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<title>Extended Convergence Analysis of Newton-Potra Solver for Equations: Extended Convergence Analysis of Newton-Potra Solver for Equations</title>
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<h1>Extended Convergence Analysis of Newton-Potra Solver for Equations</h1>
<p class="authors">
<span class="author">Ioannis K. Argyros\(^\ast \), Stepan Shakhno\(^\bullet \), Yurii Shunkin\(^\bullet \) <br />Halyna Yarmola\(^\circ \)</span>
</p>
<p class="date">April 30, 2019; accepted: July 10, 2020; published online: February 20, 2021.</p>
</div>
<div class="abstract"><p> In the paper a local and a semi-local convergence of combined iterative process for solving nonlinear operator equations is investigated. This solver is built based on Newton solver and has \(R\)-convergence order \(1.839\ldots \) The radius of the convergence ball and convergence order of the investigated solver are determined in an earlier paper. Modifications of previous conditions leads to extended convergence domain. These advantages are obtained under the same computational effort. Numerical experiments are carried out on the test examples with nondifferentiable operator. </p>
<p><b class="bf">MSC.</b> 65H10, 65J15, 49M15 </p>
<p><b class="bf">Keywords.</b> nonlinear equation, nondifferentiable operator, local and semi-local convergence, order of convergence, divided difference. </p>
</div>
<p>\(^\ast \)Department of Mathematical Sciences, Cameron University, Lawton, OK 73505, USA; <span class="tt">iargyros@cameron.edu</span> </p>
<p>\(^\bullet \)Department of Theory of Optimal Processes, <i class="it">Ivan Franko</i> National University of Lviv, 79000 Lviv, Ukraine; <span class="tt">stepan.shakhno@lnu.edu.ua; yuriy.shunkin@lnu.edu.ua</span> </p>
<p>\(^\circ \)Department of Computational Mathematics, <i class="it">Ivan Franko</i> National University of Lviv, 79000 Lviv, Ukraine; <span class="tt">halyna.yarmola@lnu.edu.ua</span> </p>
<h1 id="a0000000002">1 Introduction</h1>
<p>Consider the operator equation </p>
<div class="displaymath" id="<plasTeX.TeXFragment object at 0x0000028658D7C410>">
  \begin{align} \label{EQ_1} H(x)\equiv F(x)+Q(x)=0, \end{align}
</div>
<p> where \(F\) and \(Q\) are nonlinear operators, defined on an subset \(D\) of a Banach space \(E_1\) with values in a Banach space \(E_2\). It is known, that \(F\) is a differentiable by Frèchet operator, \(Q\) is a continuous operator, whose differentiability in general is not required. </p>
<p>A plethora of problems from diverse disciplines can be converted to equation ?? <i class="it">via</i> mathematical modelling <span class="cite">
	[
	<a href="#Rhe98" >1</a>
	, 
	<a href="#Cite_13_Kant" >2</a>
	, 
	<a href="#Cite_5" >3</a>
	, 
	<a href="#Cite_1" >4</a>
	, 
	<a href="#Cite_N_1" >5</a>
	, 
	<a href="#Cite_N_2" >6</a>
	, 
	<a href="#Cite_N_5" >7</a>
	, 
	<a href="#Cite_N_10" >8</a>
	, 
	<a href="#Cite_2" >9</a>
	, 
	<a href="#Cite_N_6" >10</a>
	, 
	<a href="#Cite_3" >11</a>
	, 
	<a href="#Cite_N_11" >12</a>
	, 
	<a href="#Cite_N_8" >13</a>
	, 
	<a href="#Cite_N_4" >14</a>
	, 
	<a href="#Cite_4" >15</a>
	, 
	<a href="#Cite_N_9" >16</a>
	, 
	<a href="#Cite_N_7" >17</a>
	, 
	<a href="#Cite_6" >18</a>
	, 
	<a href="#Cite_8" >19</a>
	, 
	<a href="#Cite_7" >20</a>
	, 
	<a href="#Cite_14_Traub" >21</a>
	, 
	<a href="#Cite_9" >22</a>
	, 
	<a href="#Cite_10" >23</a>
	, 
	<a href="#Cite_11" >24</a>
	, 
	<a href="#Cite_12_Sh" >25</a>
	, 
	<a href="#Cite_15_SYa" >26</a>
	, 
	<a href="#Cite_N_3" >27</a>
	]
</span>. Therefore, the task of computing a solution \(x_*\) is of extreme importance. We resort to iterative solvers, since closed form solutions can be obtained in rare cases. </p>
<p>The Newton solver <span class="cite">
	[
	<a href="#Cite_13_Kant" >2</a>
	]
</span> can not be used to find a solution of equation ??, because of the nondifferentiable \(Q\). However, in this case the Newton-type solver <span class="cite">
	[
	<a href="#Cite_5" >3</a>
	]
</span>, or one of combined iterative processes may be applicable <span class="cite">
	[
	<a href="#Cite_5" >3</a>
	]
</span>–<span class="cite">
	[
	<a href="#Cite_6" >18</a>
	]
</span>. </p>
<p>The special case of ?? is the equation \(F(x)=0\). Usually, to find the solution Newton’s solver is used </p>
<div class="displaymath" id="a0000000003">
  \[  x_{n+1}=x_{n} - [F'(x_n)]^{-1}F(x_n),\; n \ge 0,  \]
</div>
<p> whose convergence order is quadratic <span class="cite">
	[
	<a href="#Cite_8" >19</a>
	, 
	<a href="#Cite_7" >20</a>
	]
</span>. Hence, one can use the difference solvers. These solvers use only a nonlinear operator, and do not require analytical derivatives. One of these solvers has \(R\)-convergence order \(1.839\ldots \) </p>
<div class="displaymath" id="a0000000004">
  \[  x_{n+1}=x_n-[F(x_n,x_{n-1})+F(x_{n-2},x_n)-F(x_{n-2},x_{n-1})]^{-1}F(x_n),\;  n \ge 0,  \]
</div>
<p> where \(F(u,v)\) is a divided difference of order one. This solver was proposed by J. Traub for solving one nonlinear equation <span class="cite">
	[
	<a href="#Cite_14_Traub" >21</a>
	]
</span>, later it was generalized to Banach spaces by F. Potra <span class="cite">
	[
	<a href="#Cite_9" >22</a>
	]
</span>, and investigated under different conditions in <span class="cite">
	[
	<a href="#Cite_10" >23</a>
	, 
	<a href="#Cite_11" >24</a>
	]
</span>. </p>
<p>In the paper <span class="cite">
	[
	<a href="#Cite_12_Sh" >25</a>
	]
</span> a combined iterative process was proposed, which is built based Newton’s and Potra’s solvers </p>
<div class="displaymath" id="<plasTeX.TeXFragment object at 0x0000028658D7E6F0>">
  \begin{align} \label{EQ_2} x_{n+1}=& x_n-A_n^{-1}H(x_n),\qquad n \ge 0, \\ A_n=& F’(x_n)+Q(x_n,x_{n-1})+Q(x_{n-2},x_n)-Q(x_{n-2},x_{n-1}), \nonumber \end{align}
</div>
<p> where \(Q(x,y)\) is a divided difference of order one, to be defined later. </p>
<p>This solver was studied in <span class="cite">
	[
	<a href="#Cite_15_SYa" >26</a>
	]
</span> under weak \(\omega \)-conditions. In this work we continue the study of a local and a semi-local convergence of solver ??. It is established that the convergence order of the combined iterative process ?? is similar to the convergence order of the Potra solver. But also it is important to extend the convergence region in particular without requiring an additional hypotheses. This fact will extend the number of initial approximations. By applying a new approach we achieve fewer iterations to obtain a result with predetermined accuracy, at least as many initial points, and same or less computational cost. </p>
<p>The rest of the paper is structured as follows: In Section 2, we present the local convergence analysis of the solver ?? and a Corollary. Section 3 contains the proofs of semi-local convergence and uniqueness of solution. In Section 4, we provide the numerical example. The article ends with some conclusions. </p>
<h1 id="a0000000005">2 Local convergence of solver ??</h1>
<p> Note that we used the classic Lipschitz conditions for the derivative of first order of operator \(F\) and for divided differences of order one and two of operator \(Q\). The following theorem present the convergence radius and the convergence speed of iterative process ??. Although we assume, that \(Q\) is differentiable by Fréchet operator. </p>
<p>Set \(U=U(x_*,r_*)=\{ x: \| x-x_*\| {\lt}r_*\} \). Let \(x,y,z \in D\). <div class="definition_thmwrapper " id="Def_dd1">
  <div class="definition_thmheading">
    <span class="definition_thmcaption">
    Definition
    </span>
    <span class="definition_thmlabel">1</span>
  </div>
  <div class="definition_thmcontent">
  <p> The linear operator from \(E_1\) to \(E_2\) denoted as \(Q(x,y)\) is called divided difference of order one of \(Q\) by points \(x,y, (x \neq y)\) if it satisfies the condition </p>
<div class="displaymath" id="a0000000006">
  \[  Q(x,y)(x-y)=Q(x)-Q(y).  \]
</div>

