Solving equations with the aid of inverse rational interpolation functions

Ion Păvăloiu
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Abstract

We study the convergence of an iterative method for solving the equation \(f\left( x\right) =0,\ f:A\rightarrow B\), \(A,B\subseteq \mathbb{R}\), \(f\) assumed  bijective. Denote by \(\varphi _{a,b,\alpha,\beta}\) the set of functions of the form \[\varphi \left( x\right) =\frac{ax+b}{\alpha x+\beta},\] \(\alpha \neq0\) with \(\varphi :\mathbb{R}\backslash \{ -\textstyle\frac{\beta}{\alpha}\} \rightarrow \mathbb{R}\). If \(x_{1},x_{2},x_{3}\in I\) denote \(y_{i}=f\left( x_{i}\right) ,\ i=1,2,3\). We are interested in determining a function \(\varphi\) such that \(\varphi \left(y_{i}\right) =x_{i},i=1,2,3\), which is the inverse rational interpolation function. The value \(\varphi \left( 0\right)\) is a new approximation of the solution of equation \(f\left( x\right) =0\). We study the convergence of the iterative method generated above.

Authors

Crăciun Iancu
(Tiberiu Popoviciu Institute of Numerical Analysis)

Ion Păvăloiu
(Tiberiu Popoviciu Institute of Numerical Analysis)

Title

Original title (in French)

La resolution des équations par interpolation inverse de type Hermite

English translation of the title

Solving equations with the aid of inverse rational interpolation functions

Keywords

rational function; nonlinear equation in R; iterative method; inverse interpolation

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Cite this paper as:

C. Iancu, I. Păvăloiu, Resolution des equations à l’aide des fonctions rationnelles d’interpolation invèrse, Seminar on functional analysis and numerical methods, Preprint no. 1 (1985), pp. 71-78 (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] Imamov, A., Resenie nelineinih  urvnenii metodomi spalin-interpolirovanie. Medodi splain-funktii, Akademia Nauk  SSSR. Novosibirsk, 81 (1979) , 74-80.

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

[3] Iancu, C., Pavaloiu, I., Resolutions des equations a l’aide des fonctions spline d’interpolation  inverse.  Seminar of Functional analysis and Numerical Methods, “Babes-Bolyai” University, Faculty of Mathematics, Research Seminaries, Preprint Nr. 1, (1984), 97-1-4.

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Solving equations with the aid of inverse rational interpolation functions

