Ball Convergence of a Parametric Efficient Family of Iterative Methods for Solving Nonlinear Equations

Abstract

The goal is to extend the applicability of Newton-Traub-like methods in cases not covered in earlier articles requiring the usage of derivatives up to order seven that do not appear in the methods. The price we pay by using conditions on the first derivative that actually appear in the method is that we show only linear convergence. To find the convergence order is not our intention, however, since this is already known in the case where the spaces coincide with the multidimensional Euclidean space. Note that the order is rediscovered by using ACOC or COC, which require only the first derivative. Moreover, in earlier studies using Taylor series, no computable error distances were available based on generalized Lipschitz conditions. Therefore, we do not know, for example, in advance, how many iterates are needed to achieve a predetermined error tolerance. Furthermore, no uniqueness of the solution results is available in the aforementioned studies, but we also provide such results. Our technique can be used to extend the applicability of other methods in an analogous way, since it is so general. Finally note that local results of this type are important, since they demonstrate the difficulty in choosing initial points. Our approach also extends the applicability of this family of methods from the multi-dimensional Euclidean to the more general Banach space case. Numerical examples complement the theoretical results.

1. Introduction

Let ${\mathcal{B}}_1,$${\mathcal{B}}_2$ denote Banach spaces and $T\subseteq {\mathcal{B}}_1$ be a nonempty convex and open set. Set $\mathcal{L}B({\mathcal{B}}_1,{\mathcal{B}}_2)=\{V:{\mathcal{B}}_1\to {\mathcal{B}}_2\mathrm{is}\mathrm{by}\mathrm{a}\mathrm{bounded}\mathrm{linear}\mathrm{operator}\}.$ Many problems in mechanics, biomechanics, physics, mathematical chemistry, economics, radiative transfer, biology, ecology, medicine, engineering, and other areas [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25] are reduced to an equation (nonlinear)

$$F\left(x\right)=0,$$

(1)

with $F:T\to {\mathcal{B}}_2$ being continuously differentiable in the Fréchet sense. Therefore solving Equation (1) is an extremely important and difficult problem in general. A solution $\xi$ is very difficult to find, especially in closed or analytical form. This function forces practitioners and researchers to develop higher order and efficient methods converging to $\xi$ by starting from a point $x_0\in T$ sufficiently close to it [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28]. Motivated by Traub-like and Newton’s methods, the following methods were studied in [10]:

$$\begin{array}{cc}\hfill y_k& =x_k-F^{\prime }{\left(x_k\right)}^{-1}F\left(x_k\right)\hfill \\ \hfill z_k& =y_k-{[w_k,y_k;F]}^{-1}F\left(y_k\right)\hfill \\ \hfill x_{k+1}& =z_k-{[w_k,y_k;F]}^{-1}F\left(z_k\right),\hfill \end{array}$$

(2)

where $[·,·;F]:T\times T\to \mathcal{L}B({\mathcal{B}}_1,{\mathcal{B}}_2)$ is a divided difference of order one [18] and $w_n=w\left(x_n\right),$$w:T\to T$ is a given iteration function. To be more precise, the special choice of w given by

$$w_n=y_n+\alpha F\left(y_n\right)+\beta F{\left(y_n\right)}^2,$$

(3)

was used in [10], for parameters $\alpha$ and $\beta$ that are not zero at the same time, $f_i\left(x\right)$ the co-ordinate functions for F and

$$F{\left(x\right)}^2={(f_1^2\left(x\right),f_2^2\left(x\right),\cdots ,f_n^2\left(x\right))}^T.$$

(4)

Moreover, they used method (2) when ${\mathcal{B}}_1={\mathcal{B}}_2={\mathbb{R}}^k,$ and compared it favorably to other sixth-order methods.

However, in this article we do not necessarily assume that w is given by (3). The sixth convergence order has been verified for any value of $\alpha$ and $\beta$ using Taylor series, but hypothesizes on the derivatives up to seventh order [10]. That simply means that convergence is not guaranteed for mappings that are not up to seventh-order differentiable.

