A method for obtaining iterative formulas of higher order for roots of equations

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A. Muresan
Tiberiu Popoviciu Institute of Numerical Analysis, Romanian Academy

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A. C. Mureṣan, A method for obtaining iterative formulas of higher order for roots of equations, Rev. Anal. Numer. Theor. Approx., 26 (1997) 1-2, 131-135.

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Revue d’Analyse Numérique et de Théorie de l’Approximation

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Editura Academiei Române

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https://ictp.acad.ro/jnaat/journal/article/view/1999-vol28-no2-art3

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[1] B. Jovanović, A method for obtaining iterative formulas of higher order, Mat. Vesnik. 24, 9 (1972), pp. 365-369.
[2] I. Lazăr and A.C. Mureşan, Generalized trnasformations on rations of Fibonacci and Lucas numbers. Rev. Anal. Numér. Théorie Aprroximation 24, 1-2 (1995), pp. 169-179.
[3] A. Ralston, A First Course in Numerical Analysis, McGraw-Hill, Inc., 1965.
[4] D. M. Simeounović, On a process for obtaining iterative formulas of higher order for roots of equations. Rev. Anal. Numér. Théorie Approximation 24, 1-2 (1995), pp. 225-229.

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A METHOD FOR OBTAINING ITERATIVE FORMULAS OF HIGHER ORDER FOR ROOTS OF EQUATIONS

ADRIAN MUREŞAN

1. INTRODUCTION

Formulas of the class which use information at only one point are naturally called one-point formulas. We shall consider only stationary one-point formulas which have the form
(1) x n + 1 = F ( x n ) , (1) x n + 1 = F x n , {:(1)x_(n+1)=F(x_(n))",":}\begin{equation*} x_{n+1}=F\left(x_{n}\right), \tag{1} \end{equation*}(1)xn+1=F(xn),
with α = F ( α ) α = F ( α ) alpha=F(alpha)\alpha=F(\alpha)α=F(α), if the method converges, where α α alpha\alphaα is the root of the real or complex equation f ( x ) = 0 f ( x ) = 0 f(x)=0f(x)=0f(x)=0.
For the iterative method (1) which converges to α α alpha\alphaα, we say it is of order k k kkk if
(2)
| x n + 1 − α | = 0 ( | x n − α | k ) , n → ∞ . x n + 1 − α = 0 x n − α k , n → ∞ . |x_(n+1)-alpha|=0(|x_(n)-alpha|^(k)),n rarr oo.\left|x_{n+1}-\alpha\right|=0\left(\left|x_{n}-\alpha\right|^{k}\right), n \rightarrow \infty .|xn+1−α|=0(|xn−α|k),n→∞.
If the function F ( x ) F ( x ) F(x)F(x)F(x) is k k kkk-times differentiable in a neighborhood of the limit point x = α x = α x=alphax=\alphax=α, then [3] the iterative method (1) is of order k k kkk if and only if
(3) F ( α ) = α , F ′ ( α ) = F ′ ′ ( α ) = … = F ( k − 1 ) ( α ) = 0 , F ( k ) ( α ) ≠ 0 . (3) F ( α ) = α , F ′ ( α ) = F ′ ′ ( α ) = … = F ( k − 1 ) ( α ) = 0 , F ( k ) ( α ) ≠ 0 . {:(3)F(alpha)=alpha","F^(')(alpha)=F^('')(alpha)=dots=F^((k-1))(alpha)=0","F^((k))(alpha)!=0.:}\begin{equation*} F(\alpha)=\alpha, F^{\prime}(\alpha)=F^{\prime \prime}(\alpha)=\ldots=F^{(k-1)}(\alpha)=0, F^{(k)}(\alpha) \neq 0 . \tag{3} \end{equation*}(3)F(α)=α,F′(α)=F′′(α)=…=F(k−1)(α)=0,F(k)(α)≠0.
In Section 2 we give some results which represent the answers of the following question: If we have a method of order k k kkk, how can we obtain from it a method of order k + 1 k + 1 k+1k+1k+1 ?
In Section 3, a family of iterative functions for finding root α α alpha\alphaα is derived. The family includes the functions presented in Section 2.

