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\begin{document}4

\begin{center}
{\bf Ball  comparison for three optimal eight order methods under weak conditions}
\vskip1cm

Ioannis K. Argyros\footnote{Department of Mathematical Sciences, Cameron University, Lawton, OK 73505,
 USA, Email: iargyros@cameron.edu} and 
Santhosh George\footnote{ Department of
Mathematical and Computational Sciences, National Institute of Technology Karnataka, India-575 025, Email:sgeorge@nitk.edu.in}
\end{center}

%\maketitle
\begin{center}
\end{center}
\begin{abstract}
 We considered three optimal eighth order method for solving nonlinear equations. In earlier studies Taylors expansions and  hypotheses reaching up to the eighth derivative are used to prove the convergence of these methods. These hypotheses restrict the applicability of the methods. In our study we use  hypotheses on the first derivative.
Numerical examples  illustrating the theoretical results are also presented in this study.
\vskip .5cm
{\bf AMS} Subject Classification : 65D10, 65D99, 47J25, 47J05.
\vskip .5cm
{\bf Key Words}:   Banach space,   Fr\'{e}chet derivative, Efficiency index, ball convergence.
% \PACS{PACS code1 \and PACS code2 \and more}

\end{abstract}
%
\section{Introduction}
In this paper we are concerned with the problem of approximating a  solution $x^\ast$ of the equation
\beq\label{1.1} F(x)=0,\eeq%
%
where $F:D\subseteq S\longrightarrow T$ is a Fr\'{e}chet-differentiable operator defined on a convex set $D,$ where $S, T $ are subsets of  $\mathbb{R}$ or $\mathbb{C}.$

Equation of the form (\ref{1.1}) is used to study 
 problems in Computational Sciences and other disciplines \cite{5, 7, 15, 17, 20}.   Newton-like iterative methods \cite{2, 3, 4, 5, 6,7,8,9,10,11,12,13,14,15,16,17,18,19*,19,20, 21, 22, 23} are famous for approximating a  solution  of the equation (\ref{1.1}).
 
%  These methods are  usually studied based on: semi-local and local  convergence. The semi-local convergence matter is, based on the information around an initial point, to give conditions ensuring the convergence of the iterative procedure;  while the local one is, based on the information around a solution, to find estimates of the radii of convergence balls \cite{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20, 21}. 

 
 
 
 In this paper, we study the local convergence analysis of the  methods  defined for each $n=0, 1, 2, \cdots$ by  Siyyam et al.\cite{19}
%
\begin{eqnarray}\nonumber
y_n&=&x_n- \frac{1}{F'(x_n)}F(x_n),\\\nonumber
z_n&=&x_n+(1+\beta)\frac{1}{F'(x_n)}(F(x_n)+F(y_n)),\\\nonumber
&&-\frac{1}{F'(x_n)}F(x_n)(F(x_n)-F(y_n))^{-1}F(x_n)\\\label{1.2}
&&-\beta(\frac{1}{F'(x_n)}F(x_n)+(F'(x_n)+F^2(x_n)F'(x_n))^{-1}F(y_n))\\\nonumber
x_{n+1}&=&z_n-A_n^{-1}F(z_n),
\end{eqnarray}
%
where $x_0\in D$ is an initial point, $\beta\in S$, $A_n=F'(x_n)+([x_n,y_n,z_n;F]+[x_n,x_n,y_n;F])(z_n-x_n) + 2([x_n,y_n,z_n;F]-[x_n,x_n,y_n;F])(z_n-y_n)$ and $[.,.,.;F]$ denotes a divided difference of order two for function $F$ on $D.$ The second and third method are due to Wang et. al. \cite{23}  and are defined, respectively as 
\begin{eqnarray}\nonumber
y_n&=&x_n- \frac{1}{F'(x_n)}F(x_n),\\\nonumber
z_n&=&x_n-\frac{1}{F'(x_n)}F(x_n)(F(x_n)-2F(y_n))^{-1}(F(x_n)-F(y_n)),\\\label{1.3}
x_{n+1}&=&z_n-\frac{1}{F'(x_n)}F(z_n)\\\nonumber
&&\times\left[\frac{1}{2}+\frac{1+\frac{8F(y_n)}{5F(x_n)}+\frac{2}{5}(\frac{F(y_n)}{F(x_n)})^2}{1-\frac{12}{5}\frac{1}{F'(x_n)}F(y_n)}(1+F'(y_n)^{-1}F(z_n))\right],
\end{eqnarray}
and 
\begin{eqnarray}\nonumber
y_n&=&x_n- \frac{1}{F'(x_n)}F(x_n),\\\nonumber
z_n&=&x_n-\frac{1}{F'(x_n)}F(x_n)(F(x_n)-2F(y_n))^{-1}(F(x_n)-F(y_n)),\\\nonumber
x_{n+1}&=&z_n-F(x_n)^{-1}F(x_n)\left[\frac{1-\frac{2}{5}\frac{1}{F'(x_n)}F(y_n)+\frac{1}{5}(F(x_n)^{-1}F(y_n))^2}{1-\frac{12}{5}\frac{1}{F'(x_n)}F(y_n)}\right.\\\label{1.4}
&&\left.+(1+4\frac{1}{F'(x_n)}F(y_n))F'(y_n)^{-1}F(z_n)\right].
\end{eqnarray}

