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<title>Newton Type Iterative Methods with Higher Order of Convergence : Newton Type Iterative Methods with Higher Order of Convergence </title>
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<h1>Newton Type Iterative Methods with Higher Order of Convergence </h1>
<p class="authors">
<span class="author">Pankaj Jain\(^\ast \), Chet Raj Bhatta\(^\S \) Jivandhar Jnawali\(^{\ast \ast }\)</span>
</p>
<p class="date">August 26, 2015.</p>
</div>
<p>\(^\ast \)Department of Mathematics, South Asian University, Akbar Bhawan, Chanakya Puri, New Delhi-110021, India, e-mail: <span class="tt">pankaj.jain@sau.ac.in</span> and <span class="tt">pankajkrjain@hotmail.com</span> </p>
<p>\(^\S \)Central Department of Mathematics, Tribhuvan University, Kirtipur Kathmandu, Nepal, e-mail: <span class="tt">chetbhatta0@gmail.com</span>. </p>
<p>\(^{\ast \ast }\)Central Department of Mathematics, Tribhuvan University, Kirtipur Kathmandu, Nepal, e-mail: <span class="tt">jnawalij@gmail.com</span>. </p>

<div class="abstract"><p> Newton type iterative methods with higher order of convergence are obtained. The order of convergence is further increased by amalgamating these methods with the standard secant method. The methods are compared to the similar recent methods. </p>
<p><b class="bf">MSC.</b> 65H05 </p>
<p><b class="bf">Keywords.</b> Newton method, secant method, iterative method, nonlinear equation, order of convergence. </p>
</div>
<h1 id="a0000000002">1 Introduction</h1>
<p> Quite often, we come across numerous nonlinear equations which need to be solved. If the equation is not a polynomial equation, then it is not always easy to deal with such equations. To this end, one or the other numerical iterative method is employed. One such classical standard method is the Newton method </p>
<div class="displaymath" id="a0000000003">
  \begin{equation*}  x_{n+1}= x_n -\tfrac {f(x_n)}{f’(x_n)} \end{equation*}
</div>
<p> which is quadratically convergent. Over the years, a lot of methods have appeared, each one claims to be better than the other in some or the other aspect. We mention here the method given by Weerakoon and Fernando <span class="cite">
	[
	<a href="#wf" >8</a>
	]
</span> which is based on the Newton’s theorem </p>
<div class="displaymath" id="a0000000004">
  \begin{equation*}  f(x)=f(x_n)+ \int _{x_n}^x \  f’(\lambda )\, d{\lambda } \end{equation*}
</div>
<p>and the integral involved is approximated by the trapezoidal rule, i.e., </p>
<div class="displaymath" id="a0000000005">
  \begin{equation*}  \int _{x_n}^x \  f’(\lambda )\, d{\lambda } =\tfrac {(x-x_n)}{2}\big(f’(x)+f’(x_n)\big). \end{equation*}
</div>
<p> As a result, Weerakoon and Fernando obtained the following iterative method for solving the nonlinear equation\(f(x)=0\) : </p>
<div class="equation" id="f1.1">
<p>
  <div class="equation_content">
    \begin{equation}  x_{n+1} =x_n -\tfrac {2f(x_n)}{f’(x_n)+f’({z_{n+1}})}, \label{f1.1} \end{equation}
  </div>
  <span class="equation_label">1</span>
</p>
</div>
<p> where \({z_{n+1}} = x_n -\tfrac {f(x_n)}{f’(x_n)}.\) </p>
<p>The method so obtained is of third order. In the present paper, the aim is to modify method (<a href="#f1.1">1</a>). In fact, in (<a href="#f1.1">1</a>), \(f'\) is a function of the previously calculated iterate. In our modification, \(f'\) would be a function of some other convenient point. It is proved that the corresponding method has order of convergence 5.1925. We follow the technique of McDougall and Wotherspoon <span class="cite">
	[
	<a href="#mw" >7</a>
	]
</span> who modified Newton’s method in a similar way yielding the order of convergence of their method as \(1+\sqrt{2}.\) </p>
<p>Further, in <span class="cite">
	[
	<a href="#jaind2" >3</a>
	]
</span>, it was proved that if any method for solving nonlinear equation is used in conjunction with the standard secant method then the order of the resulting method is increased by 1. We shall show, in this paper (see Theorem 3.2), that this order can be increased by more than 1. In fact, we prove that if our own method (which is of order 5.1925) is combined with the secant method than the new method is of order 7.275. </p>
<h1 id="a0000000006">2 The Method and the Convergence</h1>
<p> We propose the following method: </p>
<p>If \(x_0\) is the initial approximation, then </p>
<div class="equation" id="f2.1">
<p>
  <div class="equation_content">
    \begin{equation} \left. \begin{aligned}  x^*_0 & =x_0\\ x_1 & =x_0^* -\tfrac {2f(x_0^*)}{f’(x_0^*)+f’({z_1})},\\ {\text{where}}\qquad {z_1}&  = x_0 -\tfrac {f(x_0)}{f’(x_0)}\\ x_1^{*} &  =x_1 -\tfrac {2f(x_1)}{f’(x_1)+f’({z_1^*})},\\ \text{with}\qquad z_1^{*} &  =x_1-\tfrac {f(x_1)}{f’[\tfrac {1}{2}(x_0+x_0^*)]}=x_1 -\tfrac {f(x_1)}{f’(x_0)}. \end{aligned}\right\rbrace \label{f2.1} \end{equation}
  </div>
  <span class="equation_label">2</span>
</p>
</div>
<p> Subsequently, for \(n\geq 1\), the iterations can be obtained as follows: </p>
<div class="equation" id="f2.2">
<p>
  <div class="equation_content">
    \begin{equation} \left. \begin{aligned}  x_n^* & =x_n -\tfrac {2f(x_n)}{f’(x_n)+f’({z_n^*})},\\ \text{where}\qquad {z_n^*}&  = x_n -\tfrac {f(x_n)}{f’[\tfrac {1}{2}(x_{n-1}+x_{n-1}^*)]}\\ x_{n+1} &  =x_n^* -\tfrac {2f(x_n^*)}{f’(x_n^*)+f’({z_{n+1}})},\\ \text{with}\qquad z_{n+1} &  =x_n-\tfrac {f(x_n)}{f’[\tfrac {1}{2}(x_n+x_n^*)]}. \end{aligned}\right\rbrace \label{f2.2} \end{equation}
  </div>
  <span class="equation_label">3</span>
</p>
</div>
<p> Below, we prove the convergence result for the method (<a href="#f2.1">2</a>)–(<a href="#f2.2">3</a>). <div class="theorem_thmwrapper " id="t2.1">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">1</span>
  </div>
  <div class="theorem_thmcontent">
  <p> Let \(\alpha \) be a simple zero of a function \( f \) which has sufficient number of smooth derivatives in a neighborhood of \(\alpha \). Then the method <span class="rm">(<a href="#f2.1">2</a>)–(<a href="#f2.2">3</a>)</span> is convergent and has the order of convergence \(5.1925\). </p>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000007">