  </div>
</div> </p>
<p><div class="definition_thmwrapper " id="Def_dd2">
  <div class="definition_thmheading">
    <span class="definition_thmcaption">
    Definition
    </span>
    <span class="definition_thmlabel">2</span>
  </div>
  <div class="definition_thmcontent">
  <p> The operator \(Q(x,y,z)\) is called divided difference of order two of \(Q\) by points \(x,y,z\) if it satisfies the condition </p>
<div class="displaymath" id="a0000000007">
  \[  Q(x,y,z)(y-z) = Q(x,y) - Q(x,z).  \]
</div>

  </div>
</div> </p>
<p><div class="theorem_thmwrapper " id="Theo1">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">3</span>
  </div>
  <div class="theorem_thmcontent">
  <p> Let \(F\) and \(Q\) are nonlinear operator, which are defined on open convex subset \(D\) of a Banach space \(E_1\) with values in a Banach space \(E_2\). Suppose, that equation ?? has a solution \(x_* \in D\) and the inverse Fréchet derivative \([H'(x_*)]^{-1}\) exists. Let \(Q(\cdot ,\cdot )\) and \(Q(\cdot ,\cdot ,\cdot )\) are the divided differences of order one and two of operator \(Q\), which are defined on the set \(D\), and the Lipschitz conditions are satisfied for each \(x,y,z \in D\) </p>
<div class="displaymath" id="<plasTeX.TeXFragment object at 0x0000028658DB5A90>">
  \begin{align} \label{A_EQ_3} \| H’(x_*)^{-1}(F’(x)-F’(x_*))\|  \le 2 l_*^0\| x-x_*\| , \end{align}
</div>
<div class="displaymath" id="<plasTeX.TeXFragment object at 0x0000028658DB6D50>">
  \begin{align} \label{A_EQ_4} \| H’(x_*)^{-1}(Q(x_*,x_*)-Q(x,x_*))\|  \le a\| x_*-x\| , \end{align}
</div>
<div class="displaymath" id="<plasTeX.TeXFragment object at 0x0000028658DB75F0>">
  \begin{align} \label{A_EQ_5} \| H’(x_*)^{-1}(Q(z,x_*)-Q(z,x))\|  \le b\| x_*-x\| , \end{align}
</div>
<div class="displaymath" id="<plasTeX.TeXFragment object at 0x0000028658DB5910>">
  \begin{align} \label{A_EQ_6} \| H’(x_*)^{-1}(Q(x,x_*,y)-Q(z,x_*,y))\|  \le q_*^0\| x-z\| , \end{align}
</div>
<p> for each \(x,y,z \in D_0=D \cap U(x_*,r_0)\), </p>
<div class="displaymath" id="<plasTeX.TeXFragment object at 0x0000028658D707D0>">
  \begin{align} \label{A_EQ_7} r_0 = \tfrac {1}{l_*^0+p_*^0+\sqrt{(l_*^0+p_*^0)^2+2q_*^0}}, \end{align}
</div>
<div class="displaymath" id="<plasTeX.TeXFragment object at 0x0000028658D71850>">
  \begin{align} \label{A_EQ_9} \| H’(x_*)^{-1}(F’(x)-F’(y))\|  \le 2l_*\| x-y\| , \end{align}
</div>
<div class="displaymath" id="<plasTeX.TeXFragment object at 0x0000028658D724B0>">
  \begin{align} \label{A_EQ_10} \| H’(x_*)^{-1}(Q(x,x_*)-Q(x,x))\|  \le c\| x_*-x\| , \end{align}
</div>
<div class="displaymath" id="<plasTeX.TeXFragment object at 0x0000028658D72990>">
  \begin{align} \label{A_EQ_11} \| H’(x_*)^{-1}(Q(x,x)-Q(x,y))\|  \le d\| x-y\| , \end{align}
</div>
<div class="displaymath" id="<plasTeX.TeXFragment object at 0x0000028658D72B10>">
  \begin{align} \label{A_EQ_12} \| H’(x_*)^{-1}(Q(x,x,y)-Q(z,x,y))\|  \le q_*\| x-z\| , \end{align}
</div>
<p> where \(2p_*^0=a+b\), \(p_*=c+d\), </p>
<div class="displaymath" id="<plasTeX.TeXFragment object at 0x0000028658D73B90>">
  \begin{align} \label{A_EQ_13} r_*=\tfrac {2}{2(l_*^0+p_*^0)+l_*+p_*+\sqrt{(2(l_*^0+p_*^0)+l_*+p_*)^2+8(q_*+q_*^0)}} \end{align}
</div>
<p> Then for each \(x_{-2},x_{-1},x_0 \in U\) iterative process ?? is well defined, and generates a sequence \(\{ x_n\} _{n \ge 0} \in U\), which converges to \(x_*\) and satisfies the estimation </p>
<div class="displaymath" id="<plasTeX.TeXFragment object at 0x0000028658D731D0>">
  \begin{align} \label{A_EQ_14} & \| x_{n+1}-x_*\|  \le \\ & \le \! \tfrac {l_*+p_*}{C_n}\| x_n-x_*\| ^2 \! +\! \tfrac {q_*}{C_n}(\| x_n-x_*\| \! +\! \| x_{n-2}-x_*\| )\| x_{n-1}-x_*\| \| x_n-x_*\| , \nonumber \end{align}
</div>
<p> where \(C_n=1-2(l_*^0+p_*^0)\| x_n-x_*\| -q_*^0(\| x_n-x_*\| +\| x_{n-2}-x_*\| )\| x_{n-1}-x_*\| \). </p>