2. HIGHER ORDER METHODS

THEOREM 1 [4]. Let (1) be an iterative method of order k ( ≥ 2 ) k ( ≥ 2 ) k( >= 2)k(\geq 2)k(≥2) and let the function F ( x ) F ( x ) F(x)F(x)F(x) be k + 1 k + 1 k+1k+1k+1-times differentiable in a neighborhood of the limit point x = α x = α x=alphax=\alphax=α. Then
(4) x n + 1 = F ( x n ) − 1 k F ′ ( x n ) ( x n − F ( x n ) ) = = x n − ( x n − F ( x n ) ) ( 1 + 1 k F ′ ( x n ) ) , n = 0 , 1 , 2 , … (4) x n + 1 = F x n − 1 k F ′ x n x n − F x n = = x n − x n − F x n 1 + 1 k F ′ x n , n = 0 , 1 , 2 , … {:[(4)x_(n+1)=F(x_(n))-(1)/(k)F^(')(x_(n))(x_(n)-F(x_(n)))=],[=x_(n)-(x_(n)-F(x_(n)))(1+(1)/(k)F^(')(x_(n)))","n=0","1","2","dots]:}\begin{gather*} x_{n+1}=F\left(x_{n}\right)-\frac{1}{k} F^{\prime}\left(x_{n}\right)\left(x_{n}-F\left(x_{n}\right)\right)= \tag{4}\\ =x_{n}-\left(x_{n}-F\left(x_{n}\right)\right)\left(1+\frac{1}{k} F^{\prime}\left(x_{n}\right)\right), n=0,1,2, \ldots \end{gather*}(4)xn+1=F(xn)−1kF′(xn)(xn−F(xn))==xn−(xn−F(xn))(1+1kF′(xn)),n=0,1,2,…
is an iterative method of order at least k + 1 k + 1 k+1k+1k+1.
THEOREM 2 [1]. Let (1) be an iterative method of order k k kkk. Let the function F ( x ) F ( x ) F(x)F(x)F(x) be k + 1 k + 1 k+1k+1k+1-times differentiable in a neighborhood of the limit point x = α x = α x=alphax=\alphax=α and let F ′ ( α ) ≠ 0 F ′ ( α ) ≠ 0 F^(')(alpha)!=0F^{\prime}(\alpha) \neq 0F′(α)≠0. Then
(5) x n + 1 = F ( x n ) − 1 k F ′ ( x n ) x n 1 − 1 k F ′ ( x n ) = x n − x n − F ( x n ) 1 − 1 k F ′ ( x n ) , n = 0 , 1 , 2 , … (5) x n + 1 = F x n − 1 k F ′ x n x n 1 − 1 k F ′ x n = x n − x n − F x n 1 − 1 k F ′ x n , n = 0 , 1 , 2 , … {:(5)x_(n+1)=(F(x_(n))-(1)/(k)F^(')(x_(n))x_(n))/(1-(1)/(k)F^(')(x_(n)))=x_(n)-(x_(n)-F(x_(n)))/(1-(1)/(k)F^(')(x_(n)))","n=0","1","2","dots:}\begin{equation*} x_{n+1}=\frac{F\left(x_{n}\right)-\frac{1}{k} F^{\prime}\left(x_{n}\right) x_{n}}{1-\frac{1}{k} F^{\prime}\left(x_{n}\right)}=x_{n}-\frac{x_{n}-F\left(x_{n}\right)}{1-\frac{1}{k} F^{\prime}\left(x_{n}\right)}, n=0,1,2, \ldots \tag{5} \end{equation*}(5)xn+1=F(xn)−1kF′(xn)xn1−1kF′(xn)=xn−xn−F(xn)1−1kF′(xn),n=0,1,2,…
is an iterative method of order at least k + 1 k + 1 k+1k+1k+1.
Theorem 3 [4]. Let (1) be an iterative method of order k k kkk. Let the function F ( x ) F ( x ) F(x)F(x)F(x) be k + 1 k + 1 k+1k+1k+1-times differentiable in a neighborhood of the limit point x = α x = α x=alphax=\alphax=α and let F ′ ( α ) ≠ 1 F ′ ( α ) ≠ 1 F^(')(alpha)!=1F^{\prime}(\alpha) \neq 1F′(α)≠1. Then
