Extending King’s Method for Finding Solutions of Equations

Abstract

King’s method applies to solve scalar equations. The local analysis is established under conditions including the fifth derivative. However, the only derivative in this method is the first. Earlier studies apply to equations containing at least five times differentiable functions. Consequently, these articles provide no information that can be used to solve equations involving functions that are less than five times differentiable, although King’s method may converge. That is why the new analysis uses only the operators and their first derivatives which appear in King’s method. The article contains the semi-local analysis for complex plane-valued functions not presented before. Numerical applications complement the theory.

1. Introduction

In this article, the function $F:\Omega \subset T⟶T$ is differentiable, where $T=\mathbb{R}$ or $T=\mathbb{C}$ and $\Omega$ is an open nonempty set.

The nonlinear equation

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

(1)

is studied in this article. An analytic form of a solution $x^{*}$ is preferred. However, this form is not always available. So, mostly iterative solution methods have been applied to approximate the solution $x^{*}.$

In particular, King’s [1] fourth-order method (KM) has been used;

$$\begin{array}{ccc}\hfill u_0\in \Omega ,v_n& =& u_n-F^{\prime }{\left(u_n\right)}^{-1}F\left(u_n\right)\hfill \\ \hfill u_{n+1}& =& v_n-A_n^{-1}(F\left(u_n\right)+\gamma F\left(v_n\right))F^{\prime }{\left(u_n\right)}^{-1}F\left(v_n\right),\hfill \end{array}$$

(2)

where $\gamma \in T$ is a parameter and $A_n=F\left(u_n\right)+(\gamma -2)F\left(v_n\right).$

As motivation consider the real function

$$\mu \left(s\right)=\left\{\begin{array}{cc}0& ifs=0\\ s^5-s^4+s^{3}log{s}^2& ifs\ne 0.\end{array}\right.$$

This definition gives

$${\mu }^{‴}\left(s\right)=6log{s}^2+60s^2-24s+22.$$

However, then, the third derivative is unbounded. So, the convergence of KM is not assured by previous analyses in [1,2,3,4,5,6,7,8].

This is the case, since Taylor series requiring derivatives of high order (not in KM) are utilized in the analysis for convergence. This is a common observation for other methods, such as Traub’s, Jarratt’s, and the Kung–Traub method to mention some [2,3,5,6,7,8,9,10]. On the top of these concerns, some other problems exist with earlier studies. No computable data are provided for distances $\parallel u_{n+1}-u_n\parallel$ or $\parallel u_n-x^{*}\parallel$ or the uniqueness and location of solution $x^{*}.$

All these concerns are addressed utilizing conditions involving only the first derivative in the method (2) [9,10,11,12,13,14,15,16].

The next four sections include semi-local analysis, local analysis, the experiment, and conclusions, respectively.

2. Semi-Local Analysis

Set $L_0,L,L_1,L_2,\delta$ and $\eta$ to be positive parameters. Set $L_3=\frac{L{L}_2}{2},$ and $L_4=\frac{\delta \left|\gamma \right|L^2}{4}.$ Let the sequence $\left\{t_n\right\}$ be given as

$$\begin{array}{ccc}\hfill t_0& =& 0,s_0=\eta ,\hfill \\ \hfill t_{n+1}& =& s_n+\frac{[L_3+L_4(s_n-t_n)]{(s_n-t_n)}^3}{(1-p_n)(1-L_0{t}_n)}\hfill \\ \hfill s_{n+1}& =& t_{n+1}+\frac{L{(t_{n+1}-t_n)}^2+2L_1(t_{n+1}-s_n)}{2(1-L_0{t}_{n+1})},\hfill \end{array}$$

(3)

where $p_n=L_2(t_n+|\gamma -2|(s_n-\eta )).$ Sequence $\left\{t_n\right\}$ shall be shown to be majorizing for KM.

Lemma 1.

Suppose

$$t_n<\frac{1}{L_0}\mathrm{and}{p}_n<1.$$

(4)

Then, the following assertions hold

$$t_n\le s_n\le t_{n+1}$$

(5)

and

$$\underset{n⟶\infty }{lim}{t}_n=t^{*}\le \frac{1}{L_0},$$

(6)

where $t^{*}$ is the unique least upper bound of sequence $\left\{t_n\right\}.$

Proof. 

Assertions (5) and (6) follow immediately by (3) and (4). □

Another result is given for the sequence $\left\{t_n\right\}$ using stronger conditions but which are easier to verify than (4). However, first, we need to introduce some concepts. Let

$$a=(L_3+L_4\eta ){\eta }^2,b=\frac{L{t}_1^2+2L_1(t_1-\eta )}{2(1-L_0{t}_1)\eta },$$

and

$$c=max\{a,b\}.$$

Develop polynomials defined on the interval $[0,1)$ as

$$f_n^{\left(1\right)}\left(t\right)=2(L_3+L_4{t}^n\eta )t^{2n-1}{\eta }^2+L_0(1+t)(1+t+\dots +t^{n-1})\eta -1,$$

$$g_n^{\left(1\right)}\left(t\right)=2(L_3+L_4{t}^{n+1}\eta )t^{n+1}\eta -2(L_3+L_4{t}^n\eta )t^{n-1}\eta +L_0(1+t),$$

$$g_1\left(t\right)=g_1^{\left(1\right)}\left(t\right),$$

and

$$\begin{array}{ccc}\hfill f_n^{\left(2\right)}\left(t\right)& =& L{[4(L_3+L_4{t}^n\eta ){\left(t^n\eta \right)}^2+1]}^2{t}^{n-1}\eta \hfill \\ & & +8L_1(L_3+L_4{t}^n\eta )t^{2n-1}{\eta }^2\hfill \\ & & +2L_0(1+t)(1+t+\dots +t^n)\eta -2.\hfill \end{array}$$

Moreover, set

$$g_2\left(t\right)=g_2^{\left(2\right)}\left(t\right).$$

Notice that polynomials $g_1$ and $g_2$ are independent of $n.$ In particular, say

$$g_1\left(t\right)=2L_{4}t{\eta }^2(t^3-1)+2L_3\eta (t^2-1)+L_0(1+t).$$

Then, condition $g_1\left(t\right)\ge 0$ needed in the next Lemma holds if

$$2L_{4}t{\eta }^2(1-t^3)+2L_3\eta (1-t^2)\le L_0(1+t).$$

The left side of this estimate is a positive multiple of $\eta .$ However, the right side of it is positive but independent of $\eta .$ So, this estimate certainly holds for sufficiently small $\eta .$ The same observation is made for polynomial $g_2$ and condition $g_2\left(t\right)\ge 0.$

An auxiliary result connects these polynomials.

