Bul.St.Univ.Baia Mare

Seria B, Matematică-Informatică, vol.VII (1991) Nr.1-2, 61-66

On the chord method

I. Păvăloiu (Cluj-Napoca)

In the paper [1], I.K. Argyros considers as divided difference of the mapping $f:X_{1}\rightarrow X_{2}$ , where $X_{1}\$ and $\ X_{2}$ are Banach spaces, a linear mapping $\left[x,y;f\right]\in\mathcal{L}\left(X_{1},X_{2}\right)$ which fulfils the following conditions:

(a)

$\left[x,y;f\right]\left(y-x\right)=f\left(y\right)-f\left(x\right),$ for every $x,y\in D,$ where $D\subseteq X_{1}$ is a subset of $X_{1};$
(b)

there exist the real constants $l_{1}\geq 0,$ $l_{2}\geq 0,$ $l_{3}\geq 0$ and $p\in(0,1]$ such that for every $x,y,u\in D$ the following inequality holds:

$\big{\|}\left[y,u;f\right]-\left[x,y;f\right]\big{\|}\leq l_{1}\left\|x-u% \right\|^{p}+l_{2}\left\|x-y\right\|^{p}+l_{3}\left\|y-u\right\|^{p}.$

In [1] the hypothesis that the equation:

(1)

f\left(x\right)=0

admits a simple solution $x^{\ast}$ in adopted, and conditions for the convergence of the sequence $\left(x_{n}\right)_{n\geq 0}$ generated by the chord method:

(2)

x_{n+1}=x_{n}-\left[x_{n-1},x_{n};f\right]^{-1}f\left(x_{n}\right),\qquad n=1,% 2,\ldots,\ x_{0},x_{1}\in D_{0}

are given.

In a recent paper [2] there is shown that, with the hypotheses considered in [1], the convergence speed of the sequence generated by (2) and the error estimation are featured by the inequality:

(3)

\left\|x^{\ast}-x_{n+1}\right\|\leq\alpha\ d_{0}^{t_{1}^{n+1}},

where $\alpha$ is a precised constant, $0<d_{0}<1$ and $t_{1}$ is the positive root of the equation $t^{2}-t-p.$

We shall admit further down that the divided difference operator fulfils the conditions (a) and (b), and search for supplementary conditions in order to make equation (1) admit a solution $x^{\ast}$ into a precised domain $D_{0}$ and the sequence $\left(x_{n}\right)_{n\geq 0}$ generated by (2) converge to this solution.

Observe firstly that the identity:

(4)

x_{n}-\left[x_{n-1},x_{n};f\right]^{-1}f\left(x_{n}\right)=x_{n-1}-\left[x_{n-% 1},x_{n};f\right]^{-1}f\left(x_{n-1}\right)

holds for every $n=1,2,\ldots$ with the hypothesis that the linear mapping $\left[x_{n-1},x_{n};f\right]$ admits an inverse mapping.

The following identity

(5)		$\displaystyle f\left(x_{n+1}\right)=$	$\displaystyle f\left(x_{n}\right)+\left[x_{n-1},x_{n};f\right]\left(x_{n+1}-x_% {n}\right)$
		$\displaystyle+\left(\left[x_{n},x_{n+1};f\right]-\left[x_{n-1},x_{n};f\right]% \right)\left(x_{n+1}-x_{n}\right),\qquad n=1,2,\ldots$

Let $B>0,\alpha>0,0<d_{0}<1,$ and $x_{0},x\in X_{1}.$ Consider the sphere

(6)

U=\Big{\{}x\in X_{1}:\|x-x_{0}\|\leq\tfrac{B\alpha d_{0}}{1-d_{0}^{t_{1}-1}}% \Big{\}}

where $t_{1}=\frac{1+\sqrt{1+4p}}{2}$ that is, the positive root, the equation:

(7)

t^{2}-t-p=0

The following theorem holds:

Theorem 1.