  <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 \(e_n\) and \(e_n^*\) denote respectively the errors in the terms \(x_n\) and \(x_n^*\). Also, we denote \(c_j=\tfrac {f^j(\alpha )}{j!f’(\alpha )},\;  j=2,3,4...\), which are constants. The error equation for the method (<a href="#f1.1">1</a>) as obtained by Weerakoon and Fernando <span class="cite">
	[
	<a href="#wf" >8</a>
	]
</span> is given by </p>
<div class="displaymath" id="a0000000008">
  \[ e_{n+1}= a e_n^3,  \]
</div>
<p> where \(a=c_2^2+\tfrac {1}{2} c_3\) and we have neglected higher power terms of \(e_n\). In particular, the error \(e_1\) in \(x_1\) in the equations (<a href="#f2.1">2</a>) is given by </p>
<div class="equation" id="f2.3">
<p>
  <div class="equation_content">
    \begin{equation}  e_1= a e_0^3.\label{f2.3} \end{equation}
  </div>
  <span class="equation_label">4</span>
</p>
</div>
<p> We now proceed to calculate the error \(e_1^*\) in \(x_1^*\). By using Taylor series expansion and binomial expansion, we get </p>
<div class="displaymath" id="a0000000009">
  \begin{equation*} \begin{aligned}  \tfrac {f(x_1)}{f’(x_0)}&  =\tfrac {f(\alpha +e_1)}{f’(\alpha +e_0)}\\ & =\big(e_1+c_2e_1^2+c_3e_1^3+{\mathcal O}(e_1^4)\big)\big(1+2c_2e_0+3c_3e_0^2+{\mathcal O}(e_0^3)\big)^{-1}\\ & = e_1- 2c_2e_0 e_1+{\mathcal O}(e_0^5) \end{aligned}\end{equation*}
</div>
<p> so that </p>
<div class="displaymath" id="a0000000010">
  \begin{equation*}  x_1-\tfrac {f(x_1)}{f’(x_0)} = \alpha + 2c_2e_0 e_1+{\mathcal O}(e_0^5). \end{equation*}
</div>
<p> Consequently, by Taylor series expansion, it can be calculated that </p>
<div class="displaymath" id="a0000000011">
  \begin{equation*}  f’(z_1^*) =f’(\alpha )\big(1+4c_2^2e_0 e_1 +{\mathcal O}(e_0^5)\big). \end{equation*}
</div>
<p> Also </p>
<div class="displaymath" id="a0000000012">
  \begin{equation*}  f’(x_1) =f’(\alpha )\big(1+2c_2^2 e_1+3c_3e_1^2 +{\mathcal O}(e_1^3)\big) \end{equation*}
</div>
<p> so that </p>
<div class="equation" id="f2.4">
<p>
  <div class="equation_content">
    \begin{equation}  f'(x_1)+f'(z_1^*) = 2f'(\alpha )\big(1+c_2 e_1+2c_2^2e_0 e_1 +{\mathcal O}(e_0^5)\big).\label{f2.4} \end{equation}
  </div>
  <span class="equation_label">5</span>
</p>
</div>
<p> Now, using (<a href="#f2.3">4</a>) and (<a href="#f2.4">5</a>), the error \(e_1^*\) in \(x_1^*\) in the equation (<a href="#f2.1">2</a>) can be calculated as </p>
<div class="displaymath" id="a0000000013">
  \begin{equation*} \begin{aligned}  e_1^* & = e_1-\big(e_1+c_2e_1^2 + {\mathcal O}(e_1^3)\big)\big(1+c_2 e_1+2c_2^2e_0 e_1 +{\mathcal O}(e_0^5)\big)^{-1}\\ &  = 2c_2^2e_0 e_1^2\\ &  = ba^2e_0^7, \end{aligned}\end{equation*}
</div>
<p> where \(b=2c_2^2\). Using \(e_1^*\), we now compute the error \(e_2\) in the term </p>
<div class="displaymath" id="a0000000014">
  \begin{equation*}  x_2 =x_1^* -\tfrac {2f(x_1^*)}{f’(x_1^*)+f’({z_2})}, \end{equation*}
</div>
<p> where </p>
<div class="displaymath" id="a0000000015">
  \begin{equation*}  {z_2} = x_1 -\tfrac {f(x_1)}{f’\big(\tfrac {x_1+x_1^*}{2}\big)}. \end{equation*}
</div>
<p> Now </p>
<div class="displaymath" id="a0000000016">
  \begin{align*}  f’\big(\tfrac {x_1+x_1^*}{2}\big) & = f’\big(\alpha +\tfrac {e_1+e_1^*}{2}\big)\\ & =f’(\alpha )\big(1+c_2 e_1+c_2e_1^* +\tfrac {3}{4}c_3e_1^2 +{\mathcal O}(e_0^9)\big) \end{align*}
</div>
<p> so that </p>
<div class="displaymath" id="a0000000017">
  \begin{align*}  \tfrac {f(x_1)}{f’\big(\frac{x_1+x_1^*}{2}\big)}& =\big( e_1+c_2e_1^2+{\mathcal O}(e_1^3)\big)\big(1+c_2 e_1+c_2e_1^* +\tfrac {3}{4}c_3e_1^2 +{\mathcal O}(e_0^9)\big)^{-1}\\ & =e_1+\tfrac {1}{4}c_3e_1^3-c_2e_1e_1^* \end{align*}
</div>
<p> and therefore </p>
<div class="displaymath" id="a0000000018">
  \begin{equation*}  z_2= \alpha -\tfrac {1}{4}c_3e_1^3+c_2e_1e_1^*, \end{equation*}
</div>
<p> where the higher power terms are neglected. Thus </p>
<div class="displaymath" id="a0000000019">
  \begin{equation*}  f’(z_2)=f’(\alpha )\big(1-\tfrac {1}{2}c_2c_3 e_1^3+2c_2^2e_1e_1^*\big) \end{equation*}
</div>
<p> and </p>
<div class="displaymath" id="a0000000020">
  \begin{equation*}  f’(x_1^*) =f’(\alpha )\big(1+2c_2 e_1^*+3c_3{e_1^*}^2\big). \end{equation*}
</div>
<p> Using the above considerations, the error \(e_2\) in \(x_2\) is given by </p>
<div class="displaymath" id="a0000000021">
  \begin{equation*} \begin{aligned}  e_2& =e_1^*-\big( e_1^*+ c_2{e_1^*}^2 +c_3e_1^{*3}\big)\big(1+c_2 e_1^*-\tfrac {1}{4}c_2c_3e_1^3\big)^{-1}\\ & =-\tfrac {1}{4}c_2c_3e_1^3 e_1^*\\ & =ce_1^3 e_1^*, \end{aligned}\end{equation*}
</div>
<p> where \(c=-\tfrac {1}{4}c_2c_3\). In fact, it can be worked out that for \(n\geq 1\), the following relation holds: </p>
<div class="equation" id="f2.5">
<p>
  <div class="equation_content">
    \begin{equation}  e_{n+1}=ce_n^3 e_n^*.\label{f2.5} \end{equation}
  </div>
  <span class="equation_label">6</span>
</p>
</div>
<p> In order to compute \(e_{n+1}\) explicitly, we need to compute \(e_n^*\). We already know \(e_1^*\). We now compute \(e_2^*\). We have </p>
<div class="displaymath" id="a0000000022">
  \begin{equation*}  x_2^* =x_2 -\tfrac {2f(x_2)}{f’(x_2)+f’({z_2^*})}, \end{equation*}
</div>
<p> where </p>
<div class="displaymath" id="a0000000023">
  \begin{equation*}  {z_2^*} = x_2 -\tfrac {f(x_2)}{f’\big(\tfrac {x_1+x_1^*}{2}\big)}. \end{equation*}
</div>
<p> Like above, it can be calculated that the error \(e_2^*\) is given by </p>
<div class="displaymath" id="a0000000024">
  \begin{equation*}  e_2^*=d e_1 e_2^2, \end{equation*}
</div>
<p> where \(d=c_2^2\) and, again, it can be checked that in general, for \(n\geq 2\), the following relation holds: </p>
<div class="equation" id="f2.6">
<p>
  <div class="equation_content">
    \begin{equation}  e_n^*=d e_{n-1} e_n^2.\label{f2.6} \end{equation}
  </div>
  <span class="equation_label">7</span>
</p>
</div>
<p> In the view of (<a href="#f2.5">6</a>) and (<a href="#f2.6">7</a>), the error at each stage in \(x_n^*\) and \(x_{n+1}\) are calculated which are tabulated below: </p>
<div class="table"  id="tab1">
   <div class="centered"> <table class="tabular">
  <tr>
    <td  style="text-align:right; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(n\) </p>