(6) x n + 1 = F ( x n ) − 1 k F ′ ( x n ) ( x n − F ( x n ) 1 − F ′ ( x n ) ) (6) x n + 1 = F x n − 1 k F ′ x n x n − F x n 1 − F ′ x n {:(6)x_(n+1)=F(x_(n))-(1)/(k)F^(')(x_(n))((x_(n)-F(x_(n)))/(1-F^(')(x_(n)))):}\begin{equation*} x_{n+1}=F\left(x_{n}\right)-\frac{1}{k} F^{\prime}\left(x_{n}\right)\left(\frac{x_{n}-F\left(x_{n}\right)}{1-F^{\prime}\left(x_{n}\right)}\right) \tag{6} \end{equation*}(6)xn+1=F(xn)−1kF′(xn)(xn−F(xn)1−F′(xn))
that is,
(7) x n + 1 = x n − ( 1 + 1 k ( F ′ ( x n ) 1 − F ′ ( x n ) ) ) ( x n − F ( x n ) ) , n = 0 , 1 , 2 , … (7) x n + 1 = x n − 1 + 1 k F ′ x n 1 − F ′ x n x n − F x n , n = 0 , 1 , 2 , … {:(7)x_(n+1)=x_(n)-(1+(1)/(k)((F^(')(x_(n)))/(1-F^(')(x_(n)))))(x_(n)-F(x_(n)))","n=0","1","2","dots:}\begin{equation*} x_{n+1}=x_{n}-\left(1+\frac{1}{k}\left(\frac{F^{\prime}\left(x_{n}\right)}{1-F^{\prime}\left(x_{n}\right)}\right)\right)\left(x_{n}-F\left(x_{n}\right)\right), n=0,1,2, \ldots \tag{7} \end{equation*}(7)xn+1=xn−(1+1k(F′(xn)1−F′(xn)))(xn−F(xn)),n=0,1,2,…
is an iterative method of order at least k + 1 k + 1 k+1k+1k+1.
Remark 1. 1. If(1)represents Newton's method for finding simple roots of the equation f ( x ) = 0 f ( x ) = 0 f(x)=0f(x)=0f(x)=0, namely,
(8) x n + 1 = x n − f ( x n ) f ′ ( x n ) , n = 0 , 1 , 2 , … , (8) x n + 1 = x n − f x n f ′ x n , n = 0 , 1 , 2 , … , {:(8)x_(n+1)=x_(n)-(f(x_(n)))/(f^(')(x_(n)))","n=0","1","2","dots",":}\begin{equation*} x_{n+1}=x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)}, n=0,1,2, \ldots, \tag{8} \end{equation*}(8)xn+1=xn−f(xn)f′(xn),n=0,1,2,…,
which means that
F ( x n ) = x n − f ( x n ) f ′ ( x n ) , F x n = x n − f x n f ′ x n , F(x_(n))=x_(n)-(f(x_(n)))/(f^(')(x_(n))),F\left(x_{n}\right)=x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)},F(xn)=xn−f(xn)f′(xn),
then from (4), (5) and (6) we obtain the following methods, respectively:
(9) x n + 1 = x n − f ( x n ) f ′ ( x n ) ⋅ 2 ( f ′ ( x n ) ) 2 + f ( x n ) f ′ ′ ( x n ) 2 ( f ′ ( x n ) ) 2 (9) x n + 1 = x n − f x n f ′ x n ⋅ 2 f ′ x n 2 + f x n f ′ ′ x n 2 f ′ x n 2 {:(9)x_(n+1)=x_(n)-(f(x_(n)))/(f^(')(x_(n)))*(2(f^(')(x_(n)))^(2)+f(x_(n))f^('')(x_(n)))/(2(f^(')(x_(n)))^(2)):}\begin{equation*} x_{n+1}=x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)} \cdot \frac{2\left(f^{\prime}\left(x_{n}\right)\right)^{2}+f\left(x_{n}\right) f^{\prime \prime}\left(x_{n}\right)}{2\left(f^{\prime}\left(x_{n}\right)\right)^{2}} \tag{9} \end{equation*}(9)xn+1=xn−f(xn)f′(xn)⋅2(f′(xn))2+f(xn)f′′(xn)2(f′(xn))2
which is known as Chebyshev's iterative method,
(10) x n + 1 = x n − f ( x n ) f ′ ( x n ) ⋅ 2 ( f ′ ( x n ) ) 2 2 ( f ′ ( x n ) ) 2 − f ( x n ) f ′ ′ ( x n ) , (10) x n + 1 = x n − f x n f ′ x n ⋅ 2 f ′ x n 2 2 f ′ x n 2 − f x n f ′ ′ x n , {:(10)x_(n+1)=x_(n)-(f(x_(n)))/(f^(')(x_(n)))*(2(f^(')(x_(n)))^(2))/(2(f^(')(x_(n)))^(2)-f(x_(n))f^('')(x_(n)))",":}\begin{equation*} x_{n+1}=x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)} \cdot \frac{2\left(f^{\prime}\left(x_{n}\right)\right)^{2}}{2\left(f^{\prime}\left(x_{n}\right)\right)^{2}-f\left(x_{n}\right) f^{\prime \prime}\left(x_{n}\right)}, \tag{10} \end{equation*}(10)xn+1=xn−f(xn)f′(xn)⋅2(f′(xn))22(f′(xn))2−f(xn)f′′(xn),