For example, consider h on $T=[-\frac{1}{2},\frac{3}{2})$ as

$$h\left(t\right)=\left\{\begin{array}{ccc}{t}^{3}ln{t}^2+t^5-t^4,\hfill & & t\ne 0\hfill \\ 0,\hfill & & t=0.\hfill \end{array}\right.$$

(5)

Then, we calculate

$$h^{\prime }\left(t\right)=3t^{2}ln{t}^2+5t^4-4t^3+2t^2,$$

$$h^{″}\left(t\right)=6tln{t}^2+20t^3+12t^2+10t$$

and

$$h^{‴}\left(t\right)=6ln{t}^2+60t^2-24t+22.$$

Notice that $h^{‴}\left(t\right)$ is unbounded on T.

Another problem is that there are no error bounds on $\parallel x_n-\xi \parallel$ or results on uniqueness of $\xi$ or how close to $\xi$ we should start, and the selection of $x_0$ is really “ a shot in the dark”. To address all these concerns about this very efficient and useful method, we only use conditions on the first derivative. Moreover, convergence radius, error estimates, and uniqueness results are computed based on these conditions. Furthermore, we rely on the computational order (COC) or approximated computational convergence order (ACOC) formulae to determine the order [8,25]. These formulae also only use the first derivative. That is how we extend the applicability of method (2). Our technique can also be used to study other methods in an analogous way.

It is worth noticing that the $\alpha =\beta =0,$ method (2) reduces to Newton-Secant-like methods, and if $\beta =0$ to Newton-Steffensen-like methods.

The layout for the rest of the article includes the convergence analysis of method (2) in Section 2 and the numerical examples in Section 3.

2. Convergence Analysis of Method (2)

Let $D=[0,\infty ).$ Let also ${\phi }_0:D\to \mathbb{R}$ be increasing and continuous, satisfying ${\phi }_0\left(0\right)=0.$ Assume

$${\phi }_0\left(t\right)=1$$

(6)

having one positive zero (at least). Let $r_0$ be the minimal zero. Set $D_0=[0,r_0).$ Assume there exists a function $\phi :D_0\to \mathbb{R}$ that is continuous and increasing satisfying $\phi \left(0\right)=0.$ Consider $g_1$ and $\overline{g_1}$ on $D_0$ as $g_1\left(t\right)=\frac{{\int }_0^1\phi \left((1-\tau )t\right)d\tau }{1-{\phi }_0\left(t\right)},$ and $\overline{g_1}\left(t\right)=g_1\left(t\right)-1.$ By these definitions $\overline{g_1}\left(0\right)=-1$ and $\overline{g_1}\left(t\right)\to \infty$ for $t\to r_0^{-}.$ The application of the intermediate value theorem on function $\overline{g_1}$ assures the existence of at least one zero in $(0,r_0).$ Denote by ${\rho }_1$ the minimal such zero.

Assume

$${\phi }_0\left(g_1\left(t\right)t\right)=1,{\phi }_2({\phi }_5\left(t\right),g_1\left(t\right)t)=1$$

(7)

have at least one positive zero, where ${\phi }_5$ is as $\phi .$ Denote by $r_1$ the minimal such zero and let $D_1=[0,{\overline{r}}_0],$ where ${\overline{r}}_0=\min \{r_0,r_1\}.$ Assume there exist functions ${\phi }_1:D_1\to \mathbb{R},$${\phi }_2:D_1\times D_1\to \mathbb{R},$ and ${\phi }_3:D_1\to \mathbb{R}$ that are continuous and increasing. Consider functions $g_2$ and $\overline{g_2}$ on $D_1$ as

$$g_2\left(t\right)=\left(\frac{{\int }_0^1\phi \left((1-\tau )g_1\left(t\right)t\right)d\tau }{1-{\phi }_0\left(g_1\left(t\right)t\right)}+\frac{{\phi }_1({\phi }_5\left(t\right)+g_1\left(t\right)t){\int }_0^1{\phi }_3\left(\tau g_1\left(t\right)t\right)d\tau }{(1-{\phi }_0\left(g_1\left(t\right)t\right))(1-{\phi }_2({\phi }_5\left(t\right),g_1\left(t\right)t))}\right)g_1\left(t\right)$$

and

$$\overline{g_2}\left(t\right)=g_2\left(t\right)-1.$$

By these definitions $\overline{g_2}\left(0\right)=-1,$ and $\overline{g_2}\left(t\right)\to \infty$ for $t\to {\overline{r}}_0^{-}.$ Denote by ${\rho }_2$ the minimal zero of function $\overline{g_2}$ on $(0,{\overline{r}}_0).$