C. Iancu and I. Păvăloiu

  1. Let's dasign by ft I → R I → R I rarr RI \rightarrow RI→Ra real function of a real variable, where I is an interval of the real axis.
Let us designate by x 0 , x 1 , x 2 x 0 , x 1 , x 2 x_(0),x_(1),x_(2)x_{0}, x_{1}, x_{2}x0,x1,x2trols distinct points of I. Consider the set φ a , ↓ , α , β φ a , ↓ , α , β varphi_(a,)darr,alpha,beta\varphi_{a,} \downarrow, \alpha, \betaφhas,↓,α,βfunctions of the form:
(1)
φ ( x ) = a x + b α x + β , α ≠ 0 , φ ( x ) = a x + b α x + β , α ≠ 0 , varphi(x)=(ax+b)/(alpha x+beta)quad,quad alpha!=0quad,\varphi(x)=\frac{a x+b}{\alpha x+\beta} \quad, \quad \alpha \neq 0 \quad,φ(x)=hasx+bαx+β,α≠0,
Or a , b , ∝ , β a , b , ∝ , β a,b,prop,betaa, b, \propto, \betahas,b,∝,βare real parameters that must be determined in such a way that the function P P P\mathcal{P}Pcoincides with the function f f fffto the points x 0 , x 1 x 0 , x 1 x_(0),x_(1)x_{0}, x_{1}x0,x1And x 2 x 2 x_(2)x_{2}x2, that is to say
(2) φ ( x 0 ) = f ( x 0 ) , φ ( x 1 ) = f ( x 1 ) , φ ( x 2 ) = f ( x 2 ) φ x 0 = f x 0 , φ x 1 = f x 1 , φ x 2 = f x 2 quad varphi(x_(0))=f(x_(0)),quad varphi(x_(1))=f(x_(1)),quad varphi(x_(2))=f(x_(2))\quad \varphi\left(x_{0}\right)=f\left(x_{0}\right), \quad \varphi\left(x_{1}\right)=f\left(x_{1}\right), \quad \varphi\left(x_{2}\right)=f\left(x_{2}\right)φ(x0)=f(x0),φ(x1)=f(x1),φ(x2)=f(x2).
We will first assume that x 0 ≠ x 1 , x 1 ≠ x 0 ≠ x 1 x 2 , let f ( x 0 ) ≠≠ f ( x 1 ) , f ( x 1 ) ≠ f ( x 2 ) , f ( x 0 ) ≠ f ( x 2 ) x 0 ≠ x 1 , x 1 ≠ x 0 ≠ x 1 x 2 ,  let  f x 0 ≠≠ f x 1 , f x 1 ≠ f x 2 , f x 0 ≠ f x 2 x_(0)!=x_(1),x_(1)!=(x_(0)!=x_(1))/(x_(2)," let ")f(x_(0))≠≠f(x_(1)),f(x_(1))!=f(x_(2)),f(x_(0))!=f(x_(2))x_{0} \neq x_{1}, x_{1} \neq \frac{x_{0} \neq x_{1}}{x_{2}, \text { let }} f\left(x_{0}\right) \neq \neq f\left(x_{1}\right), f\left(x_{1}\right) \neq f\left(x_{2}\right), f\left(x_{0}\right) \neq f\left(x_{2}\right)x0≠x1,x1≠x0≠x1x2, let f(x0)≠≠f(x1),f(x1)≠f(x2),f(x0)≠f(x2).
We deduce from (1) and (2)
(3) α x 0 + b α x 0 + β = f ( x 0 ) , a x 1 + b α x 1 + β = f ( x 1 ) , a x 2 + b α x 2 + β = f ( x 2 ) α x 0 + b α x 0 + β = f x 0 , a x 1 + b α x 1 + β = f x 1 , a x 2 + b α x 2 + β = f x 2 (alphax_(0)+b)/(alphax_(0)+beta)=f(x_(0)),(ax_(1)+b)/(alphax_(1)+beta)=f(x_(1)),(ax_(2)+b)/(alphax_(2)+beta)=f(x_(2))\frac{\alpha x_{0}+b}{\alpha x_{0}+\beta}=f\left(x_{0}\right), \frac{a x_{1}+b}{\alpha x_{1}+\beta}=f\left(x_{1}\right), \frac{a x_{2}+b}{\alpha x_{2}+\beta}=f\left(x_{2}\right)αx0+bαx0+β=f(x0),hasx1+bαx1+β=f(x1),hasx2+bαx2+β=f(x2).
System (3) contains 4 unknowns, of which only 3 are essential. If to equations (3) we add (1) and eliminate the parameters a 0 b 0 α 0 , β a 0 b 0 α 0 , β a_(0)b_(0)alpha_(0),betaa_{0} b_{0} \alpha_{0}, \betahas0b0α0,β, we get l ′ l ′ l^(')l^{\prime}L′equality
(4) φ ( x ) − f ( x 1 ) φ ( x ) − f ( x 0 ) ⋅ f ( x 1 ) − f ( x 1 ) f ( x 2 ) − f ( x 1 ) = ( x − x 1 ) ( x 2 − x 0 ) ( x − x 0 ) ( x 2 − x 1 ) φ ( x ) − f x 1 φ ( x ) − f x 0 ⋅ f x 1 − f x 1 f x 2 − f x 1 = x − x 1 x 2 − x 0 x − x 0 x 2 − x 1 (varphi(x)-f(x_(1)))/(varphi(x)-f(x_(0)))*(f(x_(1))-f(x_(1)))/(f(x_(2))-f(x_(1)))=((x-x_(1))(x_(2)-x_(0)))/((x-x_(0))(x_(2)-x_(1)))\frac{\varphi(x)-f\left(x_{1}\right)}{\varphi(x)-f\left(x_{0}\right)} \cdot \frac{f\left(x_{1}\right)-f\left(x_{1}\right)}{f\left(x_{2}\right)-f\left(x_{1}\right)}=\frac{\left(x-x_{1}\right)\left(x_{2}-x_{0}\right)}{\left(x-x_{0}\right)\left(x_{2}-x_{1}\right)}φ(x)−f(x1)φ(x)−f(x0)⋅f(x1)−f(x1)f(x2)−f(x1)=(x−x1)(x2−x0)(x−x0)(x2−x1)
from which it results
(5) φ ( x ) − f ( x 1 ) φ ( x ) − f ( x 0 ) = [ x 1 , x 2 ; f ] [ x 0 , x 2 ; f ] ⋅ x − x 1 x − x 0 (5) φ ( x ) − f x 1 φ ( x ) − f x 0 = x 1 , x 2 ; f x 0 , x 2 ; f ⋅ x − x 1 x − x 0 {:(5)(varphi(x)-f(x_(1)))/(varphi(x)-f(x_(0)))=([x_(1),x_(2);f])/([x_(0),x_(2);f])*(x-x_(1))/(x-x_(0)):}\begin{equation*} \frac{\varphi(x)-f\left(x_{1}\right)}{\varphi(x)-f\left(x_{0}\right)}=\frac{\left[x_{1}, x_{2} ; f\right]}{\left[x_{0}, x_{2} ; f\right]} \cdot \frac{x-x_{1}}{x-x_{0}} \tag{5} \end{equation*}(5)φ(x)−f(x1)φ(x)−f(x0)=[x1,x2;f][x0,x2;f]⋅x−x1x−x0
