq:213" class="eqref">2.17</a>, we get that \(\Vert {y_0-x_0}\Vert = \Vert {\mathcal F^\prime (x_0)^{-1}\mathcal F^\prime (x_0)}\Vert \le \eta = s_0-t_0 \le t^\star \). That is <a href="#eq:33" class="eqref">3.44</a> holds for \(n=0\) and \(y_0 \in \overline{U}(x_0,t^\star )\). We have by (<b class="bfseries">C</b>\(_5\)) and the choice of \(\eta _1\) that </p>
<div class="displaymath" id="a0000000022">
  \begin{align}  \Vert {\mathcal F^\prime (x_0)^{-1} \left[ \mathcal F^\prime (\tfrac {x_0+y_0}{2}) - \mathcal F^\prime (x_0) \right]}\Vert & \le \tfrac {{L}_0}{2} \Vert {y_0-x_0}\Vert \nonumber \\ & \le \tfrac {{L}_0}{2} (s_0-t_0 ) = \tfrac {{L}_0}{2} \eta {\lt} 1.\label{eq:39} \end{align}
</div>
<p> It then follows from <a href="#eq:39" class="eqref">3.49</a> and the Banach Lemma on invertible operators <span class="cite">
	[
	<a href="#18" >23</a>
	, 
	<a href="#27" >32</a>
	]
</span> that </p>
<div class="equation" id="eq:310">
<p>
  <div class="equation_content">
    \begin{equation} \label{eq:310} \begin{gathered} \mathcal F^\prime (\tfrac {x_0+y_0}{2})^{-1} \in {L}({\mathbf{Y}},\mathbf{X}),\\ \left\Vert {\mathcal F^\prime ( \tfrac {x_0+y_0}{2} )^{-1}} \mathcal F^\prime (x_{0})\right\Vert \le \tfrac {1}{1-\frac{{L}_0}{2}\eta }. \end{gathered} \end{equation}
  </div>
  <span class="equation_label">3.50</span>
</p>
</div>
<p> Using <a href="#eq:12" class="eqref">1.2</a>, <a href="#eq:213" class="eqref">2.17</a>, Lemma 3.1 and <a href="#eq:310" class="eqref">3.50</a> we obtain </p>
<div class="displaymath" id="a0000000023">
  \begin{align}  \Vert {x_1-y_0}\Vert &  \le \tfrac {1}{2} \tfrac {{L}\, \Vert {y_0-x_0}\Vert ^2}{1-\frac{{L}_0}{2}\eta }\nonumber \\ & \le \tfrac {1}{2} \tfrac {{L}\, ({s_0-t_0})^2}{1-\frac{{L}_0}{2}\eta } \label{eq:311} \end{align}
</div>
<p> and <div class="equation" id="eq:312">
<p>
  <div class="equation_content">
    \begin{equation} \label{eq:312} \Vert {x_1-y_0}\Vert \le t_1-s_0 \le \gamma (s_0-t_0). \end{equation}
  </div>
  <span class="equation_label">3.52</span>
</p>
</div>  Hence, <a href="#eq:34" class="eqref">3.45</a> holds for \(n=0\). We also get that \(\Vert {x_1-x_0}\Vert \le \Vert {x_1-y_0}\Vert + \Vert {y_0-x_0}\Vert \le t_1-s_0+ s_0-t_0 = t_1 \le t^\star ,\) which implies \(x_1\in \overline{U}(x_0,t^\star )\). Let us assume that <a href="#eq:33" class="eqref">3.44</a>, <a href="#eq:34" class="eqref">3.45</a>, \(y^\star \in \overline{U}(x_0,t^\star )\) and \(x_{k+1}\in \overline{U}(x_0,t^\star )\) hold for all \(k \le {n}\). It follows from the proof of the Lemma 2.1 and (<b class="bfseries">C</b>\(_5\)) that </p>
<div class="displaymath" id="a0000000024">
  \begin{align}  \left\Vert {\mathcal F^\prime (x_0)^{-1}\left(\mathcal F^\prime (\tfrac {x_k+y_k}{2}) - \mathcal F^\prime (x_0) \right) }\right\Vert & \le \tfrac {{L}_0}{2} \left(\Vert {x_{k} -x_0 }\Vert + \Vert {y_k-x_0}\Vert \right) \nonumber \\ & \le \tfrac {{L}_0}{2} (t_{k}+s_k) {\lt} 1 \label{eq:313} \end{align}
</div>
<p> and </p>
<div class="displaymath" id="a0000000025">
  \begin{align}  \left\Vert {\mathcal F^\prime (x_0)^{-1} \left(\mathcal F^\prime (x_{k+1}) - \mathcal F^\prime (x_0) \right) }\right\Vert & \le {L}_0 \Vert {x_{k+1}-x_0}\Vert \nonumber \\ & \le {L}_0 t_{k+1} {\lt} 1.\label{eq:314} \end{align}
</div>
<p> It then follows from <a href="#eq:313" class="eqref">3.53</a> and <a href="#eq:314" class="eqref">3.54</a> that </p>
<div class="displaymath" id="a0000000026">
  \begin{gather}  \mathcal F^\prime (\tfrac {x_k+y_k}{2})^{-1} \in {L}({\mathbf{Y}},\mathbf{X}),\nonumber \\ \mathcal F^\prime (x_{k+1})^{-1} \in {L}({\mathbf{Y}},\mathbf{X}),\nonumber \\ \left\Vert \mathcal F^\prime (\tfrac {x_k+y_k}{2})^{-1} \mathcal F^\prime (x_{0})\right\Vert \le \dfrac {1}{1-\frac{{L}_0}{2}(t_k+s_k)}, \label{eq:315}\\ \left\Vert \mathcal F^\prime (x_{k+1})^{-1} \mathcal F^\prime (x_{0})\right\Vert \le \tfrac {1}{1-{L}_0t_{k+1}}.\label{eq:316} \end{gather}
</div>
<p> Then, we have by <a href="#eq:12" class="eqref">1.2</a>, (<b class="bfseries">C</b>\(_3\)), Lemma 3.1, <a href="#eq:213" class="eqref">2.17</a>, <a href="#eq:315" class="eqref">3.55</a> and the induction hypothesis that </p>
