Extended Multi-Step Jarratt-like Schemes of High Order for Equations and Systems

Abstract

The local convergence analysis of multi-step, high-order Jarratt-like schemes is extended for solving Banach space valued systems of equations using the derivative instead of up to the ninth derivative as in previous works. Our idea expands the usage of the scheme in cases not considered earlier and can also be utilized in other schemes, too. Experiments test the theoretical results.

1. Introduction

Let $\mathsf{\Omega }\ne 0$ be an open subset of a Banach X and H a continuous operator mapping $\mathsf{\Omega }$ into Y, another Banach space.

Numerous problems from diverse disciplines turn into an equation

$$H(x)=0,$$

(1)

by mathematical modeling [1,2,3,4,5,6,7,8,9,10,11].

We want to find a solution $\gamma$ in closed form, but this is achieved just in special occasions. This is why iterative schemes are developed generating sequences converging to $\gamma$ under suitable convergence criteria [2,4,12,13,14,15,16,17,18].

We study the multi-step Jarratt-like scheme developed for $x_0\in \mathsf{\Omega }$ and all $n=0,1,2,\dots$ and for $k=1,2,\dots m-3$, ($m\ge 3$) by

$$\left\{\begin{array}{cc}\hfill & z_1=H^{\prime }{(x_n)}^{-1}H(x_n),y_1=x_n-z_1,\hfill \\ \hfill & z_2=H^{\prime }{(x_n)}^{-1}H(y_1),y_2=y_1-3z_2,\hfill \\ \hfill & z_3=H^{\prime }{(x_n)}^{-1}{H}^{\prime }(y_2)z_2,y_3=y_1-\frac{7}{4}{z}_2+\frac{1}{2}{z}_3+\frac{1}{4}{z}_4,\hfill \\ \hfill & z_4=H^{\prime }{(x_n)}^{-1}{H}^{\prime }(y_2)z_3,\hfill \\ \hfill & z_{2k+3}=H^{\prime }{(x_n)}^{-1}H(y_{k+2}),\hfill \\ \hfill & z_{2k+4}=H^{\prime }{(x_n)}^{-1}{H}^{\prime }(y_2)z_{2k+3},\hfill \\ \hfill & y_{k+3}=y_{k+2}-2z_{2k+3}+z_{2k+4}.\hfill \end{array}\right.$$

(2)

The $3m-4$ convergence order of (2) is presented in [19] for $X=Y={\mathbb{R}}^j$ from Taylor series expansions with conditions reaching the ninth derivative of H (not appearing in scheme (2)). These restrict the application of scheme (2). Consider the simple illustration for $f:\mathsf{\Omega }:=[-1/2,3/2]\to \mathbb{R}$ defined by

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

Then, it is easily seen that $f^{‴}$ is unbounded. Therefore, the work in [19] cannot guarantee convergence to a solution $\gamma$ using scheme (2). Notice that the method (2) is of order at least eight if $m=4$, which is optimum in the sense of the Kung–Traub conjecture [20]. Moreover, each step increases the convergence order by three [19]. The computational efficiency and computational cost of the method (2) have been given and discussed in detail (see Section 4 in [19]). It is also shown that it is better from the efficiency point of view when compared with other modified Jarratt-type methods. We also refer the reader to the notable relevant papers by Ignatova et al. [21], Kung et al. [20] and Petkovic [17]. Moreover, no upper bounds on $\parallel x_n-\gamma \parallel$ or results on the uniqueness of $\gamma$ are given. Motivated by all these, we develop a technique using only $H^{\prime }$ (that appears in (2)), which also gives computable upper bounds on $\parallel x_n-\gamma \parallel$ and a uniqueness result.

Hence, we extend the applicability of scheme (2). Moreover, the Computational Order of Convergence (COC) and the Approximate Computational Order of Convergence (ACOC) are utilized to compute the convergence order, which does not require usage of higher derivatives or divided differences. This is conducted in Section 2. Numerical experiments are given in Section 3. Finally, the conclusion appears in Section 4.

