Convergence of High-Order Derivative-Free Algorithms for the Iterative Solution of Systems of Not Necessarily Differentiable Equations

Abstract

In this study, we extended the applicability of a derivative-free algorithm to encompass the solution of operators that may be either differentiable or non-differentiable. Conditions weaker than the ones in earlier studies are employed for the convergence analysis. The earlier results considered assumptions up to the existence of the ninth order derivative of the main operator, even though there are no derivatives in the algorithm, and the Taylor series on the finite Euclidian space restricts the applicability of the algorithm. Moreover, the previous results could not be used for non-differentiable equations, although the algorithm could converge. The new local result used only conditions on the divided difference in the algorithm to show the convergence. Moreover, the more challenging semi-local convergence that had not previously been studied was considered using majorizing sequences. The paper included results on the upper bounds of the error estimates and domains where there was only one solution for the equation. The methodology of this paper is applicable to other algorithms using inverses and in the setting of a Banach space. Numerical examples further validate our approach.

1. Introduction

Let F indicate the mapping of a subset $D\subset E$ into itself, where E is a Banach space. In a plethora of applications, researchers have reduced the problem to finding a solution $x^{*}\in D$ of

$$F(x)=0.$$

(1)

The analytical version of the solution $x^{*}$ is difficult to determine in general. Therefore, iterative algorithms have been developed that generate sequences that converge to $x^{*}$ by means of some initial hypotheses [1,2,3,4].

The Newton’s Scheme [1,2,3] defined by

$$x_0\in D{x}_{n+1}=x_n-F^{\prime }{(x_n)}^{-1}F(x_n),n=0,1,2,\dots ,$$

(2)

is a popular quadratic-order algorithm. Recently, there has been a surge in the need to develop an algorithm with an order higher than two [5,6,7,8,9]. The Taylor series expansion provides the local order of convergence. But, there are limitations to this approach:

(C1)

The convergence analysis is usually only local and $E={\mathbb{R}}^j,$ where j is a natural number.

(C2)

The sufficient convergence hypotheses involve $F^{(d)},$ where $d=1+$ order of convergence.

(C3)

No a priori and computational error distances are available.

(C4)

The isolation of the solution $x^{*}$ is not discussed.

(C5)

The semi-local convergence, which is considered more interesting and challenging than the local convergence, is not discussed.

Our idea addresses concerns (C1)–(C5) as follows:

(C1)’

The analysis is developed in Banach space.

(C2)’

The sufficient convergence hypotheses involve only the operators on the algorithm (see Algorithm 1), i.e., the divided differences. This is in contrast with the motivational work in [10] using hypotheses on high-order derivatives in the algorithm to show the convergence of the algorithm.

(C3)’ 

Error estimates become available under the concept of $\omega -$ continuity [1,2,3,11] in the local and majorizing sequences [3,7,12,13] in the semi-local case.

(C4)’ 

The isolation of the solution $x^{*}$ is specified.

and

(C5)’ 

The semi-local convergence analysis of the algorithm is studied.

An algorithm was taken from [10] to demonstrate this idea. However, the same idea was similarly applicable in the algorithm containing the inverses of linear operators [4,11,12,13,14,15,16,17,18,19,20,21].

Let us redevelop the algorithm, but formatted in Banach space, as follows, $n=0,1,2,\dots :$ 

Algorithm 1

Step 1: Given $x_0\in D,$ solve $A_n{u}_1=F(x_n),$ for $u_1.$

Step 2: Set $y_n=x_n-u_1.$

Step 3: Solve $A_n{u}_2=F(y_n),$ for $u_2.$

Step 4: Solve $A_n{u}_3=G_n{u}_2,$ for $u_3.$

Step 5: Solve $A_n{u}_4=G_n{u}_3,$ for $u_4.$

Step 6: Ste $z_n=y_n-a_0{u}_2-(3-2a_0)u_3-(a_0-2)u_4.$

Step 7: Solve $A_n{u}_5=F(z_n),$ for $u_5.$

Step 8: Solve $A_n{u}_6=Q_n{u}_5,$ for $u_6.$

Step 9: Solve $A_n{u}_7=Q_n{u}_6,$ for $u_7.$

Step 10: Solve $A_n{u}_8=Q_n{u}_7,$ for $u_8.$

Step 11: Solve $A_n{u}_9=Q_n{u}_8,$ for $u_9.$

Step 12: Set $x_{n+1}=z_n-a_1{u}_5-a_2{u}_6-a_3{u}_7-a_4{u}_8-a_5{u}_9.$

Step 13: If $x_{n+1}=x_n,$ STOP. Otherwise, repeat the process with $n⟶n+1.$

Here, $a_0$ and $a_5$ are free real parameters, $a_1=a_5+4,$ $a_2=-4a_5-6,$ $a_3=6a_5+4$ and $a_4=-4a_5-1,$  $b_0,b_1,b_2$ are fixed real numbers, $w_n=x_n+b_{0}F(x_n),$ $h_n=y_n+b_{1}F(y_n),$ $l_n=z_n+b_{2}F(z_n),$ $A_n=[x_n,w_n;F],$ $G_n=[h_n,y_n;F]$ and $Q_n=[l_n,z_n;F].$ 

Here, $[.,.;F]:D\times D⟶L(E)$ is a divided difference of an order one for the operator F [11,18,20,22], and the notation $L(E)$ is used for the set of continuous linear operators mapping E into itself. The interesting point of the algorithm is that, because of the usage of the same coefficient operator, only one LU decomposition can be performed for solving, e.g., linear systems with multiple right-hand sides. Thus, the algorithm without memory includes three steps (see Step 2, Step 6, and Step 12, where the iterations $y_n,z_n,x_{n+1}$ are computed, respectively) and five free non-zero operator parameters. In Section 4, the parameters are further specialized. The convergence order is shown to be eight in Theorem 1 in [10]. But the existence of the ninth derivative is required for the local convergence analysis [10]. Thus, if F is not differentiable by at least that amount, the conclusions in [10] cannot assure the convergence of the algorithm to $x^{*}.$ But, the algorithm can converge. Other limitations are listed in the aforementioned concerns of (C1)–(C5). As a further motivation, consider the folowing example: If $\tilde{D}$ stands for any neighborhood containing the numbers $t=0,1$ define $F(t)=4t^{3}logt+7t^5-7t^4$ if $t\ne 0.$ Clearly, $t^{*}=1\in \tilde{D}$ solves equation $F(t)=0.$ But, the conclusions in [10] cannot assure that ${lim}_{n⟶\infty }{x}_n=1$, although the algorithm converges to $t^{*},$ as the function $F^{‴}(t)$ is not continuous at $t=0\in \tilde{D}.$ A more important semi-local analysis of the algorithm that has not yet been presented is also developed in this paper [14,15,16,17,20,22].