<div class="displaymath" id="a0000000027">
  \begin{align}  \Vert {x_{k+1}-y_k}\Vert & \le \tfrac {1}{2} \dfrac {{L}\Vert {y_{k}-x_k}\Vert ^2}{1-\frac{{L}_0}{2}(s_k+t_k)} \nonumber \\ &  \le \tfrac {{L}(s_k-t_k)^2}{ 2 \left[{1-\frac{{L}_0}{2}(s_k+t_k)}\right] } = t_{k+1} - s_k, \end{align}
</div>
<p> which shows <a href="#eq:34" class="eqref">3.45</a>. Moreover, using <a href="#eq:12" class="eqref">1.2</a>, (<b class="bfseries">C</b>\(_3\)), (<b class="bfseries">C</b>\(_4\)), <a href="#eq:213" class="eqref">2.17</a>, Lemma 3.1, we obtain in turn </p>
<div class="displaymath" id="a0000000028">
  \begin{align} & \left\Vert {\int _0^1 \mathcal F^\prime (x_0)^{-1} \left[ \mathcal F^{\prime \prime }(x_k+\theta (y_k-x_k)) (1-\theta ) - \tfrac {1}{2} \mathcal F^{\prime \prime }(x_k+\tfrac {\theta }{2}(y_k-x_k)) \right] \mathop{d\theta }}\right\Vert \nonumber \\ & \le \left\Vert {\int _0^1 \mathcal F^\prime (x_0)^{-1} \left[ \mathcal F^{\prime \prime }(x_k+\theta (y_k-x_k)) - \mathcal F^{\prime \prime }(x_k) \right] \mathop{d\theta } }\right\Vert \nonumber \\ &  + \tfrac {1}{2} \left\Vert {\int _0^1 \mathcal F^\prime (x_0)^{-1} \left[ \mathcal F^{\prime \prime }(x_k) -\mathcal F^{\prime \prime }(x_k+\tfrac {\theta }{2}(y_k-x_k)) \right] \mathop{d\theta } } \right\Vert \nonumber \\ & \le \mathcal{M}\int _0^1 \theta (1-\theta )\mathop{d\theta } \Vert {y_k-x_k}\Vert + \tfrac {\mathcal{M}}{4}\int _0^1 \theta \mathop{d\theta } \Vert {y_k-x_k}\Vert \nonumber \\ & = \tfrac {7\mathcal{M}}{24}\Vert {y_k-x_k}\Vert .\label{eq:318} \end{align}
</div>
<p> Thus, </p>
<div class="displaymath" id="eq:319">
  \begin{align} & \left\Vert { \mathcal F^\prime (x_0)^{-1} \mathcal F(x_{k+1}) }\right\Vert \label{eq:319} \\ & \le \tfrac {{L}}{2} \Vert {x_{k+1}-y_k}\Vert ^2 + \tfrac {{L}^2}{4}\tfrac {1}{1-\frac{{L}_0}{2}(s_k+t_k)} \Vert {y_{k}-x_k}\Vert ^3 + \tfrac {7\mathcal{M}}{24}\Vert {y_k-x_k}\Vert ^3\nonumber \\ & \le \tfrac {{L}}{2} (t_{k+1}-s_k)^2 + \tfrac {{L}^2(s_k-t_k)^3}{4\left[1-\frac{{L}_0}{2}(s_k+t_k)\right]} + \tfrac {7\mathcal{M}}{24} (s_k-t_k)^3\nonumber \end{align}
</div>
<p> and </p>
<div class="displaymath" id="a0000000029">
  \begin{align}  \left\Vert { y_{k+1}-x_{k+1} } \right\Vert & \le \left\Vert {\mathcal F^\prime (x_{k+1})^{-1}\mathcal F^\prime (x_{0})}\right\Vert \left\Vert {\mathcal F^\prime (x_0)^{-1}\mathcal F(x_{k+1})}\right\Vert \nonumber \\ & \le \frac{ \frac{{L}(t_{k+1}-s_k)^2}{2} + \frac{{L}^2(s_k-t_k)^3}{4\left[1-\frac{{L}_0}{2}(t_k+s_k)\right]} + \frac{7\mathcal{M}}{24}(s_k-t_k)^3}{1-{L}_0\, t_{k+1}}\nonumber \\ & =s_{k+1}-t_{k+1},\label{eq:320} \end{align}
</div>
<p> which shows <a href="#eq:33" class="eqref">3.44</a>. We also have that </p>
<div class="displaymath" id="a0000000030">
  \begin{align*}  \left\Vert { y_{k+1}-x_{0} } \right\Vert & \le \left\Vert { y_{k+1}-x_{k+1} } \right\Vert + \left\Vert { x_{k+1}-x_{0} } \right\Vert \nonumber \\ & \le s_{k+1} - t_{k+1} + t_{k+1} -t_0 = s_{k+1} \le t^\star . \end{align*}
</div>
<p> and </p>
<div class="displaymath" id="a0000000031">
  \begin{align*}  \left\Vert { x_{k+2}-x_{0} } \right\Vert & \le \left\Vert { x_{k+2}-y_{k+1} } \right\Vert + \left\Vert { y_{k+1}-x_{0} } \right\Vert \nonumber \\ & \le t_{k+2} - s_{k+1} + s_{k+1} -t_0 = t_{k+2} \le t^\star . \end{align*}
</div>
<p> Hence, \(y_{k+1}\) and \(x_{k+2}\) belong in \(\overline{U}(x_0,t^\star )\). It follows from <a href="#eq:33" class="eqref">3.44</a>, <a href="#eq:34" class="eqref">3.45</a> and the Lemma 2.1 that sequence \(\{ x_n\} \) is complete in a Banach space \(\mathbf{X}\) 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). By letting \(k\to \infty \) in <a href="#eq:319" class="eqref">3.59</a> we obtain \(\mathcal F(x^\star )=0\). Estimates <a href="#eq:36" class="eqref">3.46</a> and <a href="#eq:36" class="eqref">3.46</a> follows from <a href="#eq:33" class="eqref">3.44</a> by using standard majorization techniques <span class="cite">
	[
	<a href="#18" >23</a>
	, 
	<a href="#27" >32</a>
	]
