Ball convergence for an Aitken-Newton method

Ioannis K. Argyros$^\ast $, Munish Kansal$^\bullet $ Vinay Kanwar$^+$

March 18, 2016. Accepted: February 18, 2017. Published online: February 17, 2019.

$^\ast $Department of Mathematical Sciences, Cameron University, Lawton, OK 73505, USA, e-mail: iargyros@cameron.edu.

$^\bullet $ School of Mathematics, Thapar Institute of Engineering and Technology, Patiala, Punjab-147004, India, e-mail: mkmaths@gmail.com.

$^+$ University Institute of Engineering and Technology, Panjab University, Chandigarh-160 014, India, e-mail: vmithil@yahoo.co.in.

We present a local convergence analysis of an eighth-order Aitken-Newton method for approximating a locally unique solution of a nonlinear equation. Earlier studies have shown convergence of these methods under hypotheses up to the eighth derivative of the function although only the first derivative appears in the method. In this study, we expand the applicability of these methods using only hypotheses up to the first derivative of the function. This way the applicability of these methods is extended under weaker hypotheses. Moreover, the radius of convergence and computable error bounds on the distances involved are also given in this study. Numerical examples are also presented in this study.

MSC. 65D10, 65D99

Keywords. Nonlinear equations, Aitken-Newton method, local convergence, eighth order of convergence.

1 Introduction

Let $X$, $Y$ be Banach spaces and $D$ be a convex subset of $X.$ Let also $L(X,Y)$ denote the set of bounded linear operators from $X$ into $Y$. Many problems can be written in the form

\begin{equation} \label{I1} F(x)=0 \end{equation}

using Mathematical Modelling [ 1 ] , [ 6 ] , [ 7 ] , [ 9 ] , [ 23 ] where $F:D\subseteq X\to Y$ is a Fréchet-differentiable operator. Most solution methods for finding a solution $x^{*}$ of equation 1 are iterative, since closed form solutions can be found only in special cases [ 1 ] - [ 23 ] . In this paper, we study the local convergence of Aitken-Newton method [ 21 ] defined for each $n=0,1,2,\ldots $ by

\begin{equation} \label{M1} \left\{ \, \begin{aligned} y_n=& x_n-F’(x_n)^{-1}F(x_n),\\ z_n=& y_n-F’(y_n)^{-1}F(y_n),\\ x_{n+1}=& z_n-[z_n,y_n;F]^{-1}F(z_n) \\ & -[y_n,z_n;F]^{-1}[z_n,y_n,y_n;F][y_n,z_n;F]^{-1}F(z_n)F’(y_n)^{-1}F(y_n), \end{aligned} \right. \end{equation}

where $[\cdot ,\cdot ~ ;F],~ [\cdot ,\cdot ,\cdot ~ ;F]$ are divided differences of order one and two, respectively and

\begin{equation} \label{I2} [x,y;F]=\int _0^{1}F'(y+\theta (x-y))d\theta . \end{equation}

The above method 2 attains eighth-order of convergence using five functional evaluations, viz. $F(x_n), ~ F'(x_n), ~ F(y_n)$, $F'(y_n)$ and $F(z_n)$, per iteration. Therefore, the efficiency index [ 23 ] of the proposed method is $E=\sqrt[5]{8}\approx 1.51$, when $X=Y=\mathbb {R}$. The convergence of method 2 was shown in [ 21 ] using Taylor expansions and hypotheses reaching up to the eighth derivative of the function $F$ although only first derivative appears in the method. We will show that method 2 is well-defined and convergent using hypotheses only on the first derivative in the more general setting of a Banach space. Notice that the method 2 was not shown to be well defined in [ 21 ] . However, the eighth order of convergence was shown assuming that method 2 is well defined which may not be the case. These hypotheses limit the applicability of method 2.