  </div>
</div> <div class="proof_wrapper" id="a0000000008">
  <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 \(x,y,z \in U\). Denote \(A=F'(x)+Q(x,y)+Q(z,x)-Q(z,y)\). Then in view of conditions ??–??, we obtain </p>
<div class="displaymath" id="a0000000009">
  \begin{align*} & \| I-H’(x_*)^{-1}A\|  =\\ & \  = \| H(x_*)^{-1}[F’(x_*)-F’(x)+Q(x_*,x_*)-Q(x,y)-Q(z,x)+Q(z,y)]\|  \\ & \  \le \| H(x_*)^{-1}\left[F’(x_*)-F’(x)+Q(x_*,x_*)-Q(x,x_*) \right.\\ & \  \quad + \left. Q(z,x_*)-Q(z,x)+Q(x,x_*)-Q(x,y)+Q(z,y)-Q(z,x_*)\right]\|  \\ & \  \le \| H(x_*)^{-1}\left[F’(x_*)-F’(x)+Q(x_*,x_*)-Q(x,x_*) \right.\\ & \  \quad + \left. Q(z,x_*)-Q(z,x)+[Q(x,x_*,y)-Q(z,x_*,y)](x_*-y)\right]\|  \\ & \  \le 2l_*^0\| x-x_*\| +2p_*^0\| x-x_*\| +q_*^0\| x-z\| \| y-x_*\|  \\ & \  \le 2(l_*^0+p_*^0)\| x-x_*\| +q_*^0(\| x-x_*\| +\| z-x_*\| )\| y-x_*\| . \end{align*}
</div>
<p> By the definition of \(r_*\) ??, we get </p>
<div class="displaymath" id="<plasTeX.TeXFragment object at 0x0000028658E3CFB0>">
  \begin{align} \label{A_EQ_15} 2(l_*^0+p_*^0)r_0+2q_*^0r_0^2{\lt}1. \end{align}
</div>
<p> Then, by the Banach Lemma on invertible operators <span class="cite">
	[
	<a href="#Cite_13_Kant" >2</a>
	]
</span>, \(A\) is invertible and </p>
<div class="displaymath" id="<plasTeX.TeXFragment object at 0x0000028658E3DAF0>">
  \begin{align} \label{A_EQ_16} & \| (I-(I-H’(x_*)^{-1}A))^{-1}\|  = \| A^{-1}H’(x_*)\|  \\ & \le [1-2(l_*^0+p_*^0)\| x-x_*\| -q_*^0(\| x-x_*\| +\| z-x_*\| )\| y-x_*\| ]^{-1}. \nonumber \end{align}
</div>
<p> Suppose, that \(x_{n-2},x_{n-1},x_n \in U\). Then the operator </p>
<div class="displaymath" id="a0000000010">
  \[  A_n=F'(x_n)+Q(x_n,x_{n-1})+Q(x_{n-2},x_n)-Q(x_{n-2},x_{n-1})  \]
</div>
<p> is invertible. Next, we can write </p>
<div class="displaymath" id="<plasTeX.TeXFragment object at 0x0000028658E3E150>">
  \begin{align} \label{A_EQ_17} \| x_{n+1}-x_*\|  & = \| x_n-x_*-A_n^{-1}(H(x_n)-H(x_*))\|  \\ & \le \| A_n^{-1}H’(x_*)\| \| H’(x_*)^{-1}[H(x_n)-H(x_*)-A_n(x_n-x_*)]\| .\nonumber \end{align}
</div>
<p> In view of ??–??, we get </p>
<div class="displaymath" id="a0000000011">
  \begin{align*} & \| H’(x_*)^{-1}[H(x_n)-H(x_*)-A_n(x_n-x_*)]\|  = \\ =&  \| H’(x_*)^{-1}[F(x_n)-F(x_*)+Q(x_n)-Q(x_*)-A_n(x_n-x_*)]\|  \\ \le &  \| H’(x_*)^{-1}\int _0^1(F’(x_* +t(x_n-x_*))-F’(x_n))dt\| \| x_n-x_*\|  \\ & + \| H’(x_*)^{-1}\left[ Q(x_n,x_*)-Q(x_n,x_n)+Q(x_n,x_n)-Q(x_n,x_{n-1})-Q(x_{n-2},x_n) \right. \\ & + \left. Q(x_{n-2},x_{n-1})\right]\| \| x_n-x_*\|  \\ \le & (l_*+p_*)\| x_n-x_*\| ^2 \\ & + \|  H’(x_*)^{-1}[Q(x_n,x_n,x_{n-1})-Q(x_{n-2},x_n,x_{n-1})]\|  \| x_n-x_{n-1}\| \| x_n-x_*\|  \\ \le &  (l_*+p_*)\| x_n-x_*\| ^2+q_*\| x_n-x_{n-2}\| \| x_n-x_{n-1}\| \| x_n-x_*\| . \end{align*}
</div>
<p> Then, from ?? and ??, we obtain the estimate ??. Moreover, from inequalities ??, ?? we have in a turn </p>
<div class="displaymath" id="a0000000012">
  \begin{align*}  \| x_{n+1}-x_*\|  {\lt} \| x_n-x_*\|  {\lt} r_*, \quad n \ge 0. \end{align*}
</div>
<p> Hence, iterative process ?? is well defined, generated sequence \(\{ x_n\} _{n \ge 0}\) is in \(U\), and converges to the solution \(x_*\). Finally, by the last inequality, and estimate ?? we get, that \(\lim \limits _{n\to \infty }\| x_n-x_*\| =0\).  <div class="proof_wrapper" id="a0000000013">
  <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="corollary_thmwrapper " id="Coro1">
  <div class="corollary_thmheading">
    <span class="corollary_thmcaption">
    Corollary
    </span>
    <span class="corollary_thmlabel">4</span>
  </div>
  <div class="corollary_thmcontent">
  <p> The \(R\)-convergence order of the combined iterative solver ?? is \(1.839\ldots \) </p>

