On a Chebyshev-type method for approximating the solutions of polynomial operator equations of degree 2

1 year ago

Emil Cătinaş

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Emil Cătinaş

Emil Cătinaş

Ion Păvăloiu

Ion Păvăloiu

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E. Cătinaş, I. Păvăloiu, On a Chebyshev-type method for approximating the solutions of polynomial operator equations of degree 2, Approximation and Optimization, Proceedings of the International Congress on Approximation and Optimization (Romania)-ICAOR, Cluj-Napoca, July 29-August 1, 1996, vol. I, pp. 219-226, D.D. Stancu, Gh. Coman, W.W. Breckner and P. Blaga (eds.), Transilvania Press, 1997, ISBN 973-98180-7-2.

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Introduction

1. Introduction

For solving the nonlinear equations, in a recent paper, A. Diaconu [7] has proposed the Chebyshev-type method $(1.2)$ , for which the computation of the inverse of the derivative at each step is avoided. Similar methods have been studied by other authors [5], [6], [14], but these does not preserve the r-convergence order 3.

In this note we shall apply method $(1.2)$ for solving polynomial operator equations of degree 2. We shall show that the hypotheses for the convergence of the method take a simplified form, the convergence order remaining unaltered. We shall consider then the eigenproblem for scalar matrices, applying to it the studied method. Numerical examples are also given.

Let $X$ be a Banach space, $F : X \to X$ a nonlinear operator and consider the equation

F (x) = \bar{θ},

1.1

$\bar{θ}$ being the null element of $X$ .

For solving $(1.1)$ in [7] there are considered three sequences, ${(x_{k})}_{k \geq 0} \subset X$ and ${(B_{k})}_{k \geq 0}$ , ${(C_{k})}_{k \geq 0} \subset ℒ (X)$ given by

$C_{k}$	$= B_{k} (2 I - F^{'} (x_{k}) B_{k})$	$1.2$
$x_{k + 1}$	$= x_{k} - C_{k} F (x_{k}) - \frac{1}{2} C_{k} F^{''} (x_{k}) {(C_{k} F (x_{k}))}^{2}$
$B_{k + 1}$	$= B_{k} [3 I - 3 F^{'} (x_{k + 1}) B_{k} + {(F^{'} (x_{k + 1}) B_{k})}^{2}], k = 0, 1, \dots,$

where $x_{0} \in X$ and $B_{0} \in ℒ (X)$ , $ℒ (X)$ being the set of all linear continuous operators on $X$ and $I \in ℒ (X)$ being the identity operator.

Remark. The convergence with r-order 3 of this method is obtained despite a general principle which suggests that every newly computed unknown should be used at once in the determination of the other unknowns (e.g. the Gauss-Seidel method for linear systems). In our case we could consider $C_{k}$ instead of $B_{k}$ in the determination of $B_{k + 1} .$

2. The convergence of the method

We shall give first a lemma.

Lemma

If the sequences of real positive numbers ${(δ_{k})}_{k \geq 0}$ and ${(ρ_{k})}_{k \geq 0}$ satisfy

	$δ_{k + 1}$	$\leq {(δ_{k} + 2 ρ_{k} + 2 ρ_{k}^{2})}^{3}$
	$ρ_{k + 1}$	$\leq ρ_{k} δ_{k}^{2} + ρ_{k}^{2} δ_{k}^{2} + 2 ρ_{k}^{3} + ρ_{k}^{4}, k = 0, 1, \dots,$

where $\max {δ_{0}, ρ_{0}} \leq \frac{1}{7} d$ for some $0 < d < 1,$

then the following relation holds:

\max {δ_{k}, ρ_{k}} \leq \frac{1}{7} d^{3^{k}}, k = 0, 1, \dots .

The proof can be easily obtained by induction.

In the following we shall study the convergence of method $(1.2)$ supposing that $F$ is a polynomial operator of degree 2, i.e. $F$ is indefinite differentiable on $X$ and $F^{(i)} = θ_{i}$ for $i \geq 3,$ $θ_{i}$ being the $i$ -linear null operators.

Under this condition $F$ satisfies

F (y) = F (x) + F^{'} (x) (y - x) + \frac{1}{2} F^{''} (x) {(y - x)}^{2}, for all x, y \in X,

2.1

where, in fact, $F^{''} (x)$ is a constant bilinear operator which does not depend on $x .$

Let $x_{0} \in X$ and $r, K > 0$ be two real numbers. Denote $S = {x \in X | ‖ x - x_{0} ‖ \leq r}$ and suppose that we have the estimation

‖ F^{''} (x) ‖ \leq K, for all x \in S .

