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<title>On The Semi-Local Convergence <br />Of A Sixth Order Method In Banach Space: On The Semi-Local Convergence <br />Of A Sixth Order Method In Banach Space</title>
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<h1>On The Semi-Local Convergence <br />Of A Sixth Order Method In Banach Space</h1>
<p class="authors">
<span class="author">Ioannis K. Argyros\(^\ast \), Jinny Ann John\(^{\ast \ast }\) Jayakumar Jayaraman\(^{\ast \ast }\)</span>
</p>
<p class="date">November 8, 2022; accepted: December 18, 2022; published online: December 31, 2022.</p>
</div>
<div class="abstract"><p> High convergence order methods are important in computational mathematics, since they generate sequences converging to a solution of a non-linear equation. The derivation of the order requires Taylor series expansions and the existence of derivatives not appearing on the method. Therefore, these results cannot assure the convergence of the method in those cases when such high order derivatives do not exist. But, the method may converge. In this article, a process is introduced by which the semi-local convergence analysis of a sixth order method is obtained using only information from the operators on the method. Numerical examples are included to complement the theory. </p>
<p><b class="bf">MSC.</b> 37N30, 47J25, 49M15, 65H10, 65J15. </p>
<p><b class="bf">Keywords.</b> Non-linear equations, Fréchet derivative, Semi-Local convergence, Banach Space. </p>
</div>
<p>\(^\ast \) Department of Computing and Mathematical Sciences, Cameron University, Lawton, 73505, OK, USA, e-mail: <span class="tt">iargyros@cameron.edu</span>. </p>
<p>\(^{\ast \ast }\) Department of Mathematics, Puducherry Technological University, Pondicherry, India 605014, e-mail: <span class="tt">jinny3@pec.edu, jjayakumar@ptuniv.edu.in</span>. </p>
<h1 id="sec1">1 Introduction</h1>

<p>Finding a locally unique solution of the system of non-linear equations of the form </p>
<div class="equation" id="e:1">
<p>
  <div class="equation_content">
    \begin{equation} \label{e:1} G(x)=0 \end{equation}
  </div>
  <span class="equation_label">1</span>
</p>
</div>
<p> is a major problem with extensive applications in the field of mathematical and engineering sciences. Presently, there are numerous efficient methods to solve (<a href="#e:1">1</a>) <span class="cite">
	[
	<a href="#2" >1</a>
	, 
	<a href="#3" >9</a>
	, 
	<a href="#4" >8</a>
	, 
	<a href="#5" >17</a>
	, 
	<a href="#6" >13</a>
	, 
	<a href="#7" >16</a>
	, 
	<a href="#8" >10</a>
	, 
	<a href="#9" >14</a>
	, 
	<a href="#10" >18</a>
	]
</span>. But, in most of the cases non-linear equations and systems arising from mathematical modeling of physical systems does not have exact solutions. Because of this problem, scientists and researchers have focused on proposing iterative methods for solving non-linear systems. Newton’s method is a popular iterative process for dealing with non-linear equations. Many novel, higher-order iterative strategies for dealing with nonlinear equations have been discovered and are currently being used in recent years <span class="cite">
	[
	<a href="#2" >1</a>
	, 
	<a href="#3" >9</a>
	, 
	<a href="#4" >8</a>
	, 
	<a href="#5" >17</a>
	, 
	<a href="#6" >13</a>
	, 
	<a href="#7" >16</a>
	, 
	<a href="#8" >10</a>
	, 
	<a href="#9" >14</a>
	, 
	<a href="#10" >18</a>
	, 
	<a href="#11" >3</a>
	, 
	<a href="#12" >4</a>
	, 
	<a href="#15" >7</a>
	, 
	<a href="#18" >15</a>
	]
</span>. However, the theorems on convergence of these schemes in most of these publications are derived by applying high order derivatives. Furthermore, no results are discussed regarding the error distances, radii of convergence or the region in which the solution is unique. </p>
<p>The study of local and semi-local analysis of an iterative formula allows to estimate the convergence balls, bounds on error and uniqueness region for a solution. The results of local and semi-local convergence of efficient iterative procedures have been deduced in <span class="cite">
	[
	<a href="#11" >3</a>
	, 
	<a href="#13" >5</a>
	, 
	<a href="#14" >6</a>
	, 
	<a href="#15" >7</a>
	, 
	<a href="#16" >11</a>
	, 
	<a href="#17" >2</a>
	]
</span>. In these works, important results containing convergence radii, measurements on error estimates and expanded utility of these iterative approaches have been given. Outcomes of these type of analysis are valuable because they illustrate the complexity of selecting initial points. </p>
<p>In this article, we develop semi-local convergence theorem for a method with sixth order convergence proposed in <span class="cite">
	[
	<a href="#2" >1</a>
	]
</span>. The method can be stated as </p>
<div class="equation" id="e:2">
<p>
  <div class="equation_content">
    \begin{equation} \label{e:2} \begin{split}  y_n &  = x_n- G’(x_n)^{-1}G(x_n) \\ z_n &  = y_n - G’(y_n)^{-1}G(y_n) \\ x_{n+1} &  = z_n - G’(y_n)^{-1}G(z_n), \qquad x_0\in \Omega , \forall n = 0, 1, 2,\ldots \end{split} \end{equation}
  </div>
  <span class="equation_label">2</span>
</p>
</div>
<p> where \(G: \Omega \subset B_1 \to B_2\) is a Fréchet differentiable and continuous non-linear operator, \(B_1\) and \(B_2\) are Banach spaces and \(\Omega \neq \phi \) is a convex and open set. Let \(x^*\) represent a root of the equation (<a href="#e:1">1</a>) which is locally unique. </p>
<p>The local convergence of method (<a href="#e:2">2</a>) was established assuming that the seventh derivative (at least) of the operator \(G\) exists. As a consequence, productivity of this method is limited. To see this, we define \({G}\) on \(\Omega =[-0.5, 1.5]\) by </p>
<div class="equation" id="a0000000002">
<p>
  <div class="equation_content">
    \begin{equation}  {G}(t)= \left\{  \begin{array}{ll} t^3 \ln (t^2) + t^5 -t^4, &  \hbox{if~ ~  $ t \not= 0$} \\ 0, &  \hbox{if~ ~  $t=0$} \end{array} \right. \end{equation}
  </div>
  <span class="equation_label">6</span>
</p>
</div>
<p> Now, it is easy to find that due to the unboundedness of \({G}'''\) the results on convergence of (<a href="#e:2">2</a>) does not hold for this example. Also, previous research articles do not produce any formula for approximating the error \(\| x_n-x^*\| \), the convergence region or the uniqueness and accurate location of the root \(x^*\). This is the major motivation for developing the ball convergence theorems by considering assumptions only on \({G}'\). Our research provides important formulas for the estimation of \(\| x_n-x^*\| \) and convergence radii. This study also discusses about an exact location and the uniqueness of \(x^*\). </p>
<p>The other contents of this material can be summarized as follows: <a href="#sec2">section 2</a> discusses the development of majorizing sequence for the method (<a href="#e:2">2</a>). <a href="#sec3">Section 3</a> discusses the semi-local convergence properties of the prsented method (<a href="#e:2">2</a>). Numerical testing of convergence outcomes are placed in <a href="#sec4">section 4</a>. Concluding remarks are also stated. </p>
<h1 id="sec2">2 Majorizing Sequence</h1>