%


Convergence ball of high convergence order methods is usually very small and in general decreases as the convergence order increases. The  approach in this paper establishes the local convergence result  under  hypotheses only on the first derivative and  give a larger convergence ball than the earlier studies,  under weaker hypotheses. Notice  that in earlier studies \cite{19, 23} the convergence is shown under hypotheses on the eighth derivative.
The same technique can be used to other methods.  As a motivational example, define function $f$ on $D=[-\frac{1}{2},\frac{3}{2})$ by
% %
 \begin{equation}
 f(x)=
 \left\{
 \begin{array}{lcl}
 x^3ln x^2+x^5-x^4,& &x\not=0\\
 0, & & x=0\\
 \end{array}
 \right.
 \end{equation}
 Choose $x^*=1$. We also have that
% 
 $$
 f'(x)=3x^2lnx^2+5x^4-4x^3+2x^2,
 $$
 $$
 f''(x)=6xlnx^2+20x^3+12x^2+10x
 $$
 and
 $$
 f'''(x)=6lnx^2+60x^2-24x+22.
 $$
% 
Notice that $f'''(x)$ is unbounded on $D$. Hence, the results in \cite{19, 23}, 
 cannot apply to show the convergence of method (\ref{1.2}) (see also the numerical examples).






The rest of the paper is organized as follows. In Section 2 we present the local convergence analysis of
methods  (\ref{1.2})--(\ref{1.4}). The numerical examples are given in the concluding Section 3.
%
\setcounter{equation}{0}

\section{Local convergence}

The local convergence  of method (\ref{1.2}),  method (\ref{1.3}) and  method (\ref{1.4}) is based on  some functions and parameters.
%
Let $K_0 > 0, K >0, L_0 >0, L > 0,\,  M\geq 1$ and $\beta\in S$ be given  parameters. Let $g_1, p_1, h_{p_1}, p_2$ and $h_{p_2}$ be functions defined on the interval   $[0, \frac{1}{L_0})$ by
\begin{eqnarray*}
g_1(t)&=&\frac{Lt}{2(1-L_0t)}\\
p_1(t)&=&\frac{L_0t}{2}+Mg_1(t)\\
h_{p_1}(t)&=&p_1(t)-1,\\
p_2(t)&=&L_0t+\frac{M^2t^2}{1-L_0t}\\
h_{p_2}(t)&=&p_2(t)-1
\end{eqnarray*}
and parameter $r_1$  by 
\begin{eqnarray}\label{2.1}
r_1&=&\frac{2}{2L_0+L}.
\end{eqnarray}
We have that $g_1(r_1)=1$ and for each $t\in [0, r_1):\, 0 \leq g_1(t) < 1.$ 
We also  get  that $h_{p_1}(0)=h_{p_2}(0)=-1 <0$ 
and  $h_{p_1}(t)\longrightarrow +\infty, h_{p_2}(t)\longrightarrow +\infty$ as $t\longrightarrow \frac{1}{L_0}^-.$ It then follows from the intermediate value theorem that functions $p_1$ and $p_2$ have zeros in the interval $(0, \frac{1}{L_0}).$ Denote by $r_{p_1}$ and $r_{p_2}$ the smallest such zeros of functions $h_{p_1}$ and $r_{p_2}, $ respectively. Let $\bar{r}=\min\{r_{p_1}, r_{p_2}\}.$ Define functions $g_2$ and $h_2$ on the interval $[0, \bar{r})$ by 
\begin{eqnarray*}
 g_2(t)&=&\frac{Lt}{2(1-L_0t)}+\frac{2M^2g_1(t)}{(1-L_0t)(1-p_1(t))}\\
 &&+\frac{|1+\beta|Mg_1(t)}{1-L_0t}+\frac{M|\beta|g_1(t)}{1-p_2(t)}
\end{eqnarray*}
and $h_2(t)=g_2(t)-1.$ We have that $h_2(0)=-1 < 0$ and $h_2(t)\longrightarrow +\infty$ as $t\longrightarrow \bar{r}^-.$ Denote by $r_2$ the smallest zero of function $h_2$ in the interval $(0, \bar{r}).$  Moreover, define functions $q$ and $h_q$ on the interval $[0, \bar{r})$ by $q(t)=L_0t+(K+K_0)(1+g_2(t))t+2(K_0+K)(g_1(t)+g_2(t))t$ and $h_q(t)=q(t)-1.$ We get that $h_q(0)=-1 < 0$ and $h_q(t)\longrightarrow +\infty$ as $t\longrightarrow\bar{r}_-.$ Denote by $r_q$ the smallest zero of function $h_q$ on the interval $(0, \bar{r}).$ Let $\bar{r}_0=\min\{\bar{r}, r_q\}.$