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

    </td>
    <td  style="text-align:right" 
        rowspan=""
        colspan="">
      <p>\(e_n^*\)</p>

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

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:right; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \( e_0\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:right" 
        rowspan=""
        colspan="">
      <p>\(e_0\)</p>

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

    </td>
    <td  style="text-align:right; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(a e_0^3\) </p>

    </td>
    <td  style="text-align:right" 
        rowspan=""
        colspan="">
      <p> \(a^2 b e_0^7\)</p>

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

    </td>
    <td  style="text-align:right; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(a^5 b c e_0^{16}\) </p>

    </td>
    <td  style="text-align:right" 
        rowspan=""
        colspan="">
      <p> \(a^{11} b^2 c^2 d e_0^{35}\)</p>

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

    </td>
    <td  style="text-align:right; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(a^{26} b^5 c^6 d e_0^{83} \)</p>

    </td>
    <td  style="text-align:right" 
        rowspan=""
        colspan="">
      <p> \(a^{57} b^{11} c^{13} d^3 e_0^{182}\)</p>

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

    </td>
    <td  style="text-align:right; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>\(a^{135} b^{26} c^{32} d^6 e_0^{431}\) </p>

    </td>
    <td  style="text-align:right" 
        rowspan=""
        colspan="">
      <p> \(a^{296} b^{57} c^{70} d^{14} e_0^{945}\)</p>

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

    </td>
    <td  style="text-align:right; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(a^{701} b^{135} c^{167} d^{32} e_0^{2238}\) </p>

