A convergency theorem concerning the chord method

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Ion Păvăloiu
(Tiberiu Popoviciu Institute of Numerical Analysis)

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I. Păvăloiu, A convergency theorem concerning the chord method, Rev. Anal. Numér. Théor. Approx., 22 (1993) no. 1, pp. 83-85.

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References

[1] Argyros, K.I., The Secant Method and Point of Nonlinear Operators Mh. Math. 106 (1988), 85–94.

[2] Pavaloiu, I., Introduction in the theory of approximation of equations solutions, Dacia Ed., Cluj-Napoca 1976, (in Romanian).

[3] Pavaloiu, I., Remarks on the secant method for the solution of nonlinear operational equations, Research Seminars. Seminar on Mathematical analysis, Preprint 7, 127–132 (1991).

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A convergency theorem concerning the chord method

A convergency theorem concerning
the chord method

Ion Păvăloiu
(Cluj-Napoca)

Let X be a Banach space, and let f:XX be a mapping to solve the equation:

(1) f(x)=0,

the chord method is well known, consisting of approximating the solution of (1) by elements of the sequence (xn)n0 generated by the following relations:

(2) xn+1=xn[xn1,xn;f]1f(xn),n=1,2,,x0,x1X,

where [x,y;f](X) stands for the divided difference of f on x,yX. It is clear that to generate the elements of the sequence (xn)n0 by means of (2) we must ensure ourselves that at every iteration step the linear mapping [xn1,xn;f] is invertible. The mathematical literature dealing with the convergency of the chord method contains results which state by hypothesis that the mapping [x,y;f] admits a bounded inverse for every x,yD, where D is a subset of X.

In this note we intend to establish convergency conditions for the method (2), supposing the existence of the inverse mapping only for the divided difference [x0,x1;f].

Let r>0 be a real number, and write S(x0,r)={xX:xx0r}.

Theorem.

If the mapping f:XX, the real number r>0 and the element x1X fulfil the conditions:

  • (i)

    the mapping [x0,x1;f] admits a bounded inverse mapping, and [x0,x1;f]1B<+;

  • (ii)

    the bilinear mapping [x,y,z;f] (the second order divided difference of f on x,y,z) is bounded for every x,y,zS(x0,r), that is, [x,y,z;f]L<;

  • (iii)

    3BLr<1;

  • (iv)

    ρ0=αf(x0)<1, ρ1=αf(x1)ρ0t1, where
    α=LB2/(13BLr)2 and t1=(1+5)/2;

  • (v)

    Bρ0/[α(1ρ0t11)(13BLr)]r,

    then the following properties hold:

  • (j)

    xnS(x0,r) for every n=0,1,;

  • (jj)

    the mapping [xi1,xi;f] admits bounded inverse for every i=1,2,;

  • (jjj)

    equation (1) has at least one solution xS(x0,r);

  • (jv)

    the sequence (xn)n0 is convergent, and limxn=x;

  • (v)

    xxnBρ0t1n[α(13BLr)(1ρ0t1n(t11))].

Proof.

We shall firstly show that for every x,yS(x0,r) the following inequality holds:

(3) [x0,x1;f]1([x0,x1;f][x,y;f])3BLr<1.

Taking into account hypothesis (ii) and the definition of the second order divided difference [2], it results:

[x0,x1;f][x,y,f] [x0,x1;f][x1,x;f]+[x1,x;f][x,y;f]
Lxx0+Lyx1<3Lr.

From the above inequality and hypothesis (i) there follows (3).

Using Banach’s lemma on inverse mapping continuousness, it results from (3) that there exists [x,y;f]1, and:

[x,y;f]1B/(13BLr).

Suppose now that the following properties hold:

  • (a)

    xiS, i=0,k¯;

  • (b)

    ρi=αf(xi)ρ0t1i,i=0,k¯;

and prove that they hold for i=k+1, too.

Indeed, to prove that xn+1S we estimate the difference:

xk+1x0 i=0kxi+1xiBα113BLri=0kαf(xi)
Bρ0[α(1ρ0t11)(13BLr)]1r

To prove (b) for i=k+1 we use Newton’s identity:

(4) f(z)=f(x)+[x,y;f](zx)+[x,y,z;f](zx)(zy)

and the obvious identity:

(5) x[x,y;f]1f(x)=y[x,y;f]1f(y).

Applying (4) and taking into account (2) and (5), we deduce:

f(xk+1) =f(xk+1)f(xk)[xk1,xk;f](xk+1xk)
[xk1,xk,xk+1;f]xk+1xkxk+1xk1
LB2f(xk)f(xk1)(13BLr)2
LB2(13BLr)2α2ρkρk1,

and writing ρk+1=αf(xk+1) we obtain:

ρk+1ρkρk1<ρ0t1k+t1k1=ρ0t1k+1

that is, the property (b) holds for i=k+1, too.

From (2) one obtains the following inequalities:

xn+1xnBα1(13BLr)1ρnBρ0t1nα(13BLr)

for every n=0,1,

From these relations, for every m,n we deduce:

(6) xn+mxn i=nm+n1Bρ0t1nα(13BLr)
Bρ0t1nα1(13BLr)1(1ρ0t1n(t11))1

from which, taking into account the fact that t1>1, there follows that the sequence (xn)n0 is fundamental.

At limit (m), (6) leads to

xxn<Bρ0t1nα1(13BLr)1(1ρ0t1n(t11))1

where x=limnxn. For n=0 follows that xS(x0,r).

It is obvious that f(x)=0.

Remark.

In the conditions of the above proved theorem, it results form (3) that x is the unique solution of equation (1) in the sphere S(x0,r).

Indeed, supposing that x and y are two solutions of equation (1) in S(x0,r),xy, and using the identities:

x =x[x0,x1;f]1f(x)
y =y[x0,x1;f]1f(y)

we deduce

xy=(I[x0,x1;f]1[x,y;f])(xy)

from which, taking into account (3) it follows that:

xy3BLrxy

but, since 3BLr<1, it results that the relation xy is impossible. ∎

References


Received 1.III.1992              Institutul de Calcul

Academia Română, Filiala Cluj-Napoca

C.P.68, Cluj-Napoca

Romania

1993

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