  </div>
</div> <div class="proof_wrapper" id="a0000000014">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> By estimate ??, we have that there exist a constant \(C\), and a natural number \(N\), such that </p>
<div class="displaymath" id="a0000000015">
  \begin{align*}  \| x_{n+1}-x_*\|  \le C\| x_n-x_*\| \| x_{n-1}-x_{*}\| \| x_{n-2}-x_*\| , \quad n \ge N. \end{align*}
</div>
<p> Hence, the \(R\)-convergence order of solver ?? is the unique positive root of nonlinear equation \(t^3-t^2-t-1=0\) <span class="cite">
	[
	<a href="#Cite_9" >22</a>
	]
</span>, which is \(1.839\ldots \)  <div class="proof_wrapper" id="a0000000016">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<p><div class="remark_thmwrapper " id="Rem1">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">5</span>
  </div>
  <div class="remark_thmcontent">
  <p> The conditions used in <span class="cite">
	[
	<a href="#Cite_12_Sh" >25</a>
	]
</span> instead of ??–?? are: </p>
<p>for each \(x,y,u,v \in D\) </p>
<div class="displaymath" id="<plasTeX.TeXFragment object at 0x0000028658E9CCB0>">
  \begin{align} \label{A_EQ_18} \| H’(x_*)^{-1}(F’(x)-F’(y))\|  \le 2l_*^1\| x-y\| , \end{align}
</div>
<div class="displaymath" id="<plasTeX.TeXFragment object at 0x0000028658E9E150>">
  \begin{align} \label{A_EQ_19} \| H’(x_*)^{-1}(Q(x,y)-Q(u,v))\|  \le p_*^1(\| x-u\| +\| y-v\| ), \end{align}
</div>
<div class="displaymath" id="<plasTeX.TeXFragment object at 0x0000028658E9E8D0>">
  \begin{align} \label{A_EQ_20} \| H’(x_*)^{-1}(Q(u,x,y)-Q(v,x,y))\|  \le q_*^1\| u-v\| , \end{align}
</div>
<div class="displaymath" id="<plasTeX.TeXFragment object at 0x0000028658E9EF90>">
  \begin{align} \label{A_EQ_21} r_*^1=\tfrac {2}{3(l_*^1+p_*^1)+\sqrt{9(l_*^1+p_*^1)^2+16q_*^1}}, \end{align}
</div>
<div class="equation" id="<plasTeX.TeXFragment object at 0x0000028658E9EB70>">
<p>
  <div class="equation_content">
    \begin{equation} \label{A_EQ_22} C_n^1=1-2(l_*^1+p_*^1)\| x_n-x_*\| -q_*^1(\| x_n-x_*\| +\| x_{n-2}-x_*\| )\| x_{n-1}-x_*\| . \end{equation}
  </div>
  <span class="equation_label">21</span>
</p>
</div>
<p> But </p>
<div class="displaymath" id="a0000000017">
  \[  D_0 \subseteq D,  \]
</div>
<p> so </p>
<div class="equation" id="<plasTeX.TeXFragment object at 0x0000028658E9E330>">
<p>
  <div class="equation_content">
    \begin{equation} \label{A_EQ_23} \begin{gathered} l_*^0 \le l_*^1, \; \; \; \; \; \;  p_*^0 \le p_*^1, \; \; \; \; \; \;  q_*^0 \le q_*^1,\\ l_* \le l_*^1, \; \; \; \; \; \;  p_* \le p_*^1, \; \; \; \; \; \;  q_* \le q_*^1 \end{gathered} \end{equation}
  </div>
  <span class="equation_label">22</span>
</p>
</div>
<p> and </p>
<div class="equation" id="<plasTeX.TeXFragment object at 0x0000028658EC9B50>">
<p>
  <div class="equation_content">
    \begin{equation} \label{A_EQ_29} (C_n)^{-1} \le (C_n^1)^{-1}. \end{equation}
  </div>
  <span class="equation_label">23</span>
</p>
</div>
<p> In view of ??–??, the new results give compared to the ones in <span class="cite">
	[
	<a href="#Cite_12_Sh" >25</a>
	]
</span>. </p>
<p>At least as many initial points, and fewer iterations to achieve a predetermined accuracy. The improvements are obtained under the same or less computational cost as in <span class="cite">
	[
	<a href="#Cite_12_Sh" >25</a>
	]
</span>, since the new constants are special cases of ones in <span class="cite">
	[
	<a href="#Cite_12_Sh" >25</a>
	]
</span>. Examples where ??–?? hold as strict inequalities can be found in <span class="cite">
	[
	<a href="#Cite_N_3" >27</a>
	]
</span>. This technique is used to expand applicability of some solvers <span class="cite">
	[
	<a href="#Cite_N_5" >7</a>
	]
</span> and can be used to do the same on other solvers. </p>

