Abstract

We study the solving of nonlinear equations by an iterative method of Aitken type, which has the interpolation nodes controlled by the Newton method. We obtain a local convergence result which shows that the q-convergence order of this method is 6 and its efficiency index is \(\sqrt[5]6\), which is higher than the efficiency index of the Aitken or Newton methods. Monotone sequences are obtained for initial approximations farther from the solution, if they satisfy the Fourier condition and the nonlinear mapping satisfies monotony and convexity assumptions on the domain.

Authors

Ion Păvăloiu (Tiberiu Popoviciu Institute of Numerical Analysis, Romanian Academy)

Ion Păvăloiu

Emil Cătinaş (Tiberiu Popoviciu Institute of Numerical Analysis, Romanian Academy)

Emil Cătinaş

Keywords

Nonlinear equations in R; Aitken method; Newton method; inverse interpolatory polynomials; divided differences; monotone sequences.

Paper coordinates

I. Păvăloiu, E. Cătinaş, On an Aitken-Newton type method, Numer. Algor., 62 (2013) no. 2, pp. 253-260
10.1007/s11075-012-9577-7

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s11075-012-9577-7

On an Aitken-Newton type method

Ion Păvăloiu • Emil Cătinaş

Abstract

We study the solving of nonlinear equations by an iterative method of Aitken type, which has the interpolation nodes controlled by the Newton method. We obtain a local convergence result which shows that the q q qqq-convergence order of this method is 6 and its efficiency index is 6 5 6 5 root(5)(6)\sqrt[5]{6}65, which is higher than the efficiency index of the Aitken or Newton methods. Monotone sequences are obtained for initial approximations farther from the solution, if they satisfy the Fourier condition and the nonlinear mapping satisfies monotony and convexity assumptions on the domain.

Keywords Nonlinear equations • Aitken method • Newton method • Monotone convergence

1 Introduction

In this note we study an Aitken type method, for which the interpolation nodes are given by two iterations of Newton type. We show that this method has the q q qqq-convergence order 6 and it requires 5 function evaluations at each step. This implies that the efficiency index of this method is 6 5 6 5 root(5)(6)\sqrt[5]{6}65, which is greater than 2 2 sqrt2\sqrt{2}2 (the efficiency index of the Newton and of the Aitken method) [10, 14, 20].
Consider the equation
(1) f ( x ) = 0 (1) f ( x ) = 0 {:(1)f(x)=0:}\begin{equation*} f(x)=0 \tag{1} \end{equation*}(1)f(x)=0
where f : [ a , b ] R , a < b f : [ a , b ] R , a < b f:[a,b]rarrR,a < bf:[a, b] \rightarrow \mathbb{R}, a<bf:[a,b]R,a<b and assume
α ) α ) alpha)\alpha)α) this equation has a solution x ] a , b [ x a , b [ {:x^(**)in]a,b[\left.x^{*} \in\right] a, b[x]a,b[.
We consider two more equations
x g 1 ( x ) = 0 (2) x g 2 ( x ) = 0 x g 1 ( x ) = 0 (2) x g 2 ( x ) = 0 {:[x-g_(1)(x)=0],[(2)x-g_(2)(x)=0]:}\begin{align*} & x-g_{1}(x)=0 \\ & x-g_{2}(x)=0 \tag{2} \end{align*}xg1(x)=0(2)xg2(x)=0
g 1 , g 2 : [ a , b ] [ a , b ] g 1 , g 2 : [ a , b ] [ a , b ] g_(1),g_(2):[a,b]rarr[a,b]g_{1}, g_{2}:[a, b] \rightarrow[a, b]g1,g2:[a,b][a,b], and we assume they are equivalent to (1).
The Aitken method consists in constructing the sequence ( x n ) n 0 x n n 0 (x_(n))_(n >= 0)\left(x_{n}\right)_{n \geq 0}(xn)n0 given by [1-3, 6-8, 10-19],
(3) x n + 1 = g 1 ( x n ) f ( g 1 ( x n ) ) [ g 1 ( x n ) , g 2 ( x n ) ; f ] , n = 0 , 1 , , x 0 [ a , b ] , (3) x n + 1 = g 1 x n f g 1 x n g 1 x n , g 2 x n ; f , n = 0 , 1 , , x 0 [ a , b ] , {:(3)x_(n+1)=g_(1)(x_(n))-(f(g_(1)(x_(n))))/([g_(1)(x_(n)),g_(2)(x_(n));f])","n=0","1","dots","x_(0)in[a","b]",":}\begin{equation*} x_{n+1}=g_{1}\left(x_{n}\right)-\frac{f\left(g_{1}\left(x_{n}\right)\right)}{\left[g_{1}\left(x_{n}\right), g_{2}\left(x_{n}\right) ; f\right]}, n=0,1, \ldots, x_{0} \in[a, b], \tag{3} \end{equation*}(3)xn+1=g1(xn)f(g1(xn))[g1(xn),g2(xn);f],n=0,1,,x0[a,b],
where [ x , y ; f ] [ x , y ; f ] [x,y;f][x, y ; f][x,y;f] stands for the first order divided difference of f f fff at x x xxx and y y yyy. We suppose that f f fff is derivable on [ a , b ] [ a , b ] [a,b][a, b][a,b] and we consider the function
(4) g 1 ( x ) = x f ( x ) f ( x ) (4) g 1 ( x ) = x f ( x ) f ( x ) {:(4)g_(1)(x)=x-(f(x))/(f^(')(x)):}\begin{equation*} g_{1}(x)=x-\frac{f(x)}{f^{\prime}(x)} \tag{4} \end{equation*}(4)g1(x)=xf(x)f(x)
Denoting
g 2 ( x ) = g 1 ( g 1 ( x ) ) , g 2 ( x ) = g 1 g 1 ( x ) , g_(2)(x)=g_(1)(g_(1)(x)),g_{2}(x)=g_{1}\left(g_{1}(x)\right),g2(x)=g1(g1(x)),
we are lead to the following Aitken-type iterative method
y n = x n f ( x n ) f ( x n ) z n = y n f ( y n ) f ( y n ) (5) x n + 1 = z n f ( z n ) [ y n , z n ; f ] , n = 0 , 1 , , x 0 [ a , b ] , y n = x n f x n f x n z n = y n f y n f y n (5) x n + 1 = z n f z n y n , z n ; f , n = 0 , 1 , , x 0 [ a , b ] , {:[y_(n)=x_(n)-(f(x_(n)))/(f^(')(x_(n)))],[z_(n)=y_(n)-(f(y_(n)))/(f^(')(y_(n)))],[(5)x_(n+1)=z_(n)-(f(z_(n)))/([y_(n),z_(n);f])","n=0","1","dots","x_(0)in[a","b]","]:}\begin{align*} y_{n} & =x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)} \\ z_{n} & =y_{n}-\frac{f\left(y_{n}\right)}{f^{\prime}\left(y_{n}\right)} \\ x_{n+1} & =z_{n}-\frac{f\left(z_{n}\right)}{\left[y_{n}, z_{n} ; f\right]}, n=0,1, \ldots, x_{0} \in[a, b], \tag{5} \end{align*}yn=xnf(xn)f(xn)zn=ynf(yn)f(yn)(5)xn+1=znf(zn)[yn,zn;f],n=0,1,,x0[a,b],
which we call the Aitken-Newton method.
In Section 2 we provide a local convergence result for this method, and we show a similar result which holds for the Newton method: if f f fff maintains its monotony and convexity on a larger domain, and the initial approximation obeys the Fourier condition, then the iterates converge monotonically to the solution. These properties, together with the fact that the efficiency index of this method is shown that is higher than the efficiency index of the Newton or Aitken methods, justify the study of this method.

