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<title>The Convergence of the Euler’s method: The Convergence of the Euler’s method</title>
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<h1>The Convergence of the Euler’s method</h1>
<p class="authors">
<span class="author">Raluca Anamaria (Pomian) Salajan\(^\ast \)</span>
</p>
<p class="date">October 23, 2010.</p>
</div>
<p>\(^\ast \)Secondary School Vasile Alecsandri, Strada Păşunii, Nr. 2A, Baia Mare, Romania, e-mail: <span class="ttfamily">salajanraluca@yahoo.com</span>. </p>

<div class="abstract"><p> In this article we study the Euler’s iterative method. For this method we give a global theorem of convergence. In the last section of the paper we give a numerical example which illustrates the result exposed in this work. </p>
<p><b class="bf">MSC.</b> 37C25, 12D10. </p>
<p><b class="bf">Keywords.</b> Euler’s method, fixed point, one-point iteration method </p>
</div>
<h1 id="a0000000002">1 Introduction</h1>
<p>We consider the problem of finding a zero of the equation </p>
<div class="equation" id="a0000000003">
<p>
  <div class="equation_content">
    \begin{equation}  f(x)=0, \end{equation}
  </div>
  <span class="equation_label">1</span>
</p>
</div>
<p> where \(f:[a,b]\subset \mathbb {R}\rightarrow \mathbb {R}\) is an analytic function with simple roots. This zero can be determined as a fixed point of some iteration functions \(g:[a,b]\rightarrow [a,b]\), by means of the one-point iteration method </p>
<div class="equation" id="a0000000004">
<p>
  <div class="equation_content">
    \begin{equation}  x_{n+1}=g(x_{n}), \  x_{0} \in [a,b], \  n=0,1,..., \  n \in \mathbb {N}, \end{equation}
  </div>
  <span class="equation_label">2</span>
</p>
</div>
<p> where \(x_{0}\) is the starting value and \(g\) is a function of form </p>
<div class="displaymath" id="a0000000005">
  \begin{equation*}  g(x)=x+\varphi (x). \end{equation*}
</div>
<p>In this article we analyze the Euler’s method for approximating the solution \(x^{\ast }\in [a,b]\) of the equation (1). This method is defined by the relation </p>
<div class="equation" id="a0000000006">
<p>
  <div class="equation_content">
    \begin{equation}  x_{n+1}=x_{n}-\tfrac {2f(x_{n})}{f^{\prime }(x_{n})+\sqrt{[f^{\prime }(x_{n})]^{2}-2f(x_{n})f^{\prime \prime }(x_{n})}}, x_{0} \in [a,b],\  n\geq 0,\  n \in \mathbb {N}. \end{equation}
  </div>
  <span class="equation_label">3</span>
</p>
</div>
<p>The Euler’s method has been rediscovered by several authors, see for example [1], [2], [5], [6], [7], [8], and references therein. </p>
<h1 id="a0000000007">2 Theorems of convergence</h1>
<p>Next we will study sufficient conditions in order that the sequence \(\left\{ x_{n}\right\} _{n\geq 0}\) generated through (3) would be convergent, and if \(x^{\ast }=\lim \limits _{n\rightarrow \infty }x_{n}\), then <br />\(f(x^{\ast })=0\). </p>
<p>In order to prove the convergence of the method of form (3), we would use the next result. </p>
<p><div class="theorem_thmwrapper " id="a0000000008">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">1</span>
  </div>
  <div class="theorem_thmcontent">
  <p><span class="rm">(<span class="cite">
	[
	<a href="#3" >3</a>
	]
</span>, <span class="cite">
	[
	<a href="#4" >4</a>
	]
</span>)</span> If we consider the function \(f\),  the real number \(\delta {\gt}0\) and \(x_{0}\in \Delta ,\) where \(\  \Delta =\{ x\in \mathbb {R}:\left| x-x_{0 }\right| \leq \delta \} \subseteq [a,b]\), we could assure that the following relations hold </p>