Assume

$${\phi }_0\left(g_2\left(t\right)t\right)=1$$

(8)

having at least one positive zero at least. Let $r_2$ be the minimal zero. Set $D_2=[0,{\overline{\overline{r}}}_0),$ where ${\overline{\overline{r}}}_0=\min \{{\overline{r}}_0,r_2\}.$ Consider $g_3$ and $\overline{g_3}$ on $D_2$ as

$$g_3\left(t\right)=\left(\frac{{\int }_0^1\phi \left((1-\tau )g_2\left(t\right)t\right)d\tau }{1-{\phi }_0\left(\lambda \right)}+\frac{{\phi }_4({\phi }_5\left(t\right)+\lambda ,g_1\left(t\right)t+\lambda ){\int }_0^1{\phi }_3\left(\tau \lambda \right)d\tau }{(1-{\phi }_0\left(\lambda \right))(1-{\phi }_2({\phi }_5\left(t\right),g_1\left(t\right)t))}\right)g_2\left(t\right)$$

and

$$\overline{g_3}\left(t\right)=g_3\left(t\right)-1.$$

where

$$\lambda =g_2\left(t\right)t$$

By these definitions $\overline{g_3}\left(0\right)=-1,$ and $\overline{g_3}\left(t\right)\to \infty$ for $t\to {\overline{\overline{r}}}_0^{-}.$ Denote by ${\rho }_3$ the minimal zero of function $\overline{g_3}$ on $(0,{\overline{\overline{r}}}_0).$

Define a radius of convergence $\rho$ by

$$\rho =\min \left\{{\rho }_j\right\},j=1,2,3.$$

(9)

By the preceding we have for all $t\in [0,\rho )$

$$\begin{array}{cc}\hfill 0& \le {\phi }_0\left(t\right)<1,\hfill \end{array}$$

(10)

$$\begin{array}{cc}\hfill 0& \le {\phi }_2({\phi }_5\left(t\right),g_1\left(t\right)t)<1,\hfill \end{array}$$

(11)

$$\begin{array}{cc}\hfill 0& \le {\phi }_0\left(g_1\left(t\right)t\right)<1,\hfill \end{array}$$

(12)

$$\begin{array}{cc}\hfill 0& \le {\phi }_0\left(g_2\left(t\right)t\right)<1\hfill \end{array}$$

(13)

and

$$o\le g_j\left(t\right)<1.$$

(14)

Define $S(x,\mu )=\{y\in T:\parallel y-x\parallel <\mu \}$ and let $\overline{S}(x,\mu )$ be the closure of $S(x,\mu ).$ Next, we list the conditions (A) to be used in the convergence analysis:

$\left(a_1\right)$$F:T\to {\mathcal{B}}_2$ is continuous, differentiable, $[·,·;F]:T\times T\to \mathcal{L}B({\mathcal{B}}_1,{\mathcal{B}}_2)$ is a divided difference of order one, $F\left(\xi \right)=0,$ and $F^{\prime }{\left(\xi \right)}^{-1}\in \mathcal{L}B({\mathcal{B}}_1,{\mathcal{B}}_2)$ for some $\xi \in T.$

$\left(a_2\right)$${\phi }_0:D\to \mathbb{R}$ is increasing, continuous, ${\phi }_0\left(0\right)=0$ and for all $x\in T$

$$\parallel F^{\prime }{\left(\xi \right)}^{-1}(F^{\prime }\left(x\right)-F^{\prime }\left(\xi \right))\parallel \le {\phi }_0(\parallel x-\xi \parallel ).$$

Define $T_0=T\cap S(\xi ,r_0),$ where $r_0$ is given in (6).

