A convergency theorem concerning the chord method

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Authors

Ion Păvăloiu
(Tiberiu Popoviciu Institute of Numerical Analysis)

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Cite this paper as:

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

Ion Păvăloiu
(Cluj-Napoca)

Let X be a Banach space, and let f:X→X 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)n≥0 generated by the following relations:

(2) xn+1=xn−[xn−1,xn;f]−1⁢f⁢(xn),n=1,2,…,x0,x1∈X,

where [x,y;f]∈ℒ⁢(X) stands for the divided difference of f on x,y∈X. It is clear that to generate the elements of the sequence (xn)n≥0 by means of (2) we must ensure ourselves that at every iteration step the linear mapping [xn−1,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,y∈D, 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)={x∈X:‖x−x0‖≤r}.

Theorem.

If the mapping f:X→X, the real number r>0 and the element x1∈X fulfil the conditions:

  • (i)

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

  • (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,z∈S⁢(x0,r), that is, ‖[x,y,z;f]‖≤L<∞;

  • (iii)

    3⁢B⁢L⁢r<1;

  • (iv)

    ρ0=α⁢‖f⁢(x0)‖<1, ρ1=α⁢‖f⁢(x1)‖≤ρ0t1, where
    α=L⁢B2/(1−3⁢B⁢L⁢r)2 and t1=(1+5)/2;

  • (v)

    B⁢ρ0/[α⁢(1−ρ0t1−1)⁢(1−3⁢B⁢L⁢r)]≤r,

    then the following properties hold:

  • (j)

    xn∈S⁢(x0,r) for every n=0,1,…;

  • (jj)

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

  • (jjj)

    equation (1) has at least one solution x∗∈S⁢(x0,r);

  • (jv)

    the sequence (xn)n≥0 is convergent, and limxn=x∗;

  • (v)

    ‖x∗−xn‖≤B⁢ρ0t1n[α⁢(1−3⁢B⁢L⁢r)⁢(1−ρ0t1n⁢(t1−1))].

Proof.

We shall firstly show that for every x,y∈S⁢(x0,r) the following inequality holds:

(3) ‖[x0,x1;f]−1⁢([x0,x1;f]−[x,y;f])‖≤3⁢B⁢L⁢r<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]‖
≤L⁢‖x−x0‖+L⁢‖y−x1‖<3⁢L⁢r.

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]−1‖≤B/(1−3⁢B⁢L⁢r).

Suppose now that the following properties hold:

  • (a)

    xi∈S, i=0,k¯;

  • (b)

    ρi=α⁢‖f⁢(xi)‖≤ρ0t1i,i=0,k¯;

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

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

‖xk+1−x0‖ ≤∑i=0k‖xi+1−xi‖≤B⁢α−11−3⁢B⁢L⁢r⁢∑i=0kα⁢‖f⁢(xi)‖
≤B⁢ρ0⁢[α⁢(1−ρ0t1−1)⁢(1−3⁢B⁢L⁢r)]−1≤r

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

(4) f⁢(z)=f⁢(x)+[x,y;f]⁢(z−x)+[x,y,z;f]⁢(z−x)⁢(z−y)

and the obvious identity:

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

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

‖f⁢(xk+1)‖ =‖f⁢(xk+1)−f⁢(xk)−[xk−1,xk;f]⁢(xk+1−xk)‖
≤‖[xk−1,xk,xk+1;f]‖⋅‖xk+1−xk‖⋅‖xk+1−xk−1‖
≤L⁢B2⁢‖f⁢(xk)‖⋅‖f⁢(xk−1)‖⋅(1−3⁢B⁢L⁢r)−2
≤L⁢B2⁢(1−3⁢B⁢L⁢r)−2⋅α−2⁢ρk⁢ρk−1,

and writing ρk+1=α⁢‖f⁢(xk+1)‖ we obtain:

ρk+1≤ρk⁢ρk−1<ρ0t1k+t1k−1=ρ0t1k+1

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

From (2) one obtains the following inequalities:

‖xn+1−xn‖≤B⁢α−1⁢(1−3⁢B⁢L⁢r)−1⁢ρn≤B⁢ρ0t1nα⁢(1−3⁢B⁢L⁢r)

for every n=0,1,…

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

(6) ‖xn+m−xn‖ ≤∑i=nm+n−1B⁢ρ0t1nα⁢(1−3⁢B⁢L⁢r)
≤B⁢ρ0t1n⁢α−1⁢(1−3⁢B⁢L⁢r)−1⁢(1−ρ0t1n⁢(t1−1))−1

from which, taking into account the fact that t1>1, there follows that the sequence (xn)n≥0 is fundamental.

At limit (m→∞), (6) leads to

‖x∗−xn‖<B⁢ρ0t1n⁢α−1⁢(1−3⁢B⁢L⁢r)−1⁢(1−ρ0t1n⁢(t1−1))−1

where x∗=limn→∞xn. For n=0 follows that x∗∈S⁢(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),x∗≠y∗, and using the identities:

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

we deduce

x∗−y∗=(I−[x0,x1;f]−1⁢[x∗,y∗;f])⁢(x∗−y∗)

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

‖x∗−y∗‖≤3⁢B⁢L⁢r⁢‖x∗−y∗‖

but, since 3⁢B⁢L⁢r<1, it results that the relation x∗≠y∗ 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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