    </td>
    <td  style="text-align:right" 
        rowspan=""
        colspan="">&nbsp;</td>
  </tr>
  <tr>
    <td  style="text-align:right; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:right; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:right; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">&nbsp;</td>
  </tr>
</table> </div> <figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">1</span> 
  <span class="caption_text">Successive errors.</span> 
</figcaption>  
</div>
<p> It is observed that the powers of \(e_0\) in the errors at each iterate form a sequence </p>
<div class="equation" id="f2.7">
<p>
  <div class="equation_content">
    \begin{equation}  3,\;  16,\;  83,\;  431,\;  2238,\; ...\label{f2.7} \end{equation}
  </div>
  <span class="equation_label">8</span>
</p>
</div>
<p> and the sequence of their successive ratios is </p>
<div class="displaymath" id="a0000000025">
  \begin{equation*}  \tfrac {16}{3},\;  \tfrac {83}{16},\;  \tfrac {431}{83},\;  \tfrac {2238}{431},\;  ... \end{equation*}
</div>
<p> or,</p>
<div class="displaymath" id="a0000000026">
  \begin{equation*}  5.3334,\;  5.1875,\;  5.1927,\;  5.1925,\;  ... \end{equation*}
</div>
<p> This sequence seems to converge to the number 5.1925 approximately. Indeed, if the terms of the sequence (<a href="#f2.7">8</a>) are denoted by \({\{ \, \alpha _i}\, \} \), then it can be seen that </p>
<div class="equation" id="f2.8">
<p>
  <div class="equation_content">
    \begin{equation}  \alpha _i =5\alpha _{i-1} +\alpha _{i-2},\qquad i=2,3,4...\label{f2.8} \end{equation}
  </div>
  <span class="equation_label">9</span>
</p>
</div>
<p> If we set the limit </p>
<div class="displaymath" id="a0000000027">
  \begin{equation*}  \tfrac {\alpha _i}{\alpha _{i-1}} =\tfrac {\alpha _{i-1}}{\alpha _{i-2}}=R, \end{equation*}
</div>
<p> Then dividing (<a href="#f2.8">9</a>) by \(\alpha _{i-1}\), we obtain </p>
<div class="displaymath" id="a0000000028">
  \begin{equation*}  R^2-5R-1= 0 \end{equation*}
</div>
<p> which has its positive root as \(R=\tfrac {5+\sqrt{29}}{2}\approx 5.1925\). Hence the order of convergence of the method is at least \(5.1925\). </p>
<p>Next, we give two variants of the method (<a href="#f2.1">2</a>)–(<a href="#f2.2">3</a>). Note that, in (<a href="#f2.1">2</a>)–(<a href="#f2.2">3</a>), the arithmetic average of the points \(x_n, x_n^*, \;  n=0 ,1 ,2...\) has been used. We propose methods in which the arithmetic average is replaced by harmonic as well as geometric averages. With harmonic average, we propose the following method: If \(x_0\) is the initial approximation, then </p>
<div class="equation" id="f2.9">
<p>
  <div class="equation_content">
    \begin{equation} \left. \begin{aligned}  x^*_0 & =x_0\\ x_1 & =x_0^* -\tfrac {2f(x_0^*)}{f’(x_0^*)+f’({z_1})},\\ \text{where}\qquad {z_1}&  = x_0-\tfrac {f(x_0)}{f’\left(\tfrac {2x_0 x_0^*}{x_0+x_0^*}\right)}= x_0 -\tfrac {f(x_0)}{f’(x_0)}\\ x_1^{*} &  =x_1 -\tfrac {2f(x_1)}{f’(x_1)+f’({z_1^*})},\\ \text{with}\qquad z_1^{*} &  =x_1-\tfrac {f(x_1)}{f’\left(\tfrac {2x_0 x_0^*}{x_0+x_0^*}\right)}=x_1 -\tfrac {f(x_1)}{f’(x_0)}. \end{aligned}\right\rbrace \label{f2.9} \end{equation}
  </div>
  <span class="equation_label">10</span>
</p>
</div>
<p> Subsequently, for \(n\geq 1\), the iterations can be obtained as follows: </p>
<div class="equation" id="f2.10">
<p>
  <div class="equation_content">
    \begin{equation} \left. \begin{aligned}  x_n^* & =x_n -\tfrac {2f(x_n)}{f’(x_n)+f’({z_n^*})},\\ \text{where}\qquad {z_n^*}&  = x_n-\tfrac {f(x_n)}{f’\left(\tfrac {2x_{n-1}x_{n-1}^*}{x_{n-1}+x_{n-1}^*}\right)}\\ x_{n+1} &  =x_n^* -\tfrac {2f(x_n^*)}{f’(x_n^*)+f’({z_{n+1}})},\\ \text{with}\qquad z_{n+1} &  =x_n-\tfrac {f(x_n)}{f’\left(\tfrac {2x_n x_n^*}{x_n+x_n^*}\right)}. \end{aligned}\right\rbrace \label{f2.10} \end{equation}
  </div>
  <span class="equation_label">11</span>
</p>
</div>
<p> For the geometric average of the points \(x_n, x_n^*, \;  n=0 ,1 ,2...\), the following method is proposed: </p>
<div class="equation" id="f2.11">
<p>
  <div class="equation_content">
    \begin{equation} \left. \begin{aligned}  x^*_0 & =x_0\\ x_1 & =x_0^* -\tfrac {2f(x_0^*)}{f’(x_0^*)+f’({z_1})},\\ \text{where}\qquad {z_1}&  = x_0-\tfrac {f(x_0)}{f’\left(\sqrt{x_0 x_0^*}\, \right)}= x_0 -\tfrac {f(x_0)}{f’(x_0)}\\ x_1^{*} &  =x_1 -\tfrac {2f(x_1)}{f’(x_1)+f’({z_1^*})},\\ \text{with}\qquad z_1^{*} &  =x_1-\tfrac {f(x_1)}{f’\left(\sqrt{x_0 x_0^*}\, \right)}=x_1 -\tfrac {f(x_1)}{f’(x_0)}. \end{aligned}\right\rbrace \label{f2.11} \end{equation}
  </div>
  <span class="equation_label">12</span>
</p>
</div>
<p> Subsequently, for \(n\geq 1\), the iteration can be obtained as follows: </p>
<div class="equation" id="f2.12">
<p>
  <div class="equation_content">
    \begin{equation} \left. \begin{aligned}  x_n^* & =x_n -\tfrac {2f(x_n)}{f’(x_n)+f’({z_n^*})},\\ \text{where}\qquad {z_n^*}&  = x_n-\tfrac {f(x_n)}{f’\left(\sqrt{x_{n-1}x_{n-1}^*}\, \right)}\\ x_{n+1} &  =x_n^* -\tfrac {2f(x_n^*)}{f’(x_n^*)+f’({z_{n+1}})},\\ \text{with}\qquad z_{n+1} &  =x_n-\tfrac {f(x_n)}{f’\left(\sqrt{x_n x_n^*}\, \right)}. \end{aligned}\right\rbrace \label{f2.12} \end{equation}
  </div>
  <span class="equation_label">13</span>
</p>
</div>
<p>The convergence of the methods (<a href="#f2.9">10</a>)–(<a href="#f2.10">11</a>) and (<a href="#f2.11">12</a>)–(<a href="#f2.12">13</a>) can be proved on the similar lines as those in Theorem <a href="#t2.1">1</a>. We only state the results below: <div class="theorem_thmwrapper " id="t2.2">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">2</span>
  </div>
  <div class="theorem_thmcontent">
  <p> Let \(\alpha \) be a simple zero of a function \( f \) which has sufficient number of smooth derivatives in a neighborhood of \(\alpha \). Then for solving nonlinear equation \(f(x)= 0\), the method <span class="rm">(<a href="#f2.9">10</a>)–(<a href="#f2.10">11</a>)</span> is convergent with order of convergence \(5.1925\). </p>

  </div>
</div> <div class="theorem_thmwrapper " id="t2.3">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">3</span>
  </div>
  <div class="theorem_thmcontent">
  <p> Let \(\alpha \) be a simple zero of a function \( f \) which has sufficient number of smooth derivatives in a neighborhood of \(\alpha \). Then for solving nonlinear equation \(f(x)= 0\), the method <span class="rm">(<a href="#f2.11">12</a>)–(<a href="#f2.12">13</a>)</span> is convergent with order of convergence \(5.1925\). </p>

  </div>
</div> </p>
<h1 id="a0000000029">3 Methods with Higher Order Convergence</h1>
<p> In this section, we obtain a new iterative method by combining the iterations of method (<a href="#f2.1">2</a>)–(<a href="#f2.2">3</a>) with secant method and prove that the order of convergence is more than \(5.1925\). Precisely, we propose the following method: If \(x_0\) is the initial approximation, then </p>
<div class="equation" id="f3.1">
<p>
  <div class="equation_content">
    \begin{equation} \left. \begin{aligned}  x^*_0 & =x_0\\ x_0^{**} & =x_0^* -\tfrac {2f(x_0^*)}{f’(x_0^*)+f’({z_1})},\\ \text{where}\qquad {z_1}&  = x_0 -\tfrac {f(x_0)}{f’(x_0)}\\ x_1 &  = x_0^{**}- \tfrac {x_0^{**}-x_0^*}{f(x_0^{**})-f(x_0^*)}f(x_0^{**}). \end{aligned}\right\rbrace \label{f3.1} \end{equation}
  </div>
  <span class="equation_label">14</span>
</p>
</div>
<p> Subsequently, for \(n\geq 1\), the iterations can be obtained as follows: </p>
<div class="equation" id="f3.2">
<p>
  <div class="equation_content">
    \begin{equation} \left. \begin{aligned}  x_n^* & =x_n -\tfrac {2f(x_n)}{f’(x_n)+f’({z_n^*})},\\ \text{where}\qquad {z_n^*}&  = x_n -\tfrac {f(x_n)}{f’\big(\tfrac {x_{n-1}+x_{n-1}^*}{2}\big)}\\ x_n^{**} &  =x_n^* -\tfrac {2f(x_n^*)}{f’(x_n^*)+f’({z_{n+1}})},\\ \text{where}\qquad z_{n+1} &  =x_n-\tfrac {f(x_n)}{f’\big(\tfrac {x_{n}+x_{n}^*}{2}\big)}\\ x_{n+1} &  = x_1^{**}- \tfrac {x_n^{**}-x_n^*}{f(x_n^{**})-f(x_n^*)}f(x_n^{**}). \end{aligned}\right\rbrace \label{f3.2} \end{equation}
  </div>
  <span class="equation_label">15</span>
</p>
</div>
<p><div class="remark_thmwrapper " id="r3.1">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">4</span>
  </div>
  <div class="remark_thmcontent">
  <p> In <span class="cite">
	[
	<a href="#jaind2" >3</a>
	]
</span>, it was proved that if the iterations of any method of order \(p\) for solving nonlinear equations are used alternatively with secant method, then the new method will be of order \(p+1\). Thus, in view of that result, the method (<a href="#f3.1">14</a>)–(<a href="#f3.2">15</a>) is certainly of order at least 6.1925. However, we prove below that the order is more. </p>