</span>. Finally to show the uniqueness part. Let \(y^\star \in \overline{U}(x_0,R)\) be a solution of the equation \(\mathcal F(x)=0\). Let \(Q = \int _0^1 \mathcal F^\prime (x^\star + \theta (y^\star - x^\star )) \mathop{d\theta }\). Using (<b class="bfseries">C</b>\(_5\)), <a href="#eq:37" class="eqref">3.47</a> and <a href="#eq:38" class="eqref">3.48</a>, we get that </p>
<div class="displaymath" id="a0000000032">
  \begin{align}  \left\Vert {\mathcal F^\prime (x_0)^{-1} \left[Q-\mathcal F^\prime (x_0)\right]}\right\Vert & \le \int _0^1 \Vert {\mathcal F^\prime (x_0)^{-1} \left[\mathcal F^\prime (x^\star +\theta (y^\star -x^\star )) -\mathcal F^\prime (x_0)\right]\mathop{d\theta }}\Vert \nonumber \\ & \le {L}_0 \int _0^1 \left[ (1-\theta ) \Vert {x^\star -x_0}\Vert +\theta \Vert {y^\star -x_0}\Vert \right]\mathop{d\theta }\nonumber \\ & \le \tfrac {{L}_0}{2}(t^\star +R) {\lt} 1.\label{eq:321} \end{align}
</div>
<p> It follows from <a href="#eq:321" class="eqref">3.59</a> and the Banach lemma on invertible operators <span class="cite">
	[
	<a href="#18" >23</a>
	, 
	<a href="#27" >32</a>
	]
</span> that \(Q^{-1}\in {L}({\mathbf{Y}},\mathbf{X})\). Then, using the identity </p>
<div class="displaymath" id="a0000000033">
  \[ 0 = \mathcal F(y^\star ) - \mathcal F(x^\star ) = Q(y^\star -x^\star ) \]
</div>
<p> we deduce that \(x^\star =y^\star \). The proof of the Theorem is complete. <div class="proof_wrapper" id="a0000000034">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> <div class="remark_thmwrapper " id="a0000000035">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">3.3</span>
  </div>
  <div class="remark_thmcontent">
  <ol class="enumerate">
  <li></li>
  <li><p>The limit point \(t^\star \) can be replaced by \(t^{\star \star }\) (given in closed from by <a href="#eq:214" class="eqref">2.18</a>) in Theorem 3.2. </p>
</li>
  <li><p>The conclusions of Theorem 3.2 hold if hypotheses of Lemma 2.1 are replaced by those of Lemma 2.2. </p>
</li>
  <li><p>It follows from the (<b class="bfseries">H</b>) conditions that there exists constants \(\mathcal{K}_0\), \({L}_1\), \({L}_2\), \({L}_3\), \(\mathcal{M}_0\) satisfying </p>
<div class="displaymath" id="eq:322">
  \begin{align}  \left\Vert {\mathcal F^\prime (x_0)^{-1} \mathcal F^{\prime \prime }(x_0+\tfrac {\theta }{2}(y_0-x_0)) } \right\Vert & \le \mathcal{K}_0 \label{eq:322} \\ \left\Vert {\mathcal F^\prime (x_0)^{-1} \mathcal F^{\prime \prime }(y_0+\theta (x_1-y_0)) }\right\Vert & \le {L}_1 \label{eq:321n} \\ \left\Vert {\mathcal F^\prime (x_0)^{-1} \mathcal F^{\prime \prime }(x_0+\tfrac {\theta }{2}(y_0-x_0)) } \right\Vert & \le {L}_2 \label{eq:322n} \\ \left\Vert {\mathcal F^\prime (x_0)^{-1} \mathcal F^{\prime \prime }(\tfrac {x_0+y_0}{2}+\tfrac {\theta }{2}(y_0-x_0)) } \right\Vert & \le {L}^\prime _2 \label{eq:323} \\ \left\Vert {\mathcal F^\prime (x_0)^{-1} \left[\mathcal F^{\prime }(\tfrac {x_0+y_0}{2})-\mathcal F^\prime (x_0)\right] } \right\Vert & \le \tfrac {{L}_3}{2}\Vert {y_0-x_0}\Vert \label{eq:324} \\ \left\Vert {\mathcal F^\prime (x_0)^{-1} \left[\mathcal F^{\prime \prime }(x_0+\overline{\theta }(y_0-x_0)) - \mathcal F^{\prime \prime }(x_0) \right] } \right\Vert & \le \mathcal{M}_0\overline{\theta }\Vert {y_0-x_0}\Vert \label{eq:325} \end{align}
</div>
<p> \(\overline{\theta }=\theta \) or \(\theta /2\). For all \(\theta \in [0,1]\), where, \(y_0 = x_0 - \mathcal F^\prime (x_0)^{-1}\mathcal F(x_0)\) and \(x_1 = x_0 - \mathcal F^\prime (\frac{x_0+y_0}{2})^{-1}\mathcal F(x_0)\). Estimates <a href="#eq:322" class="eqref">3.60</a> -<a href="#eq:325" class="eqref">3.65</a> are not additional to the (<b class="bfseries">H</b>) conditions, since in practice the verification of (<b class="bfseries">C</b>\(_2\)) - (<b class="bfseries">C</b>\(_5\)) requires the computation of \(\mathcal{K}_0\), \({L}_1\), \({L}_2\), \({L}_3\) and \(\mathcal{M}_0\). Note that finding these constants only involves computations at the initial data. Moreover, these constants satisfy <a href="#eq:237" class="eqref">2.36</a>. Furthermore, according to the proof of Theorem 3.2, \(\{ r_n\} \) is a majorizing sequence for \(\{ x_n\} \) which is finer than \(\{ t_n\} \) and \(\{ v_n\} \) (see also <a href="#eq:239" class="eqref">2.38</a>-<a href="#eq:243" class="eqref">2.41</a> and the Tables in the next section). </p>
</li>
</ol>