The rest of the paper contains the following: a local analysis of the algorithm is provided in Section 2, a semi-local analysis is provided in Section 3, numerical examples are provided in Section 4, and the paper ends with a conclusion in Section 5.

2. Local Analysis

We use the symbols $U(x,a)$ and $U[x,a]$ to denote open and closed balls in $E,$ respectively, with center $x\in E$ and of radius $a>0.$ Let M denote the nonnegative axis, and NFC stands for a function that is nondecreasing and continuous on M or some subset of it. Then, the following hypotheses are required in the local analysis.

Assume:

(H1)

Nondecreasing functions and continuous (NFC) $f_1:M\to M,{\phi }_0:M\times M\to M$ exist, so that the equation

$$\begin{array}{c}\hfill {\phi }_0(f_1(t),t)-1=0\end{array}$$

admits a minimal positive solution (MPS) denoted by $\rho .$ Let $M_0=[0,\rho ).$ It follows that for each $t\in M-\left\{0\right\}$ 

$$0\le {\phi }_0(f_1(t),t)<1$$

(3)

and, consequently, the function $\lambda :M_0⟶M$ provided by

$$\lambda (t)=\frac{1}{1-{\phi }_0(f_1(t),t)}$$

is positive.

(H2)

NFC $f_2,f_3,{\phi }_1:M_0\to M$ and $\phi :M_0\times M_0⟶M$ exist, such that the equation $g_i(t)-1=0,i=1,2,3$ admits MPS denoted by ${\rho }_i,$ respectively, provided $g_i:M_0\to M$ are provided as

$$\begin{array}{ccc}\hfill g_1(t)& =& \lambda (t)\phi (f_1(t),t),\hfill \\ \hfill g_2(t)& =& [\phi (f_1(t),g_1(t)t)+|1-a_0|(1+{\int }_0^1{\phi }_1(\theta g_1(t)t)d\theta )\hfill \\ & & +|3-2a_0|\lambda (t)(1+{\phi }_0(f_2(t),g_1(t)t))\hfill \\ & & +|a_0-2|{\lambda }^2(t){(1+{\phi }_0(f_2(t),g_1(t)t))}^2]\lambda (t)g_1(t),\hfill \end{array}$$

and for

$$\begin{array}{ccc}\hfill {\lambda }_1(t)& =& \phi (f_1(t),g_2(t)t)+|1-a_1|(1+{\int }_0^1{\phi }_1(\theta g_2(t)t)d\theta )\hfill \\ & & +|a_2|\lambda (t)(1+{\phi }_0(f_3(t),g_2(t)t)\hfill \\ & & +|a_3{|\lambda (t)}^2{(1+{\phi }_0(f_3(t),g_2(t)t))}^2\hfill \\ & & +|a_4{|\lambda (t)}^3{(1+{\phi }_0(f_3(t),g_2(t)t))}^3\hfill \\ & & +|a_4{|\lambda (t)}^4{(1+{\phi }_0(f_3(t),g_2(t)t))}^4,\hfill \\ \hfill g_3(t)& =& {\lambda }_1(t)\lambda (t)g_2(t).\hfill \end{array}$$

Define parameter

$$\begin{array}{c}\hfill {\rho }^{*}=min\left\{{\rho }_i\right\},i=1,2,3\dots \end{array}$$

(4)

and the set $M^{*}=[0,{\rho }^{*}).$ These definitions imply that if $t\in M^{*}$ 

$$0\le g_i(t)<1.$$

(5)

(H3)

L is an invertible operator on E, such that for each $x\in D$ 

$$\parallel w-x^{*}\parallel \le f_1(\parallel x-x^{*}\parallel ),$$

$$\parallel L^{-1}([x,w;F]-L)\parallel \le {\phi }_0(\parallel w-x^{*}\parallel ,\parallel x-x^{*}\parallel ),w=x+b_{0}F(x).$$

Define the region $D_0=D\cap S(x^{*},{\rho }^{*})$ with $S(x^{*},{\rho }^{*})=\{y\in D:\parallel y-x^{*}\parallel <{\rho }^{*}\}.$ 

(H4)

$\parallel h-x^{*}\parallel \le f_2(\parallel y-x^{*}\parallel ),\parallel l-x^{*}\parallel \le f_3(\parallel z-x^{*}\parallel ,$ 

$$\parallel L^{-1}([x,x^{*};F]-L)\parallel \le {\phi }_1(\parallel x-x^{*}\parallel ),$$

$$\parallel L^{-1}([x,w;F]-[x,x^{*};F])\parallel \le \phi (\parallel w-x^{*}\parallel ,\parallel x-x^{*}\parallel )$$

for $x\in D_0,w=x+b_{0}F(x),h=y+b_{1}F(y)$ and $l=z+b_{2}F(z)$ and $y,z$ are provided by the last two substeps of the algorithm.

It is shown that $y,z$ exist (see Proof of Theorem 1).

and

(H5)

$S[x^{*},{\rho }^{*}]\subset D,$ where $S[x^{*},{\rho }^{*}]$ is the closure of $S(x^{*},{\rho }^{*}).$ 

A local analysis of the algorithm follows.

