Convergence analysisthe two-step Newton method of order four

Ioannis K. Argyros$^\ast $ Sanjay K. Khattri$^\S $

September 12, 2012.

$^\ast $Department of Mathematical Sciences, Cameron University, Lawton, Oklahoma 73505-6377, USA, e-mail: iargyros@cameron.edu.

$^\dag $Department of Engineering, Stord Haugesund University College, Norway, e-mail: sanjay.khattri@hsh.no.

We provide a tighter than before convergence analysis for the two-step Newton method of order four using recurrent functions. Numerical examples are also provided in this study.

MSC. 65H10; 65G99; 65J15; 47H17; 49M15

Keywords. Two-step Newton method, Newton’s method, Banach space, Kantorovich hypothesis, majorizing sequence, Lipschitz/center-Lipschitz condition.

1 Introduction

In this study, we are concerned with the problem of approximating a locally unique solution $x^\star $ of equation

\begin{equation} \label{eq:11} \mathcal{F}(x)=0, \end{equation}

where, $\mathcal{F}$ is Fréchet-differentiable operator defined on a convex subset $\mathcal{D}$ of a Banach space $\mathcal{X}$ with values in a Banach space $\mathcal{Y}$.

Many problems in computational mathematics can be brought in the form 1. The solutions of these equations are rarely found in closed form. Therefore most solution methods for these equations are iterative. Newton’s method

\begin{equation} \label{eq:12} x_{n+1} = x_n - \mathcal{F}^\prime (x_n)^{-1}\mathcal{F}(x_n)\quad (n\ge {0}), \quad (x_0\in \mathcal{D}) \end{equation}

is undoubtedly the most popular method for generating a sequence $\{ x_n\} $ converging quadratically to $x^\star $ [ 5 , 13 , 15 ] . Two-step Newton method (TSNM)

\begin{equation} \label{eq:13} \begin{split} y_n & = x_n - \mathcal{F}^\prime (x_n)^{-1}\mathcal{F}(x_n)\quad (n\ge {0}), \quad (x_0\in \mathcal{D}),\\ x_{n+1} & = y_n - \mathcal{F}^\prime (y_n)^{-1}\mathcal{F}(y_n), \end{split} \end{equation}

generates a converging sequence $\{ x_n\} $ to $x^\star $ with order four [ 5 , 9 ] . The following conditions have been used to show the semilocal convergence for the Newton’s method 2 and consequently the semilocal convergence of (TSNM) [ 5 , 13 , 15 , 17 ] ($\textbf{C}_{\text{K}}$):

\begin{align} \mathcal{F}^\prime (x_0)^{-1} & \in L(\mathcal{Y},\mathcal{X}) \quad \textrm{for some } x_0\in {\mathcal{D}}; \nonumber \\ \big\Vert {\mathcal{F}^\prime (x_0)^{-1}\mathcal{F}(x_0)}\big\Vert & \le {\nu } \nonumber \\ \big\Vert {\mathcal{F}^\prime (x_0)^{-1}\big[\mathcal{F}^\prime (x)-\mathcal{F}^\prime (y)\big]} \big\Vert & \le L \left\Vert {x-y}\right\Vert \quad \textrm{for all } x,y\in {\mathcal{D}}; \nonumber \\ h_{K} & = L{\eta } \le \tfrac {1}{2} \label{eq:14} \intertext {and} \overline{U}(x_0,\lambda )= \big\{ x \in {{{\mathcal{X}}}}\, \big\vert \, & \Vert {x-x_0}\Vert \le \lambda \big\} \subseteq \mathcal{D}, \nonumber \end{align}

for specified $\lambda \ge {0}$.

Note that ?? is the, famous for its simplicity and clarity, Kantorovich sufficient convergence hypothesis for the Newton’s method 2. A current survey on Newton-type methods can be found in [ 5 ] and the references therein (see also [1–4] and [6–17]). We have shown [ 5 ] the quadratic convergence of the Newton’s method 2 using the set of conditions ($\textbf{C}_{\text{AH}}$)

\begin{alignat}{5} \mathcal{F}^\prime (x_0)^{-1}& \in L(\mathcal{Y},\mathcal{X}) \quad & & \textrm{for some } x_0\in {\mathcal{D}}; \nonumber \\ \big\Vert {\mathcal{F}^\prime (x_0)^{-1}\mathcal{F}(x_0)}\big\Vert & \le {\eta } & {}\nonumber \\ \big\Vert {\mathcal{F}^\prime (x_0)^{-1}\big[\mathcal{F}^\prime (x)-\mathcal{F}^\prime (x_0)\big]} \big\Vert & \le L_0 \left\Vert {x-x_0}\right\Vert \quad & & \textrm{for all } x\in {\mathcal{D}}; \nonumber \\ \big\Vert {\mathcal{F}^\prime (x_0)^{-1}\big[\mathcal{F}^\prime (x)-\mathcal{F}^\prime (y)\big]} \big\Vert & \le L \left\Vert {x-y}\right\Vert \quad & & \textrm{for all } x,y\in {\mathcal{D}}; \nonumber \\ h_{AH} = \overline{L}{\eta } & \le \tfrac {1}{2} & {}\label{eq:15} \intertext {and} \overline{U}(x_0,\lambda _0) & \subseteq \mathcal{D}, & {}\nonumber \end{alignat}

for some specified $\lambda _0\ge {0}$, where

\begin{align} \overline{L} & = \tfrac {1}{8}\left(L+4L_0+\sqrt{L^2+8L_0L}\right).\label{eq:16} \end{align}