From (5) we deduce for the parameters a , b , α , β a , b , α , β a,b,alpha,betaa, b, \alpha, \betahas,b,α,βthe following expressions
(6)
a = f ( x 1 ) [ x 0 ; x 2 ; f ] − f ( x 0 ) [ x 1 ; x 2 ; f ] b = x 1 f ( x 0 ) [ x 1 , x 2 ; f ] − x 0 f ( x 1 ) ⋅ [ x 0 , x 2 ; f ] a = [ x 0 , x 2 ; f ] − [ x 1 , x 2 ; f ] β = [ x 1 , x 2 ; f ] x 1 − [ x 0 , x 2 ; f ] x 0 a = f x 1 x 0 ; x 2 ; f − f x 0 x 1 ; x 2 ; f b = x 1 f x 0 x 1 , x 2 ; f − x 0 f x 1 ⋅ x 0 , x 2 ; f a = x 0 , x 2 ; f − x 1 , x 2 ; f β = x 1 , x 2 ; f x 1 − x 0 , x 2 ; f x 0 {:[a=f(x_(1))[x_(0);x_(2);f]-f(x_(0))[x_(1);x_(2);f]],[b=x_(1)f(x_(0))[x_(1),x_(2);f]-x_(0)f(x_(1))*[x_(0),x_(2);f]],[a=[x_(0),x_(2);f]-[x_(1),x_(2);f]],[beta=[x_(1),x_(2);f]x_(1)-[x_(0),x_(2);f]x_(0)]:}\begin{aligned} & a=f\left(x_{1}\right)\left[x_{0} ; x_{2} ; f\right]-f\left(x_{0}\right)\left[x_{1} ; x_{2} ; f\right] \\ & b=x_{1} f\left(x_{0}\right)\left[x_{1}, x_{2} ; f\right]-x_{0} f\left(x_{1}\right) \cdot\left[x_{0}, x_{2} ; f\right] \\ & a=\left[x_{0}, x_{2} ; f\right]-\left[x_{1}, x_{2} ; f\right] \\ & \beta=\left[x_{1}, x_{2} ; f\right] x_{1}-\left[x_{0}, x_{2} ; f\right] x_{0} \end{aligned}has=f(x1)[x0;x2;f]−f(x0)[x1;x2;f]b=x1f(x0)[x1,x2;f]−x0f(x1)⋅[x0,x2;f]has=[x0,x2;f]−[x1,x2;f]β=[x1,x2;f]x1−[x0,x2;f]x0
where we have designated by [ x , y ; f ] [ x , y ; f ] [x,y;f][x, y ; f][x,y;f]the divided difference of the function f f fffon the points x x xxxAnd y y yyy.
In this way the function φ φ varphi\varphiφgiven by (1) which verifies conditions (2) is perfectly determined.
In what follows we designate, to fix the ideas, by F the image of the interval I I IIIby function f f fffand we will assume that it exists f − 1 : f → I f − 1 : f → I f^(-1):f rarr If^{-1}: f \rightarrow If−1:f→I.
If x ¯ ∈ I x ¯ ∈ I bar(x)in I\bar{x} \in Ix¯∈Iis a root of the equation
(7) f ( x ) = 0 ; (7) f ( x ) = 0 ; {:(7)f(x)=0quad;:}\begin{equation*} f(x)=0 \quad ; \tag{7} \end{equation*}(7)f(x)=0;
SO 0 ∈ F 0 ∈ F 0in F0 \in F0∈Fand obviously x ¯ = f − 1 ( 0 ) x ¯ = f − 1 ( 0 ) bar(x)=f^(-1)(0)\bar{x}=f^{-1}(0)x¯=f−1(0).
Let us now consider the set φ a ′ , b ′ , α ′ , β ′ φ a ′ , b ′ , α ′ , β ′ varphia^('),b^('),alpha^('),beta^(')\varphi a^{\prime}, b^{\prime}, \alpha^{\prime}, \beta^{\prime}φhas′,b′,α′,β′functions defined on R − { − β ′ α ′ } R − − β ′ α ′ R-{-(beta^('))/(alpha^('))}\mathbb{R}-\left\{-\frac{\beta^{\prime}}{\alpha^{\prime}}\right\}R−{−β′α′}with values ​​in R R R\mathbb{R}R, of the following form:
(8) φ ( y ) = a ′ y + b ′ α ′ y + β ′ , α ′ ≠ 0 (8) φ ( y ) = a ′ y + b ′ α ′ y + β ′ , α ′ ≠ 0 {:(8)varphi(y)=(a^(')y+b^('))/(alpha^(')y+beta^('))quad","quadalpha^(')!=0:}\begin{equation*} \varphi(y)=\frac{a^{\prime} y+b^{\prime}}{\alpha^{\prime} y+\beta^{\prime}} \quad, \quad \alpha^{\prime} \neq 0 \tag{8} \end{equation*}(8)φ(y)=has′y+b′α′y+β′,α′≠0
We will determine the parameters a ′ , b ′ , α ′ , β ′ a ′ , b ′ , α ′ , β ′ a^('),b^('),alpha^('),beta^(')a^{\prime}, b^{\prime}, \alpha^{\prime}, \beta^{\prime}has′,b′,α′,β′so that function (8) verifies the conditions
(?) φ ( y ) = x 0 , φ ( y 1 ) = x 1 , φ ( y 2 ) = x 2 φ ( y ) = x 0 , φ y 1 = x 1 , φ y 2 = x 2 quad varphi(y)=x_(0),varphi(y_(1))=x_(1),varphi(y_(2))=x_(2)\quad \varphi(y)=x_{0}, \varphi\left(y_{1}\right)=x_{1}, \varphi\left(y_{2}\right)=x_{2}φ(y)=x0,φ(y1)=x1,φ(y2)=x2
oi. y 1 = f ( x 1 ) , 1 = 1 , 2 , 3 y 1 = f x 1 , 1 = 1 , 2 , 3 y_(1)=f(x_(1)),1=1,2,3y_{1}=f\left(x_{1}\right), 1=1,2,3y1=f(x1),1=1,2,3.
my proceeding as cl.above we deduce for the parameters a ′ , b ′ , c ′ , β ′ a ′ , b ′ , c ′ , β ′ a^('),b^('),c^('),beta^(')a^{\prime}, b^{\prime}, c^{\prime}, \beta^{\prime}has′,b′,c′,β′the following expressions:
a ′ = x 1 [ x 1 , x 2 ; f ] − x 0 [ x 0 , x 2 ; f ] b ′ = f ( x 1 ) x 0 [ x 0 , x 2 ; f ] − f ( x 0 ) x 1 [ x 1 , x 2 ; f ] (10) α ′ = [ x 1 , x 2 ; f ] − [ x 0 , x 2 ; f ] β ′ = f ( x 1 ) [ x 0 , x 2 ; f ] − f ( x 0 ) [ x 1 , x 2 ; f ] a ′ = x 1 x 1 , x 2 ; f − x 0 x 0 , x 2 ; f b ′ = f x 1 x 0 x 0 , x 2 ; f − f x 0 x 1 x 1 , x 2 ; f (10) α ′ = x 1 , x 2 ; f − x 0 , x 2 ; f β ′ = f x 1 x 0 , x 2 ; f − f x 0 x 1 , x 2 ; f {:[a^(')=x_(1)[x_(1),x_(2);f]-x_(0)[x_(0),x_(2);f]],[b^(')=f(x_(1))x_(0)[x_(0),x_(2);f]-f(x_(0))x_(1)[x_(1),x_(2);f]],[(10)alpha^(')=[x_(1),x_(2);f]-[x_(0),x_(2);f]],[beta^(')=f(x_(1))[x_(0),x_(2);f]-f(x_(0))[x_(1),x_(2);f]]:}\begin{align*} & a^{\prime}=x_{1}\left[x_{1}, x_{2} ; f\right]-x_{0}\left[x_{0}, x_{2} ; f\right] \\ & b^{\prime}=f\left(x_{1}\right) x_{0}\left[x_{0}, x_{2} ; f\right]-f\left(x_{0}\right) x_{1}\left[x_{1}, x_{2} ; f\right] \\ & \alpha^{\prime}=\left[x_{1}, x_{2} ; f\right]-\left[x_{0}, x_{2} ; f\right] \tag{10}\\ & \beta^{\prime}=f\left(x_{1}\right)\left[x_{0}, x_{2} ; f\right]-f\left(x_{0}\right)\left[x_{1}, x_{2} ; f\right] \end{align*}has′=x1[x1,x2;f]−x0[x0,x2;f]b′=f(x1)x0[x0,x2;f]−f(x0)x1[x1,x2;f](10)α′=[x1,x2;f]−[x0,x2;f]β′=f(x1)[x0,x2;f]−f(x0)[x1,x2;f]
The function φ φ varphi\varphiφgiven by (8) where a ′ , b ′ , α ′ , β ′ a ′ , b ′ , α ′ , β ′ a^('),b^('),alpha^('),beta^(')a^{\prime}, b^{\prime}, \alpha^{\prime}, \beta^{\prime}has′,b′,α′,β′have the values ​​(10) represents the inverse interpolation rational function of the function f f fff. An approximation for the root x ¯ x ¯ bar(x)\bar{x}x¯of equation (7) is obtained as follows:
(11) x ¯ ≃ φ ( 0 ) = x 0 f ( x 1 ) [ x 0 , x 2 ; f ] − x 1 f ( x 0 ) [ x 1 , x 2 ; f ] f ( x 1 ) [ x 0 , x 2 ; f ] − f ( x 0 ) [ x 1 , x 2 ; f ] x ¯ ≃ φ ( 0 ) = x 0 f x 1 x 0 , x 2 ; f − x 1 f x 0 x 1 , x 2 ; f f x 1 x 0 , x 2 ; f − f x 0 x 1 , x 2 ; f bar(x)≃varphi(0)=(x_(0)f(x_(1))[x_(0),x_(2);f]-x_(1)f(x_(0))[x_(1),x_(2);f])/(f(x_(1))[x_(0),x_(2);f]-f(x_(0))[x_(1),x_(2);f])\bar{x} \simeq \varphi(0)=\frac{x_{0} f\left(x_{1}\right)\left[x_{0}, x_{2} ; f\right]-x_{1} f\left(x_{0}\right)\left[x_{1}, x_{2} ; f\right]}{f\left(x_{1}\right)\left[x_{0}, x_{2} ; f\right]-f\left(x_{0}\right)\left[x_{1}, x_{2} ; f\right]}x¯≃φ(0)=x0f(x1)[x0,x2;f]−x1f(x0)[x1,x2;f]f(x1)[x0,x2;f]−f(x0)[x1,x2;f]Using (11) we can obtain various iterative methods for solving equation (7).
We obtain a first method by giving ourselves the points x 0 x 0 x_(0)x_{0}x0And x 1 x 1 x_(1)x_{1}x1and taking x 2 = z 0 x 2 = z 0 x_(2)=z_(0)x_{2}=z_{0}x2=z0. It then results from (11) the following iterative method:
(12) z n + 1 = x 0 f ( x 1 ) [ x 0 , z n ; f ] − x 1 f ( x 0 ) [ x 1 , z n ; f ] f ( x 1 ) [ x 0 , x n ; f ] − f ( x 0 ) [ x 1 , z n ; f ] , z 0 = x 2 , n = 0 , 1 , … … (12) z n + 1 = x 0 f x 1 x 0 , z n ; f − x 1 f x 0 x 1 , z n ; f f x 1 x 0 , x n ; f − f x 0 x 1 , z n ; f , z 0 = x 2 , n = 0 , 1 , … … {:[(12)z_(n+1)=(x_(0)f(x_(1))[x_(0),z_(n);f]-x_(1)f(x_(0))[x_(1),z_(n);f])/(f(x_(1))[x_(0),x_(n);f]-f(x_(0))[x_(1),z_(n);f])","z_(0)=x_(2)","],[n=0","1","dots dots]:}\begin{align*} & z_{n+1}=\frac{x_{0} f\left(x_{1}\right)\left[x_{0}, z_{n} ; f\right]-x_{1} f\left(x_{0}\right)\left[x_{1}, z_{n} ; f\right]}{f\left(x_{1}\right)\left[x_{0}, x_{n} ; f\right]-f\left(x_{0}\right)\left[x_{1}, z_{n} ; f\right]}, z_{0}=x_{2}, \tag{12}\\ & n=0,1, \ldots \ldots \end{align*}(12)zn+1=x0f(x1)[x0,zn;f]−x1f(x0)[x1,zn;f]f(x1)[x0,xn;f]−f(x0)[x1,zn;f],z0=x2,n=0,1,……
We will assume that f ( x 1 ) / f ( x 0 ) ≠ [ x 1 , x ¯ ; f ] / [ x 0 , x ¯ ; f ] f x 1 / f x 0 ≠ x 1 , x ¯ ; f / x 0 , x ¯ ; f f(x_(1))//f(x_(0))!=[x_(1),( bar(x));f]//[x_(0),( bar(x));f]f\left(x_{1}\right) / f\left(x_{0}\right) \neq\left[x_{1}, \bar{x} ; f\right] /\left[x_{0}, \bar{x} ; f\right]f(x1)/f(x0)≠[x1,x¯;f]/[x0,x¯;f]and that the elements of the sequence ( z n ) cu. ∞ z n cu.  ∞ (z_(n))_("cu. ")^(oo)\left(z_{n}\right)_{\text {cu. }}^{\infty}(zn)cu. ∞can be determined. If the following ( z n ) n = 0 ∞ z n n = 0 ∞ (z_(n))_(n=0)^(oo)\left(z_{n}\right)_{n=0}^{\infty}(zn)n=0∞convergent ext and lim n → 0 z n = x ¯ lim n → 0   z n = x ¯ lim_(n rarr0)z_(n)= bar(x)\lim _{n \rightarrow 0} z_{n}=\bar{x}limn→0zn=x¯, it immediately follows that f ( x ¯ ) = 0 f ( x ¯ ) = 0 f( bar(x))=0f(\bar{x})=0f(x¯)=0, that is to say that x ¯ x ¯ bar(x)\bar{x}x¯is the root of the equation f ( x ) = 0 f ( x ) = 0 f(x)=0f(x)=0f(x)=0.
We can obtain another iterative method by giving, for example, the point x 0 x 0 x_(0)x_{0}x0and taking x 1 = z 0 x 1 = z 0 x_(1)=z_(0)x_{1}=z_{0}x1=z0And x 2 = z 1 x 2 = z 1 x_(2)=z_(1)x_{2}=z_{1}x2=z1.