<ul class="itemize">
  <li><p>the function \(f\) is of class \(C^{s}(\Delta )\), \(s\geq 2, \  s \in \mathbb {N}\) and \(\sup \limits _{x\in \Delta }\left| f^{(s)}(x)\right| = M{\lt}\infty ;\) </p>
</li>
  <li><p>we have the relation \(\left| \sum \limits _{i=0}^{s-1}\tfrac {1}{i!}f^{(i)}(x)\varphi ^{i}(x)\right| \) \(\leq \gamma \left| f(x)\right| ^{s}\)  for every <br />\(x\in \Delta ,\) where \(\gamma \in \mathbb {R,}\gamma \geq 0;\) </p>
</li>
  <li><p>the function \(\varphi \) verifies the relation  \(\left| \varphi (x)\right| \leq \eta \left| f(x)\right| ,\) for every \(x\in \Delta ,\) where \(\eta \in \mathbb {R}, \eta {\gt}0;\) </p>
</li>
  <li><p>the numbers \(\lambda ,\eta ,M\) and \(\delta \) verify the relations: </p>
<p>\(\mu _{0}=\lambda \left| f(x_{0})\right| {\lt}1\), where \(\lambda =\left( \gamma +\tfrac {M\eta ^{s}}{s!}\right) ^{\tfrac {1}{s-1}}\) and \(\tfrac {\eta \mu _{0}}{\lambda (1-\mu _{0})}\leq \delta ;\) </p>
</li>
</ul>
<p> then the sequence \(\left\{ x_{n}\right\} _{n\geq 0}\) generated by \((2)\) has the following properties: </p>
<ul class="itemize">
  <li><p>it is convergent, and if \(\  x^{\ast }=\lim \limits _{n\rightarrow \infty }x_{n} \) then \(f(x^{\ast })=0\) and \(x^{\ast }\in \Delta ;\) </p>
</li>
  <li><p>\(\left| x_{n+1}-x_{n}\right| \leq \tfrac {\eta \mu _{0}^{s^{n}}}{\lambda }\), for any \(n=0,1,..., n \in \mathbb {N};\) </p>
</li>
  <li><p>\(\left| x^{\ast }-x_{n}\right| \leq \tfrac {\eta \mu _{0}^{s^{n}}}{\lambda (1-\mu _{0}^{s^{n}})}, n=0,1,2,..., n \in \mathbb {N}.\) </p>
</li>
</ul>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000009">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>See [3], [4]. Based on Theorem 1,  in our next result we would analyze the convergence of sequence \(\{ x_{n}\} _{n\geq 0}\) given by (3). <div class="theorem_thmwrapper " id="a0000000010">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">2</span>
  </div>
  <div class="theorem_thmcontent">
  <p>If the function \(f\),  the real number \(\delta {\gt}0\) and \(x_{0}\in \Delta ,\) where \(\  \Delta =\{ x\in \mathbb {R}:\left| x-x_{0 }\right| \leq \delta \} \subseteq [a,b]\), verify the relations </p>
<ul class="itemize">
  <li><p>the function \(f\) is of class \(C^{3}(\Delta )\) and \(\  \sup \limits _{x\in \Delta }\left| f^{\prime \prime \prime }(x)\right| =M{\lt}\infty ;\) </p>
</li>
  <li><p>\(\  \  \left| \tfrac {1}{f^{\prime }(x)}\right| \leq \beta \)   for every \(x\in \Delta ,\  \beta \in \mathbb {R}, \beta {\gt}0; \) </p>
</li>
  <li><p>\(\  \tfrac {f(x)f^{\prime \prime }(x)}{[f^{\prime }(x)]^{2}}\stackrel{not}{=}L_{f}(x)\leq \tfrac {1}{2}\)  for every \(x\in \Delta ;\) </p>
</li>
  <li><p>\(\lambda =\sqrt{\tfrac {8M}{3!}\beta ^{3}}{\gt}0;\) </p>
</li>
  <li><p>\(\  \mu _{0}=\lambda \left| f(x_{0})\right| {\lt}1;\) </p>
</li>
  <li><p>\(\  \tfrac {2\beta \mu _{0}}{\lambda (1-\mu _{0})}\leq \delta ;\) </p>
</li>
</ul>
<p> then the sequence \(\{ x_{n}\} _{n\geq 0}\) generated by \((3)\) is convergent, and if \(x^{\ast }=\lim \limits _{n\rightarrow \infty }x_{n}\),  the next relations hold </p>
<ul class="itemize">
  <li><p>\(f(x^{\ast })=0\) and \(x^{\ast }\in \Delta ; \) </p>
</li>
  <li><p>\(x_{n}\in \Delta ,\) \(n=0,1,2..., n \in \mathbb {N};\) </p>
</li>
  <li><p>\(\left| f(x_{n})\right| \leq \tfrac {\mu _{0}^{3^{n}}}{\lambda },\  n=0,1,2,..., n \in \mathbb {N};\) </p>
</li>
  <li><p> \(\left| x^{\ast }-x_{n}\right| \leq \tfrac {2\beta \mu _{0}^{3^{n}}}{\lambda (1-\mu _{0}^{3^{n}})},\  n=0,1,2,..., n \in \mathbb {N}.\) </p>
</li>
</ul>