2 Convergence of the method

We obtain the following local convergence result.
Theorem 1 Assume α α alpha\alphaα ) and
β β beta\betaβ ) there exists an open interval I , x I ] a , b [ I , x I a , b [ {:I,x^(**)in I sube]a,b[\left.I, x^{*} \in I \subseteq\right] a, b[I,xI]a,b[ such that f f fff is two times differentiable on I I III, with f f f^('')f^{\prime \prime}f continuous at x x x^(**)x^{*}x.
Then there exists an interval J I J I J sube IJ \subseteq IJI such that for any initial approximation x 0 J x 0 J x_(0)in Jx_{0} \in Jx0J, the iterations (5) are well defined, remain in J J JJJ and converge to x x x^(**)x^{*}x with q q qqq-order at least 6 .
Proof The first and second relations in (5) imply the existence of θ n θ n theta_(n)\theta_{n}θn and η n η n eta_(n)\eta_{n}ηn in the interior of the intervals determined by x n x n x_(n)x_{n}xn and x x x^(**)x^{*}x, resp. y n y n y_(n)y_{n}yn and x x x^(**)x^{*}x such that
(6) x y n = f ( θ n ) 2 f ( x n ) ( x x n ) 2 , n = 0 , 1 , (7) x z n = f ( η n ) 2 f ( y n ) ( x y n ) 2 , n = 0 , 1 , (6) x y n = f θ n 2 f x n x x n 2 , n = 0 , 1 , (7) x z n = f η n 2 f y n x y n 2 , n = 0 , 1 , {:[(6)x^(**)-y_(n)=-(f^('')(theta_(n)))/(2f^(')(x_(n)))(x^(**)-x_(n))^(2)","n=0","1","dots],[(7)x^(**)-z_(n)=-(f^('')(eta_(n)))/(2f^(')(y_(n)))(x^(**)-y_(n))^(2)","n=0","1","dots]:}\begin{align*} & x^{*}-y_{n}=-\frac{f^{\prime \prime}\left(\theta_{n}\right)}{2 f^{\prime}\left(x_{n}\right)}\left(x^{*}-x_{n}\right)^{2}, n=0,1, \ldots \tag{6}\\ & x^{*}-z_{n}=-\frac{f^{\prime \prime}\left(\eta_{n}\right)}{2 f^{\prime}\left(y_{n}\right)}\left(x^{*}-y_{n}\right)^{2}, n=0,1, \ldots \tag{7} \end{align*}(6)xyn=f(θn)2f(xn)(xxn)2,n=0,1,(7)xzn=f(ηn)2f(yn)(xyn)2,n=0,1,
The third relation in (5) and the Newton identity imply that
(8) x x n + 1 = [ x , y n , z n ; f ] [ y n , z n ; f ] ( x z n ) ( x y n ) , n = 0 , 1 , (8) x x n + 1 = x , y n , z n ; f y n , z n ; f x z n x y n , n = 0 , 1 , {:(8)x^(**)-x_(n+1)=-([x^(**),y_(n),z_(n);f])/([y_(n),z_(n);f])(x^(**)-z_(n))(x^(**)-y_(n))","n=0","1","dots:}\begin{equation*} x^{*}-x_{n+1}=-\frac{\left[x^{*}, y_{n}, z_{n} ; f\right]}{\left[y_{n}, z_{n} ; f\right]}\left(x^{*}-z_{n}\right)\left(x^{*}-y_{n}\right), n=0,1, \ldots \tag{8} \end{equation*}(8)xxn+1=[x,yn,zn;f][yn,zn;f](xzn)(xyn),n=0,1,
Relations (6)-(8) lead to
(9) x x n + 1 = [ x , y n , z n ; f ] f ( η n ) ( f ( θ n ) ) 3 16 [ y n , z n ; f ] f ( y n ) ( f ( x n ) ) 3 ( x x n ) 6 , n = 0 , 1 , , (9) x x n + 1 = x , y n , z n ; f f η n f θ n 3 16 y n , z n ; f f y n f x n 3 x x n 6 , n = 0 , 1 , , {:(9)x^(**)-x_(n+1)=-([x^(**),y_(n),z_(n);f]f^('')(eta_(n))*(f^('')(theta_(n)))^(3))/(16[y_(n),z_(n);f]f^(')(y_(n))(f^(')(x_(n)))^(3))(x^(**)-x_(n))^(6)","n=0","1","dots",":}\begin{equation*} x^{*}-x_{n+1}=-\frac{\left[x^{*}, y_{n}, z_{n} ; f\right] f^{\prime \prime}\left(\eta_{n}\right) \cdot\left(f^{\prime \prime}\left(\theta_{n}\right)\right)^{3}}{16\left[y_{n}, z_{n} ; f\right] f^{\prime}\left(y_{n}\right)\left(f^{\prime}\left(x_{n}\right)\right)^{3}}\left(x^{*}-x_{n}\right)^{6}, n=0,1, \ldots, \tag{9} \end{equation*}(9)xxn+1=[x,yn,zn;f]f(ηn)(f(θn))316[yn,zn;f]f(yn)(f(xn))3(xxn)6,n=0,1,,
which shows the assertion, provided that x 0 x 0 x_(0)x_{0}x0 is sufficiently close to x x x^(**)x^{*}x.
Under supplementary conditions on f f fff we obtain the following result.
Theorem 2 If f f fff and x 0 x 0 x_(0)x_{0}x0 verify α α alpha\alphaα ), and
β ) f β f {:beta^('))quad f\left.\beta^{\prime}\right) \quad fβ)f is two times differentiable on [ a , b ] [ a , b ] [a,b][a, b][a,b], with f f f^('')f^{\prime \prime}f continuous at x x x^(**)x^{*}x;