boasts:

It then follows from (11)
(13) z n + 1 = x 0 f ( z n − 1 ) [ x 0 , z n , f ] − z n − 1 f ( x 0 ) [ z n − 1 , z n , f 1 ] f ( z n − 1 ) [ x 0 , z n , f ] − f ( x 0 ) [ z n − 1 , z n , f 1 ] z n + 1 = x 0 f z n − 1 x 0 , z n , f − z n − 1 f x 0 z n − 1 , z n , f 1 f z n − 1 x 0 , z n , f − f x 0 z n − 1 , z n , f 1 quadz_(n+1)=(x_(0)f(z_(n-1))[x_(0),z_(n),f]-z_(n-1)f(x_(0))[z_(n-1),z_(n),f1])/(f(z_(n-1))[x_(0),z_(n),f]-f(x_(0))[z_(n-1),z_(n),f1])\quad z_{n+1}=\frac{x_{0} f\left(z_{n-1}\right)\left[x_{0}, z_{n}, f\right]-z_{n-1} f\left(x_{0}\right)\left[z_{n-1}, z_{n}, f 1\right]}{f\left(z_{n-1}\right)\left[x_{0}, z_{n}, f\right]-f\left(x_{0}\right)\left[z_{n-1}, z_{n}, f 1\right]}zn+1=x0f(zn−1)[x0,zn,f]−zn−1f(x0)[zn−1,zn,f1]f(zn−1)[x0,zn,f]−f(x0)[zn−1,zn,f1]
Or z 0 = x 1 , z 1 = x 2 , n = 1 , 2 , … z 0 = x 1 , z 1 = x 2 , n = 1 , 2 , … z_(0)=x_(1),z_(1)=x_(2),n=1,2,dotsz_{0}=x_{1}, z_{1}=x_{2}, n=1,2, \ldotsz0=x1,z1=x2,n=1,2,….
It is easily proven, under conditions analogous to those of method (12) that if the sequence ( z n ∑ n = 0 ∞ z n ∑ n = 0 ∞   z_(n)sum_(n=0)^(oo)z_{n} \sum_{n=0}^{\infty}zn∑n=0∞generated by (13) is convergent and x ¯ = lim z → ∞ z n x ¯ = lim z → ∞   z n bar(x)=lim_(zrarr oo)z_(n)\bar{x}=\lim _{\mathrm{z} \rightarrow \infty} z_{n}x¯=limz→∞zn, SO f ( x ¯ ) = 0 f ( x ¯ ) = 0 f( bar(x))=0f(\bar{x})=0f(x¯)=0.
Remark 1. The approximation given by (11) for x ― x ¯ bar(x)\overline{\mathbf{x}}x―can be put in the form
(14) x ¯ ≃ x 0 − f ( x 0 ) [ x 1 , x 2 ; f ] [ x 0 , x 1 ; f ] ⋅ [ x 0 , x 2 ; f ] − f ( x 0 ) [ x 0 , x 1 , x 2 ; f ] x ¯ ≃ x 0 − f x 0 x 1 , x 2 ; f x 0 , x 1 ; f ⋅ x 0 , x 2 ; f − f x 0 x 0 , x 1 , x 2 ; f bar(x)≃x_(0)-(f(x_(0))[x_(1),x_(2);f])/([x_(0),x_(1);f]*[x_(0),x_(2);f]-f(x_(0))[x_(0),x_(1),x_(2);f])\bar{x} \simeq x_{0}-\frac{f\left(x_{0}\right)\left[x_{1}, x_{2} ; f\right]}{\left[x_{0}, x_{1} ; f\right] \cdot\left[x_{0}, x_{2} ; f\right]-f\left(x_{0}\right)\left[x_{0}, x_{1}, x_{2} ; f\right]}x¯≃x0−f(x0)[x1,x2;f][x0,x1;f]⋅[x0,x2;f]−f(x0)[x0,x1,x2;f]
Note 2. Suppose that x 0 < x 1 < x 2 x 0 < x 1 < x 2 x_(0) < x_(1) < x_(2)x_{0}<x_{1}<x_{2}x0<x1<x2And max { ( x 1 − x 0 ) max x 1 − x 0 max{(x_(1)-x_(0)):}\max \left\{\left(x_{1}-x_{0}\right)\right.max{(x1−x0),
given by (1) x 0 x 1 x 1 x 2 a , b , α , β x 0 x 1 x 1 x 2 a , b , α , β ^(x_(0))_(x_(1))^(x_(1))x_(2)_(a,b),alpha,beta{ }^{x_{0}}{ }_{x_{1}}^{x_{1}} x_{2}{ }_{a, b}, \alpha, \betax0x1x1x2has,b,α,βgiven by (6) is continuous and indefinitely differentiable in the interval [ x a , x 2 ] , c ′ x a , x 2 , c ′ [x_(a),x_(2)],c^(')\left[x_{a}, x_{2}\right], c^{\prime}[xhas,x2],c′that is to say − β α ∉ [ x 0 , x 2 ] − β α ∉ x 0 , x 2 -(beta )/(alpha)!in[x_(0),x_(2)]-\frac{\beta}{\alpha} \notin\left[x_{0}, x_{2}\right]−βα∉[x0,x2].
If x ∈ [ x 0 , x 2 ] x ∈ x 0 , x 2 x in[x_(0),x_(2)]x \in\left[x_{0}, x_{2}\right]x∈[x0,x2], and if we assume that the function f f fffadmits derivatives up to and including the 3rd order in the interval ( x 0 , x 2 x 0 , x 2 x_(0),x_(2)x_{0}, x_{2}x0,x2), then it is easy to see that there exists at least one point ξ ∈ ( x 0 , x 2 ) ξ ∈ x 0 , x 2 xi in(x_(0),x_(2))\xi \in\left(x_{0}, x_{2}\right)ξ∈(x0,x2)such as
(15) f ( x ) − φ ( x ) = f ′ ′ ′ ( 5 ) − φ v ′ ′ ( ξ ) 6 ω ( x ) f ( x ) − φ ( x ) = f ′ ′ ′ ( 5 ) − φ v ′ ′ ( ξ ) 6 ω ( x ) f(x)-varphi(x)=(f^(''')(5)-varphiv^('')(xi))/(6)omega(x)f(x)-\varphi(x)=\frac{f^{\prime \prime \prime}(5)-\varphi v^{\prime \prime}(\xi)}{6} \omega(x)f(x)−φ(x)=f′′′(5)−φv′′(ξ)6ω(x),
Or ω ( x ) = ( x − x 0 ) ( x − x 1 ) ( x − x 2 ) ω ( x ) = x − x 0 x − x 1 x − x 2 omega(x)=(x-x_(0))(x-x_(1))(x-x_(2))\omega(x)=\left(x-x_{0}\right)\left(x-x_{1}\right)\left(x-x_{2}\right)ω(x)=(x−x0)(x−x1)(x−x2)And φ φ varphi\varphiφ'' ( x ) = 6 α 2 ( α β − α b ) ( α x + β ) 4 ( x ) = 6 α 2 ( α β − α b ) ( α x + β ) 4 (x)=(6alpha^(2)(alpha beta-alpha b))/((alpha x+beta)^(4))(x)=\frac{6 \alpha^{2}(\alpha \beta-\alpha b)}{(\alpha x+\beta)^{4}}(x)=6α2(αβ−αb)(αx+β)4.
By adding to the above hypotheses the conditions f ′ ( x ) ≠ 0 f ′ ( x ) ≠ 0 f^(')(x)!=0f^{\prime}(x) \neq 0f′(x)≠0regardless of x ∈ ( x 0 , x 2 ) x ∈ x 0 , x 2 x in(x_(0),x_(2))x \in\left(x_{0}, x_{2}\right)x∈(x0,x2)And max { | f ( x 1 ) − f ( x 0 ) | , | f ( x 2 ) − f ( x 1 ) | } << | [ x 1 , x 2 ; f ] [ x 0 , x 1 ; f ] [ x 0 , x 1 , x 2 ; f ] | max f x 1 − f x 0 , f x 2 − f x 1 << x 1 , x 2 ; f x 0 , x 1 ; f x 0 , x 1 , x 2 ; f max{|f(x_(1))-f(x_(0))|,|f(x_(2))-f(x_(1))|}<<|([x_(1),x_(2);f][x_(0),x_(1);f])/([x_(0),x_(1),x_(2);f])|\max \left\{\left|f\left(x_{1}\right)-f\left(x_{0}\right)\right|,\left|f\left(x_{2}\right)-f\left(x_{1}\right)\right|\right\}< <\left|\frac{\left[x_{1}, x_{2} ; f\right]\left[x_{0}, x_{1} ; f\right]}{\left[x_{0}, x_{1}, x_{2} ; f\right]}\right|max{|f(x1)−f(x0)|,|f(x2)−f(x1)|}<<|[x1,x2;f][x0,x1;f][x0,x1,x2;f]|we obtain from (15) the following evaluation
