Abstract

Consider the nonlinear equation \(H(x):=F(x)+G(x)=0\), with \(F\) differentiable and \(G\) continuous, where \(F,G,H:X \rightarrow X\), \(X\) a Banach space. 

The Newton method for solving \(H(x)=0\) cannot be applied, and we propose an iterative method for solving the nonlinear equation, by combining the Newton method (for the differentiable part) with the chord/secant method (for the nondifferentiable part): \[x_{k+1} = \big(F^\prime(x_k)+[x_{k-1},x_k;G]\big)^{-1}(F(x_k)+G(x_k)).\]

We show that the r-convergence order of the method is the same as of the chord/secant method.

We provide some numerical examples and compare different methods for a nonlinear system in \(\mathbb{R}^2\).

Authors

E. Cătinaş, (Tiberiu Popoviciu Institute of Numerical Analysis)

Emil Cătinaş

Keywords

nonlinear equation; Banach space; Newton method; chord method; secant method; combined method; nondifferentiable mapping; nonsmooth mapping; R-convergence order.

Cite this paper as:

E. Cătinaş, On some iterative methods for solving nonlinear equations, Rev. Anal. Numér. Théor. Approx., 23 (1994) no. 1, pp. 47-53

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[1] G. Goldner, M. Balazs, On the method of the chord and on a modification of it for the solution of nonlinear operator equations, Stud. Cerc. Mat., 20 (1968), pp. 981–990 (in Romanian).

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article on journal website

[3] T. Yamamoto, A note on a posteriori error bound of Zabrejko and Nguen for Zincenko’s iteration, Numer. Funct. Anal. Optimiz., 9 (1987) 9&10, pp. 987–994.
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[4] T. Yamamoto, Ball convergence theorems and error estimates for certain iterative methods for nonlinear equations, Japan J. Appl. Math., 7 (1990) no. 1, pp. 131–143.
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[5] X. Chen, T. Yamamoto, Convergence domains of certain iterative methods for solving nonlinear equations, Numer. Funct. Anal. Optimiz., 10 (1989) 1&2, pp. 37–48.
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EMIL CĂTINAŞ
(Cluj-Napoca)

ON SOME ITERATIVE METHODS FOR SOLVING NONLINEAR EQUATIONS

1. INTRODUCTION

In the papers [3], [4] and [5] are studied nonlinear equations having the from:

f(x)+g(x)=0f(x)+g(x)=0 (1)

where, f,g:XX,Xf,g:X\rightarrow X,X is a Banach space, ff is a differentiable operator and gg is continuous but nondifferentiable. For this reason the Newton’s method, i.e. the approximation of the solution xx^{*} of the equation (1) by the sequence (xn)n0\left(x_{n}\right)_{n\geq 0} given by
(2) xn+1=xn(f(xn)+g(xn))1(f(xn)+g(xn)),n=1,2,,x0Xx_{n+1}=x_{n}-\left(f^{\prime}\left(x_{n}\right)+g^{\prime}\left(x_{n}\right)\right)^{-1}\left(f\left(x_{n}\right)+g\left(x_{n}\right)\right),\quad n=1,2,\ldots,x_{0}\in X, cannot be applied.

In the mentioned papers the following Newton-like methods are then considered:

xn+1=xnf(xn)1(f(xn)+g(xn)),n=1,2,,x0X,x_{n+1}=x_{n}-f^{\prime}\left(x_{n}\right)^{-1}\left(f\left(x_{n}\right)+g\left(x_{n}\right)\right),\quad n=1,2,\ldots,x_{0}\in X, (3)

or

xn+1=xnA(xn)1(f(xn)+g(xn)),n=1,2,,x0X,x_{n+1}=x_{n}-A\left(x_{n}\right)^{-1}\left(f\left(x_{n}\right)+g\left(x_{n}\right)\right),\quad n=1,2,\ldots,x_{0}\in X, (\prime)

where AA is a linear operator approximating ff^{\prime}. It is shown that, under certain conditions, these sequences are converging to the solution of (1).