2. Analysis

We develop the real functions ${\psi }_0$, $\psi$ and ${\psi }_1$ and real indicators. Next, these functions are connected as majorizing for the operator $H^{\prime }$ (see the conditions $(C_1)$–$(C_4)$ that follow). Set $T=[0,+\infty )$.

Suppose that there exist functions:

(a)

${\psi }_0:T\to T$ so

$${\psi }_0(s)-1=0$$

(3)

has a solution denoted by ${\rho }_0\in T-\left\{0\right\}$, which is continuous and nondecreasing (CN). Set $T_0=[0,{\rho }_0)$.

(b)

$\psi :T_0\to T$, ${\psi }_1:T_0\to T$ CN so that for

$$h_1(s)=\frac{{\int }_0^1\psi ((1-\zeta )s)d\zeta }{1-{\psi }_0(s)},$$

$$\begin{array}{ccc}\hfill h_2(s)& =& [h_1(h_1(s)s)+\frac{({\psi }_0(h_1(s)s)+{\psi }_0(s)){\int }_0^1{\psi }_1(\zeta h_1(s)s)d\zeta }{(1-{\psi }_0(s))(1-{\psi }_0(h_1(s)s))}\hfill \\ & & +\frac{2{\int }_0^1{\psi }_1(\zeta h_1(s)s)d\zeta }{1-{\psi }_0(s)}]h_1(s),\hfill \end{array}$$

$$\begin{array}{ccc}\hfill h_3(s)& =& [1+\frac{7}{4}\frac{{\int }_0^1{\psi }_1(\zeta h_1(s)s)d\zeta }{1-{\psi }_0(s)}+\frac{1}{2}\frac{{\psi }_1(h_1(s)s)h_2(s)s)}{{(1-{\psi }_0(s))}^2}\hfill \\ & & +\frac{1}{4}\frac{{\psi }_1(h_1(s)s){\psi }_1{(h_2(s)s)}^2}{{(1-{\psi }_0(s))}^3}]h_1(s),\hfill \end{array}$$

for $k=1,\dots ,n-3$

$$\begin{array}{ccc}\hfill h_{k+3}(s)& =& [1+\frac{2{\int }_0^1{\psi }_1(\zeta h_{k+2}(s)s)d\zeta }{1-{\psi }_0(s)}\hfill \\ & & +\frac{{\psi }_1(h_2(s)s){\int }_0^1{\psi }_1(\zeta h_{k+2}(s)s)d\zeta }{{(1-{\psi }_0(s))}^2}]h_{k+1}(s),\hfill \end{array}$$

and

$${\overline{h}}_i(s)=h_i(s)-1,i=1,2,\dots ,n,$$

equations

$${\overline{h}}_i(s)=0$$

(4)

have the least solutions ${\rho }_i$, respectively, in $T_0-\left\{0\right\}$.

Then, we show that $\rho$ given by

$$\rho =min\left\{{\rho }_i\right\}$$

(5)

is a convergence radius for scheme (2).

It follows by these definitions that for each $s\in [0,\rho )$

$$0\le {\psi }_0(s)<1,0\le {\psi }_0(h_1(s)s)<1$$

(6)

and

$$0\le h_i(s)<1.$$

(7)

Let $B[\gamma ,\alpha ]$ stand for the closure of ball $B(\gamma ,\alpha )$.

The conditions $(C)$ are used to provide the functions $\psi$ as given earlier:

$(C_1)$

Equation $H(x)=0$ has a simple solution $\gamma \in \mathsf{\Omega }$.

$(C_2)$

For all $u\in \mathsf{\Omega }$

$$\parallel H^{\prime }{(\gamma )}^{-1}(H^{\prime }(\gamma )-H^{\prime }(u))\parallel \le {\psi }_0(\parallel \gamma -u\parallel ).$$

Set ${\mathsf{\Omega }}_0=\mathsf{\Omega }\cap B(\gamma ,{\rho }_0)$.

$(C_3)$

For all u, $v\in {\mathsf{\Omega }}_0$

$$\parallel H^{\prime }{(\gamma )}^{-1}(H^{\prime }(u)-H^{\prime }(v))\parallel \le \psi (\parallel u-v\parallel )$$

and

$$\parallel H^{\prime }{(\gamma )}^{-1}{H}^{\prime }(v)\parallel \le {\psi }_1(\parallel v-\gamma \parallel ).$$

$(C_4)$

$B[\gamma ,\rho ]\subset \mathsf{\Omega }$.