which is known as Halley's method, and
(11) x n + 1 = x n − f ( x n ) f ′ ( x n ) ⋅ 2 ( f ′ ( x n ) ) 2 − f ( x n ) f ′ ′ ( x n ) 2 ( f ′ ( x n ) ) 2 − 2 f ( x n ) f ′ ′ ( x n ) , (11) x n + 1 = x n − f x n f ′ x n ⋅ 2 f ′ x n 2 − f x n f ′ ′ x n 2 f ′ x n 2 − 2 f x n f ′ ′ x n , {:(11)x_(n+1)=x_(n)-(f(x_(n)))/(f^(')(x_(n)))*(2(f^(')(x_(n)))^(2)-f(x_(n))f^('')(x_(n)))/(2(f^(')(x_(n)))^(2)-2f(x_(n))f^('')(x_(n)))",":}\begin{equation*} x_{n+1}=x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)} \cdot \frac{2\left(f^{\prime}\left(x_{n}\right)\right)^{2}-f\left(x_{n}\right) f^{\prime \prime}\left(x_{n}\right)}{2\left(f^{\prime}\left(x_{n}\right)\right)^{2}-2 f\left(x_{n}\right) f^{\prime \prime}\left(x_{n}\right)}, \tag{11} \end{equation*}(11)xn+1=xn−f(xn)f′(xn)⋅2(f′(xn))2−f(xn)f′′(xn)2(f′(xn))2−2f(xn)f′′(xn),
( n = 0 , 1 , 2 , … ) ( n = 0 , 1 , 2 , … ) (n=0,1,2,dots)(n=0,1,2, \ldots)(n=0,1,2,…).
The order of these methods is at least 3 but, since they do not involve derivatives of f f fff higher than the second order, their order of convergence cannot exceed 3 (see [3]).
2. In [2] it is presented a family of transformations
(12) T m ( x ) = x − f ( x ) f ′ ( x ) ⋅ ∑ k = 0 m a k [ f ′ ( x ) ] 2 m − 2 k [ f ( x ) ] k [ f ′ ′ ( x ) ] k ∑ k = 0 m b k [ f ′ ( x ) ] 2 m − 2 k [ f ( x ) ] k [ f ′ ′ ( x ) ] k (12) T m ( x ) = x − f ( x ) f ′ ( x ) ⋅ ∑ k = 0 m   a k f ′ ( x ) 2 m − 2 k [ f ( x ) ] k f ′ ′ ( x ) k ∑ k = 0 m   b k f ′ ( x ) 2 m − 2 k [ f ( x ) ] k f ′ ′ ( x ) k {:(12)T_(m)(x)=x-(f(x))/(f^(')(x))*(sum_(k=0)^(m)a_(k)[f^(')(x)]^(2m-2k)[f(x)]^(k)[f^('')(x)]^(k))/(sum_(k=0)^(m)b_(k)[f^(')(x)]^(2m-2k)[f(x)]^(k)[f^('')(x)]^(k)):}\begin{equation*} T_{m}(x)=x-\frac{f(x)}{f^{\prime}(x)} \cdot \frac{\sum_{k=0}^{m} a_{k}\left[f^{\prime}(x)\right]^{2 m-2 k}[f(x)]^{k}\left[f^{\prime \prime}(x)\right]^{k}}{\sum_{k=0}^{m} b_{k}\left[f^{\prime}(x)\right]^{2 m-2 k}[f(x)]^{k}\left[f^{\prime \prime}(x)\right]^{k}} \tag{12} \end{equation*}(12)Tm(x)=x−f(x)f′(x)⋅∑k=0mak[f′(x)]2m−2k[f(x)]k[f′′(x)]k∑k=0mbk[f′(x)]2m−2k[f(x)]k[f′′(x)]k
where m ∈ N ∗ m ∈ N ∗ m inN^(**)m \in \mathbf{N}^{*}m∈N∗ and a k , b k ∈ R a k , b k ∈ R a_(k),b_(k)inRa_{k}, b_{k} \in \mathbf{R}ak,bk∈R, which includes those of Newton and Halley, and which accelerates the convergence of the ratios of consecutive Fibonacci numbers, for some values of a k a k a_(k)a_{k}ak and b k b k b_(k)b_{k}bk to φ φ varphi\varphiφ (the golden number).