Note that

\begin{equation} L_0 \le L \label{eq:17} \end{equation}

holds in general, and $L/L_0$ can be arbitrarily large [ 4 , 5 ] . Moreover, the $L_0$ Center-Lipschitz is not an additional condition, since $L_0$ is a special case of $L$. Furthermore, we have by ??-7

\begin{equation} \label{eq:18} h_{K}\le \tfrac {1}{2}\quad \Longrightarrow \quad h_{AH} \le \tfrac {1}{2} \end{equation}

but not necessarily vice versa unless if $L_0$ = $L$. The error analysis under ?? is also tighter than ??. Hence, the applicability of Newton’s method 2 has been extended.

In this study, we provide the sufficient convergence conditions for (TSNM) corresponding to ??. The paper is organized as follows: §2 contains the semilocal convergence analysis for (TSNM), whereas the numerical examples are given in §3.

2 Semilocal Convergence Analysis for (TSNM)

We need the following result on majorizing sequence for (TSNM).

Lemma 1

Let $L_0$, $L$, $\eta $ be constants. Assume: there exist parameters $\alpha $ and $\phi $ such that

\begin{align} \tfrac {L\eta }{2(1-L_0\eta )} & \le \alpha ,\label{eq:21}\\ \tfrac {L_1\eta }{2(1-L_2\eta )} & \le \phi \le \phi _0\label{eq:22} \intertext {and} \eta & \le \eta _0\label{eq:23} \end{align}

where,

\begin{gather} L_1 = \alpha ^2\, L,\qquad L_2 = (1+\alpha )L_0,\label{eq:24}\\ \phi _1 = \tfrac {4L_0\alpha }{2(L_0+L_2)\alpha - L + \sqrt{\left[2(L_0+L)\alpha -L\right]^2+8L_0L\alpha }},\label{eq:25}\\ \phi _2 = \tfrac {2L_1}{L_1+\sqrt{L_1^2+8L_1L_2}}, \qquad \phi _3 = \tfrac {2\alpha \left[1-(L_0+L_2)\eta \right]}{L\eta },\label{eq:26}\\ \phi _0=\min \left\{ \phi _1,\phi _2,\phi _3\right\} ,\label{eq:27}\\ \eta _1 = \tfrac {2}{L_1+2L_2(1+\phi )}, \qquad \eta _2 = \tfrac {1}{L_0+L_2},\label{eq:28}\\ \eta _0 = \min \left\{ \eta _1,\eta _2\right\} . \end{gather}

Then, sequences $\{ s_n\} $, $\{ t_n\} $ generated by

\begin{equation} \label{eq:210} \begin{array}{llll} t_0=0,\quad s_0 = \eta , \quad t_{n+1} = s_n + \tfrac {L(s_n-t_n)^2}{2(1-L_0s_n)},\\ s_{n+1} = t_{n+1} + \tfrac {L(t_{n+1}-{{s_n}})^2}{2(1-L_0t_{ n+1})}, \end{array} \end{equation}

are non-decreasing, bounded from above by

\begin{equation} \label{eq:211} t^{\star \star } = \left(\tfrac {1+\alpha }{1-\phi }\right)\eta , \end{equation}

and converge to their common least upper bound $t^\star \in [0,t^{\star \star }]$. Moreover, the following estimates holds

\begin{alignat}{3} 0& \le {t_{n+1}-s_n} & \le \alpha (s_n-t_n),\label{eq:212} \intertext {and} 0& \le {s_{n+1}-t_{n+1}} & \le \phi (s_n-t_n).\label{eq:213} \end{alignat}

Proof â–¼

We shall show using induction on $k$:

\begin{alignat}{3} 0& \le \tfrac {L(s_k-t_k)}{2(1-L_0s_k)} & \le \alpha ,\label{eq:214} \intertext {and} 0& \le \tfrac {L_1(s_k-t_k)}{2(1-L_0t_{k+1})} & \le \phi .\label{eq:215} \end{alignat}

Note that estimates 19 and 19 will then follow from 20 and 20, respectively. Estimates 20 and 20 hold by the left hand side hypotheses in 9 and 10, respectively. It follows from 17, 20 and 20 that estimates 19 and 19 hold for ${{n=0}}$. Let us assume estimates 20 and 20 hold for all $k\le {{n}}$. It then follows that estimates 19 and 19 hold for $n={{k}}$. We then have:

\begin{align} 0 & \le s_k-t_k \le \phi (s_{k-1}-t_{k-1}) \le \phi \cdot \phi (s_{k-2}-t_{k-2})\le \cdots \le \phi ^k\eta ,\label{eq:216} \\ 0 & \le t_{k+1}-s_k \le \alpha (s_k-t_k) \le \alpha \phi ^k\eta ,\label{eq:217} \end{align}

and

\begin{align} t_{k+1} \le s_k + \alpha \phi ^k\eta & \le {t_k+\alpha \phi ^k\eta +\phi ^k\eta }\nonumber \\ & \le s_{k-1} + \alpha \phi ^{k-1}\eta +\alpha \phi ^k \eta +\phi ^k\eta \nonumber \\ & \le t_{k-1} + \phi ^{k-1}\eta +\alpha \phi ^{k-1} \eta +\alpha \phi ^k \eta +\phi ^k\eta \nonumber \\ & = t_{k-1} + (\phi ^{k-1}+\phi ^k)\eta +\alpha (\phi ^{k-1}+\phi ^k) \eta \nonumber \le \cdots \\ & \le s_0 + \alpha (\eta +\phi \eta +\cdots +\phi ^k\eta ) + \alpha (\phi \eta +\cdots +\phi ^k\eta )\nonumber \\ & =(1+\alpha )(1+\phi +\cdots +\phi ^k\eta ) \le t^{\star \star }.\label{eq:218} \end{align}

