Local Convergence of a Family of Weighted-Newton Methods

Abstract

This article considers the fourth-order family of weighted-Newton methods. It provides the range of initial guesses that ensure the convergence. The analysis is given for Banach space-valued mappings, and the hypotheses involve the derivative of order one. The convergence radius, error estimations, and results on uniqueness also depend on this derivative. The scope of application of the method is extended, since no derivatives of higher order are required as in previous works. Finally, we demonstrate the applicability of the proposed method in real-life problems and discuss a case where previous studies cannot be adopted.

1. Introduction

In this work, ${\mathbb{B}}_1$ and ${\mathbb{B}}_2$ denote Banach spaces, $\mathbb{A}\subseteq {\mathbb{B}}_1$ stands for a convex and open set, and $\phi :\mathbb{A}\to {\mathbb{B}}_2$ is a differentiable mapping in the Fréchet sense. Several scientific problems can be converted to the expression. This paper addresses the issue of obtaining an approximate solution $s_{*}$ of:

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

(1)

by using mathematical modeling [1,2,3,4]. Finding a zero $s_{*}$ is a laborious task in general, since analytical or closed-form solutions are not available in most cases.

We analyze the local convergence of the two-step method, given as follows:

$$\begin{array}{ccc}{y}_j\hfill & =& x_j-\delta {\phi }^{\prime }{\left(x_j\right)}^{-1}\phi \left(x_j\right),\hfill \\ x_{n+1}\hfill & =& x_j-A_j^{-1}\left(c_1\phi \left(x_j\right)+c_2\phi \left(y_j\right)\right),\hfill \end{array}$$

(2)

where $x_0\in \mathbb{A}$ is a starting point, $A_j=\alpha {\phi }^{\prime }\left(x_j\right)+\beta {\phi }^{\prime }\left(\frac{x_j+y_j}{2}\right)+\gamma {\phi }^{\prime }\left(y_j\right)$, and $\alpha ,\beta$, $\gamma ,\delta ,c_1,c_2\in \mathbb{S},$ where $\mathbb{S}=\mathbb{R}$ or $\mathbb{S}=\mathbb{C}$. The values of the parameters $\alpha ,\gamma ,\beta$, and $c_1$ are given as follows:

$$\begin{array}{c}\alpha =-\frac{1}{3}{c}_2(3{\delta }^2-7\delta +2),\beta =-\frac{4}{3}{c}_2(2\delta -1),\hfill \\ \gamma =\frac{1}{3}{c}_2(\delta -2)\mathrm{and}{c}_1=-c_2({\delta }^2-\delta +1),\mathrm{for}\delta \ne 0,c_2\ne 0.\hfill \end{array}$$

Comparisons with other methods, proposed by Cordero et al. [5], Darvishi et al. [6], and Sharma [7], defined respectively as:

$$\begin{array}{c}{w}_j=x_j-{\phi }^{\prime }{\left(x_j\right)}^{-1}\phi \left(x_j\right),\hfill \\ x_{n+1}=w_j-B_j^{-1}\phi \left(w_j\right),\hfill \end{array}$$

(3)

$$\begin{array}{c}{w}_j=x_j-{\phi }^{\prime }{\left(x_j\right)}^{-1}\phi \left(x_j\right),\hfill \\ z_j=x_j-{\phi }^{\prime }{\left(x_j\right)}^{-1}\left(\phi \left(x_j\right)+\phi \left(w_j\right)\right),\hfill \\ x_{n+1}=x_j-C_j^{-1}\phi \left(x_j\right),\hfill \end{array}$$

(4)

$$\begin{array}{c}{y}_j=x_j-\frac{2}{3}{\phi }^{\prime }{\left(x_j\right)}^{-1}\phi \left(x_j\right),\hfill \\ x_{n+1}=x_j-\frac{1}{2}{D}_j^{-1}{\phi }^{\prime }{\left(x_j\right)}^{-1}\phi \left(x_j\right),\hfill \end{array}$$

(5)

where:

$$\begin{array}{c}{B}_j=2{\phi }^{\prime }{\left(x_j\right)}^{-1}-{\phi }^{\prime }{\left(x_j\right)}^{-1}{\phi }^{\prime }\left(w_j\right){\phi }^{\prime }{\left(x_j\right)}^{-1},\hfill \\ C_j=\frac{1}{6}{\phi }^{\prime }\left(x_j\right)+\frac{2}{3}{\phi }^{\prime }\left(\frac{x_j+w_j}{2}\right)+\frac{1}{6}{\phi }^{\prime }\left(z_j\right),\hfill \\ D_j=-I+\frac{9}{4}{\phi }^{\prime }{\left(y_j\right)}^{-1}{\phi }^{\prime }\left(x_j\right)+\frac{3}{4}{\phi }^{\prime }{\left(x_j\right)}^{-1}{\phi }^{\prime }\left(y_j\right),\hfill \end{array}$$

were also reported in [8]. The local convergence of Method (2) was shown in [8] for ${\mathbb{B}}_1={\mathbb{B}}_2={\mathbb{R}}^m$ and $\mathbb{S}=\mathbb{R},$ by using Taylor series and hypotheses reaching up to the fourth Fréchet-derivative. However, the hypothesis on the fourth derivative limits the applicability of Methods (2)–(5), particularly because only the derivative of order one is required. Let us start with a simple problem. Set ${\mathbb{B}}_1={\mathbb{B}}_2=\mathbb{R}$ and $\mathbb{A}=[-\frac{5}{2},\frac{3}{2}]$. We suggest a function $\phi :\mathbb{A}\to \mathbb{R}$ as:

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

which further yield:

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

$${\phi }^{″}\left(x\right)=12x\ln x^2+20x^3-12x^2+10x,$$

$${\phi }^{‴}\left(x\right)=12\ln x^2+60x^2-12x+22,$$

where the solution is $s_{*}=1$. Obviously, the function ${\phi }^{‴}\left(x\right)$ is unbounded in the domain $\mathbb{A}$. Therefore, the results in [5,6,7,8,9] and Method (2) cannot be applicable to such problems or its special cases that require the hypotheses on the third- or higher order derivatives of $\phi$. Without a doubt, some of the iterative method in Brent [10] and Petkovíc et al. [4] are derivative free and are used to locate zeros of functions. However, there have been many developments since then. Faster iterative methods have been developed whose convergence order is determined using Taylor series or with the technique introduce in our paper. The location of the initial points is a “shot in the dark” in these references; no uniqueness results or estimates on $\parallel x_n-x_{*}\parallel$ are available. Methods on abstract spaces derived from the ones on the real line are also not addressed.

These works do not give a radius of convergence, estimations on $\parallel x_j-s_{*}\parallel$, or knowledge about the location of $s_{*}$. The novelty of this study is that it provides this information, but requiring only the derivative of order one for method (2). This expands the scope of utilization of (2) and similar methods. It is vital to note that the local convergence results are very fruitful, since they give insight into the difficult operational task of choosing the starting points/guesses.

Otherwise, with the earlier approaches: (i) use the Taylor series and high-order derivative; (ii) have no clue about the choice of the starting point $x_0$; (iii) have no estimate in advance about the number of iterations needed to obtain a predetermined accuracy; and (iv) have no knowledge of the uniqueness of the solution.

The work is laid out as follows: we give the convergence of the iterative scheme (2) with the main Theorem 1 is given in Section 2. Six numerical problems are discussed in Section 3. The final conclusions are summarized in Section 4.

2. Convergence Study

This section starts by analyzing the convergence of Scheme (2). We assume that $L>0$, $L_0>0$, $M\ge 1$ and $\gamma ,\alpha ,\beta ,\delta ,c_1,c_2\in \mathbb{S}$. We consider some maps/functions and constant numbers. Therefore, we assume the following functions $g_1$, p, and $h_p$ on the open interval $[0,\frac{1}{L_0})$ by:

$$\begin{array}{cc}\hfill g_1\left(t\right)& =\frac{1}{2(1-L_{0}t)}(Lt+2M|1-\delta \left|\right),\hfill \\ \hfill p\left(t\right)& =\frac{L_0}{|\alpha +\beta +\gamma |}\left(\left|\alpha \right|+\frac{\left|\beta \right|}{2}\left(\frac{\left|\beta \right|}{2}+\left|\gamma \right|\right)g_1\left(t\right)\right)t,\mathrm{for}\alpha +\beta +\gamma \ne 0,\hfill \\ \hfill h_p\left(t\right)& =p\left(t\right)-1,\hfill \end{array}$$

and the values of $r_1$ and $r_A$ are given as follows:

$$r_1=\frac{2\left(M\right|1-\delta |-1)}{2L_0+L},r_A=\frac{2}{L+2L_0}.$$

Consider that:

$$M|1-\delta |<1.$$

(6)

It is clear from the function $g_1$, parameters $r_1$ and $r_A$, and Equation (6), that $0<r_1\le r_A<\frac{1}{L_0}$, $g_1\left(r_1\right)=1$, and $0\le g_1\left(t\right)<1$, for each $t\in [0,r_1)$ and $h_p\left(0\right)=-1$ and $h_p\left(t\right)\to +\infty$ as $t\to {\displaystyle \frac{1^{-}}{L_0}}$. On the basis of the classical intermediate value theorem, the function $h_p$ has at least one zero in the open interval $\left(0,{\displaystyle \frac{1}{L_0}}\right)$. Let us call $r_p$ as the smallest zero. We suggest some other functions $g_2$ and $h_2$ on the interval $[0,r_p)$ by means of the expressions:

$$\begin{array}{cc}\hfill g_2\left(t\right)=& \frac{1}{2(1-L_{0}t)}[Lt+\frac{2M^2\left(|\alpha -1|+\left|\beta \right|+\left|\gamma \right|\right)\left(|1-c_1|+|c_2|g_1\left(t\right)\right)}{|\alpha +\beta +\gamma |(1-L_{0}t)(1-p\left(t\right))}\hfill \\ & +\frac{2M\left(|1-c_1|+|c_2|g_1\left(t\right)\right)}{1-L_{0}t}]\hfill \end{array}$$

and:

$$h_2\left(t\right)=g_2\left(t\right)-1.$$

Suppose that:

$$M\left(|1-c_1|+c_{2}M|1-\delta |\right)\left(1+\frac{M\left(|\alpha -1|+\left|\beta \right|+\left|\gamma \right|\right)}{|\alpha +\beta +\gamma |}\right)<1.$$

(7)

Then, we have by Equation (7) that $h_2\left(0\right)<0$ and $h_2\left(t\right)\to +\infty$ as $t\to r_p^{-}$ by the definition of $r_p$. We recall $r_2$ as the least zero of $h_2$ on $(0,r_p)$.