  </div>
</div> </p>
<p><div class="theorem_thmwrapper " id="t3.2">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">5</span>
  </div>
  <div class="theorem_thmcontent">
  <p> Let \(f\) be a function \( f \) having sufficient number of smooth derivatives in a neighborhood of \(\alpha \) which is a simple root of the equation \(f(x)= 0\). Then method <span class="rm">(<a href="#f3.1">14</a>)–(<a href="#f3.2">15</a>)</span> to approximate the root \(\alpha \) is convergent with order of convergence \(7.275\). </p>

  </div>
</div> <div class="proof_wrapper" id="a0000000030">
  <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 argue on the lines of that of Theorem <a href="#t2.1">1</a> and the error equation of the standard secant method. In particular, the errors \(e_0^*,\;  e_0^{**}\) and \(e_1\), respectively, in \(x_0^*,\;  x_0^{**}\) and \({x_1}\) in equations (<a href="#f3.1">14</a>) are given by </p>
<div class="displaymath" id="a0000000031">
  \begin{equation*} \begin{aligned}  e_0^*& = e_0\\ e_0^{**} & = ae_0^3,\qquad \text{where}\;  a=c_2^2+\tfrac {1}{2} c_3 \\ e_1& =\lambda ae_0^4,\qquad \text{where}\;  \lambda =c_2. \end{aligned}\end{equation*}
</div>
<p> Also, the errors \(e_1^*\) in \(x_1^*\) in equation (<a href="#f3.2">15</a>) is given by </p>
<div class="displaymath" id="a0000000032">
  \begin{equation*} \begin{aligned}  e_1^{*} & = 2c_2^2e_0 e_1^2\\ &  = {\lambda }^2 a^2be_0^9,\qquad \text{where}\;  b =2c_2^2\\ \end{aligned}\end{equation*}
</div>
<p> and the error \(e_1^{**}\) in \(x_1^{**}\) in equation (<a href="#f3.2">15</a>) is given by </p>
<div class="displaymath" id="a0000000033">
  \begin{equation*} \begin{aligned}  e_1^{**} & =-\tfrac {1}{4} c_2c_3 e_1^3 e_1^*\\ &  = c e_1^3 e_1^*, \end{aligned}\end{equation*}
</div>
<p> where \(c=-\tfrac {1}{4}c_2c_3\). In fact, it can be worked out that for \(n\geq 1\), the following relation holds: </p>
<div class="equation" id="f3.3">
<p>
  <div class="equation_content">
    \begin{equation}  e_n^{**} = c e_n^3 e_n^*.\label{f3.3} \end{equation}
  </div>
  <span class="equation_label">16</span>
</p>
</div>
<p> In order to compute \(e_n^{**}\) explicitly, we need to compute \(e_n\) and \(e_n^*\). We have already computed \(e_1\) and \(e_1^*\). From the proof of Theorem <a href="#t2.1">1</a> </p>
<div class="displaymath" id="a0000000034">
  \begin{equation*}  e_2^{*} = d e_1 e_2^2, \end{equation*}
</div>
<p> where \( d = c_2^2\) and, again, it can be checked that the following relation holds: </p>
<div class="equation" id="f3.4">
<p>
  <div class="equation_content">
    \begin{equation}  e_n^{*} = d e_{n-1} e_n^2.\label{f3.4} \end{equation}
  </div>
  <span class="equation_label">17</span>
</p>
</div>
<p> Also from (<a href="#f3.2">15</a>), it can be shown that </p>
<div class="displaymath" id="a0000000035">
  \begin{equation*}  e_2 = \lambda e_1^* e_2^{**}. \end{equation*}
</div>
<p> Thus, for \(n \geq 1\), it can be shown that error \(e_{n+1}\) in \(x_{n+1}\) in the method (<a href="#f3.1">14</a>)–(<a href="#f3.2">15</a>) satisfies the following recursion formula </p>
<div class="equation" id="f3.5">
<p>
  <div class="equation_content">
    \begin{equation}  e_{n+1} = \lambda e_n^* e_n^{**}\label{f3.5} \end{equation}
  </div>
  <span class="equation_label">18</span>
</p>
</div>
<p> Using the above information, the errors at each stage in \(x_n^*, \;  x_n^{**}\) and \(x_n\) are obtained and tabulated as follows: </p>
<div class="table"  id="tab2">
   <div class="centered"> <table class="tabular">
  <tr>
    <td  style="text-align:right; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(n\) </p>

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

    </td>
    <td  style="text-align:right; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>\(e_n^*\) </p>

    </td>
    <td  style="text-align:right" 
        rowspan=""
        colspan="">
      <p>\( e_n^{**}\)</p>

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

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:right; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>\( e_0\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:right; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \(e_0\) </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:right" 
        rowspan=""
        colspan="">
      <p> \( ae_0^3\)</p>

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

    </td>
    <td  style="text-align:right; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>\(\lambda a e_0^4\) </p>

    </td>
    <td  style="text-align:right; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \({\lambda }^2a^2b e_0^9\) </p>

    </td>
    <td  style="text-align:right" 
        rowspan=""
        colspan="">
      <p>\({\lambda }^5a^5bc e_0^{21}\)</p>

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

    </td>
    <td  style="text-align:right; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \({\lambda }^8a^7b^2 c e_0^{30}\) </p>

    </td>
    <td  style="text-align:right; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \({\lambda }^{17}a^{15}b^5c^2 e_0^{64}\) </p>

    </td>
    <td  style="text-align:right" 
        rowspan=""
        colspan="">
      <p>\( {\lambda }^{42}a^{36}b^{11}c^{6} e_0^{154}\)</p>

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

    </td>
    <td  style="text-align:right; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>\( {\lambda }^{60}a^{51}b^{13}c^{8} e_0^{218}\) </p>

    </td>
    <td  style="text-align:right; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \({\lambda }^{128}a^{109}b^{29}c^{17} e_0^{466}\) </p>