Next, using conditions $(C)$ together with the developed notation, we develop the local result of scheme (2).

Theorem 1.

Suppose conditions $(C)$ hold. Then, if $x_0\in B_0:=B(\gamma ,\rho )-\left\{\gamma \right\}$, iteration $\left\{x_n\right\}$ developed by scheme (2) exists, stays in $B_0$ and converges to γ.

Proof. 

Let us choose $u\in B_0$ and $t_n:=\parallel x_n-\gamma \parallel$. Then, by (5), (6) and $(C_2)$, one has

$$\parallel H^{\prime }{(\gamma )}^{-1}(H^{\prime }(u)-H^{\prime }(\gamma ))\parallel \le {\psi }_0(\parallel u-\gamma \parallel )\le {\psi }_0(\rho )<1.$$

(8)

By estimation (8) with the lemma of Banach on linear inverse mapping [5], we obtain $H^{\prime }{(u)}^{-1}\in \mathcal{L}(Y,X)$ with

$$\parallel H^{\prime }{(u)}^{-1}{H}^{\prime }(\gamma )\parallel \le \frac{1}{1-{\psi }_0(\parallel u-\gamma \parallel )}·$$

(9)

By (9) for $u=x_0$, $y_i$ exists. Using (5), (9) (for $u=x_0$ and $n=0$), (7) (for $i=1$), scheme (2) and $(C_3)$

$$\begin{array}{ccc}\hfill \parallel y_1-\gamma \parallel & =& \parallel x_0-\gamma -H^{\prime }{(x_0)}^{-1}H(x_0)\parallel \hfill \\ \hfill & \le & \parallel H^{\prime }{(x_0)}^{-1}{H}^{\prime }(\gamma )\parallel ∥{\int }_0^1{H}^{\prime }{(\gamma )}^{-1}\left(H^{\prime }(\gamma +\zeta (x_0-\gamma ))-H^{\prime }(x_0)\right)(x_0-\gamma )d\zeta ∥\hfill \\ \hfill & \le & \frac{{\int }_0^1\psi ((1-\zeta )t_0)d\zeta }{1-{\psi }_0(t_0)}{t}_0\hfill \\ \hfill & \le & h_1(t_0)t_0\le t_0<\rho ,\hfill \end{array}$$

(10)

then, $y_1\in B(\gamma ,\rho )$ and $H^{\prime }{(y_1)}^{-1}\in \mathcal{L}(Y,X)$.

Using (5)–(7) (for $i=2$), (9) (for $u=y_1$), $(C_3$), (10) and scheme (2), we obtain in turn

$$\begin{array}{ccc}\hfill \parallel y_2-\gamma \parallel & =& \parallel y_1-\gamma -H^{\prime }{(y_1)}^{-1}H(y_1)\hfill \\ \hfill & & +H^{\prime }{(y_1)}^{-1}\left(H^{\prime }(x_0)-H^{\prime }(y_1)\right)H^{\prime }{(x_0)}^{-1}H(y_1)-2H^{\prime }{(x_0)}^{-1}H(y_1)\parallel \hfill \\ \hfill & \le & [h_1(\parallel y_1-\gamma \parallel )+\frac{({\psi }_0(\parallel y_1-\gamma \parallel )+{\psi }_0(t_0)){\int }_0^1{\psi }_1(\zeta \parallel y_1-\gamma \parallel )d\zeta }{(1-{\psi }_0(\parallel y_1-\gamma \parallel ))(1-{\psi }_0(t_0))}\hfill \\ \hfill & & +\frac{2{\int }_0^1{\psi }_1(\zeta \parallel y_1-\gamma \parallel )d\zeta }{1-{\psi }_0(t_0)}]\parallel y_1-\gamma \parallel \hfill \\ \hfill & \le & h_2(t_0)t_0,\hfill \end{array}$$