Lemma 2.

The following items hold:

(i) 

$f_{n+1}^{\left(1\right)}\left(t\right)-f_n^{\left(1\right)}\left(t\right)=g_n^{\left(1\right)}\left(t\right)t^n\eta$;

(ii) 

$g_{n+1}^{\left(1\right)}\left(t\right)\ge g_n^{\left(1\right)}\left(t\right)$;

(iii) 

$g_{n+1}^{\left(1\right)}\left(t\right)-f_n^{\left(1\right)}\left(t\right)\ge g_1\left(t\right)t^n\eta ,$if$g_1\left(t\right)\ge 0$;

and

(iv) 

$f_{n+1}^{\left(2\right)}\left(t\right)-f_n^{\left(2\right)}\left(t\right)\ge g_2\left(t\right)t^{n-1}\eta ,$if$g_2\left(t\right)\ge 0.$

Proof. 

By the definition of these polynomials, we get in turn:

(i)

$$\begin{array}{ccc}\hfill f_{n+1}^{\left(1\right)}\left(t\right)& =& f_{n+1}^{\left(1\right)}\left(t\right)-f_n^{\left(1\right)}\left(t\right)+f_n^{\left(1\right)}\left(t\right)\hfill \\ & =& 2(L_3+L_4{t}^{n+1}\eta )t^{2n+1}{\eta }^2+L_0(1+t)(1+t+\dots +t^n)\eta -1\hfill \\ & & -2(L_3+L_4{t}^n\eta )t^{2n-1}{\eta }^2-L_0(1+t)(1+t+\dots +t^{n-1})\eta +f_n^{\left(1\right)}\left(t\right)+1\hfill \\ & =& f_n^{\left(1\right)}\left(t\right)+g_n^{\left(1\right)}\left(t\right)t^n\eta ;\hfill \end{array}$$

(ii)

$$\begin{array}{ccc}\hfill g_{n+1}^{\left(1\right)}\left(t\right)-g_n^{\left(1\right)}\left(t\right)& =& 2(L_3+L_4{t}^{n+2}\eta )t^{n+2}\eta \hfill \\ & & -2(L_3+L_4{t}^{n+1}\eta )t^n\eta -2(L_3+L_4{t}^{n+1})\eta )t^{n+1}\eta \hfill \\ & & +2(L_3+L_4{t}^n\eta )t^{n-1}\eta \hfill \\ & =& 2[(L_3+L_4{t}^{n+2}\eta )t^3-(L_3+L_4{t}^{n+1}\eta )t\hfill \\ & & -(L_3+L_4{t}^{n+1}\eta )t^2+(L_3+L_4{t}^n\eta )]t^{n-1}\eta \hfill \\ & =& 2{(t-1)}^2(t+1)(L_3+L_4\eta t^n(t^2+t+1))t^{n-1}\eta \ge 0.\hfill \end{array}$$

(iii)

This estimate follows immediately from the first two;

(iv)

It follows similarly from the definition of polynomials $g_2$ and $f_n^{\left(2\right)},$ since $t\in [0,1).$

□

Define the parameters

$${\beta }_1=\frac{1-L_0\eta }{1+L_0\eta },{\beta }_2=\frac{1-2L_0\eta }{1+2L_0\eta },{\beta }_3=\frac{1-2L_2\eta }{1+2L_2\eta (1+2|\gamma -2\left|\right)},$$

$$\beta =min\{{\beta }_1,{\beta }_2,{\beta }_3\}$$

and

$$M=2max\{L_0,L_2\}.$$

Notice that $\beta \in (0,1).$

Lemma 3.

Suppose:

$$L_0{t}_1<1,$$

(7)

$$M\eta <1,$$

(8)

$$c\le \alpha \le \beta ,$$

(9)

$$g_1\left(t\right)\ge 0\mathrm{at}t=\alpha$$

(10)

and

$$g_2\left(t\right)\ge 0\mathrm{at}t=\alpha$$

(11)

hold for some$\alpha \in (0,1).$Then, sequence$\left\{t_n\right\}$is convergent to$t^{*}.$Notice, criteria (7)–(11) determine the “smallness” of η to force convergence of the method.

Proof. 

Mathematical induction is used to show

$$0\le \frac{[L_3+L_4(s_m-t_m)]{(s_m-t_m)}^3}{(1-p_m)(1-L_0{t}_m)}\le \alpha ,$$

(12)

$$0\le \frac{L{(t_{m+1}-t_m)}^2+2L_1(t_{m+1}-s_m)}{2(1-L_0{t}_{m+1})}\le \alpha (s_m-t_m)$$

(13)

and

$$t_m\le s_m\le t_{m+1}.$$

(14)

These estimates are true for $m=0$ by (7) or (8) and the definition of sequence $\left\{t_m\right\}.$ Then, it follows $0\le t_1-s_0\le \alpha (s_0-t_0)=\alpha \eta$ and $0\le s_1-t_1\le \alpha (s_0-t_0)=\alpha \eta .$ Suppose:

$$0\le t_{m+1}-s_m\le \alpha (s_m-t_m)\le {\alpha }^{m+1}\eta$$

(15)

and

$$0\le s_{m+1}-t_{m+1}\le \alpha (s_m-t_m)\le {\alpha }^{m+1}\eta .$$

(16)

Then,

$$\begin{array}{ccc}\hfill t_{m+1}& \le & s_m+{\alpha }^{m+1}\eta \le t_m+{\alpha }^m\eta +{\alpha }^{m+1}\eta \hfill \\ \hfill & \le & s_{m-1}+2{\alpha }^m\eta +{\alpha }^{m+1}\eta \hfill \\ \hfill & ⋮& \\ \hfill & \le & s_1+2{\alpha }^2\eta +\dots +2{\alpha }^m\eta +{\alpha }^{m+1}\eta \hfill \\ \hfill & \le & t_1+\alpha \eta +2{\alpha }^2\eta +\dots +2{\alpha }^m\eta +{\alpha }^{m+1}\eta \hfill \\ \hfill & =& \eta +2\alpha \eta (1+\alpha +\dots +{\alpha }^{m-1})\eta +{\alpha }^{m+1}\eta \hfill \\ \hfill & =& \eta \frac{(1+\alpha )(1-{\alpha }^{m+1})}{1-\alpha }\hfill \\ \hfill & <& \eta \frac{1+\alpha }{1-\alpha }=t^{**}.\hfill \end{array}$$

(17)

