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
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
- Let's dasign by ftI rarr RI \rightarrow Ra real function of a real variable, where I is an interval of the real axis.
Let us designate byx_(0),x_(1),x_(2)x_{0}, x_{1}, x_{2}trols distinct points of I. Consider the setvarphi_(a,)darr,alpha,beta\varphi_{a,} \downarrow, \alpha, \betafunctions of the form:
(1)
(1)
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,
Ora,b,prop,betaa, b, \propto, \betaare real parameters that must be determined in such a way that the functionP\mathcal{P}coincides with the functionffto the pointsx_(0),x_(1)x_{0}, x_{1}Andx_(2)x_{2}, that is to say
(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).
(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).
We will first assume thatx_(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).
We deduce from (1) and (2)
(3)(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).
(3)(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).
System (3) contains 4 unknowns, of which only 3 are essential. If to equations (3) we add (1) and eliminate the parametersa_(0)b_(0)alpha_(0),betaa_{0} b_{0} \alpha_{0}, \beta, we getl^(')l^{\prime}equality
(4)(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)}
from which it results
(4)(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)}
from which it results
{:(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*}
From (5) we deduce for the parametersa,b,alpha,betaa, b, \alpha, \betathe following expressions
(6)
(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]],[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}
where we have designated by[x,y;f][x, y ; f]the divided difference of the functionffon the pointsxxAndyy.
In this way the functionvarphi\varphigiven by (1) which verifies conditions (2) is perfectly determined.
In what follows we designate, to fix the ideas, by F the image of the intervalIIby functionffand we will assume that it existsf^(-1):f rarr If^{-1}: f \rightarrow I.
Ifbar(x)in I\bar{x} \in Iis a root of the equation
{:(7)f(x)=0quad;:}\begin{equation*}
f(x)=0 \quad ; \tag{7}
\end{equation*}
SO0in F0 \in Fand obviouslybar(x)=f^(-1)(0)\bar{x}=f^{-1}(0).
Let us now consider the setvarphia^('),b^('),alpha^('),beta^(')\varphi a^{\prime}, b^{\prime}, \alpha^{\prime}, \beta^{\prime}functions defined onR-{-(beta^('))/(alpha^('))}\mathbb{R}-\left\{-\frac{\beta^{\prime}}{\alpha^{\prime}}\right\}with values inR\mathbb{R}, of the following form:
Let us now consider the setvarphia^('),b^('),alpha^('),beta^(')\varphi a^{\prime}, b^{\prime}, \alpha^{\prime}, \beta^{\prime}functions defined onR-{-(beta^('))/(alpha^('))}\mathbb{R}-\left\{-\frac{\beta^{\prime}}{\alpha^{\prime}}\right\}with values inR\mathbb{R}, of the following form:
{:(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*}
We will determine the parametersa^('),b^('),alpha^('),beta^(')a^{\prime}, b^{\prime}, \alpha^{\prime}, \beta^{\prime}so that function (8) verifies the conditions
(?)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}
oi.y_(1)=f(x_(1)),1=1,2,3y_{1}=f\left(x_{1}\right), 1=1,2,3.
my proceeding as cl.above we deduce for the parametersa^('),b^('),c^('),beta^(')a^{\prime}, b^{\prime}, c^{\prime}, \beta^{\prime}the following expressions:
(?)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}
oi.y_(1)=f(x_(1)),1=1,2,3y_{1}=f\left(x_{1}\right), 1=1,2,3.
my proceeding as cl.above we deduce for the parametersa^('),b^('),c^('),beta^(')a^{\prime}, b^{\prime}, c^{\prime}, \beta^{\prime}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)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*}
The functionvarphi\varphigiven by (8) wherea^('),b^('),alpha^('),beta^(')a^{\prime}, b^{\prime}, \alpha^{\prime}, \beta^{\prime}have the values (10) represents the inverse interpolation rational function of the functionff. An approximation for the rootbar(x)\bar{x}of equation (7) is obtained as follows:
(11)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]}Using (11) we can obtain various iterative methods for solving equation (7).
(11)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]}Using (11) we can obtain various iterative methods for solving equation (7).
We obtain a first method by giving ourselves the pointsx_(0)x_{0}Andx_(1)x_{1}and takingx_(2)=z_(0)x_{2}=z_{0}. 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","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*}
We will assume thatf(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]and that the elements of the sequence(z_(n))_("cu. ")^(oo)\left(z_{n}\right)_{\text {cu. }}^{\infty}can be determined. If the following(z_(n))_(n=0)^(oo)\left(z_{n}\right)_{n=0}^{\infty}convergent ext andlim_(n rarr0)z_(n)= bar(x)\lim _{n \rightarrow 0} z_{n}=\bar{x}, it immediately follows thatf( bar(x))=0f(\bar{x})=0, that is to say thatbar(x)\bar{x}is the root of the equationf(x)=0f(x)=0.