  </div>
</div> </p>
<h1 id="a0000000018">3 Semi-local convergence of solver ??</h1>
<p> Set \(U_0(x_0,r)=\{ x: \| x-x_0\| \le r\} \). The semi-local convergence of the combined Newton-Potra solver ?? is presented in what follows. </p>
<p><div class="theorem_thmwrapper " id="Theo2">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">6</span>
  </div>
  <div class="theorem_thmcontent">
  <p> Let \(F\) and \(Q\) are nonlinear operators, which are defined in open convex subset \(D\) of a Banach space \(E_1\), with values in a Banach space \(E_2\). \(Q(\cdot ,\cdot )\) and \(Q(\cdot ,\cdot ,\cdot )\) are the divided differences of order one and two of function \(Q\), which are defined on set \(D\). </p>
<p>Suppose, that the linear operator \(A_0=F'(x_0)+Q(x_0,x_{-1})+Q(x_{-2},x_0)-Q(x_{-2},x_{-1})\), where \(x_{-2},x_{-1},x_0 \in D\), is invertible fore each \(x,y,u,v \in D\) satisfies the Lipschitz conditions </p>
<div class="displaymath" id="<plasTeX.TeXFragment object at 0x0000028658ED9190>">
  \begin{align} \label{A_EQ_30} \| A_0^{-1}(F’(x)-F’(x_0))\|  \le 2l_0^0\| x-x_0\| , \end{align}
</div>
<div class="displaymath" id="<plasTeX.TeXFragment object at 0x0000028658EDA3F0>">
  \begin{align} \label{A_EQ_31} \| A_0^{-1}(Q(x_0,x_0)-Q(x,x_0))\|  \le \lambda \| x_0-x\| , \end{align}
</div>
<div class="displaymath" id="<plasTeX.TeXFragment object at 0x0000028658EDB470>">
  \begin{align} \label{A_EQ_32} \| A_0^{-1}(Q(x,x_0)-Q(x,y))\|  \le \mu \| x_0-y\| , \end{align}
</div>
<div class="displaymath" id="<plasTeX.TeXFragment object at 0x0000028658EDBB30>">
  \begin{align} \label{A_EQ_33} \| A_0^{-1}(Q(z,u)-Q(z,x))\|  \le \xi \| u-x\| , \end{align}
</div>
<div class="displaymath" id="<plasTeX.TeXFragment object at 0x0000028658EDBD10>">
  \begin{align} \label{A_EQ_34} \| A_0^{-1}(Q(x_0,x_{-1},x_0)-Q(x_{-2},x_{-1},x_0))\|  \le \overline{q}_0 \| x_0 -x_{-2}\| , \end{align}
</div>
<p> Set </p>
<div class="displaymath" id="<plasTeX.TeXFragment object at 0x0000028658EE4EF0>">
  \begin{align} \label{A_EQ_35} D_1 = D \cap U(x_0,r_0), \; \; \; \; \;  r_0 = \tfrac {1-\overline{q}_0a(a+b)}{2(l_0^0+p_0^0)} \; \;  for \; \; \;  \overline{q}_0a(a+b){\lt}1 \end{align}
</div>
<p> and \(p_0^0 = \max \{ \lambda , \mu , \xi \} \). </p>
<p>For each \(x,y,u,v \in D_1\) </p>
<div class="displaymath" id="<plasTeX.TeXFragment object at 0x0000028658EE5E50>">
  \begin{align} \label{AW_EQ_35} \| A_0^{-1}(F’(x)-F’(y))\|  \le 2l_0\| x-y\| , \end{align}
</div>
<div class="displaymath" id="<plasTeX.TeXFragment object at 0x0000028658EE6030>">
  \begin{align} \label{AW_EQ_36} \| A_0^{-1}(Q(x,y)-Q(u,v))\|  \le p_0(\| x-u\| +\| y-v\| ), \end{align}
</div>
<div class="displaymath" id="<plasTeX.TeXFragment object at 0x0000028658EE6F90>">
  \begin{align} \label{AW_EQ_37} \| A_0^{-1}(Q(u,x,y)-Q(v,x,y))\|  \le q_0\| u-v\| . \end{align}
</div>
<p> Let \(a,b\) and \(c\) are a nonnegative numbers, such that </p>
<div class="displaymath" id="<plasTeX.TeXFragment object at 0x0000028658EE5F70>">
  \begin{align} \label{AW_EQ_38} \| x_0-x_{-1}\|  \le a, \; \; \;  \| x_{-1}-x_{-2}\|  \le b, \; \; \;  \| A_0^{-1}(F(x_0)+Q(x_0))\|  \le c. \end{align}
</div>
<p> Let \(r_1\) is a nonnegative number, such that </p>
<div class="displaymath" id="a0000000019">
  \[  r_1 {\gt} \tfrac {c}{1-\gamma },  \]
</div>
<div class="displaymath" id="a0000000020">
  \[  \gamma = \tfrac {(l_0+p_0)c+q_0a(a+b)}{1-\overline{q}_0a(a+b)-2(l_0^0+p_0^0)r_1}, \; \;  0{\lt}\gamma {\lt} 1,  \]
</div>
<p> and the closed ball \(U_0(x_0,r_1)\) is included in \(D\). Then, real sequence \(\{ t_k\} _{k\geq -2}\) defined as </p>
<div class="displaymath" id="<plasTeX.TeXFragment object at 0x0000028658F114F0>">
  \begin{align} \label{EQ_R_1} \begin{gathered} t_{-2}=r_1+a+b,\; \; \; t_{-1}=r_1+a,\; \; \; t_0=r_1,\; \; \; t_1=r_1-c \end{gathered}\\ t_{n+1}-t_{n+2} = \tfrac {(l_0+p_0)(t_n-t_{n+1})+q_0(t_{n-1}-t_n)(t_{n-2}-t_n)}{1-(2l_0^0+\lambda + \xi )(t_0-t_{n+1})-(\mu + \xi )(t_0-t_n)-\overline{q}_0a(a+b)} (t_n-t_{n+1}). \end{align}
</div>

  </div>
</div> </p>
<div class="displaymath" id="a0000000021">
  \end{align}
</div>
<p> is nonnegative and decreasing converging to some \(t_* \in \mathbb {R}\), such that </p>
<div class="displaymath" id="a0000000022">
  \[  r_1 -\tfrac {c}{1-\gamma } \leq t_* \leq t_{-1}.  \]
</div>
<p> Then the iterative process ?? is well defined, remains in \(U_0(x_0,r_1)\) and converges to a solution \(x \in U_0(x_0,r_1)\) of equation \(F(x)+Q(x)=0\), Moreover, the following estimates are true </p>
<div class="displaymath" id="<plasTeX.TeXFragment object at 0x0000028658F2C5F0>">
  \begin{align} \label{EQ_R_1_5} \| x_n-x_*\|  \leq t_n-t_*. \end{align}
</div>
<p><div class="theorem_thmwrapper " id="a0000000023">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
  </div>
  <div class="theorem_thmcontent">
  