In view of 21 and 23, estimate 20 certainly holds if

\begin{gather} 0 \le \tfrac {L\phi ^k\eta }{2\left[1-L_2(1+\phi +\cdots +\phi ^{k-1})\eta -L_0t^{k-1}\eta \right]} \le \alpha ,\label{eq:219} \intertext {or} L\phi ^k\eta +2\alpha L_2(1+\phi +\cdots +\phi ^{k-1})\eta -2\alpha +2L_0\alpha t^{k-1}\eta \le {0}.\label{eq:220} \end{gather}

Estimate 24 motivates us to introduce functions $f_k$ on $[0,1)$ by

\begin{equation} \label{eq:221} f_k(t) = L\eta t^k + 2\alpha L_2(1+t+\cdots +t^{k-1})\eta +2L_0\alpha t^{k-1}\eta -2 \alpha . \end{equation}

We need a relationship between two consecutive functions $f_k$:

\begin{align} f_{k+1}(t) & = Lt^{k+1}\eta + +2\alpha L_0 t^{k}\eta +2\alpha L_2(1+t+\cdots +t^k)\eta -2\alpha -Lt^k\eta \nonumber \\ & {} \qquad -2\alpha L_2(1+t+\cdots +t^{k-1})\eta - 2L_0\alpha t^{k-1}\eta +2\alpha +f_k(t)\nonumber \\ & =f_k(t) + Lt^{k+1}\eta -Lt^k\eta +2\alpha L_2 t^k \eta + 2L_0\alpha t^{k}\eta - 2L_0\alpha t^{k-1}\eta \nonumber \\ & = f_k(t) +g(t)t^{k-1}\eta ,\label{eq:222} \intertext {where} g(t) & = Lt^2 + \left[2\alpha (L_2+L_0)-L\right]t-2L_0\alpha .\label{eq:223} \end{align}

Using 25, we see that 24 holds

\begin{align} \textrm{if } \quad f_k(\phi ) & \le {0} \label{eq:224} \\ \textrm{ or} \quad {{f_1}}(\phi ) & \le {0},\label{eq:225} \end{align}

\begin{align} \textrm{since,} \qquad g(\phi ) & \le {0} \qquad \textrm{and}\qquad f_{k+1}(\phi ) = f_k(\phi ) + g(\phi )\phi ^k\eta \le f_k(\phi ) \label{eq:226} \end{align}

where $\phi $ is chosen as in the right hand side inequality of 9. But 26 also holds by 9. Moreover, define function $f_\infty $ on $[0,1)$ by

\begin{align} f_\infty (t) & = \lim _{k\to \infty }f_k(t). \label{eq:227} \intertext {Then, we have by \eqref{eq:224}} f_\infty (\phi ) & \le {0}.\nonumber \end{align}

Hence, 19 and 20 hold for all $k$. Similarly, 20 holds if

\begin{gather} L_1\phi ^k\eta \le 2\phi \left[1-L_2(1+\phi +\cdots +\phi ^k)\eta \right]\label{eq:228} \intertext {or} L_1\phi ^k\eta +2\phi L_2(1+\phi +\cdots +\phi ^k)\eta -2\phi \le {0}.\label{eq:229} \end{gather}

As in 25 we define functions $h_k$ on $[0,1)$ by

\begin{equation} h_k(t) = L_1t^k\eta +2tL_2 (1+t+\cdots +t^k)\eta - 2\phi . \label{eq:230} \end{equation}

We need a relationship between two consecutive functions $h_k$:

\begin{align} h_{k+1}(t) & = L_1t^{k+1}\eta + 2tL_2(1+t+\cdots +t^{k+1})\eta -2\phi - L_1t^k\eta -\nonumber \\ & \quad -2tL_2(1+t+\cdots +t^k)\eta +2\phi + h_k(t)\nonumber \\ & = h_k(t) +L_1t^{k+1}\eta -L_1t^k\eta + 2L_2t^{k+2}\eta \nonumber \\ & = h_k(t) + g_1(t)t^k\eta \label{eq:231} \intertext {where} g_1(t) & = 2L_2t^2+L_1 t- L_1.\label{eq:232} \end{align}

In view of 31, estimate 30 holds if

\begin{equation} \label{eq:233} \textrm{if}\quad h_k(\phi ) \le {0} \quad \textrm{or}\quad h_1(\phi ) \le {0} \end{equation}

\begin{align} \textrm{since,}\quad g_1(\phi ) \le {0} \quad \textrm{and} \quad h_{k+1}(\phi ) = h_k(\phi ) + g_1(\phi ) \phi ^k \eta \le h_k(\phi )\label{234} \end{align}

where $\phi $ is chosen as in the right hand side of 10. Note now that 33 holds by 10. Furthermore, define functions $h_\infty $ on $[0,1)$ by

\begin{equation} h_\infty (t) =\lim _{k\to \infty }h_k(t).\label{eq:235} \end{equation}

We then have

\begin{equation} \label{eq:236} h_\infty (\phi )\le {0}. \end{equation}

That completes the induction for 19 and 20. Finally, in view of 19, 19 and 23, sequences $\{ t_n\} $, $\{ s_n\} $ converge to $t^\star $. That completes the proof of the Lemma.