<p>A scalar sequence is introduced in this section that is shown to be majorizing for the method (<a href="#e:2">2</a>) in <a href="#sec3">section 3</a>. Let \(M=\left[0,+\infty \right)\). </p>
<p>Suppose: </p>
<p>There exists a function \(\phi _0:M \to \mathbb {R}\) which is continuous and non-decreasing such that the equation </p>
<div class="displaymath" id="a0000000003">
  \begin{equation*}  \phi _0(t)-1=0 \end{equation*}
</div>
<p> has a smallest positive solution. Denote such a solution by \(\alpha \). Set \(M_0=[0,\alpha ]\). Let \(\phi :M_0 \to \mathbb {R}\) be a continuous and non-decreasing function. Let \(d \ge 0\) denote a given parameter. </p>
<p>Define the scalar sequences \(\{ t_n\} \), \(\{ s_n\} \) and \(\{ v_n\} \) for each \(n=0,1,2,\ldots \) by </p>
<div class="equation" id="2.1">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.1} \begin{split}  t_0 &  = 0, \quad s_0 = d,\\ a_n &  = \int _{0}^{1} \phi (\lambda (s_n-t_n))d\lambda (s_n-t_n), \\ v_n &  = s_n + \frac{a_n}{1-\phi _0(s_n)}, \\ b_n &  = (1+\int _{0}^{1}\phi _0(s_n+\lambda (v_n-s_n))d\lambda )(v_n-s_n) + a_n, \\ t_{n+1} &  = v_n + \frac{b_n}{1-\phi _0(s_n)}, \\ c_{n+1} &  = \int _{0}^{1}\phi (\lambda (t_{n+1}-t_n))d\lambda (t_{n+1}-t_n)+(1+\phi _0(t_n))(t_{n+1}-s_n), \\ s_{n+1} &  = t_{n+1} +\frac{c_{n+1}}{1-\phi _0(t_{n+1})}. \end{split} \end{equation}
  </div>
  <span class="equation_label">7</span>
</p>
</div>
<p>Next, we develop a general convergence result for the sequence \(\{ t_n\} \) given by the formula (<a href="#2.1">7</a>). <div class="lemma_thmwrapper " id="l1">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">1</span>
  </div>
  <div class="lemma_thmcontent">
  <p> Suppose that for each \(n=0,1,2,\ldots \) there exists a parameter \(\delta {\gt}0\) such that </p>
<div class="equation" id="2.2">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.2} \phi _0(s_n)<1, \quad \phi _0(t_{n+1})<1 \quad \text{and } t_{n+1} \le \delta . \end{equation}
  </div>
  <span class="equation_label">15</span>
</p>
</div>
<p> Then, the following assertions hold </p>
<div class="equation" id="2.3">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.3} 0\le t_n \le s_n \le v_n \le t_{n+1} \quad \text{and } \lim _{n \to \infty } t_n = t^* \le \delta . \end{equation}
  </div>
  <span class="equation_label">16</span>
</p>
</div>

  </div>
</div> <div class="proof_wrapper" id="a0000000004">
  <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 definition of the sequences \(\{ t_n\} \), \(\{ s_n\} \), \(\{ v_n\} \) and the condition (<a href="#2.2">15</a>) imply that the sequence \(\{ t_n\} \) is bounded above by \(\lambda \) and non-decreasing. Hence, it converges to its unique least upper bound \(t^*\) satisfying \(t^* \in [0,\delta ]\). <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> </p>
<p><div class="remark_thmwrapper " id="R1">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">2</span>
  </div>
  <div class="remark_thmcontent">
  
<ul class="itemize">
  <li><p>If the function \(\phi _0\) is strictly increasing, then we can set \(\delta =\phi _0^{-1}(1)\). </p>
</li>
  <li><p>Another possible choice for \(\delta \) is any number \(\gamma _0 \in [0,\delta ]\) or \(\gamma _0 \in \left[0,\alpha \right)\).<span class="qed">â–¡</span></p>
</li>
</ul>