In the present paper, for solving equation (1), we propose the following method:

xn+1=\displaystyle x_{n+1}= xn(f(xn)+[xn1,xn;g])1(f(xn)+g(xn)),n=1,2,,\displaystyle x_{n}-\left(f^{\prime}\left(x_{n}\right)+\left[x_{n-1},x_{n};g\right]\right)^{-1}\left(f\left(x_{n}\right)+g\left(x_{n}\right)\right),\quad n=1,2,\ldots, (4)
x0,x1X\displaystyle x_{0},x_{1}\in X

where by [x,y;g][x,y;g] we have denoted the first order divided difference of gg at the points x,yXx,y\in X.

So, the proposed method is obtained by combining the Newton’s method with the method of chord. The r-convergence order of this method, denoted by pp, is the same as for the method of chord (where p=1+521.618p=\frac{1+\sqrt{5}}{2}\approx 1.618 ), which is greater than the r-order of the methods (3) and (3’) (see also the numerical example), but is less than the r-order of Newton’s method (where usually p=2p=2 ).

But, unlike the method of chord, the proposed method has a better rate of convergence, because the use of f(xn)f^{\prime}\left(x_{n}\right) instead of [xn1,xn;f]\left[x_{n-1},x_{n};f\right], as it is in the method of chord, does not affect the coefficient c2c_{2} from the inequalities of the type:

xn+1xnc1xnxn12+c2xnxn1xn1xn2,\left\|x_{n+1}-x_{n}\right\|\leq c_{1}\left\|x_{n}-x_{n-1}\right\|^{2}+c_{2}\left\|x_{n}-x_{n-1}\right\|\left\|x_{n-1}-x_{n-2}\right\|,

which we shall obtain in the following.

2. THE CONVERGENCE OF THE METHOD

We shall use, as in 1 and 2 the known definitions for the divided differences of an operator.

Definition 1. An operator belonging to the space (X,X)\mathcal{L}(X,X) (the Banach space of the linear and bounded operators from XX to XX ) is called the first order divided difference of the operator g:XXg:X\rightarrow X at the points x0,y0Xx_{0},y_{0}\in X if the following properties hold:
a) [x0,y0;g](y0x0)=g(y0)g(x0)\left[x_{0},y_{0};g\right]\left(y_{0}-x_{0}\right)=g\left(y_{0}\right)-g\left(x_{0}\right), for x0y0x_{0}\neq y_{0};
b) if gg is Fréchet differentiable at x0Xx_{0}\in X, then

[x0,x0;g]=g(x0)\left[x_{0},x_{0};g\right]=g^{\prime}\left(x_{0}\right)

Definition 2. An operator belonging to the space (X,(X,X))\mathcal{L}(X,\mathcal{L}(X,X)), denoted by [x0,y0,z0;g]\left[x_{0},y_{0},z_{0};g\right] is called the second order divided difference of the operator g:XXg:X\rightarrow X at the points x0,y0,z0Xx_{0},y_{0},z_{0}\in X if the following properties hold:
a)[x0,y0,z0;g](z0x0)=[y0,z0;g][x0,y0;g]\left.\mathrm{a}^{\prime}\right)\left[x_{0},y_{0},z_{0};g\right]\left(z_{0}-x_{0}\right)=\left[y_{0},z_{0};g\right]-\left[x_{0},y_{0};g\right] for the distinct points x0,y0,z0Xx_{0},y_{0},z_{0}\in X;
b\mathrm{b}^{\prime} ) if gg is two times differentiable at x0Xx_{0}\in X, then

[x0,x0,x0;g]=12g′′(x0)\left[x_{0},x_{0},x_{0};g\right]=\frac{1}{2}g^{\prime\prime}\left(x_{0}\right)

We shall denote by Br(x1)={xXxx1<r}B_{r}\left(x_{1}\right)=\left\{x\in X\mid\left\|x-x_{1}\right\|<r\right\} the ball having the center at x1Xx_{1}\in X and the radius r>0r>0.

Concerning the convergence of the iterative process (4) we shall prove the following result.