  </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> We consider the function \(\varphi \) of form </p>
<div class="displaymath" id="a0000000012">
  \begin{equation*}  \varphi (x)=-\tfrac {2f(x)}{f^{\prime }(x)+\sqrt{ [f^{\prime }(x)]^{2}-2f(x)f^{\prime \prime }(x)}},\  x \in \Delta . \end{equation*}
</div>
<p> We’ll show that the elements of the sequence \(\{ x_{n}\} _{n\geq 0}\) generated by (3) are in \(\Delta \). </p>
<p>By conditions b), c) and f) we have </p>
<div class="displaymath" id="a0000000013">
  \begin{align*}  \left| x_{1}-x_{0}\right| & =\left| \tfrac {f(x_{0})}{f^{\prime }(x_{0})}\right| \left| \tfrac {2}{1+\sqrt{1-2L_{f}(x_{0})}}\right| \leq 2\left| \tfrac {f(x_{0})}{f^{\prime }(x_{0})}\right|\leq \\ &  \leq 2\beta \left| f(x_{0})\right| =\tfrac {2\lambda \beta \left| f(x_{0})\right| }{\lambda }{\lt} \tfrac {2\beta \mu _{0}}{\lambda (1-\mu _{0})}\leq \delta \Rightarrow x_{1}\in \Delta . \end{align*}
</div>
<p>Applying the Taylor expansion of function \(f\) around \(x_{0}\) and taking into account that </p>
<div class="displaymath" id="a0000000014">
  \begin{align*}  \varphi (x)&  = \tfrac {-f^{\prime }(x)+\sqrt{[f^{\prime }(x)]^{2}-2f(x)f^{\prime \prime }(x)}}{f^{\prime \prime }(x)}=\\ & =-\tfrac {f^{\prime }(x)-\sqrt{[f^{\prime }(x)]^{2}-2f(x)f^{\prime \prime }(x)}}{f^{\prime \prime }(x)}, \end{align*}
</div>
<p> \(\forall x\in \Delta \), and \(\varphi (x)\) is verifying the parable </p>
<div class="displaymath" id="a0000000015">
  \begin{equation*}  ax^{2}+bx+c=0, \  \textnormal{where} \  c=f(x),b=f^{\prime }(x),a=\tfrac {f^{\prime \prime }(x)}{2}, \end{equation*}
</div>
<p> we get </p>
<div class="displaymath" id="a0000000016">
  \begin{align*}  \left| f(x_{1})\right| &  \leq \left| f(x_{1})-\left(f(x_{0})+f^{\prime }(x_{0})(x_{1}-x_{0})+\tfrac {1}{2}f^{\prime \prime }(x_{0})(x_{1}-x_{0})^{2}\right)\right| + \label{ineq:formula one} \\ & \quad +\left| f(x_{0})+f^{\prime }(x_{0})(x_{1}-x_{0})+\tfrac {1}{2}f^{\prime \prime }(x_{0})(x_{1}-x_{0})^{2}\right| \leq \\ & \leq \left| \tfrac {f^{\prime \prime \prime }(\xi )}{3!}(x_{1}-x_{0})^{3}\right|+\left|f(x_{0})+f^{\prime }(x_{0})\varphi (x_{0})+\tfrac {1}{2}f^{\prime \prime }(x_{0})\varphi ^2(x_{0}) \right| \leq \\ &  \leq \tfrac {M}{3!}\left| x_{1}-x_{0}\right| ^{3}\leq \tfrac {8M\beta ^{3}}{3!}\left| f(x_{0})\right| ^{3}=\tfrac {\mu _{0}^{3}}{\lambda }, \   \xi \in \Delta . \end{align*}
</div>
<p>Because \(\left| \tfrac {1}{f^{\prime }(x_{1})}\right| \leq \beta \),  we have that <br /> \(\footnotesize {\left| x_{2}-x_{1}\right| =\left| \tfrac {f(x_{1})}{f^{\prime }(x_{1})}\right| \left| \tfrac {2}{1+\sqrt{1-2\tfrac {f(x_{1})f^{\prime \prime }(x_{1})}{[f^{\prime }(x_{1})]^{2}}}}\right| \leq 2 \left| \tfrac {f(x_{1})}{f^{\prime }(x_{1})}\right| \leq 2 \beta \left| f(x_{1})\right| \leq \tfrac {2\beta \mu _{0}^{3}}{\lambda }}\). </p>
