Inverse interpolating splines with applications to the equation solving

Ion Păvăloiu
(original), approximation of univariate functions, Approximation Theory, html, inverse interpolation, nonlinear equations in R, Numerical Analysis, paper, remainder estimation, solving F(x)=0 iteratively, splines

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Crăciun Iancu
Tiberiu Popoviciu Institute of Numerical Analysis Romania Academy, Cluj-Napoca, Romania

Tiberiu Oproiu
Romanian Academy Astronomical Observatory

Ion Păvăloiu
Tiberiu Popoviciu Institute of Numerical Analysis Romania Academy, Cluj-Napoca, Romania

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C. Iancu, T. Oproiu, I. Păvăloiu, Inverse interpolating splines with applications to the equation solving, Seminar of Functional Analysis and Numerical Method, Babes-Bolyai University, Faculty of Mathematics, Preprint nr. 1, 1986, pp. 67-82.

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[3] Creville, T.N.E. (ed), Theory and Applications of Spline Functions, Academic Press, New York, London (1969)
[4] Cuzek, J.A., Kemper, C.A., A New Error Analysis for a Cubic Spline Approximate Solution of a Class of Volterra Integro-Differential Equations, Mathematics of Computation, Vol. 27, nr. 123 (1973), 563-570.
[5] Iancu, C., On the Cubic Spline of Interpolation, Seminar of Functional Analysis and Numerical Methods, Babsc-Bolyai University, Preprint 4 (1981), 52-71
[6] Iancu, C., P[v[loiu, I., La resolution des equations par interpolation inverse de type Hermite, Seminar of Functional analysis and Numerical Methods, Babs-Bolyai University, Preprint 4 (1981), 72-84
[7] Iancu, C., Păvăloiu, I., Resolution des equations a l’aide des fonctions epline d’interpolation inverse, Seminar Functional Analysis and Numerical Methods, Babeș-Bolyai University, Preprint 1 (1984), 97-104
[8] Imamov, A., Reshenie nelinejnykh uravnenii metodom obratnogo splajn interpolirovanie, Metody splajnfunktsii, Akad. Nauk. SSSR, Novosibirsk, 81 (1979), 74-80
[9] Păvăloiu, I., Rezolvarea   ecuațiilor prin interpolare, Ed. Dacia, Cluj, 1981.
[10] Seatzu, S., Un metodo per construzione di smoothing splines naturali mono e bidimensionali, Calcolo, 12 fasc. III (1975), 259-173.
[11] Stancu, D.D., Curs și culegere de probleme de analiză numerică, vol. I, Cluj-Napoca, 1977
[12] Zavyalov, I.S., Kvasov, B.I., Miroshnichenko, V.L., Metody splajn-funktsii, Moskva, 1980.
[13] Marsden, M., Cubic Spline Interpolation of Continuous Fuctions, J. of App. Th. 10(1974), 103-111
[14] Turovicz, B., A., Sur les derives d’ordre superieur d’une function inverse, Colloq. Math., 83-87 (1959).

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1986-Iancu-Oproiu-Pavaloiu-UBB-Preprint-Inverse-interpolating-spline-with-applications-to-the-equati
INVERSE INTERPOLATING SPLINES WITH APPLICATIONS TO THE EQUATION SOLVING
C. IANCU, T. OPROIU and I. PĀVĀLOIU
  1. INTRODUCTION
Let f f fff be the function:
(1)
f : [ a , b ] → R , f : [ a , b ] → R , f:[a,b]rarrR,f:[a, b] \rightarrow \mathbb{R},f:[a,b]→R,
and:
(2) Δ x : a = x 0 < x 1 < … < x n − 1 < x n = b , n ⩾ 2 , n ∈ N , (2) Δ x : a = x 0 < x 1 < … < x n − 1 < x n = b , n ⩾ 2 , n ∈ N , {:(2)Delta_(x):a=x_(0) < x_(1) < dots < x_(n-1) < x_(n)=b","quad n >= 2","n inN",":}\begin{equation*} \Delta_{x}: a=x_{0}<x_{1}<\ldots<x_{n-1}<x_{n}=b, \quad n \geqslant 2, n \in \mathbb{N}, \tag{2} \end{equation*}(2)Δx:a=x0<x1<…<xn−1<xn=b,n⩾2,n∈N,
a partition of the interval [ a , b ] [ a , b ] [a,b][\mathrm{a}, \mathrm{b}][a,b].
We assume that the function f f fff is twice derivable on the interval [ a , b ] [ a , b ] [a,b][\mathrm{a}, \mathrm{b}][a,b], and f ′ ( x ) ≠ 0 f ′ ( x ) ≠ 0 f^(')(x)!=0\mathrm{f}^{\prime}(\mathrm{x}) \neq 0f′(x)≠0 for all x ∈ [ a , b ] x ∈ [ a , b ] xin[a,b]\mathrm{x} \in[\mathrm{a}, \mathrm{b}]x∈[a,b]. We denote F = f ( [ a , b ] ) F = f ( [ a , b ] ) F=f([a,b])F=f([a, b])F=f([a,b]).
From the above assumption it results that the function f f fff is a bijection; hence there exists f − 1 : F → [ a , b ] f − 1 : F → [ a , b ] f^(-1):F rarr[a,b]f^{-1}: F \rightarrow[a, b]f−1:F→[a,b].
Since we have assumed that the function f f fff is twice
derivable on the interval [ a , b ] [ a , b ] [a,b][a, b][a,b], and f ′ ( x ) ≠ 0 , ∀ x ∈ [ a , b ] f ′ ( x ) ≠ 0 , ∀ x ∈ [ a , b ] f^(')(x)!=0,AA x in[a,b]f^{\prime}(x) \neq 0, \forall x \in[a, b]f′(x)≠0,∀x∈[a,b], it results that the function f − 1 f − 1 f^(-1)f^{-1}f−1 will also be twice derivable on the set F F FFF, and:
( f − 1 ( f i ) ) ′ = 1 / f ′ ( x i ) f − 1 f i ′ = 1 / f ′ x i (f^(-1)(f_(i)))^(')=1//f^(')(x_(i))\left(f^{-1}\left(f_{i}\right)\right)^{\prime}=1 / f^{\prime}\left(x_{i}\right)(f−1(fi))′=1/f′(xi)
(3) ( f − 1 ( f i ) ) n = − f n ( x i ) / ( f ′ ( x i ) ) 3 , i = 0 , n ― (3) f − 1 f i n = − f n x i / f ′ x i 3 , i = 0 , n ¯ {:(3)(f^(-1)(f_(i)))^(n)=-f^(n)(x_(i))//(f^(')(x_(i)))^(3)","quad i= bar(0,n):}\begin{equation*} \left(f^{-1}\left(f_{i}\right)\right)^{n}=-f^{n}\left(x_{i}\right) /\left(f^{\prime}\left(x_{i}\right)\right)^{3}, \quad i=\overline{0, n} \tag{3} \end{equation*}(3)(f−1(fi))n=−fn(xi)/(f′(xi))3,i=0,n―
Next, we deal with the elaboration of a numerical method for solving certain equations of the form:
(4) f ( x ) = 0 , x ∈ [ a , b ] (4) f ( x ) = 0 , x ∈ [ a , b ] {:(4)f(x)=0","quad x in[a","b]:}\begin{equation*} f(x)=0, \quad x \in[a, b] \tag{4} \end{equation*}(4)f(x)=0,x∈[a,b]
using the inverse interpolating cubic splines.
Assuming that the equation (4) admits a root x ¯ ∈ [ a , b ] x ¯ ∈ [ a , b ] bar(x)in[a,b]\bar{x} \in[a, b]x¯∈[a,b], then from f ( x ¯ ) = 0 f ( x ¯ ) = 0 f( bar(x))=0f(\bar{x})=0f(x¯)=0 it follows that f − 1 ( 0 ) = x ¯ f − 1 ( 0 ) = x ¯ f^(-1)(0)= bar(x)f^{-1}(0)=\bar{x}f−1(0)=x¯, while from f ′ ( x ) ≠ 0 f ′ ( x ) ≠ 0 f^(')(x)!=0f^{\prime}(x) \neq 0f′(x)≠0 for all x ∈ [ a , b ] x ∈ [ a , b ] x in[a,b]x \in[a, b]x∈[a,b] it follows the uniqueness of x ¯ x ¯ bar(x)\bar{x}x¯.
We shall remark the following two cases for solving the equation (4):
A. The values f i = f ( x i ) , i = 0 , n ― f i = f x i , i = 0 , n ¯ f_(i)=f(x_(i)),i= bar(0,n)f_{i}=f\left(x_{i}\right), i=\overline{0, n}fi=f(xi),i=0,n―, of the function f f fff on the knots of the partition (2) are given, and f ′ ( x 0 ) == m o , f ′ ′ ( x o ) = M o f ′ x 0 == m o , f ′ ′ x o = M o f^(')(x_(0))==m_(o),f^('')(x_(o))=M_(o)f^{\prime}\left(x_{0}\right)= =m_{o}, f^{\prime \prime}\left(x_{o}\right)=M_{o}f′(x0)==mo,f′′(xo)=Mo, where m o , M o ∈ R , m o ≠ 0 m o , M o ∈ R , m o ≠ 0 m_(o),M_(o)inR,m_(o)!=0m_{o}, M_{o} \in \mathbb{R}, m_{o} \neq 0mo,Mo∈R,mo≠0, are also given.
B. The values of the function f f fff on every point of the interval [ a , b ] [ a , b ] [a,b][\mathrm{a}, \mathrm{b}][a,b] are known.
In order to solve the equation (4) in each case (A or B B BBB ), we shall construct an inverse interpolating spline.