(16) | f − 1 ( 0 ) − φ ( 0 ) | ⩽ ∣ ( f − 1 ( ξ ) ⋯ − φ ′ ′ ′ ( ξ ) ∣ 6 | f ( x 0 ) | f ( x 1 ) | f ( x 2 ) | f − 1 ( 0 ) − φ ( 0 ) ⩽ ∣ f − 1 ( ξ ) ⋯ − φ ′ ′ ′ ( ξ ) ∣ 6 f x 0 f x 1 f x 2 |f^(-1)(0)-varphi(0)| <= (∣(f^(-1)(xi)cdots-varphi^(''')(xi)∣)/(6)|f(x_(0))|f(x_(1))|f(x_(2))|\left|f^{-1}(0)-\varphi(0)\right| \leqslant \frac{\mid\left(f^{-1}(\xi) \cdots-\varphi^{\prime \prime \prime}(\xi) \mid\right.}{6}\left|f\left(x_{0}\right)\right| f\left(x_{1}\right)\left|f\left(x_{2}\right)\right||f−1(0)−φ(0)|⩽∣(f−1(ξ)⋯−φ′′′(ξ)∣6|f(x0)|f(x1)|f(x2)|Or φ ′ ′ ′ ( ζ ) = 6 α 2 ( α ′ β ′ − α ′ ℓ ′ ) ( α ′ ξ + β ′ ) 4 , ξ ∈ ( u , v ) φ ′ ′ ′ ( ζ ) = 6 α 2 α ′ β ′ − α ′ ℓ ′ α ′ ξ + β ′ 4 , ξ ∈ ( u , v ) varphi^(''')(zeta)=(6alpha^(2)(alpha^(')beta^(')-alpha^(')ℓ^(')))/((alpha^(')xi+beta^('))^(4)),xi in(u,v)quad\varphi^{\prime \prime \prime}(\zeta)=\frac{6 \alpha^{2}\left(\alpha^{\prime} \beta^{\prime}-\alpha^{\prime} \ell^{\prime}\right)}{\left(\alpha^{\prime} \xi+\beta^{\prime}\right)^{4}}, \xi \in(u, v) \quadφ′′′(ζ)=6α2(α′β′−α′ℓ′)(α′ξ+β′)4,ξ∈(u,v)Or
u = min { f ( x 0 ) , f ( x 2 ) } et v = max { f ( x 0 ) , f ( x 2 ) } . u = min f x 0 , f x 2  et  v = max f x 0 , f x 2 . u=min{f(x_(0)),f(x_(2))}" et "v=max{f(x_(0)),f(x_(2))}.u=\min \left\{f\left(x_{0}\right), f\left(x_{2}\right)\right\} \text { et } v=\max \left\{f\left(x_{0}\right), f\left(x_{2}\right)\right\} .u=min{f(x0),f(x2)} And v=max{f(x0),f(x2)}.
1 about the delimitation (16) we assumed that , the equation f ( x ) = 0 f ( x ) = 0 f(x)=0f(x)=0f(x)=0has at least one root in the interval ( x 0 , x 2 x 0 , x 2 x_(0),x_(2)x_{0}, x_{2}x0,x2).
Inequality (16) represents a delimitation of the error made in approximating the root x ¯ x ¯ bar(x)\bar{x}x¯of the equation f ( x ) = 0 f ( x ) = 0 f(x)=0f(x)=0f(x)=0by ℓ ′ p ℓ ′ p (ℓ^('))/(p)\frac{\ell^{\prime}}{p}ℓ′pgiven by the pressure (11).
We notice in (16) that l ′ ρ ′ l ′ ρ ′ (l^('))/(rho^('))\frac{l^{\prime}}{\rho^{\prime}}L′ρ′approach x ¯ x ¯ bar(x)\bar{x}x¯all the better since f ( x 0 ) , f ( x 1 ) , f ( x 2 ) f x 0 , f x 1 , f x 2 f(x_(0)),f(x_(1)),f(x_(2))f\left(x_{0}\right), f\left(x_{1}\right), f\left(x_{2}\right)f(x0),f(x1),f(x2)are closer to zero.
We now propose to determine a function of the form (8) which verifies the conditions
(17) φ ( y 0 ) = x 0 ; φ ′ ( y 0 ) = 1 / f ′ ( x 0 ) ; φ ( y 1 ) = x 1 φ y 0 = x 0 ; φ ′ y 0 = 1 / f ′ x 0 ; φ y 1 = x 1 quad varphi(y_(0))=x_(0);quadvarphi^(')(y_(0))=1//f^(')(x_(0));quad varphi(y_(1))=x_(1)quad\quad \varphi\left(y_{0}\right)=x_{0} ; \quad \varphi^{\prime}\left(y_{0}\right)=1 / f^{\prime}\left(x_{0}\right) ; \quad \varphi\left(y_{1}\right)=x_{1} \quadφ(y0)=x0;φ′(y0)=1/f′(x0);φ(y1)=x1.
We deduce from these conditions for the coefficients a', b ′ , α ′ , β ′ b ′ , α ′ , β ′ b^('),alpha^('),beta^(')b^{\prime}, \alpha^{\prime}, \beta^{\prime}b′,α′,β′the following expressions:
a ′ = x 1 [ x 0 , x 1 ; f ] − x 0 f ′ ( x 0 ) (18) b ′ = f ( x 1 ) x 0 f ′ ( x 0 ) − f ( x 0 ) x 1 [ x 0 , x 1 ; f ] α ′ = [ x 0 , x 1 f ′ ] − f ′ ( x 0 ) β ′ = f ( x 1 ) ⋅ f ′ ( x 0 ) − f ( x 0 ) [ x 0 , x 1 ; f ] a ′ = x 1 x 0 , x 1 ; f − x 0 f ′ x 0 (18) b ′ = f x 1 x 0 f ′ x 0 − f x 0 x 1 x 0 , x 1 ; f α ′ = x 0 , x 1 f ′ − f ′ x 0 β ′ = f x 1 ⋅ f ′ x 0 − f x 0 x 0 , x 1 ; f {:[a^(')=x_(1)[x_(0),x_(1);f]-x_(0)f^(')(x_(0))],[(18)b^(')=f(x_(1))x_(0)f^(')(x_(0))-f(x_(0))x_(1)[x_(0),x_(1);f]],[alpha^(')=[x_(0),x_(1)f^(')]-f^(')(x_(0))],[beta^(')=f(x_(1))*f^(')(x_(0))-f(x_(0))[x_(0),x_(1);f]]:}\begin{align*} & a^{\prime}=x_{1}\left[x_{0}, x_{1} ; f\right]-x_{0} f^{\prime}\left(x_{0}\right) \\ & b^{\prime}=f\left(x_{1}\right) x_{0} f^{\prime}\left(x_{0}\right)-f\left(x_{0}\right) x_{1}\left[x_{0}, x_{1} ; f\right] \tag{18}\\ & \alpha^{\prime}=\left[x_{0}, x_{1} f^{\prime}\right]-f^{\prime}\left(x_{0}\right) \\ & \beta^{\prime}=f\left(x_{1}\right) \cdot f^{\prime}\left(x_{0}\right)-f\left(x_{0}\right)\left[x_{0}, x_{1} ; f\right] \end{align*}has′=x1[x0,x1;f]−x0f′(x0)(18)b′=f(x1)x0f′(x0)−f(x0)x1[x0,x1;f]α′=[x0,x1f′]−f′(x0)β′=f(x1)⋅f′(x0)−f(x0)[x0,x1;f]
which lead us to the following approximation for x → x → vec(x)\vec{x}x→:
(19) x ¯ ≃ b ′ β = x 0 f ( x 1 ) f ′ ( x 0 ) − x 1 f ( x 0 ) ⋅ [ x 0 , x 1 ; f ] f ( x 1 ) f ′ ( x 0 ) − f ( x 0 ) [ x 0 , x 1 ; f ] x ¯ ≃ b ′ β = x 0 f x 1 f ′ x 0 − x 1 f x 0 ⋅ x 0 , x 1 ; f f x 1 f ′ x 0 − f x 0 x 0 , x 1 ; f quad bar(x)≃(b^('))/(beta)=(x_(0)f(x_(1))f^(')(x_(0))-x_(1)f(x_(0))*[x_(0),x_(1);f])/(f(x_(1))f^(')(x_(0))-f(x_(0))[x_(0),x_(1);f])\quad \bar{x} \simeq \frac{b^{\prime}}{\beta}=\frac{x_{0} f\left(x_{1}\right) f^{\prime}\left(x_{0}\right)-x_{1} f\left(x_{0}\right) \cdot\left[x_{0}, x_{1} ; f\right]}{f\left(x_{1}\right) f^{\prime}\left(x_{0}\right)-f\left(x_{0}\right)\left[x_{0}, x_{1} ; f\right]}x¯≃b′β=x0f(x1)f′(x0)−x1f(x0)⋅[x0,x1;f]f(x1)f′(x0)−f(x0)[x0,x1;f]
Dana 1 hypothesis f ′ ( x ) ≠ 0 f ′ ( x ) ≠ 0 f^(')(x)!=0f^{\prime}(x) \neq 0f′(x)≠0in the interval ( x 0 , x 1 ) x 0 , x 1 (x_(0),x_(1))\left(x_{0}, x_{1}\right)(x0,x1)there is the following delimitation:
(20) | x ¯ − b ′ β ′ | ⩽ | ( e − 1 ( ξ ) ) ′ ′ − φ ⋯ ( ξ ) | 6 | f ( x 0 ) | 2 ⋅ | f ( x 1 ) | x ¯ − b ′ β ′ ⩽ e − 1 ( ξ ) ′ ′ − φ ⋯ ( ξ ) 6 f x 0 2 ⋅ f x 1 |( bar(x))-(b^('))/(beta^('))| <= (|(e^(-1)(xi))^('')-varphi cdots(xi)|)/(6)|f(x_(0))|^(2)*|f(x_(1))|\left|\bar{x}-\frac{b^{\prime}}{\beta^{\prime}}\right| \leqslant \frac{\left|\left(e^{-1}(\xi)\right)^{\prime \prime}-\varphi \cdots(\xi)\right|}{6}\left|f\left(x_{0}\right)\right|^{2} \cdot\left|f\left(x_{1}\right)\right||x¯−b′β′|⩽|(e−1(ξ))′′−φ⋯(ξ)|6|f(x0)|2⋅|f(x1)|Or ξ ∈ ( u ′ , v ′ ) ξ ∈ u ′ , v ′ xi in(u^('),v^('))\xi \in\left(u^{\prime}, v^{\prime}\right)ξ∈(u′,v′), Or u ′ = min { f ( x 0 ) , f ( x 1 ) } u ′ = min f x 0 , f x 1 u^(')=min{f(x_(0)),f(x_(1))}u^{\prime}=\min \left\{f\left(x_{0}\right), f\left(x_{1}\right)\right\}u′=min{f(x0),f(x1)}And v ′ == max { f ( x 0 ) , f ( x 1 ) } v ′ == max f x 0 , f x 1 v^(')==max{f(x_(0)),f(x_(1))}v^{\prime}= =\max \left\{f\left(x_{0}\right), f\left(x_{1}\right)\right\}v′==max{f(x0),f(x1)}, and the function φ φ varphi\varphiφhas the form (8) with the coefficients given by (18).
We will now assume that we know the values ​​of the function f f fffto the points x 0 < x 1 < … < x n x 0 < x 1 < … < x n x_(0) < x_(1) < dots < x_(n)x_{0}<x_{1}<\ldots<x_{n}x0<x1<…<xn, Or x 1 ∈ I x 1 ∈ I x_(1)in Ix_{1} \in Ix1∈I, i = 0 , 1 , … , n i = 0 , 1 , … , n i=0,1,dots,ni=0,1, \ldots, ni=0,1,…,nas well as the value of the derivative of the function f f fffto the point x 0 x 0 x_(0)x_{0}x0. To clarify the ideas, we will designate by s the smallest natural number for which 1 ′ 1 ′ 1^(')1^{\prime}1′equation f ( x ) = 0 f ( x ) = 0 f(x)=0f(x)=0f(x)=0admits a root in the interval ( x g , x g + 1 x g , x g + 1 x_(g),x_(g+1)x_{g}, x_{g+1}xg,xg+1). We will assume that f ′ ( x ) > 0 , ∀ x ∈ I f ′ ( x ) > 0 , ∀ x ∈ I f^(')(x) > 0,AA x in If^{\prime}(x)>0, \forall x \in If′(x)>0,∀x∈I.
We designate by φ 0 : [ f ( x 0 ) , f ( x 1 ) ] → Z φ 0 : f x 0 , f x 1 → Z varphi_(0):[f(x_(0)),f(x_(1))]rarrZ\varphi_{0}:\left[f\left(x_{0}\right), f\left(x_{1}\right)\right] \rightarrow \mathbb{Z}φ0:[f(x0),f(x1)]→Za function of 1a form
(21) φ 0 ( y ) = a 0 y + b 0 α 0 y + β 0 φ 0 ( y ) = a 0 y + b 0 α 0 y + β 0 quadvarphi_(0)(y)=(a_(0)y+b_(0))/(alpha_(0)y+beta_(0))\quad \varphi_{0}(y)=\frac{a_{0} y+b_{0}}{\alpha_{0} y+\beta_{0}}φ0(y)=has0y+b0α0y+β0
If we demand that φ 0 φ 0 varphi_(0)\varphi_{0}φ0meets the conditions
(22) φ 0 ( y 0 ) = x 0 , φ 0 ′ ( y 0 ) = 1 / f ′ ( x 0 ) , φ 0 ( y 1 ) = x 1 φ 0 y 0 = x 0 , φ 0 ′ y 0 = 1 / f ′ x 0 , φ 0 y 1 = x 1 quadvarphi_(0)(y_(0))=x_(0),quadvarphi_(0)^(')(y_(0))=1//f^(')(x_(0)),quadvarphi_(0)(y_(1))=x_(1)\quad \varphi_{0}\left(y_{0}\right)=x_{0}, \quad \varphi_{0}^{\prime}\left(y_{0}\right)=1 / f^{\prime}\left(x_{0}\right), \quad \varphi_{0}\left(y_{1}\right)=x_{1}φ0(y0)=x0,φ0′(y0)=1/f′(x0),φ0(y1)=x1
Or y i = f ( x i ) , i = 0 , l , … , n y i = f x i , i = 0 , l , … , n y_(i)=f(x_(i)),i=0,l,dots,ny_{i}=f\left(x_{i}\right), i=0, l, \ldots, nyi=f(xi),i=0,L,…,n, then the coefficients a 0 , b 0 a 0 , b 0 a_(0),b_(0)a_{0}, b_{0}has0,b0。
α 0 , β 0 α 0 , β 0 alpha_(0),beta_(0)\alpha_{0}, \beta_{0}α0,β0have the following form is
a 0 = x 1 [ x 0 , x 1 ; f ] − x 0 f ′ ( x 0 ) b 0 = f ( x 1 ) x 0 f ′ ( x 0 ) − f ( x 0 ) x 1 [ x 0 , x 1 ; f ] a 0 = x 1 x 0 , x 1 ; f − x 0 f ′ x 0 b 0 = f x 1 x 0 f ′ x 0 − f x 0 x 1 x 0 , x 1 ; f {:[a_(0)=x_(1)[x_(0),x_(1);f]-x_(0)f^(')(x_(0))],[b_(0)=f(x_(1))x_(0)f^(')(x_(0))-f(x_(0))x_(1)[x_(0),x_(1);f]]:}\begin{aligned} & a_{0}=x_{1}\left[x_{0}, x_{1} ; f\right]-x_{0} f^{\prime}\left(x_{0}\right) \\ & b_{0}=f\left(x_{1}\right) x_{0} f^{\prime}\left(x_{0}\right)-f\left(x_{0}\right) x_{1}\left[x_{0}, x_{1} ; f\right] \end{aligned}has0=x1[x0,x1;f]−x0f′(x0)b0=f(x1)x0f′(x0)−f(x0)x1[x0,x1;f]
(23)
α 0 = [ x 0 , x 1 ; f ] − f ( x 0 ) β = f ( x 1 ) ⋅ f ′ ( x 0 ) − f ( x 0 ) [ x 0 , x 1 ; f ] α 0 = x 0 , x 1 ; f − f x 0 β = f x 1 ⋅ f ′ x 0 − f x 0 x 0 , x 1 ; f {:[alpha_(0)=[x_(0),x_(1);f]-f(x_(0))],[beta=f(x_(1))*f^(')(x_(0))-f(x_(0))[x_(0),x_(1);f]]:}\begin{aligned} & \alpha_{0}=\left[x_{0}, x_{1} ; f\right]-f\left(x_{0}\right) \\ & \beta=f\left(x_{1}\right) \cdot f^{\prime}\left(x_{0}\right)-f\left(x_{0}\right)\left[x_{0}, x_{1} ; f\right] \end{aligned}α0=[x0,x1;f]−f(x0)β=f(x1)⋅f′(x0)−f(x0)[x0,x1;f]