<p>From all that we have proved above, by using the induction, it results that the property iii) holds for every \(n\in \mathbb {N},\) </p>
<div class="equation" id="a0000000017">
<p>
  <div class="equation_content">
    \begin{equation}  \left| f(x_{n})\right| \leq \tfrac {\mu _{0}^{3^{n}}}{\lambda }. \end{equation}
  </div>
  <span class="equation_label">4</span>
</p>
</div>
<p> Analogously, from b), c) and (4) we can prove the following relation </p>
<div class="equation" id="a0000000018">
<p>
  <div class="equation_content">
    \begin{equation}  \  \left| x_{n+1}-x_{n}\right| =\left| \tfrac {f(x_{n})}{f^{\prime }(x_{n})}\right| \left| \tfrac {2}{1+\sqrt{1-2\tfrac {f(x_{n})f^{\prime \prime }(x_{n})}{[f^{\prime }(x_{n})]^{2}}}}\right| \leq \tfrac {2\beta \mu _{0}^{3^{n}}}{\lambda },\  n=0,1,..., n \in \mathbb {N}. \end{equation}
  </div>
  <span class="equation_label">5</span>
</p>
</div>
<p> From (5), e) and f) we get the relation ii) </p>
<div class="displaymath" id="ineq:formula one">
  \begin{align}  \left| x_{n+1}-x_{0}\right| &  \leq \sum \limits _{i=0}^{n}\left| x_{i+1}-x_{i}\right| \label{ineq:formula one}\leq \\ & \leq \sum \limits _{i=0}^{n}\tfrac {2\beta \mu _{0}^{3^{i}}}{\lambda }\leq \tfrac {2\beta \mu _{0}}{\lambda }(1+\mu _{0}^{3-1}+\mu _{0}^{3^{2}-1}+...+\mu _{0}^{3^{n}-1}) \nonumber \\ & {\lt} \tfrac {2\beta \mu _{0}}{\lambda (1-\mu _{0})}\leq \delta \Rightarrow x_{n+1}\in \Delta ,\  n=0,1,2,..., n \in \mathbb {N}. \nonumber \end{align}
</div>
<p>For the convergence of the sequence given by (3) we shall use the Cauchy’s theorem. By relation (5) and e) we deduce that </p>
<div class="displaymath" id="ineq:formula one">
  \begin{align}  \left| x_{n+p}-x_{n}\right| &  \leq \sum \limits _{i=n}^{n+p-1}\left| x_{i+1}-x_{i}\right| \leq \sum \limits _{i=n}^{n+p-1}\tfrac {2\beta \mu _{0}^{3^{i}}}{\lambda } \label{ineq:formula one}{\lt} \\ &  {\lt}\tfrac {2\beta \mu _{0}^{3^{n}}}{\lambda }(1+\mu _{0}^{3^{n+1}-3^{n}}+...+\mu _{0}^{3^{n+p-1}-3^{n}}) \nonumber \\ &  {\lt} \tfrac {2\beta \mu _{0}^{3^{n}}}{\lambda (1-\mu _{0}^{3^{n}})},\  p\in \mathbb {N},\  n=0,1,2,..., n \in \mathbb {N}. \nonumber \end{align}
</div>
<p>Because \(\mu _{0}{\lt}1\), it results that the sequence \(\{ x_{n}\} _{n\geq 0}\) is fundamental, so according to the Cauchy’s theorem, it is convergent. </p>
<p>If \(x^{\ast }=\lim \limits _{n\rightarrow \infty }x_{n},\) for \(p\rightarrow \infty \), from the inequality (7) we obtain the <br />relation iv) </p>
<div class="equation" id="a0000000019">
<p>
  <div class="equation_content">
    \begin{equation}  \left| x^{\ast }-x_{n}\right| \leq \tfrac {2\beta \mu _{0}^{3^{n}}}{\lambda (1-\mu _{0}^{3^{n}})},\  n=0,1,2,..., n \in \mathbb {N}. \end{equation}
  </div>
  <span class="equation_label">8</span>
</p>
</div>
<p>We show now that the relations i) hold, that is, \(x^{\ast }\) is a root of equation (1) and \(x^{\ast } \in \Delta \). </p>
<p>From the continuity of function \(f\) and from (4) for \(n\rightarrow \infty , \) it results </p>