[ d d d  _("d ")_{\text {d }}d  = 7: 2. INVERSE IMTERPOLATIMG SPLIMES
Let [ α , β ] ⊆ F [ α , β ] ⊆ F [alpha,beta]subeF[\alpha, \beta] \subseteq \mathcal{F}[α,β]⊆F be the shortest interval containing the
values f i = f ( x i ) , i = 0 , n ― f i = f x i , i = 0 , n ¯ f_(i)=f(x_(i)),i= bar(0,n)f_{i}=f\left(x_{i}\right), i=\overline{0, n}fi=f(xi),i=0,n―, of the function f f fff on the knots of the partition (2).
Since f f fff is bijective on [ a , b ] [ a , b ] [a,b][a, b][a,b], it is either increasing or decreasing on this interval.
We shall consider further that f f fff is increasing, and we shall denote by:
(5) Δ 1 : α = f 0 < f 1 < … < f n − 1 < f n = β (5) Δ 1 : α = f 0 < f 1 < … < f n − 1 < f n = β {:(5)Delta_(1):alpha=f_(0) < f_(1) < dots < f_(n-1) < f_(n)=beta:}\begin{equation*} \Delta_{1}: \alpha=f_{0}<f_{1}<\ldots<f_{n-1}<f_{n}=\beta \tag{5} \end{equation*}(5)Δ1:α=f0<f1<…<fn−1<fn=β
the partition of the interval [ α , β ] [ α , β ] [alpha,beta][\alpha, \beta][α,β] determined by the points f i , i = 0 , n ― f i , i = 0 , n ¯ f_(i),i= bar(0,n)f_{i}, i=\overline{0, n}fi,i=0,n―.
We shall construct the inverse interpolating cubic spline associated to the function f f fff.
Using the partitions (2) and (5), we construct the function H ¯ H ¯ bar(H)\bar{H}H¯, whose restrictions have the following form:
(6) H ¯ i ( y ) = a i y 3 + b i y 2 + c i y + d i , i = 0 , n − 1 ― (6) H ¯ i ( y ) = a i y 3 + b i y 2 + c i y + d i , i = 0 , n − 1 ¯ {:(6) bar(H)_(i)(y)=a_(i)y^(3)+b_(i)y^(2)+c_(i)y+d_(i)","quad i= bar(0,n-1):}\begin{equation*} \bar{H}_{i}(y)=a_{i} y^{3}+b_{i} y^{2}+c_{i} y+d_{i}, \quad i=\overline{0, n-1} \tag{6} \end{equation*}(6)H¯i(y)=aiy3+biy2+ciy+di,i=0,n−1―
for all y ∈ [ f i , f i + l ] y ∈ f i , f i + l y in[f_(i),f_(i+l)]y \in\left[f_{i}, f_{i+l}\right]y∈[fi,fi+l]. These restrictions are uniquely determined from the conditions:
(7)
H ¯ i ( f i ) = x i , H ¯ i ′ ( f i ) = D i ′ H ¯ i ′ ′ ( f i ) = D i ′ ′ H ¯ i ( f i + 1 ) = x i + 1 , i = 0 , n − 1 ― H ¯ i f i = x i , H ¯ i ′ f i = D i ′ H ¯ i ′ ′ f i = D i ′ ′ H ¯ i f i + 1 = x i + 1 , i = 0 , n − 1 ¯ {:[ bar(H)_(i)(f_(i))=x_(i)","],[ bar(H)_(i)^(')(f_(i))=D_(i)^(')],[ bar(H)_(i)^('')(f_(i))=D_(i)^('')],[ bar(H)_(i)(f_(i+1))=x_(i+1)","quad i= bar(0,n-1)]:}\begin{aligned} & \bar{H}_{i}\left(f_{i}\right)=x_{i}, \\ & \bar{H}_{i}^{\prime}\left(f_{i}\right)=D_{i}^{\prime} \\ & \bar{H}_{i}^{\prime \prime}\left(f_{i}\right)=D_{i}^{\prime \prime} \\ & \bar{H}_{i}\left(f_{i+1}\right)=x_{i+1}, \quad i=\overline{0, n-1} \end{aligned}H¯i(fi)=xi,H¯i′(fi)=Di′H¯i′′(fi)=Di′′H¯i(fi+1)=xi+1,i=0,n−1―
where:
gatwortosi ant
D 0 ′ = 1 / f ′ ( x 0 ) = 1 / m 0 (8) D 0 ′ ′ = − f ′ ′ ( x 0 ) / ( f ′ ( x 0 ) ) 3 = − M 0 / m 0 3 ; D i ′ = H ¯ i − 1 ′ ( f i ) , D i ′ ′ = H ¯ i − 1 ′ ′ ( f i ) , i = 1 , n ― D 0 ′ = 1 / f ′ x 0 = 1 / m 0 (8) D 0 ′ ′ = − f ′ ′ x 0 / f ′ x 0 3 = − M 0 / m 0 3 ; D i ′ = H ¯ i − 1 ′ f i , D i ′ ′ = H ¯ i − 1 ′ ′ f i , i = 1 , n ¯ {:[D_(0)^(')=1//f^(')(x_(0))=1//m_(0)],[(8)D_(0)^('')=-f^('')(x_(0))//(f^(')(x_(0)))^(3)=-M_(0)//m_(0)^(3);],[D_(i)^(')= bar(H)_(i-1)^(')(f_(i))","quadD_(i)^('')= bar(H)_(i-1)^('')(f_(i))","quad i= bar(1,n)]:}\begin{align*} & D_{0}^{\prime}=1 / f^{\prime}\left(x_{0}\right)=1 / m_{0} \\ & D_{0}^{\prime \prime}=-f^{\prime \prime}\left(x_{0}\right) /\left(f^{\prime}\left(x_{0}\right)\right)^{3}=-M_{0} / m_{0}^{3} ; \tag{8}\\ & D_{i}^{\prime}=\bar{H}_{i-1}^{\prime}\left(f_{i}\right), \quad D_{i}^{\prime \prime}=\bar{H}_{i-1}^{\prime \prime}\left(f_{i}\right), \quad i=\overline{1, n} \end{align*}D0′=1/f′(x0)=1/m0(8)D0′′=−f′′(x0)/(f′(x0))3=−M0/m03;Di′=H¯i−1′(fi),Di′′=H¯i−1′′(fi),i=1,n―
From (6) and (7) one obtains:
(9) a i = ( 1 − D i ′ [ x i , x i + 1 ; f ] − ( D i ′ ′ / 2 ) [ x i , x i + 1 ; f ] 2 ( x i + 1 − a i = 1 − D i ′ x i , x i + 1 ; f − D i ′ ′ / 2 x i , x i + 1 ; f 2 x i + 1 − quada_(i)=(1-D_(i)^(')[x_(i),x_(i+1);f]-(D_(i)^('')//2)[x_(i),x_(i+1);f]^(2)(x_(i+1)-:}\quad a_{i}=\left(1-D_{i}^{\prime}\left[x_{i}, x_{i+1} ; f\right]-\left(D_{i}^{\prime \prime} / 2\right)\left[x_{i}, x_{i+1} ; f\right]^{2}\left(x_{i+1}-\right.\right.ai=(1−Di′[xi,xi+1;f]−(Di′′/2)[xi,xi+1;f]2(xi+1−
− x i ) ) ( [ x i , x i + 1 ; f ] 3 ( x i + 1 − x i ) 2 ) − 1 − x i x i , x i + 1 ; f 3 x i + 1 − x i 2 − 1 {:-x_(i)))([x_(i),x_(i+1);f]^(3)(x_(i+1)-x_(i))^(2))^(-1)\left.\left.-x_{i}\right)\right)\left(\left[x_{i}, x_{i+1} ; f\right]^{3}\left(x_{i+1}-x_{i}\right)^{2}\right)^{-1}−xi))([xi,xi+1;f]3(xi+1−xi)2)−1
(10) b i = D i ′ ′ / 2 − 3 a i f i b i = D i ′ ′ / 2 − 3 a i f i quadb_(i)=D_(i)^('')//2-3a_(i)f_(i)\quad b_{i}=D_{i}^{\prime \prime} / 2-3 a_{i} f_{i}bi=Di′′/2−3aifi,
(11) c i = D i ′ − D i ′ ′ f i + 3 a i f i 2 c i = D i ′ − D i ′ ′ f i + 3 a i f i 2 quadc_(i)=D_(i)^(')-D_(i)^('')f_(i)+3a_(i)f_(i)^(2)\quad c_{i}=D_{i}^{\prime}-D_{i}^{\prime \prime} f_{i}+3 a_{i} f_{i}^{2}ci=Di′−Di′′fi+3aifi2,
(12) d i = x i − D i ′ f i + ( D i ′ ′ / 2 ) I i 2 − a i f i 3 , i = 0 , n − 1 ― d i = x i − D i ′ f i + D i ′ ′ / 2 I i 2 − a i f i 3 , i = 0 , n − 1 ¯ quadd_(i)=x_(i)-D_(i)^(')f_(i)+(D_(i)^('')//2)I_(i)^(2)-a_(i)f_(i)^(3),quad i= bar(0,n-1)\quad d_{i}=x_{i}-D_{i}^{\prime} f_{i}+\left(D_{i}^{\prime \prime} / 2\right) I_{i}^{2}-a_{i} f_{i}^{3}, \quad i=\overline{0, n-1}di=xi−Di′fi+(Di′′/2)Ii2−aifi3,i=0,n−1―,
where [ x i , x i + 1 ; f ] x i , x i + 1 ; f [x_(i),x_(i+1);f]\left[x_{i}, x_{i+1} ; f\right][xi,xi+1;f] denotes the first order divided difference of the function f f fff on the knots x i , x i + 1 x i , x i + 1 x_(i),x_(i+1)x_{i}, x_{i+1}xi,xi+1.
Taking into account the equalities:
H ¯ i ′ ( f i + 1 ) = D i + 1 ′ , (13) H ¯ i ′ ′ ( f i + 1 ) = D i + 1 ′ ′ , i = 0 , n − 1 ― , H ¯ i ′ f i + 1 = D i + 1 ′ , (13) H ¯ i ′ ′ f i + 1 = D i + 1 ′ ′ , i = 0 , n − 1 ¯ , {:[ bar(H)_(i)^(')(f_(i+1))=D_(i+1)^(')","],[(13) bar(H)_(i)^('')(f_(i+1))=D_(i+1)^('')","quad i= bar(0,n-1)","]:}\begin{align*} & \bar{H}_{i}^{\prime}\left(f_{i+1}\right)=D_{i+1}^{\prime}, \\ & \bar{H}_{i}^{\prime \prime}\left(f_{i+1}\right)=D_{i+1}^{\prime \prime}, \quad i=\overline{0, n-1}, \tag{13} \end{align*}H¯i′(fi+1)=Di+1′,(13)H¯i′′(fi+1)=Di+1′′,i=0,n−1―,
one obtains from (6), (9), (10) and (11) the following recurrence formulae:
(14)
D i + 1 ′ = D i ′ + D i ′ ′ [ x i , x i + 1 ; f ] ( x i + 1 − x i ) + + 3 a i [ x i , x i + 1 ; f ] 2 ( x i + 1 − x i ) 2 D i + 1 ′ ′ = D i ′ ′ + 6 a i [ x i , x i + 1 ; f ] ( x i + 1 − x i ) D i + 1 ′ = D i ′ + D i ′ ′ x i , x i + 1 ; f x i + 1 − x i + + 3 a i x i , x i + 1 ; f 2 x i + 1 − x i 2 D i + 1 ′ ′ = D i ′ ′ + 6 a i x i , x i + 1 ; f x i + 1 − x i {:[D_(i+1)^(')=D_(i)^(')+D_(i)^('')[x_(i),x_(i+1);f](x_(i+1)-x_(i))+],[+3a_(i)[x_(i),x_(i+1);f]^(2)(x_(i+1)-x_(i))^(2)],[D_(i+1)^('')=D_(i)^('')+6a_(i)[x_(i),x_(i+1);f](x_(i+1)-x_(i))]:}\begin{aligned} D_{i+1}^{\prime}= & D_{i}^{\prime}+D_{i}^{\prime \prime}\left[x_{i}, x_{i+1} ; f\right]\left(x_{i+1}-x_{i}\right)+ \\ & +3 a_{i}\left[x_{i}, x_{i+1} ; f\right]^{2}\left(x_{i+1}-x_{i}\right)^{2} \\ D_{i+1}^{\prime \prime}= & D_{i}^{\prime \prime}+6 a_{i}\left[x_{i}, x_{i+1} ; f\right]\left(x_{i+1}-x_{i}\right) \end{aligned}Di+1′=Di′+Di′′[xi,xi+1;f](xi+1−xi)++3ai[xi,xi+1;f]2(xi+1−xi)2Di+1′′=Di′′+6ai[xi,xi+1;f](xi+1−xi)
for the calculation of the values D i ′ D i ′ D_(i)^(')D_{i}^{\prime}Di′ and D i ′ ′ , i = 0 , n − 1 ― D i ′ ′ , i = 0 , n − 1 ¯ D_(i)^(''),i= bar(0,n-1)D_{i}^{\prime \prime}, i=\overline{0, n-1}Di′′,i=0,n−1―.
Knowing the restrictions H ¯ i H ¯ i bar(H)_(i)\bar{H}_{i}H¯i for i = 0 , n − 1 ― i = 0 , n − 1 ¯ i= bar(0,n-1)i=\overline{0, n-1}i=0,n−1―, if we have determined an interval ( x j , x j + 1 x j , x j + 1 x_(j),x_(j+1)x_{j}, x_{j+1}xj,xj+1 ) in which the root x ¯ x ¯ bar(x)\bar{x}x¯ of the given equation (4) is lying, since H ¯ j H ¯ j bar(H)_(j)\bar{H}_{j}H¯j approaches the function f − 1 f − 1 f^(-1)f^{-1}f−1 for x ∈ ( x j , x j + 1 ) x ∈ x j , x j + 1 x in(x_(j),x_(j+1))x \in\left(x_{j}, x_{j+1}\right)x∈(xj,xj+1), it results the following approximation for the root x ¯ x ¯ bar(x)\bar{x}x¯ :
(15) H ¯ j ( 0 ) = d j ≅ x ¯ . (15) H ¯ j ( 0 ) = d j ≅ x ¯ . {:(15) bar(H)_(j)(0)=d_(j)~= bar(x).:}\begin{equation*} \bar{H}_{j}(0)=d_{j} \cong \bar{x} . \tag{15} \end{equation*}(15)H¯j(0)=dj≅x¯.