Define:

$$r=\min \{r_1,r_2\}.$$

(8)

Then, notice that for all $t\in [0,r)$:

$$0<r<r_A,$$

(9)

$$0\le g_1\left(t\right)<1,$$

(10)

$$0\le p\left(t\right)<1,$$

(11)

$$0\le g_2\left(t\right)<1.$$

(12)

Assume that $Q(x,\delta )=\left\{y\in {\mathbb{B}}_1:\parallel x-y\parallel <\delta \right\}$. We can now proceed with the local convergence study of (2) adopting the preceding notations.

Theorem 1.

Let us assume that $\phi :\mathbb{A}\subset {\mathbb{B}}_1\to {\mathbb{B}}_2$ is a differentiable operator. In addition, we consider that there exist $s_{*}\in \mathbb{A}$, $L>0$, $L_0>0$, $M\ge 1$ and the parameters $\alpha ,\beta ,\gamma ,c_1,c_2\in \mathbb{S},$ with $\alpha +\beta +\gamma \ne 0,$ are such that:

$$\phi \left(s_{*}\right)=0,{\phi }^{\prime }{\left(s_{*}\right)}^{-1}\in L({\mathbb{B}}_2,{\mathbb{B}}_1),$$

(13)

$$\parallel {\phi }^{\prime }{\left(s_{*}\right)}^{-1}({\phi }^{\prime }\left(s_{*}\right)-{\phi }^{\prime }\left(x\right)\parallel \le L_0\parallel s_{*}-x\parallel ,\forall x\in \mathbb{A}.$$

(14)

Set $x,y\in {\mathbb{A}}_0=\mathbb{A}\cap Q\left(s_{*},\frac{1}{L_0}\right)$ so that:

$$\parallel {\phi }^{\prime }{\left(s_{*}\right)}^{-1}\left({\phi }^{\prime }\left(y\right)-{\phi }^{\prime }\left(x\right)\right)\parallel \le L\parallel y-x\parallel ,\forall y,x\in {\mathbb{A}}_0$$

(15)

$$\parallel {\phi }^{\prime }{\left(s_{*}\right)}^{-1}{\phi }^{\prime }\left(x\right)\parallel \le M,\forall x\in {\mathbb{A}}_0,$$

(16)

satisfies Equations (6) and (7), the condition:

$$\overline{Q}(s_{*},r)\subset \mathbb{A},$$

(17)

holds, and the convergence radius r is provided by (8). The obtained sequence of iterations $\left\{x_j\right\}$ generated for $x_0\in Q(s_{*},r)-\left\{x^{*}\right\}$ by (2) is well defined. In addition, the sequence also converges to the required root $s_{*}$, remains in $Q(s_{*},r)$ for every $n=0,1,2,\dots$, and:

$$\parallel y_j-s_{*}\parallel \le g_1(\parallel x_j-s_{*}\parallel )\parallel x_j-s_{*}\parallel \le \parallel x_j-s_{*}\parallel <r,$$

(18)

$$\parallel x_{n+1}-s_{*}\parallel \le g_2(\parallel x_j-s_{*}\parallel )\parallel x_j-s_{*}\parallel <\parallel x_j-s_{*}\parallel ,$$

(19)

where the g functions were described previously. Moreover, the limit point $s_{*}$ of the obtained sequence $\left\{x_j\right\}$ is the only root of $\phi \left(x\right)=0$ in ${\mathbb{A}}_1:=\overline{Q}(s_{*},T)\cap \mathbb{A}$, and T is defined as $T\in [r,\frac{2}{L_0})$.

Proof. 

We prove the estimates (18)–(19), by mathematical induction. Adopting the hypothesis $x_0\in Q(s_{*},r)-\left\{x^{*}\right\}$ and Equations (6) and (14), it results:

$$\parallel {\phi }^{\prime }{\left(s_{*}\right)}^{-1}({\phi }^{\prime }\left(x_0\right)-{\phi }^{\prime }\left(s_{*}\right))\parallel \le L_0\parallel x_0-s_{*}\parallel <L_{0}r<1.$$

(20)

Using Equation (20) and the results on operators by [1,2,3] that ${\phi }^{\prime }\left(x_0\right)\ne 0$, we get:

$$\parallel {\phi }^{\prime }{\left(x_0\right)}^{-1}{\phi }^{\prime }\left(s_{*}\right)\parallel \le {\displaystyle \frac{1}{1-L_0\parallel x_0-s_{*}\parallel }}.$$

(21)

Therefore, it is clear that $y_0$ exists. Then, by using Equations (8), (10), (15), (16), and (21), we obtain:

$$\begin{array}{ccc}\parallel y_0-s_{*}\parallel \hfill & =& \parallel \left(x_0-s_{*}-{\phi }^{\prime }{\left(x_0\right)}^{-1}\phi \left(x_0\right)\right)+(1-\delta ){\phi }^{\prime }{\left(x_0\right)}^{-1}\phi \left(x_0\right)\parallel \hfill \\ & \le & \parallel {\phi }^{\prime }{\left(x_0\right)}^{-1}{\phi }^{\prime }\left(s_{*}\right)\parallel \parallel {\int }_0^1{\phi }^{\prime }{\left(x^{*}\right)}^{-1}\left[{\phi }^{\prime }(s_{*}+\theta (x_0-s_{*}))-{\phi }^{\prime }\left(x_0\right)\right](x_0-s_{*})d\theta \parallel \hfill \\ & & +\parallel {\phi }^{\prime }{\left(x_0\right)}^{-1}{\phi }^{\prime }\left(s_{*}\right)\parallel \parallel {\int }_0^1{\phi }^{\prime }{\left(x^{*}\right)}^{-1}{\phi }^{\prime }(s_{*}+\theta (x_0-s_{*}))(x_0-s_{*})d\theta \parallel \hfill \\ & \le & {\displaystyle \frac{L\parallel x_0-x^{*}{\parallel }^2}{2(1-L_0\parallel x_0-s_{*}\parallel )}}+{\displaystyle \frac{M|1-\delta |\parallel x_0-s_{*}\parallel }{1-L_0\parallel x_0-s_{*}\parallel }}\hfill \\ & =& g_1(\parallel x_0-s_{*}\parallel )\parallel x_0-s_{*}\parallel <\parallel x_0-s_{*}\parallel <r,\hfill \end{array}$$

