# Three-Step Derivative-Free Method of Order Six

^{1}

^{2}

^{3}

^{4}

^{*}

## Abstract

**:**

## 1. Introduction

**Motivational Limitations**

- (a)
- The existence of high-order derivatives not present in the method.
- (b)
- $B=\mathbb{R}.$
- (c)
- A priori error analysis is not provided for $\parallel {x}_{n}-{x}^{*}\parallel .$
- (d)
- Results on the isolation of the solution case not present, either.
- (e)
- The more challenging and important semi-local analysis (SLA) is not given.

**Novelty of the paper**

- (a)′
- (b)′
- The analysis of convergence is carried out in the more general setting of Banach space valued operators, not only on $\mathbb{R}$.
- (c)′
- An a priori error analysis is provided to determine upper error bounds on $\parallel {x}_{n}-{x}^{*}\parallel .$ This allows the determination of the number of iterations in advance to be carried out in order to achieve a predecided error tolerance.
- (d)′
- Computational results on the isolation of solutions are developed based on generalized continuity conditions [3,4,5,6,7,8] on the divided differences (see conditions $\left({C}_{1}\right)$ and $\left({C}_{2}\right)$ in Section 2).It is also worth noting that the usual conditions in the convergence analysis of this and the other methods mentioned in the aforementioned references require that ${F}^{\prime}\left({x}^{*}\right)$ is invertible. That is, ${x}^{*}$ must be a simple solution of equation $F\left(x\right)=0,$ although derivative ${F}^{\prime}\left({x}^{*}\right)$ is not present in Method (3). Thus, the earlier results in [13] cannot assure the convergence of Method (3) in cases operator F is a nondifferentiable operator, although the method may converge. But conditions $\left({C}_{1}\right)$ and $\left({C}_{2}\right)$ under our approach do not require ${F}^{\prime}\left({x}^{*}\right)$ to exist or be invertible.Thus, our approach can be utilized to solve equations like (1) in cases the operator is nondifferentiable.
- (e)′