As a motivational example, define function $F$ on $D=[-\frac{1}{2},\frac{5}{2}]$ by

\begin{equation} \nonumber F(x)=\begin{cases} x^3\ln {x^2}+x^5-x^4,\qquad x\neq 0,\\ 0,~ ~ ~ \qquad \qquad \qquad \qquad x=0. \end{cases} \end{equation}

We have $x^{*}=1$,

\begin{equation} \label{I2} \begin{aligned} & F’(x)=3x^2\ln {x^2}+5x^4-4x^3+2x^2,\\ & F”(x)=6x\ln {x^2}+20x^3-12x^2+10x,\\ & F”’(x)=6\ln {x^2}+60x^2-24x+22.\\ \end{aligned} \end{equation}

Clearly, function $F'''(x)$ is unbounded on $D$. Hence, the results in [ 21 ] cannot be applied to solve equation $F(x)=0,$ where $F$ is given by 6. Moreover, the results in [ 21 ] do not provide computable convergence radii, error bounds on the distances $|x_n-x^{*}|$ and uniqueness of the solution results. We address all these problems using only hypotheses on the first derivative. We use the computational order of convergence (COC) to approximate the convergence order (which does not depend upon the solution $x^{*}$). Moreover, we present the results in a more general setting of a Banach space.

The rest of the paper is organized as follows: In Section 2, we present the local convergence of method 2. The numerical examples are presented in the concluding Section 3.

2 Local convergence analysis

We present the local convergence analysis of method 2 in this section using some scalar functions and parameters.

Let $L_0{\gt}0,~ L{\gt}0,~ K{\gt}0,$ and $M\geq 1 $ be given parameters. Define function $g_1$ on the interval $[0,\frac{1}{L_0})$ by

\begin{equation} \nonumber \begin{aligned} g_1(t)& =\tfrac {Lt}{2(1-L_0t)},\\ \end{aligned} \end{equation}

and parameter $r_1$ by

\[ r_1=\tfrac {2}{2L_0+L}. \]

We have that $g_1(r_1)=1$ and $0\leq g_1(t){\lt}1$ for each $t\in [0,r_1)$. Define functions $p_1$ and $h_{p_1}$ on the interval $[0,\frac{1}{L_0})$ by

\[ p_1(t)=L_0g_1(t)t \]

and $h_{p_1}(t)={p_1}(t)-1.$

We get that $h_{p_1}(0)=-1{\lt}0$ and $h_{p_1}(t)\to +\infty $ as $t\to \frac{1^{-}}{L_0}$. It follows from the intermediate value theorem that function $h_{p_1}$ has zeros in the interval $(0,\frac{1}{L_0}).$ Denote by $r_{p_1}$ the smallest such zero. Moreover, define functions $g_2$ and $h_{2}$ on the interval $[0,r_{p_1})$ by

\begin{equation} \nonumber \begin{aligned} g_2(t)=\tfrac {Lg_1^2(t)t^3}{2(1-p(t))} \end{aligned} \end{equation}

and $h_{2}(t)=g_2(t)-1.$

We get that $h_{2}(0)=-1{\lt}0$ and $h_{2}(t)\to +\infty $ as $t\to r_{p_1}^{-}.$ Denote by $r_{2}$ the smallest zero of function $h_2$ in the interval $(0,r_2)$. Furthermore, define functions $p_2$ and $h_{p_2}$ on the interval $[0,r_{p_1})$ by

\[ p_2(t)=\tfrac {L_0}{2}\Big(g_1(t)+g_2(t)\Big)t \]

and $h_{p_2}(t)={p_2}(t)-1.$

We have that $h_{p_2}(0)=-1{\lt}0$ and $h_{p_2}(t)\to +\infty $ as $t\to r_{p_1}^{-}$. Denote by $r_{p_2}$ the smallest such zero of function $h_{p_2}$ in the interval $(0,r_{p_1}).$ Finally, define functions $g_3$ and $h_{3}$ on the interval $[0,r_{p_2})$ by

\begin{equation} \nonumber \begin{aligned} & g_3(t)=\left(1+\tfrac {M}{1-p_2(t)}+\tfrac {KM^2 g_1(t)t}{(1-p_2(t))^2(1-p_1(t))}\right)g_2(t) \end{aligned} \end{equation}