  </div>
</div> </p>
<h1 id="sec3">3 Convergence Analysis</h1>

<p>The semi-local convergence analysis of the method (<a href="#e:2">2</a>) is developed in this section. For this, we need an auxiliary Ostrowski-type result for the method (<a href="#e:2">2</a>), which is stated below. <div class="lemma_thmwrapper " id="l2">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">3</span>
  </div>
  <div class="lemma_thmcontent">
  <p> Suppose that the iterates \(\{ x_n\} \), \(\{ y_n\} \) and \(\{ z_n\} \) appearing in the method <a href="#e:2" class="eqref">2</a> exist for each \(n=0,1,2,\ldots \). Then, the following assertions hold: </p>
<div class="displaymath" id="3.1">
  \begin{align}  G(y_n) &  =p_n= \int _{0}^{1}(G’(x_n+\lambda (y_n-x_n))d\lambda -G’(x_n))(y_n-x_n),\label{3.1}\\ z_n-y_n &  = -G(y_n)^{-1} p_n, \label{3.2}\\ q_n &  = \int _{0}^{1}G’(y_n+\lambda (z_n-y_n))d\lambda (z_n-y_n), \label{3.3}\\ G(z_n)&  =q_n+p_n, \label{3.4}\\ x_{n+1}-z_n &  =-G’(y_n)^{-1}(q_n+p_n), \label{3.5}\\ G(x_{n+1}) &  = \delta _{n+1}= \! \! \int _{0}^{1}\! \! (G’(x_n\! +\! \lambda (x_{n+1}\! -\! x_n))\! -\! G’(x_n))d\lambda (x_{n+1}-x_n) \\ &  \qquad + G’(x_n)(x_{n+1}-y_n) \  \  \text{and} \label{3.6}\\ y_{n+1}-x_{n+1} &  =-G’(x_{n+1})\delta _{n+1}. \label{3.7} \end{align}
</div>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000006">
  <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 write by the second sub-step of the method (<a href="#e:2">2</a>) in turn that </p>
<div class="displaymath" id="a0000000007">
  \begin{equation*} \begin{split}  G(y_n) &  = G(y_n) - G(x_n)-G’(x_n)(y_n-x_n) \\ &  = \int _{0}^{1} (G’(x_n+\lambda (y_n-x_n))d\lambda -G’(x_n))(y_n-x_n)=p_n \end{split}\end{equation*}
</div>
<p> showing (<a href="#3.1">17</a>) and consequently (<a href="#3.2">18</a>). </p>
<p>Moreover, by using (<a href="#3.1">17</a>), (<a href="#3.3">19</a>) and the following identity, </p>
<div class="displaymath" id="a0000000008">
  \begin{equation*} \begin{split}  G(z_n) &  = G(z_n)-G(y_n)+G(y_n) \\ &  = \int _{0}^{1}G’(y_n+\lambda (z_n-y_n))d\lambda (z_n-y_n) + G(y_n) \end{split}\end{equation*}
</div>
<p> we obtain (<a href="#3.4">20</a>). </p>
<p>Hence, the identity (<a href="#3.5">21</a>) holds by the third sub-step of the method (<a href="#e:2">2</a>) and (<a href="#3.4">20</a>). </p>
<p>Furthermore, by the first sub-step of the method (<a href="#e:2">2</a>), we get (<a href="#3.6">23</a>) as follows: </p>
<div class="displaymath" id="a0000000009">
  \begin{equation*} \begin{split} & G(x_{n+1}) = \\ &  = G(x_{n+1})\! -\! G(x_n)\! -\! G’(x_n)(y_n\! -\! x_n) \! -\!  G’(x_n)(x_{n+1}\! -\! x_n)\! +\! G’(x_n)(x_{n+1}\! -\! x_n)\\ &  = (G(x_{n+1})-G(x_n)-G’(x_n)(x_{n+1}-x_n)) + G’(x_n)(x_{n+1}-y_n) \\ &  = \int _{0}^{1}(G’(x_n+\lambda (x_{n+1}-x_n))d\lambda - G’(x_n))(x_{n+1}-x_n)+ G’(x_n)(x_{n+1}-y_n) \\ &  = \delta _{n+1}. \end{split}\end{equation*}
</div>
<p> Finally, the identity (<a href="#3.7">24</a>) follows from the first sub-step of the method (<a href="#e:2">2</a>) and (<a href="#3.6">23</a>). <div class="proof_wrapper" id="a0000000010">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<p>The following conditions are required to prove the semi-local convergence analysis of the method (<a href="#e:2">2</a>). Further, these conditions that guarantee the existence of the iterates \(\{ x_n\} \), \(\{ y_n\} \) and \(\{ z_n\} \). Suppose: </p>
<ul class="itemize">
  <li></li>
  <li> </li>
  <li><p>There exists an element \(x_0 \in \Omega \) and a parameter \(d\ge 0\) such that \(G'(x_0)^{-1} \in \mathscr {L}(B_2,B_1)\) and \(\| G'(x_0)^{-1}G(x_0)\|  \le d\). </p>
</li>
  <li><p>\(\| G'(x_0)^{-1}(G'(u)-G'(x_0))\|  \le \phi _0(\| u-x_0\| )\) for each \(u \in \Omega \).<br />Set \(\Omega _0 = U(x_0,\alpha ) \cap \Omega \). </p>
</li>
  <li><p>\(\| G'(x_0)^{-1}(G'(u_2)-G'(u_1))\|  \le \phi (\| u_2-u_1\| )\) for each \(u_1,u_2 \in \Omega _0\). </p>
</li>
  <li><p>The conditions (<a href="#2.2">15</a>) hold and </p>
</li>
  <li><p>\(U[x_0,t^*] \subset \Omega \), where the parameter \(t^*\) is given in the <a href="#l1">lemma 1</a>. </p>
</li>
</ul>
<p>Next, we state and prove the semi-local convergence result for method (<a href="#e:2">2</a>). </p>
<p><div class="theorem_thmwrapper " id="th1">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">4</span>
  </div>
  <div class="theorem_thmcontent">
  <p> Suppose that the conditions (\(H_1\))–(\(H_5\)) hold. Then, the sequences \(\{ x_n\} \), \(\{ y_n\} \), \(\{ z_n\} \) generated by the method <a href="#e:2" class="eqref">2</a> are well-defined, remain in \(U[x_0,t^*]\) and converge to a solution \(x^* \in U[x_0,t^*]\) of the equation \(G(x)=0\). Moreover, the following assertions hold </p>
<div class="displaymath" id="3.8">
  \begin{align}  \| y_n-x_n\|  &  \le s_n-t_n, \label{3.8} \\ \| z_n-y_n\|  &  \le v_n-s_n, \label{3.9} \\ \| x_{n+1}-z_n\|  &  \le t_{n+1}-v_n \label{3.10} \quad \textrm{and }\\ \| x^* - x_n\|  &  \le t^*-t_n. \label{3.11} \end{align}
</div>