    </td>
    <td  style="text-align:right" 
        rowspan=""
        colspan="">
      <p>\( {\lambda }^{308}a^{260}b^{68}c^{42} e_0^{1120}\)</p>

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

    </td>
    <td  style="text-align:right; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \({\lambda }^{437}a^{369}b^{97}c^{59} e_0^{1586}\) </p>

    </td>
    <td  style="text-align:right; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> \({\lambda }^{934}a^{789}b^{208}c^{126} e_0^{3390}\) </p>

    </td>
    <td  style="text-align:right" 
        rowspan=""
        colspan="">
      <p>\( {\lambda }^{2245}a^{1896}b^{499}c^{304} e_0^{8148}\)</p>

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

    </td>
    <td  style="text-align:right; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>\({\lambda }^{3180}a^{2685}b^{707}c^{430} e_0^{11538}\) </p>

    </td>
    <td  style="text-align:right; border-right:1px solid black" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:right" 
        rowspan=""
        colspan="">&nbsp;</td>
  </tr>
  <tr>
    <td  style="text-align:right; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:right; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:right; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">&nbsp;</td>
    <td  style="text-align:right; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">&nbsp;</td>
  </tr>
</table> </div> <figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">2</span> 
  <span class="caption_text">Successive errors.</span> 
</figcaption>  
</div>
<p>We do the analysis of Table <a href="#tab2">2</a> as done in the proof of Theorem <a href="#t2.1">1</a> for Table <a href="#tab1">1</a>. Note that the powers of \(e_0\) in the error at each iterate from the sequence </p>
<div class="equation" id="f3.6">
<p>
  <div class="equation_content">
    \begin{equation}  4 ,\;  30,\;  218,\;  1586,\;  11538,\;  ....\label{f3.6}\end{equation}
  </div>
  <span class="equation_label">19</span>
</p>
</div>
<p> and the sequence of their successive ratios is </p>
<div class="displaymath" id="a0000000036">
  \[ \tfrac {30}{4},\;  \tfrac {218}{30},\;  \tfrac {1586}{218},\;  \tfrac {11538}{1586},\;  ... \]
</div>
<p> or </p>
<div class="displaymath" id="a0000000037">
  \[ 7.5,\;  7.2667,\;  7.2752,\;  7.2749,\;  .... \]
</div>
<p> If the terms of the sequence (<a href="#f3.6">19</a>) are denoted by \(\{ N_i\} \), then it can be seen that </p>
<div class="displaymath" id="a0000000038">
  \[ N_i = 7N_{i-1}+ 2N_{i-2}, \quad i= 2,\, 3,\, 4,\,  .... \]
</div>
<p> Thus, as in Theorem <a href="#t2.1">1</a>, the rate of convergence of method (<a href="#f3.1">14</a>)–(<a href="#f3.2">15</a>) is at least 7.275. <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>It is natural to consider the variants of the method (<a href="#f3.1">14</a>)–(<a href="#f3.2">15</a>), where in the expression of \(z_n\) and \(z_n^*\), the arithmetic mean is replaced by harmonic mean as well as geometric mean as done in methods (<a href="#f2.9">10</a>)–(<a href="#f2.10">11</a>) and (<a href="#f2.11">12</a>)–(<a href="#f2.12">13</a>), respectively. Precisely, with harmonic mean, we propose the following method: </p>
<div class="equation" id="f3.7">
<p>
  <div class="equation_content">
    \begin{equation} \left. \begin{aligned}  x^*_0 & =x_0\\ x_0^{**} & =x_0^* -\tfrac {2f(x_0^*)}{f’(x_0^*)+f’({z_1})},\\ \text{where}\qquad {z_1}&  = x_0-\tfrac {f(x_0)}{f’\left(\tfrac {2x_0 x_0^*}{x_0+x_0^*}\right)}= x_0 -\tfrac {f(x_0)}{f’(x_0)}\\ x_1 &  = x_0^{**}- \tfrac {x_0^{**}-x_0^*}{f(x_0^{**})-f(x_0^*)}f(x_0^{**}) \end{aligned}\right\rbrace \label{f3.7} \end{equation}
  </div>
  <span class="equation_label">20</span>
</p>
</div>
<p> followed by (for \( n \geq 1\)) </p>
<div class="equation" id="f3.8">
<p>
  <div class="equation_content">
    \begin{equation} \left. \begin{aligned}  x_n^* & =x_n -\tfrac {2f(x_n)}{f’(x_n)+f’({z_n^*})},\\ \text{where}\qquad {z_n^*}&  = x_n-\tfrac {f(x_n)}{f’\left(\tfrac {2x_{n-1}x_{n-1}^*}{x_{n-1}+x_{n-1}^*}\right)}\\ x_n^{**} &  =x_n^* -\tfrac {2f(x_n^*)}{f’(x_n^*)+f’({z_{n+1}})},\\ \text{where}\qquad z_{n+1} &  =x_n-\tfrac {f(x_n)}{f’\left(\tfrac {2x_n x_n^*}{x_n+x_n^*}\right)}\\ x_{n+1} &  = x_1^{**}- \tfrac {x_n^{**}-x_n^*}{f(x_n^{**})-f(x_n^*)}f(x_n^{**}) \end{aligned}\right\rbrace \label{f3.8} \end{equation}
  </div>
  <span class="equation_label">21</span>
</p>
</div>
<p> and with the geometric mean, we we propose the following : </p>
<div class="equation" id="f3.9">
<p>
  <div class="equation_content">
    \begin{equation} \left. \begin{aligned}  x^*_0 & =x_0\\ x_0^{**} & =x_0^* -\tfrac {2f(x_0^*)}{f’(x_0^*)+f’({z_1})},\\ \text{where}\qquad {z_1}&  = x_0-\tfrac {f(x_0)}{f’\left(\sqrt{x_0 x_0^*}\, \right)}= x_0 -\tfrac {f(x_0)}{f’(x_0)}\\ x_1 &  = x_0^{**}- \tfrac {x_0^{**}-x_0^*}{f(x_0^{**})-f(x_0^*)}f(x_0^{**}) \end{aligned}\right\rbrace \label{f3.9} \end{equation}
  </div>
  <span class="equation_label">22</span>
</p>
</div>
<p> followed by (for \( n \geq 1\)) </p>
<div class="equation" id="f3.10">
<p>
  <div class="equation_content">
    \begin{equation} \left. \begin{aligned}  x_n^* & =x_n -\tfrac {2f(x_n)}{f’(x_n)+f’({z_n^*})},\\ \text{where}\qquad {z_n^*}&  = x_n-\tfrac {f(x_n)}{f’\left(\sqrt{x_{n-1}x_{n-1}^*}\, \right)}\\ x_n^{**} &  =x_n^* -\tfrac {2f(x_n^*)}{f’(x_n^*)+f’({z_{n+1}})},\\ \text{where}\qquad z_{n+1} &  =x_n-\tfrac {f(x_n)}{f’\left(\sqrt{x_n x_n^*}\, \right)}\\ x_{n+1} &  = x_1^{**}- \tfrac {x_n^{**}-x_n^*}{f(x_n^{**})-f(x_n^*)}f(x_n^{**}). \end{aligned}\right\rbrace \label{f3.10} \end{equation}
  </div>
  <span class="equation_label">23</span>
</p>
</div>
<p> The convergence of the methods (<a href="#f3.7">20</a>)–(<a href="#f3.8">21</a>) and (<a href="#f3.9">22</a>)–(<a href="#f3.10">23</a>) can be proved by using the arguments as used in the proof of Theorem <a href="#t3.2">5</a>. We skip the details for conciseness. </p>
<h1 id="a0000000040">4 Algorithms and Numerical Examples</h1>
<p> We give below an algorithm to implement the method (<a href="#f2.1">2</a>)–(<a href="#f2.2">3</a>): </p>
<p><div class="algorithm_thmwrapper " id="a4.1">
  <div class="algorithm_thmheading">
    <span class="algorithm_thmcaption">
    Algorithm
    </span>
    <span class="algorithm_thmlabel">6</span>
  </div>
  <div class="algorithm_thmcontent">
  <p> <i class="it">Step</i> 1 : For the given tolerance \(\varepsilon {\gt}0\) and iteration \(N\), choose the initial approximation \(x_0\) and set \(n=0\). </p>
<p><i class="it">Step</i> 2 : Follow the following sequence of expressions: </p>
<div class="displaymath" id="a0000000041">
  \begin{align*}  x^*_0 & =x_0\\ x_1 & =x_0^* -\tfrac {2f(x_0^*)}{f’(x_0^*)+f’({z_1})},\\ \text{where}\qquad {z_1}&  = x_0 -\tfrac {f(x_0)}{f’(x_0)}\\ x_1^{*} &  =x_1 -\tfrac {2f(x_1)}{f’(x_1)+f’({z_1^*})},\\ \text{where}\qquad z_1^{*} &  =x_1-\tfrac {f(x_1)}{f’\big(\tfrac {x_{0}+x_{0}^*}{2}\big)}=x_1 -\tfrac {f(x_1)}{f’(x_0)} \end{align*}
</div>
<p> <i class="it">Step</i> 3 : For \(n=1,2,3,\hdots \), calculate \(x_2,x_3,x_4,\hdots \) by the following sequence of expressions: </p>
<div class="displaymath" id="a0000000042">
  \begin{align*}  x_n^* & =x_n -\tfrac {2f(x_n)}{f’(x_n)+f’({z_n^*})},\\ \text{where}\qquad {z_n^*}&  = x_n -\tfrac {f(x_n)}{f’\big(\tfrac {x_{n-1}+x_{n-1}^*}{2}\big)}\\ x_{n+1} &  =x_n^* -\tfrac {2f(x_n^*)}{f’(x_n^*)+f’({z_{n+1}})},\\ \text{where}\qquad z_{n+1} &  =x_n-\tfrac {f(x_n)}{f’\big(\tfrac {x_{n}+x_{n}^*}{2}\big)} \end{align*}
</div>
<p><i class="it">Step</i> 4 : Stop if either \(|x_{n+1}-x_n|{\lt}\varepsilon \) or \(n{\gt}N\). </p>
<p><i class="it">Step</i> 5 : Set \(n=n+1\) and repeat Step 3. </p>