3. ANOTHER FORM FOR THE INVERSE INTERPOLATIEG SPLINE

In this section we shall give a new form for the inverae interpolating spline (6). For this purpose, using the model proposed in [5], we shall construct the cubic interpolating apline for f − 1 f − 1 f^(-1)f^{-1}f−1 on the knots of the partition (5), whom restriction on the interval [ f i − 1 , f i ] f i − 1 , f i [f_(i-1),f_(i)]\left[f_{i-1}, f_{i}\right][fi−1,fi] has the following form:
(16) H ¯ i − 1 ( y ) = ( ( D i ′ ′ − D i − 1 ′ ′ ) / ( 6 k i ) ) ( y − f i − 1 ) 3 + ( D i − 1 ′ ′ / 2 ) . ⋅ ( y − f i − 1 ) 2 + D i − 1 ′ ( y − f i − 1 ) + x i − 1 (16) H ¯ i − 1 ( y ) = D i ′ ′ − D i − 1 ′ ′ / 6 k i y − f i − 1 3 + D i − 1 ′ ′ / 2 . ⋅ y − f i − 1 2 + D i − 1 ′ y − f i − 1 + x i − 1 {:[(16) bar(H)_(i-1)(y)=((D_(i)^('')-D_(i-1)^(''))//(6k_(i)))(y-f_(i-1))^(3)+(D_(i-1)^('')//2).],[*(y-f_(i-1))^(2)+D_(i-1)^(')(y-f_(i-1))+x_(i-1)]:}\begin{align*} \bar{H}_{i-1}(y)= & \left(\left(D_{i}^{\prime \prime}-D_{i-1}^{\prime \prime}\right) /\left(6 k_{i}\right)\right)\left(y-f_{i-1}\right)^{3}+\left(D_{i-1}^{\prime \prime} / 2\right) . \tag{16}\\ & \cdot\left(y-f_{i-1}\right)^{2}+D_{i-1}^{\prime}\left(y-f_{i-1}\right)+x_{i-1} \end{align*}(16)H¯i−1(y)=((Di′′−Di−1′′)/(6ki))(y−fi−1)3+(Di−1′′/2).⋅(y−fi−1)2+Di−1′(y−fi−1)+xi−1
where:
y ∈ [ f i − 1 , f i ] , D i n ≠ D i − 1 n , k i = f i − f i − 1 , i = 1 , n ― , y ∈ f i − 1 , f i , D i n ≠ D i − 1 n , k i = f i − f i − 1 , i = 1 , n ¯ , y in[f_(i-1),f_(i)],quadD_(i)^(n)!=D_(i-1)^(n),quadk_(i)=f_(i)-f_(i-1),quad i= bar(1,n),y \in\left[f_{i-1}, f_{i}\right], \quad D_{i}^{n} \neq D_{i-1}^{n}, \quad k_{i}=f_{i}-f_{i-1}, \quad i=\overline{1, n},y∈[fi−1,fi],Din≠Di−1n,ki=fi−fi−1,i=1,n―,
and:
D i = H ¯ i ( f i ) , D i ′ = H ¯ i ′ ′ ( f i ) , i = 0 , n ― . D i = H ¯ i f i , D i ′ = H ¯ i ′ ′ f i , i = 0 , n ¯ . D_(i)= bar(H)_(i)(f_(i)),quadD_(i)^(')= bar(H)_(i)^('')(f_(i)),quad i= bar(0,n).D_{i}=\bar{H}_{i}\left(f_{i}\right), \quad D_{i}^{\prime}=\bar{H}_{i}^{\prime \prime}\left(f_{i}\right), \quad i=\overline{0, n} .Di=H¯i(fi),Di′=H¯i′′(fi),i=0,n―.
Denoting by H ( 3 , Δ f ) H 3 , Δ f H(3,Delta_(f))H\left(3, \Delta_{f}\right)H(3,Δf) the set of all the cubic interpolating splines corresponding to the partition (5), the following proposition holds:
Proposition 1. Every spline H ¯ ∈ H ( 3 , Δ f ) H ¯ ∈ H 3 , Δ f bar(H)in H(3,Delta_(f))\bar{H} \in H\left(3, \Delta_{f}\right)H¯∈H(3,Δf) which has the form (16) is uniquely determined if it satisfies the following conditions:
H ¯ i − 1 ( f i ) = x i H ¯ i − 1 f i = x i bar(H)_(i-1)(f_(i))=x_(i)\bar{H}_{i-1}\left(f_{i}\right)=x_{i}H¯i−1(fi)=xi
(17)
H ¯ i − 1 ( f i ) = D i ′ , D 0 ′ = a 1 , D 0 ′ ′ = a 2 , H ¯ i − 1 f i = D i ′ , D 0 ′ = a 1 , D 0 ′ ′ = a 2 , {:[ bar(H)_(i-1)(f_(i))=D_(i)^(')","],[D_(0)^(')=a_(1)","quadD_(0)^('')=a_(2)","]:}\begin{aligned} & \bar{H}_{i-1}\left(f_{i}\right)=D_{i}^{\prime}, \\ & D_{0}^{\prime}=a_{1}, \quad D_{0}^{\prime \prime}=a_{2}, \end{aligned}H¯i−1(fi)=Di′,D0′=a1,D0′′=a2,
where a 1 , a 2 ∈ R a 1 , a 2 ∈ R a_(1),a_(2)inRa_{1}, a_{2} \in \mathbb{R}a1,a2∈R are given.
Proof. Using the formula (16), the first two conditions
D i ′ ′ = ( 6 / k i 2 ) ( x i − x i − 1 ) − ( 6 / k i ) D i − 1 ′ − 2 D i − 1 ′ ′ , (18) D i ′ = ( 3 / k i ) ( x i − x i − 1 ) − 2 D i − 1 ′ − ( k i / 2 ) D i − 1 ′ ′ , i = I , n ― . D i ′ ′ = 6 / k i 2 x i − x i − 1 − 6 / k i D i − 1 ′ − 2 D i − 1 ′ ′ , (18) D i ′ = 3 / k i x i − x i − 1 − 2 D i − 1 ′ − k i / 2 D i − 1 ′ ′ , i = I , n ¯ . {:[D_(i)^('')=(6//k_(i)^(2))(x_(i)-x_(i-1))-(6//k_(i))D_(i-1)^(')-2D_(i-1)^('')","],[(18)D_(i)^(')=(3//k_(i))(x_(i)-x_(i-1))-2D_(i-1)^(')-(k_(i)//2)D_(i-1)^('')","quad i= bar(I,n).]:}\begin{align*} & D_{i}^{\prime \prime}=\left(6 / k_{i}^{2}\right)\left(x_{i}-x_{i-1}\right)-\left(6 / k_{i}\right) D_{i-1}^{\prime}-2 D_{i-1}^{\prime \prime}, \\ & D_{i}^{\prime}=\left(3 / k_{i}\right)\left(x_{i}-x_{i-1}\right)-2 D_{i-1}^{\prime}-\left(k_{i} / 2\right) D_{i-1}^{\prime \prime}, \quad i=\overline{I, n} . \tag{18} \end{align*}Di′′=(6/ki2)(xi−xi−1)−(6/ki)Di−1′−2Di−1′′,(18)Di′=(3/ki)(xi−xi−1)−2Di−1′−(ki/2)Di−1′′,i=I,n―.
Since, in the problem we have studied, we have 'D' = = a 1 = 1 / f ′ ( x o ) = a 1 = 1 / f ′ x o =a_(1)=1//f^(')(x_(o))=a_{1}=1 / f^{\prime}\left(x_{o}\right)=a1=1/f′(xo) and D o ′ ′ = a 2 = − f ′ ′ ( x o ) / ( f ′ ( x o ) ) 3 D o ′ ′ = a 2 = − f ′ ′ x o / f ′ x o 3 D_(o)^('')=a_(2)=-f^('')(x_(o))//(f^(')(x_(o)))^(3)D_{o}^{\prime \prime}=a_{2}=-f^{\prime \prime}\left(x_{o}\right) /\left(f^{\prime}\left(x_{o}\right)\right)^{3}Do′′=a2=−f′′(xo)/(f′(xo))3 are known, it follows that the system (18) is compatible determined and provides the values D 1 ′ , D 2 ′ , … , D n ′ , D 1 ′ ′ , D 2 ′ ′ , … , D n ′ ′ D 1 ′ , D 2 ′ , … , D n ′ , D 1 ′ ′ , D 2 ′ ′ , … , D n ′ ′ D_(1)^('),D_(2)^('),dots,D_(n)^('),D_(1)^(''),D_(2)^(''),dots,D_(n)^('')D_{1}^{\prime}, D_{2}^{\prime}, \ldots, D_{n}^{\prime}, D_{1}^{\prime \prime}, D_{2}^{\prime \prime}, \ldots, D_{n}^{\prime \prime}D1′,D2′,…,Dn′,D1′′,D2′′,…,Dn′′ which determine the cubic spline whose restrictions are of the form (16).
Note. If the function f f fff is decreasing, one proceeds analogously, performing a rearrangement of the knots from right to left and taking the values D n ′ D n ′ D_(n)^(')D_{n}^{\prime}Dn′ and D n ′ ′ D n ′ ′ D_(n)^('')D_{n}^{\prime \prime}Dn′′ instead of D o ′ D o ′ D_(o)^(')D_{o}^{\prime}Do′ and D o ′ ′ D o ′ ′ D_(o)^('')D_{o}^{\prime \prime}Do′′, respectively.