  </div>
</div> </p>
<h1 id="a0000000036">4 Numerical examples</h1>
<p> <div class="example_thmwrapper " id="a0000000037">
  <div class="example_thmheading">
    <span class="example_thmcaption">
    Example
    </span>
    <span class="example_thmlabel">4.1</span>
  </div>
  <div class="example_thmcontent">
  <p>Let \({\mathbf{X}} = {{\mathbf{Y}}}=\mathbb {R}\) be equipped with the max-norm, \(x_0=1\), \({\mathbf{D}}=[\psi ,2-\psi ]\). Let us define \({\mathcal F}\) on \({\mathbf{D}}\) by </p>
<div class="equation" id="eq:411">
<p>
  <div class="equation_content">
    \begin{equation} \label{eq:411} {\mathcal F}(x) = x^m - \psi . \end{equation}
  </div>
  <span class="equation_label">4.66</span>
</p>
</div>
<p> Here, \(a\in (0,1.0)\). Through some algebraic manipulations, for the conditions (<b class="bfseries">H</b>), we obtain </p>
<div class="displaymath" id="a0000000038">
  \begin{gather*}  \eta = \frac{1-\psi }{m}, \quad {L}= { \left( 2-\psi \right) ^{m-2} \left( {m}-1\right) }, \quad {L}_0= {\tfrac { \left( 2-\psi \right) ^{m-1}-1}{1-\psi }}\\ \textrm{and}\quad \mathcal{M}= (m-1)(m-2)(2-\psi )^{m-3}. \end{gather*}
</div>
<p> Furthermore, we see that for \(m=8\) and \(\psi =0.79\) the criteria <a href="#eq:14" class="eqref">1.3</a> and <a href="#eq:14" class="eqref">1.3</a> yield </p>
<div class="displaymath" id="a0000000039">
  \begin{equation*}  {0.02625000000}\le {0.02128483707}\quad \textrm{and} \quad {0.02625000000}\le {0.02024888538} \end{equation*}
</div>
<p> respectively. Thus we observe that the criteria <a href="#eq:14" class="eqref">1.3</a> and <a href="#eq:14" class="eqref">1.3</a> are not satisfied. Even though the criteria <a href="#eq:14" class="eqref">1.3</a> and <a href="#eq:14" class="eqref">1.3</a> fall short but Midpoint method, starting at \(x_0=1\), converges for \(m=8\) and \(a=0.79\) as reported in Table <a href="#table:ourtable01">4.</a>. </p>
<div class="table"  id="table:ourtable01">
    <div class="centered"><table class="tabular">
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px" 
        rowspan=""
        colspan="">
      <p> \(n\) </p>