  </div>
</div> </p>
<p><div class="example_thmwrapper " id="e4.2">
  <div class="example_thmheading">
    <span class="example_thmcaption">
    Example
    </span>
    <span class="example_thmlabel">7</span>
  </div>
  <div class="example_thmcontent">
  <p> We apply method (<a href="#f2.1">2</a>)–(<a href="#f2.2">3</a>) on the nonlinear equation </p>
<div class="equation" id="tab3">
<p>
  <div class="equation_content">
    \begin{equation}  \cos x -x e^x + x^2 =0.\label{f4.1} \end{equation}
  </div>
  <span class="equation_label">24</span>
</p>
</div>
<p> This equation has a simple root in the interval \((0,1)\). Taking initial approximation as \(x_0=1\), Table <a href="#tab3">24</a> shows the iterations of McDougall-Wotherspoon method, a third order method (<a href="#f1.1">1</a>) and our method (<a href="#f2.1">2</a>)–(<a href="#f2.2">3</a>). </p>

  </div>
</div> </p>
<div class="table"  id="tab3">
    <div class="centered"> <table class="tabular">
  <tr>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>\(n\) </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> W-F Method (<a href="#f1.1">1</a>) </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> M-W method </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> (<a href="#f2.1">2</a>)–(<a href="#f2.2">3</a>) method</p>

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

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">1.1754860092539474</small> </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.89033621746836966</small> </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.64406452481689269</small></p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p><small class="small">2.</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.7117526001461193</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.66469560530044569</small> </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915407608296659</small></p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p><small class="small">3.</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63945030188514695</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63928150457301036</small> </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915411559451774</small></p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p><small class="small">4.</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915408656045591</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915408990276223</small> </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.6391540955014231</small></p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p><small class="small">5.</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915410631623149</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915410965853769</small> </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915407540832936</small></p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p><small class="small">6.</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915412607200606</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.6391540698096656</small> </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.6391541149198805</small></p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p><small class="small">7.</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915408622313585</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915408956544117</small> </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915409482678587</small></p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p><small class="small">8.</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915410597891142</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915410932121663</small> </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915407473369212</small></p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p><small class="small">9.</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.639154125734686</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915406947234454</small> </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915411424524327</small></p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p><small class="small">10.</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915408588581579</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915408922812</small> </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915409415214863</small></p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p><small class="small">11.</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915410564159136</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915410898389557</small> </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915407405905489</small></p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p><small class="small">12.</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915412539736594</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915406913502348</small> </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915411357060603</small></p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p><small class="small">13.</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915408554849573</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915408889079894</small> </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.6391540934775114</small></p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p><small class="small">14.</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915410530427119</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915410864657451</small> </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915407338441765</small></p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p><small class="small">15.</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915412506004576</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915406879770231</small> </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.6391541128959688</small></p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p><small class="small">16.</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915408521117556</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915408855347788</small> </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915409280287416</small></p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p><small class="small">17.</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915410496695113</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915410830925345</small> </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915407270978042</small></p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p><small class="small">18.</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.6391541247227257</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915406846038125</small> </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915411222133156</small></p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p><small class="small">19.</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.6391540848738555</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915408821615682</small> </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915409212823693</small></p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p><small class="small">20.</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915410462963107</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915410797193239</small> </p>

    </td>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915407203514318</small></p>