  </div>
</div> </p>
<p><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> Mathematical induction on \(n\) is applied to first show that the iterates \(\{ x_k\} \), \(\{ y_k\} \), \(\{ z_k\} \) exist and the estimates (<a href="#3.8">34</a>)–(<a href="#3.10">36</a>) hold. </p>
<p>It follows by the condition \((H_1)\) and the first sub-step of the method (<a href="#e:2">2</a>) for \(n=0\) that </p>
<div class="displaymath" id="a0000000012">
  \begin{equation*}  \| y_0-x_0\|  = \| G’(x_0)^{-1}G(x_0)\|  \le d = s_0-t_0 = s_0 {\lt}t^*, \end{equation*}
</div>
<p> so, the estimate (<a href="#3.8">34</a>) holds for \(n=0\). Let \(w \in U(x_0,t^*)\). Then, by applying the condition (\(H_2\)), we get in turn that </p>
<div class="equation" id="3.12">
<p>
  <div class="equation_content">
    \begin{equation} \label{3.12} \| G'(x_0)^{-1}(G'(w)-G'(x_0))\|  \le \phi _0(\| w-x_0\| ) \le \phi _0(t^*) <1. \end{equation}
  </div>
  <span class="equation_label">38</span>
</p>
</div>
<p>The estimate (<a href="#3.12">38</a>) together with the Banach lemma on linear operators that have inverses <span class="cite">
	[
	<a href="#1" >12</a>
	]
</span> imply that \(G'(w) \in \mathscr {L}(B_2,B_1)\) and </p>
<div class="equation" id="3.13">
<p>
  <div class="equation_content">
    \begin{equation} \label{3.13} \| G'(w)^{-1}G'(x_0)\|  \le \tfrac {1}{1-\phi _0(\| w-x_0\| )}. \end{equation}
  </div>
  <span class="equation_label">39</span>
</p>
</div>
<p>In particular, for \(w=y_0, G'(y_0)^{-1} \in \mathscr {L}(B_2,B_1)\). Consequently, the iterates \(z_0\) and \(x_1\) exist by the second and the third sub-step of the method (<a href="#e:2">2</a>). Then, by <a href="#l1">lemma 1</a>, (\(H_3\)), (<a href="#3.13">39</a>) for \(w=y_0\) and (<a href="#2.1">7</a>), we obtain in turn </p>
<div class="displaymath" id="a0000000013">
  \begin{align*}  \| z_k-y_k\|  &  \le \| G’(y_k)^{-1}G’(x_0)\| \| G’(x_0)^{-1}G(y_k)\|  \\ &  \le \frac{\int _{0}^{1}\phi (\lambda \| y_k-x_k\| )d\lambda \| y_k-x_k\| }{1-\phi _0(\| y_k-x_0\| )} \\ &  \le \frac{\int _{0}^{1}\phi (\lambda (s_k-t_k))d\lambda (s_k-t_k)}{1-\phi _0(s_k)} = v_k-s_k, \\ \| z_k-x_0\|  &  \le \| z_k-y_k\| +\| y_k-x_0\|  \\ &  \le v_k-s_k+s_k-t_0= v_k \le t^*, \\ \| x_{k+1}-z_k\|  & \le \| G’(y_k)^{-1}G’(x_0)\| \| G’(x_0)^{-1}G(z_k)\|  \\ &  \quad \times \bigg( \frac{(1+\int _{0}^{1}\phi _0(\| y_k-x_0\| +\lambda \| z_k-y_k\| )d\lambda )\| z_k-y_k\| }{1-\phi _0(s_k)} \\ &  \quad + \frac{\int _{0}^{1}\phi (\lambda \| y_k-x_k\| )d\lambda \| y_k-x_k\| }{1-\phi _0(s_k)} \bigg) \\ &  \le \tfrac {b_k}{1-\phi _0(s_k)}, \\ \| x_{k+1}-x_0\|  &  \le \| x_{k+1}-y_k\| +\| y_k-x_0\|  \\ &  \le t_{k+1}-s_k+s_k-t_0 = t_{k+1} \le t^*, \\ \| y_{k+1}-x_{k+1}\|  &  \le \| G’(y_k+1)^{-1}G’(x_0)\| \| G’(x_0)^{-1}G(x_{k+1})\|  \\ &  \le \tfrac {c_{k+1}}{1-\phi _0(t_{k+1})}= s_{k+1}-t_{k+1}, \\ \| y_{k+1}-x_0\|  &  \le \| y_{k+1}-x_{k+1}\|  + \| x_{k+1}-x_0\| \\ &  \le s_{k+1}-t_{k+1}+t_{k+1}-t_0 = s_{k+1} \le t^*, \end{align*}
</div>
<p> where we also used </p>
<div class="displaymath" id="a0000000014">
  \begin{align*} &  \bigg\| \int _{0}^{1}G’(x_0)^{-1}(G’(y_k+\lambda (z_k-y_k))d\lambda )\bigg\|  = \\ &  = \bigg\| G’(x_0)^{-1}(G’(y_k+\lambda (z_k-y_k))d\lambda - G’(x_0)+G’(x_0)) \bigg\|  \\ &  \le 1+ \int _{0}^{1}\phi _0(\| y_k-x_0\| +\lambda \| z_k-y_k\| )d\lambda & \\ &  \le 1+ \int _{0}^{1}\phi _0(s_k+\lambda (v_k-s_k))d\lambda & \end{align*}
</div>
<p> and </p>
<div class="displaymath" id="a0000000015">
  \begin{equation*} \begin{split}  \| G’(x_0)^{-1}G(x_{k+1})\|  &  \le \bigg\| \| \int _{0}^{1}G’(x_0)^{-1}(G’(x_k+\lambda (x_{k+1}-x_k))d\lambda \\ &  - G’(x_k))(x_{k+1}-x_k)\bigg\| \|  \\ &  + \| G’(x_0)^{-1}(G’(x_k)-G(x_0)+G’(x_0))(x_{k+1}-y_k)\| \\ &  \le \int _{0}^{1}\phi (\lambda \| x_{k+1}-x_k\| )d\lambda \| x_{k+1}-x_k\|  \\ &  \qquad + (1+\phi _0(\| x_k-x_0\| ))\| x_{k+1}-y_k\|  \\ &  \le \delta _{k+1}. \end{split}\end{equation*}
</div>
<p>Therefore, the assertions (<a href="#3.8">34</a>)–(<a href="#3.10">36</a>) hold and the iterates \(\{ x_k\} \), \(\{ y_k\} , \{ z_k\}  \in U[x_0,t^*]\). Moreover, it follows that the sequence \(\{ x_k\} \) is fundamental in a Banach space \(B_1\) and as such it converges to some \(x^* \in U[x_0,t^*]\). </p>
<p>By letting \(k \to \infty \) in the estimate \(\| G'(x_0)^{-1}G(x_{k+1})\|  \le \delta _{k+1}\) and using the continuity of \(G\), we deduce that \(G(x^*)=0\). Let \(m\ge 0\) be an integer. Then, we can write in turn that </p>
<div class="displaymath" id="a0000000016">
  \begin{align*}  \| x_{n+m}-x_n\|  &  \le \| x_{n+m}-x_{n+m-1}\| +\| x_{n+m-1}-x_n\|  \\ &  \le t_{n+m}-t_{n+m-1}+t_{n+m-1}-t_n \\ &  \le \ldots \le t_{n+m}-t_n. \end{align*}
</div>
<p>By letting \(m\to \infty \) in the preceding estimate we show the assertion (<a href="#3.11">37</a>). <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> </p>
<p>Next, the uniqueness of the solution \(x^*\) in a certain region is determined. </p>
<p><div class="proposition_thmwrapper " id="p1">
  <div class="proposition_thmheading">
    <span class="proposition_thmcaption">
    Proposition
    </span>
    <span class="proposition_thmlabel">5</span>
  </div>
  <div class="proposition_thmcontent">
  <p> Suppose: </p>
<ul class="itemize">
  <li><p>There exists a solution \(\Lambda \in U(x_0,\rho )\) of the equation \(G(x)=0\) for some \(\rho {\gt}0\). </p>
</li>
  <li><p>The condition (\(H_2\)) holds on \(U(x_0,\rho )\). </p>
</li>
  <li><p>There exists \(R\ge \rho \) such that </p>
<div class="displaymath" id="a0000000018">
  \begin{equation*}  \int _{0}^{1}\phi _0((1-\lambda )\rho +\lambda R)d\lambda {\lt}1. \end{equation*}
</div>
<p> Set \(\Omega _1= U[x_0,R] \cap \Omega \). </p>
</li>
</ul>
<p>Then, the equation \(G(x)=0\) is uniquely solvable by \(\Lambda \) in the region \(\Omega _1\). </p>