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

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(\Vert {\mathcal F(x_n)}\Vert \) </p>

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

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px" 
        rowspan=""
        colspan="">
      
<p>\(0\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px" 
        rowspan=""
        colspan="">
      <p> \({2.8793487895751115e-02}\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \({2.0999999999999996e-01}\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \({1.0000000000000000e+00}\) </p>

    </td>
  </tr>
  <tr>
    <td  style="None" 
        rowspan=""
        colspan="">
      <p>\(1\) </p>

    </td>
    <td  style="None" 
        rowspan=""
        colspan="">
      <p> \({2.4193435090165668e-04}\) </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \({1.5761236253916969e-03}\) </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> \({9.7120651210424889e-01}\) </p>

    </td>
  </tr>
  <tr>
    <td  style="None" 
        rowspan=""
        colspan="">
      <p>\(2\) </p>

    </td>
    <td  style="None" 
        rowspan=""
        colspan="">
      <p> \({1.5756220982003125e-10}\) </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \({1.0255710554280571e-09}\) </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> \({9.7096457775334723e-01}\) </p>

    </td>
  </tr>
  <tr>
    <td  style="None" 
        rowspan=""
        colspan="">
      <p>\(3\) </p>

    </td>
    <td  style="None" 
        rowspan=""
        colspan="">
      <p> \({4.3565100027748919e-29}\) </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \({2.8356485759462416e-28}\) </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> \({9.7096457759578502e-01}\) </p>

    </td>
  </tr>
  <tr>
    <td  style="None" 
        rowspan=""
        colspan="">
      <p>\(4\) </p>

    </td>
    <td  style="None" 
        rowspan=""
        colspan="">
      <p> \({9.2087073799154864e-85}\) </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \({5.9939396332225704e-84}\) </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> \({9.7096457759578502e-01}\) </p>

    </td>
  </tr>
  <tr>
    <td  style="None" 
        rowspan=""
        colspan="">
      <p>\(5\) </p>

    </td>
    <td  style="None" 
        rowspan=""
        colspan="">
      <p> \({8.6971817284824960e-252}\) </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \({5.6609880311094046e-251}\) </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> \({9.7096457759578502e-01}\) </p>

    </td>
  </tr>
  <tr>
    <td  style="None" 
        rowspan=""
        colspan="">
      <p>\(6\) </p>

    </td>
    <td  style="None" 
        rowspan=""
        colspan="">
      <p> \({7.3268645115540533e-753}\) </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \({4.7690497451183881e-752}\) </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> \({9.7096457759578502e-01}\) </p>

    </td>
  </tr>
  <tr>
    <td  style="border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>\(7\) </p>

    </td>
    <td  style="border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \({0.0000000000000000e+00}\) </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \({1.1979236472187808e-2023}\) </p>

    </td>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \({9.7096457759578502e-01}\) </p>