4. SOLUTION OF THE EQUATION f ( x ) = 0 f ( x ) = 0 f(x)=0f(x)=0f(x)=0

The sign of the values f 1 , i = 0 , n ― f 1 , i = 0 , n ¯ f_(1),i= bar(0,n)f_{1}, i=\overline{0, n}f1,i=0,n―, of the partition (5) points out the interval ( x j , x j + 1 ) ⊂ ( a , b ) x j , x j + 1 ⊂ ( a , b ) (x_(j),x_(j+1))sub(a,b)\left(x_{j}, x_{j+1}\right) \subset(a, b)(xj,xj+1)⊂(a,b) in which the root x ¯ x ¯ bar(x)\bar{x}x¯ of the equation (4) is lying.
The method we present in this paper is given by the formulae (15) and (16), obtaining the following approximation for the root x ¯ x ¯ bar(x)\bar{x}x¯ :
(19) x ¯ ≅ H ¯ j ( 0 ) = ( ( D j + 1 ′ ′ − D j ′ ′ ) / ( 6 k j + 1 ) ) ( − f j ) 3 + x ¯ ≅ H ¯ j ( 0 ) = D j + 1 ′ ′ − D j ′ ′ / 6 k j + 1 − f j 3 + quad bar(x)~= bar(H)_(j)(0)=((D_(j+1)^('')-D_(j)^(''))//(6k_(j+1)))(-f_(j))^(3)+\quad \bar{x} \cong \bar{H}_{j}(0)=\left(\left(D_{j+1}^{\prime \prime}-D_{j}^{\prime \prime}\right) /\left(6 k_{j+1}\right)\right)\left(-f_{j}\right)^{3}+x¯≅H¯j(0)=((Dj+1′′−Dj′′)/(6kj+1))(−fj)3+
+ ( D j ′ ′ / 2 ) ( − f j ) 2 + D j ′ ( − f j ) + x j + D j ′ ′ / 2 − f j 2 + D j ′ − f j + x j +(D_(j)^('')//2)(-f_(j))^(2)+D_(j)^(')(-f_(j))+x_(j)+\left(D_{j}^{\prime \prime} / 2\right)\left(-f_{j}\right)^{2}+D_{j}^{\prime}\left(-f_{j}\right)+x_{j}+(Dj′′/2)(−fj)2+Dj′(−fj)+xj.