We need an Ostrowski-type relationship between iterates $\{ x_n\} $ and $\{ y_n\} $ [ 5 , 14 ] .

Lemma 2

Let us assume iterates $\{ x_n\} $ and $\{ y_n\} $ in (TSNM) are well defined for all $n\ge {0}$. Then, the following identities hold:

\begin{align} \mathcal{F}(x_{n+1}) & = \int _{0}^1 \big[ \mathcal{F}^\prime (y_n + \theta (x_{n+1} - y_n) )- \mathcal{F}^\prime (y_n) \big](x_{n+1}-y_n)\mathrm{d}\theta ,\label{eq:237} \intertext {and} \mathcal{F}(y_n) & = \int _{0}^1 \big[ \mathcal{F}^\prime (x_n + \theta (y_{n} - x_n) )- \mathcal{F}^\prime (x_n) \big] (y_n-x_n)\mathrm{d}\theta .\label{eq:238} \end{align}

Proof â–¼

Identity 37 follows from the Taylor’s theorem and the first iteration in (TSNM), whereas 37 follows from Taylor’s theorem and the second iteration in (TSNM). That completes the proof of the Lemma.

We can show the following semilocal convergence result for (TSNM).

Lemma 3

Let $\mathcal{F}:\mathcal{D} \subset \mathcal{X}\rightarrow \mathcal{Y}$ be Fréchet-differentiable operator. Assume: there exist $x_0\in {\mathcal{D}}$, $L_0{\gt}{0}$, $L{\gt}0$ and $\eta \ge 0$ such that for all $x,y\in {\mathcal{D}}$:

\begin{align} \mathcal{F}^\prime (x_0)^{-1} & \in L(\mathcal{Y},\mathcal{X}), \label{eq:239} \\ \big\Vert {\mathcal{F}^\prime \left(x_0\right)^{-1} \mathcal{F}(x_0)}\big\Vert & \le \eta , \label{eq:240}\\ \big\Vert {\mathcal{F}^\prime \left(x_0\right)^{-1} \left(\mathcal{F}^\prime (x)-\mathcal{F}^\prime (x_0)\right)}\big\Vert & \le L_0\left\Vert {x-x_0}\right\Vert ,\label{eq:241} \\ \big\Vert {\mathcal{F}^\prime \left(x_0\right)^{-1} \left(\mathcal{F}^\prime (x)-\mathcal{F}^\prime (y)\right)}\big\Vert & \le L\left\Vert {x-y}\right\Vert ,\label{eq:242}\\ \overline{U}(x_0,t^\star ) & \subseteq {\mathcal{D}};\label{eq:243} \end{align}

hypotheses of Lemma 2.1 hold, where $t^\star $ is given in Lemma 2.1. Then, sequences $\{ x_n\} $ and $\{ y_n\} $ generated by (TSNM) are well defined, remain in $\overline{U}(x_0,t^\star )$ for all $n\ge {0}$ and converge to a solution $x^\star \in \overline{U}(x_0,t^\star )$ of equation $\mathcal{F}(x)=0.$ Moreover, the following estimates hold

\begin{align} \left\Vert {y_{n} - x_n }\right\Vert & \le s_{n}-t_n, \label{eq:244} \\ \left\Vert {x_{n+1} - y_n }\right\Vert & \le t_{n+1}-s_n, \label{eq:245} \\ \left\Vert {x_{n+1} - x_n }\right\Vert & \le t_{n+1}-t_n, \label{eq:246} \\ \left\Vert {y_{n+1} - y_n }\right\Vert & \le s_{n+1}-s_n, \label{eq:247} \\ \left\Vert {x_{n} - x^\star }\right\Vert & \le t^\star -t_n, \label{eq:248} \\ \left\Vert {y_{n} - x^\star }\right\Vert & \le t^\star -s_n. \label{eq:249} \end{align}

Furthermore, if there exists $R\ge {t^\star }$ such that

\begin{align} \overline{U}(x_0,R) & \subseteq {\mathcal{D}} \label{eq:250} \intertext {and} L_0(t^\star +R) & {\lt} 2,\label{eq:251} \end{align}

then, $x^\star $ is the only solution of $\mathcal{F}(x)=0$ in $\overline{U}(x_0,R)$

Proof â–¼

We shall show using induction on $k$ that (TSNM) is well defined, the iterates remain in $\overline{U}(x_0,t^\star )$ for all $n\ge {0}$ and estimates 43 and 44 hold for all $n\ge {0}$. Iterate $y_0$ is well defined by the first equation in (TSNM) for $n=0$ and 38. We also have by 13 and 39

\begin{equation} \nonumber \left\Vert {y_0-x_0}\right\Vert = \left\Vert {\mathcal{F}^\prime (x_0)^{-1}\mathcal{F}(x_0)}\right\Vert \le \eta = {{s_0 = s_0-t_0}} \le t^{\star }. \end{equation}

