Remarks on the secant method for the solution of nonlinear operator equations

Abstract

Let \(X_{1},X_{2}\) be two Banach spaces and \(f:X_{1}\rightarrow X_{2}\) a nonlinear equation. We study the chord method for solving the equation \(f\left( x\right) =0\). Assuming the first order divided differences of \(f\) satisfy a Holder type condition, we obtain sufficient convergence conditions and error estimations at each step.

Authors

Ion Păvăloiu
(Tiberiu Popoviciu Institute of Numerical Analysis)

Keywords

chord method; divided differences; Holder condition

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Cite this paper as:

I. Păvăloiu, 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.

About this paper

Journal

Seminar on mathematical analysis,
Preprint

Publisher Name

“Babes-Bolyai” University,
Faculty of Mathematics,
Research Seminars

DOI

Not available yet.

References

[1] Argyros, I.K., The secant method and fixed points of nonlinear operators, Mh. Math. 106, 85 94 (1988).

[2] Dennis, J.E., Toward a unified convergence theory for Newton like methods, Nonlinear Functional analysis and Applications (Ed. by L.B. Rall), pp. 425–472, New York, John Wiley (1986).

[3] Pavaloiu, I., Introduction to the Theory of Approximation of Equations Solutions, Dacia Ed., Cluj-Napoca, 1976 (in Romanian)

Paper (preprint) in HTML form

Remarks on the secant method for the solution of nonlinear operatorial equations

"Babeş-Bolyai" University

Faculty of Mathematics and Physics

Research Seminars

Seminar on Mathematical Analysis

Preprint Nr.7, 1991, pp.127-132

Remarks on the secant method for the solution of nonlinear operatorial equations

Ion Păvăloiu

This note has for purpose some refinements of the convergence conditions and error delimitations obtained by I.K. Argyros in [2] with respect to the secant method for the solution of the equation:

(1) f(x)=0,

where f:X1X2 is a nonlinear operator, while X1 and X2 are Banach spaces.

If we denote by [x,y;f] the divided difference of the mapping f on the point x and y, then for fixed x,y we have [x,y;f](X1,X2). It is known that in certain conditions the sequence (xn)n0 generated by the secant method:

(2) xn+1=xn[xn1,xn;f]1f(xn),x0,x1X1,n=1,2,

converges to the solution x of equation (1).

1. Generalizing a result on J.E. Dennis [3], I.K. Argyros [2] studies the convergence of the method (2) with the assumptions that the operator f is Fréchet differentiable, while the derivative f(x) fulfils a Hölder-like condition on a set DX1,namely there exist a constant C>0 and number p(0,1] such that the inequality:

(3) f(x)f(y)Cxyp

holds for every x,yD. In this case we shall say that f()HD(C,p).

In the quoted paper I.K. Argyros defines the divided difference operator [x,y;f] as a linear operator which fulfils the conditions:

(4) [x,y;f](yx)=f(y)f(x),x,yD,

and, in addition, for every x,y,uD the following inequality holds:

(5) [x,y;f][y,u;f]l1xup+l2xyp+l2yup,

where l10,l20 are constants which do not depend on x,y and u, while p(0,1].

Let x be a simple solution of (1). We mean by that the mapping f(x) admits a bounded inverse mapping, and if [x,x;f]=f(x) then [x,x;f] admits a bounded inverse mapping. Thus the continuity of the mapping [x,y;f] with respect to the variable x and y ensures the existence of a number ε>0 such that the mapping [x,y;f] admits a bounded inverse mapping for every x,yU(x,ε), where U(x,ε)={xX1:xx<ε} that is, the set B(x,y)=[x,y;f]1 is uniformly bounded in U(x,ε)={xX1:xxε}.

Theorem 1.

[2] Let f:X1X2 and let DX1 an open set. The following conditions are fulfilled:

  • (a)

    xD is a simple solution of the equation (1);

  • (b)

    there exist ε0,b>0 such that [x,y;f]1b for every x,yU(x,ε);

  • (c)

    there exists a convex set D0D such that xD0, and there exists ε1>0, with 0<ε1<ε such that f()HD0(C,p) for every x,yD0 and U(x,ε1)D0.

Let r>0 such that:

(6) 0<r<min{ε1,(q(p))1/p}

where:

(7) q(p)=bp+1[2p(l1+l2)(1+p)+C].