Finally, define functions $g_3$ and $h_3$ on the interval $[0, \bar{r}_0)$ by 
$$g_3(t)=(1+\frac{M}{1-q(t)})g_2(t)$$
and $h_3(t)=g_3(t)-1.$ We get that $h_3(0)=-1 < 0$ and $h_3(t)\longrightarrow+\infty$ as $t\longrightarrow \bar{r}_0^-.$ Denote by $r_3$ the smallest zero of function $h_3$ on the interval $(0, \bar{r}_0).$ Define the radius of convergence $r$ by 
\beq\label{2.1}
r=\min\{r_i\},\, i=1,2,3.
\eeq 

%
Then, we have that 
\beq\label{2.2}
0 < r < r_1 < \frac{1}{L_0}
\eeq 
and  for each $t\in [0, r)$
\beq\label{2.3}
0\leq g_i(t) < 1,\, i=1,2,3
\eeq 

\beq\label{2.4}
0\leq p_j(t) < 1,\, j=1,2
\eeq 
and 

\beq\label{2.5}
0\leq q(t) < 1.
\eeq 
Let us denote by  $U(v,\rho),\,\overline{U}(v, \rho)$  the open and closed balls in $S$  with center $v\in S$  and of radius $\rho > 0.$


Next,  we present the local convergence analysis of method (\ref{1.2}) using the preceding notation. 
%%
%%
%%
\bt\label{T2.1}
Let $F:D\subset S\longrightarrow T$ be a differentiable function. Let also $[.,.,.;F]$ denote a divided difference of order two for function $F$ on $D.$ Suppose that there exist $x^\ast\in D$
%
\beq\label{2.6}
 F(x^\ast)=0,\,\, F'(x^\ast)\neq 0
 \eeq%
 and 
 \beq\label{2.7}
 \|F'(x^\ast)^{-1}(F'(x)-F'(x^\ast))\|\leq L_0\|x-x^\ast\|.
 \eeq%
Moreover, suppose that there exist $L > 0$ and $M \geq 1$ and $ K > 0$  such that for each $x, y, z \in D_0=D\cap U(x^\ast, \frac{1}{L_0})$
 \beq\label{2.8}
 \|F'(x^\ast)^{-1}(F'(x)-F'(y))\|\leq L\|x-y\|,
 \eeq%
 %
 \beq\label{2.9}
 \|F'(x^\ast)^{-1}F'(x)\|\leq M,
 \eeq%
 \beq\label{2.10}
 \|F'(x^*)^{-1}[x,x,y;F]\|\leq K_0,\, \|F'(x^*)^{-1}[x,y,z;F]\|\leq K
\eeq 
and 
\beq\label{2.11}
\bar{U}(x^\ast, r)\subseteq D,
\eeq
where  the radius of convergence $r$ is defined by (\ref{2.1}). 
Then, the sequence $\{x_n\}$ generated  for $x_0\in U(x^\ast, r)-\{x^\ast\}$ is well defined  in $U(x^\ast, r),$ remains in $U(x^*,r)$  for each $n=0, 1, 2,\cdots,$ and converges to $x^\ast.$ Moreover, the following estimates hold
\beq\label{2.12}
\|y_n-x^\ast\|\leq g_1(\|x_n-x^*\|)\|x_n-x^\ast\| \leq \|x_n-x^\ast\| < r,
\eeq%
\beq\label{2.13}
\|z_n-x^\ast\|\leq g_2(\|x_n-x^*\|)\|x_n-x^\ast\| \leq \|x_n-x^\ast\| 
\eeq
and
\beq\label{2.14}
\|x_{n+1}-x^\ast\|\leq g_3(\|x_n-x^\ast\|)\|x_n-x^\ast\| \leq \|x_n-x^\ast\|,
\eeq
where the ``$g$'' functions are defined previously. 
Furthermore, for  $T\in[r,\frac{2}{L_0})$  the  limit point $x^\ast$ is the only solution of the equation $F(x)=0$ in $D_1=D\cap\bar{U}(x^\ast,T).$
%
\et
%
\noindent