  </div>
</div> <div class="proof_wrapper" id="a0000000024">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> Using mathematical induction, we show that the iterative process ?? is well defined </p>
<div class="displaymath" id="<plasTeX.TeXFragment object at 0x0000028658F2D670>">
  \begin{align} \label{EQ_R_3} t_{k+1} - t_{k+2}\leq & \gamma (t_k-t_{k+1}), \\ \label{EQ_R_4} t_{k+1} \geq & t_{k+2}\geq r_1 - \tfrac {c}{1-\gamma }. \end{align}
</div>
<p>Using ?? and \(k=0\), we obtain </p>
<div class="displaymath" id="a0000000025">
  \begin{align*}  t_1-t_2 =&  \tfrac {(l_0+p_0)(t_0-t_1)+q_0(t_{-1}-t_0)(t_{-2}-t_0)}{1-(2l_0^0+\lambda + \xi )(t_0-t_1)-\overline{q}_0a(a+b)}(t_0-t_1) \\ \leq &  \tfrac {(l_0+p_0)c+q_0a(a+b)}{1-\overline{q}_0a(a+b)-2(l_0^0+p_0^0)r_1}(t_0-t_1), \\ t_0 \geq & t_1,\;  t_1 \geq t_2 \geq t_1 - \gamma (t_0-t_1) = r_1 - (1-\gamma )c = r_1 - \tfrac {(1-\gamma ^2)c}{1-\gamma } \\ \geq &  r_1 - \tfrac {c}{1-\gamma } \geq 0, \end{align*}
</div>
<p> so ??–?? are true for \(k=0\). </p>
<p>Suppose, that estimates ??–?? are satisfied for each \(k \le n\). Then, for \(k = n\) we have the following </p>
<div class="displaymath" id="a0000000026">
  \begin{align*}  t_{n+1}-t_{n+2} =&  \tfrac {(l_0+p_0)(t_n-t_{n+1})+q_0(t_{n-1}-t_n)(t_{n-2}-t_n)}{1-(2l_0^0+\lambda + \xi )(t_0-t_{n+1})-(\mu + \xi )(t_0-t_n)-\overline{q}_0a(a+b)} (t_n-t_{n+1}) \\ \leq & \tfrac {(l_0+p_0)c+q_0a(a+b)}{1-\overline{q}_0a(a+b)-2(l_0^0+p_0^0)r_1}(t_n-t_{n+1}) = \gamma (t_n-t_{n+1} ), \\ t_{n+1}\geq & t_{n+2} \geq t_{n+1}-\gamma (t_n-t_{n+1})\geq r_1 - \tfrac {(1-\gamma ^2)c}{1-\gamma } \geq r_1 - \tfrac {c}{1-\gamma } \geq 0. \end{align*}
</div>
<p>Hence, that \(\{ t_n\} _{n \geq -2}\) is decreasing, nonnegative sequence which converges to some \(t_*\geq 0\). Next, we show, that iterative process ?? is well defined, and following estimate is true for each \(n\geq -2\) </p>
<div class="displaymath" id="<plasTeX.TeXFragment object at 0x0000028658F5D2B0>">
  \begin{align} \label{EQ_R_5} \| x_n-x_{n+1}\|  \leq t_n - t_{n+1}, \;  n\geq -2. \end{align}
</div>
<p> In view of Lipschitz conditions ??–??, for \(k=n+1\), we obtain </p>
<div class="displaymath" id="a0000000027">
  \begin{align*} & \| I-A_0^{-1}A_{n+1}\| = \\ & =\| A_0^1(A_0-A_{n+1})\|  \\ & \le \| A_0^{-1}[F’(x_0)-F’(x_{n+1})]\|  \\ & \quad + \| A_0^{-1}\left[ Q(x_0,x_{-1})-Q(x_0,x_0)+Q(x_{-2},x_0)-Q(x_{-2},x_{-1})+Q(x_0,x_0) \right. \\ & \quad -\!  \left. Q(x_{n+1},x_0)\! +\! Q(x_{n+1},x_0)\! -\! Q(x_{n+1},x_n)\! +\! Q(x_{n-1},x_n)\! -\! Q(x_{n-1},x_{n+1}) \right] \|  \\ & = \| A_0^{-1}[F’(x_0)-F’(x_{n+1})]\| +\| A_0^{-1}[(Q(x_0,x_{-1},x_0)-Q(x_{-2},x_{-1},x_0)) \\ & \quad \times (x_{-1}-x_0) + Q(x_0,x_0)-Q(x_{n+1},x_0)+Q(x_{n+1},x_0)-Q(x_{n+1},x_n) \\ & \quad +Q(x_{n-1},x_n) -Q(x_{n-1},x_{n+1})]\|  \\ & \le 2l_0^0\| x_0-x_{n+1}\| +\overline{q}_0a(a+b) +\lambda \| x_0-x_{n+1}\|  +\mu \| x_0-x_n\|  + \xi \| x_n-x_{n+1}\|  \\ & \le 2(l_0^0+p_0^0)(t_0-t_{n+1})+\overline{q}_0a(a+b) \le 2(l_0^0+p_0^0)t_0+\overline{q}_0a(a+b) \\ & \le 2(l_0^0+p_0^0)r_1+\overline{q}_0a(a+b) {\lt} 1. \end{align*}
</div>
<p>Hence, \(A_{n+1}\) is invertible and </p>
<div class="displaymath" id="a0000000028">
  \begin{align*} & \| A_{n+1}^{-1}A_0\|  \le \\ &  [1\! -\! \overline{q}_0a(a+b)\! -\! 2l_0^0\| x_0\! -\! x_{n+1}\|  \! -\! \lambda \| x_0\! -\! x_{n+1}\| \! -\! \mu \| x_0\! -\! x_n\| \! -\! \xi \| x_n\! -\! x_{n+1}\| ]^{-1}. \end{align*}
</div>
<p>Taking into account the definition of the divided difference and conditions ??–?? we get in a turn </p>
<div class="displaymath" id="a0000000029">
  \begin{align*} & \| A_0^{-1}[F(x_{n+1})+Q(x_{n+1})]\| = \\ =&  \| A_0^{-1}[F(x_{n+1})+Q(x_{n+1})-F(x_n)-Q(x_n)-A_n(x_{n+1}-x_n)]\|  \\ \le &  \Big\| A_0^{-1} \int _0^1\left( F’(x_n+t(x_{n+1}-x_n))-F’(x_n)\right)dt\Big\| \| x_n-x_{n+1}\|  \\ & \quad + \| A_0^{-1}[Q(x_n,x_{n+1})\! -\! Q(x_n,x_n)\! +\! (Q(x_n,x_n,x_{n-1})\! -\! Q(x_{n-2},x_n,x_{n-1})) \\ & \quad \times (x_n\! -\! x_{n-1})]\|  \| x_n-x_{n+1}\|  \\ \le &  (l_0+p_0)\| x_n-x_{n+1}\| ^2 + q_0\| x_{n-2}-x_n\| \| x_{n-1}-x_n\| \| x_n-x_{n+1}\| . \end{align*}
</div>
<p> Hence </p>
<div class="displaymath" id="a0000000030">
  \begin{align*} & \| x_{n+1}-x_{n+2}\| =\| A_0^{-1}H(x_{n+1})\|  \le \| A_0^{-1}A_0\| \| A_0^{-1}[F(x_{n+1})+Q(x_{n+1})]\|  \\ & \le \tfrac {(l_0+p_0)\| x_n-x_{n+1}\| ^2+q_0\| x_{n-1}-x_n\| \| x_{n-2}-x_n\| \| x_n-x_{n+1}\| }{1-(2l_0^0+\lambda )\| x_0-x_{n+1}\| -\mu \| x_0-x_n\| -\xi \| x_n-x_{n+1}\| -\overline{q}_0a(a+b)} \\ & \le \tfrac {(l_0+p_0)(t_n-t_{n+1})+q_0(t_{n-1}-t_n)(t_{n-2}-t_n)}{1-(2l_0^0+\lambda + \xi )(t_0-t_{k+1})-(\mu + \xi )(t_0-t_n)-\overline{q}_0a(a+b)}(t_n-t_{n+1}) \\ & = t_{n+1}-t_{n+2}. \end{align*}
</div>
<p> That is, iterative process ?? is well defined for each \(n\). Moreover </p>
<div class="displaymath" id="<plasTeX.TeXFragment object at 0x0000028658FD19D0>">
  \begin{align} \label{AW_EQ_45} \| x_n-x_k\|  \le t_n-t_k,\; \; \; -2\le n \le k, \end{align}
</div>
<p> so the sequence \(\{ x_n\} _{n \ge 0}\) is fundamental, and as such convergent in the Banach space \(E_1\). By letting \(k \rightarrow \infty \) in ??, we get ??. </p>
<p>Let us show, that \(x_*\) is a root of equation \(F(x)+Q(x)=0\). </p>
<div class="displaymath" id="a0000000031">
  \begin{align*} & \| A_0^{-1}H(x_{n+1})\|  \le \\ & \le (l_0+p_0)\| x_n-x_{n+1}\| ^2 + q_0 \| x_n-x_{n-2}\| \| x_n-x_{n-1}\| \| x_n-x_{n+1}\|  \rightarrow 0 \end{align*}
</div>
<p> for \(n \rightarrow \infty \). Hence, \(F(x_*)+Q(x_*)=0\). </p>
<p>Next we will show the uniqueness of solution \(x_*\). Suppose, that \(x_{**} \in U_0(x_0,r_1)\), exists \(x_{**} \neq x_*\) and \(H(x_{**})=0\). Denote </p>
<div class="displaymath" id="a0000000032">
  \[ P=\int _0^1 F'(x_*+t(x_{**}-x_*))dt + Q(x_{**},x_*). \]
</div>
<p> Then \(P(x_{**}-x_*)=H(x_{**})-H(x_*)\). In case operator \(P^{-1}\) is invertible, we obtain, that \(x_{**}=x_*\). </p>
<div class="displaymath" id="a0000000033">
  \begin{align*} & \| I-A_0^{-1}P\|  = \| A_0^{-1}(A_0-P)\|  \leq \\ & \le \| A_0^{-1} \int _0^1 (F’(x_0)-F’(x_*+t(x_{**}-x_*)))dt \|  \\ & \quad + \| A_0^{-1}[Q(x_0,x_{-1})+Q(x_{-2},x_0)-Q(x_{-2},x_{-1})-Q(x_{**},x_*)]\|  \\ & \le (l_0^0+p_0^0)(\| x_0-x_*\| +\| x_0-x_{**}\| )+\overline{q}_0a(a+b) \\ & \le 2(l_0^0+p_0^0)r_1 + \overline{q}_0a(a+b) {\lt} 1. \end{align*}
</div>
<p> Hence, \(P^{-1}\) exists.  <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> </p>
<p><div class="theorem_thmwrapper " id="Teo3">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">7</span>
  </div>
  <div class="theorem_thmcontent">
  <p> Let conditions of <a href="#Theo2">theorem 6</a> are true. Then for each \(n\ge 1\) the following estimate is true </p>
<div class="displaymath" id="<plasTeX.TeXFragment object at 0x000002865901D8B0>">
  \begin{align} \label{AW_EQ_46} \| x_n-x_*\|  \le \tfrac {(l_0+p_0)(t_{n-1}-t_n)+q_0(t_{n-3}-t_{n-1})(t_{n-2}-t_{n-1})}{1- \overline{q}_0a(a+b)-(l_0^0+p_0^0)(2t_0-t_n)}(t_{n-1}-t_n). \end{align}
</div>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000035">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> Taking into account estimates ??–??, we get </p>
<div class="displaymath" id="a0000000036">
  \begin{align*} & \Big\| I-A_0^{-1}\Big(\int _0^1F’(x_*+t(x_n-x_*))dt+Q(x_n,x_*)\Big)\Big\|  \le \\ & \le \Big\| A_0^{-1}\int _0^1 (F’(x_0)-F’(x_*+t(x_n-x_*)))dt\Big\|  + \big\| A_0^{-1}[Q(x_0,x_{-1}) \\ & \quad - Q(x_0,x_0)+Q(x_{-2},x_0)-Q(x_{-2},x{-1})+Q(x_0,x_0)-Q(x_n,x_*)]\big\|  \\ & \le (l_0+p_0)(\| x_0-x_n\| +\| x_0-x_*\| )+q_0a(a+b) \\ & \le (l_0^0+p_0^0)(2t_0-t_n)+\overline{q}_0a(a+b) {\lt} 1. \end{align*}
</div>
<p>Hence, \(\int _0^1F'(x_*+t(x_n-x_*))dt+Q(x_n,x_*)\) is invertible and </p>
<div class="displaymath" id="a0000000037">
  \begin{align*} & \Big\| \Big(\int _0^1F’(x_*+t(x_n-x_*)\Big)dt+Q(x_n,x_*))^{-1}A_0\Big\| \le \\ & \le (1-\overline{q}_0a(a+b)-(l_0^0+p_0^0)(\| x_0-x_n\| +\| x_0-x_*|))^{-1}. \end{align*}
</div>
<p> Using the estimation </p>
<div class="displaymath" id="a0000000038">
  \begin{align*}  \| x_n-x_*\| =& \Big\| \Big(\int _0^1F’(x_*+t(x_n-x_*))dt+Q(x_n,x_*)\Big)^{-1}(H(x_n)-H(x_*))\Big\|  \\ \le &  \| (\int _0^1F’(x_*+t(x_n-x_*))dt+Q(x_n,x_*))^{-1}A_0\| \| A_0^{-1}H(x_n)\| , \end{align*}
</div>
<p> we obtain estimate ??.  <div class="proof_wrapper" id="a0000000039">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<p><div class="remark_thmwrapper " id="Rem2">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">8</span>
  </div>
  <div class="remark_thmcontent">
  <p> The corresponding conditions in <span class="cite">
	[
	<a href="#Cite_12_Sh" >25</a>
	]
</span> are given for each \(x,y,u,v \in D\) by </p>
<div class="displaymath" id="a0000000040">
  \begin{align*}  \| A_0^{-1}(F’(x)-F’(y))\|  \le & 2l_0^1\| x-y\| , \\ \| A_0^{-1}(Q(x,y)-Q(u,v))\|  \le & p_0^1(\| x-u\| +\| y-v\| ), \\ \| A_0^{-1}(Q(u,x,y)-Q(v,x,y))\|  \le &  q_0^1\| u-v\| , \\ q_0a(a+b){\lt}& 1, \\ \overline{r}_1 {\gt} & \tfrac {c}{1-\overline{\gamma }}, \\ \end{align*}
</div>
<p> where </p>
<div class="displaymath" id="a0000000041">
  \[  \overline{\gamma } = \tfrac {(l_0^1+p_0^1)c+q_0a(a+b)}{1-q_0^1a(a+b)-2(l_0^1+p_0^1)r_1}, \; \;  0{\lt} \overline{\gamma } {\lt} 1.  \]
</div>
<p> We have that \(D_1 \subseteq D\), so as in the local convergence case </p>
<div class="displaymath" id="a0000000042">
  \begin{align*}  l_0^0 \le l_0^1, \; \; \; \; \; \;  l_0 \le l_0^1, \\ p_0^0 \le p_0^1, \; \; \; \; \; \;  p_0 \le p_0^1, \\ q_0^0 \le q_0^1, \; \; \; \; \; \;  q_0 \le q_0^1 \end{align*}
</div>
<p> and the old majorizing sequence call it \(\{ s_n\} \) (using \(l_0^1,p_0^1,q_0^1)\) is less tight than \(t_n\) <span class="cite">
	[
	<a href="#Cite_12_Sh" >25</a>
	]
</span>. Hence, the applicability of solver ?? has been extended in the semilocal convergence too.<span class="qed">â–¡</span></p>