Then, if x0x1U¯(x,r), the iterates xn,n=2,3,, generated by (2) are well defined and belong to the set U¯(x,r), while the sequence (xn)n0 converges to the unique solution x of equation (1).

Moreover, the following estimation:

(8) xn+1xγ1xn1xpxnx+γ2xnxp+1

holds for sufficiently great n, where:

(9) γ1=b(l1+l2)2p,
(10) γ2=bC1+p

while l1,l2 and p were precised by the relation (5).

In order to prove this theorem the author uses the following two lemmas:

Lemma 1.

[2]. Let f:X1X2 and DX1. Suppose that D is an open set and f() does exist in every point of D. If, for a convex set D0D,f()HD0(C,p), then for every x,yD0 the following inequality holds:

f(x)f(y)f(x)(yx)C1+pxy1+p.
Lemma 2.

[2]. If [x,y;f] fulfils the conditions (4) and (5), the following relations hold:

  • (a)

    [x,x;f]=f(x) for every xD0;

  • (b)

    f()HD0(2(l1+l2),p).

From the proof of Theorem 1 follows, for the error estimation and for the convergence speeds of the sequence (xn)n0, the inequality:

(11) xn+1x(M(r))n+1x0x

where one shows that M(r)(0,1).

2. We shall make further down some remarks upon the above exposed results, showing that the hypotheses imposed in [2] can lead to more rich conclusions with respect to both the convergency order of the secant method and the error estimation.

Suppose that x0 and x1 fulfil the conditions:

  • (a’)

    xx0αd0;

  • (b’)

    xx1min{αd0t1,xx0}

where 0<d0<1,α=(q(p))1, while t1 is the positive root of the equation:

(12) t2tp =0
namely t1 =1+(1+4p)1/22.

Using the condition (4) and (5), Lemmas 1 and 2, and the hypotheses of 1, it results easily from (2),for n=1, the inequality [2]:

(13) x2xγ1x0xpx1x+γ2x1xp+1

from which, using (a’) and (b’) and the fact that t1 is a root of equation (12), we obtain:

x2x γ1αpd0pαd0t1+γ2α1+pd0t1(1+p)
=α1+p(γ1d0t1+p+γ2d0t1(1+p))
=α1+pd0t1+p(γ1+γ2d0p(t11))
=αd0t12(γ1+γ2d0p(t11))αp.

But

(γ1+γ2d0p(t11))αp=γ1+γ2d0p(t11)γ1+γ2<1,

then the following inequality holds

x2xαd0t12.

We prove now that x2xx1x. From the inequality (13) we obtain:

x2x(γ1αpd0p+γ2αpd0t1p)x1x
αpd0p(γ1+γ2d0p(t11))x1x<x1x

since d0p<1 and, as we saw above, αp(γ1+γ2d0p(t11))<1.

Assume now that for n, n2, the following relations hold:

  • (a”)

    xn1xαd0t1n1;

  • (b”)

    xnxmin{αd0t1n,xn1x}

Proceeding as in the case of x2, and taking into account (a”), (b”) and (8), we obtain:

xn+1x α1+pd0t1n+1(γ1+γ2d0pt1n1(t1+1))=
=αd0t1n+1αp(γ1+γ2d0pt1n1(t11))αd0t1n+1,

since, as previously, it is easy to show that:

αp(γ1+γ2d0pt1n1(t11))<1

In order to complete the proof, we shall show that:

xn+1xxnx

Indeed, form (8) we deduce:

xn+1x(γ1αpd0pt1n1+γ2αpd0pt1n)xnx.

But d0<1 and αp(γ1+γ2d0p(t11))<1, therefore:

xn+1xxnx.

We proved in this way the following theorem:

Theorem 2.

If the conditions of Theorem 1 are fulfilled, with the difference that x0 and x1 are chosen in such a manner to verify the relations (a’) and (b’), where α=(q(p))1/p and d0(0,1), then, for every n,xnU={xX1|xx<α} and the following inequality holds:

(14) xn+1xαd0t1n+1,n=0,1,
Remark.

The inequality (14) contains in its right-hand side a number substantially smaller than that yielded by relation (11).

References


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This paper is in final form and no version of it is or will be submitted for publication elsewhere.

1991

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