{\bf Proof.} We shall  show   that method (\ref{1.2}) is well defined in  $U(x^\ast, r)$  remains in $U(x^*,r)$ for each $n=0,1,2,\ldots,$ and  converges to $x^*$ so that estimates (\ref{2.12})--(\ref{2.14}) are satisfied. Using hypothesis $x_0\in U(x^*,r)-\{x^*\},$  (\ref{2.2}) and (\ref{2.7}), we have that 
\beq\label{2.15}
\|F'(x^\ast)^{-1}(F'(x_0)-F'(x^\ast))\|\leq L_0\|x_0-x^\ast\|\leq L_0 r < 1.
\eeq%
It follows from   (\ref{2.15}) and the Banach Lemma on invertible functions \cite{5, 7,  15}  that $F'(x_0)\neq 0$ and
\beq\label{2.16}
\|F'(x_0)^{-1}F'(x^\ast)\|\leq \frac{1}{1-L_0\|x_0-x^\ast\|}.
\eeq%
Hence,   $y_0$  is   well defined. By the first sub-step of method (\ref{1.2}) for $n=0,$ (\ref{2.2}), (\ref{2.3}), (\ref{2.6}), (\ref{2.8}) and (\ref{2.16}), we get in turn that
\begin{eqnarray}\nonumber
 \|y_0-x^\ast\|&=&
\|x_0-x^\ast-F'(x_0)^{-1}F'(x_0) \|\\\nonumber
&\leq& \|F'(x_0)^{-1}F'(x^\ast)\|\|\\\nonumber
&&\times\int_0^1F'(x^\ast)^{-1}(F'(x^\ast+\theta(x_0-x^\ast))-F'(x_0))(x_0-x^\ast)d\theta\|\\\nonumber
&\leq& \frac{L\|x_0-x^\ast\|^2}{2(1-L_0\|x_0-x^\ast\|)}\\\label{2.17}
&\leq&g_1(\|x_0-x^\ast\|)\|x_0-x^\ast\| <  \|x_0-x^\ast\| < r,
\end{eqnarray}
which shows (\ref{2.12}) for $n=0$ and $y_0\in U(x^\ast, r).$
%
We can write by (\ref{2.6}) that 
\beq\label{2.18}
F(y_0)=F(y_0)-F(x^*)=\int_0^1F'(x^*+\theta(y_0-x^*))(y_0-x^*)d\theta.
\eeq 
Notice that $\|x^*+\theta(y_0-x^*)-x^*\|=\theta\|y_0-x^*\| < r,$ so $x^*+\theta(y_0-x^*)\in U(x^*, r)$ for each $\theta\in [0,1].$ Then, by (\ref{2.9}), (\ref{2.17}) and (\ref{2.18}), we get that 
\beq\label{2.19}
\|F(y_0)F'(x^*)^{-1}\|\leq M\|y_0-x^*\|\leq Mg_1(\|x_0-x^*\|)\|x_0-x^*\|.
\eeq 
We must show in turn that  $F(x_0)-F(y_0)\neq0$ and $F'(x_0)+\frac{F^2(x_0)}{F'(x_0)}\neq0.$
%
We have by (\ref{2.2}), (\ref{2.4}), (\ref{2.7}) and (\ref{2.19}) that 
\begin{eqnarray}\nonumber
&&\|(F'(x^*)(x_0-x^*))^{-1}(F(x)-F(x^*)-F'(x^*)(x_0-x^*)-F(y_0))\|\\nonumber
&\leq&\|x_0-x^*\|^{-1}(\frac{L_0}{2}\|x_0-x^*\|^2+M\|y_0-x^*\|)\\\label{2.20}
 &\leq&p_1(\|x_0-x^*\|) < p_1(r) < 1,
\end{eqnarray}
so
\beq\label{2.21}
\|(F(x_0)-F(y_0))^{-1}F'(x^*)\|\leq\frac{1}{\|x_0-x^*\|(1-p_1(\|x_0-x^*\|)}.
\eeq 
Sinilarly, by (\ref{2.2}), (\ref{2.4}), (\ref{2.7}) and (\ref{2.19}) (for $x_0=y_0$) that 
\begin{eqnarray}\nonumber
&&\|F'(x^*)^{-1}(F'(x_0)-F'(x^*)+\frac{F^2(x_0)}{F'(x_0)})\|\\\nonumber
 &\leq&L_0\|x_0-x^*\|+\frac{M^2\|x_0-x^*\|^2}{1-L_0\|x_0-x^*\|}=p_2(\|x_0-x^*\|)\\\label{2.22}
 & <& p_2(r) < 1,
\end{eqnarray}
so
\beq\label{2.23}
\|(F'(x_0)+\frac{F^2(x_0)}{F'(x_0)})^{-1}F'(x^*)\|\leq \frac{1}{1-p_2(\|x_0-x^*\|)}.
\eeq 