(22)

illustrating that $y_0\in Q(s_{*},r)$ and Equation (18) is true for $j=0$.

Now, we demonstrate that the linear operator $A_0$ is invertible. By Equations (8), (10), (14), and (22), we obtain:

$$\begin{array}{c}\parallel {\left((\alpha +\beta +\gamma ){\phi }^{\prime }\left(s_{*}\right)\right)}^{-1}\left(A_0-(\alpha +\beta +\gamma ){\phi }^{\prime }\left(s_{*}\right)\right)\parallel \hfill \\ \le \frac{L_0}{|\alpha +\beta +\gamma |}\left[\left|\alpha \right|\parallel x_0-s_{*}\parallel +\frac{\left|\beta \right|}{2}\left(\parallel x_0-s_{*}\parallel +\parallel y_0-s_{*}\parallel \right)+\left|\gamma \right|\parallel y_0-s_{*}\parallel \right]\hfill \\ \le \frac{L_0}{|\alpha +\beta +\gamma |}\left[\left|\alpha \right|+\frac{\left|\beta \right|}{2}\left(\frac{\left|\beta \right|}{2}+\left|\gamma \right|\right)g_1\left(\parallel x_0-s_{*}\parallel \right)\parallel x_0-s_{*}\parallel \right]\hfill \\ =p(\parallel x_0-s_{*}\parallel )<p\left(r\right)<1.\hfill \end{array}$$

(23)

Hence, $A_0^{-1}\in L({\mathbb{B}}_2,{\mathbb{B}}_1),$

$$\parallel A_0^{-1}{\phi }^{\prime }\left(s_{*}\right)\parallel \le \frac{1}{|\alpha +\beta +\gamma |(1-p(\parallel x_0-s_{*}\parallel \left)\right)},$$

(24)

and $x_1$ exists. Therefore, we need the identity:

$$\begin{array}{cc}\hfill x_1-s_{*}=& x_0-s_{*}-{\phi }^{\prime }{\left(x_0\right)}^{-1}\phi \left(x_0\right)-{\phi }^{\prime }{\left(x_0\right)}^{-1}\left((1-c_1)\phi \left(x_0\right)+c_2\phi \left(y_0\right)\right)\hfill \\ & +{\phi }^{\prime }{\left(x_0\right)}^{-1}\left(A_0-{\phi }^{\prime }\left(x_0\right)\right)A_0^{-1}\left(c_1\phi \left(x_0\right)+c_2\phi \left(y_0\right)\right).\hfill \end{array}$$

(25)

Further, we have:

$$\begin{array}{cc}\hfill \parallel x_1-s_{*}\parallel & \le \parallel x_0-s_{*}-{\phi }^{\prime }{\left(x_0\right)}^{-1}\phi \left(x_0\right)\parallel +\parallel {\phi }^{\prime }{\left(x_0\right)}^{-1}\left((1-c_1)\phi \left(x_0\right)+c_2\phi \left(y_0\right)\right)\parallel \hfill \\ & +\parallel {\phi }^{\prime }{\left(x_0\right)}^{-1}{\phi }^{\prime }\left(s_{*}\right)\parallel \parallel {\phi }^{\prime }{\left(s_{*}\right)}^{-1}\left(A_0-{\phi }^{\prime }\left(x_0\right)\right)\parallel \parallel A_0^{-1}{\phi }^{\prime }\left(s_{*}\right)\parallel \parallel {\phi }^{\prime }{\left(s_{*}\right)}^{-1}\left(c_1\phi \left(x_0\right)+c_2\phi \left(y_0\right)\right)\parallel \hfill \\ & \le \frac{L\parallel x_0-s_{*}{\parallel }^2}{2\left(1-L_0\parallel x_0-s_{*}\parallel \right)}+\frac{M\left(|1-c_1|\parallel x_0-s_{*}\parallel +|c_2|\parallel y_0-s_{*}\parallel \right)}{1-L_0\parallel x_0-s_{*}\parallel }\hfill \\ & +\frac{M^2\left(|\alpha -1|+\left|\beta \right|+\left|\gamma \right|\right)\left(|1-c_1|+|c_2|g_1(\parallel x_0-s_{*}\parallel )\right)\parallel x_0-s_{*}\parallel }{|\alpha +\beta +\gamma |(1-L_0\parallel x_0-s_{*}\parallel )\left(1-p(\parallel x_0-s_{*}\parallel )\right)}\hfill \\ & \le g_2(\parallel x_0-s_{*}\parallel )\parallel x_0-s_{*}\parallel <\parallel x_0-s_{*}\parallel <r,\hfill \end{array}$$