Concerning the convergence of method $(1.2),$ the following result holds:

Theorem

If the operator $F$ and the elements $x_{0} \in X$ , $B_{0} \in ℒ (X)$ satisfy:

a) there exists $F^{'} {(x_{0})}^{- 1}$ and $‖ F^{'} {(x_{0})}^{- 1} ‖ \leq b_{0}$ for some $b_{0} > 0;$

b) $q = K b_{0} r < 1;$

c) $\max {δ_{0}, ρ_{0}} \leq \frac{1}{7} d$ for some $0 < d < 1$ , where

ρ_{0} = \frac{K a^{2}}{2} ‖ F (x_{0}) ‖, δ_{0} = ‖ I - F^{'} (x_{0}) B_{0} ‖, a = \frac{64}{49} b and b = \frac{b_{0}}{1 - q};

d) $\frac{16 d}{49 K a (1 - d^{2})} \leq r,$

then the following properties hold:

1) the sequences ${(x_{k})}_{k \geq 0}$ , ${(B_{k})}_{k \geq 0}$ , ${(C_{k})}_{k \geq 0}$ converge and ${(x_{k})}_{k \geq 0} \subset S;$

2) denoting $x^{*} = lim x_{k}, B = lim B_{k}, C = lim C_{k}$ , then $F (x^{*}) = \bar{θ}$ and $B = C = F^{'} {(x^{*})}^{- 1};$

3) the following estimations are true:

	$‖ x^{*} - x_{k} ‖$	$\leq \frac{16 d^{3^{k}}}{49 K a (1 - d^{2 \cdot 3^{k}})};$
	$‖ B - B_{k} ‖$	$\leq \frac{1656 a d^{3^{k}}}{2401 (1 - d^{2 \cdot 3^{k}})}, k = 0, 1, \dots$

Demonstration Proof

We shall prove firstly that the first order derivative of $F$ is invertible on $S$ . By a) and b) we have

‖ I - F^{'} {(x_{0})}^{- 1} F^{'} (x) ‖ \leq ‖ F^{'} {(x_{0})}^{- 1} ‖ ‖ F^{'} (x_{0}) - F^{'} (x) ‖ \leq b_{0} K r = q < 1,

for all $x \in S$ . It follows that the operator $T (x) = F^{'} {(x_{0})}^{- 1} F^{'} (x)$ is invertible on $S$ and $T {(x)}^{- 1} = F^{'} {(x)}^{- 1} F^{'} (x_{0}),$ whence $F^{'} {(x)}^{- 1} = T {(x)}^{- 1} F^{'} {(x_{0})}^{- 1}$ and

‖ F^{'} {(x)}^{- 1} ‖ \leq \frac{b_{0}}{1 - q} = b .

For the norms of $B_{0}$ and $C_{0}$ , taking into account the hypotheses, we get

	$‖ B_{0} ‖ \leq$	$‖ B_{0} - F^{'} {(x_{0})}^{- 1} ‖ + ‖ F^{'} {(x_{0})}^{- 1} ‖$
	$\leq$	$‖ F^{'} {(x_{0})}^{- 1} ‖ (1 + ‖ I - F^{'} (x_{0}) B_{0} ‖)$
	$\leq$	$b_{0} (1 + δ_{0}) \leq \frac{8}{7} b_{0} < \frac{8}{7} b$

and

‖ C_{0} ‖ \leq ‖ B_{0} ‖ + ‖ I - F^{'} (x_{0}) B_{0} ‖ \cdot ‖ B_{0} ‖ \leq ‖ B_{0} ‖ (1 + δ_{0}) \leq \frac{64}{49} b = a,

so $‖ B_{0} ‖ \leq a$ and $‖ C_{0} ‖ \leq a$ .

From $(1.2)$ we have

	$‖ x_{1} - x_{0} ‖ \leq$	$a (1 + \frac{K a^{2}}{2} ‖ F (x_{0}) ‖) ‖ F (x_{0}) ‖$
	$\leq$	$a (1 + ρ_{0}) ‖ F (x_{0}) ‖$
	$<$	$\frac{8}{7} a ‖ F (x_{0}) ‖ = \frac{2 ρ_{0}}{K a^{2}} \cdot \frac{8}{7} a = \frac{16 d}{49 K a},$

whence, taking into account d) it follows that $x_{1} \in S$ .