Evidently, (12) holds if

$$2(L_3+L_4{\alpha }^m\eta ){\left({\alpha }^m\eta \right)}^2+L_0\alpha \frac{(1+\alpha )}{1-\alpha }(1-{\alpha }^m)\eta -\alpha \le 0$$

(18)

or

$$f_m^{\left(1\right)}\left(t\right)\le 0\mathrm{at}t=\alpha .$$

(19)

Define

$$f_{\infty }^{\left(1\right)}\left(t\right)=\underset{m⟶\infty }{lim}{f}_m^{\left(1\right)}\left(t\right).$$

(20)

It can be shown instead from Lemma 2 that

$$f_{\infty }^{\left(1\right)}\left(t\right)\le 0att=\alpha .$$

(21)

However, by (15) and (20),

$$f_{\infty }^{\left(1\right)}\left(t\right)=\frac{L_0(1+t)\eta }{1-t}-1.$$

(22)

Then, (21) holds by (10) and (22). Moreover, instead of (13), we can show

$$\frac{[L{(t_{n+1}-s_n+s_n-t_n)}^2+2L_1\left(\frac{(L_3+L_4(s_n-t_n)]{(s_n-t_n)}^3}{(1-p_n)(1-L_0{t}_n)}\right)}{2(1-L_0{t}_{m+1})}\le \alpha (s_n-t_n),$$

(23)

since

$$\frac{1}{1-L_0{t}_m}\le 2,$$

(24)

$$\frac{1}{1-p_m}\le 2$$

(25)

and

$$\begin{array}{ccc}\hfill 0& \le & t_{m+1}-t_m\hfill \\ \hfill & \le & (1+\alpha )(s_m-t_m)\hfill \end{array}$$

(26)

hold. Indeed, (24) holds if

$$2L_2{t}_m\le 2L_2\frac{(1+\alpha )\eta }{1-\alpha }\le 1$$

or

$$\alpha \le \frac{1-2L_0\eta }{1+2L_0\eta }.$$

However, this holds because of the choice of ${\beta }_2$ and (9). Moreover, estimate (25) holds if

$$2L_2\left(|\gamma -2|[\frac{(1+\alpha )\eta }{1-\alpha }-\eta ]+\frac{(1+\alpha )\eta }{1-\alpha }\right)\le 1,$$

which is true by the choice of ${\beta }_3$ and (9). Then, (23) holds if

$$\begin{array}{ccc}& & L{[4(L_3+L_4(s_n-t_n)){(s_n-t_n)}^2+1]}^2(s_n-t_n)\hfill \\ & & +8L_1(L_3+L-4(s_n-t_n)){(s_n-t_n)}^2\le \alpha \hfill \end{array}$$

or

$$\begin{array}{ccc}\hfill & & L[4(L_3+L_4{\alpha }^n\eta ){\left({\alpha }^n\eta \right)}^2+1]{\alpha }^{n-1}\eta \hfill \\ \hfill & & +8L_1(L_3+L-4{\alpha }^n\eta ){\alpha }^{2n-1}{\eta }^2-1\le 0\hfill \end{array}$$

(27)

or

$$f_m^{\left(2\right)}\left(t\right)\le 0\mathrm{at}t=\alpha .$$

(28)

or

$$f\left(t\right)\le 0\mathrm{at}t=\alpha .$$

However, this holds by (11). By sequence $\left\{t_m\right\},$ (12) and (13), the estimate (14) also holds. Therefore, the induction for estimates (12)–(14) is terminated. Hence, $\left\{t_m\right\}$ is bounded by $t^{**},$ which is non-decreasing. Hence, it converges to $t^{*}$. □

The semi-local convergence analysis of KM uses conditions (H). Suppose that there exist:

(H1)

$u_0\in \Omega ,\eta \ge 0,\delta \ge 0:$$F^{\prime }\left(u_0\right)\ne 0,A_0\ne 0,$$\parallel F^{\prime }{\left(u_0\right)}^{-1}F\left(u_0\right)\parallel \le \eta$ and $\parallel A_0^{-1}{F}^{\prime }\left(u_0\right)\parallel \le \delta$;

(H2)

$L_0>0:$$\parallel F^{\prime }{\left(u_0\right)}^{-1}(F^{\prime }\left(v\right)-F^{\prime }\left(u_0\right))\parallel \le L_0\parallel v-u_0\parallel$ for all $v\in \Omega .$ Set ${\Omega }_0=U(u_0,\frac{1}{L_0})\cap \Omega$;

(H3)

$L>0,L_1>0,$$L_2>0:$$\parallel F^{\prime }{\left(u_0\right)}^{-1}(F^{\prime }\left(v\right)-F^{\prime }\left(u\right))\parallel \le L\parallel v-u\parallel ,$

$$\parallel F^{\prime }{\left(u_0\right)}^{-1}{F}^{\prime }\left(v\right)\parallel \le L_1$$

and

$$\parallel A_0^{-1}{F}^{\prime }\left(v\right)\parallel \le L_2,$$

for all $v,w\in {\Omega }_0$;

(H4)

The conditions in Lemma 1 or in Lemma 3 are true;

(H5)

$U[u_0,t^{*}]\subset \Omega .$

Theorem 1.

Assume conditions H hold. Then, KM is well defined in $U(u_0,t^{*}),$ lies in $U(u_0,t^{*}),$ for all $n=0,1,2,\dots$ and converges to a solution $x^{*}\in U[u_0,t^{*}]$ of Equation (1), so

$$\parallel v_m-u_m\parallel \le s_m-t_m$$

(29)

and

$$\parallel u_{m+1}-v_m\parallel \le t_{m+1}-s_m.$$

(30)

Proof. 