(26)

which demonstrates that $x_1\in Q\left(s_{*}r\right)$ and (19) is true for $j=0$, where we used (15) and (21) for the derivation of the first fraction in the second inequality. By means of Equations (21) and (16), we have:

$$\begin{array}{cc}\hfill \parallel \phi {\left(s_{*}\right)}^{-1}\phi \left(x_0\right)\parallel & =\parallel {\phi }^{\prime }{\left(s_{*}\right)}^{-1}\left(\phi \left(x_0\right)-\phi \left(s_{*}\right)\right)\parallel \hfill \\ & =\parallel {\int }_0^1{\phi }^{\prime }{\left(s_{*}\right)}^{-1}{\phi }^{\prime }(s_{*}+\theta (x_0-s_{*}))d\theta \parallel \le M\parallel x_0-s_{*}\parallel .\hfill \end{array}$$

In the similar fashion, we obtain $\parallel {\phi }^{\prime }{\left(s_{*}\right)}^{-1}\phi \left(y_0\right)\parallel \le M\parallel y_0-s_{*}\parallel \le M{g}_1(\parallel x_0-s_{*}\parallel )\parallel x_0-s_{*}\parallel$ (by (22)) and the definition of $\mathbb{A}$ to arrive at the second section. We reach (18) and (19), just by changing $x_0$, $z_0,y_0$, and $x_1$ by $x_j$, $z_j,y_j$, and $x_{j+1}$, respectively. Adopting the estimates $\parallel x_{j+1}-s_{*}\parallel \le q\parallel x_j-s_{*}\parallel <r,$ where $q=g_2(\parallel x_0-s_{*}\parallel )\in [0,1)$, we conclude that $x_{j+1}\in Q(s_{*},r)$ and ${\displaystyle \underset{j\to \infty }{\lim }{x}_j=s_{*}}$. To illustrate the unique solution, we assume that $y_{*}\in {\mathbb{A}}_1$, satisfying $\phi \left(y_{*}\right)=0$ and $U={\int }_0^1{\phi }^{\prime }(y_{*}+\theta (s_{*}-y_{*}))d\theta$. From Equation (14), we have:

$$\begin{array}{c}\parallel {\phi }^{\prime }{\left(s_{*}\right)}^{-1}(U-{\phi }^{\prime }\left(s_{*}\right))\parallel \le \parallel {\int }_0^1{L}_0|y_{*}+\theta (s_{*}-y_{*})-s_{*}\parallel d\theta \hfill \\ \le {\int }_0^1(1-t)\parallel y_{*}-s_{*}\parallel d\theta \le {\displaystyle \frac{L_0}{2}}T<1.\hfill \end{array}$$

(27)

It follows from Equation (27) that U is invertible. Therefore, the identity $0=\phi \left(y_{*}\right)-\phi \left(s_{*}\right)=U(y_{*}-s_{*})$ leads to $y_{*}=s_{*}$. □

3. Numerical Experiments

Herein, we illustrate the previous theoretical results by means of six examples. The first two are standard test problems. The third is a counter problem where we show that the previous results are not applicable. The remaining three examples are real-life problems considered in several disciplines of science.

Example 1.

We assume that ${\mathbb{B}}_1={\mathbb{B}}_2={\mathbb{R}}^3,\mathbb{A}=\overline{Q}(0,1)$. Then, the function φ is defined on $\mathbb{A}$ for $u={(x_1,x_2,x_3)}^T$ as follows:

$$\phi \left(u\right)={\left(e_1^x-1,x_2-\frac{1}{2}(1-e)x_2^2,x_3\right)}^T.$$

(28)

We yield the following Fréchet-derivative:

$${\phi }^{\prime }\left(u\right)=\left[\begin{array}{ccc}{e}^{x_1}& 0& 0\\ 0& (e-1)x_2+1& 0\\ 0& 0& 1\end{array}\right].$$

It is important to note that we have $s_{*}={(0,0,0)}^T,L_0=e-1<L=e^{\frac{1}{L_0}}$, $\delta =1$, $M=2$, $c_1=1$, and ${\phi }^{\prime }\left(s_{*}\right)={\phi }^{\prime }{\left(s_{*}\right)}^{-1}=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$. By considering the parameter values that were defined in Theorem 1, we get the different radii of convergence that are depicted in

Table 1. Radii of convergence for Example 1, where $L_0<L$.

Cases	Different Values of Parameters That Are Defined in Theorem 1
	$\mathit{\alpha }$	$\mathit{\beta }$	$\mathit{\gamma }$	${\mathit{c}}_{\mathbf{2}}$	${\mathit{r}}_{\mathbf{1}}$	${\mathit{r}}_{\mathbf{2}}$	$\mathit{r}=\mathbf{min}\{{\mathit{r}}_{\mathbf{1}},{\mathit{r}}_{\mathbf{2}}\}$
1	$-\frac{2}{3}$	$\frac{4}{3}$	$\frac{1}{3}$	$-1$	0.382692	0.0501111	0.0501111
2	$-\frac{2}{3}$	$\frac{4}{3}$	$-100$	$\frac{1}{100}$	0.382692	0.334008	0.334008
3	1	1	1	0	0.382692	0.382692	0.382692
4	1	1	1	$\frac{1}{100}$	0.382692	0.342325	0.342325
5	10	$\frac{1}{10}$	$\frac{1}{10}$	$\frac{1}{100}$	0.382692	0.325413	0.325413

and

Table 2. Radii of convergence for Example 1, where $L_0=L=e$ by [3,11].