Further, by $(1.2)$ and $(2.1)$ ,

	$‖ F (x_{1}) ‖ \leq$	$‖ I - F^{'} (x_{0}) C_{0} ‖ (1 + \frac{1}{2} ‖ F^{''} (x_{0}) ‖ {‖ C_{0} ‖}^{2} ‖ F^{'} (x_{0}) ‖) ‖ F (x_{0}) ‖$
		$+ \frac{1}{2} {‖ F^{''} (x_{0}) ‖}^{2} {‖ C_{0} ‖}^{4} {‖ F (x_{0}) ‖}^{3} + \frac{1}{8} {‖ F^{''} (x_{0}) ‖}^{3} {‖ C_{0} ‖}^{6} {‖ F (x_{0}) ‖}^{4},$

whence

	$\frac{K a^{2}}{2} ‖ F (x_{1}) ‖ \leq$	$\frac{K a^{2}}{2} ‖ F (x_{0}) ‖ \cdot ‖ I - F^{'} (x_{0}) C_{0} ‖ (1 + \frac{K a^{2}}{2} ‖ F (x_{0}) ‖)$
		$+ 2 {(\frac{K a^{2}}{2} ‖ F (x_{0}) ‖)}^{3} + {(\frac{K a^{2}}{2} ‖ F (x_{0}) ‖)}^{4} .$

Denoting $ρ_{1} = \frac{K a^{2}}{2} ‖ F (x_{1}) ‖$ and taking into account the inequality

‖ I - F^{'} (x_{0}) C_{0} ‖ \leq {‖ I - F^{'} (x_{0}) B_{0} ‖}^{2} = δ_{0}^{2}

it follows

ρ_{1} \leq ρ_{0} δ_{0}^{2} + ρ_{0}^{2} δ_{0}^{2} + 2 ρ_{0}^{3} + ρ_{0}^{4} .

2.2

From the third relation of $(1.2)$ we get

	$‖ I - F^{'} (x_{1}) B_{1} ‖ =$	$‖ {(I - F^{'} (x_{1}) B_{0})}^{3} ‖$
	$\leq$	${‖ I - F^{'} (x_{1}) B_{0} ‖}^{3} \leq$
	$\leq$	${(‖ I - F^{'} (x_{0}) B_{0} ‖ + ‖ B_{0} ‖ K ‖ x_{1} - x_{0} ‖)}^{3}$
	$\leq$	${(δ_{0} + 2 ρ_{0} + 2 ρ_{0}^{2})}^{3},$

i.e.,

δ_{1} \leq {(δ_{0} + 2 ρ_{0} + 2 ρ_{0}^{2})}^{3} .

2.3

By $(2.2), (2.3)$ and hypothesis b), we obtain

\begin{matrix} ρ_{1} & \leq & \frac{1}{7} d^{3} \\ δ_{1} & \leq & \frac{1}{7} d^{3} . \end{matrix}

Suppose now that the following properties hold:

$𝜶$ $)$ $x_{0}, x_{1}, \dots, x_{k} \in S;$

$𝜷)$ $ρ_{i} := \frac{K a^{2}}{2} ‖ F (x_{i}) ‖ \leq \frac{1}{7} d^{3^{i}}$ and $δ_{i} := ‖ I - F^{'} (x_{i}) B_{i} ‖ \leq \frac{1}{7} d^{3^{i}}, i = \bar{0, k} .$

It easily follows that $‖ B_{k} ‖ \leq a$ , $‖ C_{k} ‖ \leq a$ and

$‖ x_{k + 1} - x_{k} ‖ \leq$	$a (1 + \frac{K a^{2}}{2} ‖ F (x_{k}) ‖) ‖ F (x_{k}) ‖$	$2.4$
$\leq$	$a (1 + ρ_{k}) ‖ F (x_{k}) ‖$
$\leq$	$\frac{16 ρ_{k}}{7 K a} \leq \frac{16 d^{3}}{49 K a} .$

From the above formula it follows that $x_{k + 1} \in S$ :

‖ x_{k + 1} - x_{0} ‖ \leq \frac{16 d}{49 K a} \sum_{i = 0}^{k} d^{3^{i} - 1} \leq \frac{16 d}{49 K a (1 - d^{2})} \leq r .