    </td>
  </tr>
</table> <figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">4.</span> 
  <span class="caption_text">Midpoint method applied to <a href="#eq:411" class="eqref">4.66</a>.</span> 
</figcaption> </div> 
</div>
<p> Moreover from equations <a href="#eq:24" class="eqref">2.10</a>–<a href="#eq:26" class="eqref">2.11</a> we obtain </p>
<div class="displaymath" id="a0000000040">
  \begin{equation*}  \eta _1 = {0.03842120732},\quad \eta _2 = {0.02841309975},\quad \eta _3 ={0.02710183588}. \end{equation*}
</div>
<p> From <a href="#eq:27" class="eqref">2.12</a>, we get \(\eta _0=\eta _3 = {0.02710183588}\). We notice that the assumption <a href="#eq:212" class="eqref">2.16</a>, of Lemma 2.1, holds. That is \(\eta = {0.02625} {\lt} \eta _0 = {0.02710183588}\). From <a href="#eq:322" class="eqref">3.60</a>-<a href="#eq:324" class="eqref">3.64</a>, we obtain </p>
<div class="displaymath" id="a0000000041">
  \begin{gather*}  \mathcal{K}_0 = 7,\quad {L}_1 = 7\Big(\tfrac {7+\psi }{8}\Big)^6, \quad {L}_2 = 7,\quad {L}_3 = 6\tfrac {\left\Vert \left( 2/3+1/3\, \psi - \left( -1/3+1/3\, \psi \right) ^{2} \right) ^{2}-1\right\Vert }{\left\Vert -1/3+1/3\, \psi - \left( -1/3+1/3\, \psi \right) ^{2}\right\Vert },\\ \mathcal{M}_0 = 7, \quad L_2^\prime = 6\Big(\tfrac {5+\psi }{3}\Big). \end{gather*}
</div>
<p> We can verify that the conditions <a href="#eq:237" class="eqref">2.36</a> are fulfilled. Additionally, for the sequences \(\{ t_n\} \) (given by <a href="#eq:212" class="eqref">2.16</a>), \(\{ r_n\} \) (given by <a href="#eq:236" class="eqref">2.35</a>) and \(\{ v_n\} \) (given by <a href="#eq:238" class="eqref">2.37</a>), we produce the Table <a href="#table:ourtable20">4.</a>.  </p>
<div class="table"  id="table:ourtable20">
    <div class="centered"><table class="tabular">
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px" 
        rowspan=""
        colspan="">
      <p> \(n\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px" 
        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; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \({r}_{n+1}-{r}_{n}\) </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" 
        rowspan=""
        colspan="">
      <p>\(0\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px" 
        rowspan=""
        colspan="">
      <p> \({3.08364074880371e-02}\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \({2.90926725239779e-02}\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \({3.54228149760742e-02}\) </p>

    </td>
  </tr>
  <tr>
    <td  style="None" 
        rowspan=""
        colspan="">
      <p>\(1\) </p>

    </td>
    <td  style="None" 
        rowspan=""
        colspan="">
      <p> \({6.19878272557754e-03}\) </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \({6.51490076943387e-04}\) </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> \({2.81817144632719e-02}\) </p>

    </td>
  </tr>
  <tr>
    <td  style="None" 
        rowspan=""
        colspan="">
      <p>\(2\) </p>

    </td>
    <td  style="None" 
        rowspan=""
        colspan="">
      <p> \({1.02193763653189e-04}\) </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \({7.47329177011811e-08}\) </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> \({-4.56679057964336e-03}\) </p>

    </td>
  </tr>
  <tr>
    <td  style="None" 
        rowspan=""
        colspan="">
      <p>\(3\) </p>

    </td>
    <td  style="None" 
        rowspan=""
        colspan="">
      <p> \({5.69217297753260e-10}\) </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \({1.16187879922849e-19}\) </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> \({6.40020812159250e-04}\) </p>

    </td>
  </tr>
  <tr>
    <td  style="None" 
        rowspan=""
        colspan="">
      <p>\(4\) </p>

    </td>
    <td  style="None" 
        rowspan=""
        colspan="">
      <p> \({9.87611999985374e-26}\) </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \({4.36625392841961e-55}\) </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> \({-4.79847100295654e-07}\) </p>

    </td>
  </tr>
  <tr>
    <td  style="None" 
        rowspan=""
        colspan="">
      <p>\(5\) </p>

    </td>
    <td  style="None" 
        rowspan=""
        colspan="">
      <p> \({5.15835437807280e-73}\) </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \({2.31714641078614e-161}\) </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> \({2.20353540551787e-16}\) </p>

    </td>
  </tr>
  <tr>
    <td  style="None" 
        rowspan=""
        colspan="">
      <p>\(6\) </p>

    </td>
    <td  style="None" 
        rowspan=""
        colspan="">
      <p> \({7.34997098358734e-215}\) </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \({3.46327466154480e-480}\) </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> \({-2.13374334510703e-44}\) </p>

    </td>
  </tr>
  <tr>
    <td  style="None" 
        rowspan=""
        colspan="">
      <p>\(7\) </p>

    </td>
    <td  style="None" 
        rowspan=""
        colspan="">
      <p> \({2.12622379402663e-640}\) </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \({1.15634469982568e-1436}\) </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> \({1.93735237524841e-128}\) </p>

    </td>
  </tr>
  <tr>
    <td  style="None" 
        rowspan=""
        colspan="">
      <p>\(8\) </p>

    </td>
    <td  style="None" 
        rowspan=""
        colspan="">
      <p> \({5.14729456000340e-1917}\) </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \({0.00000000000000e+00}\) </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> \({-1.45013334596713e-380}\) </p>

    </td>
  </tr>
  <tr>
    <td  style="border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>\(9\) </p>

    </td>
    <td  style="border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \({0.00000000000000e+00}\) </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \({0.00000000000000e+00}\) </p>

    </td>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \({6.08143261073181e-1137}\) </p>

    </td>
  </tr>
</table> <figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">4.</span> 
  <span class="caption_text">Sequences \(\{ t_n\} \), \(\{ {r}_n\} \) and \(\{ v_n\} \).</span> 
</figcaption> </div> 
</div>
<p> In Table <a href="#table:ourtable20">4.</a>, we observe that the sequence \(\{ {r}_n\} \) provides tighter error bounds than the sequence \(\{ t_n\} \). The convergence of the sequence \(\{ v_n\} \) is not expected, since <a href="#eq:14" class="eqref">1.3</a> or <a href="#eq:14" class="eqref">1.3</a> are not satisfied. Note also that \(\{ v_n\} \) was essentially used as a majorizing sequence for the Midpoint method in <span class="cite">
	[
	<a href="#1" >1</a>
	, 
	<a href="#3" >3</a>
	, 
	<a href="#4" >4</a>
	, 
	<a href="#5" >10</a>
	, 
	<a href="#6" >11</a>
	, 
	<a href="#7" >12</a>
	, 
	<a href="#8" >13</a>
	, 
	<a href="#9" >14</a>
	, 
	<a href="#10" >15</a>
	, 
	<a href="#11" >16</a>
	, 
	<a href="#19" >24</a>
	, 
	<a href="#20" >25</a>
	, 
	<a href="#21" >26</a>
	, 
	<a href="#23" >28</a>
	]
</span>. <span class="qed">â–¡</span></p>