and $h_3(t)=g_3(t)-1.$

We obtain that $h_3(0)=-1{\lt}0$ and $h_3(t)\to +\infty $ as $t \to r_{p_2}^{-}$. Denote by $r_{3}$ the smallest zero of function $h_3$ in the interval $(0,r_{p_2}).$ Define the radius of convergence $r$ by

\begin{equation} \label{L1} r=\min \{ r_i\} ,~ i=1,2,3. \end{equation}

Then, we have that

\begin{align} \label{L2} 0{\lt}r\leq r_1{\lt}\tfrac {1}{L_0} \end{align}

and for each $t\in [0,r)$

\begin{equation} \label{L3} 0\leq g_i(t)<1,~ i=1,2,3 \end{equation}

and

\begin{equation} \label{L4} 0\leq p_j(t)<1, ~ j=1,2. \end{equation}

Let $U(z,\rho )~ \text{and}~ \bar{U}(z,\rho )$ stand respectively for the open and closed balls in $X$ with center at $z\in X$ and of radius $\rho {\gt}0.$ Next, the local convergence analysis of method 2 shall be presented using previous notations.

Theorem 1

Let $F:{D} \subseteq X \to Y$ be a Fréchet-differentiable operator. Let $[\cdot ,\cdot ~ ;F],~ [\cdot ,\cdot ,\cdot ~ ;F]$ are divided differences of order one and two on $D$, respectively. Suppose there exist $x^{*}\in D$ and $L_0{\gt}0,$ such that for each $x\in D$

\begin{equation} \label{L5} F(x^{*})=0,~ F'(x^{*})^{-1}\in L(Y,X)\quad \end{equation}

and

\begin{equation} \label{L6} \lVert F'(x^{*})^{-1}(F'(x)-F'(x^{*}))\rVert \leq L_0\lVert x-x^{*}\rVert . \end{equation}

Moreover, suppose there exist $L{\gt}0,~ K{\gt}0$ and $M\geq 1$ such that for each $x,y\in D_0:=D\cap U\left(x^{*},\frac{1}{L_0}\right)$, we have

\begin{equation} \label{L7} \lVert F'(x^{*})^{-1}(F'(x)-F'(y))\rVert \leq L\lVert x-y \rVert ,\quad , \end{equation}

\begin{equation} \label{L8} \lVert F'(x^{*})^{-1}F'(x)\rVert \leq M, \end{equation}

\begin{equation} \label{L9} \lVert F'(x^{*})^{-1}[x,y,y;F] \rVert \leq K, \end{equation}

and

\begin{equation} \label{L10} \bar{U}(x^{*},r)\subseteq D, \end{equation}

where radius of convergence $r$ is defined by 7. Then, the sequence $\{ x_n\} $ generated for $x_0\in U(x^{*},r)$-$\{ x^{*}\} $ by method 2 is well defined, remains in $U(x^{*},r)$ for each $n=0,1,2,\ldots $ and converges to the solution $x^{*}$. Moreover, the following error estimates hold

\begin{align} \label{L11} \lVert y_n-x^{*} \rVert \leq g_1(\lVert x_n-x^{*}\rVert )\lVert x_n-x^{*}\rVert \leq \lVert x_n-x^{*}\rVert {\lt}r, \end{align}

\begin{equation} \label{L12} \lVert z_n-x^{*} \rVert \leq g_2(\lVert x_n-x^{*}\rVert )\lVert x_n-x^{*}\rVert \leq \lVert x_n-x^{*}\rVert \end{equation}

and

\begin{equation} \label{L13} \lVert x_{n+1}-x^{*} \rVert \leq g_3(\lVert x_n-x^{*}\rVert )\lVert x_n-x^{*}\rVert \leq \lVert x_n-x^{*}\rVert , \end{equation}

where the $``g_i,~ i=1,2,3"$ functions are defined previously. For $T\in [r,\frac{2}{L_0}),$ the limit point $x^{*}$ is the only solution of equation $F(x)=0$ in $D_1:=D\cap \bar{U}(x^{*},T).$