  </div>
</div> <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> Let \(A=\int _{0}^{1}G'(\Lambda +\lambda (\Lambda _1-\Lambda ))d\lambda \) for some \(\Lambda _1 \in \Omega _1\) with \(G(\Lambda _1)=0\). Then, in view of (\(ii\)) and (\(iii\)), we have in turn that </p>
<div class="displaymath" id="a0000000020">
  \begin{equation*} \begin{split}  \| G’(x_0)^{-1}(A-G’(x_0))\|  &  \le \int _{0}^{1}\phi _0((1-\lambda )\| \Lambda -x_0\| +\lambda \| \Lambda _1-x_0\| )d\lambda \\ &  \le \int _{0}^{1}\phi _0((1-\lambda )\rho +\lambda \rho _1)d\lambda \\ &  {\lt} 1. \end{split}\end{equation*}
</div>
<p> It follows that the linear operator \(A\) is invertible. Therefore, we can write </p>
<div class="displaymath" id="a0000000021">
  \begin{equation*} \begin{split}  \Lambda _1-\Lambda &  = A^{-1}(G(\Lambda _1)-G(\Lambda )) \\ &  = A^{-1}(0-0) = A^{-1}(0) \\ &  = 0, \end{split}\end{equation*}
</div>
<p> showing that \(\Lambda _1=\Lambda \). <div class="proof_wrapper" id="a0000000022">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<p><div class="remark_thmwrapper " id="R2">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">6</span>
  </div>
  <div class="remark_thmcontent">
  
<ol class="enumerate">
  <li><p>The conditions (\(H_1-H_5\)) were not used in the Proposition <a href="#p1">5</a>. But if they were used, then we can set \(\rho =t^*\). </p>
</li>
  <li><p>The limit point \(t^*\) in the condition (\(H_5\)) can be replaced by \(\delta \).<span class="qed">â–¡</span></p>
</li>
</ol>

  </div>
</div> </p>
<h1 id="sec4">4 Numerical Examples</h1>

<p><div class="example_thmwrapper " id="ex1">
  <div class="example_thmheading">
    <span class="example_thmcaption">
    Example
    </span>
    <span class="example_thmlabel">7</span>
  </div>
  <div class="example_thmcontent">
  <p> We reconsider the motivational example from the introduction part of this work. Select \(x_0=0.9955\). Conditions <span class="rm">(\(H_1\))–(\(H_3\))</span> are verified for </p>
<div class="displaymath" id="a0000000023">
  \begin{equation*}  \| G’(x_0)^{-1}G(x_0)\| =0.00456182=d, \end{equation*}
</div>
<p> \(\phi _0(t)=12.8089t\), \(\alpha =0.0780704\), \(\Omega _0=U(x_0,\alpha ) \cap \Omega \), \(\phi (t)=1.12091t\). Let \(\delta = 0.06\). The conditions <a href="#2.2" class="eqref">15</a> are tested and the results are given in <a href="#t1">table 1</a>. </p>
<div class="table"  id="t1">
   <div class="centered"> <small class="footnotesize"> <table class="tabular">
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> <i class="it">n</i> </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> \(1\) </p>