Denoting $ρ_{k + 1} = \frac{K a^{2}}{2} ‖ F (x_{k + 1}) ‖$ and $δ_{k + 1} = ‖ I - F^{'} (x_{k + 1}) B_{k + 1} ‖$ , the following relations are obtained in the same manner as for $ρ_{1}$ and $δ_{1}$ :

	$ρ_{k + 1}$	$\leq ρ_{k} δ_{k}^{2} + ρ_{k}^{2} δ_{k}^{2} + 2 ρ_{k}^{3} + ρ_{k}^{4}$
	$δ_{k + 1}$	$\leq {(δ_{k} + 2 ρ_{k} + 2 ρ_{k}^{2})}^{3},$

whence, by the Lemma, we get

\begin{matrix} ρ_{k + 1} \leq \frac{1}{7} d^{3^{k + 1}} \\ δ_{k + 1} \leq \frac{1}{7} d^{3^{k + 1}} . \end{matrix}

2.5

Then properties $𝜶)$ , $𝜷),$ $(2.4)$ and $(2.5)$ hold for all $k \in ℕ .$ We shall prove now that ${(x_{k})}_{k \geq 0}$ is a Cauchy sequence.

Indeed,

‖ x_{k + s} - x_{k} ‖ \leq \frac{16 d^{3^{k}}}{49 K a} \sum_{i = k}^{k + s - 1} d^{3^{s} - 3^{k}} \leq \frac{16 d^{3^{k}}}{49 K a (1 - d^{2 \cdot 3^{k}})},

for all $s, k \in ℕ$ , which proves that ${(x_{k})}_{k \geq 0}$ converges. Denoting $x^{*} = lim_{k \to \infty} x_{k}$ and passing to limit for $s \to \infty$ in the above inequality, we obtain

‖ x^{*} - x_{k} ‖ \leq \frac{16 d^{3^{k}}}{49 K a (1 - d^{2 \cdot 3^{k}})}, k = 0, 1, \dots

The convergence of ${(B_{k})}_{k \geq 0}$ is obtained from the third relation of $(1.2)$ :

	$‖ B_{k + 1} - B_{k} ‖ \leq$	$‖ B_{k} ‖ \cdot ‖ 2 I - F^{'} (x_{k + 1}) B_{k} ‖ \cdot ‖ I - F^{'} (x_{k + 1}) B_{k} ‖$
	$\leq$	$a (1 + δ_{k} + 2 ρ_{k} + 2 ρ_{k}^{2}) (ρ_{k} + 2 ρ_{k} + 2 ρ_{k}^{2}) \leq a \frac{1656}{2401} d^{3^{k}} .$

Denoting $B = lim B_{k}$ it easily follows that

‖ B - B_{k} ‖ \leq \frac{1656 a d^{3^{k}}}{2401 (1 - d^{2 \cdot 3^{k}})}, k = 0, 1, \dots ∎

Remark. The inequalities from the hypothesis of the Lemma can be replaced by $δ_{0} \leq α d, ρ_{0} \leq β d$ for $0 < d < 1$ and $α, β > 0$ satisfying ${(α + 2 β + 2 β^{2} d)}^{3} \leq α$ and $β α^{2} + β^{2} α^{2} + 2 β^{3} + β^{4} d \leq β .$

We shall obviously obtain in the conclusion that $δ_{k} \leq α d^{3^{k}}, ρ_{k} \leq β d^{3^{k}}, k = 0, 1, \dots .$ The theorem can be reformulated then accordingly. ∎

3. The eigenproblem for linear continuous operators

3.1. The infinite dimensional case

Let $V$ be a Banach space over the field $𝕂$ (where $𝕂 = ℂ$ or $𝕂 = ℝ$ ) and $A : V \to V$ a linear continuous operator. The scalar $λ \in 𝕂$ is an eigenvalue of $A$ iff the equation

A v - λ v = θ

3.1

has at least a solution $v^{*} \neq θ$ , called an eigenvector of $A$ corresponding to $λ$ , where $θ$ is the null element of $V$ .

For the simultaneous determination of a $v^{*}$ and a $λ$ we attach to equation (3.1) another equation

G (v) = 1,

3.2

where $G : V \to 𝕂$ is a polynomial functional of degree two for which $G (0) \neq 1$ (a norming function).

Remark. The functional $G$ may also be taken as a polynomial functional of degree one, i.e. a linear continuous functional, but then $dim Ker G = dim V - dim Im G = n - 1,$ for the finite dimensional case, and $dim Ker G = \infty$ otherwise. So, there exist eigenvectors which do not fulfill equation (3.2). ∎

Denote the Banach space $X = V \times 𝕂$ and for $x = (\binom{v}{λ})$ , with $v \in V$ and $λ \in 𝕂$ , define

‖ x ‖ = \max {‖ v ‖, | λ |} .