We have by $\left\{t_n\right\}$ and (H1)

$$\parallel v_0-u_0\parallel =\parallel F^{\prime }{\left(u_0\right)}^{-1}F\left(u_0\right)\parallel \le \eta =s_0-t_0<t^{*}.$$

So, (29) is true if $m=0$ and $v_0\in U(u_0,t^{*}).$ Pick $u\in U(u_0,t^{*}).$ By (H1), (H2) and $t^{*},$ then

$$\parallel F^{\prime }{\left(u_0\right)}^{-1}(F^{\prime }\left(u_0\right)-F^{\prime }\left(u\right))\parallel \le L_0\parallel u_0-u\parallel \le L_0{t}^{*}<1.$$

That is $F^{\prime }\left(u\right)\ne 0$ with

$$\parallel F^{\prime }{\left(u\right)}^{-1}{F}^{\prime }\left(u_0\right)\parallel \le \frac{1}{1-L_0\parallel u-u_0\parallel }.$$

(31)

By the Banach lemma on functions [11,12,13], iteration $u_1$ is well-defined. Suppose $u_k,v_k\in U(u_0,t^{*}).$ Then, we can write

$$u_{k+1}-v_k=A_k^{-1}(F\left(u_k\right)+\gamma F\left(v_k\right))F^{\prime }{\left(u_k\right)}^{-1}F\left(v_k\right).$$

(32)

By (H1), (H3), we get

$$\begin{array}{ccc}\hfill \parallel A_0^{-1}(A_k-A_0)\parallel & \le & \parallel A_0^{-1}(F\left(u_0\right)-F\left(u_k\right))\parallel \hfill \\ \hfill & & +|\gamma -2|\parallel A_0^{-1}(F\left(v_0\right)-F\left(v_k\right))\parallel \hfill \\ \hfill & \le & \parallel {\int }_0^1{A}_0^{-1}{F}^{\prime }(u_0+\theta (u_k-u_0))d\theta \parallel \parallel u_k-v_0\parallel \hfill \\ \hfill & & +|\gamma -2|\parallel {\int }_0^1{A}_0^{-1}(F^{\prime }(v_0+\theta (v_k-v_0))d\theta \parallel \parallel v_k-v_0\parallel \hfill \\ \hfill & \le & L_2(\parallel u_k-u_0\parallel +|\gamma -2|\parallel v_k-v_0\parallel )\hfill \\ \hfill & \le & {\overline{p}}_k\le p_k=L_2(t_k+|\gamma -2|(s_k-\eta ))<1,\hfill \end{array}$$

so $A_k\ne O$ and

$$\parallel A_k^{-1}{A}_0\parallel \le \frac{1}{1-p_k}.$$

(33)

Then, by (H3), (3), (31) (for $u=u_0$), (32) and (33), we obtain

$$\begin{array}{ccc}\hfill \parallel u_{k+1}-v_k\parallel & \le & \parallel A_k^{-1}{A}_0\parallel [\parallel A_0^{-1}F\left(u_k\right)\parallel +\left|\gamma \right|\parallel A_0^{-1}{F}^{\prime }\left(u_0\right)\parallel \hfill \\ \hfill & & \times \parallel F^{\prime }{\left(u_0\right)}^{-1}F\left(v_k\right)\parallel ]\parallel F^{\prime }{\left(u_k\right)}^{-1}{F}^{\prime }\left(u_0\right)\parallel \parallel F^{\prime }{\left(u_0\right)}^{-1}F\left(v_k\right)\parallel \hfill \\ \hfill & \le & \frac{[L_2((s_k-t_k)+\delta \left|\gamma \right|\frac{L}{2}{(s_k-t_k)}^2]\frac{L}{2}{(s_k-t_k)}^2}{(1-p_k)(1-L_0{t}_k)}\hfill \\ \hfill & =& t_{k+1}-s_k,\hfill \end{array}$$

(34)

so (30) holds, where we also used that (29) and (30) hold for all k smaller than $n-1.$ We also get

$$\begin{array}{ccc}\hfill \parallel F^{\prime }{\left(u_0\right)}^{-1}F\left(u_k\right)\parallel & \le & \parallel F^{\prime }{\left(u_0\right)}^{-1}{F}^{\prime }\left(u_k\right)(v_k-u_k)\parallel \hfill \\ \hfill & \le & L_1\parallel v_k-u_k\parallel \le L_1(s_k-t_k),\hfill \end{array}$$

(35)

$$\begin{array}{ccc}\hfill F\left(v_k\right)& =& F\left(v_k\right)-F\left(u_k\right)+F\left(u_k\right)\hfill \\ \hfill & =& {\int }_0^1{F}^{\prime }(u_k+\theta (v_k-u_k))d\theta (v_k-u_k)-F^{\prime }\left(u_k\right)(v_k-u_k),\hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill \parallel F^{\prime }{\left(u_0\right)}^{-1}F\left(v_k\right)\parallel & \le & \frac{L}{2}{(s_k-t_k)}^2\hfill \end{array}$$

(36)

We also have

$$\parallel u_{k+1}-u_0\parallel \le \parallel u_{k+1}-v_k\parallel +\parallel v_k-u_0\parallel \le (t_{k+1}-s_k)+(s_k-t_0)=t_{K+1}<t^{*},$$

so $u_{k+1}\in U(u_0,t^{*}).$ Then, we write

$$\begin{array}{ccc}\hfill F\left(u_{k+1}\right)& =& F\left(u_{k+1}\right)-F\left(u_k\right)+F\left(u_k\right)\hfill \\ \hfill & =& F\left(u_{k+1}\right)-F\left(u_k\right)-F^{\prime }\left(u_k\right)(v_k-u_k)\hfill \\ \hfill & =& F\left(u_{k+1}\right)-F\left(u_k\right)-F^{\prime }\left(u_k\right)(u_{k+1}-u_k)+F^{\prime }\left(u_k\right)(u_{k+1}-v_k)\hfill \\ \hfill & =& {\int }_0^1(F^{\prime }(u_k+\theta (u_{k+1}-u_k))d\theta -F^{\prime }\left(u_k\right))(u_{k+1}-u_k)\hfill \\ \hfill & & +F^{\prime }\left(u_k\right)(u_{k+1}-v_k).\hfill \end{array}$$

(37)

By (H3), we get

$$\begin{array}{ccc}\hfill \parallel F^{\prime }{\left(u_0\right)}^{-1}F\left(u_{k+1}\right)\parallel & \le & \frac{L}{2}\parallel u_{k+1}-u_k{\parallel }^2+L_1\parallel u_{k+1}-v_k\parallel \hfill \\ \hfill & \le & \frac{L}{2}{(t_{k+1}-t_k)}^2+L_1(t_{k+1}-s_k).\hfill \end{array}$$

(38)

Then, by the first substep of KM

$$\begin{array}{ccc}\hfill \parallel v_{k+1}-u_{k+1}\parallel & \le & \parallel F^{\prime }{\left(u_{k+1}\right)}^{-1}{F}^{\prime }\left(u_0\right)\parallel \parallel F^{\prime }{\left(u_0\right)}^{-1}F\left(u_{k+1}\right)\parallel \hfill \\ \hfill & \le & \frac{\frac{L}{2}{(t_{k+1}-t_k)}^2+L_1(t_{k+1}-s_k)}{1-L_0{t}_{k+1}}\hfill \\ \hfill & =& s_{k+1}-t_{k+1},\hfill \end{array}$$