Cases	Different Values of Parameters That Are Defined in Theorem 1
	$\mathit{\alpha }$	$\mathit{\beta }$	$\mathit{\gamma }$	${\mathit{c}}_{\mathbf{2}}$	${\mathit{r}}_{\mathbf{1}}$	${\mathit{r}}_{\mathbf{2}}$	$\mathit{r}=\mathbf{min}\{{\mathit{r}}_{\mathbf{1}},{\mathit{r}}_{\mathbf{2}}\}$
1	$-\frac{2}{3}$	$\frac{4}{3}$	$\frac{1}{3}$	$-1$	$0.245253$	0.0326582	0.0326582
2	$-\frac{2}{3}$	$\frac{4}{3}$	$-100$	$\frac{1}{100}$	0.245253	0.213826	0.213826
3	1	1	1	0	0.245253	0.245253	0.245253
4	1	1	1	$\frac{1}{100}$	0.245253	0.219107	0.219107
5	10	$\frac{1}{10}$	$\frac{1}{10}$	$\frac{1}{100}$	0.245253	0.208097	0.208097

.

Example 2.

Let us consider that ${\mathbb{B}}_1={\mathbb{B}}_2=C[0,1],\mathbb{A}=\overline{Q}(0,1)$ and introduce the space of continuous maps in $[0,1]$ having the max norm. We consider the following function φ on $\mathbb{A}$:

$$\phi \left(\varphi \right)\left(x\right)=\phi \left(x\right)-5{\int }_0^{1}x\tau \varphi {\left(\tau \right)}^{3}d\tau ,$$

(29)

which further yields:

$${\phi }^{\prime }\left(\varphi \left(\mu \right)\right)\left(x\right)=\mu \left(x\right)-15{\int }_0^{1}x\tau \varphi {\left(\tau \right)}^2\mu \left(\tau \right)d\tau ,foreach\mu \in \mathbb{A}.$$

We have $s_{*}=0,L=15,L_0=7.5,$ $M=2$, $\delta =1$, and $c_1=1$. We will get different radii of convergence on the basis of distinct parametric values as mentioned in

Table 3. Radii of convergence for Example 2, where $L_0<L$.

Cases	Different Values of Parameters That Are Defined in Theorem 1
	$\mathit{\alpha }$	$\mathit{\beta }$	$\mathit{\gamma }$	${\mathit{c}}_{\mathbf{2}}$	${\mathit{r}}_{\mathbf{1}}$	${\mathit{r}}_{\mathbf{2}}$	$\mathit{r}=\mathbf{min}\{{\mathit{r}}_{\mathbf{1}},{\mathit{r}}_{\mathbf{2}}\}$
1	$-\frac{2}{3}$	$\frac{4}{3}$	$\frac{1}{3}$	$-1$	$0.0666667$	0.00680987	0.00680987
2	$-\frac{2}{3}$	$\frac{4}{3}$	$-100$	$\frac{1}{100}$	0.0666667	0.0594212	0.0594212
3	1	1	1	0	0.0666667	0.0666667	0.0666667
4	1	1	1	$\frac{1}{100}$	0.0666667	0.0609335	0.0609335
5	10	$\frac{1}{10}$	$\frac{1}{10}$	$\frac{1}{100}$	0.0666667	0.0588017	0.0588017

and

Table 4. Radii of convergence for Example 2, where $L_0=L=15$ by [3,11].

Cases	Different Values of Parameters That Are Defined in Theorem 1
	$\mathit{\alpha }$	$\mathit{\beta }$	$\mathit{\gamma }$	${\mathit{c}}_{\mathbf{2}}$	${\mathit{r}}_{\mathbf{1}}$	${\mathit{r}}_{\mathbf{2}}$	$\mathit{r}=\mathbf{min}\{{\mathit{r}}_{\mathbf{1}},{\mathit{r}}_{\mathbf{2}}\}$
1	$-\frac{2}{3}$	$\frac{4}{3}$	$\frac{1}{3}$	$-1$	0.0444444	0.00591828	0.00591828
2	$-\frac{2}{3}$	$\frac{4}{3}$	$-100$	$\frac{1}{100}$	0.0444444	0.0387492	0.0387492
3	1	1	1	0	0.0444444	0.0444444	0.0444444
4	1	1	1	$\frac{1}{100}$	0.0444444	0.0397064	0.0397064
5	10	$\frac{1}{10}$	$\frac{1}{10}$	$\frac{1}{100}$	0.0444444	0.0377112	0.0377112

.

Example 3.

Let us return to the problem from the Introduction. We have $s_{*}=1,L=L_0=96.662907$, $M=2$, $\delta =1$, and $c_1=1$. By substituting different values of the parameters, we have distinct radii of convergence listed in

Table 5. Radii of convergence for Example 3.