  </div>
</div> </p>
<h1 id="a0000000043">4 Numerical experiments</h1>
<p> In order to demonstrate the results of iterative solver ??, we carried out numerical experiments on test cases with nondifferentiable operator. The calculations are performed for different initial approximations with accuracy \(\varepsilon = 10^{-10}\). The iterative process was performed until following conditions are satisfied: </p>
<div class="displaymath" id="a0000000044">
  \[  \| x_{n+1}-x_n\| _{\infty } \le \varepsilon , \; \; \;  \| H(x_{n+1})\| _{\infty } \le \varepsilon .  \]
</div>
<p> Additional initial approximations were chosen by the following formula: </p>
<div class="displaymath" id="a0000000045">
  \[  x_{-1}=x_0-10^{-4}, \;  x_{-2}=x_0-2\cdot 10^{-4}.  \]
</div>
<p> To compare the convergence speed of the combined Newton-Potra solver with a basic solvers, the number of iterations, required to obtain a solution of systems of nonlinear equations, are presented in a table. The Newton-type solver for equation ?? has the form <span class="cite">
	[
	<a href="#Cite_5" >3</a>
	]
</span>: </p>
<div class="displaymath" id="<plasTeX.TeXFragment object at 0x0000028659046C90>">
  \begin{align} \label{EQ_20} x_{n+1}=x_n-[F’(x_n)]^{-1}H(x_n),\; n \ge 0, \end{align}
</div>
<p> and the Potra solver <span class="cite">
	[
	<a href="#Cite_9" >22</a>
	]
</span>: </p>
<div class="displaymath" id="<plasTeX.TeXFragment object at 0x0000028659078EF0>">
  \begin{align} \label{EQ_21} x_{n+1}=x_n-[H(x_n,x_{n-1})+H(x_{n-2},x_n)-H(x_{n-2},x_{n-1})]^{-1}H(x_n),\; n \ge 0. \end{align}
</div>
<p>Consider the system of two equations <div class="example_thmwrapper " id="a0000000046">
  <div class="example_thmheading">
    <span class="example_thmcaption">
    Example
    </span>
    <span class="example_thmlabel">9</span>
  </div>
  <div class="example_thmcontent">
  <div class="displaymath" id="a0000000047">
  \[ \left\{ \begin{array}{l} {4xy^{2} - x^{3} + y^{3} - 1 +|x|=0,} \\ {2y^{2} - x^{2}y^{2} +1+|x+y|=0.} \end{array}\right.  \]
</div>