We can obtain another iterative method by giving, for example, the pointx_(0)x_{0}and takingx_(1)=z_(0)x_{1}=z_{0}Andx_(2)=z_(1)x_{2}=z_{1}.
boasts:
It then follows from (11)
(13)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]}
Orz_(0)=x_(1),z_(1)=x_(2),n=1,2,dotsz_{0}=x_{1}, z_{1}=x_{2}, n=1,2, \ldots.
It is easily proven, under conditions analogous to those of method (12) that if the sequence (z_(n)sum_(n=0)^(oo)z_{n} \sum_{n=0}^{\infty}generated by (13) is convergent andbar(x)=lim_(zrarr oo)z_(n)\bar{x}=\lim _{\mathrm{z} \rightarrow \infty} z_{n}, SOf( bar(x))=0f(\bar{x})=0.
(13)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]}
Orz_(0)=x_(1),z_(1)=x_(2),n=1,2,dotsz_{0}=x_{1}, z_{1}=x_{2}, n=1,2, \ldots.
It is easily proven, under conditions analogous to those of method (12) that if the sequence (z_(n)sum_(n=0)^(oo)z_{n} \sum_{n=0}^{\infty}generated by (13) is convergent andbar(x)=lim_(zrarr oo)z_(n)\bar{x}=\lim _{\mathrm{z} \rightarrow \infty} z_{n}, SOf( bar(x))=0f(\bar{x})=0.
Remark 1. The approximation given by (11) forbar(x)\overline{\mathbf{x}}can be put in the form
(14)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]}
(14)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]}
Note 2. Suppose thatx_(0) < x_(1) < x_(2)x_{0}<x_{1}<x_{2}Andmax{(x_(1)-x_(0)):}\max \left\{\left(x_{1}-x_{0}\right)\right.,
given by (1)^(x_(0))_(x_(1))^(x_(1))x_(2)_(a,b),alpha,beta{ }^{x_{0}}{ }_{x_{1}}^{x_{1}} x_{2}{ }_{a, b}, \alpha, \betagiven by (6) is continuous and indefinitely differentiable in the interval[x_(a),x_(2)],c^(')\left[x_{a}, x_{2}\right], c^{\prime}that is to say-(beta )/(alpha)!in[x_(0),x_(2)]-\frac{\beta}{\alpha} \notin\left[x_{0}, x_{2}\right].
given by (1)^(x_(0))_(x_(1))^(x_(1))x_(2)_(a,b),alpha,beta{ }^{x_{0}}{ }_{x_{1}}^{x_{1}} x_{2}{ }_{a, b}, \alpha, \betagiven by (6) is continuous and indefinitely differentiable in the interval[x_(a),x_(2)],c^(')\left[x_{a}, x_{2}\right], c^{\prime}that is to say-(beta )/(alpha)!in[x_(0),x_(2)]-\frac{\beta}{\alpha} \notin\left[x_{0}, x_{2}\right].Ifx in[x_(0),x_(2)]x \in\left[x_{0}, x_{2}\right], and if we assume that the functionffadmits derivatives up to and including the 3rd order in the interval (x_(0),x_(2)x_{0}, x_{2}), then it is easy to see that there exists at least one pointxi in(x_(0),x_(2))\xi \in\left(x_{0}, x_{2}\right)such as
(15)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),
Oromega(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)Andvarphi\varphi''(x)=(6alpha^(2)(alpha beta-alpha b))/((alpha x+beta)^(4))(x)=\frac{6 \alpha^{2}(\alpha \beta-\alpha b)}{(\alpha x+\beta)^{4}}.
By adding to the above hypotheses the conditionsf^(')(x)!=0f^{\prime}(x) \neq 0regardless ofx in(x_(0),x_(2))x \in\left(x_{0}, x_{2}\right)Andmax{|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|we obtain from (15) the following evaluation
(16)|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|Orvarphi^(''')(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) \quadOr
(15)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),
Oromega(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)Andvarphi\varphi''(x)=(6alpha^(2)(alpha beta-alpha b))/((alpha x+beta)^(4))(x)=\frac{6 \alpha^{2}(\alpha \beta-\alpha b)}{(\alpha x+\beta)^{4}}.
By adding to the above hypotheses the conditionsf^(')(x)!=0f^{\prime}(x) \neq 0regardless ofx in(x_(0),x_(2))x \in\left(x_{0}, x_{2}\right)Andmax{|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|we obtain from (15) the following evaluation
(16)|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|Orvarphi^(''')(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) \quadOr
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\} .
1 about the delimitation (16) we assumed that , the equationf(x)=0f(x)=0has at least one root in the interval (x_(0),x_(2)x_{0}, x_{2}).
Inequality (16) represents a delimitation of the error made in approximating the rootbar(x)\bar{x}of the equationf(x)=0f(x)=0by(ℓ^('))/(p)\frac{\ell^{\prime}}{p}given by the pressure (11).