We now determine the functions
φ i : [ f ( x i ) , f ( x i + 1 ) ] → R φ i : f x i , f x i + 1 → R varphi_(i):[f(x_(i)),f(x_(i+1))]rarrR\varphi_{i}:\left[f\left(x_{i}\right), f\left(x_{i+1}\right)\right] \rightarrow \mathbb{R}φi:[f(xi),f(xi+1)]→R, Or
(24) φ i ( y ) = a i y + b i α i y + β i , 1 = 1 , 2 , … , n . (24) φ i ( y ) = a i y + b i α i y + β i , 1 = 1 , 2 , … , n . {:(24)varphi_(i)(y)=(a_(i)y+b_(i))/(alpha_(i)y+beta_(i))","1=1","2","dots","n.:}\begin{equation*} \varphi_{i}(y)=\frac{a_{i} y+b_{i}}{\alpha_{i} y+\beta_{i}}, 1=1,2, \ldots, n . \tag{24} \end{equation*}(24)φi(y)=hasiy+biαiy+βi,1=1,2,…,n.
in such a way that they satisfy the conditions:
(25) φ i ( y i ) = x i , φ i ′ ( y i ) = φ i − 1 ′ ( y i ) , φ i ( y i + 1 ) = x i + 1 , (25) φ i y i = x i , φ i ′ y i = φ i − 1 ′ y i , φ i y i + 1 = x i + 1 , {:(25)varphi_(i)(y_(i))=x_(i)","quadvarphi_(i)^(')(y_(i))=varphi_(i-1)^(')(y_(i))","quadvarphi_(i)(y_(i+1))=x_(i+1)",":}\begin{equation*} \varphi_{i}\left(y_{i}\right)=x_{i}, \quad \varphi_{i}^{\prime}\left(y_{i}\right)=\varphi_{i-1}^{\prime}\left(y_{i}\right), \quad \varphi_{i}\left(y_{i+1}\right)=x_{i+1}, \tag{25} \end{equation*}(25)φi(yi)=xi,φi′(yi)=φi−1′(yi),φi(yi+1)=xi+1,
We then consider as an approximation for the root x ¯ x ¯ bar(x)\bar{x}x¯from equation (7) the number
x ¯ p = φ s ( 0 ) x ¯ p = φ s ( 0 ) bar(x)_(p)=varphi_(s)(0)\bar{x}_{p}=\varphi_{s}(0)x¯p=φs(0)
If we write T i − 1 = 1 / φ i − 1 ′ ( y i ) T i − 1 = 1 / φ i − 1 ′ y i T_(i-1)=1//varphi_(i-1)^(')(y_(i))T_{i-1}=1 / \varphi_{i-1}^{\prime}\left(y_{i}\right)Ti−1=1/φi−1′(yi), then we obtain for the coefficients a i , b i , α i , β i a i , b i , α i , β i a_(i),b_(i),alpha_(i),beta_(i)a_{i}, b_{i}, \alpha_{i}, \beta_{i}hasi,bi,αi,βifrom (24) the following expressions:
(26)
a i = x i + 1 [ x i , x i + 1 ; f ] − x i T i − 1 b i = f ( x i + 1 ) x i T i − 1 − f ( x i ) x i + 1 [ x i , x i + 1 ; f ] a i = [ x i , x i + 1 ; f ] − T i − 1 β i = f ( x i + 1 ) T i − 1 − f ( x i ) [ x i , x i + 1 ; f ] a i = x i + 1 x i , x i + 1 ; f − x i T i − 1 b i = f x i + 1 x i T i − 1 − f x i x i + 1 x i , x i + 1 ; f a i = x i , x i + 1 ; f − T i − 1 β i = f x i + 1 T i − 1 − f x i x i , x i + 1 ; f {:[a_(i)=x_(i+1)[x_(i),x_(i+1);f]-x_(i)T_(i-1)],[b_(i)=f(x_(i+1))x_(i)T_(i-1)-f(x_(i))x_(i+1)[x_(i),x_(i+1);f]],[a_(i)=[x_(i),x_(i+1);f]-T_(i-1)],[beta_(i)=f(x_(i+1))T_(i-1)-f(x_(i))[x_(i),x_(i+1);f]]:}\begin{aligned} & a_{i}=x_{i+1}\left[x_{i}, x_{i+1} ; f\right]-x_{i} T_{i-1} \\ & b_{i}=f\left(x_{i+1}\right) x_{i} T_{i-1}-f\left(x_{i}\right) x_{i+1}\left[x_{i}, x_{i+1} ; f\right] \\ & a_{i}=\left[x_{i}, x_{i+1} ; f\right]-T_{i-1} \\ & \beta_{i}=f\left(x_{i+1}\right) T_{i-1}-f\left(x_{i}\right)\left[x_{i}, x_{i+1} ; f\right] \end{aligned}hasi=xi+1[xi,xi+1;f]−xiTi−1bi=f(xi+1)xiTi−1−f(xi)xi+1[xi,xi+1;f]hasi=[xi,xi+1;f]−Ti−1βi=f(xi+1)Ti−1−f(xi)[xi,xi+1;f]
It results from the condition x ¯ ≃ φ β ( 0 ) x ¯ ≃ φ β ( 0 ) bar(x)≃varphi_(beta)(0)\bar{x} \simeq \varphi_{\beta}(0)x¯≃φβ(0)the following expression for the root approximation x ¯ x ¯ bar(x)\bar{x}x¯:
(27) x ¯ ≃ b s β p s = f ( x s + 1 ) x s T s − 1 − f ( x s ) x s + 1 [ x s , x s + 1 ; f ] f ( x s + 1 ) T s − 1 − f ( x s ) [ x s , x s + 1 ; f ] (27) x ¯ ≃ b s β p s = f x s + 1 x s T s − 1 − f x s x s + 1 x s , x s + 1 ; f f x s + 1 T s − 1 − f x s x s , x s + 1 ; f {:(27) bar(x)≃(b_(s))/(betap_(s))=(f(x_(s+1))x_(s)T_(s-1)-f(x_(s))x_(s+1)[x_(s),x_(s+1);f])/(f(x_(s+1))T_(s-1)-f(x_(s))[x_(s),x_(s+1);f]):}\begin{equation*} \bar{x} \simeq \frac{b_{s}}{\beta p_{s}}=\frac{f\left(x_{s+1}\right) x_{s} T_{s-1}-f\left(x_{s}\right) x_{s+1}\left[x_{s}, x_{s+1} ; f\right]}{f\left(x_{s+1}\right) T_{s-1}-f\left(x_{s}\right)\left[x_{s}, x_{s+1} ; f\right]} \tag{27} \end{equation*}(27)x¯≃bsβps=f(xs+1)xsTs−1−f(xs)xs+1[xs,xs+1;f]f(xs+1)Ts−1−f(xs)[xs,xs+1;f]

BIBLIOGRAPHY

  1. IMAMOV, A., Resenie nelineinih uravnenii metodomi splain-- interpolirovanie, Metodi splain-funktii, Akademia Nauk SSSR. Novosibirsk, 81 (1979), 74 − 80 74 − 80 74-8074-8074−80,
  2. PAVALOIU, I., Rezolvarea ecuatillor prin interpolare, Editura Dacia, Cluj, 1981.
  3. IANCU, C., PAVATOTU, I.: Solving equations using inverse interpolation spline functions, Seminar of Functional Analysis and Numerical Methods, "Babes-Bolyai" University, Faculty of Mathematics, Research Seminaries, Preprint Nr.1, (1984), 97-104.
1985

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