Theorem 1. 

Under conditions (H1)–(H5), the sequence $\left\{x_n\right\}$ is convergent to $x^{*}$ provided that $x_0\in U(x^{*},{\rho }^{*})-\left\{x^{*}\right\}.$ Moreover, the following assertions hold

$$\begin{array}{cc}\hfill \left\{x_n\right\}& \subset S(x^{*},{\rho }^{*}),\hfill \end{array}$$

(6)

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

(7)

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

(8)

and

$$\begin{array}{cc}\hfill \parallel x_{n+1}-x^{*}\parallel & \le g_3(\parallel x_n-x^{*}\parallel )\parallel x_n-x^{*}\parallel \le \parallel x_n-x^{*}\parallel ,\hfill \end{array}$$

(9)

where the functions $g_i$ are as previously provided and the radius ${\rho }^{*}$ is defined by the Formula (4).

Proof. 

From the hypothesis $x_0\in S(x^{*},{\rho }^{*}).$ If $F(x_0)\ne 0$ and $b_0\ne 0,$ then $w_0\ne x_0.$ It follows that the divided difference $A_0=[x_0,w_0;F]$ is well defined. Then, by (H1), (4) and (H3), we obtain

$$\begin{array}{ccc}\hfill \parallel L^{-1}(A_0-L)\parallel & \le & {\phi }_0(\parallel w_0-x^{*}\parallel ,\parallel x_0-x^{*}\parallel )\hfill \\ \hfill & \le & {\phi }_0(f_1(\parallel x_0-x^{*}\parallel ,\parallel x_0-x^{*}\parallel )\hfill \\ \hfill & \le & {\phi }_0(f_1({\rho }^{*}),{\rho }^{*})<1.\hfill \end{array}$$

(10)

Thus, by the Banach perturbation Lemma on the linear operators with inverses [1,2,6,18] $A_0^{-1}$  exists,

$$\parallel A_0^{-1}L\parallel \le \frac{1}{1-{\phi }_0(f_1(\parallel x_0-x^{*}\parallel ,\parallel x_0-x^{*}\parallel )}$$

(11)

and the iterate $y_0$ exists in the first substep of the algorithm.

Then, we can write

$$\begin{array}{ccc}\hfill y_0-x^{*}& =& x_0-x^{*}-A_0^{-1}F(x_0)\hfill \\ \hfill & =& A_0^{-1}(A_0-[x_0,x^{*};F])(x_0-x^{*}).\hfill \end{array}$$

(12)

Using (4), (5) (for $i=1$ ), (11) and (12)

$$\begin{array}{ccc}\hfill \parallel y_0-x^{*}\parallel & \le & \parallel A_0^{-1}L\parallel \parallel L^{-1}(A_0-[x_0,x^{*};F])\parallel \parallel x_0-x^{*}\parallel \hfill \\ \hfill & \le & \frac{\phi (\parallel w_0-x^{*}\parallel ,\parallel x_0-x^{*}\parallel )\parallel x_0-x^{*}\parallel }{1-{\phi }_0(f_1(\parallel x_0-x^{*}\parallel ,\parallel x_0-x^{*}\parallel )}\hfill \\ \hfill & \le & g_1(\parallel x_0-x^{*}\parallel )\parallel x_0-x^{*}\parallel \le \parallel x_0-x^{*}\parallel <{\rho }^{*}.\hfill \end{array}$$

(13)

Thus, in the second substep of the algorithm, the following happens

$$\begin{array}{ccc}\hfill z_0-x^{*}& =& y_0-x^{*}-A_0^{-1}F(y_0)\hfill \\ \hfill & & +(1-a_0)A_0^{-1}F(y_0)-(3-2a_0)A_0^{-1}{G}_0{A}_0^{-1}{G}_0{A}_0^{-1}F(y_0)\hfill \\ & & -(a_0-2)A_0^{-1}{G}_0{A}_0^{-1}{G}_0{A}_0^{-1}F(y_0)\hfill \end{array}$$

leading to

$$\begin{array}{ccc}\hfill \parallel z_0-x^{*}\parallel & \le & [\phi (\parallel w_0-x^{*}\parallel ,\parallel y_0-x^{*}\parallel )\lambda (\parallel x_0-x^{*}\parallel )\hfill \\ \hfill & & +|1-a_0|\lambda (\parallel x_0-x^{*}\parallel )(1+{\int }_0^1{\phi }_0(\theta \parallel y_0-x^{*}\parallel )d\theta )\hfill \\ \hfill & & +|3-2a_0|{\lambda }^2(\parallel x_0-x^{*}\parallel )(1+{\phi }_0(\parallel h_0-x^{*}\parallel ,\parallel y_0-x^{*}\parallel ))\hfill \\ \hfill & & +|a_0-2|{\lambda }^3(\parallel x_0-x^{*}\parallel )(1+{\phi }_0(\parallel h_0-x^{*}\parallel ,\parallel y_0-x^{*}{\parallel ))}^2]\parallel y_0-x^{*}\parallel \hfill \\ \hfill & \le & g_2(\parallel x_0-x^{*}\parallel )\parallel x_0-x^{*}\parallel \le \parallel x_0-x^{*}\parallel .\hfill \end{array}$$

(14)

Hence, the iterate $z_0\in S(x^{*},{\rho }^{*})$ and the assertion (8) holds if $n=0.$ 