(11)

hence, $y_2\in B(\gamma ,\rho )$. Similarly, by (5), (7) (for $i=3$), (10), (11) and scheme (2), we have in turn that

$$\begin{array}{ccc}\hfill \parallel y_3-\gamma \parallel & =& \parallel y_1-\gamma -\frac{7}{4}{H}^{\prime }{(x_0)}^{-1}H(y_1)+\frac{1}{2}{H}^{\prime }{(x_0)}^{-1}{H}^{\prime }(y_2)H^{\prime }{(x_0)}^{-1}H(y_1)\hfill \\ \hfill & & +\frac{1}{4}{H}^{\prime }{(x_0)}^{-1}{H}^{\prime }(y_2)H^{\prime }{(x_0)}^{-1}{H}^{\prime }(y_2)H^{\prime }{(x_0)}^{-1}H(y_1)\parallel \hfill \\ \hfill & \le & [1+\frac{7}{4}\frac{{\int }_0^1{\psi }_1(\zeta \parallel y_1-\gamma \parallel )d\zeta }{1-{\psi }_0(t_0)}+\frac{1}{2}\frac{{\psi }_1(\parallel y_1-\gamma \parallel ){\psi }_1(\parallel y_2-\gamma \parallel )}{{(1-{\psi }_0(t_0))}^2}\hfill \\ \hfill & & +\frac{1}{4}\frac{{\psi }_1(\parallel y_1-\gamma \parallel ){\psi }_1(\parallel y_2-\gamma {\parallel )}^2}{{(1-{\psi }_0(t_0))}^3}]\parallel y_1-\gamma \parallel \hfill \\ & \le & h_3(t_0)t_0\le t_0,\hfill \end{array}$$

(12)

therefore, $y_3\in B(\gamma ,\rho )$.

Then, for the rest of the substeps, similarly, we obtain for $k=1,\dots ,m-3$

$$\begin{array}{ccc}\hfill \parallel y_{k+3}-\gamma \parallel & =& \parallel y_{k+2}-\gamma -2H^{\prime }{(x_0)}^{-1}H(y_{k+2})+H^{\prime }{(x_0)}^{-1}{H}^{\prime }(y_2)H^{\prime }{(x_0)}^{-1}H(y_{k+2})\parallel \hfill \\ \hfill & \le & [1+\frac{2{\int }_0^1{\psi }_1(\zeta \parallel y_{k+2}-\gamma \parallel )d\zeta }{1-{\psi }_0(t_0)}\hfill \\ \hfill & & +\frac{{\psi }_1(\parallel y_2-\gamma \parallel ){\int }_0^1{\psi }_1(\zeta \parallel y_{k+2}-\gamma \parallel )d\zeta }{{(1-{\psi }_0(t_0))}^2}]\parallel y_{k+2}-\gamma \parallel \hfill \\ \hfill & \le & h_{k+3}(t_0)t_0\le t_0.\hfill \end{array}$$

(13)

By the definition of $x_1=y_m$ and (10)–(13), we deduce

$$t_1\le c{t}_0,$$

(14)

where

$$c=h_1(t_0)h_2(t_0)\dots h_m(t_0)\in [0,1).$$

Simply replace $x_0$ by $x_j$ in the preceding calculations to obtain

$$t_{j+1}\le c{t}_j\le c^{j+1}{t}_0<\rho ,$$

(15)

so, ${lim}_{j\to +\infty }{x}_j=\gamma$ and $x_{j+1}\in B(\gamma ,\rho )$. □

We have the uniqueness result:

Proposition 1.

Suppose:

(i) 

Point γ is a simple solution in $B(\gamma ,{\rho }_0)\subset \mathsf{\Omega }$ of (1) and

(ii) 

$${\int }_0^1{\psi }_0(\zeta \overline{\rho })d\zeta <1\mathit{for}\mathit{some}\overline{\rho }\ge {\rho }_0.$$

(16)

Then, γ is unique in ${\mathsf{\Omega }}_1=\mathsf{\Omega }\cap \overline{B}(\gamma ,\overline{\rho })$ as a solution of (1).

Proof. 