5. ESTTILATE OF THE ERROR

Theorem 1. Let the function (1) be increasing. If the inverse function f − 1 f − 1 f^(-1)f^{-1}f−1 is approximated on the interval [ f i − 1 f i − 1 [f_(i-1):}\left[f_{i-1}\right.[fi−1, f i ] , i = I , n ― f i , i = I , n ¯ {:f_(i)],i= bar(I,n)\left.f_{i}\right], i=\overline{I, n}fi],i=I,n―, by a cubic apline of the form (16), observing the conditions (17) and D 1 % − D 1 − 1 # > 0 D 1 % − D 1 − 1 # > 0 D_(1)^(%)-D_(1-1)^(#) > 0D_{1}^{\%}-D_{1-1}^{\#}>0D1%−D1−1#>0, then the following inequalities hold:
(20) | H ¯ i − 1 ( y ) − f − 1 ( y ) | < { δ x , for D i − 1 ′ ′ > 0 , D i − 1 ′ > 0 ; δ x − 2 D i − 1 ′ k i , for D i − 1 ′ ′ > 0 , δ x − D i − 1 ′ ′ k i 2 , for D i − 1 ′ ′ < 0 ; D i − 1 > 0 ; δ x − 2 D i − 1 k i − D i − 1 ′ ′ k i 2 , for D i − 1 ′ ′ < 0 , D i − 1 < 0 , H ¯ i − 1 ( y ) − f − 1 ( y ) < δ x ,       for  D i − 1 ′ ′ > 0 , D i − 1 ′ > 0 ; δ x − 2 D i − 1 ′ k i ,       for  D i − 1 ′ ′ > 0 , δ x − D i − 1 ′ ′ k i 2 ,       for  D i − 1 ′ ′ < 0 ; D i − 1 > 0 ;      δ x − 2 D i − 1 k i − D i − 1 ′ ′ k i 2 ,  for  D i − 1 ′ ′ < 0 , D i − 1 < 0 , quad| bar(H)_(i-1)(y)-f^(-1)(y)| < {[delta_(x)","," for "D_(i-1)^('') > 0","D_(i-1)^(') > 0;],[delta_(x)-2D_(i-1)^(')k_(i)","," for "D_(i-1)^('') > 0","],[delta_(x)-D_(i-1)^('')k_(i)^(2)","," for "D_(i-1)^('') < 0;],[D_(i-1) > 0;,],[delta_(x)-2D_(i-1)k_(i)-D_(i-1)^('')k_(i)^(2)","" for "],[D_(i-1)^('') < 0","D_(i-1) < 0","]:}\quad\left|\bar{H}_{i-1}(y)-f^{-1}(y)\right|<\left\{\begin{array}{rr}\delta_{x}, & \text { for } D_{i-1}^{\prime \prime}>0, D_{i-1}^{\prime}>0 ; \\ \delta_{x}-2 D_{i-1}^{\prime} k_{i}, & \text { for } D_{i-1}^{\prime \prime}>0, \\ \delta_{x}-D_{i-1}^{\prime \prime} k_{i}^{2}, & \text { for } D_{i-1}^{\prime \prime}<0 ; \\ D_{i-1}>0 ; & \\ \delta_{x}-2 D_{i-1} k_{i}-D_{i-1}^{\prime \prime} k_{i}^{2}, \text { for } \\ D_{i-1}^{\prime \prime}<0, D_{i-1}<0,\end{array}\right.|H¯i−1(y)−f−1(y)|<{δx, for Di−1′′>0,Di−1′>0;δx−2Di−1′ki, for Di−1′′>0,δx−Di−1′′ki2, for Di−1′′<0;Di−1>0;δx−2Di−1ki−Di−1′′ki2, for Di−1′′<0,Di−1<0,
where δ x = x i − 2 x i − 1 + x , x ∈ [ x i − 1 , x i ] , i = 1 , n ― δ x = x i − 2 x i − 1 + x , x ∈ x i − 1 , x i , i = 1 , n ¯ delta_(x)=x_(i)-2x_(i-1)+x,quad x in[x_(i-1),x_(i)],quad i= bar(1,n)\delta_{x}=x_{i}-2 x_{i-1}+x, \quad x \in\left[x_{i-1}, x_{i}\right], \quad i=\overline{1, n}δx=xi−2xi−1+x,x∈[xi−1,xi],i=1,n―.
Proof. It results from (16):
H ¯ i − 1 ( y ) − f − 1 ( y ) = ( ( D i n − D i − 1 n ) / ( 6 k i ) ) ( y − f i − 1 ) 3 + + ( D i − 1 n / 2 ) ( y − f i − 1 ) 2 + D i − 1 ′ ( y − f i − 1 ) + x i − 1 − f − 1 ( y ) . H ¯ i − 1 ( y ) − f − 1 ( y ) = D i n − D i − 1 n / 6 k i y − f i − 1 3 + + D i − 1 n / 2 y − f i − 1 2 + D i − 1 ′ y − f i − 1 + x i − 1 − f − 1 ( y ) . {:[ bar(H)_(i-1)(y)-f^(-1)(y)=((D_(i)^(n)-D_(i-1)^(n))//(6k_(i)))(y-f_(i-1))^(3)+],[quad+(D_(i-1)^(n)//2)(y-f_(i-1))^(2)+D_(i-1)^(')(y-f_(i-1))+x_(i-1)-f^(-1)(y).]:}\begin{aligned} & \bar{H}_{i-1}(y)-f^{-1}(y)=\left(\left(D_{i}^{n}-D_{i-1}^{n}\right) /\left(6 k_{i}\right)\right)\left(y-f_{i-1}\right)^{3}+ \\ & \quad+\left(D_{i-1}^{n} / 2\right)\left(y-f_{i-1}\right)^{2}+D_{i-1}^{\prime}\left(y-f_{i-1}\right)+x_{i-1}-f^{-1}(y) . \end{aligned}H¯i−1(y)−f−1(y)=((Din−Di−1n)/(6ki))(y−fi−1)3++(Di−1n/2)(y−fi−1)2+Di−1′(y−fi−1)+xi−1−f−1(y).
and then:
(21) | H ¯ i − 1 ( y ) − f − 1 ( y ) | < | ( D i ′ ′ − D i − 1 ′ ′ ) / ( 6 k i ) | | y − f i − 1 | 3 + H ¯ i − 1 ( y ) − f − 1 ( y ) < D i ′ ′ − D i − 1 ′ ′ / 6 k i y − f i − 1 3 + | bar(H)_(i-1)(y)-f^(-1)(y)| < |(D_(i)^('')-D_(i-1)^(''))//(6k_(i))||y-f_(i-1)|^(3)+\left|\bar{H}_{i-1}(y)-f^{-1}(y)\right|<\left|\left(D_{i}^{\prime \prime}-D_{i-1}^{\prime \prime}\right) /\left(6 k_{i}\right)\right|\left|y-f_{i-1}\right|^{3}+|H¯i−1(y)−f−1(y)|<|(Di′′−Di−1′′)/(6ki)||y−fi−1|3+
+ | D i − 1 n / 2 | | y ˙ − f i − 1 | 2 + | D i − 1 | | y − f i − 1 | + | x i − 1 − x | + D i − 1 n / 2 y ˙ − f i − 1 2 + D i − 1 y − f i − 1 + x i − 1 − x +|D_(i-1)^(n)//2||(y^(˙))-f_(i-1)|^(2)+|D_(i-1)||y-f_(i-1)|+|x_(i-1)-x|+\left|D_{i-1}^{n} / 2\right|\left|\dot{y}-f_{i-1}\right|^{2}+\left|D_{i-1}\right|\left|y-f_{i-1}\right|+\left|x_{i-1}-x\right|+|Di−1n/2||y˙−fi−1|2+|Di−1||y−fi−1|+|xi−1−x|.
From the conditions of the theorem we have | y − p i − 1 | < y − p i − 1 < |y-p_(i-1)| <\left|y-p_{i-1}\right|<|y−pi−1|<
⟨ | f i − f i − 1 ∣= k i ⟨ | f i − f i − 1 ∣= k i (:|f_(i)-f_(i-1)∣=k_(i)\langle | f_{i}-f_{i-1} \mid=k_{i}⟨|fi−fi−1∣=ki, and then the inequality (21) can be put in the form:
(22) | H ¯ i − 1 ( y ) − f − 1 ( y ) | < ( ( D i n − D i − 1 n ) / ( 6 k i ) ) k i 3 + | D i − 1 n / 2 | k i 2 + H ¯ i − 1 ( y ) − f − 1 ( y ) < D i n − D i − 1 n / 6 k i k i 3 + D i − 1 n / 2 k i 2 + | bar(H)_(i-1)(y)-f^(-1)(y)| < ((D_(i)^(n)-D_(i-1)^(n))//(6k_(i)))k_(i)^(3)+|D_(i-1)^(n)//2|k_(i)^(2)+\left|\bar{H}_{i-1}(y)-f^{-1}(y)\right|<\left(\left(D_{i}^{n}-D_{i-1}^{n}\right) /\left(6 k_{i}\right)\right) k_{i}^{3}+\left|D_{i-1}^{n} / 2\right| k_{i}^{2}+|H¯i−1(y)−f−1(y)|<((Din−Di−1n)/(6ki))ki3+|Di−1n/2|ki2+
+ | D i − 1 | k i + x − x i − 1 . + D i − 1 k i + x − x i − 1 . +|D_(i-1)|k_(i)+x-x_(i-1).+\left|D_{i-1}\right| k_{i}+x-x_{i-1} .+|Di−1|ki+x−xi−1.
If D i − 1 n > 0 D i − 1 n > 0 D_(i-1)^(n) > 0D_{i-1}^{n}>0Di−1n>0, one obtains from (22):
(23) | H ¯ i − 1 ( y ) − f − 1 ( y ) | < ( D i n / 6 ) k i 2 − ( D i − 1 n / 6 ) k i 2 + ( D i − 1 n / 2 ) k i 2 + H ¯ i − 1 ( y ) − f − 1 ( y ) < D i n / 6 k i 2 − D i − 1 n / 6 k i 2 + D i − 1 n / 2 k i 2 + | bar(H)_(i-1)(y)-f^(-1)(y)| < (D_(i)^(n)//6)k_(i)^(2)-(D_(i-1)^(n)//6)k_(i)^(2)+(D_(i-1)^(n)//2)k_(i)^(2)+\left|\bar{H}_{i-1}(y)-f^{-1}(y)\right|<\left(D_{i}^{n} / 6\right) k_{i}^{2}-\left(D_{i-1}^{n} / 6\right) k_{i}^{2}+\left(D_{i-1}^{n} / 2\right) k_{i}^{2}+|H¯i−1(y)−f−1(y)|<(Din/6)ki2−(Di−1n/6)ki2+(Di−1n/2)ki2+
+ | D i − 1 | k i + x − x i − 1 = ( D i n / 6 ) k i 2 + 2 ( D i − 1 n / 6 ) k i 2 + + D i − 1 k i + x − x i − 1 = D i n / 6 k i 2 + 2 D i − 1 n / 6 k i 2 + +|D_(i-1)|k_(i)+x-x_(i-1)=(D_(i)^(n)//6)k_(i)^(2)+2(D_(i-1)^(n)//6)k_(i)^(2)++\left|D_{i-1}\right| k_{i}+x-x_{i-1}=\left(D_{i}^{n} / 6\right) k_{i}^{2}+2\left(D_{i-1}^{n} / 6\right) k_{i}^{2}++|Di−1|ki+x−xi−1=(Din/6)ki2+2(Di−1n/6)ki2+
+ | D i − 1 | k i + x − x i − 1 + D i − 1 k i + x − x i − 1 +|D_(i-1)|k_(i)+x-x_(i-1)+\left|D_{i-1}\right| k_{i}+x-x_{i-1}+|Di−1|ki+x−xi−1.
Taking into account the first equality (18), the inequality (23) will acquire the form:
(24) | H ¯ i − 1 ( y ) − f − 1 ( y ) | < ( k i 2 / 6 ) ( D i n + 2 D i − 1 n ) + | D i − 1 | k i + x − H ¯ i − 1 ( y ) − f − 1 ( y ) < k i 2 / 6 D i n + 2 D i − 1 n + D i − 1 k i + x − | bar(H)_(i-1)(y)-f^(-1)(y)| < (k_(i)^(2)//6)(D_(i)^(n)+2D_(i-1)^(n))+|D_(i-1)|k_(i)+x-\left|\bar{H}_{i-1}(y)-f^{-1}(y)\right|<\left(k_{i}^{2} / 6\right)\left(D_{i}^{n}+2 D_{i-1}^{n}\right)+\left|D_{i-1}\right| k_{i}+x-|H¯i−1(y)−f−1(y)|<(ki2/6)(Din+2Di−1n)+|Di−1|ki+x−
− x i − 1 = ( k i 2 / 6 ) ( 6 ( x i − x i − 1 ) / k i 2 − 6 D i − 1 / k i ) + | D i − 1 | k i + x − x i − 1 = x i − x i − 1 − D i − 1 k i + | D i − 1 | k i + x − x i − 1 = = x i − 2 x i − 1 + x − D i − 1 k i + | D i − 1 | k i . − x i − 1 = k i 2 / 6 6 x i − x i − 1 / k i 2 − 6 D i − 1 / k i + D i − 1 k i + x − x i − 1 = x i − x i − 1 − D i − 1 k i + D i − 1 k i + x − x i − 1 = = x i − 2 x i − 1 + x − D i − 1 k i + D i − 1 k i . {:[-x_(i-1)=(k_(i)^(2)//6)(6(x_(i)-x_(i-1))//k_(i)^(2)-6D_(i-1)//k_(i))+|D_(i-1)|k_(i)],[+x-x_(i-1)=x_(i)-x_(i-1)-D_(i-1)k_(i)+|D_(i-1)|k_(i)+x-x_(i-1)=],[=x_(i)-2x_(i-1)+x-D_(i-1)k_(i)+|D_(i-1)|k_(i).]:}\begin{aligned} & -x_{i-1}=\left(k_{i}^{2} / 6\right)\left(6\left(x_{i}-x_{i-1}\right) / k_{i}^{2}-6 D_{i-1} / k_{i}\right)+\left|D_{i-1}\right| k_{i} \\ & +x-x_{i-1}=x_{i}-x_{i-1}-D_{i-1} k_{i}+\left|D_{i-1}\right| k_{i}+x-x_{i-1}= \\ & =x_{i}-2 x_{i-1}+x-D_{i-1} k_{i}+\left|D_{i-1}\right| k_{i} . \end{aligned}−xi−1=(ki2/6)(6(xi−xi−1)/ki2−6Di−1/ki)+|Di−1|ki+x−xi−1=xi−xi−1−Di−1ki+|Di−1|ki+x−xi−1==xi−2xi−1+x−Di−1ki+|Di−1|ki.
For D i − 1 > 0 D i − 1 > 0 D_(i-1) > 0D_{i-1}>0Di−1>0, we obtain from the inequality (24):
(25) | H ¯ i − 1 ( y ) − f − 1 ( y ) | < x i − 2 x i − 1 + x = δ x H ¯ i − 1 ( y ) − f − 1 ( y ) < x i − 2 x i − 1 + x = δ x | bar(H)_(i-1)(y)-f^(-1)(y)| < x_(i)-2x_(i-1)+x=delta_(x)\left|\bar{H}_{i-1}(y)-f^{-1}(y)\right|<x_{i}-2 x_{i-1}+x=\delta_{x}|H¯i−1(y)−f−1(y)|<xi−2xi−1+x=δx,
while for D i − 1 < 0 D i − 1 < 0 D_(i-1) < 0D_{i-1}<0Di−1<0 we have:
(26) | H ¯ i − 1 ( y ) − p − 1 ( y ) | < δ x − 2 D i − 1 k i H ¯ i − 1 ( y ) − p − 1 ( y ) < δ x − 2 D i − 1 k i | bar(H)_(i-1)(y)-p^(-1)(y)| < delta_(x)-2D_(i-1)k_(i)\left|\bar{H}_{i-1}(y)-p^{-1}(y)\right|<\delta_{x}-2 D_{i-1} k_{i}|H¯i−1(y)−p−1(y)|<δx−2Di−1ki.