Considering the operator $F : X \to X$ given by

F (x) = (\begin{matrix} A v - λ v \\ G (v) - 1 \end{matrix}) = (\begin{matrix} (A - λ I) v \\ G (v) - 1 \end{matrix}),

and denoting by $\bar{θ} = (\binom{θ}{0})$ the null element of $X$ , then the operatorial equation for which the solution yields an eigenvalue and an eigenvector of $A$ is

F (x) = \bar{θ} .

3.3

It is known that the Fréchet derivatives of $F$ are (see [4]):

	$F^{'} (x_{0}) h_{1} =$	$(\begin{matrix} A - λ_{0} I & - v_{0} \\ G^{'} (v_{0}) & 0 \end{matrix}) (\begin{matrix} u_{1} \\ α_{1} \end{matrix}) = (\begin{matrix} A u_{1} - λ_{0} u_{1} - α_{1} v_{0} \\ G^{'} (v_{0}) u_{1} \end{matrix}),$
	$F^{''} (x_{0}) h_{1} h_{2} =$	$(\begin{matrix} - α_{1} u_{2} - α_{2} u_{1} \\ G^{''} (v_{0}) u_{1} u_{2} \end{matrix}),$

where $x_{0} = (\binom{v_{0}}{λ_{0}}),$ $h_{1} = (\binom{u_{1}}{α_{1}}), h_{2} = (\binom{u_{2}}{α_{2}})$ with $v_{0},$ $u_{1}, u_{2} \in V$ and $λ_{0}, α_{1}, α_{2} \in 𝕂$ .

It is obvious that $F^{(i)} (x_{0}) h_{1} \dots h_{i} = \bar{θ},$ for all $i \geq 3$ and $x_{0}$ , $h_{1}, \dots, h_{i} \in X .$

Our theorem can be applied for this function $F$ and we can take $K = \max {2, ‖ G^{''} ‖} .$

3.2. The eigenproblem for complex matrices

In the following we shall apply the studied method for the approximation of the eigenvalues and eigenvectors of complex matrices.

Let $A = {(a_{i j})}_{i, j = 1, n} \in ℳ_{n} (𝕂)$ be a square matrix with the elements $a_{i j} \in 𝕂 .$ Consider $V = 𝕂^{n}$ and $X = V \times 𝕂 = 𝕂^{n + 1} .$ In this case the equation (3.3) is written

F_{i} (x) = F_{i} (x^{(1)}, \dots, x^{(n + 1)}) = 0, i = \bar{1, n + 1},

where for $i = \bar{1, n}$ we have

F_{i} (x) = a_{i 1} x^{(1)} + \dots + a_{i, i - 1} x^{(i - 1)} + (a_{i i} - x^{(n + 1)}) x^{(i)} + a_{i, i + 1} x^{(i + 1)} + \dots + a_{i n} x^{(n)},

and for the norming function $G$ we can take

F_{n + 1} (x^{(1)}, \dots, x^{(n + 1)}) = \frac{1}{2} \sum_{i = 1}^{n} {(x^{(i)})}^{2} - 1 .

3.4

A solution $x^{*}$ of equation $F (x) = \bar{θ}$ yields an eigenvalue $λ = x_{*}^{(n + 1)}$ and a corresponding eigenvector $v^{*} = (x_{*}^{(1)}, \dots, x_{*}^{(n)})$ of the matrix $A$ .

The first and second order derivatives of $F$ are given by

F^{'} (x) h = (\begin{matrix} a_{11} - x^{(n + 1)} & a_{12} & \dots & a_{1 n} & - x^{(1)} \\ a_{21} & a_{22} - x^{(n + 1)} & \dots & a_{2 n} & - x^{(2)} \\ ⋮ & ⋮ & ⋮ & ⋮ \\ a_{n 1} & a_{n 2} & \dots & a_{n n} - x^{(n + 1)} & - x^{(n)} \\ x^{(1)} & x^{(2)} & \dots & x^{(n)} & 0 \end{matrix}) (\begin{matrix} h^{(1)} \\ h^{(2)} \\ ⋮ \\ h^{(n)} \\ h^{(n + 1)} \end{matrix}),

and

F^{''} (x) h k = (\begin{matrix} - k^{(n + 1)} & 0 & \dots & 0 & - k^{(1)} \\ 0 & - k^{(n + 1)} & \dots & 0 & - k^{(2)} \\ ⋮ & ⋮ & ⋮ & ⋮ \\ 0 & 0 & \dots & - k^{(n + 1)} & - k^{(n)} \\ k^{(1)} & k^{(2)} & \dots & k^{(n)} & 0 \end{matrix}) (\begin{matrix} h^{(1)} \\ h^{(2)} \\ ⋮ \\ h^{(n)} \\ h^{(n + 1)} \end{matrix}) .