  </div>
</div> <div class="example_thmwrapper " id="a0000000042">
  <div class="example_thmheading">
    <span class="example_thmcaption">
    Example
    </span>
    <span class="example_thmlabel">4.2</span>
  </div>
  <div class="example_thmcontent">
  <p>In this example, we provide an application of our results to a special nonlinear Hammerstein integral equation of the second kind. Consider the integral equation </p>
<div class="equation" id="eq:45">
<p>
  <div class="equation_content">
    \begin{equation} \label{eq:45} x(s) = 1 + \tfrac {4}{5} \int _0^1 G(s,t)x(t)^3\mathop{dt}, \quad s\in [0,1], \end{equation}
  </div>
  <span class="equation_label">4.67</span>
</p>
</div>
<p> where, \(G\) is the Green kernel on \([0,1]\times [0,1]\) defined by </p>
<div class="equation" id="eq:46">
<p>
  <div class="equation_content">
    \begin{equation} \label{eq:46} G(s,t) = \left\{  \begin{aligned}  t(1-s), \quad t \le {s};\\ s(1-t), \quad s \le t. \end{aligned} \right. \end{equation}
  </div>
  <span class="equation_label">4.68</span>
</p>
</div>
<p> Let \(\mathbf{X}={\mathbf{Y}}=\mathcal{C}[0,1]\) and \(\mathbf{D}\) be a suitable open convex subset of \(\mathbf{X}_1:=\{ x\in \mathbf{X}: x(s){\gt}0,s\in [0,1]\} \), which will be given below. Define \(\mathcal F:\mathbf{D}\rightarrow {\mathbf{Y}}\) by </p>
<div class="equation" id="eq:47">
<p>
  <div class="equation_content">
    \begin{equation} \label{eq:47} [\mathcal F(x)](s) = x(s)-1-\tfrac {4}{5}\int _0^1 G(s,t)x(t)^3\mathop{dt},\quad s\in [0,1]. \end{equation}
  </div>
  <span class="equation_label">4.69</span>
</p>
</div>
<p> The first and second derivatives of \(\mathcal F\) are given by </p>
<div class="displaymath" id="eq:49">
  \begin{align}  [\mathcal F(x)^\prime y](s) = y(s)-\tfrac {12}{5}\int _0^1 G(s,t)x(t)^2y(t)\mathop{dt},\quad s\in [0,1],\label{eq:48} \intertext {and} [\mathcal F(x)^{\prime \prime }yz](s) = \tfrac {24}{5}\int _0^1 G(s,t)x(t)y(t)z(t)\mathop{dt},\quad s\in [0,1],\label{eq:49} \end{align}
</div>
<p> respectively. We use the max-norm. Let \(x_0(s)=1\) for all \(s\in [0,1]\). Then, for any \(y\in \mathbf{D}\), we have </p>
<div class="displaymath" id="a0000000043">
  \begin{gather}  [(I-\mathcal F^\prime (x_0))(y)](s) = \tfrac {12}{5}\int _0^1G(s,t)y(t)\mathop{dt},\quad s\in [0,1], \intertext {which means} \Vert {I-\mathcal F^\prime (x_0)}\Vert \le \tfrac {12}{5}\max _{s\in [0,1]}\int _0^1G(s,t)\mathop{dt} = \tfrac {12}{5\times 8} = \tfrac {3}{10}{\lt}1. \end{gather}
</div>
<p> It follows from the Banach theorem that \(\mathcal F^\prime (x_0)^{-1}\) exists and </p>
<div class="equation" id="a0000000044">
<p>
  <div class="equation_content">
    \begin{equation}  \Vert {\mathcal F^\prime (x_0)^{-1}}\Vert \le \tfrac {1}{1-\frac{3}{10}} = \tfrac {10}{7}. \end{equation}
  </div>
  <span class="equation_label">4.72</span>
</p>
</div>
<p> On the other hand, we have from <a href="#eq:47" class="eqref">4.69</a> that </p>
<div class="displaymath" id="a0000000045">
  \begin{equation*}  \Vert {\mathcal F(x_0)}\Vert = \tfrac {4}{5} \max _{s\in [0,1]}\int _0^1G(s,t)\mathop{dt} = \tfrac {1}{10}. \end{equation*}
</div>
<p> Then, we get \(\eta =1/7\). Note that \(\mathcal F^{\prime \prime }(x)\) is not bounded in \(\mathbf{X}\) or its subset \(\mathbf{X}_1\). Take into account that a solution \(x^\star \) of equation <a href="#eq:11" class="eqref">1.1</a> with \(\mathcal F\) given by <a href="#eq:46" class="eqref">4.68</a> must satisfy </p>
<div class="equation" id="a0000000046">
<p>
  <div class="equation_content">
    \begin{equation}  \Vert {x^\star }\Vert -1 - \tfrac {1}{10} \Vert {x^\star }\Vert ^3 \le {0}, \end{equation}
  </div>
  <span class="equation_label">4.73</span>
</p>
</div>
<p> i.e., \(\Vert {x^\star }\Vert \le \rho _1 = 1.153467305\) and \(\Vert {x^\star }\Vert \ge \rho _2 = 2.423622140,\) where \(\rho _1\) and \(\rho _2\) are the positive roots of the real equation \(z-1-z^3/10=0\). Consequently, if we look for a solution such that \(x^\star {\lt} \rho _1\in \mathbf{X}_1\), we can consider \(\mathbf{D}:=\{ x:x\in \mathbf{X}_1\quad \textrm{and}\quad \Vert {x}\Vert {\lt}r\} ,\) with \(r\in (\rho _1,\rho _2),\) as a nonempty open convex subset of \(\mathbf{X}\). For example, choose \(r=1.7\). Using <a href="#eq:37" class="eqref">3.47</a> and <a href="#eq:38" class="eqref">3.48</a>, we have that for any \(x,y,z\in \mathbf{D}\) </p>
<div class="displaymath" id="a0000000047">
  \begin{align} & \left\Vert {\left[(\mathcal F^\prime (x)- \mathcal F^\prime (x_0))y\right](s)} \right\Vert = \\ & = \tfrac {12}{5}\left\Vert \int _0^1 G(s,t) (x(t)^2-x_0(t)^2)y(t)\mathop{dt}\right\Vert \nonumber \\ & \le \tfrac {12}{5}\int _0^1 G(s,t) \Vert x(t)-x_0(t)\Vert \, \Vert {x(t)+x_0(t)} \Vert y(t)\mathop{dt}\nonumber \\ & \le \tfrac {12}{5}\int _0^1 G(s,t)\, (r+1) \Vert {x(t) - x_0(t)}\Vert y(t)\mathop{dt}, \quad s\in [0,1]\nonumber \intertext {and} \Vert {(F^{\prime \prime }(x)yz)(s)}\Vert & = \tfrac {24}{5} \int _0^1 G(s,t)x(t)y(t)z(t)\mathop{dt}, \quad s\in [0,1]. \end{align}
</div>
<p> Then, we get </p>
<div class="displaymath" id="a0000000048">
  \begin{align}  \Vert {\mathcal F^\prime (x)-\mathcal F^\prime (x_0)} \Vert &  \le \tfrac { 12}{5} \tfrac {1}{8}(r+1)\Vert {x-x_0}\Vert = \tfrac {81}{100}\Vert {x-x_0}\Vert ,\\ \Vert {F^{\prime \prime }(x)}\Vert &  \le \tfrac {24}{5}\times \tfrac {r}{8} = \tfrac {51}{50} \end{align}
</div>
<p> and </p>
<div class="displaymath" id="a0000000049">
  \begin{align}  \left\Vert \left[\left[ {F^{\prime \prime }(x) - \mathcal F^{\prime \prime }(\overline{x}) } \right] yz\right](s) \right\Vert & = \tfrac {24}{5} \left\Vert { \int _0^1 {G(s,t)} \left( x(t)-\overline{x}(t) \right))y(t)z(t) }\right\Vert \mathop{dt}\\ & \le \tfrac {24}{5} \tfrac {1}{8}\Vert {x-\overline{x}}\Vert = \tfrac {3}{5 }\Vert {x-\overline{x}}\Vert . \end{align}
</div>
<p> Now we can choose constants as follows: </p>
<div class="displaymath" id="a0000000050">
  \begin{equation*}  \mathcal{M}= \tfrac {6}{7},\quad {L}= \tfrac {51}{35}, \quad {L}_0 =\tfrac {81}{70},\quad \textrm{and}\quad \eta = \tfrac {1}{7}. \end{equation*}
</div>
<p> From equations <a href="#eq:24" class="eqref">2.10</a> – <a href="#eq:26" class="eqref">2.11</a>, we obtain </p>
<div class="displaymath" id="a0000000051">
  \begin{equation*}  \eta _1= {0.5775706579},\quad \eta _2 = {0.4268353213}, \quad \eta _3 = {0.4076031248}. \end{equation*}
</div>
<p> Thus from <a href="#eq:27" class="eqref">2.12</a> </p>
<div class="displaymath" id="a0000000052">
  \begin{equation*}  \eta _0 = \eta _2 = {0.4268353213}. \end{equation*}
</div>
<p> Since \(\eta _0 \neq \eta _1\). Thus from <a href="#eq:212" class="eqref">2.16</a>, we get </p>
<div class="displaymath" id="a0000000053">
  \begin{equation*}  \tfrac {1}{7}\le 0.3473064574. \end{equation*}
</div>
<p> Thus, the assumption <a href="#eq:212" class="eqref">2.16</a> holds. Furthermore, it can be checked that the criteria <a href="#eq:14" class="eqref">1.3</a> (\({0.1428571429}{\lt}{0.3070646192}\)) and <a href="#eq:14" class="eqref">1.3</a> (\({0.1428571429}{\lt}{0.3036094577}\)) also hold. Likewise we select the constants </p>
<div class="displaymath" id="a0000000054">
  \begin{equation*}  \mathcal{K}_0 = \tfrac {51}{50},\quad \mathcal{M}_0 = \tfrac {4}{9}, \quad {L}_1 = \tfrac {51}{50}, \quad {L}_2 = \tfrac {52}{55}, \quad {L}_2^\prime = \tfrac {50}{45}. \end{equation*}
</div>
<p> <p>We can verify that the conditions <a href="#eq:237" class="eqref">2.36</a> are fulfilled. Additionally, to verify the criteria <a href="#eq:233" class="eqref">2.33</a> and check the convergence of the sequences \(\{ t_n\} \) (given by <a href="#eq:212" class="eqref">2.16</a>), \(\{ r_n\} \) (given by <a href="#eq:236" class="eqref">2.35</a>) and \(\{ v_n\} \) (given by <a href="#eq:238" class="eqref">2.37</a>), we produce the Table <a href="#table:ourtable3">4.</a>. </p>