  </div>
</div> The solution of this system is \((x_{*} ,y_{*} ) = (2, -1)\). The numerical results are presented in <a href="#tab:Tab1">table 1</a>. </p>
<div class="table"  id="tab:Tab1">
   <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:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p> Newton type </p>

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

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

    </td>
  </tr>
  <tr>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(x_0\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> solver ?? </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> solver ?? </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> solver ?? </p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>(1.1,0.1) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p> 23 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p> 19 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p> 14 </p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>(5,\(-\)5) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> 24 </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> 25 </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> 18 </p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>(1.85, \(-\)0.85) </p>

    </td>
    <td  style="text-align:center; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 13 </p>

    </td>
    <td  style="text-align:center; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 11 </p>

    </td>
    <td  style="text-align:center; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 7 </p>

    </td>
  </tr>
</table> </small> <figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">1</span> 
  <span class="caption_text">Number of iteration made to solve the problem, for initial approximation \(x_0\).</span> 
</figcaption>  </div>
</div>
<p>Consider the system of three equations. <div class="example_thmwrapper " id="a0000000048">
  <div class="example_thmheading">
    <span class="example_thmcaption">
    Example
    </span>
    <span class="example_thmlabel">10</span>
  </div>
  <div class="example_thmcontent">
  <div class="displaymath" id="a0000000049">
  \[ \left\{ \begin{array}{l} {z^2(1-y)-xy+|y-z^2|=0,} \\ {z^2(x^3-x)-y^2+|3y^2-z^2+1|=0,} \\ {6xy^3+y^2z^2-xy^2z+|x+z-y|=0} \end{array}\right.  \]
</div>

  </div>
</div> It is known, that one of solutions of the system is \((x_*,y_*,z_*)^T=(-1,2,3)^T\). The results of solvers ??, ??, ?? are presented in <a href="#tab:Tab2">table 2</a>. </p>
<div class="table"  id="tab:Tab2">
   <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:center" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p> Newton type </p>

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

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

    </td>
  </tr>
  <tr>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>\(x_0\) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> solver ?? </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> solver ?? </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> solver ?? </p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p>(\(-\)0.5, 2.3, 3.5) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p> 142 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p> 11 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center" 
        rowspan=""
        colspan="">
      <p> 10 </p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p>(\(-\)1.5, 2.5, 3.5) </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> 131 </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> 10 </p>

    </td>
    <td  style="text-align:center" 
        rowspan=""
        colspan="">
      <p> 8 </p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:center; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>(\(-\)10, 20, 30) </p>

    </td>
    <td  style="text-align:center; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 128 </p>

    </td>
    <td  style="text-align:center; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 23 </p>

    </td>
    <td  style="text-align:center; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 17 </p>

    </td>
  </tr>
</table> </small> <figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">2</span> 
  <span class="caption_text">Number of iteration made to solve the problem.</span> 
</figcaption>  </div>
</div>
<h1 id="a0000000050">5 Conclusions</h1>
<p> Based on the obtained results we showed the advantages of combined solver ?? over basic solvers, in particular, over Potra solver ??, even the theoretical convergence order of both solvers are the same. Moreover the convergence region of iterative solvers in general is small, which limits the choice of initial points. So by using the new Lipschitz constants we get at least as many initial points and fewer iterations to achieve predetermined accuracy, without any additional cost. This technique can be applied to extend the applicability of other iterative solvers. Therefore, the proposed combined solver ?? is an effective alternative for solving nonlinear equations with nondifferentiable operator. </p>
<p><small class="footnotesize">  </small></p>
<div class="bibliography">
<h1>Bibliography</h1>
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</dd>
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</dd>
  <dt><a name="Cite_11">24</a></dt>
  <dd><p><i class="sc">S.M. Shakhno, O.M. Makukh</i>, <i class="it">About iterative methods in conditions of Hölder continuity of the divided differences of the second order</i>, Matematychni Metody ta Fizyko-Mekhanichni Polya, <b class="bf">49</b> (2006) no. 2, pp. 90–98 (in Ukrainian). </p>
</dd>
  <dt><a name="Cite_12_Sh">25</a></dt>
  <dd><p><i class="sc">S.M. Shakhno, A.-V.I. Babjak, H.P. Yarmola</i> <i class="it">Combined Newton-Potra method for solving nonlinear operator equations</i>, Journal of Computational and Applied Mathematics, Kyiv, <b class="bf">3</b> (2015) 120, pp. 170–178 (in Ukrainian). </p>
</dd>
  <dt><a name="Cite_15_SYa">26</a></dt>
  <dd><p><i class="sc">S.M. Shakhno, H.P. Yarmola</i>, <i class="it">On convergence of Newton-Potra method under weak conditions</i>, Visnyk&#160;Lviv Univ. Ser. Appl. Math. Inform., <b class="bf">25</b> (2017), pp. 49–55 (in Ukrainian). </p>
</dd>
  <dt><a name="Cite_N_3">27</a></dt>
  <dd><p><a href ="https://doi.org/10.1201/9781315153469"> <i class="sc">I.K. Argyros, A.A. Magréñan</i>, <i class="it">Iterative Methods and Their Dynamics with Applications</i>, CRC Press, New York, NY, USA, 2017. <img src="img-0001.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
</dl>


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