That is 43 holds for $n=0$ and $y_0\in \overline{U}(x_0,t^\star )$. Using (TSNM) for $n=0$, we see that $x_1$ is well defined. Let $w\in \overline{U}(x_0,t^\star )$. Then, we have by Lemma 2.1 and 40:

\begin{equation} \big\Vert {\mathcal{F}^\prime \left(x_0\right)^{-1} \big[\mathcal{F}^\prime (w)-\mathcal{F}^\prime (x_0)\big]}\big\Vert \le L_0\left\Vert {w-x_0}\right\Vert \le L_0t^\star < 1. \label{eq:252} \end{equation}

It follows from 50 and the Banach lemma on invertible operators [ 5 , 13 , 15 ] that $\mathcal{F}^\prime \left(w\right)^{-1}$ exists and

\begin{align} \big\Vert {\mathcal{F}^\prime \left(w\right)^{-1} \mathcal{F}^\prime (x_0)}\big\Vert & \le \tfrac {1}{1-L_0\Vert {w-x_0}\Vert }.\label{eq:253} \end{align}

In particular, for $x_1\in \overline{U}(x_0,t^\star )$, we have

\begin{align} \big\Vert {\mathcal{F}^\prime \left(x_1\right)^{-1} \mathcal{F}^\prime (x_0)}\big\Vert & \le \tfrac {1}{1-L_0\Vert {x_1-x_0}\Vert } \le \tfrac {1}{1-L_0(t_1-t_0)} = \tfrac {1}{1-L_0t_1}.\label{eq:254} \end{align}

Moreover, in view of 37 for $n=0$, (TSNM), 13 and 39-41, we get

\begin{align} & \left\Vert {x_{1} - y_0 }\right\Vert =\\ & =\left\Vert {\int _0^1{ {\left[\mathcal{F}^\prime (y_0)^{-1} \mathcal{F}^\prime (x_0)\right]}\mathcal{F}^\prime (x_0)^{-1}\left[ \mathcal{F}^\prime (x_0\! +\! \theta (y_0-x_0))\! -\! \mathcal{F}^\prime (x_0)\right] }}\textrm{d}\theta (y_0-x_0)\! \! \ \right\Vert \nonumber \\ & \le \tfrac {L_0}{1-L_0\Vert {y_0-x_0}\Vert } \int _0^1 {\theta \Vert {y_0-x_0}\Vert ^2} \textrm{d}{\theta }\nonumber \\ & =\tfrac {L_0}{2({1-L_0\Vert {y_0-x_0}\Vert })} \Vert {y_0-x_0}\Vert ^2\nonumber \\ & \le \tfrac {L_0}{2(1-L_0s_0)} (s_0-t_0)^2 = t_1-s_0,\nonumber \end{align}

which shows 44 for $n=0$. We also have

\begin{equation} \nonumber \left\Vert {x_{1} - x_0 }\right\Vert \le \left\Vert {x_{1} - y_0 }\right\Vert + \left\Vert {y_{0} - x_0 }\right\Vert \le t_1 - s_0 + s_0 - t_0 = t_1-t_0 \le t^\star , \end{equation}

which implies 45 holds for $n=0$ and $x_1\in \overline{U}(x_0,t^\star )$.

Using (TSNM), 13, 37 (for $n=0$) and 52, we get

\begin{align} \left\Vert {y_1-x_1}\right\Vert & = \left\Vert {\left[\mathcal{F}^\prime (x_1)^{-1}\mathcal{F}^\prime (x_0)\right] \left[\mathcal{F}^\prime (x_0)^{-1}\mathcal{F}(x_1)\right] }\right\Vert \nonumber \\ & \le \left\Vert {\mathcal{F}^\prime (x_1)^{-1}\mathcal{F}^\prime (x_0)}\right\Vert \left\Vert {\mathcal{F}^\prime (x_0)^{-1}\mathcal{F}(x_1) }\right\Vert \nonumber \\ & \le \tfrac {1}{1-L_0t_1}\left\Vert \int _0^1{\mathcal{F}^\prime (x_0)^{-1} \left[\mathcal{F}^\prime (y_0+\theta (x_1-y_0)) - \mathcal{F}^\prime (y_0)\right]\textrm{d}\theta (x_1-y_0) }\right\Vert \nonumber \\ & \le \tfrac {L_0}{1-L_0t_1}\int _0^1{ \theta \Vert {x_1-y_0}\Vert \textrm{d}\theta \Vert {x_1-y_0}\Vert }\nonumber \\ & \le \tfrac {{L}}{1-L_0t_1}{ \tfrac {1}{2}(t_1-s_0)(t_1-s_0) } = s_1-t_1,\nonumber \end{align}

which implies 43 for $n=1$. We then have:

\begin{gather} \Vert {y_1-y_0}\Vert \le \Vert {y_1-x_1}\Vert + \Vert {x_1-y_0}\Vert \le s_1-t_1 + t_1-s_0 = s_1-s_0,\nonumber \\ \Vert {y_1-x_0}\Vert \le \Vert {y_1-y_0}\Vert + \Vert {y_0-x_0}\Vert \le s_1-s_0 + s_0-t_0 = s_1 \le t^\star ,\nonumber \end{gather}

which imply 46 for $n=0$ and $y_1\in \overline{U}(x_0,t^\star )$. Let us now assume 43-46, $y_n,x_k\in \overline{U}(x_0,t^\star )$ for all $n\le {k}$. Using (TSNM), 13, 37, 37, 41, 51 and the induction hypotheses, we have in turn:

\begin{align} \Vert {x_{k+1}-x_0}\Vert & \le \Vert {x_{k+1}-x_k}\Vert + \Vert {x_{k}-x_{k-1}}\Vert +\cdots + \Vert {x_{1}-x_0}\Vert \nonumber \\ & \le t_{k+1}-t_k + t_k-t_{k-1}+\cdots + t_1-t_0 = t_{k+1} \le t^\star , \label{eq:255}\\ \Vert {y_{k}-x_0}\Vert & \le \Vert {y_{k}-x_k}\Vert + \Vert {x_{k}-x_{0}}\Vert \label{eq:256}\\ & \le s_k-t_k + t_k - t_0 \nonumber \\ & = s_k \le t^\star \nonumber \end{align}

\begin{align} & \left\Vert {y_{k+1}-x_{k+1}}\right\Vert = \label{eq:257}\\ & =\left\Vert { \big[ \mathcal{F}^\prime (x_{k+1})^{-1} \mathcal{F}^\prime (x_0)\big] \big[ \mathcal{F}^\prime (x_{0})^{-1} \mathcal{F}(x_{k+1})\big]}\right\Vert \nonumber \\ & \le \left\Vert { \mathcal{F}^\prime (x_{k+1})^{-1} \mathcal{F}^\prime (x_0)} \right\Vert \left\Vert { \mathcal{F}^\prime (x_{0})^{-1} \mathcal{F}(x_{k+1})}\right\Vert \nonumber \\ & \le \tfrac {1}{1-L_0\Vert {x_{k+1}\! -\! x_0}\Vert }\! \int _0^1\! \big\Vert {\mathcal{F}^\prime (x_{0})^{-1}} \left[\mathcal{F}^\prime (y_k\! +\! \theta (x_{k+1}-y_k)) - \mathcal{F}^\prime (y_k)\right]\textrm{d}\theta (x_{k+1}\! -\! y_k)\big\Vert \nonumber \\ & \le \tfrac {L}{1-L_0t_{k+1}} \int _0^1\theta \Vert {x_{k+1}-y_k}\Vert ^2\textrm{d}\theta \nonumber \\ & \le \tfrac {L}{1-L_0t_{k+1}} \tfrac {1}{2}(t_{k+1}-s_k)^2\nonumber \\ & = s_{k+1}-t_{k+1}, \nonumber \end{align}

\begin{align} & \left\Vert {x_{k+2}-y_{k+1}}\right\Vert = \label{eq:258} \left\Vert { \big[ \mathcal{F}^\prime (y_{k+1})^{-1} \mathcal{F}^\prime (x_0)\big] \big[ \mathcal{F}^\prime (x_{0})^{-1} \mathcal{F}(y_{k+1})\big]}\right\Vert \leq \end{align}

\begin{align} & \le \tfrac {1}{1\! -\! L_0s_{k+1}}\! \! \int _0^1\! \! \big\Vert {\mathcal{F}^\prime (x_{0})^{-1}} \! \! \! \left[\mathcal{F}^\prime (x_{k+1}\! +\! \theta (y_{k+1}\! -\! x_{k+1}))\! -\! \mathcal{F}^\prime (x_{k+1})\right]\! \! \textrm{d}\theta (y_{k+1}\! -\! x_{k+1})\big\Vert \nonumber \\ & \le \tfrac {L}{1-L_0s_{k+1}} \int _0^1 \theta {\Vert {y_{k+1}-x_{k+1}}\Vert ^2}\textrm{d}\theta \nonumber \\ & \le \tfrac {L}{2(1-L_0s_{k+1})}(s_{k+1}-t_{k+1})^2 = t_{k+2}-s_{k+1}, \nonumber \end{align}

\begin{align} \Vert {y_{k+2}-y_{k+1}}\Vert & \le \Vert {y_{k+2}-x_{k+2}}\Vert + \Vert {x_{k+2}-y_{k+1}}\Vert \nonumber \\ & \le s_{k+2}-t_{k+2}+t_{k+2}-s_{k+1} = s_{k+2}-s_{k+1}, \label{eq:259} \\ \Vert {x_{k+2}-x_{k+1}}\Vert & \le \Vert {x_{k+2}-y_{k+1}}\Vert + \Vert {y_{k+1}-x_{k+1}}\Vert \nonumber \\ & \le t_{k+2}-s_{k+1}+s_{k+1}-t_{k+1} = {{t_{k+2}-t_{k+1}}}\label{eq:260} \end{align}

which show 43-46 hold for all $n\ge {0}$. Estimates 47 and 48 follow from 45 and 46, respectively by using standard majorization technique [ 5 , 13 , 15 ] . It follows from Lemma 2.1 and 43-47 that (TSNM) is Cauchy in a Banach space $\mathcal{X}$ and as such it converges to some $x^\star \in \overline{U}(x_0,t^\star )$ (since $\overline{U}(x_0,t^\star )$ is a closed set). Moreover, we have by 56

\begin{align} \big\Vert {\mathcal{F}^\prime (x_{0})^{-1} \mathcal{F}(x_{k+1}) }\big\Vert & \le \tfrac {L}{2}\Vert {x_{k+1}-y_k}\Vert \Vert {x_{k+1}-y_k}\Vert \rightarrow {0}, \quad \textrm{as} \quad k\to \infty .\label{eq:261} \end{align}