**Definition**

**1.**

**Definition**

**2.**

## 2. Local Analysis

- $\left({C}_{1}\right)$
- There exist continuous as well as nondecreasing functions ${f}_{1}:M=[0,\infty )\to M,$${f}_{2}:M\to M,$ and ${\phi}_{0}:M\times M\to M$ such that equation$${\phi}_{0}({f}_{1}\left(t\right),{f}_{2}\left(t\right))-1=0$$
- $\left({C}_{2}\right)$
- There exists an invertible linear operator L and ${x}^{*}\in \mathrm{\Omega}$ with $F\left({x}^{*}\right)=0$ so that for $x\in \mathrm{\Omega}$,$$\begin{array}{cc}\hfill \phantom{\rule{1.em}{0ex}}& \phantom{\rule{4pt}{0ex}}\left|\right|{L}^{-1}([{u}_{1},{u}_{2};F]-L)\left|\right|\le {\phi}_{0}\left(\right||{u}_{1}-{x}^{*}\left|\right|,\left|\right|{u}_{2}-{x}^{*}\left|\right|),\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& \phantom{\rule{4pt}{0ex}}\left|\right|{u}_{1}-{x}^{*}\left|\right|\le {f}_{1}\left(\right|\left|d\right|\left|\right),\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& \phantom{\rule{4pt}{0ex}}\left|\right|{u}_{2}-{x}^{*}\left|\right|\le {f}_{2}\left(\right|\left|d\right|\left|\right),\hfill \end{array}$$
- $\left({C}_{3}\right)$
- There exist continuous as well as nondecreasing functions ${\phi}_{1}:{M}_{1}\to M,$$\phi :{M}_{1}\times {M}_{1}\times {M}_{1}\to M,$ and ${\phi}_{2}:{M}_{1}\times {M}_{1}\times {M}_{1}\times {M}_{1}\to M$ for each $x\in {\mathrm{\Omega}}_{0}$,$$\begin{array}{cc}\hfill \phantom{\rule{1.em}{0ex}}& \phantom{\rule{4pt}{0ex}}\left|\right|{L}^{-1}([x,{x}^{*};F]-L)\left|\right|\le {\phi}_{1}\left(\right|\left|d\right|\left|\right),\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& \phantom{\rule{4pt}{0ex}}\left|\right|{L}^{-1}([{u}_{1},{u}_{2};F]-[x,{x}^{*};F])\left|\right|\le \phi \left(\right|\left|d\right||,||{u}_{1}-{x}^{*}\left|\right|,\left|\right|{u}_{2}-{x}^{*}\left|\right|),\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& \phantom{\rule{4pt}{0ex}}\left|\right|{L}^{-1}([{u}_{1},{u}_{2};F]-[y,x;F])\left|\right|\le {\phi}_{2}\left(\right|\left|d\right||,||y-{x}^{*}\left|\right|,\left|\right|{u}_{1}-{x}^{*}\left|\right|,\left|\right|{u}_{2}-{x}^{*}\left|\right|).\hfill \end{array}$$
- $\left({C}_{4}\right)$
- Equation ${g}_{1}\left(t\right)-1=0$ has a PSS denoted by ${r}_{1}\in M-\left\{0\right\},$ where ${g}_{1}:M\times M\times M\u27f6M,$$${g}_{1}\left(t\right)=\frac{\phi (t,{f}_{1}\left(t\right),{f}_{2}\left(t\right))}{1-{\phi}_{0}({f}_{1}\left(t\right),{f}_{2}\left(t\right))}.$$We define function $P:{M}_{1}\times {M}_{1}\times {M}_{1}\u27f6M$ by$$p\left(t\right)=\frac{\phi (t,{g}_{1}\left(t\right),{f}_{2}\left(t\right))}{1-{\phi}_{0}({f}_{1}\left(t\right),{f}_{2}\left(t\right))}.$$
- $\left({C}_{5}\right)$
- Equations ${g}_{k}\left(t\right)-1=0$, $k=2,3$ have PSS denoted by ${r}_{2},{r}_{3}\in M-\left\{0\right\},$ respectively, where ${g}_{2}:{M}_{1}\u27f6M,$${g}_{3}:{M}_{1}\u27f6M$ are$$\begin{array}{cc}\hfill {g}_{2}\left(t\right)=& \phantom{\rule{4pt}{0ex}}\left[1+(1+2p\left(t\right)+2(c-2){p}^{2}\left(t\right))\frac{1+{\int}_{0}^{1}{\phi}_{1}\left(\theta {g}_{1}\left(t\right)t\right)d\theta}{1-{\phi}_{0}({f}_{1}\left(t\right),{f}_{2}\left(t\right))}\right]{g}_{1}\left(t\right)\hfill \end{array}$$$$\begin{array}{cc}\hfill {g}_{3}\left(t\right)=& \phantom{\rule{4pt}{0ex}}\left[1+(1+2p\left(t\right)+2(c-2){p}^{2}\left(t\right))\frac{1+{\int}_{0}^{1}{\phi}_{1}\left(\theta {g}_{2}\left(t\right)t\right)d\theta}{1-{\phi}_{0}({f}_{1}\left(t\right),{f}_{2}\left(t\right))}\right]{g}_{2}\left(t\right).\hfill \end{array}$$
- $\left({C}_{6}\right)$
- $u[{x}^{*},r]\subset \mathrm{\Omega},$ with $r=\mathit{min}\left\{{r}_{i}\right\}$, $i=1,2,3.$It is implied, if $t\in [0,r)$, that$$0\le {\phi}_{0}({f}_{1}\left(t\right),{f}_{2}\left(t\right))<1$$$$0\le {g}_{i}\left(t\right)<1.$$