If D i − 1 n < 0 D i − 1 n < 0 D_(i-1)^(n) < 0D_{i-1}^{n}<0Di−1n<0, then we obtain from (22):
(27) | H ¯ i − 1 ( y ) − f − 1 ( y ) | < ( D i n / 6 ) k i 2 − ( D i − 1 n / 6 ) k i 2 − ( D i − 1 n / 2 ) k i 2 + H ¯ i − 1 ( y ) − f − 1 ( y ) < D i n / 6 k i 2 − D i − 1 n / 6 k i 2 − D i − 1 n / 2 k i 2 + | bar(H)_(i-1)(y)-f^(-1)(y)| < (D_(i)^(n)//6)k_(i)^(2)-(D_(i-1)^(n)//6)k_(i)^(2)-(D_(i-1)^(n)//2)k_(i)^(2)+\left|\bar{H}_{i-1}(y)-f^{-1}(y)\right|<\left(D_{i}^{n} / 6\right) k_{i}^{2}-\left(D_{i-1}^{n} / 6\right) k_{i}^{2}-\left(D_{i-1}^{n} / 2\right) k_{i}^{2}+|H¯i−1(y)−f−1(y)|<(Din/6)ki2−(Di−1n/6)ki2−(Di−1n/2)ki2+
+ | D i − 1 | k i + x i − x i − 1 = ( D i ′ ′ / 6 ) k i 2 − 4 ( D i − 1 ′ ′ / 6 ) k i 2 + + | D i − 1 | k i + x − x i − 1 . + D i − 1 k i + x i − x i − 1 = D i ′ ′ / 6 k i 2 − 4 D i − 1 ′ ′ / 6 k i 2 + + D i − 1 k i + x − x i − 1 . {:[+|D_(i-1)|k_(i)+x_(i)-x_(i-1)=(D_(i)^('')//6)k_(i)^(2)-4(D_(i-1)^('')//6)k_(i)^(2)+],[+|D_(i-1)|k_(i)+x-x_(i-1).]:}\begin{aligned} & +\left|D_{i-1}\right| k_{i}+x_{i}-x_{i-1}=\left(D_{i}^{\prime \prime} / 6\right) k_{i}^{2}-4\left(D_{i-1}^{\prime \prime} / 6\right) k_{i}^{2}+ \\ & +\left|D_{i-1}\right| k_{i}+x-x_{i-1} . \end{aligned}+|Di−1|ki+xi−xi−1=(Di′′/6)ki2−4(Di−1′′/6)ki2++|Di−1|ki+x−xi−1.
Taking into account the first equality (18), the inequality (27) will acquire the following form:
(28) | H ¯ i − 1 ( y ) − p − 1 ( y ) | < ( k i 2 / 6 ) ( 6 ( x i − x i − 1 ) / k i 2 − 6 D i − 1 / k i − H ¯ i − 1 ( y ) − p − 1 ( y ) < k i 2 / 6 6 x i − x i − 1 / k i 2 − 6 D i − 1 / k i − | bar(H)_(i-1)(y)-p^(-1)(y)| < (k_(i)^(2)//6)(6(x_(i)-x_(i-1))//k_(i)^(2)-6D_(i-1)//k_(i)-:}\left|\bar{H}_{i-1}(y)-p^{-1}(y)\right|<\left(k_{i}^{2} / 6\right)\left(6\left(x_{i}-x_{i-1}\right) / k_{i}^{2}-6 D_{i-1} / k_{i}-\right.|H¯i−1(y)−p−1(y)|<(ki2/6)(6(xi−xi−1)/ki2−6Di−1/ki−
− 2 D i − 1 ′ ′ ) − 4 ( D i − 1 ′ ′ / 6 ) k i 2 + | D i − 1 ′ | k i + x − x i − 1 = x i − x i − 1 − D i − 1 k i − 2 ( D i − 1 ′ ′ / 6 ) k i 2 − 4 ( D i − 1 ′ ′ / 6 ) k i 2 + | D i − 1 ′ | k i + x − − x i − 1 = δ x − D i − 1 ′ ′ k i 2 − D i − 1 k i + | D i − 1 | k i . − 2 D i − 1 ′ ′ − 4 D i − 1 ′ ′ / 6 k i 2 + D i − 1 ′ k i + x − x i − 1 = x i − x i − 1 − D i − 1 k i − 2 D i − 1 ′ ′ / 6 k i 2 − 4 D i − 1 ′ ′ / 6 k i 2 + D i − 1 ′ k i + x − − x i − 1 = δ x − D i − 1 ′ ′ k i 2 − D i − 1 k i + D i − 1 k i . {:[{:-2D_(i-1)^(''))-4(D_(i-1)^('')//6)k_(i)^(2)+|D_(i-1)^(')|k_(i)+x-x_(i-1)=x_(i)-x_(i-1)],[-D_(i-1)k_(i)-2(D_(i-1)^('')//6)k_(i)^(2)-4(D_(i-1)^('')//6)k_(i)^(2)+|D_(i-1)^(')|k_(i)+x-],[-x_(i-1)=delta_(x)-D_(i-1)^('')k_(i)^(2)-D_(i-1)k_(i)+|D_(i-1)|k_(i).]:}\begin{aligned} & \left.-2 D_{i-1}^{\prime \prime}\right)-4\left(D_{i-1}^{\prime \prime} / 6\right) k_{i}^{2}+\left|D_{i-1}^{\prime}\right| k_{i}+x-x_{i-1}=x_{i}-x_{i-1} \\ & -D_{i-1} k_{i}-2\left(D_{i-1}^{\prime \prime} / 6\right) k_{i}^{2}-4\left(D_{i-1}^{\prime \prime} / 6\right) k_{i}^{2}+\left|D_{i-1}^{\prime}\right| k_{i}+x- \\ & -x_{i-1}=\delta_{x}-D_{i-1}^{\prime \prime} k_{i}^{2}-D_{i-1} k_{i}+\left|D_{i-1}\right| k_{i} . \end{aligned}−2Di−1′′)−4(Di−1′′/6)ki2+|Di−1′|ki+x−xi−1=xi−xi−1−Di−1ki−2(Di−1′′/6)ki2−4(Di−1′′/6)ki2+|Di−1′|ki+x−−xi−1=δx−Di−1′′ki2−Di−1ki+|Di−1|ki.
For D i − 1 > 0 D i − 1 > 0 D_(i-1) > 0D_{i-1}>0Di−1>0, the inequality (28) provides the following result:
(29) | H ¯ i − 1 ( y ) − p − 1 ( y ) | < δ x − D i − 1 n k i 2 H ¯ i − 1 ( y ) − p − 1 ( y ) < δ x − D i − 1 n k i 2 | bar(H)_(i-1)(y)-p^(-1)(y)| < delta_(x)-D_(i-1)^(n)k_(i)^(2)\left|\bar{H}_{i-1}(y)-p^{-1}(y)\right|<\delta_{x}-D_{i-1}^{n} k_{i}^{2}|H¯i−1(y)−p−1(y)|<δx−Di−1nki2,
while for D − 1 < 0 D − 1 < 0 D_(-1) < 0D_{-1}<0D−1<0 we obtain from the same inequality (28) the result:
(30) | H ¯ i − 1 ( y ) − f − 1 ( y ) | < δ x − 2 D i − 1 k i − D i − 1 n k i 2 H ¯ i − 1 ( y ) − f − 1 ( y ) < δ x − 2 D i − 1 k i − D i − 1 n k i 2 | bar(H)_(i-1)(y)-f^(-1)(y)| < delta_(x)-2D_(i-1)k_(i)-D_(i-1)^(n)k_(i)^(2)\left|\bar{H}_{i-1}(y)-f^{-1}(y)\right|<\delta_{x}-2 D_{i-1} k_{i}-D_{i-1}^{n} k_{i}^{2}|H¯i−1(y)−f−1(y)|<δx−2Di−1ki−Di−1nki2.
So, theorem 1 is proved.
Theorem 2. In the conditions of theorem 1, if the function f f fff is decreasing, then the following inequalities hold:
(31) | H ¯ i − 1 ( y ) − f − 1 ( y ) | < { α x , for D i ′ ′ < 0 , D i ′ < 0 ; α x − 2 D i ′ ⋅ k i , for D i ′ ′ < 0 , D i ′ > 0 ; α x + D i ′ ′ ⋅ k i 2 , for D i ′ ′ > 0 , D i ′ < 0 ; α x + D i ′ ′ ⋅ k i 2 − 2 D i ⋅ k i , for D i ′ ′ > 0 , D i > 0 , H ¯ i − 1 ( y ) − f − 1 ( y ) < α x ,  for  D i ′ ′ < 0 , D i ′ < 0 ; α x − 2 D i ′ ⋅ k i ,  for  D i ′ ′ < 0 , D i ′ > 0 ; α x + D i ′ ′ ⋅ k i 2 ,  for  D i ′ ′ > 0 , D i ′ < 0 ; α x + D i ′ ′ ⋅ k i 2 − 2 D i ⋅ k i ,  for  D i ′ ′ > 0 , D i > 0 , | bar(H)_(i-1)(y)-f^(-1)(y)| < {[alpha_(x)","" for "D_(i)^('') < 0","D_(i)^(') < 0;],[alpha_(x)-2D_(i)^(')*k_(i)","" for "D_(i)^('') < 0","D_(i)^(') > 0;],[alpha_(x)+D_(i)^('')*k_(i)^(2)","" for "D_(i)^('') > 0","D_(i)^(') < 0;],[alpha_(x)+D_(i)^('')*k_(i)^(2)-2D_(i)*k_(i)","" for "D_(i)^('') > 0","],[D_(i) > 0","]:}\left|\bar{H}_{i-1}(y)-f^{-1}(y)\right|<\left\{\begin{array}{l}\alpha_{x}, \text { for } D_{i}^{\prime \prime}<0, D_{i}^{\prime}<0 ; \\ \alpha_{x}-2 D_{i}^{\prime} \cdot k_{i}, \text { for } D_{i}^{\prime \prime}<0, D_{i}^{\prime}>0 ; \\ \alpha_{x}+D_{i}^{\prime \prime} \cdot k_{i}^{2}, \text { for } D_{i}^{\prime \prime}>0, D_{i}^{\prime}<0 ; \\ \alpha_{x}+D_{i}^{\prime \prime} \cdot k_{i}^{2}-2 D_{i} \cdot k_{i}, \text { for } D_{i}^{\prime \prime}>0, \\ D_{i}>0,\end{array}\right.|H¯i−1(y)−f−1(y)|<{αx, for Di′′<0,Di′<0;αx−2Di′⋅ki, for Di′′<0,Di′>0;αx+Di′′⋅ki2, for Di′′>0,Di′<0;αx+Di′′⋅ki2−2Di⋅ki, for Di′′>0,Di>0,
where:
(32) α x = 2 x i − x − x i − 1 ⩽ 2 h i , i = 1 , n ― . (32) α x = 2 x i − x − x i − 1 ⩽ 2 h i , i = 1 , n ¯ . {:(32)alpha_(x)=2x_(i)-x-x_(i-1) <= 2h_(i)","quad i= bar(1,n).:}\begin{equation*} \alpha_{x}=2 x_{i}-x-x_{i-1} \leqslant 2 h_{i}, \quad i=\overline{1, n} . \tag{32} \end{equation*}(32)αx=2xi−x−xi−1⩽2hi,i=1,n―.
The proof of theorem 2 is similar to that of theorem 1. We notice that:
(33) δ x = x i − x i − 1 + x − x i − 1 ⩽ 2 h i , i = 1 , n ― δ x = x i − x i − 1 + x − x i − 1 ⩽ 2 h i , i = 1 , n ¯ quaddelta_(x)=x_(i)-x_(i-1)+x-x_(i-1) <= 2h_(i),quad i= bar(1,n)\quad \delta_{x}=x_{i}-x_{i-1}+x-x_{i-1} \leqslant 2 h_{i}, \quad i=\overline{1, n}δx=xi−xi−1+x−xi−1⩽2hi,i=1,n―.
Using the formulae (32) and (33), one obtains from the-- orems 1 and 2 the following results:
Theorem 3. In the conditions of theorem 1, the following estimate holds:
(34) ‖ H ¯ − f − 1 ‖ ⩽ max { 2 h i , 2 h i + p 1 , 2 h i + p 2 , 2 h i + p 1 + p 2 } H ¯ − f − 1 ⩽ max 2 h i , 2 h i + p 1 , 2 h i + p 2 , 2 h i + p 1 + p 2 ||( bar(H))-f^(-1)|| <= max{2h_(i),2h_(i)+p_(1),2h_(i)+p_(2),2h_(i)+p_(1)+p_(2)}\left\|\bar{H}-f^{-1}\right\| \leqslant \max \left\{2 h_{i}, 2 h_{i}+p_{1}, 2 h_{i}+p_{2}, 2 h_{i}+p_{1}+p_{2}\right\}‖H¯−f−1‖⩽max{2hi,2hi+p1,2hi+p2,2hi+p1+p2},
1 ⩽ i ⩽ n 1 ⩽ i ⩽ n 1 <= i <= n1 \leqslant i \leqslant n1⩽i⩽n
where:
p 1 = − 2 min 1 ⩽ i ⩽ n D i − 1 k i p 2 = − min 1 ⩽ i ⩽ n D i − 1 ′ ′ k i 2 p 1 = − 2 min 1 ⩽ i ⩽ n   D i − 1 k i p 2 = − min 1 ⩽ i ⩽ n   D i − 1 ′ ′ k i 2 {:[p_(1)=-2min_(1 <= i <= n)D_(i-1)k_(i)],[p_(2)=-min_(1 <= i <= n)D_(i-1)^('')k_(i)^(2)]:}\begin{aligned} & p_{1}=-2 \min _{1 \leqslant i \leqslant n} D_{i-1} k_{i} \\ & p_{2}=-\min _{1 \leqslant i \leqslant n} D_{i-1}^{\prime \prime} k_{i}^{2} \end{aligned}p1=−2min1⩽i⩽nDi−1kip2=−min1⩽i⩽nDi−1′′ki2
Theorem 4. In the conditions of theorem 2, the following estimate holds:
(35) ‖ H ¯ − f − 1 ‖ ⩽ max 1 ⩽ i ⩽ n { 2 h i , 2 h i + p ¯ 1 , 2 h i + p ¯ 2 , 2 h i + p ¯ 1 + p ¯ 2 } H ¯ − f − 1 ⩽ max 1 ⩽ i ⩽ n   2 h i , 2 h i + p ¯ 1 , 2 h i + p ¯ 2 , 2 h i + p ¯ 1 + p ¯ 2 ||( bar(H))-f^(-1)|| <= max_(1 <= i <= n){2h_(i),2h_(i)+ bar(p)_(1),2h_(i)+ bar(p)_(2),2h_(i)+ bar(p)_(1)+ bar(p)_(2)}\left\|\bar{H}-\mathrm{f}^{-1}\right\| \leqslant \max _{1 \leqslant \mathrm{i} \leqslant \mathrm{n}}\left\{2 h_{i}, 2 h_{i}+\bar{p}_{1}, 2 h_{i}+\bar{p}_{2}, 2 h_{i}+\bar{p}_{1}+\bar{p}_{2}\right\}‖H¯−f−1‖⩽max1⩽i⩽n{2hi,2hi+p¯1,2hi+p¯2,2hi+p¯1+p¯2}.
1 ⩽ i ⩽ n 1 ⩽ i ⩽ n 1 <= i <= n1 \leqslant i \leqslant n1⩽i⩽n
  1. NUMERICAL APPLICATION
Based on the theory of Section 3, a calculation program written in BASIC (accuracy: 11 significant digits) was elaborated. This program allows to determine the root of an
equation f ( x ) = 0 f ( x ) = 0 f(x)=0f(x)=0f(x)=0, where the analytical expression of f ( x ) f ( x ) f(x)f(x)f(x) is known (case B, Section 1).