That is $\mathcal{F}(x^\star ) = 0.$ Finally to show uniqueness, let $y^\star \in \overline{U}(x_0,R)$ be a solution of equation $\mathcal{F}(x)=0.$ Let us define linear operator $M$ by

\begin{equation} \label{eq:262} M = \int _{0}^1\mathcal{F}^\prime (y^\star +\theta (x^\star -y^\star )){\rm d\theta }. \end{equation}

Then using 40, 49 and 49, we get in turn

\begin{align} \left\Vert {\mathcal{F}^\prime (x_0)\left[M-\mathcal{F}^\prime (x_0)\right]}\right\Vert & \le L_0\int _{0}^1\left\Vert {y^\star +\theta (x^\star -y^\star ) -x_0}\right\Vert {\rm d\theta }\nonumber \\ & \le L_0\int _{0}^1\left[(1-\theta )\Vert {y^\star -x_0}\Vert +\theta \Vert {x^\star -x_0}\Vert \right] {\rm d\theta }\nonumber \\ & \le \tfrac {L_0}{2}(R+t^\star ) {\lt} 1.\label{eq:263} \end{align}

It follows from 58 and the Banach Lemma on invertible operators that $M^{-1}$ exists. Then, in view of the identity

\begin{equation} \label{eq:264} 0 = \mathcal{F}(x^\star ) -\mathcal{F}(y^\star ) = M(x^\star -y^\star ), \end{equation}

we conclude that $x^\star =y^\star $. That completes the proof of the Theorem.

Remark 4

1) Limit point $t^\star $ can be replaced by $t^{\star \star }$, given in closed form by 14, in hypotheses 39 and 47.

2) The verification of conditions 9-10 require simple algebra (see also Example 3.1).

3) If $L_0 = L$, then scalar sequences $\{ s_n\} $, $\{ t_n\} $ given by 13 reduce essentially to the ones used in [ 9 ] . In particular, we have in this case

\begin{equation} \label{eq:265} \begin{array}{llll} \overline{t}_0 =0,\quad \overline{s}_0 = \eta , \quad \overline{t}_{n+1} = \overline{s}_n + \tfrac {L(\overline{s}_n-\overline{t}_n)^2}{2(1-L\overline{s}_n)},\\ \overline{s}_{n+1} = \overline{t}_{n+1} + \tfrac {L( \overline{t}_{n+1}-\overline{s}_n)^2 }{ 2 (1-L\overline{t}_{n+1})} \end{array} \end{equation}

If $L_0 {\lt} L$ iteration 13 is tighter than 60. Moreover, in view of the proof of the Theorem 2.3, we note that sequence

\begin{equation} \label{eq:266} \begin{array}{llll} \overline{\overline{t}}_0 =0,\quad \overline{\overline{s}}_0 = \eta , \quad \overline{\overline{t}}_{n+1} = \overline{\overline{s}}_n + \tfrac {L^\star (\overline{\overline{s}}_n-\overline{\overline{t}}_n)^2}{2(1-L_0\overline{ \overline{s}} _n) } , \\ \overline{\overline{s}}_{n+1} = \overline{\overline{t}}_{n+1} + \tfrac {L^\star (\overline{\overline{t}}_{n+1}-\overline{\overline{s}}_n)^2 }{ 2 (1-L_0\overline{\overline{t}}_{n+1})}, \end{array} \end{equation}

is also majorizing for (TSNM), where

\begin{equation} \nonumber L^\star = \left\{ \begin{array}{lll} {L_0}, \quad & \textrm{if}\quad n=0 \\ {L}, \quad & \textrm{if}\quad n>0. \end{array}\right. \end{equation}

In case $L_0{\lt}L,$ 29 is even a tighter majorizing sequence than 60. Furthermore, $L, L_1$ can be replaced by $L_0,L_1^\star = \alpha ^2L_0$ at the left hand sides of 9 and 10, respectively.

4) If $\alpha =0$, define $L_1=L$, then it is simple algebra to show that conditions of Lemma 2.1 reduce to ??. Moreover, if $L_0=L$, these conditions reduce to ??. That is we have Newton’s method 2 and iteration 13 reduces to

\begin{equation} \label{eq:267} t_0 = 0, \quad t_1=\eta , \quad t_{n+2} = t_{n+1} + \tfrac {L(t_{n+1}-t_n)^2}{2(1-L_0t_{n+1})}. \end{equation}

In the case of Newton’s method for $L_0=L$, we have the well-known Kantorovich majorizing sequence.

\begin{equation} \label{eq:268} \nu _0 = 0, \quad \nu _1=\eta , \quad \nu _{n+2} = \nu _{n+1} + \tfrac {L(\nu _{n+1}-\nu _n)^2}{2(1-L_0\nu _{n+1})}. \end{equation}

Note that if $L_0{\lt}L$, $\{ t_n\} $ is a tighter majorizing sequence than $\{ \nu _n\} $ for the Newton’s method [ 5 , 13 , 15 ] .â–¡

3 Numerical Examples

Let $\mathcal{X}=\mathcal{Y}=\mathbb {R}^2$ be equipped with the max-norm, $x_0=(1,1)^T$, $\mathcal{D}=\overline{U}(x_0,1-p)$, $p\in [0,1)$ and define $\mathcal{F}$ on $\mathcal{D}$ by

\begin{equation} \label{eq:31} \mathcal{F}(x) = \left(\xi _1^3-p,\xi _2^3-p\right)^T, \quad x=\left(\xi _1,\xi _2\right)^T. \end{equation}