(39)

and

$$\begin{array}{ccc}\hfill \parallel v_{k+1}-u_0\parallel & \le & \parallel v_{k+1}-u_{k+1}\parallel +\parallel u_{k+1}-u_0\parallel \hfill \\ & \le & s_{k+1}-t_{k+1}+t_{k+1}-t_0=s_{k+1}<t^{*}.\hfill \end{array}$$

Therefore, (29) holds and $v_{k+1}\in U[u_0,t^{*}].$ The induction is finished. So, $\left\{u_k\right\}$ is Cauchy in $T.$ Hence, there exists $x^{*}\in U[u_0,t^{*}]$ such that ${lim}_{k⟶\infty }{x}_n=x^{*}.$ By letting k approach ∞ in (35), $F\left(x^{*}\right)=0.$ □

Notice that $\frac{1}{L_0}$ under conditions of Lemma 1 or $\frac{(1+\alpha )\eta }{1-\alpha }$ under conditions of Lemma 3 provided in closed form may be used for $t^{*}$ in Theorem 1.

Proposition 1.

Suppose

(1) 

The point $b\in U[u_0,r_0]\subset \Omega$ is a solution of Equation (1) with $F^{\prime }\left(b\right)\ne 0,$ and condition (H2) holds;

(2) 

Point $r\ge r_0$ exists:

$$L_0(r+r_0)<2.$$

(40)

Set ${\Omega }_1=U[u_0,r]\cap \Omega .$ Then, b uniquely solves Equation (1) in ${\Omega }_1.$

Proof. 

Let $\xi \in {\Omega }_1$ satisfy $F\left(\xi \right)=0.$ Set $B={\int }_0^1{F}^{\prime }(b+q(\xi -b))dq.$ Then, by (H2) and (40), we obtain in turn that

$$\begin{array}{ccc}\hfill \parallel F^{\prime }{\left(u_0\right)}^{-1}(B-F^{\prime }\left(u_0\right))\parallel & \le & L_0{\int }_0^1((1-q)\parallel u_0-b\parallel +q\parallel u_0-\xi \parallel )dq\hfill \\ & \le & \frac{L_0}{2}(r_0+r)<1.\hfill \end{array}$$

Therefore, $\xi =b$ follows from $B\ne 0$ and $B(\xi -b)=F\left(\xi \right)-F\left(b\right)=0-0=0.$ □

3. Local Convergence

Set $K_0,K$, and $K_1$ to be positive parameters. Define function $g_1:[0,\frac{1}{K_0})⟶\mathbb{R}$ by

$$g_1\left(t\right)=\frac{Kt}{2(1-K_{0}t)}.$$

Notice that

$${\rho }_0=\frac{2}{2K_0+K}<\frac{1}{K_0}$$

(41)

is a radius of convergence for Newton’s method provided by us in [11,12,13]. This point ${\rho }_0$ also solves the equation

$$G_1\left(t\right)=g_1\left(t\right)-1=0.$$

Develop $q:[0,\frac{1}{K_0})⟶\mathbb{R},Q:[0,\frac{1}{K_0})⟶\mathbb{R}$ by

$$q\left(t\right)=|\gamma -2|K_1{g}_1\left(t\right)+\frac{K}{2}t$$

and

$$Q\left(t\right)+1=q\left(t\right).$$

Then, we have $Q\left(0\right)=-1$ and $Q\left({\rho }_0\right)=\frac{K}{2}{\rho }_0+|\gamma -2|K_1>0.$ The intermediate value theorem assures Q has zeros in $(0,{\rho }_0).$ Let ${\rho }_Q$ stand for the smallest zero in $(0,{\rho }_0).$ Define functions $g_2:[0,{\rho }_Q)⟶\mathbb{R}$ and $G_2:[0,{\rho }_Q)⟶\mathbb{R}$ by

$$g_2\left(t\right)=g_1\left(t\right)\left(1+\frac{K_1^2(1+|\gamma |g_1\left(t\right))}{(1-q\left(t\right))(1-K_{0}t)}\right)$$

and

$$G_2\left(t\right)=g_2\left(t\right)-1.$$

It follows $G_2\left(0\right)=-1$ and $G_2\left(t\right)⟶\infty$ as $t⟶{\rho }_Q^{-}.$ Let $\rho$ be the smallest such zero of $G_2$ on $(0,{\rho }_Q).$ Set $I=[0,\rho ).$ Then, the definition of $\rho$ implies that for all $t\in I$

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

(42)

$$0\le q\left(t\right)<1$$

(43)

and

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

(44)

The local convergence of KM uses conditions (C). Suppose that there is

(C1)

a solution $x^{*}\in \Omega$ of Equation (1) with $F^{\prime }\left(x^{*}\right)\ne 0$;

(C2)

$K_0>0,$ so that

$$\parallel F^{\prime }{\left(x^{*}\right)}^{-1}(F^{\prime }\left(x^{*}\right)-F^{\prime }\left(w\right))\parallel \le K_0\parallel x^{*}-w\parallel$$

for all $w\in \Omega .$ Define ${\Omega }_2=U(x^{*},\frac{1}{K_0})\cap \Omega$;

(C3)

There exist $K>0,K_1>0$ such that

$$\parallel F^{\prime }{\left(x^{*}\right)}^{-1}(F^{\prime }\left(w\right)-F^{\prime }\left(v\right))\parallel \le K\parallel w-v\parallel$$

and

$$\parallel F^{\prime }{\left(x^{*}\right)}^{-1}{F}^{\prime }\left(v\right)\parallel \le K_1\parallel x^{*}-v\parallel$$

for all $v,w\in {\Omega }_2$;

(C4)

$U[x^{*},\rho ]\subset \Omega .$

Theorem 2.

Choose $u_0\in U(x^{*},\rho )-\left\{x^{*}\right\}.$ Then, under conditions (C), sequence $\left\{u_n\right\}$ generated by KM converges to $x^{*},$ so that

$$\parallel v_n-x^{*}\parallel \le g_1\left(d_n\right)d_n\le d_n<\rho$$

(45)

and

$$d_{n+1}\le g_2\left(d_n\right)d_n\le d_n,$$

(46)

where $d_n=\parallel u_n-x^{*}\parallel ,$ and the functions $g_1$, $g_2$ were previously defined.

Proof. 