As input data one considers: n = n = n=n=n= the number of knots, and the knots ' 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. The values y i = f ( x i ) , i = 0 , n ― y i = f x i , i = 0 , n ¯ y_(i)=f(x_(i)),i= bar(0,n)y_{i}=f\left(x_{i}\right), i=\overline{0, n}yi=f(xi),i=0,n―, and those of the first and second order derivatives of these ones are computed by means of the numerical-type functions FWA(X), FNB(X), FNC(X), defined by the programmer for f ( x ) , f ′ ( x ) f ( x ) , f ′ ( x ) f(x),f^(')(x)f(x), f^{\prime}(x)f(x),f′(x) and f ′ ′ ( x ) f ′ ′ ( x ) f^('')(x)f^{\prime \prime}(x)f′′(x), respectively.
The values f ( x i ) , i = 0 , n ― f x i , i = 0 , n ¯ f(x_(i)),i= bar(0,n)f\left(x_{i}\right), i=\overline{0, n}f(xi),i=0,n―, are increasingly ordered (the lines 230 − 310 230 − 310 230-310230-310230−310 of the program); after that, the values D 0 ′ = l / f ′ ( x 0 ) D 0 ′ = l / f ′ x 0 D_(0)^(')=l//f^(')(x_(0))D_{0}^{\prime}=l / f^{\prime}\left(x_{0}\right)D0′=l/f′(x0) and D 0 ′ ′ = − f ′ ′ ( x 0 ) / ( f ′ ( x 0 ) ) 3 D 0 ′ ′ = − f ′ ′ x 0 / f ′ x 0 3 D_(0)^('')=-f^('')(x_(0))//(f^(')(x_(0)))^(3)D_{0}^{\prime \prime}=-f^{\prime \prime}\left(x_{0}\right) /\left(f^{\prime}\left(x_{0}\right)\right)^{3}D0′′=−f′′(x0)/(f′(x0))3 are calculated. Then one determines D i ′ D i ′ D_(i)^(')D_{i}^{\prime}Di′ and D i ′ ′ , i = l , n ― D i ′ ′ , i = l , n ¯ D_(i)^(''),i= bar(l,n)D_{i}^{\prime \prime}, i=\overline{l, n}Di′′,i=l,n―, with the recurrent relations (18).
After these calculations, the interval [ y i , y i + 1 ] , i == 0 , n − 1 ― y i , y i + 1 , i == 0 , n − 1 ¯ [y_(i),y_(i+1)],i== bar(0,n-1)\left[y_{i}, y_{i+1}\right], i= =\overline{0, n-1}[yi,yi+1],i==0,n−1―, in which the root is lying, is tested, and then, by means of the formula (19), one estimates the value of the root x ¯ ≅ H ¯ ( 0 ) x ¯ ≅ H ¯ ( 0 ) bar(x)~= bar(H)(0)\bar{x} \cong \bar{H}(0)x¯≅H¯(0) and f ( x ¯ ) f ( x ¯ ) f( bar(x))f(\bar{x})f(x¯).
The program was especially applied to the case when only three knots, 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,x2, are given (see Table 1), the root lying into one of the intervals ( x 0 , x 1 ) , ( x 1 , x 2 ) x 0 , x 1 , x 1 , x 2 (x_(0),x_(1)),(x_(1),x_(2))\left(x_{0}, x_{1}\right),\left(x_{1}, x_{2}\right)(x0,x1),(x1,x2).
In order to determine as accurately as possible the value of the root, the program contains a sequence which restricts the interval in which the knots are lying, as follows: if f ( x 0 ) ⋅ f ( x ¯ ) > 0 f x 0 ⋅ f ( x ¯ ) > 0 f(x_(0))*f( bar(x)) > 0f\left(x_{0}\right) \cdot f(\bar{x})>0f(x0)⋅f(x¯)>0, then one removes x 0 x 0 x_(0)x_{0}x0, while in the opposite case x n x n x_(n)x_{n}xn is removed; in both cases the value x ¯ x ¯ bar(x)\bar{x}x¯ found at the previous step is inserted instead of the removed knot. This successive removal process stops when | f ( x ¯ ) | | f ( x ¯ ) | |f( bar(x))||f(\bar{x})||f(x¯)|
becomes smaller than 10 − 10 10 − 10 10^(-10)10^{-10}10−10 (this restriction being imposed by the restriction of the microcomputer we used).
Table 1 lists the results obtained by means of this program in the case of five equations. The listing of the program is given further down.
Table 1.
Equation f ( x ) = 0 f ( x ) = 0 f(x)=0f(x)=0f(x)=0 Interval in which the root is lying Knots used at start Root ( x ¯ x ¯ bar(x)\bar{x}x¯ ) at every step | f ( x ¯ ) | | f ( x ¯ ) | |f( bar(x))||f(\bar{x})||f(x¯)|
4 x 3 + 3 x 2 + + 3 x − 1 = 0 4 x 3 + 3 x 2 + + 3 x − 1 = 0 4x^(3)+3x^(2)++3x-1=04 x^{3}+3 x^{2}+ +3 \mathrm{x}-1=04x3+3x2++3x−1=0 (0.2, 0.4) 0.2, 0.3, 0.4
0.2499800875
0.2500000081
0.2500000000
0.2499800875 0.2500000081 0.2500000000| 0.2499800875 | | :--- | | 0.2500000081 | | 0.2500000000 |
1.05E-04
4.23E-08
0.00E+00
1.05E-04 4.23E-08 0.00E+00| 1.05E-04 | | :--- | | 4.23E-08 | | 0.00E+00 |
x 2 − 10 ln ⁡ x x 2 − 10 ln ⁡ x x^(2)-10 ln xx^{2}-10 \ln xx2−10ln⁡x
− 3 = 0 − 3 = 0 -3=0-3=0−3=0
x^(2)-10 ln x -3=0| $x^{2}-10 \ln x$ | | :--- | | $-3=0$ |
 ( 4 , 6 ) ( 4 , 6 ) (4,6)(4,6)(4,6) 4,5,6
4.1512952567
4.1514567631
4.1514567195
4.1512952567 4.1514567631 4.1514567195| 4.1512952567 | | :--- | | 4.1514567631 | | 4.1514567195 |
9.52E-04
2.57E-07
0.00E+00
9.52E-04 2.57E-07 0.00E+00| 9.52E-04 | | :--- | | 2.57E-07 | | 0.00E+00 |
ln ⁡ x − 4 + + x 2 = 0 ln ⁡ x − 4 + + x 2 = 0 ln x-4++x^(2)=0\ln x-4+ +x^{2}=0ln⁡x−4++x2=0 ( 1 , 3 ) ( 1 , 3 ) (1,3)(1,3)(1,3) 1,2,3
1.8448743194
1.8412032474
1.8411000557
1.8410971431
1.8410970608
1.8410970585
1.8410970585
1.8448743194 1.8412032474 1.8411000557 1.8410971431 1.8410970608 1.8410970585 1.8410970585| 1.8448743194 | | :--- | | 1.8412032474 | | 1.8411000557 | | 1.8410971431 | | 1.8410970608 | | 1.8410970585 | | 1.8410970585 |
1.60E-02
4.49E-04
1.27E-05
3.58E-07
1.01E-08
2.84E-10
7.28E-12
1.60E-02 4.49E-04 1.27E-05 3.58E-07 1.01E-08 2.84E-10 7.28E-12| 1.60E-02 | | :--- | | 4.49E-04 | | 1.27E-05 | | 3.58E-07 | | 1.01E-08 | | 2.84E-10 | | 7.28E-12 |
x - ( 1 / 10 1 / 10 1//101 / 101/10 ).
⋅ sin ⁡ x − 1 == 0 ⋅ sin ⁡ x − 1 == 0 *sin x-1==0\cdot \sin x-1= =0⋅sin⁡x−1==0
x - ( 1//10 ). *sin x-1==0| x - ( $1 / 10$ ). | | :--- | | $\cdot \sin x-1= =0$ |
 (0.5, 2) 0.5,1.5,2
1.0890477291
1.0885982411
1.0885977529
1.0885977524
1.0890477291 1.0885982411 1.0885977529 1.0885977524| 1.0890477291 | | :--- | | 1.0885982411 | | 1.0885977529 | | 1.0885977524 |
4.29E-04
4.66E-07
5.06E-10
1.82E-12
4.29E-04 4.66E-07 5.06E-10 1.82E-12| 4.29E-04 | | :--- | | 4.66E-07 | | 5.06E-10 | | 1.82E-12 |
x − 0.2 x − 0.2 x-0.2x-0.2x−0.2.
- sin ⁡ x − sin ⁡ x − sin x-\sin x-sin⁡x−
− 5 = 0 − 5 = 0 -5=0-5=0−5=0
x-0.2. - sin x- -5=0| $x-0.2$. | | :--- | | - $\sin x-$ | | $-5=0$ |
 (4.5, 6.5) 4.5, 5.5, 6.5
4.8006851220
4.8007808072
4.8007808029
4.8006851220 4.8007808072 4.8007808029| 4.8006851220 | | :--- | | 4.8007808072 | | 4.8007808029 |
8.05E-05
4.28E-09
7.28E-09
8.05E-05 4.28E-09 7.28E-09| 8.05E-05 | | :--- | | 4.28E-09 | | 7.28E-09 |
Equation f(x)=0 Interval in which the root is lying Knots used at start Root ( bar(x) ) at every step |f( bar(x))| 4x^(3)+3x^(2)++3x-1=0 (0.2, 0.4) 0.2, 0.3, 0.4 "0.2499800875 0.2500000081 0.2500000000" "1.05E-04 4.23E-08 0.00E+00" "x^(2)-10 ln x -3=0" (4,6) 4,5,6 "4.1512952567 4.1514567631 4.1514567195" "9.52E-04 2.57E-07 0.00E+00" ln x-4++x^(2)=0 (1,3) 1,2,3 "1.8448743194 1.8412032474 1.8411000557 1.8410971431 1.8410970608 1.8410970585 1.8410970585" "1.60E-02 4.49E-04 1.27E-05 3.58E-07 1.01E-08 2.84E-10 7.28E-12" "x - ( 1//10 ). *sin x-1==0" (0.5, 2) 0.5,1.5,2 "1.0890477291 1.0885982411 1.0885977529 1.0885977524" "4.29E-04 4.66E-07 5.06E-10 1.82E-12" "x-0.2. - sin x- -5=0" (4.5, 6.5) 4.5, 5.5, 6.5 "4.8006851220 4.8007808072 4.8007808029" "8.05E-05 4.28E-09 7.28E-09"| Equation $f(x)=0$ | Interval in which the root is lying | Knots used at start | Root ( $\bar{x}$ ) at every step | $\|f(\bar{x})\|$ | | :--- | :--- | :--- | :--- | :--- | | $4 x^{3}+3 x^{2}+ +3 \mathrm{x}-1=0$ | (0.2, 0.4) | 0.2, 0.3, 0.4 | 0.2499800875 <br> 0.2500000081 <br> 0.2500000000 | 1.05E-04 <br> 4.23E-08 <br> 0.00E+00 | | $x^{2}-10 \ln x$ <br> $-3=0$ | $(4,6)$ | 4,5,6 | 4.1512952567 <br> 4.1514567631 <br> 4.1514567195 | 9.52E-04 <br> 2.57E-07 <br> 0.00E+00 | | $\ln x-4+ +x^{2}=0$ | $(1,3)$ | 1,2,3 | 1.8448743194 <br> 1.8412032474 <br> 1.8411000557 <br> 1.8410971431 <br> 1.8410970608 <br> 1.8410970585 <br> 1.8410970585 | 1.60E-02 <br> 4.49E-04 <br> 1.27E-05 <br> 3.58E-07 <br> 1.01E-08 <br> 2.84E-10 <br> 7.28E-12 | | | | | | | | | | | | | | | | | | | | | | | | | | x - ( $1 / 10$ ). <br> $\cdot \sin x-1= =0$ | (0.5, 2) | 0.5,1.5,2 | 1.0890477291 <br> 1.0885982411 <br> 1.0885977529 <br> 1.0885977524 | 4.29E-04 <br> 4.66E-07 <br> 5.06E-10 <br> 1.82E-12 | | $x-0.2$. <br> - $\sin x-$ <br> $-5=0$ | (4.5, 6.5) | 4.5, 5.5, 6.5 | 4.8006851220 <br> 4.8007808072 <br> 4.8007808029 | 8.05E-05 <br> 4.28E-09 <br> 7.28E-09 |
100 REM Inverse interpolating cubic spline for eq. solving.
110. DEF FHA ( x ) = 4 ( x ) = 4 (x)=4(x)=4(x)=4. W x ∧ 3 + 3.7 x ∧ 2 + 3 . ∗ x − 1 x ∧ 3 + 3.7 x ∧ 2 + 3 . ∗ x − 1 x^3+3.7x^2+3.**x-1x^{\wedge} 3+3.7 x^{\wedge} 2+3 . * x-1x∧3+3.7x∧2+3.∗x−1.
130 DEF FNB ( X ) = 12.3 X 4 2 + 6.3 X + 3 ( X ) = 12.3 X 4 2 + 6.3 X + 3 (X)=12.3X^(4)2+6.3 X+3(X)=12.3 X^{4} 2+6.3 X+3(X)=12.3X42+6.3X+3.
30 DE FNC ( X ) = 24 . ; X + 6 ( X ) = 24 . ; X + 6 (X)=24.;X+6(X)=24 . ; X+6(X)=24.;X+6.
150 READ N LPEINTSPC (10)" Input Data" ILPRINT
160 DIM
170 F 0 X ( N + 1 ) , F ( N + 1 ) , BP ( N + 1 ) , BS ( N + 1 ) 170 F 0 X ( N + 1 ) , F ( N + 1 ) , BP ( N + 1 ) , BS ( N + 1 ) 170F0X(N+1),F(N+1),BP(N+1),BS(N+1)170 \mathrm{~F} 0 \mathrm{X}(\mathrm{N}+1), \mathrm{F}(\mathrm{N}+1), \mathrm{BP}(\mathrm{N}+1), \mathrm{BS}(\mathrm{N}+1)170 F0X(N+1),F(N+1),BP(N+1),BS(N+1)
180 FOR I=0 T0 H:READ X(I): NEXT I
180 FOR I = 0 I = 0 I=0I=0I=0 TO N:AA=FNA(X(I)):F(I)=AA