3.5

where $x = (x^{(1)}, \dots, x^{(n + 1)}), h = (h^{(1)}, \dots, h^{(n + 1)}), k = (k^{(1)}, \dots, k^{(n + 1)}) \in 𝕂^{n + 1} .$

Suppose that $𝕂^{n + 1}$ is equipped with the norm

‖ x ‖ = \max_{1 \leq i \leq n + 1} | x^{(i)} |, x = (x^{(1)}, \dots, x^{(n + 1)}) \in 𝕂^{n + 1} .

Then, from (3.5) it follows that we can take $K = ‖ F^{''} ‖ = n .$ Our theorem can be stated accordingly.

Another possible choice for the definition of $F_{n + 1},$ is:

F_{n + 1} (x^{(1)}, \dots, x^{(n + 1)}) = \frac{1}{2 n} \sum_{i = 1}^{n} {(x^{(i)})}^{2} - 1,

3.6

in which case we can take $K = ‖ F^{''} ‖ = 2$ . In this case $K$ does not depend on $n$ .

Remark. In [15] it is proved the following result. If $(v, λ)$ is an eigenpair then $F^{'} (v, λ)$ is nonsingular iff $λ$ is simple. Hence our results apply only for simple eigenvalues.

4. Numerical examples

Consider

A = (\begin{matrix} 1 & 1 & 1 & 1 \\ 1 & 1 & - 1 & - 1 \\ 1 & - 1 & 1 & - 1 \\ 1 & - 1 & - 1 & 1 \end{matrix}),

having the eigenvalues and the corresponding eigenvectors $λ_{1, 2, 3} = 2,$ $v_{1} = (1, 0, 0, 1),$ $v_{2} = (1, 0, 1, 0),$ $v_{3} = (1, 1, 0, 0)$ , $λ_{4} = - 2,$ $v_{4} = (- 1, 1, 1, 1) .$

Taking $x_{0} = (- 1, 3, 1, 3, - 1.5),$ $B_{0} = F^{'} {(x_{0})}^{- 1}$ and using formula (3.4) for $F_{n + 1}$ we get:

\begin{matrix} k & x_{k}^{(1)} & x_{k}^{(2)} & x_{k}^{(3)} & x_{k}^{(4)} & x_{k}^{(5)} \\ 0 & - 0.3000000000 & 1.000000000 & 0.3000000000 & 1.000000000 & - 1.700000000 \\ 1 & - 0.6344924627 & 0.6757048140 & 0.6344924627 & 0.6757048140 & - 1.901115297 \\ 2 & - 0.7066264375 & 0.7068982234 & 0.7066264375 & 0.7068982234 & - 1.998773347 \\ 3 & - 0.7071067705 & 0.7071067768 & 0.7071067705 & 0.7071067768 & - 1.999999976 \\ 4 & - 0.7071067812 & 0.7071067812 & 0.7071067812 & 0.7071067812 & - 2.000000000 \end{matrix}

Using formula (3.6) for $F_{n + 1}$ and taking $x_{0} = (- 1, 1.8, 1, 1.8, - 1.6), B_{0} = F^{'} {(x_{0})}^{- 1},$ we obtain

\begin{matrix} k & x_{k}^{(1)} & x_{k}^{(2)} & x_{k}^{(3)} & x_{k}^{(4)} & x_{k}^{(5)} \\ 0 & - 1.000000000 & 1.800000000 & 1.000000000 & 1.800000000 & - 1.600000000 \\ 1 & - 1.382930642 & 1.405515579 & 1.382930642 & 1.405515579 & - 1.972622248 \\ 2 & - 1.414185197 & 1.414209835 & 1.414185197 & 1.414209835 & - 1.999975824 \\ 3 & - 1.414213562 & 1.414213562 & 1.414213562 & 1.414213562 & - 2.000000000 \end{matrix}

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