3. A NEW METHOD

THEOREM 4. Let (1) be an iterative method of order k ( ≥ 2 ) k ( ≥ 2 ) k( >= 2)k(\geq 2)k(≥2). Let the function F ( x ) F ( x ) F(x)F(x)F(x) be k + 1 k + 1 k+1k+1k+1-times differentiable in a neighborhood of the limit point x = α x = α x=alphax=\alphax=α and let s s sss be a finite parameter such that 1 − F ′ ( α ) ( s + 1 k ) ≠ 0 1 − F ′ ( α ) s + 1 k ≠ 0 1-F^(')(alpha)(s+(1)/(k))!=01-F^{\prime}(\alpha)\left(s+\frac{1}{k}\right) \neq 01−F′(α)(s+1k)≠0. Then
(13) x n + 1 = x n − ( x n − F ( x n ) ) ⋅ 1 − s F ′ ( x n ) 1 − ( s + 1 k ) F ′ ( x n ) , n = 0 , 1 , 2 , … (13) x n + 1 = x n − x n − F x n ⋅ 1 − s F ′ x n 1 − s + 1 k F ′ x n , n = 0 , 1 , 2 , … {:(13)x_(n+1)=x_(n)-(x_(n)-F(x_(n)))*(1-sF^(')(x_(n)))/(1-(s+(1)/(k))F^(')(x_(n)))","n=0","1","2","dots:}\begin{equation*} x_{n+1}=x_{n}-\left(x_{n}-F\left(x_{n}\right)\right) \cdot \frac{1-s F^{\prime}\left(x_{n}\right)}{1-\left(s+\frac{1}{k}\right) F^{\prime}\left(x_{n}\right)}, n=0,1,2, \ldots \tag{13} \end{equation*}(13)xn+1=xn−(xn−F(xn))⋅1−sF′(xn)1−(s+1k)F′(xn),n=0,1,2,…
is an iterative method of order at least k + 1 k + 1 k+1k+1k+1.
Proof. In the method (13) the iterative function is
(14) G ( x ) = x − ( x − F ( x ) ) ⋅ 1 − s F ′ ( x n ) 1 − ( s + 1 k ) F ′ ( x n ) = = F ( x ) − 1 k F ′ ( x ) ⋅ ( x − F ( x ) 1 − ( s + 1 k ) F ′ ( x n ) ) (14) G ( x ) = x − ( x − F ( x ) ) ⋅ 1 − s F ′ x n 1 − s + 1 k F ′ x n = = F ( x ) − 1 k F ′ ( x ) ⋅ x − F ( x ) 1 − s + 1 k F ′ x n {:[(14)G(x)=x-(x-F(x))*(1-sF^(')(x_(n)))/(1-(s+(1)/(k))F^(')(x_(n)))=],[=F(x)-(1)/(k)F^(')(x)*((x-F(x))/(1-(s+(1)/(k))F^(')(x_(n))))]:}\begin{align*} & G(x)=x-(x-F(x)) \cdot \frac{1-s F^{\prime}\left(x_{n}\right)}{1-\left(s+\frac{1}{k}\right) F^{\prime}\left(x_{n}\right)}= \tag{14}\\ & =F(x)-\frac{1}{k} F^{\prime}(x) \cdot\left(\frac{x-F(x)}{1-\left(s+\frac{1}{k}\right) F^{\prime}\left(x_{n}\right)}\right) \end{align*}(14)G(x)=x−(x−F(x))⋅1−sF′(xn)1−(s+1k)F′(xn)==F(x)−1kF′(x)⋅(x−F(x)1−(s+1k)F′(xn))
For the function G ( x ) G ( x ) G(x)G(x)G(x) we shall prove that