**Theorem**

**1.**

**Proof.**

**Remark**

**1.**

- (i)
- The real functions ${f}_{1}$ and ${f}_{2}$ are left uncluttered in Theorem 1. But some choices are motivated by calculations$$\begin{array}{cc}\hfill {u}_{1}-{x}^{*}& \phantom{\rule{4pt}{0ex}}=d-bF\left(x\right)=(I-a[x,{x}^{*};F])\left(d\right)\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& \phantom{\rule{4pt}{0ex}}=[(1-a\phantom{\rule{4pt}{0ex}}L)-aL{L}^{-1}([x,{x}^{*};F]-L)]\left(d\right),\hfill \\ \hfill \parallel {u}_{1}-{x}^{*}\parallel & \phantom{\rule{4pt}{0ex}}\le [\parallel I-a\phantom{\rule{4pt}{0ex}}L\parallel +|a|\parallel L\parallel {\phi}_{1}(\parallel \left(d\right)\parallel )]\parallel \left(d\right)\parallel .\hfill \end{array}$$$${f}_{1}\left(t\right)=\left(\parallel I-a\phantom{\rule{4pt}{0ex}}L\parallel +\left|a\right|\parallel L\parallel {\phi}_{1}\left(t\right)\right)t,$$$${f}_{2}\left(t\right)=\left(\parallel I+b\phantom{\rule{4pt}{0ex}}L\parallel +\left|b\right|\parallel L\parallel {\phi}_{1}\left(t\right)\right)t.$$
- (ii)
- Conditions can be expressed without ${u}_{1}$ and ${u}_{2}$ like, for example,$$\parallel {L}^{-1}([x,y;F]-L)\parallel \le {\overline{\phi}}_{0}(\parallel d\parallel ,\parallel y-{x}^{*}\parallel ).$$$$\overline{r}=\mathit{max}\{r,{f}_{1}\left(r\right),{f}_{2}\left(r\right)\}.$$
- (iii)
- Linear operator L is chosen so that functions $\u201c\phi \u201d$ are as tight as possible. Some popular choices are: $L={F}^{T}\left({x}^{*}\right)$ (the differentiable case) or $L=[{x}_{-1},{x}_{0};F]$, ${x}_{-1},{x}_{0}\in \mathrm{\Omega}$ (the non-differentiable case). It is worth noticing that the invertibility of ${F}^{\prime}\left({x}^{*}\right)$ is not assumed or implied.