and $z_0$ is well defined. Using the second substep of method (\ref{1.2}), (\ref{2.2}),  (\ref{2.16}), (\ref{2.17}),  (\ref{2.19}), (\ref{2.21}) and (\ref{2.23}) we obtain in turn that 
%
\begin{eqnarray}\nonumber
z_0-x^*&=&x_0-x^*-F'(x_0)^{-1}F(x_0)+(2+\beta)F'(x_0)^{-1}F(x_0)\\\nonumber
&&+(1+\beta)F'(x_0)^{-1}F(y_0)-2\frac{F^2(x_0)}{F'(x_0)(F(x_0)-F(y_0))}\\\nonumber
&&-\beta F'(x_0)F(x_0)-\beta\frac{F(y_0)}{F'(x_0)+\frac{F^2(x_0)}{F'(x_0)}}\\\nonumber
&=&y_0-x^*-2[F'(x^*)^{-1}F(x_0)][F'(x_0)^{-1}F'(x^*)]\\\nonumber
&&\times[(F(x_0)-F(y_0))^{-1}F'(x^*)][F'(x^*)^{-1}F(y_0)]\\\nonumber
&&+(1+\beta)[F'(x_0)^{-1}F'(x^*)][F'(x^*)^{-1}F(y_0)]\\\label{2.24}
&&-\beta[F'(x^*)^{-1}F(y_0)][(F'(x_0)+\frac{F^2(x_0)}{F'(x_0)})^{-1}F'(x^*)], 
\end{eqnarray}
so
\begin{eqnarray}\nonumber
\|z_0-x^\ast\|&\leq&\|y_0-x^\ast\|\\\nonumber
&&+\frac{2M^2\|y_0-x^*\|\|x_0-x^*\|}{\|x_0-x^*\|(1-L_0\|x_0-x^\ast\|)(1-p_1(\|x_0-x^\ast\|)}\\\nonumber
&&+\frac{|1+\beta|M\|y_0-x^*\|}{1-L_0\|x_0-x^\ast\|}+\frac{|\beta|M\|y_0-x^\ast\|}{1-p_2(\|x_0-x^\ast\|)}
\\\label{2.25}
&=&g_2(\|x_0-x^\ast\|)\|x_0-x^\ast\| \leq \|x_0-x^\ast\| < r,
\end{eqnarray}
which shows (\ref{2.13}) for r $n=0$ and $z_0\in U(x^\ast, r).$ Next, we must show that $A_0\neq0.$ Using
%
%
 (\ref{2.2}), (\ref{2.5}),  (\ref{2.7}), (\ref{2.10}), (\ref{2.17}) and (\ref{2.25}),  we get in turn that  
%
\begin{eqnarray}\nonumber
&&\|F'(x^*)^{-1}(A_0-F'(x^*))\|\\\nonumber
&\leq&L_0\|x_0-x^*\|\\\nonumber
&&+(K_0+K)[\|z_0-x^*\|+\|x_0-x^*\|]+2(K_0+K)[\|z_0-x^*\|+\|y_0-x^*\|]\\\nonumber
&\leq&L_0\|x_0-x^*\|+(K_0+K)(1+g_2(\|x_0-x^*\|)\|x_0-x^*\|\\\nonumber
&&2(K_0+K)(g_1(\|x_0-x^*\|)+g_2(\|x_0-x^*\|))\|x_0-x^*\|\\\nonumber
&=&q(\|x_0-x^*\|) < q(r) < 1,
\end{eqnarray}
so
\beq\label{2.26}
\|A_0^{-1}F'(x^*)\|\leq\frac{1}{1-q(\|x_0-x^*\|)}
\eeq 
and $x_1$ is well defined. Then, from   (\ref{2.2}), (\ref{2.3}),  (\ref{2.17}), 
(\ref{2.19}) (for $y_0=z_0$),  (\ref{2.26}),  and the last substep of method  (\ref{1.2}) for $n=0,$ we have   that
\begin{eqnarray}\nonumber
\|x_1-x^\ast\|&\leq&\|z_0-x^\ast\|+\frac{M\|z_0-x^*\|}{1-q(\|x_0-x^\ast\|)}
\\\label{2.27}
&=&g_3(\|x_0-x^\ast\|)\|x_0-x^\ast\| \leq \|x_0-x^\ast\| < r,
\end{eqnarray}
which implies  (\ref{2.14}) holds  for $n=0$ and $x_1\in U(x^\ast, r).$ By simply replacing $x_0, y_0, z_0, x_1$ by $x_k,  y_k, z_k,  x_{k+1}$ in the preceding estimates we arrive at  (\ref{2.12})--(\ref{2.14}). Using the estimate
$\|x_{k+1}-x^\ast\|\leq c \|x_k-x^\ast\|, c=g_3(\|x_0-x^\ast\|)\in [0,1),$
 we deduce  that $\lim_{k\rightarrow \infty}x_k=x^\ast$ and $x_{k+1}\in U(x^\ast,r)$.
%
The proof of the uniqueness part is  standard \cite{8}.
%
\hfill $\Box$