    </td>
    <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> \(3\) </p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p><b class="bf">\(\phi _0(s_n)\)</b> </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \(0.0584319\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \(0.0596158\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \(0.0596159\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \(0.0596159\) </p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p><b class="bf">\(\phi _0(t_{n+1})\)</b> </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> \(0.0589276\) </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> \(0.0596158\) </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> \(0.0596159\) </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> \(0.0596159\) </p>

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

    </td>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(0.00460052\) </p>

    </td>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(0.00465425\) </p>

    </td>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(0.00465426\) </p>

    </td>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(0.00465426\) <sup></sup> </p>

    </td>
  </tr>
</table>  </small></div> <figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">1</span> 
  <span class="caption_text">Estimates for <a href="#ex1">example 7</a> (\(t^*=0.00465426\))</span> 
</figcaption>  
</div>
<p> Hence, we can observe that conditions <span class="rm">(\(H_4\))</span> and <span class="rm">(\(H_5\))</span> hold and therefore we can conclude that the sequence \(\{ x_n\} \) generated by the method <a href="#e:2" class="eqref">2</a> converges to a solution \(x^*\) of the equation \(G(x)=0\) in \(U[x_0,t^*]\). </p>

  </div>
</div> </p>
<p><div class="example_thmwrapper " id="ex2">
  <div class="example_thmheading">
    <span class="example_thmcaption">
    Example
    </span>
    <span class="example_thmlabel">8</span>
  </div>
  <div class="example_thmcontent">
  <p> The applicability of our work in the real world can be demonstrated by considering the quartic equation for fractional conversion which represents the fraction of the nitrogen-hydrogen feed that gets converted to ammonia. At \(500^{\circ }\)C and \(250\) atm, this equation can be formulated as </p>
<div class="displaymath" id="a0000000024">
  \begin{equation*}  G(x)= x^4-7.79075x^3+14.7445x^2+2.511x-1.674. \end{equation*}
</div>
<p> Let \(\Omega =(0.3,0.4)\) and \(x_0=0.3\). Then, the conditions (\(H_1\)) - (\(H_3\)) are valid if </p>
<div class="displaymath" id="a0000000025">
  \begin{equation*}  \| G’(x_0)^{-1}G(x_0)\| =0.0217956=d, \end{equation*}
</div>
<p> \(\phi _0(t)=\phi (t)=1.56036\)t, \(\alpha =0.640877\) and \(\Omega _0= U[x_0,\alpha ] \cap \Omega \). Choose \(\delta = 0.3\). Conditions (<a href="#2.2">15</a>) are tested and the outcomes are given in <a href="#t2">table 2</a>. </p>
<div class="table"  id="t2">
   <div class="centered"> <small class="footnotesize"> <table class="tabular">
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> <i class="it">n</i> </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> \(1\) </p>

    </td>
    <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> \(3\) </p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p><b class="bf">\(\phi _0(s_n)\)</b> </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \(0.034009\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \(0.0384204\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \(0.0384459\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \(0.0384459\) </p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p><b class="bf">\(\phi _0(t_{n+1})\)</b> </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> \(0.0358473\) </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> \(0.038431\) </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> \(0.0384459\) </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> \(0.0384459\) </p>

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

    </td>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(0.0229737\) </p>

    </td>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(0.0246296\) </p>

    </td>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(0.0246391\) </p>

    </td>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(0.0246391\) <sup></sup> </p>

    </td>
  </tr>
</table> </small></div> <figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">2</span> 
  <span class="caption_text">Estimates for <a href="#ex2">example 8</a> (\(t^*=0.0246391\))</span> 
</figcaption>  
</div>
<p> Thus, we can observe that conditions <span class="rm">(\(H_4\))</span> and <span class="rm">(\(H_5\))</span> hold and therefore we can conclude that the sequence \(\{ x_n\} \) generated by the method <a href="#e:2" class="eqref">2</a> converges to a solution \(x^*\) of the equation \(G(x)=0\) in \(U[x_0,t^*]\). </p>

  </div>
</div> </p>
<p><div class="example_thmwrapper " id="ex3">
  <div class="example_thmheading">
    <span class="example_thmcaption">
    Example
    </span>
    <span class="example_thmlabel">9</span>
  </div>
  <div class="example_thmcontent">
  <p> Define the cubic polynomial </p>
<div class="displaymath" id="a0000000026">
  \begin{equation*}  G(x)=x^3-a. \end{equation*}
</div>
<p> on the open ball \(\Omega =U(x_0,1-a)\) for some \(a \in \left[0,1\right)\). If one chooses \(x_0=1\), then the conditions (\(A_1\))-(\(A_3\)) are verified for \(d = \frac{1-a}{3}\), \(\phi _0(t)=(3-a)t\), \(\alpha =\delta =\frac{1}{3-a}\), \(\Omega _0= U(x_0,\frac{1}{3-a})\) and \(\phi (t)=2(1+\frac{1}{3-a})t\). Let \(a=0.95\). <a href="#t3">Table 3</a> depicts the outcomes on testing of conditions <a href="#2.2" class="eqref">15</a>. </p>
<div class="table"  id="t3">
   <div class="centered"> <small class="footnotesize"> <table class="tabular">
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> <i class="it">n</i> </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> \(1\) </p>

    </td>
    <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> \(3\) </p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p><b class="bf">\(\phi _0(s_n)\)</b> </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \(0.0341667\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \(0.0358153\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \(0.0358166\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \(0.0358166\) </p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p><b class="bf">\(\phi _0(t_{n+1})\)</b> </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> \(0.0348607\) </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> \(0.0358158\) </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> \(0.0358166\) </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> \(0.0358166\) </p>