Similarly, in the last substep of the algorithm, we write

$$\begin{array}{ccc}\hfill x_1-x^{*}& =& z_0-x^{*}-A_0^{-1}F(z_0)\hfill \\ \hfill & & +(1-a_1)u_5-a_2{u}_6-a_3{u}_7-a_4{u}_8-a_5{u}_9,\hfill \end{array}$$

leading to

$$\begin{array}{ccc}\hfill \parallel x_1-x^{*}\parallel & \le & [\phi (\parallel w_0-x^{*}\parallel ,\parallel z_0-x^{*}\parallel )\lambda (\parallel x_0-x^{*}\parallel )\hfill \\ \hfill & & +|1-a_0|\lambda (\parallel x_0-x^{*}\parallel )(1+{\int }_0^1{\phi }_1(\theta \parallel z_0-x^{*}\parallel )d\theta )\hfill \\ \hfill & & +|a_2|{\lambda }^2(\parallel x_0-x^{*}\parallel )(1+{\phi }_0(\parallel l_0-x^{*}\parallel ,\parallel z_0-x^{*}\parallel ))\hfill \\ \hfill & & +|a_3|{\lambda }^3(\parallel x_0-x^{*}\parallel )(1+{\phi }_0(\parallel l_0-x^{*}\parallel ,\parallel z_0-x^{*}{\parallel ))}^2\hfill \\ \hfill & & +|a_4|{\lambda }^4(\parallel x_0-x^{*}\parallel )(1+{\phi }_0(\parallel l_0-x^{*}\parallel ,\parallel z_0-x^{*}{\parallel ))}^3\hfill \\ \hfill & & +|a_5|{\lambda }^5(\parallel x_0-x^{*}\parallel )(1+{\phi }_0(\parallel l_0-x^{*}\parallel ,\parallel z_0-x^{*}{\parallel ))}^4]\parallel z_0-x^{*}\parallel \hfill \\ \hfill & \le & g_3(\parallel x_0-x^{*}\parallel )\parallel x_0-x^{*}\parallel \le \parallel x_0-x^{*}\parallel .\hfill \end{array}$$

Thus, assertion (6) holds if $n=1$ and (9) if $n=0.$ Repeat the preceding calculations with $x_m,y_m,z_m,x_{m+1}$ replacing $x_0,y_0,z_0,x_1$ to complete the induction for assertions (7)–(9). Then, from the estimate

$$\parallel x_{m+1}-x^{*}\parallel \le c\parallel x_m-x^{*}\parallel <{\rho }^{*},$$

where $c=g_3(\parallel x_0-x^{*}\parallel )\in [0,1),$ we conclude that ${lim}_{m⟶\infty }{x}_m=x^{*}$ and the iterate $x_{m+1}\in S(x^{*},{\rho }^{*}).$    □

The uniqueness region can be determined.

Proposition 1. 

Suppose:

A solution of $\overline{x}\in S(x^{*},{\rho }_4)$ exists for the equation $F(x)=0$ for ${\rho }_4>0;$ the condition (H3) holds on the ball $S(x^{*},{\rho }_4)$ and ${\rho }_5\ge {\rho }_4$ exists, such that

$${\int }_0^1{\phi }_1(\theta {\rho }_5)d\theta <1.$$

(15)

Define the region $D_1=D\cap S[x^{*},{\rho }_5].$ Then, the equation $F(x)=0$ is uniquely solvable by $x^{*}$ in the region $D_1.$ 

Proof. 

Let us consider the divided difference $[x^{*},\overline{x},F]$ provided that $x\ne \overline{x}.$ Then, it follows by (H3) and (15) that

$$\parallel L^{-1}(M-L)\parallel \le {\int }_0^1{\phi }_1(\theta \parallel \overline{x}-x^{*}\parallel )d\theta \le {\int }_0^1{\phi }_1(\theta {\rho }_5)d\theta <1.$$

Thus, the linear operator M is invertible. It follows from $\overline{x}-x^{*}=M^{-1}(F\overline{x})-F(x^{*}))=M^{-1}(0)=0$ that $\overline{x}=x^{*}.$    □

Remark 1. 

 

(i) 

We can certainly choose ${\rho }_4={\rho }^{*}$ in Proposition 1.

(ii) 

Possible choice for the uncluttered functions $f_i$ can be obtained as follows:

$$\begin{array}{ccc}\hfill w_0-x^{*}& =& x_0-x^{*}+b_{0}F(x_0)\hfill \\ & =& (I+b_0[x_0,x^{*};F])(x_0-x^{*})\hfill \\ & =& (I+b_{0}L{L}^{-1}([x_0,x^{*};F]-L+L))(x_0-x^{*})\hfill \\ & =& ((I+b_{0}L)+b_{0}L{L}^{-1}([x_0,x^{*};F]-L))(x_0-x^{*}),\hfill \\ \hfill \parallel w_0-x^{*}\parallel & \le & [\parallel I+b_{0}L\parallel +|b_0|\parallel L\parallel {\phi }_1(\parallel x_0-x^{*}\parallel )]\parallel x_0-x^{*}\parallel .\hfill \end{array}$$

Thus, we can define

$$f_1(t)=(\parallel I+b_{0}L\parallel +|b_0|\parallel L\parallel {\phi }_1(t))t.$$

Similarly, we set

$$f_2(t)=(\parallel I+b_{1}L\parallel +|b_1|\parallel L\parallel {\phi }_1(t))t$$

and

$$f_3(t)=(\parallel I+b_{2}L\parallel +|b_2|\parallel L\parallel {\phi }_1(t))t.$$

(iii) 

A possible choice for L in local convergence studies may be $L=F^{\prime }(x^{*})$, or $L=I$, or any other linear operator satisfying the conditions (H1)–(H5) (see also the Example 1 in the Section 4).