    </td>
  </tr>
</table> <figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">3</span> 
  <span class="caption_text">Numerical results for different methods.</span> 
</figcaption>  </div> 
</div>
<p><div class="example_thmwrapper " id="e4.3">
  <div class="example_thmheading">
    <span class="example_thmcaption">
    Example
    </span>
    <span class="example_thmlabel">8</span>
  </div>
  <div class="example_thmcontent">
  <p> We consider the same equation (<a href="#tab3">24</a>) but now implement method (<a href="#f3.1">14</a>)–(<a href="#f3.2">15</a>) and compare with other methods. Table <a href="#tab4">4</a>, shows the corresponding iterates. One can also compare the last columns of Table <a href="#tab3">3</a> and Table <a href="#tab4">4</a> which correspond to methods (<a href="#f2.1">2</a>)–(<a href="#f2.2">3</a>) and (<a href="#f3.1">14</a>)–(<a href="#f3.2">15</a>), respectively. This clearly indicates the fast convergence of (<a href="#f3.1">14</a>)–(<a href="#f3.2">15</a>). </p>

  </div>
</div> </p>
<div class="table"  id="tab4">
   <div class="centered"> <table class="tabular">
  <tr>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>\(n\) </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> W-F Method (<a href="#f1.1">1</a>) </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> M-W method </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> (<a href="#f3.1">14</a>)–(<a href="#f3.2">15</a>) method</p>

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

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">1.1754860092539474</small> </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.89033621746836966</small> </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63919747126530391</small></p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p><small class="small">2.</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.7117526001461193</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.66469560530044569</small> </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915410580338361</small></p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p><small class="small">3.</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63945030188514695</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63928150457301036</small> </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915409891807362</small></p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p><small class="small">4.</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915408656045591</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915408990276223</small> </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915409203276374</small></p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p><small class="small">5.</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915410631623149</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915410965853769</small> </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915408514745375</small></p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p><small class="small">6.</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915412607200606</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.6391540698096656</small> </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915411145121981</small></p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p><small class="small">7.</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915408622313585</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915408956544117</small> </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> <small class="small">division by zero</small></p>

    </td>
  </tr>
  <tr>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p><small class="small">8.</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915410597891142</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915410932121663</small> </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">&nbsp;</td>
  </tr>
  <tr>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p><small class="small">9.</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.639154125734686</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915406947234454</small> </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">&nbsp;</td>
  </tr>
  <tr>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p><small class="small">10.</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915408588581579</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915408922812</small> </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">&nbsp;</td>
  </tr>
  <tr>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p><small class="small">11.</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915410564159136</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915410898389557</small> </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">&nbsp;</td>
  </tr>
  <tr>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p><small class="small">12.</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915412539736594</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915406913502348</small> </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">&nbsp;</td>
  </tr>
  <tr>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p><small class="small">13.</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915408554849573</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915408889079894</small> </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">&nbsp;</td>
  </tr>
  <tr>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p><small class="small">14.</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915410530427119</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915410864657451</small> </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">&nbsp;</td>
  </tr>
  <tr>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p><small class="small">15.</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915412506004576</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915406879770231</small> </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">&nbsp;</td>
  </tr>
  <tr>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p><small class="small">16.</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915408521117556</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915408855347788</small> </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">&nbsp;</td>
  </tr>
  <tr>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p><small class="small">17.</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915410496695113</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915410830925345</small> </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">&nbsp;</td>
  </tr>
  <tr>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p><small class="small">18.</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.6391541247227257</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915406846038125</small> </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">&nbsp;</td>
  </tr>
  <tr>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p><small class="small">19.</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.6391540848738555</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915408821615682</small> </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">&nbsp;</td>
  </tr>
  <tr>
    <td  style="text-align:left; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p><small class="small">20.</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915410462963107</small> </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> <small class="small">0.63915410797193239</small> </p>

    </td>
    <td  style="text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">&nbsp;</td>
  </tr>
</table> <figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">4</span> 
  <span class="caption_text">Numerical results for different methods.</span> 
</figcaption>  </div> 
</div>
<p><small class="footnotesize">  </small></p>
<div class="bibliography">
<h1>Bibliography</h1>
<dl class="bibliography">
  <dt><a name="bradie">1</a></dt>
  <dd><p><i class="sc">B. Bradie</i>, <i class="it">A Friendly Introduction to Numerical Analysis</i>, Pearson, 2007. </p>
</dd>
  <dt><a name="jaind1">2</a></dt>
  <dd><p><a href ="http://dx.doi.org/10.1080/00207160.2012.746677"> <i class="sc">D. Jain</i>, <i class="it">Families of Newton-like methods with fourth-order convergence</i>, Int. J. Comp. Math., <b class="bf">90</b> (2013), pp.&#160;1072–1082. <img src="img-0001.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
  <dt><a name="jaind2">3</a></dt>
  <dd><p><i class="sc">D. Jain</i>, <i class="it">Newton and Steffensen type methods with flexible order of convergence</i>, Jordanian J. Math. Stat., <b class="bf">8</b> (2015), pp.&#160;43–57. </p>
</dd>
  <dt><a name="jg">4</a></dt>
  <dd><p><i class="sc">D. Jain</i> and <i class="sc">B. Gupta</i>, <i class="it">Two step Newton and Steffensen type methods for solving nonlinear equations</i>, Tbiliai Math. J., <b class="bf">5</b> (2012), pp.&#160;17–29. </p>
</dd>
  <dt><a name="jainp">5</a></dt>
  <dd><p><a href ="http://dx.doi.org/10.1016/j.amc.2007.04.087"> <i class="sc">P. Jain</i>, <i class="it">Steffensen type methods for solving non-linear equations</i>, Appl. Math. Comput., <b class="bf">194</b> (2007), pp.&#160;527–533. <img src="img-0001.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
  <dt><a name="kast">6</a></dt>
  <dd><p><a href ="http://dx.doi.org/10.1080/00207390210131786"> <i class="sc">A.B. Kasturiarachi</i>, <i class="it">Leap frogging Newton’s method</i>, Int. J. Math. Educ. Sci. Technol., <b class="bf">33</b> (2002), pp.&#160;521–527. <img src="img-0001.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
  <dt><a name="mw">7</a></dt>
  <dd><p><a href ="http://dx.doi.org/10.1016/j.aml.2013.10.008"> <i class="sc">T. J. McDougall</i> and <i class="sc">S. J. Wotherspoon</i>, <i class="it">A simple modification of Newton’s method to achieve convergence of order \(1+\sqrt2\)</i>, Appl. Math. Lett., <b class="bf">29</b> (2014), pp.&#160;20–25. <img src="img-0001.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
  <dt><a name="wf">8</a></dt>
  <dd><p><a href ="http://dx.doi.org/10.1016/S0893-9659(00)00100-2"> <i class="sc">S. Weerakoon</i> and <i class="sc">T.G.I. Fernando</i>, <i class="it">A variant of Newton’s method with accelerated third-order convergence</i>, Appl. Math. Lett., <b class="bf">13</b> (2002), pp.&#160;87–93. <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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