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

    </td>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(0.0170052\) </p>

    </td>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(0.0174711\) </p>

    </td>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(0.0174715\) </p>

    </td>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(0.0174715\) <sup></sup> </p>

    </td>
  </tr>
</table>  </small></div> <figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">3</span> 
  <span class="caption_text">Estimates for <a href="#ex3">example 9</a> (\(t^*=0.0174715\))</span> 
</figcaption>  
</div>
<p> Thus, we find that conditions <span class="rm">(\(H_4\))</span> and <span class="rm">(\(H_5\))</span> also hold and therefore we can conclude that the sequence \(\{ x_n\} \) generated by the method <a href="#e:2" class="eqref">2</a> converges to a solution \(x^*\) of the equation \(G(x)=0\) in \(U[x_0,t^*]\). </p>

  </div>
</div> </p>
<p><div class="example_thmwrapper " id="ex4">
  <div class="example_thmheading">
    <span class="example_thmcaption">
    Example
    </span>
    <span class="example_thmlabel">10</span>
  </div>
  <div class="example_thmcontent">
  <p> Consider the non-linear system </p>
<div class="displaymath" id="a0000000027">
  \begin{equation*}  G(w_1,w_2)= \big(w_1+e^{w_2}-\cos w_2, 3w_1-w_2-\sin w_2\big)^T \end{equation*}
</div>
<p> defined on \(\Omega =U(0,1)\) for \(w=(w_1,w_2)^T\). We obtain the operator \(G'\) as follows </p>
<div class="displaymath" id="a0000000028">
  \begin{equation*}  G’(w)= \begin{bmatrix}  1 

&  e^{w_2}+\sin w_2 

\\ 3 

&  -1-\cos w_2 

\end{bmatrix}. \end{equation*}
</div>
<p> We find \(x^*=(0,0)^T\) is a solution. Choose \(x_0=(0.1,0.1)^T\). Then, by applying method <a href="#e:2" class="eqref">2</a>, we find that the above system converges to \(x^*\) and the error estimates are given <a href="#t4">table 4</a>. </p>
<div class="table"  id="t4">
   <div class="centered"> 0.95 <small class="footnotesize"> <table class="tabular">
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \(\| x_0-x^*\| \) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \(\| x_1-x_*\| \) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \(\| x_2-x_*\| \) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \(\| x_3-x_*\| \) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \(\| x_4-x_*\| \) </p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>\(0.100\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(6.28455\cdot 10^{-3}\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(3.99276\cdot 10^{-10}\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(2.62582\cdot 10^{-53}\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(2.12431\cdot 10^{-312}\) </p>

    </td>
  </tr>
</table>  </small></div> <figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">4</span> 
  <span class="caption_text">Error Estimates for <a href="#ex4">example 10</a></span> 
</figcaption>  
</div>

  </div>
</div> </p>
<p><div class="example_thmwrapper " id="ex5">
  <div class="example_thmheading">
    <span class="example_thmcaption">
    Example
    </span>
    <span class="example_thmlabel">11</span>
  </div>
  <div class="example_thmcontent">
  <p> Consider the \(4\times 4\) system of non-linear equations defined on the open ball \(\Omega = U(0,1)\) given by </p>
<div class="displaymath" id="a0000000029">
  \begin{equation*}  v_i-\cos \Big(2v_i- \sum _{i=1}^{4}v_i\Big)=0, \quad i=1,2,3,4. \end{equation*}
</div>
<p> \(x^*=(0.5149,\ldots ,0.5149)^T\) is a solution. We choose the initial approximation as \(x_0=(0.5,\ldots ,0.5)^T\). Then, the method <a href="#e:2" class="eqref">2</a> is applied to this system and is found to be convergent to \(x^*\) and the error estimates obtained are given in <a href="#t5">table 5</a>. </p>
<div class="table"  id="t5">
   <div class="centered"> 0.95 <small class="footnotesize"> <table class="tabular">
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \(\| x_0-x^*\| \) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \(\| x_1-x_*\| \) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \(\| x_2-x_*\| \) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \(\| x_3-x_*\| \) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> \(\| x_4-x_*\| \) </p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>\(0.01490\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(1.21463\cdot 10^{-3}\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(3.87516\cdot 10^{-10}\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(4.08658\cdot 10^{-49}\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> \(5.62059\cdot 10^{-283}\) </p>

    </td>
  </tr>
</table>  </small></div> <figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">5</span> 
  <span class="caption_text">Error Estimates for <a href="#ex5">example 11</a></span> 
</figcaption>  
</div>

  </div>
</div> </p>
<p><div class="example_thmwrapper " id="ex6">
  <div class="example_thmheading">
    <span class="example_thmcaption">
    Example
    </span>
    <span class="example_thmlabel">12</span>
  </div>
  <div class="example_thmcontent">
  <p> Let us solve the non-linear system of logarithmic type where for \(w=(q_1,q_2,\ldots ,q_m)^T\) and \(Q(w)= \ln (q_i+1)- \frac{q_i}{20}\), \(i=1,2,\ldots ,m\), </p>
<div class="displaymath" id="a0000000030">
  \begin{equation*}  Q(w)=0 \end{equation*}
</div>
<p> Select \(m=100\) and \(x_0=(\frac{1}{2},\frac{1}{2},\ldots ,\frac{1}{2})^T\) (\(50\) times). Then, by applying the method <a href="#e:2" class="eqref">2</a>, the solution \(x^*=(1,1,\ldots ,1)^T\) (\(50\) times) is obtained after three iterations. </p>