<div class="table"  id="table:ourtable3">
    <div class="centered"><table class="tabular">
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px" 
        rowspan=""
        colspan="">
      <p> \(n\) </p>

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

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px" 
        rowspan=""
        colspan="">
      
<p>\(0\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px" 
        rowspan=""
        colspan="">
      <p> \({1.50961386e-01}\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \({1.54802014e-01}\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \({1.59065628e-01}\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \({1.42857143e-01}\) </p>

    </td>
  </tr>
  <tr>
    <td  style="None" 
        rowspan=""
        colspan="">
      <p>\(1\) </p>

    </td>
    <td  style="None" 
        rowspan=""
        colspan="">
      <p> \({2.98912541e-03}\) </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \({1.68915209e-03}\) </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> \({3.40423355e-03}\) </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> \({2.98518380e-03}\) </p>

    </td>
  </tr>
  <tr>
    <td  style="None" 
        rowspan=""
        colspan="">
      <p>\(2\) </p>

    </td>
    <td  style="None" 
        rowspan=""
        colspan="">
      <p> \({2.89675336e-08}\) </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \({2.33411706e-09}\) </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> \({4.63079794e-08}\) </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> \({2.89675332e-08}\) </p>

    </td>
  </tr>
  <tr>
    <td  style="None" 
        rowspan=""
        colspan="">
      <p>\(3\) </p>

    </td>
    <td  style="None" 
        rowspan=""
        colspan="">
      <p> \({2.64962951e-23}\) </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \({6.18331262e-27}\) </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> \({1.17578438e-22}\) </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> \({2.64962951e-23}\) </p>

    </td>
  </tr>
  <tr>
    <td  style="None" 
        rowspan=""
        colspan="">
      <p>\(4\) </p>

    </td>
    <td  style="None" 
        rowspan=""
        colspan="">
      <p> \({2.02771106e-68}\) </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \({1.14952103e-79}\) </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> \({1.92461253e-66}\) </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> \({2.02771106e-68}\) </p>

    </td>
  </tr>
  <tr>
    <td  style="None" 
        rowspan=""
        colspan="">
      <p>\(5\) </p>

    </td>
    <td  style="None" 
        rowspan=""
        colspan="">
      <p> \({9.08801065e-204}\) </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \({7.38591110e-238}\) </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> \({8.44092751e-198}\) </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> \({9.08801065e-204}\) </p>

    </td>
  </tr>
  <tr>
    <td  style="None" 
        rowspan=""
        colspan="">
      <p>\(6\) </p>

    </td>
    <td  style="None" 
        rowspan=""
        colspan="">
      <p> \({8.18195521e-610}\) </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \({1.95914029e-712}\) </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> \({7.12083230e-592}\) </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> \({8.18195521e-610}\) </p>

    </td>
  </tr>
  <tr>
    <td  style="None" 
        rowspan=""
        colspan="">
      <p>\(7\) </p>

    </td>
    <td  style="None" 
        rowspan=""
        colspan="">
      <p> \({5.97065407e-1828}\) </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \({0.00000000e+00}\) </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> \({4.27516145e-1774}\) </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> \({5.97065407e-1828}\) </p>

    </td>
  </tr>
  <tr>
    <td  style="None" 
        rowspan=""
        colspan="">
      <p>\(8\) </p>

    </td>
    <td  style="None" 
        rowspan=""
        colspan="">
      <p> \({0.00000000e+00}\) </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \({0.00000000e+00}\) </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> \({0.00000000e+00}\) </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> \({0.00000000e+00}\) </p>

    </td>
  </tr>
  <tr>
    <td  style="border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>\(9\) </p>

    </td>
    <td  style="border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \({0.00000000e+00}\) </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \({0.00000000e+00}\) </p>

    </td>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \({0.00000000e+00}\) </p>

    </td>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \({0.00000000e+00}\) </p>

    </td>
  </tr>
</table> <figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">4.</span> 
  <span class="caption_text">Sequences \(\{ t_n\} \), \(\{ {r}_n\} \) and \(\{ v_n\} \).</span> 
</figcaption> </div> 
</div>
 </p>
<p>In the Table <a href="#table:ourtable3">4.</a>, we observe that the sequence \(\{ {r}_n\} \) provides tighter error bounds than sequences \(\{ t_n\} \) and \(\{ v_n\} \). This is also true by <a href="#eq:239" class="eqref">2.38</a>. Additionally, we notice in Table <a href="#table:ourtable3">4.</a> that the criterion <a href="#eq:233" class="eqref">2.33</a> holds. That is \(s_n-t_n \le \eta _0\). </p>
<p>Concerning the uniqueness balls, let us denote the radii corresponding to <a href="#eq:321" class="eqref">3.59</a>, <a href="#eq:14" class="eqref">1.3</a><span class="cite">
	[
</span>]1,3,4,18,19,20,21 and <a href="#eq:14" class="eqref">1.3</a> <span class="cite">
	[
</span>]7,9 by \(\gamma _1\), \(\gamma _2\) and \(\gamma _3\), respectively. These are given as the smallest positive roots of the polynomials </p>
<div class="displaymath" id="a0000000055">
  \begin{align}  p_1(t) & = {L}_0\, t-1 \quad (\textrm{for}\quad t^\star = R),\\ p_2(t) & = \tfrac {\mathcal{K}}{2}t^2 -t + \eta , \intertext {and} p_3(t) & = \tfrac {\mathcal{M}}{3}t^3 + \tfrac {{L}}{2}t^2-t+\eta \end{align}
</div>
<p> respectively. Using the values of \({L}_0\), \({L}\), \(\mathcal{M}\) and \(\eta \) we get </p>
<div class="equation" id="a0000000056">
<p>
  <div class="equation_content">
    \begin{equation}  \gamma _1 = {0.8641975309},\quad {{\gamma _2= {0.1677113044}}},\quad \gamma _3 = {0.1636113702}. \end{equation}
  </div>
  <span class="equation_label">4.81</span>
</p>
</div>
<p> Here, \(\mathcal{K}= {1.767276239}\). Note that \(\overline{U}(x_0,r-1)\subseteq \mathbf{D}\), \({L}_0{\lt}{L}\) and \(\gamma _3{\lt}\gamma _2{\lt}\gamma _1\). Therefore, the new approach provides the largest uniqueness ball and since \(r-1{\lt}\gamma _1\), we deduce that \(x^\star \) is unique in \(\overline{U}(x_0,r-1)=\overline{U}(1,0.7)\subseteq \mathbf{D}\). </p>

  </div>
</div> </p>
<p><small class="footnotesize">  </small></p>
<div class="bibliography">
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</dd>
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</dd>
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</a> </p>
</dd>
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</a> </p>
</dd>
  <dt><a name="25">30</a></dt>
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</a> </p>
</dd>
  <dt><a name="26">31</a></dt>
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</dd>
  <dt><a name="27">32</a></dt>
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</dd>
  <dt><a name="28">33</a></dt>
  <dd><p><a href ="http://dx.doi.org/10.1016/0377-0427(88)90389-5"><i class="sc">T. Yamamoto,</i> <i class="it">On the method of tangent hyperbolas in Banach spaces,</i> J. Comput. Appl. Math., <b class="bf">21</b> (1988) no. 1, 75–86 <img src="img-0002.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
</dl>


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