γ) x 0 [ a , b ] x 0 [ a , b ] quadx_(0)in[a,b]\quad x_{0} \in[a, b]x0[a,b] verifies the Fourier condition: f ( x 0 ) f ( x 0 ) > 0 f x 0 f x 0 > 0 f(x_(0))f^('')(x_(0)) > 0f\left(x_{0}\right) f^{\prime \prime}\left(x_{0}\right)>0f(x0)f(x0)>0 (see [10]),
and, moreover,
i 1 . f ( x ) > 0 , x [ a , b ] ; i 1 . f ( x ) > 0 , x [ a , b ] ; i_(1).quadf^(')(x) > 0,AA x in[a,b];\mathrm{i}_{1} . \quad f^{\prime}(x)>0, \forall x \in[a, b] ;i1.f(x)>0,x[a,b];
ii. f ( x ) 0 , x [ a , b ] f ( x ) 0 , x [ a , b ] quadf^('')(x) >= 0,AA x in[a,b]\quad f^{\prime \prime}(x) \geq 0, \forall x \in[a, b]f(x)0,x[a,b],
then the sequences ( x n ) n 0 , ( y n ) n 0 x n n 0 , y n n 0 (x_(n))_(n >= 0),(y_(n))_(n >= 0)\left(x_{n}\right)_{n \geq 0},\left(y_{n}\right)_{n \geq 0}(xn)n0,(yn)n0 and ( z n ) n 0 z n n 0 (z_(n))_(n >= 0)\left(z_{n}\right)_{n \geq 0}(zn)n0 generated by (5), remain in [ a , b a , b a,ba, ba,b ] and obey
j 1 . x n > y n > z n > x n + 1 > x , n = 0 , 1 , j 1 . x n > y n > z n > x n + 1 > x , n = 0 , 1 , j_(1).quadx_(n) > y_(n) > z_(n) > x_(n+1) > x^(**),n=0,1,dots\mathrm{j}_{1} . \quad x_{n}>y_{n}>z_{n}>x_{n+1}>x^{*}, n=0,1, \ldotsj1.xn>yn>zn>xn+1>x,n=0,1,;
jj 1 . lim x n = lim y n = lim z n = x jj 1 . lim x n = lim y n = lim z n = x jj_(1).quad limx_(n)=limy_(n)=limz_(n)=x^(**)\mathrm{jj}_{1} . \quad \lim x_{n}=\lim y_{n}=\lim z_{n}=x^{*}jj1.limxn=limyn=limzn=x.
Proof By α α alpha\alphaα ) and i 1 i 1 i_(1)i_{1}i1 it follows that x x x^(**)x^{*}x is the unique solution of (1). Let x n ] a , b [ x n a , b [ {:x_(n)in]a,b[\left.x_{n} \in\right] a, b[xn]a,b[ be an approximation which verifies the relation f ( x n ) f ( x n ) > 0 f x n f x n > 0 f(x_(n))f^('')(x_(n)) > 0f\left(x_{n}\right) f^{\prime \prime}\left(x_{n}\right)>0f(xn)f(xn)>0. Then by i i 1 i i 1 ii_(1)i i_{1}ii1 it follows that f ( x n ) > 0 f x n > 0 f(x_(n)) > 0f\left(x_{n}\right)>0f(xn)>0, which, together with i 1 i 1 i_(1)i_{1}i1 lead to x n > x x n > x x_(n) > x^(**)x_{n}>x^{*}xn>x.
From i 1 , i i 1 i 1 , i i 1 i_(1),ii_(1)i_{1}, i i_{1}i1,ii1 and relation (6) we have x y n 0 x y n 0 x^(**)-y_(n) <= 0x^{*}-y_{n} \leq 0xyn0, i.e. y n x y n x y_(n) >= x^(**)y_{n} \geq x^{*}ynx. Analogously, from (7) and (8) it follows z n x z n x z_(n) >= x^(**)z_{n} \geq x^{*}znx and x n + 1 x x n + 1 x x_(n+1) >= x^(**)x_{n+1} \geq x^{*}xn+1x.
The first relation in (5) and f ( x n ) > 0 , f ( x n ) > 0 f x n > 0 , f x n > 0 f(x_(n)) > 0,f^(')(x_(n)) > 0f\left(x_{n}\right)>0, f^{\prime}\left(x_{n}\right)>0f(xn)>0,f(xn)>0 imply that y n < x n y n < x n y_(n) < x_(n)y_{n}<x_{n}yn<xn. Analogously, the second relation in (5) leads to z n < y n z n < y n z_(n) < y_(n)z_{n}<y_{n}zn<yn, while the third relation in (5) to x n + 1 < z n x n + 1 < z n x_(n+1) < z_(n)x_{n+1}<z_{n}xn+1<zn. Conclusion j 1 j 1 j_(1)j_{1}j1 is therefore proved. Moreover, it is clear that the elements of the sequences ( x n ) n 0 , ( y n ) n 0 x n n 0 , y n n 0 (x_(n))_(n >= 0),(y_(n))_(n >= 0)\left(x_{n}\right)_{n \geq 0},\left(y_{n}\right)_{n \geq 0}(xn)n0,(yn)n0 and ( z n ) n 0 z n n 0 (z_(n))_(n >= 0)\left(z_{n}\right)_{n \geq 0}(zn)n0 remain in the interval [ x , x 0 ] [ a , b ] x , x 0 [ a , b ] [x^(**),x_(0)]sub[a,b]\left[x^{*}, x_{0}\right] \subset[a, b][x,x0][a,b]. By j 1 j 1 j_(1)j_{1}j1 it follows that these three sequences are convergent.
Let lim x n = lim x n = limx_(n)=ℓ\lim x_{n}=\elllimxn=. Relation j 1 j 1 j_(1)j_{1}j1 implies that lim y n = lim z n = lim y n = lim z n = limy_(n)=limz_(n)=ℓ\lim y_{n}=\lim z_{n}=\elllimyn=limzn=. The first relation in (5) attracts that = f ( ) f ( ) = f ( ) f ( ) ℓ=ℓ-(f(ℓ))/(f^(')(ℓ))\ell=\ell-\frac{f(\ell)}{f^{\prime}(\ell)}=f()f(), so f ( ) = 0 f ( ) = 0 f(ℓ)=0f(\ell)=0f()=0, i.e., = x = x ℓ=x^(**)\ell=x^{*}=x. \square
The following immediate consequence holds.