(15) G ( α ) = α , G ′ ( α ) = G ′ ′ ( α ) = … = G ( k ) ( α ) = 0 . (15) G ( α ) = α , G ′ ( α ) = G ′ ′ ( α ) = … = G ( k ) ( α ) = 0 . {:(15)G(alpha)=alpha","G^(')(alpha)=G^('')(alpha)=dots=G^((k))(alpha)=0.:}\begin{equation*} G(\alpha)=\alpha, G^{\prime}(\alpha)=G^{\prime \prime}(\alpha)=\ldots=G^{(k)}(\alpha)=0 . \tag{15} \end{equation*}(15)G(α)=α,G′(α)=G′′(α)=…=G(k)(α)=0.
(3) hold.
By hypothesis, (1) is an iterative method of order k k kkk and, therefore, relations
We obtain from (14)
and
(16) G ( α ) = α (16) G ( α ) = α {:(16)G(alpha)=alpha:}\begin{equation*} G(\alpha)=\alpha \tag{16} \end{equation*}(16)G(α)=α
(17) ( x − F ( x ) 1 − ( s + 1 k ) F ′ ( x ) ) ′ + ( r 2 ) ( r − 1 ) ( x ) ( x − F ( x ) 1 − ( s + 1 k ) F ′ ( x ) ) + … + x − F ( x ) 1 − s + 1 k F ′ ( x ) ′ + ( r 2 ) ( r − 1 ) ( x ) x − F ( x ) 1 − s + 1 k F ′ ( x ) + … + quad((x-F(x))/(1-(s+(1)/(k))F^(')(x)))^(')+((r)/(2))^((r-1))(x)((x-F(x))/(1-(s+(1)/(k))F^(')(x)))+dots+\quad\left(\frac{x-F(x)}{1-\left(s+\frac{1}{k}\right) F^{\prime}(x)}\right)^{\prime}+\binom{r}{2}^{(r-1)}(x)\left(\frac{x-F(x)}{1-\left(s+\frac{1}{k}\right) F^{\prime}(x)}\right)+\ldots+(x−F(x)1−(s+1k)F′(x))′+(r2)(r−1)(x)(x−F(x)1−(s+1k)F′(x))+…+
+ F ′ ( x ) ( x − F ( x ) 1 − ( s + 1 k ) F ′ ( x ) ) ( r ) ] + F ′ ( x ) x − F ( x ) 1 − s + 1 k F ′ ( x ) ( r ) {:+F^(')(x)((x-F(x))/(1-(s+(1)/(k))F^(')(x)))^((r))]\left.+F^{\prime}(x)\left(\frac{x-F(x)}{1-\left(s+\frac{1}{k}\right) F^{\prime}(x)}\right)^{(r)}\right]+F′(x)(x−F(x)1−(s+1k)F′(x))(r)]
As regards relations (3), we obtain from (17)
(18) G ( r ) ( α ) = 0 for r = 1 , 2 , … , k − 1 . (18) G ( r ) ( α ) = 0  for  r = 1 , 2 , … , k − 1 .  {:(18)G^((r))(alpha)=0" for "r=1","2","dots","k-1". ":}\begin{equation*} G^{(r)}(\alpha)=0 \text { for } r=1,2, \ldots, k-1 \text {. } \tag{18} \end{equation*}(18)G(r)(α)=0 for r=1,2,…,k−1. 
Since
