On the convergence order of the iterative methods

Abstract

We study the connection between the convergence order of two sequences. We show that the exist sequences that do not have exact convergence order.

We apply the obtained results to the study of the convergence order of the iterative methods.

Authors

Ion Păvăloiu
(Tiberiu Popoviciu Institute of Numerical Analysis)

Title

Original title (in French)

Sur l’ordre de convergence des méthodes d’itération

English translation of the title

On the convergence order of the iterative methods

Keywords

convergence order; nonlinear equation; iterative method

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Notes

Cite this paper as:

I. Păvăloiu, Sur l’ordre de convergence des méthodes d’itération, Mathematica, 23(46) (1981) no. 1, pp. 261-272 (in French).

About this paper

Journal

Mathematica

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DOI

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References

[1] Brezinski, C., Comparation de suites convergentes, Revue francaise d’informatique et de recherches operationnelles, Nr. R-2, 95–99, (1971).

[2] Pavaloiu, I., Introducere în teoria aproximarii solutiilor ecuatiilor, Editura Dacia, Cluj-Napoca, (1976).

[3] Pavaloiu, I., Sur les procedes iteratifs a un ordre eleve de convergence, Mathematica, Cluj, 12(35), 2 309–324, (1970).

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On the convergence order of the iterative methods

by
Ion Pavaloiu

In the work [ 1 ] C. Brezinski defined the notion of order of convergence of a sequence and a notion of comparison of the speeds of convergence of two sequences.

In numerical analysis, a set of methods for solving equations leads to the approximation of the solutions of the equations considered by elements of a suitably constructed sequence.

If we can construct a sequence whose terms approximate the solution of the given equation, then it is very important to know the speed of convergence of this sequence towards the solution of the equation considered.

In this note we will look for a relationship between the order of convergence of two sequences that have the same speed of convergence. We will apply the results to a problem of convergence of numerical methods for solving equations.

Either(X,r)a metric space and(inn)n=0∞a sequence of elements of space X. We assume that the following(inn)n=0∞ converges to the elementin∈X.

We designate byrany real and positive number.

Definition 1 .

[ 1 ] . We say that the following(inn)n=0∞has the order of convergencerif there are two positive constantsC1AndC2in such a way that for each n=0,1,…,inequalities take place

(1) C2⁢r⁢(inn,in)r≤r⁢(inn+1,in)≤C1⁢r⁢(inn,in)r,n=0,1,…

Concerning the order of convergence of a sequence, in the work [ 1 ] we demonstrate the following theorem.

Theorem 1 .

Yes, please.(inn)n=0∞has the order of convergencer, SOris unique.

Be it now(inn)n=0∞,inn ∈X,n=0,1,…,And(inn)n=0∞,inn∈X,n=0,1,…,two sequences that converge to the same limitin∈X.

Definition 2 .

[ 1 ] . We say that the following(inn)n=0∞converges at least as quickly as(inn)n=0∞if the following inequalities are verified:

(2) r⁢(inn,in)≤C⁢r⁢(inn,in),n=0,1,…,

OrCis a positive constant independent ofn.

Definition 3 .

[ 1 ] We say that the suites(inn)n=0∞, inn∈X,n=1,2,…,And(inn)n=0∞,inn∈X,n=1,2,…,have the same speed of convergence if there are two positive constantsC1And C2such that, for eachn=0,1,…,the inequalities are satisfied:

(3) C1⁢r⁢(inn,in)≤r⁢(inn,in)≤C2⁢r⁢(inn,in)
Note 1 .

It is easy to show that if the inequalities ( 3 ) hold, then obviously, for eachn=0,1,…,inequalities take place

(4) 1C2⁢r⁢(inn,in)≤r⁢(inn,in)≤1C1⁢r⁢(inn,in)
Note 2 .

Concerning the order of convergence of a sequence, we will now show, by a simple example, that there exist sequences without the order of convergence.

EitherX=RAnd(an)n=0∞a sequence of real numbers defined using equalities.

(5) {a2⁢k+1=12⁢k+1,k=0,1,…,a2⁢k=1kk,k=1,2,…

The sequel(xn)n=0∞defined above is convergent verso0.

We will show that this sequence has no order of convergence in the sense of definition 1 .

From inequalities ( 1 ) we deduce the following inequalities:

C2≤r⁢(inn+1,in)r⁢(inn,in)r≤C1,n=0,1,…,

from which it follows that for the continuation(inn)n=0∞has an order of convergencerit is necessary that there exists arin such a way that the following(dn)n=0∞given by equalitydn=r⁢(inn+1,in)r⁢(inn,in)r,n=0,1,…,be limited.

But, in the case of the sequence ( 5 ) we have

dn=an+1anr,n=0,1,…,

sequence which is not bounded, because its subsequence(d2⁢k)k=1∞is not limited for anyr>0because we have:

d2⁢k=a2⁢k+1(a2⁢k)k=kr⁢k2⁢k+1,k=0,1,…∎

Now we will prove the following theorem:

Theorem 2 .