$\left(a_3\right)$$\phi :D_0\to \mathbb{R}$ is continuous, increasing, $\phi \left(0\right)=0$ and for $x,y\in T_0$

$$\parallel F^{\prime }{\left(\xi \right)}^{-1}(F^{\prime }\left(y\right)-F^{\prime }\left(x\right))\parallel \le \phi (\parallel y-x\parallel ).$$

$\left(a_4\right)$${\phi }_1:D_1\to \mathbb{R},$${\phi }_3:D_1\to \mathbb{R},$${\phi }_2:D_1\times D_1\to \mathbb{R},$${\phi }_5:D_1\to \mathbb{R},$ are increasing, continuous, $w:T\to T$ is continuous and for all $x,y\in T_1:=T\cap S(\xi ,{\overline{r}}_0)$

$$\begin{array}{cc}\hfill \parallel F^{\prime }{\left(\xi \right)}^{-1}([w,y;F]-F^{\prime }\left(y\right))\parallel & \le {\phi }_1(\parallel w-y\parallel ),\hfill \\ \hfill \parallel F^{\prime }{\left(\xi \right)}^{-1}([w,y;F]-F^{\prime }\left(\xi \right))\parallel & \le {\phi }_2(\parallel w-\xi \parallel ,\parallel y-\xi \parallel ),\hfill \\ \hfill \parallel F^{\prime }{\left(\xi \right)}^{-1}{F}^{\prime }\left(x\right)\parallel & \le {\phi }_3(\parallel x-\xi \parallel ),\hfill \end{array}$$

and

$$\parallel w-\xi \parallel \le {\phi }_5(\parallel x-\xi \parallel ),$$

where ${\overline{r}}_0$ is given in (7).

$\left(a_5\right)$${\phi }_4:D_2\times D_2\to \mathbb{R}$ is increasing, continuous and for all $y,z\in T_2:=T\cap S(x_0,{\overline{\overline{r}}}_0)$

$$\parallel F^{\prime }{\left(\xi \right)}^{-1}([w,y;F]-F^{\prime }\left(z\right))\parallel \le {\phi }_4(\parallel w-z\parallel ,\parallel y-z\parallel ),$$

$\left(a_6\right)$$\overline{S}(\xi ,\rho )\subset T,$$r_0,{\overline{r}}_0,{\overline{\overline{r}}}_0$ given by (6), (7), (8) exist and $\rho$ is defined in (9).

$\left(a_7\right)$ There exist $\overline{\rho }\ge \rho$ such that

$${\int }_0^1{\phi }_0\left(\tau \overline{\rho }\right)d\tau <1.$$

Define $T_3=T\cap \overline{S}(\xi ,\overline{\rho }).$

Theorem 1.

Assume conditions (A) hold. Then, for $x_0\in S(\xi ,\rho )-\left\{\xi \right\},$$\left\{x_n\right\}$ produced by (2) is such that $\left\{x_n\right\}\subseteq S(\xi ,\rho )$, ${lim}_{n\to \infty }{x}_n=\xi$ so that

$$\begin{array}{cc}\hfill \parallel y_n-\xi \parallel & \le g_1(\parallel x_n-\xi \parallel )\parallel x_n-\xi \parallel \le \parallel x_n-\xi \parallel <\rho \hfill \end{array}$$

(15)

$$\begin{array}{cc}\hfill \parallel z_n-\xi \parallel & \le g_2(\parallel x_n-\xi \parallel )\parallel x_n-\xi \parallel \le \parallel x_n-\xi \parallel ,\hfill \end{array}$$

(16)

and

$$\parallel x_{n+1}-\xi \parallel \le g_3(\parallel x_n-\xi \parallel )\parallel x_n-\xi \parallel \le \parallel x_n-\xi \parallel ,$$

(17)

where functions $g_j$ were given previously. The only solution of Equation (1) is $\xi ,$ in $T_3,$ where $T_3$ is given in $\left(a_7\right).$

Proof. 