Using 35-37, we get

\begin{alignat}{5} \eta & = \tfrac {1-p}{3},\quad & L_0 & = 3-p \quad & & \textrm{and}\quad & L & = 2(2-p){\gt}L_0. \nonumber \intertext {Let $p = 0.7$. Then, we get} \eta & = 0.1, & L_0 & = 2.3 & & \textrm{and} & L & = 2.6. \nonumber \end{alignat}

The Newton-Kantorovich hypothesis ?? is satisfied, since

\begin{equation} \nonumber \tfrac {2}{3}(1-p)(2-p) = 0.26 < 1\quad \textrm{for all}\quad {p \in [0,{1}/{2})}. \end{equation}

Using Lemma 2.1, for $\alpha = 0.17 \quad \textrm{and}\quad \phi = 0.0052$, we get

\begin{gather*} \begin{alignedat} \end{alignedat}{6} L_1 & = 0.07514, & \qquad L_2 & = 2.691 & \qquad & \phi & =& & 0.756703694,\\ \phi _2 & = 0.111383518, & \phi _3 & = 0.666923077, & \phi _0 & =& & \phi _2 \\ \eta _1 & = 0.364622409, & \eta _2 & = 0.200360649, & \eta _0 & =& & \eta _2, \end{gather*}

Lη/[2(1-L_0η)] = 0.168831169 and L_1η/[2(1-L_2η)] = 0.005140238.

\end{gather*}

Hence, the hypotheses of Lemma 2.1 are satisfied. Moreover, we have by 18 that

\begin{equation} \nonumber t^{\star \star } =0.11761158 < 1-p = 0.3. \end{equation}

Furthermore, using 47 (for $t^\star $ replaced by $t^{\star \star }$), we get

\begin{equation} \nonumber t^{\star \star } < R < \tfrac {2}{L_0}-t^{\star \star } = 0.751953637. \end{equation}

So, we can choose $R=0.3$. Hence, hypotheses of Theorem 2.3 hold, and (TSNM) converges to

\begin{equation*} x^\star = \left(\sqrt[3]{0.7},\sqrt[3]{0.7}\right)^T = \left(0.887904002, 0.887904002\right)^T. \end{equation*}

We compare 13 to 60.

$n$	$s_n\! -\! t_n$	$t_{n+1}\! -\! s_n$	$\overline{s}_n\! -\! \overline{t}_n$	$\overline{t}_{n+1}\! -\! \overline{s}_n$	$\overline{\overline{s}}_n\! -\! \overline{\overline{t}}_n$	$\overline{\overline{t}}_{n+1}\! -\! \overline{\overline{s}}_n$
$0$	${1.00e-01}$	${1.69e-02}$	${1.00e-01}$	${1.76e-02}$	${1.00e-01}$	${1.49e-02}$
$1$	${5.07e-04}$	${4.57e-07}$	${5.78e-04}$	${6.27e-07}$	${3.49e-04}$	${2.15e-07}$
$2$	${3.73e-13}$	${2.47e-25}$	${5.37e-13}$	${1.02e-24}$	${8.19e-14}$	${1.19e-26}$
$3$	${1.09e-49}$	${2.11e-98}$	${1.94e-48}$	${1.09e-96}$	${2.49e-52}$	${1.09e-103}$
$4$	${7.91e-196}$	${1.11e-390}$	${9.44e-191}$	${1.67e-380}$	${2.11e-206}$	${7.88e-412}$
$5$	${2.21e-780}$	${8.70e-1560}$	${5.24e-760}$	${5.15e-1519}$	${1.09e-822}$	${2.13e-1644}$

Table 1 Comparison among 13, 64 and 65

As expected from the theoretical results iteration 13 is faster than 64.

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\(n\)	\(s_n\! -\! t_n\)	\(t_{n+1}\! -\! s_n\)	\(\overline{s}_n\! -\! \overline{t}_n\)	\(\overline{t}_{n+1}\! -\! \overline{s}_n\)	\(\overline{\overline{s}}_n\! -\! \overline{\overline{t}}_n\)	\(\overline{\overline{t}}_{n+1}\! -\! \overline{\overline{s}}_n\)
\(0\)	\({1.00e-01}\)	\({1.69e-02}\)	\({1.00e-01}\)	\({1.76e-02}\)	\({1.00e-01}\)	\({1.49e-02}\)
\(1\)	\({5.07e-04}\)	\({4.57e-07}\)	\({5.78e-04}\)	\({6.27e-07}\)	\({3.49e-04}\)	\({2.15e-07}\)
\(2\)	\({3.73e-13}\)	\({2.47e-25}\)	\({5.37e-13}\)	\({1.02e-24}\)	\({8.19e-14}\)	\({1.19e-26}\)
\(3\)	\({1.09e-49}\)	\({2.11e-98}\)	\({1.94e-48}\)	\({1.09e-96}\)	\({2.49e-52}\)	\({1.09e-103}\)
\(4\)	\({7.91e-196}\)	\({1.11e-390}\)	\({9.44e-191}\)	\({1.67e-380}\)	\({2.11e-206}\)	\({7.88e-412}\)
\(5\)	\({2.21e-780}\)	\({8.70e-1560}\)	\({5.24e-760}\)	\({5.15e-1519}\)	\({1.09e-822}\)	\({2.13e-1644}\)