**Proposition**

**1.**

**Proof.**

## 3. Semi-Local Analysis

- $\left({H}_{1}\right)$
- There exist continuous as well as nondecreasing functions ${\psi}_{0}:M\times M\to M,$${f}_{3}:M\to M$ and ${f}_{4}:M\to M$ such that equation$${\psi}_{0}({f}_{3}\left(t\right),{f}_{4}\left(t\right))-1=0$$
- $\left({H}_{2}\right)$
- There exist an initial point ${x}_{0}\in \mathrm{\Omega}$ and a linear operator L which is invertible such that$$\begin{array}{cc}\hfill \parallel {L}^{-1}([{u}_{1},{u}_{2};F]-L)\parallel & \phantom{\rule{4pt}{0ex}}\le {\phi}_{0}(\parallel {u}_{1}-{x}_{0}\parallel ,\parallel {u}_{2}-{x}_{0}\parallel ),\hfill \\ \hfill \parallel {u}_{1}-{x}_{0}\parallel & \phantom{\rule{4pt}{0ex}}\le {f}_{3}(\parallel x-{x}_{0}\parallel )\hfill \end{array}$$$$\begin{array}{cc}\hfill \parallel {u}_{2}-{x}_{0}\parallel & \phantom{\rule{4pt}{0ex}}\le {f}_{4}(\parallel x-{x}_{0}\parallel )\hfill \end{array}$$Notice that conditions $\left({H}_{1}\right)$ and $\left({H}_{2}\right)$ offer, for $x={x}_{0}$,$$\parallel {L}^{-1}([{u}_{1},{u}_{2};F]-L)\parallel \le {\phi}_{0}({f}_{3}\left(0\right),{f}_{4}\left(0\right))<1.$$Thus, ${[{u}_{1},{u}_{2};F]}^{-1}$ is invertible and we can set $\parallel {[{u}_{1},{u}_{2};F]}^{-1}F\left({x}_{0}\right)\parallel \le {b}_{0}$ for some ${b}_{0}\ge 0.$
- $\left({H}_{3}\right)$
- There exists continuous as well as nondecreasing function $\psi :{M}_{2}\times {M}_{2}\times {M}_{2}\times {M}_{2}\to M,$ so for $x,y\in {\mathrm{\Omega}}_{2}$;$$\parallel {L}^{-1}([{u}_{1},{u}_{2};F]-[x,y;F])\parallel \le \psi (\parallel x-{x}_{0}\parallel ,\parallel y-{x}_{0}\parallel ,\parallel {u}_{1}-{x}_{0}\parallel ,\parallel {u}_{2}-{x}_{0}\parallel ).$$We define the scalar sequence $\left\{{a}_{r}\right\}$, $\left\{{b}_{r}\right\}$ and $\left\{{c}_{r}\right\}$ for ${a}_{0}=0,$${b}_{0}\in [0,{\delta}_{3})$ and each $r=0,1,2,\dots $ by$$\begin{array}{cc}\hfill {q}_{r}=& \phantom{\rule{4pt}{0ex}}\frac{\psi ({a}_{r},{b}_{r},{f}_{3}\left({a}_{r}\right),{f}_{4}\left({a}_{r}\right))}{1-{\psi}_{0}({f}_{3}\left({a}_{r}\right),{f}_{4}\left({a}_{r}\right))},\hfill \\ \hfill {\lambda}_{r}=& \phantom{\rule{4pt}{0ex}}\psi ({a}_{r},{b}_{r},{f}_{3}\left({a}_{r}\right),{f}_{4}\left({a}_{r}\right)),\hfill \\ \hfill {c}_{r}=& \phantom{\rule{4pt}{0ex}}{b}_{r}+\frac{(1+2{q}_{r}+2|c-2|{q}_{r}^{2}){\lambda}_{r}}{1-{\psi}_{0}({f}_{3}\left({a}_{r}\right),{f}_{4}\left({a}_{r}\right))},\hfill \\ \hfill {\mu}_{r}=& \phantom{\rule{4pt}{0ex}}{\int}_{0}^{1}(1+{\psi}_{0}({b}_{r}+\theta ({c}_{r}-{b}_{r}))d\theta )({c}_{r}-{b}_{r})+{\lambda}_{r},\hfill \\ \hfill {a}_{r+1}=& {c}_{r}+\frac{(1+2{q}_{r}+2|c-2|{q}_{r}^{2}){\mu}_{r}}{1-{\psi}_{0}({f}_{3}\left({a}_{r}\right),{f}_{4}\left({a}_{r}\right))}\hfill \\ \hfill {\delta}_{r+1}=& \phantom{\rule{4pt}{0ex}}\psi ({a}_{r},{a}_{r+1},{f}_{3}\left({a}_{r}\right),{f}_{4}\left({a}_{r}\right))({a}_{r+1}-{a}_{r})+(1+{\psi}_{0}({f}_{3}\left({a}_{r}\right),{f}_{4}\left({a}_{r}\right))({a}_{r+1}-{a}_{r})\hfill \end{array}$$$$\begin{array}{cc}\hfill {b}_{r+1}=& \phantom{\rule{4pt}{0ex}}{a}_{r+1}+\frac{{\delta}_{r+1}}{1-{\psi}_{0}({f}_{3}\left({a}_{r+1}\right),{f}_{4}\left({a}_{r+1}\right))}.\hfill \end{array}$$
- $\left({H}_{4}\right)$
- There exists $\overline{\delta}\in [0,{\delta}_{3})$, so$${\psi}_{0}({f}_{3}\left({a}_{r}\right),{f}_{4}\left({a}_{r}\right))<1\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\mathrm{and}\phantom{\rule{4pt}{0ex}}{a}_{r}\le \overline{\delta}\phantom{\rule{4pt}{0ex}}\mathrm{for}\phantom{\rule{4.pt}{0ex}}\mathrm{all}\phantom{\rule{4pt}{0ex}}r=0,1,2,\dots .