Pick $z\in U(x^{*},\rho )-\left\{x^{*}\right\}.$ Then, by (C1) and (C2)

$$\parallel F^{\prime }{\left(x^{*}\right)}^{-1}(F^{\prime }\left(z\right)-F^{\prime }\left(x^{*}\right))\parallel \le K_0\parallel z-x^{*}\parallel \le K_0\rho <1.$$

(47)

So, we have $F^{\prime }\left(z\right)\ne 0$ and

$$\parallel F^{\prime }{\left(z\right)}^{-1}{F}^{\prime }\left(x^{*}\right)\parallel \le \frac{1}{1-K_0\parallel z-x^{*}\parallel }.$$

(48)

If $z=u_0,$ we see that iterate $v_0$ is well-defined by KM for $n=0.$ Moreover, we can write

$$\begin{array}{ccc}\hfill v_0-x^{*}& =& u_0-x^{*}-F^{\prime }{\left(u_0\right)}^{-1}F\left(u_0\right)\hfill \\ \hfill & =& F^{\prime }{\left(u_0\right)}^{-1}\left[{\int }_0^1(F^{\prime }(x^{*}+\theta (u_0-x^{*}))-F^{\prime }\left(u_0\right))d\theta (u_0-x^{*})\right].\hfill \end{array}$$

(49)

By (42), (48) (for $z=u_0$), (C3) and (46), we have in turn that

$$\begin{array}{ccc}\hfill \parallel v_0-x^{*}\parallel & \le & \frac{K\parallel u_0-x^{*}{\parallel }^2}{2(1-K)\parallel u_0-x^{*}\parallel }\hfill \\ \hfill & =& g_1(\parallel u_0-x^{*}\parallel )\parallel u_0-x^{*}\parallel \hfill \\ \hfill & \le & \parallel u_0-x^{*}\parallel <\rho .\hfill \end{array}$$

(50)

Hence, iterate $v_0\in U(x^{*},\rho )$ and (42) holds if $n=0.$ Next, we show that $A_0\ne 0.$ If $u_0\ne x^{*},$ we obtain by (C1), (C2), and (46)

$$\begin{array}{ccc}& & \parallel \left(F^{\prime }\left(x^{*}\right){(u_0-x^{*})}^{-1}\right)[A_0-F^{\prime }\left(x^{*}\right)(u_0-x^{*})]\parallel \hfill \\ & \le & \frac{1}{\parallel u_0-x^{*}\parallel }[\parallel F^{\prime }{\left(x^{*}\right)}^{-1}(F\left(u_0\right)-F\left(x^{*}\right)-F^{\prime }\left(x^{*}\right)(u_0-x^{*}))\parallel \hfill \\ & & +|\gamma -2|\parallel F^{\prime }{\left(x^{*}\right)}^{-1}F\left(v_0\right)\parallel ]\hfill \\ & \le & \frac{1}{\parallel u_0-x^{*}\parallel }[\frac{K}{2}\parallel u_0-x^{*}{\parallel }^2+|\gamma -2|K_1\parallel v_0-x^{*}\parallel ]\hfill \\ & \le & \frac{K}{2}\parallel u_0-x^{*}\parallel +|\gamma -2|K_1{g}_1(\parallel u_0-x^{*}\parallel )\hfill \\ & =& q(\parallel u_0-x^{*}\parallel )\le q\left(\rho \right)<1.\hfill \end{array}$$

It follows that $A_0\ne 0,$ and

$$\parallel A_0^{-1}{F}^{\prime }\left(x^{*}\right)\parallel \le \frac{1}{\parallel u_0-x^{*}\parallel (1-q(\parallel u_0-x^{*}\parallel \left)\right)}.$$

(51)

Then, using (44), (C3), (48), (50), and (51)

$$\begin{array}{ccc}\hfill \parallel u_1-x^{*}\parallel & \le & \parallel v_0-x^{*}\parallel +\parallel A_0^{-1}{F}^{\prime }\left(x^{*}\right)\parallel (\parallel F^{\prime }{\left(x^{*}\right)}^{-1}F\left(u_0\right)\parallel \hfill \\ \hfill & & +\left|\gamma \right|\parallel F^{\prime }{\left(x^{*}\right)}^{-1}F\left(v_0\right)\parallel )\parallel F^{\prime }{\left(u_0\right)}^{-1}{F}^{\prime }\left(x^{*}\right)\parallel \parallel F^{\prime }{\left(x^{*}\right)}^{-1}F\left(v_0\right)\parallel \hfill \\ \hfill & \le & \left(1+\frac{K_1^2(\parallel u_0-x^{*}\parallel )+\left|\gamma \right|\parallel v_0-x^{*}\parallel }{\parallel u_0-x^{*}\parallel (1-q(\parallel u_0-x^{*}\parallel \left)\right)(1-K_0\parallel u_0-x^{*}\parallel )}\right)\parallel v_0-x^{*}\parallel \hfill \\ \hfill & \le & \left(1+\frac{K_1^2(\parallel u_0-x^{*}\parallel )+\left|\gamma \right|g_1(\parallel u_0-x^{*}\parallel )}{(1-q(\parallel u_0-x^{*}\parallel \left)\right)(1-K_0\parallel u_0-x^{*}\parallel )}\right)g_1(\parallel u_0-x^{*}\parallel )\parallel u_0-x^{*}\parallel \hfill \\ \hfill & =& g_2(\parallel u_0-x^{*}\parallel )\parallel u_0-x^{*}\parallel \le \parallel u_0-x^{*}\parallel <\rho .\hfill \end{array}$$

(52)

That is iterate $u_1\in U(u_0,x^{*})$ and (43) holds for $n=0.$ Simply switch $u_0,v_0,u_1$ by $u_k,v_k,u_{k+1}$ in the above calculations to terminate the induction for (42) and (43). Then, it follows from the estimate

$$\parallel u_{k+1}-x^{*}\parallel \le \lambda \parallel u_k-x^{*}\parallel <\rho ,$$

(53)

where $\lambda =g_2(\parallel u_0-x^{*}\parallel )\in [0,1)$. We conclude ${lim}_{k⟶\infty }{u}_k=x^{*}$ and $u_{k+1}\in U(x^{*},\rho ).$ □

A uniqueness of the solution result follows next.

Proposition 2.

Suppose

(1) 

Element $\lambda \in U(x^{*},{\rho }_0)\subseteq \Omega$ solves Equation (1), $F\left(\lambda \right)=0$, and (C2) holds;

(2) 

There exists ${\rho }^{*}\ge {\rho }_0$ such that

$$K_0{\rho }^{*}<2.$$

(54)

Set ${\Omega }_3=U[\lambda ,{\rho }^{*}]\cap \Omega .$ Then, element λ uniquely solves Equation (1) in ${\Omega }_3$.

Proof. 