Proof â–¼

We shall show using mathematical induction that sequence $\{ x_n\} $ of iterates generated by 2 is well defined, remains in $U(x^{*},r)$ and satisfies estimations 17–19. By hypothesis, $x_0\in U(x^{*},r)-x^{*}$, 8 and 12, we have that

\begin{align} \label{L14} \lVert F’(x^{*})^{-1}(F’(x_0)-F’(x^{*}))\rVert \leq L_0\lVert x_0-x^{*}\rVert {\lt}L_0r{\lt}1. \end{align}

It follows from estimate 20 and the Banach lemma on invertible operators [ 6 , 9 , 20 , 22 , 23 ] that $F'(x_0)^{-1}\in L(Y,X)$ and

\begin{equation} \label{L15} \lVert F'(x_0)^{-1}F'(x^{*})\rVert \leq \tfrac {1}{1-L_0\Vert x_0-x^{*}\rVert }. \end{equation}

Hence, $y_0$ is well defined. Using 8, 9, 13, 21 and the first substep of method 2 for $n=0$, we get in turn that

\begin{align} & \lVert y_0-x^{*}\rVert \leq \label{L16} \\ & \leq \lVert x_0-x^{*}-F’(x_0)^{-1}F’(x_0)\rVert \nonumber \\ & \leq \lVert F’(x_0)^{-1}F’(x^{*})\rVert \Big\Vert \int _{0}^{1}F’(x^{*})^{-1}[F’(x^{*}+\theta (x_0-x^{*}))-F’(x_0))](x_0-x^{*})d\theta \Big\Vert \nonumber \\ & \leq \dfrac {L\lVert x_0-x^{*}\rVert ^2}{2(1-L_0\lVert x_0-x^{*}\rVert )}= g_1(\lVert x_0-x^{*}\rVert )\lVert x_0-x^{*}\rVert \leq \lVert x_0-x^{*}\rVert {\lt}r, \nonumber \end{align}

which shows 17 for $n=0$ and $y_0\in U(x^{*},r)$. Then, we have as in 20 that

\begin{align} \label{L17} \lVert F’(x^{*})^{-1}(F’(y_0)-F’(x^{*}))\rVert \leq L_0\lVert y_0-x^{*} \rVert & \leq L_0g_1(\lVert x_0-x^{*}\rVert )\lVert x_0-x^{*}\rVert \\ & =p_1(\lVert x_0-x^{*}\rVert ){\lt}p_1(r){\lt}1, \end{align}

so $F'(y_0)^{-1}\in L(Y,X)$. Therefore, we have

\begin{equation} \label{L18} \lVert F'(y_0)^{-1}F'(x^{*})\rVert \leq \tfrac {1}{1-p_1\Vert x_0-x^{*}\rVert } \end{equation}

and as in 22

\begin{align} \label{L19} \lVert z_0-x^{*}\rVert & \leq \tfrac {L\lVert y_0-x^{*}\rVert ^2}{2(1-L_0\lVert y_0-x^{*}\rVert )}\\ & \leq g_2(\lVert x_0-\xi \rVert ) \lVert x_0-x^{*}\rVert \leq \lVert x_0-x^{*}\rVert {\lt}r, \end{align}

which implies that 18 holds for $n=0$ and $z_0\in U(x^{*},r).$

Next, we show that $[y_0,z_0;F]^{-1}$ exists. In view of the definition of divided difference $[\cdot ,\cdot ;F]$, 8, 9, 12, 22 and 25 that

\begin{align} \label{L20} & \lVert F’(x^{*})^{-1}\left([z_0,y_0;F]-F’(x^{*})\right)\rVert \leq \\ & \leq \tfrac {L_0}{2}\left(\lVert z_0-x^{*}\rVert +\lVert y_0-x^{*}\rVert \right) \nonumber \\ & \leq \tfrac {L_0}{2}\Big(g_1(\lVert x_0-x^{*}\rVert )\! +\! g_2(\lVert x_0-x^{*}\rVert )\Big)\lVert x_0-x^{*}\rVert \nonumber \\ & =p_2(\lVert x_0-x^{*}\rVert ){\lt}p_2(r){\lt}1, \nonumber \end{align}

\begin{equation} \label{L21} \lVert [z_0,y_0;F]^{-1}-F'(x^{*})\rVert \leq \tfrac {1}{1-p_2\Vert x_0-x^{*}\rVert }. \end{equation}