If $v\in S(\xi ,\rho ),$ then (6), (9), (11) and $\left(a_2\right)$ give

$$\parallel F^{\prime }{\left(\xi \right)}^{-1}(F^{\prime }\left(\xi \right)-F^{\prime }\left(v\right))\parallel \le {\phi }_0(\parallel \xi -v\parallel )\le {\phi }_0\left(r_0\right)\le {\phi }_0\left(\rho \right)<1.$$

(18)

This estimation together with the lemma by Banach for operators that are invertible [5,22] assures $F^{\prime }{\left(v\right)}^{-1}\in \mathcal{L}B({\mathcal{B}}_1,{\mathcal{B}}_2)$ with

$$\parallel F^{\prime }{\left(v\right)}^{-1}{F}^{\prime }\left(\xi \right)\parallel \le \frac{1}{1-{\phi }_0(\parallel v-\xi \parallel )}.$$

(19)

It also follows that iterate $y_0$ exists by (19), and the first substep of method (2). Using (2) (the first substep) (9), (11), (14) (for $j=1$), (19) (for $v=x_0$), and $\left(a_3\right)$

$$\begin{array}{cc}\parallel y_0-\xi \parallel \hfill & =\parallel x_0-\xi -F^{\prime }{\left(x_0\right)}^{-1}F\left(x_0\right)\parallel \hfill \\ \hfill & \le \parallel F^{\prime }{\left(x_0\right)}^{-1}{F}^{\prime }\left(\xi \right)\parallel \hfill \\ \hfill & \parallel {\int }_0^1{F}^{\prime }{\left(\xi \right)}^{-1}(F^{\prime }(\xi +\tau (x_0-\xi ))-F^{\prime }\left(x_0\right))d\tau (x_0-\xi )\parallel \hfill \\ \hfill & \le \frac{{\int }_0^1\phi ((1-\tau )\parallel x_0-\xi \parallel )d\tau \parallel x_0-\xi \parallel }{1-{\phi }_0(\parallel x_0-\xi \parallel )}=g_1(\parallel x_0-\xi \parallel )\parallel x_0-\xi \parallel \hfill \\ \hfill & \le \parallel x_0-\xi \parallel <\rho ,\hfill \end{array}$$

(20)

showing that $y_0\in S(\xi ,\rho )$ and (15) is true for $n=0.$ By $\left(a_1\right)$ and $\left(a_4\right)$ we have

$$\begin{array}{cc}\hfill \parallel F^{\prime }{\left(\xi \right)}^{-1}F\left(v\right)\parallel & =\parallel F^{\prime }{\left(\xi \right)}^{-1}(F\left(v\right)-F\left(\xi \right))\parallel \hfill \\ \hfill & =\parallel {\int }_0^1{F}^{\prime }{\left(\xi \right)}^{-1}{F}^{\prime }(\xi +\tau (v-\xi ))d\tau (v-\xi )\parallel \hfill \\ \hfill & \le {\int }_0^1{\phi }_3(\tau \parallel v-\xi \parallel )d\tau \parallel v-\xi \parallel .\hfill \end{array}$$

(21)

We get the estimates by (9), (11), and $\left(a_3\right)$

$$\parallel F^{\prime }{\left(\xi \right)}^{-1}([w_0,y_0;F]-F^{\prime }\left(\xi \right))\parallel \le {\phi }_2(\parallel w_0-\xi \parallel ,\parallel y_0-\xi \parallel )\le {\phi }_0\left(\rho \right)<1$$

(22)

leading to

$$\parallel {[w_0,y_0;F]}^{-1}{F}^{\prime }\left(\xi \right)\parallel \le \frac{1}{1-{\phi }_2(\parallel w_0-\xi \parallel ,\parallel y_0-\xi \parallel )},$$