Corollary 3 If f f fff and x 0 [ a , b ] x 0 [ a , b ] x_(0)in[a,b]x_{0} \in[a, b]x0[a,b] verify α α alpha\alphaα ), β β beta^(')\beta^{\prime}β ), γ γ gamma\gammaγ ) and, moreover
i. f ( x ) < 0 , x [ a , b ] ; f ( x ) < 0 , x [ a , b ] ; quadf^(')(x) < 0,AA x in[a,b];\quad f^{\prime}(x)<0, \forall x \in[a, b] ;f(x)<0,x[a,b];
ii i 2 f ( x ) 0 , x [ a , b ] 2 f ( x ) 0 , x [ a , b ] _(2)quadf^('')(x) <= 0,AA x in[a,b]_{2} \quad f^{\prime \prime}(x) \leq 0, \forall x \in[a, b]2f(x)0,x[a,b],
then the elements of the sequences ( x n ) n 0 , ( y n ) n 0 x n n 0 , y n n 0 (x_(n))_(n >= 0),(y_(n))_(n >= 0)\left(x_{n}\right)_{n \geq 0},\left(y_{n}\right)_{n \geq 0}(xn)n0,(yn)n0, and ( z n ) n 0 z n n 0 (z_(n))_(n >= 0)\left(z_{n}\right)_{n \geq 0}(zn)n0 generated by (5) remain in the interval [ a , b ] [ a , b ] [a,b][a, b][a,b] and satisfy the conclusions j 1 j 1 j_(1)j_{1}j1 and j j 1 j j 1 jj_(1)j j_{1}jj1 of Theorem 2.
It is easy to see that if instead of (1) we consider
(10) f ( x ) = 0 (10) f ( x ) = 0 {:(10)-f(x)=0:}\begin{equation*} -f(x)=0 \tag{10} \end{equation*}(10)f(x)=0
then function h : [ a , b ] R h : [ a , b ] R h:[a,b]rarrRh:[a, b] \rightarrow \mathbb{R}h:[a,b]R given by relation
h ( x ) = f ( x ) h ( x ) = f ( x ) h(x)=-f(x)h(x)=-f(x)h(x)=f(x)
verifies hypothesis of Theorem 2.
Theorem 4 If f f fff and x 0 [ a , b ] x 0 [ a , b ] x_(0)in[a,b]x_{0} \in[a, b]x0[a,b] verify α ) , β ) , γ α , β , γ {: alpha),beta^(')),gamma\left.\left.\alpha\right), \beta^{\prime}\right), \gammaα),β),γ ) and, moreover,
i 3 . f ( x ) > 0 , x [ a , b ] ; i 3 . f ( x ) > 0 , x [ a , b ] ; i_(3).quadf^(')(x) > 0,AA x in[a,b];\mathrm{i}_{3} . \quad f^{\prime}(x)>0, \forall x \in[a, b] ;i3.f(x)>0,x[a,b];
ii. f ( x ) 0 , x [ a , b ] f ( x ) 0 , x [ a , b ] f^('')(x) <= 0,AA x in[a,b]f^{\prime \prime}(x) \leq 0, \forall x \in[a, b]f(x)0,x[a,b],
then the elements of the sequences ( x n ) n 0 , ( y n ) n 0 x n n 0 , y n n 0 (x_(n))_(n >= 0),(y_(n))_(n >= 0)\left(x_{n}\right)_{n \geq 0},\left(y_{n}\right)_{n \geq 0}(xn)n0,(yn)n0 and ( z n ) n 0 z n n 0 (z_(n))_(n >= 0)\left(z_{n}\right)_{n \geq 0}(zn)n0 generated by (5) remain in [ a , b ] [ a , b ] [a,b][a, b][a,b] and, moreover, obey
j 3 . x n < y n < z n < x n + 1 < x , n = 0 , 1 , j 3 . x n < y n < z n < x n + 1 < x , n = 0 , 1 , j_(3).quadx_(n) < y_(n) < z_(n) < x_(n+1) < x^(**),n=0,1,dots\mathrm{j}_{3} . \quad x_{n}<y_{n}<z_{n}<x_{n+1}<x^{*}, n=0,1, \ldotsj3.xn<yn<zn<xn+1<x,n=0,1,, ;
jj 3 . lim x n = lim y n = lim z n = x jj 3 . lim x n = lim y n = lim z n = x jj_(3).quad limx_(n)=limy_(n)=limz_(n)=x^(**)\mathrm{jj}_{3} . \quad \lim x_{n}=\lim y_{n}=\lim z_{n}=x^{*}jj3.limxn=limyn=limzn=x.
The proof of this result is similar to the proof of Theorem 2.
If we replace (1) by (10) then we obtain:
Corollary 5 If f f fff and x 0 [ a , b ] x 0 [ a , b ] x_(0)in[a,b]x_{0} \in[a, b]x0[a,b] verify hypothesis α α alpha\alphaα ), β β beta^(')\beta^{\prime}β ), γ γ gamma\gammaγ ) and, moreover,
i n . f ( x ) < 0 , x [ a , b ] ; i n . f ( x ) < 0 , x [ a , b ] ; i_(n).quadf^(')(x) < 0,AA x in[a,b];\mathrm{i}_{n} . \quad f^{\prime}(x)<0, \forall x \in[a, b] ;in.f(x)<0,x[a,b];
ii n . f ( x ) 0 , x [ a , b ] ii n . f ( x ) 0 , x [ a , b ] ii_(n).quadf^('')(x) >= 0,AA x in[a,b]\mathrm{ii}_{n} . \quad f^{\prime \prime}(x) \geq 0, \forall x \in[a, b]iin.f(x)0,x[a,b],
then the elements of the sequences ( x n ) n 0 , ( y n ) n 0 x n n 0 , y n n 0 (x_(n))_(n >= 0),(y_(n))_(n >= 0)\left(x_{n}\right)_{n \geq 0},\left(y_{n}\right)_{n \geq 0}(xn)n0,(yn)n0 and ( z n ) n 0 z n n 0 (z_(n))_(n >= 0)\left(z_{n}\right)_{n \geq 0}(zn)n0 generated by (5) remain in [ a , b ] [ a , b ] [a,b][a, b][a,b] and satisfy the statements j 3 j 3 j_(3)j_{3}j3 and j j 3 j j 3 jj_(3)j j_{3}jj3 of Theorem 4.