Theorem 3. If there exist the elements x0,x1Xx_{0},x_{1}\in X and the positive real numbers r,l,M,Kr,l,M,K and ε\varepsilon such that the conditions
i) the operator ff is Fréchet differentiable on Br(x1)B_{r}\left(x_{1}\right) and ff^{\prime} satisfies

f(x)f(y)lxy,x,yBr(x1);\left\|f^{\prime}(x)-f^{\prime}(y)\right\|\leq l\|x-y\|,\quad\forall x,y\in B_{r}\left(x_{1}\right);

ii) the operator gg is continuous on Br(x1)B_{r}\left(x_{1}\right),
iii) for any distinct points x,yBr(x1)x,y\in B_{r}\left(x_{1}\right) there exists the application (f(y)+[x,y;g])1\left(f^{\prime}(y)+[x,y;g]\right)^{-1} and the inequality

(f(y)+[x,y;g])1M\left\|\left(f^{\prime}(y)+[x,y;g]\right)^{-1}\right\|\leq M

is true;
iv) for any distinct points x,y,zBr(x1)x,y,z\in B_{r}\left(x_{1}\right) we have the inequality

[x,y,z;g]K\|[x,y,z;g]\|\leq K

v) the elements x0,x1x_{0},x_{1} satisfy

x1x0Mε, where ε=f(x1)+g(x1);\left\|x_{1}-x_{0}\right\|\leq M\varepsilon,\quad\text{ where }\varepsilon=\left\|f\left(x_{1}\right)+g\left(x_{1}\right)\right\|;

vi) the following relations hold:

x2x1\displaystyle\left\|x_{2}-x_{1}\right\| x1x0, with x2 given by (4) for n=1\displaystyle\leq\left\|x_{1}-x_{0}\right\|,\quad\text{ with }x_{2}\text{ given by (4) for }n=1
q\displaystyle q =M2ε(l2+2K)<1, and\displaystyle=M^{2}\varepsilon\left(\frac{l}{2}+2K\right)<1,\text{ and }
r\displaystyle r =Mεqk=1quk\displaystyle=\frac{M\varepsilon}{q}\sum_{k=1}^{\infty}q^{u_{k}}

where (uk)k0\left(u_{k}\right)_{k\geq 0} is the Fibonacci’s sequence uk+1=uk+uk1,k1u_{k+1}=u_{k}+u_{k-1},k\geq 1, u0=u1=1;u_{0}=u_{1}=1;
are fulfilled, then the sequence (xn)n0\left(x_{n}\right)_{n\geq 0} generated by (4) is well defined, all its terms belonging to Br(x1)B_{r}\left(x_{1}\right).

Moreover, the following properties are true:
j) the sequence (xn)n0\left(x_{n}\right)_{n\geq 0} is convergent;
jj) let x=limnxnx^{*}=\lim_{n\rightarrow\infty}x_{n}. Then xx^{*} is a solution of the equation (1);
jjj) we have the a priori error estimates:

xxnMεq(1qpn(p1)5)(q15)pn,n1,p=1+52\left\|x^{*}-x_{n}\right\|\leq\frac{M\varepsilon}{q\left(1-q^{\frac{p^{n}(p-1)}{\sqrt{5}}}\right)}\left(q^{\frac{1}{\sqrt{5}}}\right)^{p^{n}},\quad n\geq 1,p=\frac{1+\sqrt{5}}{2}

Proof. We shall prove first by induction that, for any n2n\geq 2,

xn\displaystyle x_{n} Br(x1)\displaystyle\in B_{r}\left(x_{1}\right) (5)
xnxn1\displaystyle\left\|x_{n}-x_{n-1}\right\| xn1xn2, and\displaystyle\leq\left\|x_{n-1}-x_{n-2}\right\|,\text{ and } (6)
xnxn1\displaystyle\left\|x_{n}-x_{n-1}\right\| qun11Mε.\displaystyle\leq q^{u_{n-1}-1}M\varepsilon. (7)

For n=2n=2, from vv ) and vi)vi) we infer the above relations.
Let us suppose now that relations (5), (6) and (7) hold for n=2,3,,kn=2,3,\ldots,k, where k2k\geq 2. Since xk,xk1Br(x1)x_{k},x_{k-1}\in B_{r}\left(x_{1}\right), we can construct xk+1x_{k+1} from (4), whence, using iii)iii), we have
xk+1xk=(f(xk)+[xk1,xk;g])1(f(xk)+g(xk))Mf(xk)+g(xk)\left\|x_{k+1}-x_{k}\right\|=\left\|\left(f^{\prime}\left(x_{k}\right)+\left[x_{k-1},x_{k};g\right]\right)^{-1}\left(f\left(x_{k}\right)+g\left(x_{k}\right)\right)\right\|\leq M\left\|f\left(x_{k}\right)+g\left(x_{k}\right)\right\|.
For the estimation of f(xk)+g(xk)\left\|f\left(x_{k}\right)+g\left(x_{k}\right)\right\| we shall rely on the equality

