An algorithm in the solving of equations by interpolation

9 years ago

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.

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"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 by $f : I \to ℝ$ a function where $I$ is an interval of the real axis. We consider the equation

(1)

f (x) = i,

and we assume that this equation has only one solution $\bar{x} \in I$ . We designate by $F = f (I)$ the set of velors of the function $f$ For $x \in I$ . We will assume that the function $f$ admits an inverse function $f^{- 1} : F \to I$ .

We designate by $x_{i} \in I, i = 1, 2, \dots, n + 1, x_{i} \neq x_{j}$ For $i \neq j$ , $n + 1$ different approximations of the solution $\bar{x}$ from equation ( 1 ) and we write ${and}_{i} = f (x_{i}), i = 1, 2, \dots, n + 1$ .

The polynomial

(2)

L_{n} ({and}_{1}, {and}_{2}, \dots, {and}_{n + 1}; f^{- 1} | and) = \sum_{i = 1}^{n + 1} x_{i} \frac{oh (and)}{(and - {and}_{i}) {oh}^{'} ({and}_{i})}

(3)

oh (and) = \prod_{i = 1}^{n + 1} (and - {and}_{i})

check the conditions

(4)

L_{n} ({and}_{1}, {and}_{2}, \dots, {and}_{n + 1}; f^{- 1} | {and}_{i}) = x_{i}, i = 1, 2, \dots, n + 1 .

And $\bar{x} \in I,$ SO $0 \in F$ And $f^{- 1} (0) = \bar{x}$ . In this case, make them $and = 0$ an ( 2 ) we obtain an approximation for $\bar{x}$ , that's to say

(5)

\bar{x} ≃ L_{n} ({and}_{1}, {and}_{2}, \dots, {and}_{n + 1}; f^{- 1} | 0) .

Taking into account the inequality

(6)			$f^{- 1} (0) - L_{n} ({and}_{1}, {and}_{2}, \dots, {and}_{n + 1}; f^{- 1} \| 0) =$
		$= {(- 1)}^{n + 1} [{and}_{1}, {and}_{2}, \dots, {and}_{n + 1}, 0; f^{- 1} \| 0] {and}_{1} {and}_{2} \dots {and}_{n + 1}$

and write about it $x_{n + 2} = L_{n} ({and}_{1}, {and}_{2}, \dots, {and}_{n + 1}; f^{- 1} | 0),$ we get

(7)

\bar{x} - x_{n + 2} = {(- 1)}^{n + 1} [{and}_{1}, {and}_{2}, \dots, {and}_{n + 1}, 0; f^{- 1}] {and}_{1} {and}_{2} \dots {and}_{n + 1} .

We deduce that, ignoring the factor

{(- 1)}^{n + 1} [{and}_{1}, {and}_{2}, \dots, {and}_{n + 1}, 0; f^{- 1}]

$x_{n + 2}$ will be an approximation for $\bar{x}$ , all the better as the numbers ${and}_{1}, {and}_{2}, \dots, {and}_{n + 1}$ are closer to $0$ .

Let us now designate by $x_{i + 1}, x_{i + 2}, \dots, x_{i + n + 1} \in I, n + 1$ approximations of the solution $\bar{x},$ so the numbers

(8)

x_{i + n + 2} = L_{n} ({and}_{i + 1}, {and}_{i + 2}, \dots, {and}_{i + n + 1}; f^{- 1} | 0)

can also be considered as approximations for $\bar{x}$ .

We will now assume that for all ${in}_{1}, \dots, {in}_{n + 1} \in F,$

| [{in}_{1}, {in}_{2}, \dots, {in}_{n + 1}, 0; f^{- 1}] | \leq M < + \infty .