Remark 6 Relations (9) show us that the Aitken-Newton method has the q q qqq-convergence order at least 6 (it is exactly 6 if f ( x ) 0 f x 0 f^('')(x^(**))!=0f^{\prime \prime}\left(x^{*}\right) \neq 0f(x)0, see [9]). In order to obtain x n + 1 x n + 1 x_(n+1)x_{n+1}xn+1 from x n x n x_(n)x_{n}xn in (5) we need to perform the following function evaluations: f ( x n ) , f ( x n ) , f ( y n ) , f ( y n ) f x n , f x n , f y n , f y n f(x_(n)),f^(')(x_(n)),f(y_(n)),f^(')(y_(n))f\left(x_{n}\right), f^{\prime}\left(x_{n}\right), f\left(y_{n}\right), f^{\prime}\left(y_{n}\right)f(xn),f(xn),f(yn),f(yn) and f ( z n ) f z n f(z_(n))f\left(z_{n}\right)f(zn), i.e., 5 function evaluations. This shows that the efficiency index of this method is 6 5 6 5 root(5)(6)\sqrt[5]{6}65 which is greater than of Aitken or Newton method.
Remark 7 Under additional information on the bounds of the size of derivatives, one can obtain obtain some a posteriori error estimations of the error:
(11) | x x n + 1 | M 2 m | x n + 1 y n | | x n + 1 z n | , n = 0 , 1 , (11) x x n + 1 M 2 m x n + 1 y n x n + 1 z n , n = 0 , 1 , {:(11)|x^(**)-x_(n+1)| <= (M)/(2m)|x_(n+1)-y_(n)||x_(n+1)-z_(n)|","n=0","1","dots:}\begin{equation*} \left|x^{*}-x_{n+1}\right| \leq \frac{M}{2 m}\left|x_{n+1}-y_{n}\right|\left|x_{n+1}-z_{n}\right|, n=0,1, \ldots \tag{11} \end{equation*}(11)|xxn+1|M2m|xn+1yn||xn+1zn|,n=0,1,
where
m min x [ a , b ] | f ( x ) | , M max x [ a , b ] | f ( x ) | m min x [ a , b ] f ( x ) , M max x [ a , b ] f ( x ) m <= min_(x in[a,b])|f^(')(x)|,M >= max_(x in[a,b])|f^('')(x)|m \leq \min _{x \in[a, b]}\left|f^{\prime}(x)\right|, M \geq \max _{x \in[a, b]}\left|f^{\prime \prime}(x)\right|mminx[a,b]|f(x)|,Mmaxx[a,b]|f(x)|
In order to prove them, we consider the Newton identity,
f ( x n + 1 ) = f ( y n ) + [ y n , z n ; f ] ( x n + 1 y n ) + + [ x n + 1 , y n , z n ; f ] ( x n + 1 y n ) ( x n + 1 z n ) , n = 0 , 1 , f x n + 1 = f y n + y n , z n ; f x n + 1 y n + + x n + 1 , y n , z n ; f x n + 1 y n x n + 1 z n , n = 0 , 1 , {:[f(x_(n+1))=f(y_(n))+[y_(n),z_(n);f](x_(n+1)-y_(n))+],[+[x_(n+1),y_(n),z_(n);f](x_(n+1)-y_(n))(x_(n+1)-z_(n))","n=0","1","dots]:}\begin{aligned} f\left(x_{n+1}\right)= & f\left(y_{n}\right)+\left[y_{n}, z_{n} ; f\right]\left(x_{n+1}-y_{n}\right)+ \\ & +\left[x_{n+1}, y_{n}, z_{n} ; f\right]\left(x_{n+1}-y_{n}\right)\left(x_{n+1}-z_{n}\right), n=0,1, \ldots \end{aligned}f(xn+1)=f(yn)+[yn,zn;f](xn+1yn)++[xn+1,yn,zn;f](xn+1yn)(xn+1zn),n=0,1,
whence, taking into acount (5), we get
f ( x n + 1 ) f ( x ) = [ x n + 1 , y n , z n ; f ] ( x n + 1 y n ) ( x n + 1 z n ) , n = 0 , 1 , , f x n + 1 f x = x n + 1 , y n , z n ; f x n + 1 y n x n + 1 z n , n = 0 , 1 , , f(x_(n+1))-f(x^(**))=[x_(n+1),y_(n),z_(n);f](x_(n+1)-y_(n))(x_(n+1)-z_(n)),n=0,1,dots,f\left(x_{n+1}\right)-f\left(x^{*}\right)=\left[x_{n+1}, y_{n}, z_{n} ; f\right]\left(x_{n+1}-y_{n}\right)\left(x_{n+1}-z_{n}\right), n=0,1, \ldots,f(xn+1)f(x)=[xn+1,yn,zn;f](xn+1yn)(xn+1zn),n=0,1,,
or
x n + 1 x = [ x n + 1 , y n , z n ; f ] [ x , x n + 1 ; f ] ( x n + 1 y n ) ( x n + 1 z n ) , n = 0 , 1 , x n + 1 x = x n + 1 , y n , z n ; f x , x n + 1 ; f x n + 1 y n x n + 1 z n , n = 0 , 1 , x_(n+1)-x^(**)=([x_(n+1),y_(n),z_(n);f])/([x^(**),x_(n+1);f])(x_(n+1)-y_(n))(x_(n+1)-z_(n)),n=0,1,dotsx_{n+1}-x^{*}=\frac{\left[x_{n+1}, y_{n}, z_{n} ; f\right]}{\left[x^{*}, x_{n+1} ; f\right]}\left(x_{n+1}-y_{n}\right)\left(x_{n+1}-z_{n}\right), n=0,1, \ldotsxn+1x=[xn+1,yn,zn;f][x,xn+1;f](xn+1yn)(xn+1zn),n=0,1,
The mean value formulas for divided differences lead to (11).
These estimations can be applied in connection to any of the results proved above.