Cases	Different Values of Parameters That Are Defined in Theorem 1
	$\mathit{\alpha }$	$\mathit{\beta }$	$\mathit{\gamma }$	${\mathit{c}}_{\mathbf{2}}$	${\mathit{r}}_{\mathbf{1}}$	${\mathit{r}}_{\mathbf{2}}$	$\mathit{r}=\mathbf{min}\{{\mathit{r}}_{\mathbf{1}},{\mathit{r}}_{\mathbf{2}}\}$
1	$-\frac{2}{3}$	$\frac{4}{3}$	$\frac{1}{3}$	$-1$	$0.00689682$	0.000918389	0.000918389
2	$-\frac{2}{3}$	$\frac{4}{3}$	$-100$	$\frac{1}{100}$	$0.00689682$	$0.00601304$	$0.00601304$
3	1	1	1	0	$0.00689682$	$0.00689682$	$0.00689682$
4	1	1	1	$\frac{1}{100}$	$0.00689682$	$0.00616157$	$0.00616157$
5	10	$\frac{1}{10}$	$\frac{1}{10}$	$\frac{1}{100}$	$0.00689682$	$0.0133132$	$0.0133132$

.

Example 4.

The chemical reaction [12] illustrated in this case shows how $W_1$ and $W_2$ are utilized at rates $q_{*}-Q_{*}$ and $Q_{*}$, respectively, for a tank reactor (known as CSTR), given by:

$$\begin{array}{c}\hfill W_2+W_1\to W_3\\ \hfill W_3+W_1\to W_4\\ \hfill W_4+W_1\to W_5\\ \hfill W_5+W_1\to W_6\end{array}$$

Douglas [13] analyzed the CSTR problem for designing simple feedback control systems. The following mathematical formulation was adopted:

$$K_C\frac{2.98(x+2.25)}{(x+1.45){(x+2.85)}^2(x+4.35)}=-1,$$

where the parameter $K_C$ has a physical meaning and is described in [12,13]. For the particular value of choice $K_C=0$, we obtain the corresponding equation:

$$\phi \left(x\right)=x^4+11.50x^3+47.49x^2+83.06325x+51.23266875.$$

(30)

The function φ has four zeros $s_{*}=\left(-1.45,-2.85,-2.85,-4.35\right)$. Nonetheless, the desired zero is $s_{*}=-4.35$ for Equation (30). Let us also consider $\mathbb{A}=[-4.5,-4]$.

Then, we obtain:

$$L_0=1.2547945,L=29.610958,M=2,\delta =1,c_1=1.$$

Now, with the help of different values of the parameters, we get different radii of convergence displayed in

Table 6. Radii of convergence for Example 4.

Cases	Different Values of Parameters That Are Defined in Theorem 1
	$\mathit{\alpha }$	$\mathit{\beta }$	$\mathit{\gamma }$	${\mathit{c}}_{\mathbf{2}}$	${\mathit{r}}_{\mathbf{1}}$	${\mathit{r}}_{\mathbf{2}}$	$\mathit{r}=\mathbf{min}\{{\mathit{r}}_{\mathbf{1}},{\mathit{r}}_{\mathbf{2}}\}$
1	$-\frac{2}{3}$	$\frac{4}{3}$	$\frac{1}{3}$	$-1$	$0.0622654$	0.00406287	0.00406287
2	$-\frac{2}{3}$	$\frac{4}{3}$	$-100$	$\frac{1}{100}$	$0.0622654$	$0.0582932$	$0.0582932$
3	1	1	1	0	$0.0622654$	$0.0622654$	$0.0622654$
4	1	1	1	$\frac{1}{100}$	$0.0622654$	$0.0592173$	$0.0592173$
5	10	$\frac{1}{10}$	$\frac{1}{10}$	$\frac{1}{100}$	$0.0622654$	$0.0585624$	$0.0585624$

.

Example 5.

Here, we assume one of the well-known Hammerstein integral equations (see pp. 19–20, [14]) defined by:

$$x\left(s\right)=1+\frac{1}{5}{\int }_0^{1}F(s,t)x{\left(t\right)}^{3}dt,x\in C[0,1],s,t\in [0,1],$$

(31)

where the kernel F is:

$$F(s,t)=\left\{\begin{array}{c}s(1-t),s\le t,\hfill \\ (1-s)t,t\le s.\hfill \end{array}\right.$$

We obtain (31) by using the Gauss–Legendre quadrature formula with ${\displaystyle {\int }_0^1\varphi \left(t\right)dt\simeq \sum _{k=1}^8{w}_k\varphi \left(t_k\right),}$ where $t_k$ and $w_k$ are the abscissas and weights, respectively. Denoting the approximations of $x\left(t_i\right)$ with $x_i(i=1,2,3,\dots ,8)$, then it yields the following $8\times 8$ system of nonlinear equations:

$${\displaystyle 5x_i-5-\sum _{k=1}^8{a}_{ik}{x}_k^3=0,i=1,2,3\dots ,8,}$$

$$a_{ik}=\left\{\begin{array}{c}\hfill w_k{t}_k(1-t_i),k\le i,\\ \hfill w_k{t}_i(1-t_k),i<k.\end{array}\right.$$

The values of $t_k$ and $w_k$ can be easily obtained from the Gauss–Legendre quadrature formula when $k=8$. The required approximate root is:

$$\begin{array}{c}{s}_{*}=(1.002096\dots ,1.009900\dots ,1.019727\dots ,1.026436\dots ,1.026436\dots ,\\ {1.019727\dots ,1.009900\dots ,1.002096\dots )}^T.\end{array}$$

Then, we have:

$$L_0=L=\frac{3}{40},M=2,\delta =1,c_1=1$$

and $\mathbb{A}=Q(s_{*},0.11).$ By using the different values of the considered disposable parameters, we have different radii of convergence displayed in

Table 7. Radii of convergence for Example 5.