Let $q\in {\mathsf{\Omega }}_1$ be such that $H(q)=0$. Define $Q={\int }_0^1{H}^{\prime }(\gamma +\zeta (q-\gamma ))d\zeta$. Then, in view of $(C_2)$ and (16), we obtain in turn

$$\parallel H^{\prime }{(\gamma )}^{-1}(Q-H^{\prime }(\gamma ))\parallel \le {\int }_0^1{\psi }_0(\zeta \parallel \gamma -q\parallel )d\zeta \le {\int }_0^1{\psi }_0(\zeta \overline{\rho })d\zeta <1,$$

so, $\gamma =q$, since $Q^{-1}\in \mathcal{L}(Y,X)$ and $0=H(q)-H(\gamma )=Q(q-\gamma )$. □

Definition 1.

The COC is defined as

$$a=ln\left(\frac{\parallel s_{n+1}-\gamma \parallel }{\parallel s_n-\gamma \parallel }\right)/ln\left(\frac{\parallel s_n-\gamma \parallel }{\parallel s_{n-1}-\gamma \parallel }\right)$$

and, for ${\beta }_n:=\parallel s_n-s_{n-1}\parallel$, the ACOC [16] by

$$b=ln\left(\frac{{\beta }_{n+1}}{{\beta }_n}\right)/ln\left(\frac{{\beta }_n}{{\beta }_{n-1}}\right).$$

3. Applications

We test the conditions $(C)$.

Example 1.

If $X=Y=\mathsf{\Omega }=\mathbb{R}$, let function $H(s)=sins$. The conditions $(C)$ must be verified. Clearly, $\gamma =0$ solves the equation $H(s)=0$. Hence, we choose $\gamma =0$ in conditions $(C)$. In particular, the choice $\gamma =0$ satisfies the condition $(C_1)$. Concerning the condition $(C_2)$, we have, in turn, by the definition of H and, since $H^{\prime }(\gamma )=1$ and $\parallel H^{\prime }{(\gamma )}^{-1}(H^{\prime }(\gamma )-H^{\prime }(u))\parallel \le {\psi }_0(\parallel \gamma -u\parallel )$, that

$$\parallel H^{\prime }{(\gamma )}^{-1}(H^{\prime }(\gamma )-H^{\prime }(u))\parallel =|cos\gamma -cosu|\le {\psi }_0(|\gamma -u|)$$

provided that ${\psi }_0(s)=s$, since there exists $c\in \mathbb{R}$ such that

$$|cos\gamma -cosu|=|sinc(\gamma -u)|\le |sinc||\gamma -u|=|\gamma -u|.$$

That is the condition $(C_1)$ that is satisfied for this choice of ${\psi }_0$. Moreover, by solving the equation ${\psi }_0(s)-1=0$, we obtain ${\rho }_0=1$, and we can set ${\mathsf{\Omega }}_0=B(0,1)$.

Similarly, we have

$$\parallel H^{\prime }{(\gamma )}^{-1}(H^{\prime }(u)-H^{\prime }(v))\parallel \le |sind(u-v)|\le |sind||u-v|\le |u-v|$$

for some, $d\in \mathbb{R}$, so we can choose $\psi (s)=s$. Then, the first condition in $(C_2)$ is satisfied. Concerning the second condition in $(C_2)$, we obtain in turn that

$$\parallel H^{\prime }{(\gamma )}^{-1}{H}^{\prime }(v)\parallel =|1sinv|\le 1,$$

so, we can choose ${\psi }_1(s)=1$. Then, we have for $k=3$ the resutls of

Table 1. Radius for Example 1.

$\mathit{\rho }$	${\mathit{\rho }}_0$	${\mathit{\rho }}_1$	${\mathit{\rho }}_2$	${\mathit{\rho }}_3$
0.306688…	1	0.666667…	0.343549…	0.306688…

.

Example 2.

If $\mathsf{\Omega }=B[0,1]$, and considering the continuous operator $H:\mathsf{\Omega }\subset S\to S$ (with the maximum norm) as

$$H(\psi )(s)=\psi (s)-5{\int }_0^{1}s\zeta \psi {(\zeta )}^{3}d\zeta ,$$

where space S contains continuous functions defined on the interval $[0,1]$, we have that

$$H^{\prime }(\psi (\lambda ))(s)=\lambda (s)-15{\int }_0^{1}s\zeta \psi {(\zeta )}^2\lambda (\zeta )d\zeta ,\mathit{for}\mathit{each}\lambda \in \mathsf{\Omega }.$$

Then, we obtain that $\gamma =0$, and for $k=3$ we obtain the functions ${\psi }_0(s)=\psi (s)=\frac{15}{2}s$, ${\psi }_1(s)=2$. This way, we obtain the

Table 2. Radius for Example 2.