3. Semi-Local Analysis

The role of $x^{*},{\phi }_0,\phi$ is exchanged with $x_0,{\psi }_0$ and $\psi$ as follows:

Assume:

(E1)

NFC $f_4:M\to M,{\psi }_0:M\times M\to M$ exists, such that equation

$${\psi }_0(f_4(t),t)-1=0$$

has MPS ${\rho }_6\in M-\left\{0\right\}.$ Let $M_1=[0,{\rho }_6).$ Define $\psi :M_1⟶M$ of the sequence $\left\{{\alpha }_n\right\}$ for ${\alpha }_0=0,$  ${\beta }_0\in [0,{\rho }_6)$ and each $n=0,1,2,\dots$ by

$$\begin{array}{ccc}\hfill q_n& =& \frac{1}{1-{\psi }_0(f_4({\alpha }_n),{\alpha }_n)},\hfill \\ \hfill p_n& =& \psi ({\alpha }_n,{\beta }_n,f_4({\alpha }_n))({\beta }_n-{\alpha }_n),\hfill \\ \hfill {\gamma }_n& =& {\beta }_n+|a_0|q_n{p}_n+|3-2a_0|q_n^2(1+{\psi }_0(f_5({\beta }_n),{\beta }_n))\hfill \\ & & +|a_0-2|q_n^3(1+{\psi }_0(f_5({\beta }_n),{\beta }_n)),\hfill \\ \hfill d_n& =& (1+{\psi }_0({\alpha }_n,{\gamma }_n))({\gamma }_n-{\alpha }_n)+p_n,\hfill \end{array}$$

$$\begin{array}{ccc}\hfill {\alpha }_{n+1}& =& {\gamma }_n+|a_1|q_n{d}_n+|a_2|q_n^2{d}_n(1+{\psi }_0(f_5({\gamma }_n),{\gamma }_n))\hfill \\ & & +|a_3|q_n^3{d}_n{(1+{\psi }_0(f_5({\gamma }_n),{\gamma }_n))}^2\hfill \\ & & +|a_4|q_n^4{d}_n{(1+{\psi }_0(f_5({\gamma }_n),{\gamma }_n))}^3\hfill \\ & & +|a_5|q_n^5{d}_n{(1+{\psi }_0(f_5({\gamma }_n),{\gamma }_n))}^4,\hfill \\ \hfill {\delta }_{n+1}& =& \psi ({\alpha }_n,{\alpha }_{n+1},f_4({\alpha }_n))({\alpha }_{n+1}-{\alpha }_n)\hfill \\ & & +(1+{\psi }_0(f_4({\alpha }_n),{\alpha }_n))({\alpha }_{n+1}-{\beta }_n),\hfill \end{array}$$

and

$${\beta }_{n+1}={\alpha }_{n+1}+\frac{{\delta }_{n+1}}{1-{\psi }_0(f_4({\alpha }_{n+1}),{\alpha }_{n+1})}.$$

(E2)

${\rho }_7\in [0,{\rho }_6)$ exists, such that for each $n=0,1,2,\dots$ 

$${\psi }_0(f_4({\alpha }_n),{\alpha }_n)<1\mathrm{and}{\alpha }_n\le {\rho }_7.$$

It follows that $0\le {\alpha }_n\le {\beta }_n\le {\gamma }_n\le {\alpha }_{n+1}<{\rho }_7$ and ${\rho }_8\in [0,{\rho }_7)$ exists, such that ${lim}_{n⟶\infty }{\alpha }_n={\rho }_8.$ 

(E3)

An invertible linear operator of L and $x_0\in D$ exists, such that for each $x\in D$ 

$$\parallel L^{-1}([w,x;F]-L)\parallel \le {\psi }_0(\parallel w-x_0\parallel ,\parallel x-x_0\parallel )$$

and

$$\parallel w-x_0\parallel \le f_4(\parallel x-x_0\parallel ).$$

Notice that by condition (E1)

$$\parallel L^{-1}([w_0,x_0;F]-L)\parallel \le {\psi }_0(\parallel w_0-x_0\parallel ,0)\le {\psi }_0(f_4(0),0)<1.$$

Thus, the linear operator $A_0=[w_0,x_0;F]$ is invertible and we can take

$$\parallel A_0^{-1}F(x_0)\parallel \le {\beta }_0.$$

(E4)

Let $D_1=D\cap S(x_0,{\rho }_6)$ 

$$\parallel L^{-1}([y,x;F]-[w,x;F])\parallel \le \psi (\parallel x-x_0\parallel ,\parallel y-x_0\parallel ,\parallel w-x_0\parallel )$$

for each $x,y\in D_1.$ 

and

(E5)

$S[x_0,{\rho }_8]\subset D.$ 

Then, using induction, as in the local case, we obtain the estimates

$$\begin{array}{ccc}\hfill \parallel y_0-x_0\parallel & =& \parallel A_0^{-1}F(x_0)\parallel \le {\beta }_0={\beta }_0-{\alpha }_0<{\rho }_8,\hfill \\ \hfill F(y_k)& =& F(y_n)-F(x_n)-[w_n,x_n;F](y_n-x_n)\hfill \\ & =& ([y_n,x_n;F]-[w_n,x_n;F])(y_n-x_n),\hfill \\ \hfill \parallel L^{-1}F(y_n)\parallel & \le & \psi (\parallel x_n-x_0\parallel ,\parallel y_n-x_0\parallel ,\parallel w_n-x_0\parallel )\parallel y_n-x_n\parallel \hfill \\ & \le & \psi ({\alpha }_n,{\beta }_n,f_4({\alpha }_n))({\beta }_n-{\alpha }_n)=p_n,\hfill \end{array}$$

$$\begin{array}{ccc}\hfill \parallel A_n^{-1}L\parallel & \le & \frac{1}{1-{\psi }_0(\parallel w_n-x_0\parallel ,\parallel x_n-x_0\parallel )}\hfill \\ & \le & \frac{1}{1-{\psi }_0(f_4({\alpha }_n),{\alpha }_n)}=q_n,\hfill \\ \hfill \parallel z_n-y_n\parallel & \le & |a_0|q_n{p}_n+|3-2a_0|q_n^2(1+{\psi }_0(\parallel h_n-x_0\parallel ,\parallel y_n-x_0\parallel )\hfill \\ & & +|a_0-2|q_n^3(1+{\psi }_0(\parallel h_n-x_0\parallel ,\parallel y_0-x_0{\parallel )}^2\hfill \\ & \le & |a_0|q_n{p}_n+|3-2a_0|q_n^2(1+{\psi }_0(f_5({\beta }_n),{\beta }_n))\hfill \\ & & +|a_0-2|q_n^{3}2{(1+{\psi }_0(f_5({\beta }_n),{\beta }_n))}^2\hfill \\ & =& {\gamma }_n-{\beta }_n,\hfill \end{array}$$