Cases	Different Values of Parameters That Are Defined in Theorem 1
	$\mathit{\alpha }$	$\mathit{\beta }$	$\mathit{\gamma }$	${\mathit{c}}_{\mathbf{2}}$	${\mathit{r}}_{\mathbf{1}}$	${\mathit{r}}_{\mathbf{2}}$	$\mathit{r}=\mathbf{min}\{{\mathit{r}}_{\mathbf{1}},{\mathit{r}}_{\mathbf{2}}\}$
1	$-\frac{2}{3}$	$\frac{4}{3}$	$\frac{1}{3}$	$-1$	$8.88889$	1.18366	1.18366
2	$-\frac{2}{3}$	$\frac{4}{3}$	$-100$	$\frac{1}{100}$	$8.88889$	$7.74984$	$7.74984$
3	1	1	1	0	$8.88889$	$8.88889$	$8.88889$
4	1	1	1	$\frac{1}{100}$	$8.88889$	$7.94127$	$7.94127$
5	10	$\frac{1}{10}$	$\frac{1}{10}$	$\frac{1}{100}$	$8.88889$	$7.54223$	$7.54223$

.

Example 6.

One can find the boundary value problem in [14], given as:

$$y^{″}=\frac{1}{2}{y}^3+3y^{\prime }-\frac{3}{2-x}+\frac{1}{2},y\left(0\right)=0,y\left(1\right)=1.$$

(32)

We suppose the following partition of $[0,1]$:

$$x_0=0<x_1<x_2<x_3<\cdots <x_j,\mathit{where}{x}_{i+1}=x_i+h,h=\frac{1}{j}.$$

In addition, we assume that $y_0=y\left(x_0\right)=0,y_1=y\left(x_1\right),\dots ,y_{j-1}=y\left(x_{j-1}\right)$ and $y_j=y\left(x_j\right)=1$. Now, we can discretize this problem (32) relying on the first- and second-order derivatives, which is given by:

$$y_k^{\prime }=\frac{y_{k+1}-y_{k-1}}{2h},y_k^{″}=\frac{y_{k-1}-2y_k+y_{k+1}}{h^2},k=1,2,\dots ,j-1.$$

Hence, we find the following general $(j-1)\times (j-1)$ nonlinear system:

$$y_{k+1}-2y_k+y_{k-1}-\frac{h^2}{2}{y}_k^3-\frac{3}{2-x_k}{h}^2-\frac{1}{h^2}=0,k=1,2,\dots ,j-1.$$

We choose the particular value of $j=7$ that provides us a $6\times 6$ nonlinear systems. The roots of this nonlinear system are $s_{*}=(0.07654393\dots ,0.1658739\dots ,$$0.2715210\dots ,0.3984540\dots ,0.5538864\dots ,0.7486878\dots {)}^T$, and the results are mentioned in

Table 8. Radii of convergence for Example 6.

Cases	Different Values of Parameters That Are Defined in Theorem 1
	$\mathit{\alpha }$	$\mathit{\beta }$	$\mathit{\gamma }$	${\mathit{c}}_{\mathbf{2}}$	${\mathit{r}}_{\mathbf{1}}$	${\mathit{r}}_{\mathbf{2}}$	$\mathit{r}=\mathbf{min}\{{\mathit{r}}_{\mathbf{1}},{\mathit{r}}_{\mathbf{2}}\}$
1	$-\frac{2}{3}$	$\frac{4}{3}$	$\frac{1}{3}$	$-1$	$0.00904977$	0.00119169	0.00119169
2	$-\frac{2}{3}$	$\frac{4}{3}$	$-100$	$\frac{1}{100}$	$0.00904977$	$0.00789567$	$0.00789567$
3	1	1	1	0	$0.00904977$	$0.00904977$	$0.00904977$
4	1	1	1	$\frac{1}{100}$	$0.00904977$	$0.00809175$	$0.00809175$
5	10	$\frac{1}{10}$	$\frac{1}{10}$	$\frac{1}{100}$	$0.00904977$	$0.00809175$	$0.00809175$

.

Then, we get that:

$$L_0=73,L=75,M=2,\delta =1,c_1=1,$$

and $\mathbb{A}=Q(s_{*},0.15).$

With the help of different values of the parameters, we have the different radii of convergence listed in Table 8.

Remark 1.

It is important to note that in some cases, the radii $r_i$ are larger than the radius of $Q(s_{*},r)$. A similar behavior for Method (2) was noticed in Table 7. Therefore, we have to choose all $r_i=0.11$ because Expression (17) must be also satisfied.

4. Concluding Remarks

The local convergence of the fourth-order scheme (2) was shown in earlier works [5,6,8,15] using Taylor series expansion. In this way, the hypotheses reach to four-derivative of the function $\phi$ in the particular case when ${\mathbb{B}}_1={\mathbb{B}}_2={\mathbb{R}}^m$ and $S=\mathbb{R}$. These hypotheses limit the applicability of methods such (2). We analyze the local convergence using only the first derivative for Banach space mapping. The convergence order can be found using the computational order of convergence $\left(COC\right)$ or the approximate computational order of convergence $\left(ACOC\right)$ (Appendix A), avoiding the computation of higher order derivatives. We found also computable radii and error bounds not given before using Lipschitz constants, expanding, therefore, the applicability of the technique. Six numerical problems were proposed for illustrating the feasibility of the new approach. Our technique can be used to study other iterative methods containing inverses of mapping such as (3)–(5) (see also [1,2,3,4,5,6,7,8,9,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45]) and to expand their applicability along the same lines.