Hence, $x_1$ is well defined. Notice that $\lVert x^{*}+\theta (x_0-x^{*})-x^{*}\rVert \leq \theta \lVert x_0-x^{*}\rVert {\lt}r$ for each $\theta \in [0,1],$ so $x^{*}+\theta (x_0-x^{*})\in U(x^{*},r)$ for each $\theta \in [0,1].$ Then, by 11 and 14, we get that

\begin{align} \label{L22} & \lVert F’(x^{*})^{-1} F(x_0) \rVert = \\ & =\lVert F’(x^{*})^{-1}( F(x_0)-F(x^{*}) \rVert \nonumber \\ & =\lVert \int _0^1 F’(x^{*})^{-1} F’(x^{*}+\theta (x_0-x^{*}))(x_0-x^{*}) \rVert \leq M \lVert x_0-x^{*} \rVert \nonumber \end{align}

and similarly

\begin{equation} \label{L23} \begin{aligned} \lVert F’(x^{*})^{-1} F(y_0) \rVert & \leq M \lVert y_0-x^{*} \rVert \leq M g_1(\lVert x_0-x^{*} \rVert )\lVert x_0-x^{*} \rVert . \end{aligned} \end{equation}

By the last substep of method 2 for $n=0$, 7, 8, 22, 25, 26, 29–31, we obtain in turn that

\begin{align} \label{L24} & \lVert x_1-x^{*} \rVert \leq \\ & \leq \lVert z_0-x^{*}\rVert +\lVert [z_0,y_0;F]^{-1}F’(x^{*})\rVert \lVert F’(x^{*})^{-1}F’(z_0)\rVert \nonumber + \\ & \quad +\lVert [z_0,y_0;F]^{-1}F’(x^{*})\rVert \lVert [z_0,y_0;F]F’(x^{*})^{-1}\rVert \lVert [z_0,y_0;F]^{-1}F’(x^{*})\rVert \cdot \nonumber \\ & \quad \cdot \lVert F’(x^{*})^{-1} [z_0,y_0,y_0;F]^{-1}F’(x^{*})\rVert \lVert F’(y_0)^{-1}F’(x^{*})\rVert \lVert F’(x^{*})^{-1}F(y_0)\rVert \nonumber \\ & \leq \lVert z_0-x^{*} \rVert +\tfrac {M\lVert z_0-x^{*} \rVert }{1-p_2(\lVert x_0-x^{*} \rVert )}+\tfrac {KM^2\lVert z_0-x^{*} \rVert \lVert y_0-x^{*} \rVert }{(1-p_2(\lVert x_0-x^{*} \rVert ))^2(1-p_1(\lVert x_0-x^{*} \rVert ))}\nonumber \\ & =\Big(1+\tfrac {M}{1-p_2(\lVert x_0-x^{*} \rVert )}+\tfrac {KM^2\lVert y_0-x^{*} \rVert }{(1-p_2(\lVert x_0-x^{*} \rVert ))^2(1-p_1(\lVert x_0-x^{*} \rVert ))}\Big)\lVert z_0-x^{*} \rVert \nonumber \\ & \leq g_3(\lVert x_0-x^{*}\rVert )\lVert x_0-x^{*}\rVert \leq \lVert x_0-x^{*}\rVert {\lt}r, \nonumber \end{align}

which implies that 19 holds for $n=0$ and $x_1\in U(x^{*},r).$ By simply replacing $y_0,z_0,x_1$ by $y_n,z_n,x_{n+1}$ in the preceding estimates, we complete the induction for estimates 17–19. Then, in view of the estimate $\lVert x_{n+1}-x^{*}\rVert \leq c \lVert x_{n}-x^{*}\rVert {\lt}r,~ c=g_3(\lVert x_{0}-x^{*}\rVert )\in [0,1),$ we deduce that $\lim _{n\to \infty }x_n=x^{*}$ and $x_{n+1}\in U(x^{*},r).$ The proof of the uniqueness follows using standard arguments [ 11 ] .