<div class="displaymath" id="a0000000020">
  \begin{equation*}  0\leq \left| f(x^{\ast })\right| \leq \lim \limits _{n\rightarrow \infty }\tfrac {\mu _{0}^{3^{n}}}{\lambda }=0, \end{equation*}
</div>
<p> that is, \(f(x^{\ast })=0\). </p>
<p>From f) and the inequality (8) for \(n=0\), we obtain </p>
<div class="displaymath" id="a0000000021">
  \begin{equation*}  \left| x^{\ast }-x_{0}\right| \leq \tfrac {2\beta \mu _{0}^{3^{0}}}{\lambda (1-\mu _{0}^{3^{0}})}\leq \delta , \end{equation*}
</div>
<p> so, \(x^{\ast }\in \Delta \). <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> It is evidently that all the assumptions of Theorem 1 are verified for \(s=3\),  \(\gamma =0\) and \(\eta =2\beta \). </p>
<h1 id="a0000000023">3 Numerical example</h1>
<p> We shall present a numerical example, which illustrates the result exposed in Theorem 2. </p>
<p><div class="example_thmwrapper " id="Tab: one">
  <div class="example_thmheading">
    <span class="example_thmcaption">
    Example
    </span>
    <span class="example_thmlabel">3</span>
  </div>
  <div class="example_thmcontent">
  <p>We used the following test functions and display the zeros \(x^{\ast }\) found. </p>
<div class="displaymath" id="a0000000024">
  \begin{align*} &  f_{1}(x)={\rm ln}(x^2-3), \  x^{\ast }=-2, \  x \in [-2.35,-1.9],\\ &  f_{2}(x)=x^{3}-3x^{2}-13x+15, \  x^{\ast }=5, \  x \in [4.5,5.5],\\ &  f_{3}(x)=x^5-1, \  x^{\ast }=1, \  x \in [0.88,1.38]. \end{align*}
</div>
<p>For the derivatives of order 1, 2 and 3 of \(f_{i},\  i=1,2,3\), we have the relations </p>
<div class="displaymath" id="a0000000025">
  \begin{align*} & f^{\prime }_{1}(x)=\tfrac {2x}{-3+x^2}, \  f^{\prime \prime }_{1}(x)=\tfrac {-4x^2}{(-3+x^2)^2}+\tfrac {2}{-3+x^2}, \\ & f^{\prime \prime \prime }_{1}(x)=\tfrac {16x^3}{(-3+x^2)^3}-\tfrac {12x}{(-3+x^2)^2}, \end{align*}
</div>
<p> from which we get \(\beta =0.536702\) and \(M=422.22\); </p>
<div class="displaymath" id="a0000000026">
  \begin{align*}  f^{\prime }_{2}(x)=3x^2-6x-13, \  f^{\prime \prime }_{2}(x)=6x-6, \  f^{\prime \prime \prime }_{2}(x)=6, \end{align*}
</div>
<p> from which we get \(\beta =0.0481928\) and \(M=6\); </p>
<div class="displaymath" id="a0000000027">
  \begin{align*}  f^{\prime }_{3}(x)=5x^4, \  f^{\prime \prime }_{3}(x)=20x^3, \  f^{\prime \prime \prime }_{3}(x)=60x^2, \end{align*}
</div>
<p> from which we get \(\beta =0.333503\) and \(M=114.264\). </p>
<p>In the Table 1 are listed the values for \(x_{0}\), \(M\), \(\beta \), \(\lambda \), \(\mu _{0}\), \(\delta \) and \(\tfrac {2\beta \mu _{0}}{\lambda (1-\mu _{0})}\), for each test functions. </p>
<div class="table"  id="a0000000028">
   <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; border-right:1px solid black; border-left:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(i\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(x_{0}\) </p>