(23)

so $z_0$ and $x_1$ exist. Then, by (19) (for $v=x_0,y_0$), (9), (14) (for $j=2$), (20), (21), $\left(a_3\right)$ method (2) (second substep) we obtain

$$\begin{array}{ll}\parallel z_0-\xi \parallel & =\parallel (y_0-\xi -F^{\prime }{\left(y_0\right)}^{-1}F\left(y_0\right))\\ & +F^{\prime }{\left(y_0\right)}^{-1}([w_0,y_0;F]-F^{\prime }\left(y_0\right)){[w_0,y_0;F]}^{-1}F\left(y_0\right)\parallel \\ & \le \left\{\frac{{\int }_0^1\phi ((1-\tau )\parallel y_0-\xi \parallel d\tau }{1-{\phi }_0(\parallel y_0-\xi \parallel )}\right.\\ & \left.+\frac{{\phi }_1(\parallel (w_0-\xi )+(\xi -y_0)\parallel ){\int }_0^1{\phi }_3(\tau \parallel y_0-\xi \parallel )d\tau }{(1-{\phi }_0(\parallel y_0-\xi \parallel \left)\right)(1-{\phi }_2(\parallel w_0-\xi \parallel ,\parallel y_0-\xi \parallel \left)\right)}\right\}\parallel y_0-\xi \parallel \\ & \le g_2(\parallel x_0-\xi \parallel )\parallel x_0-\xi \parallel \le \parallel x_0-\xi \parallel <\rho ,\end{array}$$

(24)

showing that $z_0\in S(\xi ,\rho )$ and (15) is true for $n=0.$ In view of (9), (14), (19) (for $v=z_0$), (20)–(24), and the last substep of method (2), we obtain the estimations

$$\begin{array}{cc}\parallel x_1-\xi \parallel \hfill & =\parallel (z_0-\xi -F^{\prime }{\left(z_0\right)}^{-1}F\left(z_0\right))\hfill \\ \hfill & +F^{\prime }{\left(z_0\right)}^{-1}([w_0,y_0;F]-F^{\prime }\left(z_0\right)){[w_0,y_0;F]}^{-1}F\left(z_0\right)\parallel \hfill \\ \hfill & \le \left\{\frac{{\int }_0^1\phi ((1-\tau )\parallel z_0-\xi \parallel d\tau }{1-{\phi }_0(\parallel z_0-\xi \parallel )}\right.\hfill \\ \hfill & \left.+\frac{{\phi }_4(\parallel (w_0-\xi )+(\xi -z_0)\parallel ,\parallel (y_0-\xi )+(\xi -z_0)\parallel ){\int }_0^1{\phi }_3(\tau \parallel z_0-\xi \parallel )d\tau }{(1-{\phi }_0(\parallel z_0-\xi \parallel \left)\right)(1-{\phi }_2(\parallel w_0-\xi \parallel ,\parallel z_0-\xi \parallel \left)\right)}\right\}\hfill \\ \hfill & \times \parallel z_0-\xi \parallel \hfill \\ \hfill & \le g_3(\parallel x_0-\xi \parallel )\parallel x_0-\xi \parallel \le \parallel x_0-\xi \parallel <\rho ,\hfill \end{array}$$

(25)

which completes the induction for estimations (15)–(17) if $n=0$ and $x_1\in S(\xi ,\rho ).$ By repeating the previous estimations for $x_m,y_m,z_m,x_{m+1}$ replacing $x_0,y_0,z_0,x_1$, respectively, the induction for items (15)–(17) is completed. Moreover, by the estimation

$$\parallel x_{m+1}-\xi \parallel \le \gamma \parallel x_m-\xi \parallel \le \rho ,\gamma =g_3(\parallel x_0-\xi \parallel )\in [0,1),$$

(26)

we arrive at ${lim}_{m\to \infty }{x}_m=\xi ,$ and $x_{m+1}\in S(\xi ,\rho ).$ Let $G={\int }_0^1{F}^{\prime }(\xi +\tau ({\xi }_0-\xi ))d\tau$ for ${\xi }_0\in T_3$ and $F\left({\xi }_0\right)=0.$ By $\left(a_2\right),\left(a_7\right),$ we get that

$$\parallel F^{\prime }{\left(\xi \right)}^{-1}(G-F^{\prime }\left(\xi \right))\parallel \le {\int }_0^1{\phi }_0((1-\tau )\parallel {\xi }_0-\xi \parallel )d\tau \le {\int }_0^1{\phi }_0\left(\tau \overline{\rho }\right)d\tau <1,$$

(27)

so $G^{-1}\in \mathcal{L}B({\mathcal{B}}_1,{\mathcal{B}}_2),$ leading to $\xi ={\xi }_0,$ where we also used the estimation $0=F\left({\xi }_0\right)-F\left(\xi \right)=G({\xi }_0-\xi ).$ □

Remark 1. 