$$\begin{array}{ccc}\hfill \parallel z_n-x_0\parallel & \le & \parallel z_n-y_n\parallel +\parallel y_n-x_0\parallel \hfill \\ & \le & {\gamma }_n-{\beta }_n+{\beta }_n-{\alpha }_0\hfill \\ & =& {\gamma }_n<{\rho }_8,\hfill \\ \hfill \parallel x_{n+1}-z_n\parallel & \le & |a_1|q_n{d}_n+|a_2|q_n^2(1+{\psi }_0(\parallel l_n-x_0\parallel ,\parallel z_n-x_0\parallel )\hfill \\ & & +|a_3|q_n^3{d}_n{(1+{\psi }_0(f_6({\gamma }_n),{\gamma }_n))}^2\hfill \\ & & +|a_4|q_n^4{d}_n{(1+{\psi }_0(f_6({\gamma }_n),{\gamma }_n))}^3\hfill \\ & & +|a_5|q_n^5{d}_n{(1+{\psi }_0(f_6({\gamma }_n),{\gamma }_n))}^4\hfill \\ & =& {\alpha }_{n+1}-{\gamma }_n,\hfill \\ \hfill \parallel x_{n+1}-x_0\parallel & \le & \parallel x_{n+1}-z_n\parallel +\parallel z_n-x_0\parallel \hfill \\ & \le & {\alpha }_{n+1}-{\gamma }_n+{\gamma }_n-{\alpha }_0\hfill \\ & =& {\alpha }_{n+1}<{\rho }_8,\hfill \end{array}$$

$$\begin{array}{ccc}\hfill F(x_{n+1})& =& F(x_{n+1})-F(x_n)-M_n(y_n-x_n)\hfill \\ & =& (F(x_{n+1})-F(x_n)-M_n(x_{n+1}-x_n))+M_n(x_{n+1}-y_n)\hfill \\ & =& ([x_{n+1},x_n;F]-M_n)(x_{n+1}-x_n)+M_n(x_{n+1}-y_n),\hfill \end{array}$$

$$\begin{array}{ccc}\hfill \parallel L^{-1}F(x_{n+1})\parallel & \le & \psi (\parallel x_n-x_0\parallel ,\parallel x_{n+1}-x_0\parallel ,\parallel w_n-x_0\parallel )\parallel x_{n+1}-x_n\parallel \hfill \\ \hfill & & +(1+{\psi }_0(\parallel w_n-x_0\parallel ,\parallel x_n-x_0\parallel ))\parallel x_{n+1}-y_n\parallel \hfill \\ \hfill & \le & \psi ({\alpha }_n,{\alpha }_{n+1},f_4({\alpha }_n))({\alpha }_{n+1}-{\alpha }_n)\hfill \\ \hfill & & +(1+{\psi }_0(f_4({\alpha }_n),{\alpha }_n))({\alpha }_{n+1}-{\beta }_n)\hfill \\ \hfill & =& {\delta }_{n+1}\hfill \end{array}$$

(16)

$$\begin{array}{ccc}\hfill \parallel y_{n+1}-x_{n+1}\parallel & \le & \parallel A_{n+1}^{-1}L\parallel \parallel L^{-1}F(x_{n+1})\parallel \hfill \\ & \le & \frac{{\delta }_{n+1}}{1-{\psi }_0(f_4(\parallel w_{n+1}-x_0\parallel ,\parallel x_{n+1}-x_0\parallel )}\hfill \\ & \le & \frac{{\delta }_{n+1}}{1-{\psi }_0(f_4({\alpha }_{n+1}),{\alpha }_{n+1})}\hfill \\ & =& {\beta }_{n+1}-{\alpha }_{n+1},\hfill \end{array}$$

$$\begin{array}{ccc}\hfill \parallel y_{n+1}-x_0\parallel & \le & \parallel y_{n+1}-x_{n+1}\parallel +\parallel x_{n+1}-x_0\parallel \hfill \\ & \le & {\beta }_{n+1}-{\alpha }_{n+1}+{\alpha }_{n+1}-{\alpha }_0\hfill \\ & =& {\beta }_{n+1}<{\rho }_8.\hfill \end{array}$$

Thus, $\left\{x_n\right\}\subset S(x_0,{\rho }_8)$ and is Cauchy in a Banach space $E.$ Hence, $x^{*}\in S[x_0,{\rho }_8]$ exists, such that ${lim}_{n⟶\infty }{x}_n=x^{*}.$ 

By letting $n⟶\infty$ in (16), we obtain $F(x^{*})=0.$ Notice that from the estimate,

$$\parallel x_{n+j}-x_n\parallel \le {\alpha }_{n+j}-{\alpha }_n.$$

(17)

If $j⟶\infty$ in (17), we obtain

$$\parallel x^{*}-x_n\parallel \le {\alpha }^{*}-{\alpha }_n.$$

Therefore, we arrive at the semi-local result for the Algorithm 1:

Theorem 2. 

Suppose that conditions (E1)–(E5) hold. Then, the following assertions hold

$$\left\{x_n\right\}\subset S(x_0,{\rho }_8),$$

$$\parallel y_n-x_n\parallel \le {\beta }_n-{\alpha }_n,$$

$$\parallel x_{n+1}-z_n\parallel \le {\alpha }_{n+1}-{\gamma }_n$$

and $x^{*}\in S[x_0,{\rho }_8]$ exists, solving the equation $F(x)=0.$ 

The uniqueness of the solution for equation of $F(x)=0$ is specified.

Proposition 2. 

Suppose: A solution of $\tilde{x}\in S(x_0,{\rho }_9)$ exists for the equation $F(x)=0$ for some ${\rho }_9>0;$ condition (E3) holds on the ball $S(x_0,{\rho }_9)$ and ${\rho }_{10}\ge {\rho }_9$ exists, such that

$${\psi }_0({\rho }_9,{\rho }_{10})<1.$$

(18)

Define the region $D_2=D\cap S[x_0,{\rho }_{10}].$ Then, equation $F(x)=0$ is uniquely solvable by $\tilde{x}$ in the region $D_2.$ 

Proof. 