Let $\overline{x}\in {\Omega }_3$ with $F\left(\overline{x}\right)=0.$ Set $E={\int }_0^1{F}^{\prime }(\lambda +\tau (\overline{x}-\lambda ))d\tau .$ Then, using (C2) and (54), we get in turn that

$$\begin{array}{ccc}\hfill \parallel F^{\prime }{\left(\lambda \right)}^{-1}(E-F^{\prime }\left(\lambda \right))\parallel & \le & K_0{\int }_0^1(1-\tau )\parallel \lambda -\overline{x}\parallel d\tau \hfill \\ & \le & \frac{K_0}{2}{\rho }^{*}<1.\hfill \end{array}$$

Hence, $\overline{x}=\lambda$ follows from $E\ne 0$ and $E(\lambda -\overline{x})=F\left(\lambda \right)-F\left(\overline{x}\right)=0-0=0.$ □

Next, the fourth-order convergence is shown using only the first derivative. Suppose:

$$\parallel A_n^{-1}{F}^{\prime }\left(z\right)\parallel \le \omega$$

(55)

and

$$\parallel F^{\prime }{\left(x\right)}^{-1}(F^{\prime }\left(x\right)-F^{\prime }\left(y\right))\parallel \le {\omega }_0\parallel x-y\parallel$$

(56)

hold for all $x,y,z\in \Omega ,$ for some constants $\omega >0$ and ${\omega }_0>0.$ Further, suppose

$$\frac{\theta {\omega }_0^2}{2}(\frac{3}{2}+\frac{{\omega }_0}{4}+\frac{\left|\gamma \right|{\omega }_0}{2}(1+\frac{{\omega }_0}{4}))-1>0.$$

(57)

Let $\psi \left(t\right)=\phi \left(t\right)-1=0,$ where $\phi \left(t\right)=\frac{\omega {\omega }_0^2}{2}(\frac{3}{2}+\frac{{\omega }_0}{4}t+\frac{\left|\gamma \right|{\omega }_0}{2}(t+\frac{{\omega }_0}{4}{t}^2)t^3.$ Then, $\psi \left(0\right)=-1<0$ and $\psi \left(1\right)=\frac{\theta {\omega }_0^2}{2}(\frac{3}{2}+\frac{{\omega }_0}{4}+\frac{\left|\gamma \right|{\omega }_0}{2}(1+\frac{{\omega }_0}{4}))-1>0.$ Hence, by the intermediate value theorem, $\psi \left(t\right)=0$ has positive solutions. Let $r_o$ be the smallest such solution.

Theorem 3.

Suppose conditions (55)–(57) hold. Then, sequence $\left\{u_n\right\}$ given in (2) is convergent to $x^{*}$ with order four, i.e.,

$$\parallel u_{n+1}-x^{*}\parallel \le \varrho \left(r_o\right)d_n^4,$$

where $\varrho \left(r_o\right)=\frac{\theta {\omega }_0^2}{2}(\frac{3}{2}+\frac{{\omega }_0}{4}{r}_0+\frac{\left|\gamma \right|{\omega }_0}{2}(r_o+\frac{{\omega }_0}{4}{r}_o^2).$

Proof. 

The first substep of (2) and (56) gives

$$\begin{array}{ccc}\hfill \parallel v_n-x^{*}\parallel & \le & \parallel F^{\prime }{\left(u_n\right)}^{-1}{\int }_0^1[F^{\prime }\left(u_n\right)-F^{\prime }(x^{*}+\theta (u_n-x^{*}))]d\theta (u_n-x^{*})\parallel \hfill \\ & \le & \frac{{\omega }_0}{2}{d}_n^2.\hfill \end{array}$$

Note

$$\begin{array}{ccc}\hfill u_{n+1}-x^{*}& =& v_n-x^{*}-A_n^{-1}(F\left(u_n\right)+\gamma F\left(v_n\right))F^{\prime }{\left(u_n\right)}^{-1}F\left(v_n\right)\hfill \\ & =& A_n^{-1}[A_n-(F\left(u_n\right)+\gamma F\left(v_n\right))F^{\prime }{\left(u_n\right)}^{-1}{\int }_0^1{F}^{\prime }(x^{*}+\theta (v_n-x^{*}))d\theta ](v_n-x^{*})\hfill \\ & =& A_n^{-1}F\left(u_n\right)F^{\prime }{\left(u_n\right)}^{-1}[F^{\prime }\left(u_n\right)-{\int }_0^1{F}^{\prime }(x^{*}+\theta (v_n-x^{*}))d\theta ](v_n-x^{*})\hfill \\ & & +A_n^{-1}F\left(v_n\right)F^{\prime }{\left(u_n\right)}^{-1}\gamma [F^{\prime }\left(u_n\right)-{\int }_0^1{F}^{\prime }(x^{*}+\theta (v_n-x^{*}))d\theta ](v_n-x^{*})\hfill \\ & & -2A_n^{-1}F\left(v_n\right)(v_n-x^{*}),\hfill \end{array}$$

so, since $F\left(u_n\right)={\int }_0^1{F}^{\prime }(x^{*}+u(u_n-x^{*}))du(u_n-x^{*})$ and $F\left(v_n\right)={\int }_0^1{F}^{\prime }(x^{*}+u(v_n-x^{*}))du(v_n-x^{*})$

$$\begin{array}{ccc}\hfill d_{n+1}& \le & \parallel A_n^{-1}{\int }_0^1{F}^{\prime }(x^{*}+u(u_n-x^{*}))du{F}^{\prime }{\left(u_n\right)}^{-1}[F^{\prime }\left(u_n\right)\hfill \\ & & -{\int }_0^1{F}^{\prime }(x^{*}+\theta (v_n-x^{*}))d\theta ](u_n-x^{*})(v_n-x^{*})\parallel \hfill \\ & & +\parallel A_n^{-1}{\int }_0^1{F}^{\prime }(x^{*}+u(v_n-x^{*}))du{F}^{\prime }{\left(u_n\right)}^{-1}\gamma [F^{\prime }\left(u_n\right)\hfill \\ & & -{\int }_0^1{F}^{\prime }(x^{*}+\theta (v_n-x^{*}))d\theta ]{(v_n-x^{*})}^2\parallel \hfill \\ & & +2\parallel A_n^{-1}{\int }_0^1{F}^{\prime }(x^{*}+u(v_n-x^{*}))du{(v_n-x^{*})}^2\parallel \hfill \end{array}$$