We then have

(9)

| \bar{x} - x_{i + n + 2} | \leq M \cdot | {and}_{i + 1} | \cdot | {and}_{i + 2} | \dots | {and}_{i + n + 1} |, i = 0, 1, \dots,

If the first order divided difference of the function $f$ is limited, that is to say $| [x, and; f] | \leq b$ for all $x, and \in I,$ then we get from ( 9 )

(10)

| \bar{x} - x_{i + n + 2} |

\leq M b^{n + 1} | \bar{x} - x_{i + 1} | | \bar{x} - x_{i + 2} | \dots | \bar{x} - x_{i + n + 1} |, i = 1, 2, \dots

We will write $a = M b^{n + 1}$ And $r_{i} = a^{\frac{1}{n}} | \bar{x} - x_{i} |, i = 1, 2, \dots$

It then follows from ( 10 )

(11)

r_{i + n + 2} \leq r_{i + 1} \cdot r_{i + 2} \dots r_{i + n + 1}, i = 0, 1, 2, \dots

We now consider the equation to differences

(12)

c_{i + n + 2} = c_{i + 1} + c_{i + 2} + \dots + c_{i + n + 1}, i = 0, 1, \dots

and we assume that there exists a number $d < 1$ such as $r_{s} \leq d^{c_{s}}$ For $s = 1, 2, \dots, n + 1$ .

We associate to equation ( 12 ) the equation

(13)

t^{n + 1} = t^{n} + t^{n - 1} + \dots + t + 1 .

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

(14)

c_{i + k} = C_{1} t_{1}^{i + k} + C_{2} t_{2}^{i + k} + \dots + C_{n + 1} t_{n + 1}^{i + k}

Or $t_{1}, t_{2}, \dots, t_{n + 1}$ are the roots of equation ( 13 ) and $C_{1}, C_{2}, \dots, C_{n + 1}$ are determined by the initial conditions $c_{s} = c_{s}^{0}, s = 1, 2, \dots, n + 1 .$

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

If we admit that

(15)

r_{s} \leq d^{{oh}^{s}}; s = 1, 2, \dots, n + 1

then we easily deduce from ( 11 ) that the inequalities ( 15 ) hold for all $s \in ℕ$ , which means that

(16)

lim_{n \to \infty} r_{n} = 0

that is to say that the continuation ${(x_{n})}_{n = 1}^{\infty}$ converges to the solution of equation ( 1 ).

We will subsequently present an algorithm for calculating the values of polynomials. ${oh}_{k} (0)$ And ${oh}_{k}^{'} ({and}_{i}), i = k, k + 1, \dots, k + n,$ Or

(17)

{oh}_{k} (and) = \prod_{i = k}^{k + n} (and - {and}_{i}), k = 1, 2, \dots

The polynomials ( 17 ) are used to construct the sequence of inverse Lagrange interpolation polynomials, which lead to the sequence of approximations ${(x_{n})}_{n = 1}^{\infty}$ , whose elements are given by the equality ( 8 ).

It follows from ( 17 )

{oh}_{k + 1} (and) = {oh}_{k} (and) \cdot \frac{and - {and}_{n + k + 1}}{and - {and}_{k}}

from which we deduce the relationship

(18)

{oh}_{k + 1} (0) = {oh}_{k} (0) \cdot \frac{{and}_{n + k + 1}}{{and}_{k}}

which provides us with a recurrence formula for calculating the values of polynomials ${oh}_{k}$ For $and = 0$ .

For ${oh}_{k + 1}^{'} (and)$ We have

{oh}_{k + 1} (and) = \frac{[{oh}_{k} (and) (and - {and}_{n + k + 1}) + {oh}_{k} (and)] (and - {and}_{k}) - {oh}_{k} (and) (and - {and}_{n + k + 1})}{{(and - {and}_{k})}^{2}} .

We deduce the following recurrence formulas:

(19)

{oh}_{k + 1}^{'} ({and}_{i}) = {\begin{matrix} {oh}_{k}^{'} ({and}_{i}) \frac{{and}_{i} - {and}_{n + k + 1}}{{and}_{i} - {and}_{k}}, & For i = \bar{k + 1, k + n} \\ {oh}_{k} \frac{{and}_{i}}{{and}_{i} - {and}_{k}}, & For i = n + k + 1 \end{matrix}

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.