Next, we introduce the needed functions as the corresponding ones above Theorem \ref{T2.1} but for method (\ref{1.3}). Define  functions $\varphi_1, \varphi_2, \varphi_3, h_{\varphi_1}, h_{\varphi_2}, h_{\varphi_3}$ on the interval $[0, \frac{1}{L_0})$ by 
$$\varphi_1(t)=\frac{12}{5}\frac{Mg_1(t)}{1-\frac{L_0}{2}t},\, h_{\varphi_1}(t)=\varphi_1(t)-1,$$
$$\varphi_2(t)=\frac{L_0}{2}t+2Mg_1(t),\, h_{\varphi_2}(t)=\varphi_2(t)-1,$$
$$\varphi_3(t)=\frac{L_0}{2}g_1(t)t\,\, \textnormal{and}\,\, h_{\varphi_3}(t)=\varphi_3(t)-1.$$

We have that $h_{\varphi_1}(0)=h_{\varphi_2}(0)=h_{\varphi_3}(0)=-1 < 0$ and $h_{\varphi_1}(t)\longrightarrow +\infty,$ $h_{\varphi_2}(t)\longrightarrow+\infty, h_{\varphi_3}(t)\longrightarrow+\infty$ as $t\longrightarrow\frac{1}{L_0}^-.$ Denote by $r_{\varphi_1}, r_{\varphi_2}, r_{\varphi_3}$ the smallest zero of functions $h_{\varphi_1}, h_{\varphi_2}, h_{\varphi_3},$ respectively on the interval $(0, \frac{1}{L_0}).$ Moreover, define functions $g_2$ and $h_2$ on the interval $[0, r_{\varphi_2})$ by 
$$g_2(t)=(1+\frac{M^2}{(1-L_0t)(1-\varphi_2(t))})g_1(t)$$
and 
$h_2(t)=g_2(t)-1.$ We get that $h_2(0)=-1 <0$ and $h_2(t)\longrightarrow +\infty$ as $t\longrightarrow r_{\varphi_2}.$ Denote by $r_2$ the smallest such zero. Finally, for 
$$\bar{r}=\min\{r_{\varphi_1}, r_{\varphi_2}, r_{\varphi_3}\}$$
define functions $g_3$ and $h_3$ on the interval $[0, \bar{r})$ by 
\begin{eqnarray*}
 g_3(t)&=&[1+\frac{M}{1-L_0t}(\frac{1}{2}+\frac{1+\frac{8Mg_1(t)}{5(1-\frac{L_0}{2}t)}+\frac{2}{5}(\frac{Mg_1(t)}{1-\frac{L_0}{2}t})^2}{1-\varphi_1(t)})\\
 &&\times(\frac{1}{2}+\frac{M\bar{g}_2(t)}{1-\frac{L_0}{2}t}]g_2(t),
\end{eqnarray*}
$$h_3(t)=g_3(t)-1$$
and 
$$\bar{g}_2(t)=1+\frac{M62}{(1-L_0t)(1-\varphi_2(t))}.$$
We have that $h_3(0)=-1 < 0$ and $h_3(t)\longrightarrow +\infty$ as $t\longrightarrow\bar{r}^-.$ Denote by $r_3$ the smallest zero of function $g_3$ on the interval $(0, \bar{r}).$ Define the radius of convergence $\rho_1$ by 
\beq\label{2.28}
\rho_1=\min\{r_i\},\, i=1,2,3.
\eeq 