If the divided difference $\left[x,y;f\right]$ fulfils the conditions (a) and (b) for every $x,y\in U$ and the following hypotheses:

(1)

the mapping $\left[x,y;f\right]$ admits a bounded inverse mapping for every $x,y\in U,$ namely there exists a constant $B>0$ such that $\big{\|}\left[x,y;f\right]^{-1}\big{\|}\leq B$
(ii)

$\alpha=\tfrac{1}{B^{\left(1+p\right)/p}\left(l_{1}+l_{2}+l_{3}\right)^{1/p}};$
(iii)

$\left\|x_{1}-x_{0}\right\|\leq B\alpha d_{0},\quad\left\|f\left(x_{0}\right)% \right\|\leq\alpha d_{0},\quad\left\|f\left(x_{1}\right)\right\|\leq\alpha d_{% 0}^{t_{1}}$

are also fulfilled, then equation (1) has at least one solution $x^{\ast}\in U$ and the sequence $\left(x_{n}\right)_{n\geq 0}$ generated by (2) converges to $x^{\ast},$ the convergence speed and the error estimation being featured by the inequality:

$\left\|x^{\ast}-x_{n}\right\|\leq\tfrac{B\alpha d_{0}^{t_{1}^{n}}}{1-d_{0}^{t_% {1}^{n}\left(t_{1}-1\right)}}.$

Proof.

From (2) for $n=1$ we deduce:

\left\|x_{2}-x_{1}\right\|\leq B\left\|f\left(x_{1}\right)\right\|\leq B\alpha d% _{0}^{t_{1}}

from which, taking also into account iii. it follows

	$\displaystyle\left\\|x_{2}-x_{0}\right\\|$	$\displaystyle\leq\left\\|x_{2}-x_{1}\right\\|+\left\\|x_{1}-x_{0}\right\\|$
		$\displaystyle\leq B\alpha d_{0}^{t_{1}}+B\alpha d_{0}$
		$\displaystyle\leq B\alpha d_{0}\left(1+d_{0}^{t_{1}-1}\right)$
		$\displaystyle<\frac{B\alpha d_{0}}{1-d_{0}^{t_{1}-1}}$

from which it results that $x_{2}\in U$ .

Using the fact that $x_{2}\in U,$ the identities (4) and (5), and the inequality a), we obtain

	$\displaystyle\left\\|f\left(x_{2}\right)\right\\|$	$\displaystyle\leq\left\\|x_{2}-x_{1}\right\\|\left(l_{1}\left\\|x_{2}-x_{0}\right% \\|^{p}+l_{2}\left\\|x_{1}-x_{0}\right\\|^{p}+l_{3}\left\\|x_{2}-x_{1}\right\\|^{p}\right)$
		$\displaystyle\leq B\left\\|f\left(x_{1}\right)\right\\|\left(l_{1}B^{p}\left\\|f% \left(x_{0}\right)\right\\|^{p}+l_{2}\left\\|x_{1}-x_{0}\right\\|^{p}+l_{3}B^{p}% \left\\|f\left(x_{1}\right)\right\\|^{p}\right)$
		$\displaystyle\leq B\alpha d_{0}^{t_{1}}\left(l_{1}B^{p}\alpha^{p}d_{0}^{p}+l_{% 2}B^{p}\alpha^{p}d_{0}^{p}+l_{3}B^{p}\alpha^{p}d_{0}^{t_{1}}\right)$
		$\displaystyle\leq B^{p+1}\alpha^{p+1}d_{0}^{t_{1}+p}\left(l_{1}+l_{2}+l_{3}d_{% 0}^{p\left(t_{1}-1\right)}\right)$
		$\displaystyle=B^{p+1}\alpha^{p+1}\left(l_{1}+l_{2}+l_{3}d_{0}^{p\left(t_{1}-1% \right)}\right)d_{0}^{t_{1}^{2}-p}\leq\alpha d_{0}^{t_{1}^{2}}$

since $\alpha^{p}B^{p+1}\big{(}l_{1}+l_{2}+l_{3}d_{0}^{p\left(t_{1}-1\right)}\big{)}% \leq\alpha^{p}B^{p+1}\left(l_{1}+l_{2}+l_{3}\right)<1.$