3 Numerical examples

Example 8 Consider the equation
f ( x ) = e x + sin x 2 , x [ 0 , 1 ] f ( x ) = e x + sin x 2 , x [ 0 , 1 ] f(x)=e^(x)+sin x-2,x in[0,1]f(x)=e^{x}+\sin x-2, x \in[0,1]f(x)=ex+sinx2,x[0,1]
The derivatives of f f fff are given by
f ( x ) = e x + cos x > 0 , x [ 0 , 1 ] f ( x ) = e x sin x f ( x ) = e x + cos x > 0 , x [ 0 , 1 ] f ( x ) = e x sin x {:[f^(')(x)=e^(x)+cos x > 0","quad x in[0","1]],[f^('')(x)=e^(x)-sin x]:}\begin{aligned} f^{\prime}(x) & =e^{x}+\cos x>0, \quad x \in[0,1] \\ f^{\prime \prime}(x) & =e^{x}-\sin x \end{aligned}f(x)=ex+cosx>0,x[0,1]f(x)=exsinx
Some elementary considerations on f f f^('')f^{\prime \prime}f show that f ( x ) > 0 , x [ 0 , 1 ] f ( x ) > 0 , x [ 0 , 1 ] f^('')(x) > 0,x in[0,1]f^{\prime \prime}(x)>0, x \in[0,1]f(x)>0,x[0,1]. Since f f fff is continuous, f ( x ) > 0 , x [ 0 , 1 ] f ( x ) > 0 , x [ 0 , 1 ] f^(')(x) > 0,x in[0,1]f^{\prime}(x)>0, x \in[0,1]f(x)>0,x[0,1] and f ( 0 ) = 1 , f ( 1 ) = e + sin 1 2 > 0 f ( 0 ) = 1 , f ( 1 ) = e + sin 1 2 > 0 f(0)=-1,f(1)=e+sin 1-2 > 0f(0)=-1, f(1)=e+\sin 1-2>0f(0)=1,f(1)=e+sin12>0, it follows that f f fff has a unique solution on [ 0 , 1 ] [ 0 , 1 ] [0,1][0,1][0,1].
Taking x 0 = 1 x 0 = 1 x_(0)=1x_{0}=1x0=1, the hypotheses of Theorem 2 are satisfied.
Table 1 Numerical results when solving e x + sin x 2 = 0 e x + sin x 2 = 0 e^(x)+sin x-2=0e^{x}+\sin x-2=0ex+sinx2=0
n n nnn x n x n x_(n)x_{n}xn y n y n y_(n)y_{n}yn z n z n z_(n)z_{n}zn f ( x n ) f x n f(x_(n))f\left(x_{n}\right)f(xn)
0 1.000000000000000 e + 0 1.000000000000000 e + 0 1.000000000000000e+01.000000000000000 \mathrm{e}+01.000000000000000e+0 5.213403278939761 e 1 5.213403278939761 e 1 5.213403278939761e-15.213403278939761 \mathrm{e}-15.213403278939761e1 4.498799895489901 e 1 4.498799895489901 e 1 4.498799895489901e-14.498799895489901 \mathrm{e}-14.498799895489901e1 1.5 e + 0 1.5 e + 0 1.5e+01.5 \mathrm{e}+01.5e+0
1 4.486920253023863 e 1 4.486920253023863 e 1 4.486920253023863e-14.486920253023863 \mathrm{e}-14.486920253023863e1 4.486719164440748 e 1 4.486719164440748 e 1 4.486719164440748e-14.486719164440748 \mathrm{e}-14.486719164440748e1 4.486719163512726 e 1 4.486719163512726 e 1 4.486719163512726e-14.486719163512726 \mathrm{e}-14.486719163512726e1 4.9 e 5 4.9 e 5 4.9e-54.9 \mathrm{e}-54.9e5
2 4.486719163512727 e 1 4.486719163512727 e 1 4.486719163512727e-14.486719163512727 \mathrm{e}-14.486719163512727e1 0
n x_(n) y_(n) z_(n) f(x_(n)) 0 1.000000000000000e+0 5.213403278939761e-1 4.498799895489901e-1 1.5e+0 1 4.486920253023863e-1 4.486719164440748e-1 4.486719163512726e-1 4.9e-5 2 4.486719163512727e-1 0| $n$ | $x_{n}$ | $y_{n}$ | $z_{n}$ | $f\left(x_{n}\right)$ | | :--- | :--- | :--- | :--- | :--- | | 0 | $1.000000000000000 \mathrm{e}+0$ | $5.213403278939761 \mathrm{e}-1$ | $4.498799895489901 \mathrm{e}-1$ | $1.5 \mathrm{e}+0$ | | 1 | $4.486920253023863 \mathrm{e}-1$ | $4.486719164440748 \mathrm{e}-1$ | $4.486719163512726 \mathrm{e}-1$ | $4.9 \mathrm{e}-5$ | | 2 | $4.486719163512727 \mathrm{e}-1$ | | | 0 |
In Table 1 we present the results obtained in double precision using MATLAB.
One can easily verify that min x [ 0 , 1 ] | f ( x ) | = 2 min x [ 0 , 1 ] f ( x ) = 2 min_(x in[0,1])|f^(')(x)|=2\min _{x \in[0,1]}\left|f^{\prime}(x)\right|=2minx[0,1]|f(x)|=2 and max x [ 0 , 1 ] | f ( x ) | e max x [ 0 , 1 ] f ( x ) e max_(x in[0,1])|f^('')(x)| <= e\max _{x \in[0,1]}\left|f^{\prime \prime}(x)\right| \leq emaxx[0,1]|f(x)|e.
Taking into account (11) we get
| x x 2 | e 4 | x 2 y 1 | | x 2 z 1 | . x x 2 e 4 x 2 y 1 x 2 z 1 . |x^(**)-x_(2)| <= (e)/(4)|x_(2)-y_(1)|*|x_(2)-z_(1)|.\left|x^{*}-x_{2}\right| \leq \frac{e}{4}\left|x_{2}-y_{1}\right| \cdot\left|x_{2}-z_{1}\right| .|xx2|e4|x2y1||x2z1|.
The quantity in the right hand side is majorized by 7.0 e 27 7.0 e 27 7.0e-277.0 \mathrm{e}-277.0e27, which, together with the fact that f ( x 2 ) = 0 f x 2 = 0 f(x_(2))=0f\left(x_{2}\right)=0f(x2)=0, shows that in this particular case, x x x^(**)x^{*}x can be computed with accuracy higher than the machine epsilon.

