An algorithm in the solving of equations by interpolation

Abstract

Consider the nonlinear equation in \(R\), \(f\left( x\right) =0\), where \(f:A\rightarrow B \), \((A,B\subseteq \mathbb{R})\) which is assumed bijective. The Lagrange inverse interpolation polynomial leads to an iterative method of the form \[x_{i+n+2}=L_{n}[ y_{i+1},y_{i+2},…,y_{i+n+1}:f_{|0}^{-1}],\ \quad x_{1},x_{2},…,x_{n+1}\in I.\] At each iterative step we need to compute the values \(\omega_{k}\left( x\right) \) and \(\omega_{k}^{\prime}\left( y_{i}\right) ,\ i=k,\ldots,k+n\), where \(\omega_{k}\left( y\right) =\Pi_{i=k}^{k_{n}}\left( y-y_{k}\right) \). We give an algorithm for computing \(\omega_{k+1}\left( 0\right) \) and \(\omega_{k+1}^{\prime}\left( y_{i}\right) \) using \(\omega_{k}\left(0\right) \) and \(\omega_{k}^{\prime}\left( y\right) \).

Authors

Ion Păvăloiu
(Tiberiu Popoviciu Institute of Numerical Analysis)

Title

Original title (in French)

Un algorythme de calcul dans la résolution des equations par interpolation

English translation of the title

An algorithm in the solving of equations by interpolation

Keywords

nonlinear equation in \(\mathbb{R}\); inverse interpolation; iterative method; Lagrange polynomial; inverse Lagrange interpolation polynomial

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Cite this paper as:

I. Păvăloiu, Un algorythme de calcul dans la résolution des equations par interpolation, Seminar on functional analysis and numerical methods. Preprint no. 1 (1987), pp. 130-134 (in French).

About this paper

Journal

Seminar on functional analysis and numerical methods,
Preprint

Publisher Name

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

DOI

Not available yet.

References

[1] I. Pavaloiu, Rezolvarea ecuatiilor prin interpolare, Editura Dacia, Cluj-Napoca, 1981.

Paper (preprint) in HTML form

"Babeş-Bolyai" University

Faculty of Mathematics and Physics

Research Seminars

Seminar on Functional Analysis and Numerical Methods

Preprint Nr.1, 1987, pp.130-134


An algorithm in the solving of equations by interpolation

Ion Pavaloiu

Let us designate byf:Ia function whereIis an interval of the real axis. We consider the equation

(1) f(x)=i,

and we assume that this equation has only one solutionx¯I. We designate byF=f(I)the set of velors of the functionfForxI. We will assume that the functionfadmits an inverse functionf1:FI.

We designate byxiI,i=1,2,,n+1,xixjFor ij, n+1different approximations of the solutionx¯from equation ( 1 ) and we writeandi=f(xi),i=1,2,,n+1.

The polynomial

(2) Ln(and1,and2,,andn+1;f1|and)=i=1n+1xioh(and)(andandi)oh(andi)

Or

(3) oh(and)=i=1n+1(andandi)

check the conditions

(4) Ln(and1,and2,,andn+1;f1|andi)=xi,i=1,2,,n+1.

Andx¯I,SO0FAndf1(0)=x¯. In this case, make themand=0an ( 2 ) we obtain an approximation for x¯, that's to say

(5) x¯Ln(and1,and2,,andn+1;f1|0).

Taking into account the inequality

(6) f1(0)Ln(and1,and2,,andn+1;f1|0)=
=(1)n+1[and1,and2,,andn+1,0;f1|0]and1and2andn+1

and write about itxn+2=Ln(and1,and2,,andn+1;f1|0),we get

(7) x¯xn+2=(1)n+1[and1,and2,,andn+1,0;f1]and1and2andn+1.

We deduce that, ignoring the factor

(1)n+1[and1,and2,,andn+1,0;f1]

xn+2will be an approximation forx¯, all the better as the numbersand1,and2,,andn+1are closer to0.

Let us now designate byxi+1,xi+2,,xi+n+1I,n+1 approximations of the solutionx¯,so the numbers

(8) xi+n+2=Ln(andi+1,andi+2,,andi+n+1;f1|0)

can also be considered as approximations forx¯.

We will now assume that for allin1,,inn+1F,

|[in1,in2,,inn+1,0;f1]|M<+.

We then have

(9) |x¯xi+n+2|M|andi+1||andi+2||andi+n+1|,i=0,1,,

If the first order divided difference of the functionfis limited, that is to say|[x,and;f]|bfor allx,andI,then we get from ( 9 )

(10) |x¯xi+n+2| Mbn+1|x¯xi+1||x¯xi+2||x¯xi+n+1|,i=1,2,

We will writea=Mbn+1Andri=a1n|x¯xi|,i=1,2,

It then follows from ( 10 )

(11) ri+n+2ri+1ri+2ri+n+1,i=0,1,2,

We now consider the equation to differences

(12) ci+n+2=ci+1+ci+2++ci+n+1,i=0,1,

and we assume that there exists a numberd<1such asrsdcsFors=1,2,,n+1.

We associate to equation ( 12 ) the equation

(13) tn+1=tn+tn1++t+1.

In this case, the solution to equation ( 12 ) can be expressed as follows:

(14) ci+k=C1t1i+k+C2t2i+k++Cn+1tn+1i+k

Ort1,t2,,tn+1are the roots of equation ( 13 ) andC1,C2,,Cn+1are determined by the initial conditionscs=cs0,s=1,2,,n+1.

It is easily seen that equation ( 13 ) has only one real rootoh; 1<oh<2.

If we admit that

(15) rsdohs;s=1,2,,n+1

then we easily deduce from ( 11 ) that the inequalities ( 15 ) hold for alls, which means that

(16) limnrn=0

that is to say that the continuation(xn)n=1converges to the solution of equation ( 1 ).

We will subsequently present an algorithm for calculating the values ​​of polynomials.ohk(0)Andohk(andi),i=k,k+1,,k+n,Or

(17) ohk(and)=i=kk+n(andandi),k=1,2,

The polynomials ( 17 ) are used to construct the sequence of inverse Lagrange interpolation polynomials, which lead to the sequence of approximations(xn)n=1, whose elements are given by the equality ( 8 ).

It follows from ( 17 )

ohk+1(and)=ohk(and)andandn+k+1andandk

from which we deduce the relationship

(18) ohk+1(0)=ohk(0)andn+k+1andk

which provides us with a recurrence formula for calculating the values ​​of polynomialsohkForand=0.

Forohk+1(and)We have

ohk+1(and)=[ohk(and)(andandn+k+1)+ohk(and)](andandk)ohk(and)(andandn+k+1)(andandk)2.

We deduce the following recurrence formulas:

(19) ohk+1(andi)={ohk(andi)andiandn+k+1andiandk,For i=k+1,k+n¯ohkandiandiandk,For i=n+k+1

The recurrence relations ( 18 ) and ( 19 ) offer us the possibility of obtaining the sequence of inverse Lagrange interpolation polynomials, using at any iteration step certain elements which have already been calculated during the previous step.

Bibliography



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

1987

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