$\mathit{\rho }$	${\mathit{\rho }}_0$	${\mathit{\rho }}_1$	${\mathit{\rho }}_2$	${\mathit{\rho }}_3$
0.024324…	0.133333…	0.088888…	0.032839…	0.024324…

.

Example 3.

Let $\mathsf{\Omega }=B(0,1)$, $X=Y={\mathbb{R}}^3$ and $\gamma ={(0,0,0)}^T$. Let H defined on Ω be

$$H(w)=H(w_1,w_2,w_3)={\left(e^{w_1}-1,\frac{e-1}{2}{w}_2^2+w_2,w_3\right)}^T.$$

For the point $w={(w_1,w_2,w_3)}^T$, the derivative $H^{\prime }$ is

$$H^{\prime }(w)=\left(\begin{array}{ccc}{e}^{w_1}& 0& 0\\ 0& (e-1)w_2+1& 0\\ 0& 0& 1\end{array}\right).$$

Taking into account rows with their max. norm and

$$H^{\prime }(\gamma )=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right),$$

for $k=3$ we obtain through conditions $(C)$${\psi }_0(s)=(e-1)s$, $\psi (s)=e^{\frac{1}{e-1}}s$, ${\psi }_1(s)=e^{\frac{1}{e-1}}$ and the radius in

Table 3. Radius for Example 3.

$\mathit{\rho }$	${\mathit{\rho }}_0$	${\mathit{\rho }}_1$	${\mathit{\rho }}_2$	${\mathit{\rho }}_3$
0.115169…	0.581977…	0.382692…	0.148762…	0.115169…

.

Example 4.

Taking into account the academic example in the first section of this work, for $k=3$, we obtain ${\psi }_0(s)=\psi (s)=96.662907s$, ${\psi }_1(s)=1.0631$. The corresponding radius are in

Table 4. Radius for Example 4.

$\mathit{\rho }$	${\mathit{\rho }}_0$	${\mathit{\rho }}_1$	${\mathit{\rho }}_2$	${\mathit{\rho }}_3$
0.003149…	0.010345…	0.006896…	0.003463…	0.003149…

.

Example 5.

We consider the conventional form of Kepler’s equation

$$G(s)=s-\sigma sin(s)-\tau =0,$$

where $0\le \tau \le \pi$ and $0\le \sigma <1$. In [22], we can find, for distinct values of τ and σ, a numerical study. In this case, we select value $\tau =0.1$ and $\sigma =0.27$, so the solution obtained is $\gamma \approx 0.136828\dots$. Since

$$G^{\prime }(s)=1-\sigma cos(s),$$

we obtain

$$\begin{array}{ccc}\hfill \parallel G^{\prime }{(\gamma )}^{-1}(G^{\prime }(s)-G^{\prime }(t))\parallel & =& \frac{\parallel \sigma (cos(s)-cos(t))\parallel }{\parallel 1-\sigma cos(\gamma )\parallel }\hfill \\ \hfill & =& \frac{2\sigma \parallel sin\left(\frac{s+t}{2}\right)sin\left(\frac{s-t}{2}\right)\parallel }{\parallel 1-\sigma cos(\gamma )\parallel }\hfill \\ \hfill & \le & \frac{\sigma }{\parallel 1-\sigma cos(\gamma )\parallel }\parallel s-t\parallel \hfill \end{array}$$

and

$$\parallel G^{\prime }{(\gamma )}^{-1}{G}^{\prime }(s)\parallel =\frac{\parallel 1-\sigma cos(s)\parallel }{\parallel 1-\sigma cos(\gamma )\parallel }\le \frac{1+\sigma }{\parallel 1-\sigma cos(\gamma )\parallel }·$$

Consequently, we have ${\psi }_0(s)=\psi (s)=0.3685888t$ and ${\psi }_1(s)=1.7337327$. The calculated value parameters for $k=3$ are given in

Table 5. Radius for Example 5.