  </div>
</div> </p>
<h1 id="sec5">5 Conclusion</h1>

<p>The semi-local convergence analysis for the method (<a href="#e:2">2</a>) is established by applying generalized Lipschitz condition only on the first derivative. Estimates on convergence balls, measurable error distances and the existence-uniqueness regions for the solution are deduced. At the end, the suggested theoretical outcomes are verified on numerical examples including one application problem. The major advantage is that the technique does not really depend on the method (<a href="#e:2">2</a>). Hence, it can be extended on other single and multi-step methods in the same manner. </p>
<p><small class="footnotesize">  </small></p>
<div class="bibliography">
<h1>Bibliography</h1>
<dl class="bibliography">
  <dt><a name="2">1</a></dt>
  <dd><p>Hameer&#160;Akhtar Abro and Muhammad&#160;Mujtaba Shaikh. A new time-efficient and convergent nonlinear solver. <em>Applied Mathematics and Computation</em>, 355:516–536, 2019. </p>
</dd>
  <dt><a name="17">2</a></dt>
  <dd><p>Christopher&#160;I Argyros, Ioannis&#160;K Argyros, Samundra Regmi, Jinny&#160;Ann John, and Jayakumar Jayaraman. Semi-Local Convergence of a Seventh Order Method with One Parameter for Solving Non-Linear Equations. <em>Foundations</em>, 2(4):827–838, 2022. </p>
</dd>
  <dt><a name="11">3</a></dt>
  <dd><p>Ioannis&#160;K Argyros. Unified convergence criteria for iterative Banach space valued methods with applications. <em>Mathematics</em>, 9(16):1942, 2021. </p>
</dd>
  <dt><a name="12">4</a></dt>
  <dd><p>Ioannis&#160;K Argyros. <em>The Theory and Applications of Iteration Methods</em>. 2nd edition, CRC Press/Taylor and Francis Publishing Group Inc., Boca Raton,Florida,USA, 2022. </p>
</dd>
  <dt><a name="13">5</a></dt>
  <dd><p>Ioannis&#160;K Argyros, Christopher Argyros, Johan Ceballos, and Daniel González. Extended Comparative Study between Newton’s and Steffensen-like Methods with Applications. <em>Mathematics</em>, 10(16):2851, 2022. </p>
</dd>
  <dt><a name="14">6</a></dt>
  <dd><p>Ioannis&#160;K Argyros, Debasis Sharma, Christopher&#160;I Argyros, Sanjaya&#160;Kumar Parhi, and Shanta&#160;Kumari Sunanda. Extended iterative schemes based on decomposition for nonlinear models. <em>Journal of Applied Mathematics and Computing</em>, 68(3):1485–1504, 2022. </p>
</dd>
  <dt><a name="15">7</a></dt>
  <dd><p>Michael&#160;I Argyros, Ioannis&#160;K Argyros, Samundra Regmi, and Santhosh George. Generalized Three-Step Numerical Methods for Solving Equations in Banach Spaces. <em>Mathematics</em>, 10(15):2621, 2022. </p>
</dd>
  <dt><a name="4">8</a></dt>
  <dd><p>Diyashvir Kreetee&#160;Rajiv Babajee. On the Kung-Traub conjecture for iterative methods for solving quadratic equations. <em>Algorithms</em>, 9(1):1, 2015. </p>
</dd>
  <dt><a name="3">9</a></dt>
  <dd><p>Alicia Cordero, Esther Gómez, and Juan&#160;R Torregrosa. Efficient high-order iterative methods for solving nonlinear systems and their application on heat conduction problems. <em>Complexity</em>, 2017, 2017. </p>
</dd>
  <dt><a name="8">10</a></dt>
  <dd><p>Djordje Herceg and Dragoslav Herceg. A family of methods for solving nonlinear equations. <em>Applied Mathematics and Computation</em>, 259:882–895, 2015. </p>
</dd>
  <dt><a name="16">11</a></dt>
  <dd><p>Jinny&#160;Ann John, Jayakumar Jayaraman, and Ioannis&#160;K Argyros. Local Convergence of an Optimal Method of Order Four for Solving Non-Linear System. <em>International Journal of Applied and Computational Mathematics</em>, 8(4):1–8, 2022. </p>
</dd>
  <dt><a name="1">12</a></dt>
  <dd><p>L&#160;V Kantorovich and G&#160;P Akilov. <em>Functional Analysis in Normed Spaces</em>. Pergamon Press, Oxford, 1964. </p>
</dd>
  <dt><a name="6">13</a></dt>
  <dd><p>HT&#160;Kung and Joseph&#160;Frederick Traub. Optimal order of one-point and multipoint iteration. <em>Journal of the ACM (JACM)</em>, 21(4):643–651, 1974. </p>
</dd>
  <dt><a name="9">14</a></dt>
  <dd><p>Taher Lotfi, Parisa Bakhtiari, Alicia Cordero, Katayoun Mahdiani, and Juan&#160;R Torregrosa. Some new efficient multipoint iterative methods for solving nonlinear systems of equations. <em>International Journal of Computer Mathematics</em>, 92(9):1921–1934, 2015. </p>
</dd>
  <dt><a name="18">15</a></dt>
  <dd><p>Kalyanasundaram Madhu, DKR Babajee, and Jayakumar Jayaraman. An improvement to double-step Newton method and its multi-step version for solving system of nonlinear equations and its applications. <em>Numerical Algorithms</em>, 74(2):593–607, 2017. </p>
</dd>
  <dt><a name="7">16</a></dt>
  <dd><p>Muhammad&#160;Aslam Noor, Khalida&#160;Inayat Noor, Eisa Al-Said, and Muhammad Waseem. Some new iterative methods for nonlinear equations. <em>Mathematical Problems in Engineering</em>, 2010, 2010. </p>
</dd>
  <dt><a name="5">17</a></dt>
  <dd><p>Janak&#160;Raj Sharma, Rajni Sharma, and Nitin Kalra. A novel family of composite Newton–Traub methods for solving systems of nonlinear equations. <em>Applied Mathematics and Computation</em>, 269:520–535, 2015. </p>
</dd>
  <dt><a name="10">18</a></dt>
  <dd><p>Muhammad Waseem, Muhammad&#160;Aslam Noor, and Khalida&#160;Inayat Noor. Efficient method for solving a system of nonlinear equations. <em>Applied Mathematics and Computation</em>, 275:134–146, 2016. </p>
</dd>
</dl>


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