If the sequels(inn)n=0∞And (inn)n=0∞⁢inn,inn∈X,n=0,1,…,have the same speed of convergence, and if one of the two sequences has an order of convergence, then the other sequence has the same order of convergence.

Demonstration..

To clarify the ideas we designate byrthe order of convergence of the sequence(inn)n=0∞. Because the two sequences have the same speed of convergence, the following inequalities result:

(6) C1⁢r⁢(inn,in)≤r⁢(inn,in)≤C2⁢r⁢(inn,in),n=0,1,…,

and taking into account that the following(inn)n=0∞has the order of convergencerresults:

(7) Cr′⁢r⁢(inn,in)r≤r⁢(inn+1,in)≤C2′⁢r⁢(inn,in)r,n=0,1,…,

OrC1,C2,C1′,C2′are positive constants, independent ofn.

From ( 6 ) and ( 7 ) we deduce:

(8) r⁢(inn+1,in)≤C2⁢r⁢(inn+1,in)≤C2⁢C2′⁢r⁢(inn,in)r≤C2⁢C2′C1r⁢r⁢(inn,in)r,n=0,1,…

Still from ( 6 ) and ( 7 ) we deduce:

(9) r⁢(inn+1,in)≥C1⁢r⁢(inn+1,in)≥C1⁢C1′⁢r⁢(inn,in)r≥C1⁢C1′C2r⁢r⁢(inn,in)r,n=0,1,…

From inequalities ( 8 ) and ( 9 ) the double inequality results:

(10) C1⁢C1′C2r⁢r⁢(inn,in)r≤r⁢(inn+1,in)≤C2⁢C2′C1r⁢r⁢(inn,in)r,n=0,1,…

where the constantsC1⁢C1′C2rAndC2⁢C2′C1rdo not depend onn.

From Theorem 1 it follows that the real and positive numberrfor which the inequalities ( 10 ) take place is unique.

Subsequently, we will apply the results presented above to specify the order of convergence of the numerical methods for solving operational equations.

LetXAndANDtwo normed linear spaces and either

(11) P⁢(x)=i

an equation, whereP:X→ANDis an application andi is the zero element of the spaceAND.

We designate byx¯∈Xa solution of equation ( 11 ) and we assume that we have constructed a sequence(xn)n=0∞of elements of spaceX, which converges towardsx¯.

In practice we do not always manage to obtain the exact solution.x¯of equation ( 11 ) but we can approximate the elementx¯by a suitable element of the sequence (xn)n=0∞.

It is also important to know what is the convergence speed of the sequence(xn)n=0∞towardsx¯ and to evaluate what is the number of steps such thatxn′ considered as an approximation of the solutionx¯can replace this solution with a precision given in advance.

But we don't know the elementx¯and this is why it is very difficult to apply the results presented in this note to study the speed of convergence of the sequence(xn)n=0∞.

A simple criterion to verify thatxn,approachx¯is to see what the distance is between the real number‖P⁢(xn)‖And0. From this remark we propose to study the speed of convergence of the sequence(P⁢(xn))n=0∞.

We now assume that the applicationPis continuous and then from the fact thatx¯=limn→∞xnit follows that limn→∞‖P⁢(xn)‖=0.

Using notions similar to those presented in the first part of this work, we can clarify the notion of order of convergence for the following(xn)n=0∞which approaches the solution x¯of equation ( 11 ). ∎

Definition 4 .

We say that the sequel(xn)n=0∞has the order of convergencerrelative to equation ( 11 ), if there are two positive constants,r1,r2in such a way that for eachn=0,1,…,we have:

(12) r1⁢‖P⁢(xn)‖r≤‖P⁢(xn+1)‖≤r2⁢‖P⁢(xn)‖r,

It obviously follows from Theorem 1  that if (xn)n=0∞has the order of convergencer, SO ris unique.

We now consider two sequences(xn)n=0∞And(andn)n=0∞which we assume have the same limitx¯∈XOrx¯is the solution to equation ( 11 ).

By analogy with the notions specified in definitions 2 and 3 we present the following definitions:

Definition 5 .

We say that the sequel(xn)n=0∞converges at least as quickly as(andn)n=0∞relative to equation ( 11 ) if the inequalities are verified:

(13) ‖P⁢(xn)‖≤r⁢‖P⁢(andn)‖,n=0,1,…

Orr>1is a constant independent ofn.

Definition 6 .

We say that the sequels(xn)n=0∞And(andn)n=0∞have the same speed of convergence relative to equation ( 11 ) if there are two positive constantsr1Andr2in such a way that for eachn=0,1,…,we have:

r1⁢‖P⁢(xn)‖≤‖P⁢(andn)‖≤r2⁢‖P⁢(xn)‖.

A consequence of Theorem 2 is the following.

Theorem 3 .

If the sequels(xn)n=0∞And (andn)n=0∞have the same speed of convergence relative to equation ( 11 )  and if one of the two sequences has an order of convergence relative to equation ( 11 ) then the other sequence has the same order of convergence relative to equation ( 11 ).


Bibliography

Received, 20.IV.1980


Institute of Mathematics,

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PO Box 68 3400 Cluj-Napoca

Romania

1981

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