We notice in (16) that(l^('))/(rho^('))\frac{l^{\prime}}{\rho^{\prime}}approachbar(x)\bar{x}all the better sincef(x_(0)),f(x_(1)),f(x_(2))f\left(x_{0}\right), f\left(x_{1}\right), f\left(x_{2}\right)are closer to zero.
We now propose to determine a function of the form (8) which verifies the conditions
(17)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} \quad.
(17)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} \quad.
We deduce from these conditions for the coefficients a',b^('),alpha^('),beta^(')b^{\prime}, \alpha^{\prime}, \beta^{\prime}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]],[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*}
which lead us to the following approximation forvec(x)\vec{x}:
(19)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]}
(19)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]}
Dana 1 hypothesisf^(')(x)!=0f^{\prime}(x) \neq 0in the interval(x_(0),x_(1))\left(x_{0}, x_{1}\right)there is the following delimitation:
(20)|( 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|Orxi in(u^('),v^('))\xi \in\left(u^{\prime}, v^{\prime}\right), Oru^(')=min{f(x_(0)),f(x_(1))}u^{\prime}=\min \left\{f\left(x_{0}\right), f\left(x_{1}\right)\right\}Andv^(')==max{f(x_(0)),f(x_(1))}v^{\prime}= =\max \left\{f\left(x_{0}\right), f\left(x_{1}\right)\right\}, and the functionvarphi\varphihas the form (8) with the coefficients given by (18).
(20)|( 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|Orxi in(u^('),v^('))\xi \in\left(u^{\prime}, v^{\prime}\right), Oru^(')=min{f(x_(0)),f(x_(1))}u^{\prime}=\min \left\{f\left(x_{0}\right), f\left(x_{1}\right)\right\}Andv^(')==max{f(x_(0)),f(x_(1))}v^{\prime}= =\max \left\{f\left(x_{0}\right), f\left(x_{1}\right)\right\}, and the functionvarphi\varphihas the form (8) with the coefficients given by (18).
We will now assume that we know the values of the functionffto the pointsx_(0) < x_(1) < dots < x_(n)x_{0}<x_{1}<\ldots<x_{n}, Orx_(1)in Ix_{1} \in I,i=0,1,dots,ni=0,1, \ldots, nas well as the value of the derivative of the functionffto the pointx_(0)x_{0}. To clarify the ideas, we will designate by s the smallest natural number for which1^(')1^{\prime}equationf(x)=0f(x)=0admits a root in the interval (x_(g),x_(g+1)x_{g}, x_{g+1}). We will assume thatf^(')(x) > 0,AA x in If^{\prime}(x)>0, \forall x \in I.
We designate byvarphi_(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}a function of 1a form
(21)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}}
(21)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}}
If we demand thatvarphi_(0)\varphi_{0}meets the conditions
(22)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}
Ory_(i)=f(x_(i)),i=0,l,dots,ny_{i}=f\left(x_{i}\right), i=0, l, \ldots, n, then the coefficientsa_(0),b_(0)a_{0}, b_{0}。
alpha_(0),beta_(0)\alpha_{0}, \beta_{0}have the following form is
(22)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}
Ory_(i)=f(x_(i)),i=0,l,dots,ny_{i}=f\left(x_{i}\right), i=0, l, \ldots, n, then the coefficientsa_(0),b_(0)a_{0}, b_{0}。
alpha_(0),beta_(0)\alpha_{0}, \beta_{0}have 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]]:}\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}
(23)
{:[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}
We now determine the functions
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}, Or
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}, Or
{:(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*}
in such a way that they satisfy the conditions:
{:(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*}
We then consider as an approximation for the rootbar(x)\bar{x}from equation (7) the number
bar(x)_(p)=varphi_(s)(0)\bar{x}_{p}=\varphi_{s}(0)
If we writeT_(i-1)=1//varphi_(i-1)^(')(y_(i))T_{i-1}=1 / \varphi_{i-1}^{\prime}\left(y_{i}\right), then we obtain for the coefficientsa_(i),b_(i),alpha_(i),beta_(i)a_{i}, b_{i}, \alpha_{i}, \beta_{i}from (24) the following expressions:
(26)
(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)],[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}
It results from the conditionbar(x)≃varphi_(beta)(0)\bar{x} \simeq \varphi_{\beta}(0)the following expression for the root approximationbar(x)\bar{x}:
{:(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*}
BIBLIOGRAPHY
- IMAMOV, A., Resenie nelineinih uravnenii metodomi splain-- interpolirovanie, Metodi splain-funktii, Akademia Nauk SSSR. Novosibirsk, 81 (1979),74-8074-80,
- PAVALOIU, I., Rezolvarea ecuatillor prin interpolare, Editura Dacia, Cluj, 1981.
- 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.