Example 9

f ( x ) = ln ( x 2 + x + 2 ) x + 1 = 0 , x [ 4 , 5 ] . f ( x ) = ln x 2 + x + 2 x + 1 = 0 , x [ 4 , 5 ] . f(x)=ln(x^(2)+x+2)-x+1=0,x in[4,5].f(x)=\ln \left(x^{2}+x+2\right)-x+1=0, x \in[4,5] .f(x)=ln(x2+x+2)x+1=0,x[4,5].
We have
f ( x ) = x 2 + x 1 x 2 + x + 2 < 0 , for x [ 4 , 5 ] ; f ( x ) = 2 x 2 2 x + 1 ( x 2 + x + 2 ) 2 < 0 , for x [ 4 , 5 ] ; f ( x ) = x 2 + x 1 x 2 + x + 2 < 0 ,  for  x [ 4 , 5 ] ; f ( x ) = 2 x 2 2 x + 1 x 2 + x + 2 2 < 0 ,  for  x [ 4 , 5 ] ; {:[f^(')(x)=(-x^(2)+x-1)/(x^(2)+x+2) < 0","" for "x in[4","5];],[f^('')(x)=(-2x^(2)-2x+1)/((x^(2)+x+2)^(2)) < 0","" for "x in[4","5];]:}\begin{aligned} & f^{\prime}(x)=\frac{-x^{2}+x-1}{x^{2}+x+2}<0, \text { for } x \in[4,5] ; \\ & f^{\prime \prime}(x)=\frac{-2 x^{2}-2 x+1}{\left(x^{2}+x+2\right)^{2}}<0, \text { for } x \in[4,5] ; \end{aligned}f(x)=x2+x1x2+x+2<0, for x[4,5];f(x)=2x22x+1(x2+x+2)2<0, for x[4,5];
f ( 4 ) = ln 22 3 > 0 f ( 4 ) = ln 22 3 > 0 f(4)=ln 22-3 > 0f(4)=\ln 22-3>0f(4)=ln223>0 and f ( 5 ) = ln 32 4 < 0 f ( 5 ) = ln 32 4 < 0 f(5)=ln 32-4 < 0f(5)=\ln 32-4<0f(5)=ln324<0.
We take x 0 = 5 x 0 = 5 x_(0)=5x_{0}=5x0=5, so hypotheses of Corollary 3 are verified. The obtained results are presented in Table 2.
In this case we have min x [ 4 , 5 ] | f ( x ) | > 1 3 min x [ 4 , 5 ] f ( x ) > 1 3 min_(x in[4,5])|f^(')(x)| > (1)/(3)\min _{x \in[4,5]}\left|f^{\prime}(x)\right|>\frac{1}{3}minx[4,5]|f(x)|>13 and max x [ 4 , 5 ] | f ( x ) | 1 8 max x [ 4 , 5 ] f ( x ) 1 8 max_(x in[4,5])|f^('')(x)| <= (1)/(8)\max _{x \in[4,5]}\left|f^{\prime \prime}(x)\right| \leq \frac{1}{8}maxx[4,5]|f(x)|18, whence, by (11) we get
| x x 2 | 3 16 | x 2 y 1 | | x 2 z 1 | . x x 2 3 16 x 2 y 1 x 2 z 1 . |x^(**)-x_(2)| <= (3)/(16)|x_(2)-y_(1)|*|x_(2)-z_(1)|.\left|x^{*}-x_{2}\right| \leq \frac{3}{16}\left|x_{2}-y_{1}\right| \cdot\left|x_{2}-z_{1}\right| .|xx2|316|x2y1||x2z1|.
The quantity in the right hand side is majorized by 1.4 e 31 1.4 e 31 1.4e-311.4 \mathrm{e}-311.4e31, which, together with the fact that f ( x 2 ) = 0 f x 2 = 0 f(x_(2))=0f\left(x_{2}\right)=0f(x2)=0 shows that in this example too x x x^(**)x^{*}x is computed with accuracy higher than the machine epsilon.
Table 2 Numerical results when solving ln ( x 2 + x + 2 ) x + 1 = 0 ln x 2 + x + 2 x + 1 = 0 ln(x^(2)+x+2)-x+1=0\ln \left(x^{2}+x+2\right)-x+1=0ln(x2+x+2)x+1=0
n n nnn x n x n x_(n)x_{n}xn y n y n y_(n)y_{n}yn z n z n z_(n)z_{n}zn f ( x n ) f x n f(x_(n))f\left(x_{n}\right)f(xn)
0 5.000000000000000 e + 0 5.000000000000000 e + 0 5.000000000000000e+05.000000000000000 \mathrm{e}+05.000000000000000e+0 4.185883280456726 e + 0 4.185883280456726 e + 0 4.185883280456726e+04.185883280456726 \mathrm{e}+04.185883280456726e+0 4.152656878948953 e + 0 4.152656878948953 e + 0 4.152656878948953e+04.152656878948953 \mathrm{e}+04.152656878948953e+0 5.3 e 1 5.3 e 1 -5.3e-1-5.3 \mathrm{e}-15.3e1
1 4.152590868900850 e + 0 4.152590868900850 e + 0 4.152590868900850e+04.152590868900850 \mathrm{e}+04.152590868900850e+0 4.152590736757159 e + 0 4.152590736757159 e + 0 4.152590736757159e+04.152590736757159 \mathrm{e}+04.152590736757159e+0 4.152590736757158 e + 0 4.152590736757158 e + 0 4.152590736757158e+04.152590736757158 \mathrm{e}+04.152590736757158e+0 7.9 e 8 7.9 e 8 -7.9e-8-7.9 \mathrm{e}-87.9e8
2 4.152590736757158 e + 0 4.152590736757158 e + 0 4.152590736757158e+04.152590736757158 \mathrm{e}+04.152590736757158e+0 0
n x_(n) y_(n) z_(n) f(x_(n)) 0 5.000000000000000e+0 4.185883280456726e+0 4.152656878948953e+0 -5.3e-1 1 4.152590868900850e+0 4.152590736757159e+0 4.152590736757158e+0 -7.9e-8 2 4.152590736757158e+0 0| $n$ | $x_{n}$ | $y_{n}$ | $z_{n}$ | $f\left(x_{n}\right)$ | | :--- | :--- | :--- | :--- | :--- | | 0 | $5.000000000000000 \mathrm{e}+0$ | $4.185883280456726 \mathrm{e}+0$ | $4.152656878948953 \mathrm{e}+0$ | $-5.3 \mathrm{e}-1$ | | 1 | $4.152590868900850 \mathrm{e}+0$ | $4.152590736757159 \mathrm{e}+0$ | $4.152590736757158 \mathrm{e}+0$ | $-7.9 \mathrm{e}-8$ | | 2 | $4.152590736757158 \mathrm{e}+0$ | | | 0 |