$$\begin{array}{ccc}& \le & \parallel {\int }_0^1{A}_n^{-1}{F}^{\prime }(x^{*}+u(u_n-x^{*}))du\hfill \\ & & {\int }_0^1{F}^{\prime }{\left(u_n\right)}^{-1}[F^{\prime }\left(u_n\right)-F^{\prime }(x^{*}+\theta (v_n-x^{*}))d\theta ](u_n-x^{*})(v_n-x^{*})\parallel \hfill \\ & & +\left|\gamma \right|\parallel {\int }_0^1{A}_n^{-1}{F}^{\prime }(x^{*}+u(v_n-x^{*}))du\hfill \\ & & +{\int }_0^1{F}^{\prime }{\left(u_n\right)}^{-1}[F^{\prime }\left(u_n\right)-F^{\prime }(x^{*}+\theta (v_n-x^{*}))d\theta ]{(v_n-x^{*})}^2\parallel \hfill \\ & & +2\parallel {\int }_0^1{A}_n^{-1}{F}^{\prime }(x^{*}+u(v_n-x^{*}))du{(v_n-x^{*})}^2\parallel .\hfill \end{array}$$

Therefore, (55) and (56) give

$$\begin{array}{ccc}\hfill d_{n+1}& \le & \omega {\omega }_0\left[d_n+\frac{\parallel v_n-x^{*}\parallel }{2}\right]d_n\parallel v_n-x^{*}\parallel \hfill \\ & & +\left|\gamma \right|\omega {\omega }_0\left[d_n+\frac{\parallel v_n-x^{*}\parallel }{2}\right]{\parallel v_n-x^{*}\parallel }^2\hfill \\ & & +2\omega \parallel v_n-x^{*}{\parallel }^2\hfill \\ & \le & \omega \frac{{\omega }_0^2}{2}\left[1+\frac{{\omega }_0}{4}{d}_n\right]d_n^4\hfill \\ & & +\left|\gamma \right|\omega \frac{{\omega }_0^3}{4}\left[d_n+\frac{{\omega }_0}{4}{d}_n^2\right]d_n^4\hfill \\ & & +\frac{{\omega }_0^2}{4}\omega d_n^4\hfill \\ & \le & \phi \left(d_n\right)d_n\hfill \\ & \le & \varrho \left(r_o\right)d_n^4.\hfill \end{array}$$

□

4. Numerical Example

We verify convergence criteria using KM.

Example 1.

Let us consider a scalar function F defined on the set $\Omega =U[u_0,1-s]$ for $s\in (0,\frac{1}{2})$ by

$$F\left(x\right)=x^3-s.$$

Choose $\gamma =2$ and $u_0=1.$ Then, we obtain the estimates $\eta =\frac{1-s}{3},$

$$\begin{array}{ccc}\hfill |F^{\prime }{\left(u_0\right)}^{-1}(F^{\prime }\left(x\right)-F^{\prime }\left(u_0\right))|& =& |x^2-u_0^2|\hfill \\ & \le & |x+u_0\left|\right|x-u_0|\le (|x-u_0|+2|u_0\left|\right)|x-u_0|\hfill \\ & =& (1-s+2)|x-u_0|=(3-s)|x-u_0|,\hfill \end{array}$$

for each $x\in \Omega ,$ so $L_0=3-s,$${\Omega }_0=U(u_0,\frac{1}{L_0})\cap \Omega =U(u_0,\frac{1}{L_0}),$

$$\begin{array}{ccc}\hfill |F^{\prime }{\left(u_0\right)}^{-1}(F^{\prime }\left(y\right)-F^{\prime }\left(x\right)|& =& |y^2-x^2|\hfill \\ & \le & |y+x||y-x|\le \left(\right|y-u_0+x-u_0+2u_0\left)\right||y-x|\hfill \\ & =& \left(\right|y-u_0|+|x-u_0|+2|u_0\left|\right)|y-x|\hfill \\ & \le & (\frac{1}{L_0}+\frac{1}{L_0}+2)|y-x|=2(1+\frac{1}{L_0})|y-x|,\hfill \end{array}$$

for each $x,y\in \Omega ,$ and so $L=2(1+\frac{1}{L_0}),$

$$\begin{array}{ccc}\hfill \left|F^{\prime }{\left(u_0\right)}^{-1}(F^{\prime }\left(y\right)-F^{\prime }\left(x\right)\right|& =& \left(\right|y-u_0|+|x-u_0|+2|u_0\left|\right)|y-x|\hfill \\ & \le & (1-s+1-s+2)|y-x|=2(2-s)|y-x|,\hfill \end{array}$$

for each $x,y\in \Omega ,$ so $L_1={(2-s)}^2$ and $L_2=\frac{3{(2-s)}^2}{F\left(1\right)+|\gamma -2|{(1-\frac{F\left(1\right)}{F^{\prime }\left(1\right)})}^2-s}.$

Then, for $s=0.95,\gamma =0.5$, we have $\frac{1}{L_0}=0.4878$.

According to the information taken from

Table 1. Sequence (3) and condition (4).

n	1	2	3	4	5	6
$p_n$	0	0.1004	0.1033	0.1033	0.1033	0.1033
$t_n$	0.0167	0.0172	0.0172	0.0172	0.0172	0.0172
$s_n$	0.0167	0.0172	0.0172	0.0172	0.0172	0.0172

, the conditions of Lemma 1 hold. Consequently, the sequence converges and the interval of initial points has been further extended.

Example 2.

Set function $F:I=[-1,1]⟶\mathbb{R}$ as

$$F\left(x\right)=e^x-1.$$

Notice that $x^{*}=0$ solves equation $F\left(x\right)=0.$ Choose $\gamma =2.$ Then, conditions of Theorem 3 hold for $\omega ={\omega }_0=e^2.$ Then, the radius is $r_o=0.1381.$

Example 3.

The example used in the introduction gives $\omega ={\omega }_0=96.6629073.$ Then, for $\gamma =2,$ the radius is

$$r_o=0.0092.$$

Recall that it was shown in the Introduction that earlier articles cannot be used to solve this problem. The method used is a specialization of KM for $\gamma =2.$

5. Conclusions

In this article, the extension of KM is presented. The convergence of KM has been shown by assuming the existence of a fifth derivative which was not considered before. This observation holds true for other high-convergence order methods such as Traub’s and Jarratt’s method. Other such methods can be found in [1,2,3,4,5,6,7,8] and the references therein. Therefore, these results cannot assure convergence. However, these methods may converge. Other concerns involve the absence of error estimates or uniqueness results that can be computed. This is our motivation for presenting a convergence analysis based on the first derivative used in KM. The generality of the technique allows its usage in other methods mentioned previously. This can be a fruitful direction of future research.