(a) 

Let ${\phi }_0\left(t\right)=K_{0}t$ and $\phi \left(t\right)=Kt.$ The radius $r_1=\frac{2}{2K_0+K}$ was obtained by Argyros in [1] as the convergence radius for Newton’s method under conditions (17)–(19). Notice that the convergence radius for Newton’s method given independently by Rheinboldt [23] and Traub [25] is given by

$$\rho =\frac{2}{3K_1}<r_1,$$

where $K_1$ is the Lipschitz constant on $T,$ so $K_0\le K_1$ and $K\le K_1.$ Define $f\left(x\right)=e^x-1$ and $T=\overline{S}(0,1)$. Then, we find $K_0=e-1<K=e^{\frac{1}{e-1}}<K=e$, so $\rho =0.24252961<r_1=0.3827$.

Moreover, the new error bounds [1,2,3,4] are:

$$\parallel x_{n+1}-x^{*}\parallel \le \frac{L}{1-L_0\parallel x_n-x^{*}\parallel }{\parallel x_n-x^{*}\parallel }^2,$$

whereas the old ones [12,14] are

$$\parallel x_{n+1}-x^{*}\parallel \le \frac{L_1}{1-L_1\parallel x_n-x^{*}\parallel }{\parallel x_n-x^{*}\parallel }^2.$$

Therefore the new bounds are tighter if $L_0<L$ or $L<L_1.$ Clearly, we do not expect the radius of convergence of method (2) given by r to be larger than $r_1$ (see (8)).

(b) 

By (a2) and

$$\begin{array}{ccc}\hfill \parallel F^{\prime }{\left(x^{*}\right)}^{-1}{F}^{\prime }\left(x\right)\parallel & =& \parallel F^{\prime }{\left(x^{*}\right)}^{-1}(F^{\prime }\left(x\right)-F^{\prime }\left(x^{*}\right))+I\parallel \hfill \\ & \le & 1+\parallel F^{\prime }{\left(x^{*}\right)}^{-1}(F^{\prime }\left(x\right)-F^{\prime }\left(x^{*}\right))\parallel \le 1+{\phi }_0(\parallel x-x^{*}\parallel ),\hfill \end{array}$$

the condition on ${\phi }_3$ can be dropped and we can set

$${\phi }_3\left(t\right)=1+{\phi }_0\left(t\right).$$

3. Numerical Examples

We use $[x,y;F]={\int }_0^1{F}^{\prime }(y+\tau (x-y))d\tau$ and w as given in (3) for $\alpha =\frac{1}{10}\parallel F^{\prime }\left(\xi \right)\parallel ,$ and $\beta =\frac{1}{10}{\parallel F^{\prime }\left(\xi \right)\parallel }^2$. In view of the definition of the divided difference, conditions (A) and the estimation (for $x\in S(\xi ,\rho )$)

$$\begin{array}{cc}\hfill \parallel w\left(x\right)-\xi \parallel & \le \parallel y\left(x\right)-\xi \parallel +\alpha \parallel F^{\prime }{\left(\xi \right)}^{-1}F\left(y\left(x\right)\right)\parallel +\beta \parallel F^{\prime }{\left(\xi \right)}^{-1}{F\left(y\left(x\right)\right)\parallel }^2\hfill \\ \hfill & \le g_1\left(t\right)t+\alpha {\int }_0^1{\phi }_3\left(\tau g_1\left(t\right)t\right)d\tau +\beta {\left({\int }_0^1{\phi }_3\left(\tau g_1\left(t\right)t\right)d\tau \right)}^2={\phi }_5\left(t\right).\hfill \end{array}$$

Then, we can choose functions ${\phi }_i,i=1,2,4$ in terms of functions ${\phi }_0,\phi ,{\phi }_5$ as follows

$$\begin{array}{cc}\hfill {\phi }_1\left(t\right)& =\frac{\phi \left({\phi }_5\left(t\right)\right)+\phi \left(g_1\left(t\right)t\right)}{2}\hfill \\ \hfill {\phi }_2(s,t)& =\frac{{\phi }_0\left({\phi }_5\left(s\right)\right)+{\phi }_0\left(g_1\left(t\right)t\right)}{2}\hfill \end{array}$$

and

$${\phi }_4(s,t)={\phi }_2(s,t)+{\phi }_0\left(t\right)$$

in all examples.