Remark 2

1) It follows from 12 that condition 14 can be dropped, if we set

\[ M(t)=1+L_0t \]

\[ M(t)=M=2,~ \text{since}~ ~ t\in \Big[0,\tfrac {1}{L_0}\Big). \]

2) The results obtained here can also be used for operators $F$ satisfying autonomous differential equations [ 6 , 9 ] of the form:

\[ F'(x)=P(F(x)), \]

where $P$ is a continuous operator. Then, since $F'(x^{*})=P(F(x^{*}))=P(0)$, we can apply the results without actually knowing $x^{*}$. For example, let $f(x) = e^x-1$. Then, we can choose $P(x) =x +1.$
3) The radius $\bar{r_1}=\frac{2}{2L_0+L_1}$ was shown by Argyros [ 6 ] to be the convergence radius of Newton’s method

\begin{equation} \label{V31} x_{n+1}=x_n-F'(x_n)^{-1}F(x_n),~ \text{for each}~ n=0,1,2,\ldots \end{equation}

under the conditions 11–13 on $D$, where $L_1$ is the Lipschitz constant on $D.$ We have that $L\leq L_1$ and $L_0\leq L_1,$ so $\bar{r_1}\leq r_1.$ It follows that the convergence radius $r$ of the method 2 cannot be larger than the convergence radius $r_1$ of the second order Newton’s method 33. As already noted in [ 6 ] , $\bar{r_1}$ is at least as large as the convergence ball given by Rheinboldt [ 22 ]

\begin{equation} \nonumber r_R=\tfrac {2}{3L_1}. \end{equation}

In particular, for $L_0{\lt}L_1,$ we have that

\[ r_R{\lt}\bar{r_1} \]

and

\[ \tfrac {r_R}{\bar{r_1}}\to \tfrac {1}{3}~ \text{as}~ \tfrac {L_0}{L_1}\to 0. \]

That is our convergence ball $\bar{r_1}$ is at most three times larger than Rheinboldt’s. The same value of $r_R$ was given by Traub [ 23 ] . 4) It is worth noticing that method 2 is not changing when we use the conditions of Theorem 1 instead of stronger conditions used in previous studies. Moreover, we can consider the semi-computational order of convergence defined by

\begin{align*} Q_{L}(k)= & \tfrac {\ln |x_{n+1}-x^{*}| } {\ln |x_n-x^{*}| } \\ Q_\Lambda (k)= & \ln \Big(\tfrac {|x_{n+1}-x^{*}|}{| x_{n}-x^{*}|}\Big)/\ln \Big(\tfrac {| x_{n}-x^{*}|}{| x_{n-1}-x^{*}|}\Big), \end{align*}

for which some interesting properties were obtained in [ 19 ] and [ 12 ] . We can even compute the (full) computational order of convergence defined by

\begin{align*} Q^\prime _{L}(k)= & \tfrac {\ln |x_{n}-x_{n-1}| } {\ln |x_{n-1}-x_{n-2}| } \\ Q^\prime _\Lambda (k)=& \ln \Big(\tfrac {| x_{n}-x_{n-1}|}{| x_{n-1}-x_{n-2}|}\Big)/\ln \Big(\tfrac {| x_{n-1}-x_{n-2}|}{| x_{n-2}-x_{n-3}|}\Big), \end{align*}

for which interesting properties were obtained in [ 12 ] . We also recommend to the motivated reader the excellent survey on these notions [ 13 ] , containing full proofs and historical aspects.

This way we obtain in practice the order of convergence in a way that avoids the bounds involving estimates using estimates higher than the first Fréchet derivative of operator $F.$ Notice also that the computation of $\xi ^{*}$ does not require knowledge of $x^{*}.$ 5) Also, condition 12 can be replaced by

\begin{equation} \label{L6'} \lVert F'(x^{*})^{-1} ([x,y;F]-F'(x^{*}))\rVert \leq \tfrac {\bar{L_0}}{2} (\lVert x-x^{*}\rVert +\lVert y-x^{*}\rVert ). \end{equation}

In this case, the choice of the divided difference given by 6 can be dropped. Moreover, $\bar{L_0}$ and 34 can replace $L_0$ and 12, respectively in the proof of Theorem 1. â–¡

3 Numerical examples

We present numerical examples in this section.