We make the following comments regarding the convergence order of the obtained sequences. Since they are monotonic in these two examples, the quotient convergence factors [9, ch. 9] can be determined by (9), taking into account the mean value formulas for divided differences, as
(12) Q 6 { x k } = lim k | x k + 1 x | | x k x | 6 = ( f ( x ) ) 5 32 ( f ( x ) ) 5 (12) Q 6 x k = lim k x k + 1 x x k x 6 = f x 5 32 f x 5 {:(12)Q_(6){x_(k)}=lim_(k rarr oo)(|x_(k+1)-x^(**)|)/(|x_(k)-x^(**)|^(6))=((f^('')(x^(**)))^(5))/(32(f^(')(x^(**)))^(5)):}\begin{equation*} Q_{6}\left\{x_{k}\right\}=\lim _{k \rightarrow \infty} \frac{\left|x_{k+1}-x^{*}\right|}{\left|x_{k}-x^{*}\right|^{6}}=\frac{\left(f^{\prime \prime}\left(x^{*}\right)\right)^{5}}{32\left(f^{\prime}\left(x^{*}\right)\right)^{5}} \tag{12} \end{equation*}(12)Q6{xk}=limk|xk+1x||xkx|6=(f(x))532(f(x))5
Therefore, since in both examples f ( x ) 0 f x 0 f^('')(x^(**))!=0f^{\prime \prime}\left(x^{*}\right) \neq 0f(x)0, the q q qqq-convergence order is exactly 6 . However, the above quantity requires the knowledge of the solution x x x^(**)x^{*}x. In [4] and [5], the asymptotical constant Q 6 { x k } Q 6 x k Q_(6){x_(k)}Q_{6}\left\{x_{k}\right\}Q6{xk} was approximated by some quantities computable at each step:
(13) | x k + 1 x k | | x k x k 1 | 6 , k = 1 , 2 , (13) x k + 1 x k x k x k 1 6 , k = 1 , 2 , {:(13)(|x_(k+1)-x_(k)|)/(|x_(k)-x_(k-1)|^(6))","k=1","2","dots:}\begin{equation*} \frac{\left|x_{k+1}-x_{k}\right|}{\left|x_{k}-x_{k-1}\right|^{6}}, k=1,2, \ldots \tag{13} \end{equation*}(13)|xk+1xk||xkxk1|6,k=1,2,
In Example 8, formula (12) (with x x x^(**)x^{*}x considered as x 2 x 2 x_(2)x_{2}x2 ) yields the value 6.3e-4, while formula (13) for k = 1 k = 1 k=1k=1k=1 yields 7.1 e 4 7.1 e 4 7.1e-47.1 \mathrm{e}-47.1e4, i.e., two close quantities. These values are also close in Example 9, where we obtain 1.0e-6, respectively 3.5e-7.
Some other formulas to determine the convergence order were considered in [21]:
p ln | ( x k + 1 x ) / ( x k x ) | ln | ( x k x ) / ( x k 1 x ) | , k = 1 , 2 , p ln x k + 1 x / x k x ln x k x / x k 1 x , k = 1 , 2 , p~~(ln|(x_(k+1)-x^(**))//(x_(k)-x^(**))|)/(ln|(x_(k)-x^(**))//(x_(k-1)-x^(**))|),k=1,2,dotsp \approx \frac{\ln \left|\left(x_{k+1}-x^{*}\right) /\left(x_{k}-x^{*}\right)\right|}{\ln \left|\left(x_{k}-x^{*}\right) /\left(x_{k-1}-x^{*}\right)\right|}, k=1,2, \ldotspln|(xk+1x)/(xkx)|ln|(xkx)/(xk1x)|,k=1,2,
subsequently approximated in [5] by
p ln | ( x k + 1 x k ) / ( x k x k 1 ) | ln | ( x k x k 1 ) / ( x k 1 x k 2 ) | , k = 2 , 3 , p ln x k + 1 x k / x k x k 1 ln x k x k 1 / x k 1 x k 2 , k = 2 , 3 , p~~(ln|(x_(k+1)-x_(k))//(x_(k)-x_(k-1))|)/(ln|(x_(k)-x_(k-1))//(x_(k-1)-x_(k-2))|),k=2,3,dotsp \approx \frac{\ln \left|\left(x_{k+1}-x_{k}\right) /\left(x_{k}-x_{k-1}\right)\right|}{\ln \left|\left(x_{k}-x_{k-1}\right) /\left(x_{k-1}-x_{k-2}\right)\right|}, k=2,3, \ldotspln|(xk+1xk)/(xkxk1)|ln|(xkxk1)/(xk1xk2)|,k=2,3,
but since in the presented examples the solution was approximated in only three steps (the convergence order is high), we cannot use these formulas.

References

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  17. Păvăloiu, I.: Aitken-Steffensen-type method for nonsmooth functions (III). Rev. Anal. Numér. Théor. Approx. 32(1), 73-78 (2003)
  18. Quan, Z., Peng, Z., Li, Z., Wenchao, M.: Variants of Steffensen secant method and applications. Appl. Math. Comput. 216(12), 3486-3496 (2010)
  19. Sharma, R.J.: A composite thid order Newton-Steffensen method for solving nonlinear equations. Appl. Math. Comput. 169, 242-246 (2005)
  20. Traub, Y.F.: Iterative Method for Solution of Equations. Prentice-Hall, Englewood Cliffs, NJ (1964)
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  1. I. Păvăloiu ⋅ E. Cătinaş ( \boxtimes )
    "Tiberiu Popoviciu" Institute of Numerical Analysis, Romanian Academy, P.O. Box 68-1, Cluj-Napoca, Romania
    e-mail: ecatinas@ictp.acad.ro
    I. Păvăloiu
    e-mail: pavaloiu@ictp.acad.ro
2013

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