Finally, for method (\ref{1.4}), define functions $g_1$ and $g_2$ as in method (\ref{1.3}) but define function $g_3$ and $h_3$ by 
\begin{eqnarray*}
 g_3(t)&=&[1+\frac{M}{1-L_0t}\frac{1+\frac{2Mg_1(t)}{5(1-\frac{L_0}{2}t)}+\frac{1}{5}(\frac{Mg_1(t)}{1-\frac{L_0}{2}t})^2}{1-\varphi_1(t)}\\
 &&\times(1+\frac{4Mg_1(t)}{1-\frac{L_0}{2}t})\frac{M\bar{g}_2(t)}{1-\varphi_3(t)}]g_2(t),
\end{eqnarray*}
$$h_3(t)=g_3(t)-1$$
and  radius of convergence $\rho_2$ by 
\beq\label{2.29}
\rho_2=\min\{r_i\}, i=1,2,3.
\eeq 
Next, drop the hypotheses on the divided differences and $K$ from Theorem \ref{T2.1} and exchange the ``$g$'' functions and $r$ with the corresponding ``$g$'' functions for method (\ref{1.3}), $\rho_1$ and method (\ref{1.4}), $\rho_2.$ Call the resulting hypotheses (C) and (H), respectively. Then, we obtain the corresponding results.

\bt\label{T2.2}
Under the (C) hypotheses the conclusions of Theorem \ref{T2.1} hold for method (\ref{1.3}) with $\rho_1$ replacing $r.$
\et

\bt\label{T2.3}
Under the (H) hypotheses the conclusions of Theorem \ref{T2.1} hold for method (\ref{1.4}) with $\rho_2$ replacing $r.$
\et


 \brem
 \label{R2.4}
  \begin{enumerate}
 \item[(a)]  
 The radius $r_1$ was obtained by Argyros in \cite{4} as the convergence radius for Newton's method under  condition (\ref{2.12})-(\ref{2.14}). Notice that the convergence radius for Newton's method given  independently by
 Rheinboldt \cite{19*} and Traub \cite{21} is given by
 \beq\nonumber
 \rho=\frac{2}{3L}<r_1.
 \eeq%
 As an example, let us consider the function $f(x)=e^x-1$. Then $x^*=0$. Set $D=U(0,1)$. Then, we have that $L_0=e-1<l=e$, so $\rho=0.24252961<r_1=0.3827$.
 