$\mathit{\rho }$	${\mathit{\rho }}_0$	${\mathit{\rho }}_1$	${\mathit{\rho }}_2$	${\mathit{\rho }}_3$
0.550956…	2.713050…	1.808700…	0.720969…	0.550956…

.

Example 6.

We analyze the following system of nonlinear equations

$$\left\{\begin{array}{ccc}\hfill {\nu }_2{\nu }_3+{\nu }_4({\nu }_2+{\nu }_3)& =& 0\hfill \\ \hfill {\nu }_1{\nu }_3+{\nu }_4({\nu }_1+{\nu }_3)& =& 0\hfill \\ \hfill {\nu }_1{\nu }_2+{\nu }_4({\nu }_1+{\nu }_2)& =& 0\hfill \\ \hfill {\nu }_1{\nu }_2+{\nu }_1{\nu }_3+{\nu }_2{\nu }_3-1& =& 0\hfill \end{array}\right.$$

(17)

with ${\nu }_1,{\nu }_2,{\nu }_3,{\nu }_4\in \mathbb{R}$. Then, the solution of the previous scheme is showed for $\gamma ={({\nu }_1,{\nu }_2,{\nu }_3,{\nu }_4)}^T$ by function $H:=({\nu }_1,{\nu }_2,{\nu }_3,{\nu }_4):{\mathbb{R}}^4\to {\mathbb{R}}^4$ defined by

$$H({\nu }_1,{\nu }_2,{\nu }_3,{\nu }_4)=\left(\begin{array}{c}{\nu }_2{\nu }_3+{\nu }_4({\nu }_2+{\nu }_3)\\ {\nu }_1{\nu }_3+{\nu }_4({\nu }_1+{\nu }_3)\\ {\nu }_1{\nu }_2+{\nu }_4({\nu }_1+{\nu }_2)\\ {\nu }_1{\nu }_2+{\nu }_1{\nu }_3+{\nu }_2{\nu }_3-1\end{array}\right)$$

The Fréchet derivative is given by

$$H^{\prime }({\nu }_1,{\nu }_2,{\nu }_3,{\nu }_4)=\left(\begin{array}{cccc}0& {\nu }_3+{\nu }_4& {\nu }_2+{\nu }_4& {\nu }_2+{\nu }_3\\ {\nu }_3+{\nu }_4& 0& {\nu }_1+{\nu }_4& {\nu }_1+{\nu }_3\\ {\nu }_2+{\nu }_4& {\nu }_1+{\nu }_4& 0& {\nu }_1+{\nu }_2\\ {\nu }_2+{\nu }_3& {\nu }_1+{\nu }_3& {\nu }_1+{\nu }_2& 0\end{array}\right)$$

Then, we obtain

$$\begin{array}{cc}\hfill & {\nu }_1=0.57735026\dots \hfill \\ \hfill & {\nu }_2=0.57735026\dots \hfill \\ \hfill & {\nu }_3=0.57735026\dots \hfill \\ \hfill & {\nu }_4=-0.28867513\dots \hfill \end{array}$$

(18)

Then, we have that ${\psi }_0(s)=0.28867513s$, $\psi (s)=1.55198152s$, ${\psi }_1(s)=1.06466589s$ and the radii are in

Table 6. Radius for Example 6.

$\mathit{\rho }$	${\mathit{\rho }}_0$	${\mathit{\rho }}_1$	${\mathit{\rho }}_2$	${\mathit{\rho }}_3$
0.391641…	0.775990…	0.484339…	0.413569…	0.391641…

.

4. Conclusions

A technique is introduced involving only $H^{\prime }$, which also gives computable upper error bounds on $\parallel x_n-\gamma \parallel$. This way, the method (2) becomes applicable in cases not possible before, since higher hypotheses on $H^{\prime }$ were required. The COC or ACOC are used to determine the convergence order. The technique is not based on method (2) and is very general. Hence, it can be used to extend the applicability of other methods [11,17,20,21]. This will be the focus of our future work.