( x − F ( x ) 1 − ( s + 1 k ) F ′ ( x ) ) ′ = ( 1 − F ′ ( x ) ) [ 1 − ( s + 1 k ) F ′ ( x ) ] + ( x − F ( x ) ) ( s + 1 k ) F ′ ′ ( x ) [ 1 − ( s + 1 k ) F ′ ( x ) ] 2 x − F ( x ) 1 − s + 1 k F ′ ( x ) ′ = 1 − F ′ ( x ) 1 − s + 1 k F ′ ( x ) + ( x − F ( x ) ) s + 1 k F ′ ′ ( x ) 1 − s + 1 k F ′ ( x ) 2 ((x-F(x))/(1-(s+(1)/(k))F^(')(x)))^(')=((1-F^(')(x))[1-(s+(1)/(k))F^(')(x)]+(x-F(x))(s+(1)/(k))F^('')(x))/([1-(s+(1)/(k))F^(')(x)]^(2))\left(\frac{x-F(x)}{1-\left(s+\frac{1}{k}\right) F^{\prime}(x)}\right)^{\prime}=\frac{\left(1-F^{\prime}(x)\right)\left[1-\left(s+\frac{1}{k}\right) F^{\prime}(x)\right]+(x-F(x))\left(s+\frac{1}{k}\right) F^{\prime \prime}(x)}{\left[1-\left(s+\frac{1}{k}\right) F^{\prime}(x)\right]^{2}}(x−F(x)1−(s+1k)F′(x))′=(1−F′(x))[1−(s+1k)F′(x)]+(x−F(x))(s+1k)F′′(x)[1−(s+1k)F′(x)]2
we obtain
G ( k ) ( a ) = F ( k ) ( a ) − 1 k [ k F ( k ) ( a ) ] = 0 G ( k ) ( a ) = F ( k ) ( a ) − 1 k k F ( k ) ( a ) = 0 G^((k))(a)=F^((k))(a)-(1)/(k)[kF^((k))(a)]=0G^{(k)}(a)=F^{(k)}(a)-\frac{1}{k}\left[k F^{(k)}(a)\right]=0G(k)(a)=F(k)(a)−1k[kF(k)(a)]=0
hence conditions (15) are fulfilled.
Remark 2. For s = − 1 k s = − 1 k s=-(1)/(k)s=-\frac{1}{k}s=−1k we obtain the iterative method (4), for s = 0 s = 0 s=0s=0s=0 we obtain the iterative method (5) and for s = 1 − 1 k s = 1 − 1 k s=1-(1)/(k)s=1-\frac{1}{k}s=1−1k we obtain the iterative method (6).

REFERENCES

  1. B. Jovanović, A method for obtaining iterative formulas of higher order, Mat. Vesnik. 24, 9 (1972), 365-369.
  2. I. Lazăr and A. C. Mureşan, Generalized transformations on ratios of Fibonacci and Lucas numbers, Rev. Anal. Numér. Théorie Approximation 24, 1-2 (1995), 169-179.
  3. A. Ralston, A First Course in Numerical Analysis, McGraw-Hill, Inc., 1965.
  4. D. M. Simeunović, On a process for obtaining iterative formulas of higher order for roots of equations, Rev. Anal. Numér. Théorie Approximation 24, 1-2 (1995), 225-229.
Received January 15, 1997
Romanian Academy of Sciences
"Tiberiu Popoviciu" Institute of Numerical Analysis
P.O. Box 68
3400 Cluj-Napoca
Romania
1997

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