Example 1.

Consider ${\mathcal{B}}_1$ and ${\mathcal{B}}_2$ to be the space of continuous functions and defined on interval $[0,1]$ with $T=\overline{S}(0,1).$ Let G on T be

$$G\left(\psi \right)\left(z\right)=\psi \left(z\right)-5{\int }_0^{1}z\tau \psi {\left(\tau \right)}^{3}d\tau .$$

(28)

Then, we find

$$G^{\prime }\left(\psi \right)(\left(\xi \left(x\right)\right)=\xi \left(x\right)-15{\int }_0^{1}z\tau \psi {\left(\tau \right)}^2\xi \left(\tau \right)d\tau ,\mathrm{for}\mathrm{each}\xi \in T.$$

Notice that we can take $\xi =0,$ giving $\phi \left(t\right)=15t,{\phi }_0\left(t\right)=7.5t,{\phi }_3\left(t\right)=15,\alpha =\beta =\frac{1}{10}.$ This way, we have that

$${\rho }_1=0.066667,{\rho }_2=0.00097005,\rho ={\rho }_3=0.000510334.$$

Example 2.

By the motivational example, represented in Figure 1, we choose ${\phi }_0\left(t\right)=\phi \left(t\right)=96.662907t$, ${\phi }_3\left(t\right)=1.0631$, $\alpha =\frac{3}{10},$ and $\beta =\frac{9}{10}.$ Then, the parameters for method (2) are for $\xi =1$

$${\rho }_1=0.00689682,{\rho }_2=0.00000221412,\rho ={\rho }_3=0.00000121412.$$

Example 3.

Let ${\mathbb{B}}_1={\mathbb{B}}_2={\mathbb{R}}^3$, $T=S(0,1),$ and let F on T be

$$F\left(x\right)=F(x_1,x_2,x_3)={(e^{x_1}-1,\frac{e-1}{2}{x_2}^2+x_2,x_3)}^T.$$

(29)

For the points $u={(u_1,u_2,u_3)}^T$, we find

$$F^{\prime }\left(u\right)=\left(\begin{array}{ccc}{e}^{u_1}& 0& 0\\ 0& (e-1)u_2+1& 0\\ 0& 0& 1\end{array}\right).$$

Then, for $x^{*}={(0,0,0)}^T$ we get $F^{\prime }\left(\xi \right)=diag(1,1,1),$ $\phi \left(t\right)=e^{\frac{1}{e-1}}t$, ${\phi }_0\left(t\right)=(e-1)t$, ${\phi }_3\left(t\right)=e^{\frac{1}{e-1}}$, $\alpha =\beta =\frac{1}{10}.$

Then, we obtain that

$${\rho }_1=0.382692,{\rho }_2=0.96949,\rho ={\rho }_3=0.154419.$$

4. Conclusions

In this article we extended the applicability of Newton-Traub-like methods in cases not covered before requiring the usage of derivatives up to order seven that do not appear in the methods. The price we pay by using conditions on the first derivative that actually appears on the method is that we show only linear convergence. To find the convergence order is not, however, our intention, since this is already known in the case of spaces that coincide with the multidimensional Euclidean space. Notice that the order is rediscovered by using ACOC or COC, which require only the first derivative. Moreover, in earlier studies using Taylor series, no computable error distances were available based on generalized Lipschitz conditions. Therefore, we do not know, for example, in advance, how many iterates are needed to achieve a predetermined error tolerance. Furthermore, no uniqueness of the solution results is available in the aforementioned studies, but we also provide such results. Our technique can be used to extend the applicability of other methods in an analogous way, since it is so general. Finally notice that local results of this type are important, since they demonstrate the difficulty in choosing initial points.