Let $\tilde{y}\in D_2$ with $F(\tilde{y})=0$ and $\tilde{y}\ne \tilde{x}.$ Then, the divided difference $Q=[\tilde{y},\tilde{x};F]=[\tilde{w},\tilde{x};F]$ is well defined. It then follows from (18) that

$$\begin{array}{ccc}\hfill \parallel L^{-1}(Q-L)\parallel & \le & {\psi }_0(\parallel \tilde{y}-x_0\parallel ,\parallel \tilde{x}-x_0\parallel )\hfill \\ & \le & {\psi }_0({\rho }_9,{\rho }_{10})<1.\hfill \end{array}$$

Therefore, we deduce $\tilde{y}=\tilde{x}.$    □

Remark 2. 

 

(i) 

The limit point ${\rho }_8$ can be replaced by ${\rho }_6$ in (E5) (provided in the condition (E1)).

(ii) 

Suppose that all conditions (E1)–(E5) hold in Proposition 1. Then, set ${\rho }_9={\rho }_8$ and $\tilde{x}=x^{*}.$ 

(iii) 

Functions $f_i,i=4,5,6$ can be specified as in the local case by the following estimates:

$$\begin{array}{ccc}\hfill w-x_0& =& [(I+b_{0}L)+b_{0}L{L}^{-1}([x,x_0;F]-L)](x-x_0)+b_{0}F(x_0),\hfill \\ \hfill \parallel w-x_0\parallel & \le & (\parallel I+b_{0}L\parallel +|b_0|\parallel L\parallel {\psi }_1(\parallel x-x_0\parallel )\parallel x-x_0\parallel \hfill \\ & & +|b_0|\parallel F(x_0)\parallel .\hfill \end{array}$$

Hence, we can define

$$f_4({\alpha }_n)=(\parallel I+b_{0}L\parallel +|b_0|\parallel L\parallel {\psi }_1({\alpha }_n)){\alpha }_n+|b_0|\parallel F(x_0)\parallel .$$

Similarly, we choose

$$f_5({\beta }_n)=(\parallel I+b_{1}L\parallel +|b_1|\parallel L\parallel {\psi }_1({\beta }_n)){\beta }_n+|b_1|\parallel F(x_0)\parallel$$

and

$$f_6({\gamma }_n)=(\parallel I+b_{2}L\parallel +|b_2|\parallel L\parallel {\psi }_2({\gamma }_n)){\gamma }_n+|b_2|\parallel F(x_0)\parallel .$$

(iv) 

A possible choice for L may be $L=A_0$, provided that the operator $A_0$ is invertible or $L=I.$ Other choices are possible, as long as conditions (E1)–(E4) are validated.

4. Numerical Examples

In this Section, we chose $a_0=3,a_5=0,b_0=-1,b_1=1$, and $b_2=-1$ for all of the examples to obtain the specialization of the algorithm, which is defined by Algorithm 2.

Algorithm 2

Step1: Given $x_0\in D,$ Solve $A_n{u}_1=F(x_n),$ for $u_1.$

Step 2: Set $y_n=x_n-u_1.$

Step 3: Solve $A_n{u}_2=F(y_n),$ for $u_2.$

Step 4: Solve $A_n{u}_3=G_n{u}_2,$ for $u_3.$

Step 5: Solve $A_n{u}_4=G_n{u}_3,$ for $u_4.$

Step 6: Set $z_n=y_n-3u_2+3u_3-u_4.$

Step 7: Solve $A_n{u}_5=F(z_n),$ for $u_5.$

Step 8: Solve $A_n{u}_6=Q_n{u}_5,$ for $u_5.$

Step 9: Solve $A_n{u}_7=Q_n{u}_6,$ for $u_6.$

Step 10: Solve $A_n{u}_8=Q_n{u}_7,$ for $u_8.$

Step 11: Set $x_{n+1}=z_n-4u_5+6u_6-4u_7+u_8.$

Step 12: If $x_{n+1}=x_n,$ STOP. Otherwise, repeat the process with $n⟶n+1.$

Here $w_n=x_n-F(x_n),h_n=y_n+F(y_n),$ $l_n=z_n-F(z_n),A_n=[x_n,w_n;F],$ $G_n=[h_n,y_n;F]$, $Q_n=[l_n,z_n;F]$.

Also, we considered the choice of the divided difference $[x,y;F]={\int }_0^1{F}^{\prime }(x+\theta (y-x))d\theta$ and $L=F^{\prime }(x^{*}).$ 

In Example 1, we provided the choice of the operator L as well as the functions ${\phi }_0,\phi ,\mathrm{and} {\phi }_1$ to validate the local convergence conditions (H1)–(H5). Notice that functions $f_1,f_2$ and $f_3$ were chosen, as in Remark 1 (ii). There was no need to choose the operator L in the rest of the examples as the convergence of the aforementioned Algorithm was established (semi-local convergence). The stopping criterion is $\parallel x_n-x_{n-1}\parallel <ϵ,$ where $ϵ$ is the desired error tolerance.

Example 1. 