$$Then, by Formula (18) and this condition that$$\begin{array}{cc}\hfill F\left({y}_{r}\right)& \phantom{\rule{4pt}{0ex}}=F\left({y}_{r}\right)-F\left({x}_{r}\right)-[{u}_{1},{u}_{2};F]({y}_{r}-{x}_{r}),\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& \phantom{\rule{4pt}{0ex}}=([{y}_{r},{x}_{r};F]-[{u}_{1},{u}_{2};F])({y}_{r}-{x}_{r}),\hfill \end{array}$$$$\begin{array}{cc}\hfill \parallel {L}^{-1}F\left({y}_{r}\right)\parallel & \phantom{\rule{4pt}{0ex}}\le \psi (\parallel {x}_{r}-{x}_{0}\parallel ,\parallel {y}_{r}-{x}_{0}\parallel ,\parallel {u}_{1}-{x}_{0}\parallel ,\parallel {u}_{2}-{x}_{0}\parallel )\parallel {y}_{r}-{x}_{r}\parallel ,\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& \phantom{\rule{4pt}{0ex}}\le \psi ({a}_{r},{b}_{r},{f}_{3}\left({a}_{r}\right),{f}_{4}\left({a}_{r}\right))({b}_{r}-{a}_{r})={\lambda}_{r}.\hfill \end{array}$$Moreover,$$\begin{array}{cc}\hfill F\left({z}_{r}\right)& \phantom{\rule{4pt}{0ex}}=F\left({z}_{r}\right)-F\left({y}_{r}\right)+F\left({y}_{r}\right),\hfill \\ \hfill \parallel {L}^{-1}F\left({z}_{r}\right)\parallel & \phantom{\rule{4pt}{0ex}}\le \left(1+{\int}_{0}^{1}{\psi}_{0}(\parallel {y}_{r}-{x}_{0}\parallel +\theta \parallel {z}_{r}-{y}_{r}\parallel )d\theta \right)\parallel {z}_{r}-{y}_{r}\parallel +{\lambda}_{r},\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& \phantom{\rule{4pt}{0ex}}\le \left(1+{\int}_{0}^{1}{\psi}_{0}({b}_{r}+\theta \parallel {c}_{r}-{b}_{r}\parallel )d\theta \right)\parallel {c}_{r}-{b}_{r}\parallel +{\lambda}_{r}={\mu}_{r},\hfill \\ \hfill \parallel {x}_{r+1}-{z}_{r}\parallel & \phantom{\rule{4pt}{0ex}}\le \parallel {A}_{r}L\parallel \parallel {L}^{-1}F\left({z}_{r}\right)\parallel ,\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& \phantom{\rule{4pt}{0ex}}\le \frac{(1+2{q}_{r}+2|c-2|{q}_{r}^{2}){\mu}_{r}}{1-{\psi}_{0}({f}_{3}\left({a}_{r}\right),{f}_{4}\left({a}_{r}\right))},\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& \phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}{a}_{r+1}-{a}_{r},\hfill \\ \hfill \parallel {x}_{r+1}-{x}_{0}\parallel & \phantom{\rule{4pt}{0ex}}\le \parallel {x}_{r+1}-{z}_{r}\parallel +\parallel {z}_{r}-{x}_{0}\parallel ,\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& \phantom{\rule{4pt}{0ex}}\le {a}_{r+1}-{c}_{r}+{c}_{r}-{a}_{0}={a}_{r+1}<{a}^{*},\hfill \end{array}$$$$\begin{array}{cc}\hfill F\left({x}_{r+1}\right)& \phantom{\rule{4pt}{0ex}}=F\left({x}_{r+1}\right)-F\left({x}_{r}\right)-[{u}_{1},{u}_{2};F]({y}_{r}-{x}_{r}),\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& \phantom{\rule{4pt}{0ex}}=([{x}_{r+1},{x}_{r};F]-[{u}_{1},{u}_{2};F])({x}_{r+1}-{x}_{r})+[{u}_{1},{u}_{2};F]({x}_{r+1}-{y}_{r}),\hfill \end{array}$$$$\begin{array}{cc}\hfill \parallel {L}^{-1}F\left({x}_{r+1}\right)\parallel & \phantom{\rule{4pt}{0ex}}\le \psi ({a}_{r},{a}_{r+1},{f}_{3}\left({a}_{r}\right),{f}_{4}\left({a}_{r}\right))({a}_{r+1}-{a}_{r})\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& \phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}+(1+{\psi}_{0}({f}_{3}\left({a}_{r}\right),{f}_{4}\left({a}_{r}\right)))({a}_{r+1}-{b}_{r})={\delta}_{r+1}.\hfill \end{array}$$Consequently, we obtain$$\begin{array}{cc}\hfill \parallel {y}_{r+1}-{x}_{r+1}\parallel & \phantom{\rule{4pt}{0ex}}\le \parallel {[{u}_{1},{u}_{2};F]}^{-1}L\parallel \parallel {L}^{-1}F\left({a}_{r+1}\right)\parallel \hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& \phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\le \frac{{\delta}_{r+1}}{1-{\psi}_{0}({f}_{3}\left({a}_{r+1}\right),{f}_{4}\left({a}_{r+1}\right))}={b}_{r+1}-{a}_{r+1}\hfill \end{array}$$$$\begin{array}{cc}\hfill \parallel {y}_{r+1}-{x}_{0}\parallel & \phantom{\rule{4pt}{0ex}}\le \parallel {y}_{r+1}-{x}_{r+1}\parallel +\parallel {x}_{r+1}-{x}_{0}\parallel \hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& \phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\le {b}_{r+1}-{a}_{r+1}+{a}_{r+1}-{a}_{0}={b}_{r+1}<{a}^{*}.\hfill \end{array}$$