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

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

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

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

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

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(\tfrac {2\beta \mu _{0}}{\lambda (1-\mu _{0})}{\lt}\delta \) </p>

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

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> -1.989 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 422.22 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 0.53670 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 9.32908 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 0.41860 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 0.089 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 0.08284 </p>

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

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 4.875 </p>

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

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 0.04819 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 0.02992 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 0.11414 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 0.625 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 0.41503 </p>

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

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 1.035 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 114.264 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 0.33350 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 2.37724 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 0.33873 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 0.353 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 0.14372 </p>

    </td>
  </tr>
</table></small><figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">1</span> 
  <span class="caption_text"></span> 
</figcaption></div>
</div>
<p>The implementations were done in Mathematica 7.0 with double precision. From the Table 1 we can conclude that all the assumptions a)–f) of Theorem 2 are verified. </p>
<p>In the next Table 2 we can observe that, the convergence is faster and the method (3) converges at \(x^{\ast }\). </p>
<div class="table"  id="a0000000029">
   <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; border-right:1px solid black; border-left:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(i\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(x_{0}\) </p>

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

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(x_{2}\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(x_{3}=x^{\ast }\) </p>

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

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> -1.989 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> -2.0000063482540900 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> -1.9999999999999990 </p>

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

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

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 4.875 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 5.0000611233335144 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 4.999999999999993 </p>

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

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

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 1.027 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 0.999958832170524 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:center; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 1.000000000000139 </p>

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

    </td>
  </tr>
</table></small><figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">2</span> 
  <span class="caption_text"></span> 
</figcaption></div>
</div>

  </div>
</div> </p>
<p><small class="footnotesize">  </small></p>
<div class="bibliography">
<h1>Bibliography</h1>
<dl class="bibliography">
  <dt><a name="1">1</a></dt>
  <dd><p><i class="sc">Amat, S.</i>, <i class="sc">Busquier, S.</i> and <i class="sc">Plaza, S.</i>, <i class="it">Review of some iterative root-finding methods from a dynamical point of view</i>, Scientia, Series A: Mathematical Sciences, <b class="bf">10</b>, pp.&#160;3–35, 2004. </p>
</dd>
  <dt><a name="2">2</a></dt>
  <dd><p><i class="sc">Osada, N.</i>, <i class="it">A one parameter family of locally quartically convergent zero-finding methods</i>, J. Comput. Appl. Math., <b class="bf">205</b>, pp.&#160;116–128, 2007. </p>
</dd>
  <dt><a name="3">3</a></dt>
  <dd><p><i class="sc">Păvăloiu, I.</i>, <i class="it">Sur les procedées itérative à un order élevé de convergence</i>, Mathématica, <b class="bf">12(35)</b>, no. 2, pp.&#160;309–324, 1970. </p>
</dd>
  <dt><a name="4">4</a></dt>
  <dd><p><i class="sc">Păvăloiu, I.</i> and <i class="sc">Pop, N.</i>, <i class="it">Interpolare şi aplicaţii </i>, Editura Risoprint, Cluj-Napoca, 2005. </p>
</dd>
  <dt><a name="5">5</a></dt>
  <dd><p><i class="sc">Petković, L. D.</i>, <i class="sc">Petković, M. S.</i> and <i class="sc">Z̆iviković, D.</i>, <i class="it">Hansen-Patrick’s family is of Laguerre’s type</i>, Novi Sad J. Math., <b class="bf">33</b>, no. 1, pp.&#160;109–115, 2003. </p>
</dd>
  <dt><a name="6">6</a></dt>
  <dd><p><i class="sc">Petrović, M.</i>, <i class="sc">Tričković, S.</i> and <i class="sc">Herceg, D.</i>, <i class="it">Higher order Euler-like methods</i>, Novi Sad J. Math., <b class="bf">28</b>, no. 3, pp.&#160;129–136, 1998. </p>
</dd>
  <dt><a name="7">7</a></dt>
  <dd><p><i class="sc">Varona, J.</i>, <i class="it">Graphic and numerical comparison between iterative methods</i>, The Mathematical Intelligencer, <b class="bf">24</b>, no. 1, pp.&#160;37–46, 2002. </p>
</dd>
  <dt><a name="8">8</a></dt>
  <dd><p><i class="sc">Ye, X.</i> and <i class="sc">Li, C.</i>, <i class="it">Convergence of the family of the deformed Euler-Halley iterations under the Hölder condition of the second derivative</i>, Journal of Computational and Applied Mathematics, <b class="bf">194</b>, pp.&#160;294–308, 2006. </p>
</dd>
</dl>


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