NU (耓) =+訳."###"; I, F (I)
210 NEXT 1
220 LPRINT:LPRINT. SPC (4) "H(0)"; SPC(12)"f(H(0))"
230 REM Increasing arrangement of the values y ( i ) y ( i ) y(i)y(i)y(i).
240 IV=0
250 FOR I = 0 I = 0 I=0I=0I=0 TO N-1
260 IF F ( I ) <= F ( I + 1 ) F ( I ) <= F ( I + 1 ) F(I)<=F(I+1)F(I)<=F(I+1)F(I)<=F(I+1) THEN GOTO 300
280 T = X ( I ) : F ( I ) = F ( I + 1 ) : F ( I + 1 ) = T T 280 T = X ( I ) : F ( I ) = F ( I + 1 ) : F ( I + 1 ) = T T 280 T=X(I):F(I)=F(I+1):F(I+1)=TT280 T=X(I): F(I)=F(I+1): F(I+1)=T T280T=X(I):F(I)=F(I+1):F(I+1)=TT
290 T = X ( I ) : X ( I ) = X ( I + 1 ) : X ( I + 1 ) = TT 290 T = X ( I ) : X ( I ) = X ( I + 1 ) : X ( I + 1 ) = TT 290T=X(I):X(I)=X(I+1):X(I+1)=TT290 \mathrm{~T}=\mathrm{X}(\mathrm{I}): \mathrm{X}(\mathrm{I})=\mathrm{X}(\mathrm{I}+1): \mathrm{X}(\mathrm{I}+1)=\mathrm{TT}290 T=X(I):X(I)=X(I+1):X(I+1)=TT
300 NEXT I
310 IF N U <> 0 N U <> 0 NU<>0N U<>0NU<>0 THEN GOTO 240
320 REM Values of DS(i) and DP (I).
330 D P ( 0 ) = 1 . / F N B ( X ( 0 ) ) 330 D P ( 0 ) = 1 . / F N B ( X ( 0 ) ) 330 DP(0)=1.//FNB(X(0))330 D P(0)=1 . / F N B(X(0))330DP(0)=1./FNB(X(0))
340 DS ( 0 ) = − F = − F =-F=-F=−F NC ( X ( 0 ) X ( 0 ) X(0)X(0)X(0) )/ (FNB ( X ( 0 ) X ( 0 ) X(0)X(0)X(0) )^3)
350 FOR I = 0 I = 0 I=0I=0I=0 TO N − 1 N − 1 N-1N-1N−1
360 K = F ( I + 1 ) − F ( I ) : H = X ( I + 1 ) − X ( I ) 360 K = F ( I + 1 ) − F ( I ) : H = X ( I + 1 ) − X ( I ) 360K=F(I+1)-F(I):H=X(I+1)-X(I)360 \mathrm{~K}=\mathrm{F}(I+1)-\mathrm{F}(I): \mathrm{H}=X(I+1)-X(I)360 K=F(I+1)−F(I):H=X(I+1)−X(I)
370 DS ( I + 1 ) = 6 370 DS ( I + 1 ) = 6 370DS(I+1)=6370 \mathrm{DS}(\mathrm{I}+1)=6370DS(I+1)=6. ∗ ∗ ***∗ H/ ( K ∗ K ) − 2 ( K ∗ K ) − 2 (K**K)-2(\mathrm{K} * \mathrm{~K})-2(K∗ K)−2. ∗ ∗ ***∗ DS ( I ) − 6 ( I ) − 6 (I)-6(\mathrm{I})-6(I)−6. ∗ ∗ ***∗ DP ( I ) / K ( I ) / K (I)//K(\mathrm{I}) / \mathrm{K}(I)/K
380 DP ( I + 1 ) = 3 ( I + 1 ) = 3 (I+1)=3(I+1)=3(I+1)=3, H/K-KDS(I)/2.-2. *DP (I)
390 NEXT I
400 FOR I = 0 I = 0 I=0I=0I=0 TO N − 1 N − 1 N-1N-1N−1
410 IF F ( I ) ≈ F ( I + 1 ) < 0 F ( I ) ≈ F ( I + 1 ) < 0 F(I)~~F(I+1) < 0F(I) \approx F(I+1)<0F(I)≈F(I+1)<0. THEN GOTO 430
420 NEXT I
430 Y = 0 . 430 Y = 0 . 430quad Y=0.quad430 \quad Y=0 . \quad430Y=0.
440 AA=Y-F(I)
450 BB = ( D S ( I + 1 ) − D S ( I ) ) ∗ A A ∧ 3 / ( F ( I + 1 ) − F ( I ) ) / 6 450 BB = ( D S ( I + 1 ) − D S ( I ) ) ∗ A A ∧ 3 / ( F ( I + 1 ) − F ( I ) ) / 6 450BB=(DS(I+1)-DS(I))**AA^3//(F(I+1)-F(I))//6450 \mathrm{BB}=(D S(I+1)-D S(I)) * A A^{\wedge} 3 /(F(I+1)-F(I)) / 6450BB=(DS(I+1)−DS(I))∗AA∧3/(F(I+1)−F(I))/6.
470 (I)
470 HY = BB + CC 470 HY = BB + CC 470HY=BB+CC470 \mathrm{HY}=\mathrm{BB}+\mathrm{CC}470HY=BB+CC
490 LPRINT USING " ∗ ∗ ^(**){ }^{*}∗ ####### " ; HY ;
500 IF NN (HYN)
510 IF F (O) 540
520 FOR I = 0 I = 0 I=0I=0I=0 TO (O). THEN X ( N ) = H Y X ( N ) = H Y X(N)=HYX(N)=H YX(N)=HY ELSE X ( O ) = H Y X ( O ) = H Y X(O)=HYX(O)=H YX(O)=HY
530 GOTB 230 N:AA=FNA(X(I)):F(I)=AA: NEXT I
S40 LPRINT:LP
550 LPRINT USING USING" ROOt: H ( 0 ) = + # H ( 0 ) = + # H(0)=+#\mathrm{H}(0)=+\#H(0)=+#. ########"; HY
560 STOP
580 BATA 0.2 , 0.3 , 0.4 0.2 , 0.3 , 0.4 0.2,0.3,0.40.2,0.3,0.40.2,0.3,0.4
Note. We mention that a program was elaborated also for the inverse interpolating spline presented in Section 2 , but in many concrete cases, when the knots got near to the root (in the meaning of the above removal), undeterminacies appeared because of the small values of the denominator in formula (9). That is why the inverse interpolating spline proposed in Section 3 (which has not created such problems) has been chosen.

REFERENCES

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1986

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