From the above inequality follows therefore:

\left\|f\left(x_{2}\right)\right\|\leq\alpha d_{0}^{t_{1}^{2}}

Suppose by induction that:

(a’)

$x_{i}\in U,\ i=0,1,\ldots,k;$
(b’)

$\left\|f\left(x_{i}\right)\right\|\leq\alpha d_{0}^{t_{1}^{i}},\ i=1,2,\ldots,k.$

Then, for $x_{k+1}$ we have:

	$\displaystyle\left\\|x_{k+1}-x_{0}\right\\|\leq\left\\|x_{k+1}-x_{k}\right\\|+% \left\\|x_{k}-x_{k-1}\right\\|+\ldots+\left\\|x_{1}-x_{0}\right\\|$
	$\displaystyle\leq B\left\\|f\left(x_{k}\right)\right\\|+B\left\\|f\left(x_{k-1}% \right)\right\\|+\ldots+B\alpha d_{0}$
	$\displaystyle\leq B\alpha d_{0}^{t_{1}^{k}}+B\alpha d_{0}^{t_{1}^{k-1}}+\ldots% +B\alpha d_{0}$
	$\displaystyle=B\alpha d_{0}\left(1+d_{0}^{t_{1}-1}+d_{0}^{t_{1}^{2}-1}+\ldots+% d_{0}^{t_{1}^{k}-1}\right)$
	$\displaystyle\leq B\alpha d_{0}\left(1+d_{0}^{t_{1}-1}+d_{0}^{2\left(t_{1}-1% \right)}+\ldots+d_{0}^{k\left(t_{1}-1\right)}\right)\leq\tfrac{B\alpha d_{0}}{% 1-d_{0}^{t_{1}-1}}$

from which follows that $x_{k+1}\in U.$ Proceeding now for $x_{k+1},$ as in the case of $x_{2},$ we obtain:

	$\displaystyle\left\\|f\left(x_{k+1}\right)\right\\|$	$\displaystyle\leq B^{p+1}\alpha^{p+1}\left(l_{1}+l_{2}+l_{3}d_{o}^{pt_{1}^{k-1% }\left(t_{1}-1\right)}\right)d_{0}^{t_{1}^{k-1}\left(t_{1}+p\right)}$
		$\displaystyle\leq B^{p+1}\alpha^{p+1}\left(l_{1}+l_{2}+l_{3}\right)d_{0}^{t_{1% }^{k+1}}\leq\alpha d_{0}^{t_{1}^{k+1}}$

It results therefore that the relations (a’) and (b’) hold for every $i\in\mathbb{N}.$

Now we shall show that the sequence $\left(x_{n}\right)_{n\geq 0}$ is fundamental.

Indeed, for every $n,s\in\mathbb{N}$ we have:

	$\displaystyle\left\\|x_{n+s}-x_{n}\right\\|$	$\displaystyle\leq\sum_{k=n}^{n+s-1}\left\\|x_{k+1}-x_{k}\right\\|\leq\sum_{k=n}^% {n+s-1}B\left\\|f\left(x_{k}\right)\right\\|\leq B\alpha\sum_{k=n}^{n+s-1}d_{0}^% {t_{1}^{k}}$
		$\displaystyle=B\alpha d_{0}^{t_{1}^{n}}\sum_{k=n}^{n+s-1}d_{0}^{t_{1}^{k}-t_{1% }^{n}}=B\alpha d_{0}^{t_{1}^{n}}\sum_{k=n}^{n-s-1}d_{0}^{t_{1}^{n}\left(t_{1}^% {k-n}-1\right)}$
		$\displaystyle\leq B\alpha d_{0}^{t_{1}^{n}}\sum_{k=n}^{n+s-1}d_{0}^{t_{1}^{n}% \left(k-n\right)\left(t_{1}-1\right)}=B\alpha d_{0}^{t_{1}^{n}}\sum_{k=n}^{n+s% -1}\left(d_{0}^{t_{1}^{n}\left(t_{1}-1\right)}\right)^{k-n}$
		$\displaystyle\leq\tfrac{B\alpha d_{0}^{t_{1}^{n}}}{1-d_{0}^{t_{1}^{n}\left(t_{% 1}-1\right)}}.$