**Theorem**

**2.**

**Proof.**

**Remark**

**2.**

**Proposition**

**2.**

**Proof.**

## 4. Numerical Tests

**Example**

**1.**

**Example**

**2.**

**Example**

**3.**

**Example**

**4.**

**Example**

**5.**

## 5. Conclusions

## Author Contributions

## Funding

## Institutional Review Board Statement

## Informed Consent Statement

## Data Availability Statement

## Conflicts of Interest

## Correction Statement

## References

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**MDPI and ACS Style**

Kumar, S.; Sharma, J.R.; Argyros, I.K.; Regmi, S.
Three-Step Derivative-Free Method of Order Six. *Foundations* **2023**, *3*, 573-588.
https://doi.org/10.3390/foundations3030034

**AMA Style**

Kumar S, Sharma JR, Argyros IK, Regmi S.
Three-Step Derivative-Free Method of Order Six. *Foundations*. 2023; 3(3):573-588.
https://doi.org/10.3390/foundations3030034

**Chicago/Turabian Style**

Kumar, Sunil, Janak Raj Sharma, Ioannis K. Argyros, and Samundra Regmi.
2023. "Three-Step Derivative-Free Method of Order Six" *Foundations* 3, no. 3: 573-588.
https://doi.org/10.3390/foundations3030034