Let $\mathcal{M}=\mathbb{R}\times \mathbb{R}\times \mathbb{R}$ and $\Omega =S[x^{*},1].$ The mapping F is defined on Ω for $a={(a_1,a_2,a_3)}^{tr}\in \mathbb{R}$ as

$$\begin{array}{c}\hfill F(a)={(a_1,e^{a_2}-1,\frac{e^{-1}}{2}{a}_3^2+a_3)}^{tr}.\end{array}$$

Then, $F^{\prime }$ is calculated to be

$$F^{\prime }(a)=\left[\begin{array}{ccc}1& 0& 0\\ 0& e^{a_2}& 0\\ 0& 0& (e-1)a_3+1\end{array}\right]$$

Then, $x^{*}={(0,0,0)}^{tr}$ solves equation $F(a)=0$ . Moreover, the definition of $F^{\prime }$ provides $F^{\prime }(x^{*})=I$ . Take $L=I.$ 

Then, conditions (H3)–(H5) are valid if we define for $f_i$ as provided in Remark 1.

$$\begin{array}{ccc}\hfill {\phi }_1(t)& =& \frac{e-1}{2}t,\hfill \\ \hfill f_1(t)& =& f_2(t)=f_3(t)={\phi }_1(t)t,\hfill \\ \hfill {\phi }_0(s,t)& =& \frac{1}{2}(e-1)(t+f_1(t)),\hfill \end{array}$$

(19)

$$\begin{array}{ccc}\hfill \phi (s,t)& =& \frac{1}{2}(e-1)f_1(t).\hfill \end{array}$$

(20)

These choices of scalar functions validate the conditions of Theorem 1. This assures the convergence of the sequence $\left\{x_n\right\}$ to solution $x^{*}.$ 

Then, from Formula (4), we deduce

$${\rho }_1=0.59619959522338323554725671786752,$$

$${\rho }_2=0.24652697702590073500524094190487$$

and

$${\rho }_3={\rho }^{*}=0.10739579951416893362926171861238.$$

Example 2. 

The solution sought for the nonlinear system

$$\begin{array}{ccc}\hfill 3{\theta }_1^2{\theta }_2+{\theta }_2^2-1+|{\theta }_1-1|& =& 0\hfill \\ \hfill {\theta }_1^4+{\theta }_1{\theta }_2^3-1+|{\theta }_2|& =& 0\hfill \end{array}$$

Let $F=(Q_1,Q_2)$ for $({\theta }_1,{\theta }_2)\in \mathbb{R}\times \mathbb{R}$, where

$$\begin{array}{ccc}\hfill Q_1& =& 3{\theta }_1^2{\theta }_2+{\theta }_2^2-1+|{\theta }_1-1|=0and\hfill \\ \hfill Q_2& =& {\theta }_1^4+{\theta }_1{\theta }_2^3-1+|{\theta }_2|=0.\hfill \end{array}$$

Then, the system becomes

$$\begin{array}{c}\hfill F(s)=0fors={({\theta }_1,{\theta }_2)}^T.\end{array}$$

The divided difference $L=[.,.;F]$ belongs in the space $M_{2\times 2}(\mathbb{R})$ and is the standard $2\times 2$ matrix in ${\mathbb{R}}^2$ [11,18]. Let us choose $x_0={(5,5)}^T$ . It turns out that the algorithm converges to the solution of $x^{*}$, as the initial guess of $x_0$ is close enough to it. Hence, there is no need to validate the conditions of Theorem 2, which are sufficient. Then, the application of the algorithm provides the solution ${\theta }^{*}$ after three iterations. The solution ${\theta }^{*}={({\theta }_1^{*},{\theta }_2^{*})}^T$, where

$$\begin{array}{c}\hfill {\theta }_1^{*}=0.894655373334687\end{array}$$

and

$$\begin{array}{c}\hfill {\theta }_2^{*}=3.27826421746298.\end{array}$$

Example 3. 

Consider the system of 100 equations defined by

$$F(x)=0,$$

where

$$F(x)(i)=\{\begin{array}{cc}{x}_i^{2}sin(x_{i+1})-1,& it1\le i\le 99\\ x_i^{2}sin(x_i)-1,& iti=100.\end{array}$$

The results obtained for the initial point $(2,2,\dots ,2)$ are provided in Table 1.

Table 1. Iterated solutions of Example 3.

Iteration	Solution	$|\mathit{F}(\mathit{x})(\mathit{i})|$	Time
$\mathit{n}$	${\mathit{x}}_{\mathit{i}}$		s
1	0.52465745776846004734532218993115	0.862114977251371	9.196941
2	1.0666417888794666022247900197049	0.003827358620836	19.437201
3	1.0682235441972490182834127193622	1.110223024625157 × ${10}^{-16}$	28.369249
4	1.0682235441972490182834711142631	1.110223024625157 × ${10}^{-16}$	37.619249

Example 4. 

In this example, we consider a system of five equations defined by

$$F(x)=0,$$

where

$$F(x)(i)=\sum _{j=1,j\ne i}^5{x}_i-e^{x_i}.$$

The results obtained for the initial points $(1,1,1,1,1)$ are provided in Table 2.

Table 2. Iterated solutions of Example 4.

Iteration	Solution	$|\mathit{F}(\mathit{x})(\mathit{i})|$	Time
$\mathit{n}$	${\mathit{x}}_{\mathit{i}}$		s
1	0.20391080591998655968666298576863	1.081148328339054 × ${10}^{-4}$	0.002347
2	0.20388835470224017654139458954887	1.110223024625157 × ${10}^{-16}$	0.002501
3	0.20388835470224017654139458954887	1.110223024625157 × ${10}^{-16}$	0.002839

We used a 4-core 64-bit Windows machine with 11th Gen Intel(R) Core (TM) i5-1135G7 CPU @ 2.40 GHz for all our computations using MATLAB R2023b.

5. Conclusions

The step eighth-order Algorithm without derivatives of the operator was studied in this paper using assumptions only on the first divided difference of the operator. Earlier studies using the Taylor series expansion algorithm made assumptions up to a ninth order derivative not on the algorithm [10].

We provided sufficient convergence conditions involving only the operators on the algorithm, computable error upper bounds on $\parallel x_n-x^{*}\parallel ,$ and presented the uniqueness of the solution results. It is worth noticing that the methodology of this study was not dependent on the convergence order of the iterative algorithm as the convergence conditions did not make use of it. Moreover, the assumption that the solution was simple was not made or implied by the convergence conditions. Thus, in case convergence conditions were satisfied, the methods also found solutions of multiplicity greater than one. The approach in this paper was applied to other algorithms with inverses to obtain the same benefits [4,20,21,22]. This will be the focus of our future research.