Example 3

Let $X=Y=\mathbb {R}^3$, $D=\bar{U}(0,1),~ x^{*}=(0,0,0)^T$. Define function $F$ on $D$ for $w=(x,y,z)^T$ by

\[ F(w)=(e^x-1,\tfrac {e-1}{2}y^2+y,z)^T. \]

Then, the Fréchet derivative is given by

\[ F'(w)=\begin{bmatrix} {} e^x & 0 & 0 \\ 0 & (e-1)y+1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}. \]

We have that $L_0=e-1$, $L=e^{\frac{1}{L_0}}$, $M=e^{\frac{1}{L_0}}$, $K=\tfrac {e^{\frac{1}{L_0}}}{2}$ and $L_1=e$. The parameters using method 2 are:

\begin{align*} & r_1=0.382692,~ r_2=0.45738,~ r_3=0.38726, \\ & r=0.38726,~ \bar{r_1}=0.324947,~ r_R=0.245253. \end{align*}

Example 4

Let $X=Y=C[0,1],$ the space of continuous functions defined on $[0,1]$ and be equipped with the max norm. Let $D=\bar{U}(0,1)$ and $B(x)=F''(x)$ for each $x\in D$. Define function $F$ on $D$ by

\begin{equation} F(\phi )(x)=\phi (x)-5\int _0^1 x\theta \phi (\theta )^3 d\theta . \end{equation}

We have that

\begin{equation} F'(\phi (\xi ))(x)=\xi (x)-15\int _0^1 x\theta \phi (\theta )^2 \xi (\theta ) d\theta , ~ \text{for each}~ \xi \in D. \end{equation}

Then, we get that $x^{*}=0,~ L_0=7.5,~ L_1=15,~ L=15,~ K=15$ and $M=2.$ The parameters using method 2 are:

$r_1=0.0666667,~ r_2=0.114372,~ r_3=0.0783549,~ r=0.0783549,$
$\bar{r_1}=0.0666667,~ r_R=0.0444444.$ â–¡

Example 5

Returning back to the motivational example at the introduction of this paper, we have that $L=L_0=146.6629073$, $M=2$, $L_1=L$ and $K=48.3315.$ The parameters using method 2 are:

\begin{align*} & r_1=0.00454557,~ r_2=.006787,~ r_3=0.004991287,\\ & \bar{r_1}=0.00454557,~ r_R=0.00454557,~ \xi ^{*}=7.8403. \end{align*}

Example 6

Let $X=Y=\mathbb {R}$ and define function $F$ on $D=\mathbb {R}$ by

\begin{equation} \label{N1} F(x)=\beta x-\gamma \sin {(x)}-\delta ,\end{equation}

where $\beta ,~ \gamma ,~ \delta $ are given real numbers. Suppose that there exists a solution $x^{*}$ of $F(x)=0$ with $F'(x^{*})\neq 0.$ Then, we have

\[ L_1=L_0=L=\tfrac {|\gamma |}{|\beta -\gamma \cos {x^{*}}|},~ M=\tfrac {|\gamma |+|\beta |}{|\beta -\gamma \cos {x^{*}}|}~ \text{and}~ K=\tfrac {|\gamma |}{2|\beta -\gamma \cos {x^{*}}|}. \]

Then one can find the convergence radii for different values of $\beta ,~ \gamma $ and $\delta $. As a specific example, let us consider Kepler’s equation 37 with $\beta =1$, $0\leq \gamma {\lt}1$ and $0\leq \delta \leq \pi .$ A numerical study was presented in [ 14 ] for different values of $\gamma $ and $\delta .$

Let us take $\gamma =0.9$ and $\delta =0.1$. Then the solution is given by $x^{*}=0.6308435.$ Hence, for method 2 the parameters are:

\begin{equation} \begin{aligned} & r_1=0.202387,~ r_2=0.261858,~ r_3=0.196578,\\ & r=0.196578, \bar{r_1}=0.202387, r_R=0.202387,~ \xi ^{*}=8.0353. \end{aligned}\end{equation}

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