 Moreover, the new error bounds \cite{4,5,6,7} are:
 $$
 \|x_{n+1}-x^*\|\leq \frac{L}{1-L_0\|x_n-x^*\|}\|x_n-x^*\|^2,
 $$
 whereas the old ones \cite{15,17}
 $$
 \|x_{n+1}-x^*\|\leq \frac{L}{1-L\|x_n-x^*\|}\|x_n-x^*\|^2.
 $$
% 
 Clearly, the new error bounds are more precise,  if $L_0<L$.
 Clearly, we do not expect the radius of convergence of method (\ref{1.2}) given by $r$ to be larger than $r_1$ (see (\ref{2.3})) .
 \item[(b)] The local results can be used for projection methods such as Arnoldi's method, the  generalized minimum residual method(GMREM), the generalized conjugate method(GCM) for combined Newton/finite projection methods and in connection to the mesh independence principle in order to develop the cheapest and  most efficient mesh refinement strategy \cite{4, 5,6,7}.
 \item[(c)] The results can be also be used to solve equations where the operator $F'$ satisfies the autonomous differential equation \cite{5, 7, 15, 17}:
 $$F'(x)=p(F(x)),$$ 
 where $p$ is a known continuous operator. Since $F'(x^\ast)= p(F(x^\ast))=p(0),$ we can apply the results  without actually knowing the solution $x^\ast.$ Let as an example $F(x)=e^x-1.$ Then, we can choose $p(x)=x+1$  and $x^\ast=0.$
 \item[(d)] It is worth noticing that method (\ref{1.2}) are not changing if we use the  new  instead of the old conditions \cite{23}. Moreover, for the error bounds in practice we can use the computational order of convergence (COC)
 $$
 \xi=\frac{ln\frac{\|x_{n+2}-x_{n+1}\|}{\|x_{n+1}-x_{n}\|}}{ln\frac{\|x_{n+1}-x_{n}\|}{\|x_{n}-x_{n-1}\|}},\quad \textnormal{ for each}\,\, n=1,2,\ldots
 $$
 or the approximate computational order of convergence (ACOC)
 $$
 \xi^*=\frac{ln\frac{\|x_{n+2}-x^*\|}{\|x_{n+1}-x^*\|}}{ln\frac{\|x_{n+1}-x^*\|}{\|x_{n}-x^*\|}},\quad \textnormal{ for each }\,\, n=0,1,2,\ldots
 $$
 instead of the error bounds obtained in Theorem 2.1. Related work on convergence orders can be found in \cite{9}.
 \item[(e)]In view of  (\ref{2.8}) and the estimate 
  \begin{eqnarray*}
   \|F'(x^\ast)^{-1}F'(x)\|&=&\|F'(x^\ast)^{-1}(F'(x)-F'(x^\ast))+I\|\\
   &\leq& 1+\|F'(x^\ast)^{-1}(F'(x)-F'(x^\ast))\| \leq 1+L_0\|x-x^\ast\|
  \end{eqnarray*}
 condition (\ref{2.10}) can be dropped and $M$ can be replaced by 
 $$M(t)=1+L_0 t$$
 or
 $$M(t)=M=2,$$
 since $t\in [0, \frac{1}{L_0}).$
% 
 \end{enumerate}
 \erem
%
%
\setcounter{equation}{0}
\section{Numerical Example}
We present a numerical example in this section.
%%%%%%%%%%%%
% %%%%%%%%%%%%
% \bexample
% Let $X=Y=\mathbb{R}^2,\,\,D=\overline{U}(0,1).$ Define $F$ on $D$ for $v= (x, y)^T$ by
% \beq\label{3.2}
% F(v)=(e^x-1, \frac{e-1}{2}y^2+y)^T.
% \eeq%
% Then, the Fr\'{e}chet-derivative is given by
% $$F'(v)=\left[
% \begin{array}{cc}
%   e^x&0\\
% 0&(e-1)y+1
%    \end{array}\right].
% $$
% Notice that $x^\ast=(0,0)^T,\,F'(x^\ast)=F'(x^\ast)^{-1}=diag\{1,1\},\, L_0=e-1,  L= e^{1/L_0}= M = 1.7896, K=K_0=\frac{L}{2}, \beta=0.5.$
% Then, the parameters for  method (\ref{1.2}) are
% $$r_1=0.3827,\, r_2=0.0717=r,\, r_3=0.0792.$$
% \eexample
% %
% 

\bexample
Returning back to the motivation example at the introduction on this paper, we have $L_0=L=96.662907,\,M=1.0631,  K=K_0=\frac{L}{2}, \beta=-1.$ 
Then,  the parameters for method (\ref{1.2}) are
$$r_1=0.0069,\, r_2=0.0051=r ,\, r_3=0.1217.$$
We have 
$ACOC =    1.7960$
 and $COC = 1.8371.$


\eexample


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\end{document} 