By the last inequality and the fact that $0<d_{0}<1$ and $t_{1}>1$ follows that the sequence $\left(x_{n}\right)_{n\geq 2}$ is fundamental. For $s\rightarrow\infty,$ from the inequality:

\left\|x_{n+s}-x_{n}\right\|\leq\tfrac{B\alpha d_{0}^{t_{1}^{n}}}{1-d_{0}^{t_{% 1}^{n}\left(t_{1}-1\right)}}

follows the inequality:

\left\|x^{\ast}-x_{n}\right\|\leq\tfrac{B\alpha d_{0}^{t_{1}^{n}}}{1-d_{0}^{t_% {1}^{n}\left(t_{1}-1\right)}}.

In [1] Argyros showed that if the divided difference $\left[x,y;f\right]$ fulfils the conditions (a) and (b) then $f$ is Fréchet differentiable and $\left[x,x;f\right]=f^{\prime}\left(x\right).$ From this fact follows that the mapping $f$ is continuous on $B$ : hence at limit for $n\rightarrow\infty$ in the inequality:

\left\|f\left(x_{n}\right)\right\|\leq\alpha d_{0}^{t_{1}^{n}},

one obtains:

\left\|f\left(x^{\ast}\right)\right\|\leq 0

from which results $f\left(x^{\ast}\right)=0.$ With this the theorem is entirely proved ∎

Remark 2.

In [5], [6] Schmidt imposes in the divided difference conditions similar to the conditions (a) and (b) given by Argyros in [1], but for $p=1$ . The same conditions are reproduced in [2], too.

References

[1] Argyros, I. K., The secant method and fixed points on nonlinear operators, Mh. Math., 106 (1988), 85–94.
[2] Balázs, M. şi Goldner, G., ^†^†margin: clickable $\rightarrow$ Observaţii asupra diferenţelor divizate şi asupra metodei coardei. Revista de analiză numerică şi teoria aproximaţiei vol. 3 (1974) fasc. 1, 19–30.
[3] ^†^†margin: clickable $\rightarrow$ Păvăloiu, I., Remarks on the secant method for the solution of nonlinear operatorial equations, Research Seminars, Seminar on Mathematical Analysis, Preprint no. 7, (1991), pp. 127–132.
[4] Păvăloiu, I., ^†^†margin: clickable $\rightarrow$ Introducere în teoria aproximării soluţiilor ecuaţiilor. Ed. Dacia, Cluj-Napoca, 1976.
[5] Schmidt, I. W., Eine Ubertagungen der Regula Falsi auf Gleichungen, in ”Banachräumen” I ZAMM, 48, 1–8 (1963).
[6] Schmidt, I. W., Eine Ubertagungen der Regula Falsi auf Gleichungen, in ”Banachräumen” II 97–110 (1963).

Institutul de Calcul

Academia Română

Cluj-Napoca

On the chord method

Abstract

Authors

Keywords

PDF

Cite this paper as:

About this paper

Journal

Publisher Name

Stable JSTOR

Print ISSN

Online ISSN

References

Paper (preprint) in HTML form

On the chord method

Theorem 1.

Proof.

Remark 2.

References

Related Posts