g(xk)g(xk1)[xk2,xk1;g](xkxk1)=\displaystyle g\left(x_{k}\right)-g\left(x_{k-1}\right)-\left[x_{k-2},x_{k-1};g\right]\left(x_{k}-x_{k-1}\right)=
=[xk2,xk1,xk;g](xkxk1)(xkxk2)\displaystyle=\left[x_{k-2},x_{k-1},x_{k};g\right]\left(x_{k}-x_{k-1}\right)\left(x_{k}-x_{k-2}\right)

(easily obtained from Definition 1 and Definition 2), which imply, using iv),

g(xk)g(xk1)[xk2,xk1;g](xkxk1)\displaystyle\left\|g\left(x_{k}\right)-g\left(x_{k-1}\right)-\left[x_{k-2},x_{k-1};g\right]\left(x_{k}-x_{k-1}\right)\right\|\leq (8)
Kxkxk1(xkxk1+xk1xk2)\displaystyle\leq K\left\|x_{k}-x_{k-1}\right\|\left(\left\|x_{k}-x_{k-1}\right\|+\left\|x_{k-1}-x_{k-2}\right\|\right)

and on the inequality

f(xk)f(xk1)f(xk1)(xkxk1)l2xkxk12\left\|f\left(x_{k}\right)-f\left(x_{k-1}\right)-f^{\prime}\left(x_{k-1}\right)\left(x_{k}-x_{k-1}\right)\right\|\leq\frac{l}{2}\left\|x_{k}-x_{k-1}\right\|^{2} (9)

valid because of the assumptions i)i) concerning ff.
For n=k1n=k-1, by (4), we get

(f(xk1)+[xk2,xk1;g])(xkxk1)f(xk1)g(xk1)=0,-\left(f^{\prime}\left(x_{k-1}\right)+\left[x_{k-2},x_{k-1};g\right]\right)\left(x_{k}-x_{k-1}\right)-f\left(x_{k-1}\right)-g\left(x_{k-1}\right)=0,

whence

f(xk)+g(xk)=\displaystyle f\left(x_{k}\right)+g\left(x_{k}\right)= f(xk)f(xk1)f(xk1)(xkxk1)+g(xk)g(xk1)\displaystyle f\left(x_{k}\right)-f\left(x_{k-1}\right)-f^{\prime}\left(x_{k-1}\right)\left(x_{k}-x_{k-1}\right)+g\left(x_{k}\right)-g\left(x_{k-1}\right)-
[xk2,xk1;g](xkxk1)\displaystyle-\left[x_{k-2},x_{k-1};g\right]\left(x_{k}-x_{k-1}\right)

The above relation, together with (8), (9) and (6) for n=kn=k imply

xk+1xk\displaystyle\left\|x_{k+1}-x_{k}\right\|\leq
Mf(xk)+g(xk)\displaystyle\leq M\left\|f\left(x_{k}\right)+g\left(x_{k}\right)\right\|
Ml2xkxk12+MKxkxk1(xkxk1+xk1xk2)\displaystyle\leq\frac{Ml}{2}\left\|x_{k}-x_{k-1}\right\|^{2}+MK\left\|x_{k}-x_{k-1}\right\|\left(\left\|x_{k}-x_{k-1}\right\|+\left\|x_{k-1}-x_{k-2}\right\|\right)
Mxkxk1(l2xk1xk2+2Kxk1xk2)\displaystyle\leq M\left\|x_{k}-x_{k-1}\right\|\left(\frac{l}{2}\left\|x_{k-1}-x_{k-2}\right\|+2K\left\|x_{k-1}-x_{k-2}\right\|\right)
=M(l2+2K)xkxk1xk1xk2.\displaystyle=M\left(\frac{l}{2}+2K\right)\left\|x_{k}-x_{k-1}\right\|\left\|x_{k-1}-x_{k-2}\right\|.

From the hypothesis of the induction we have on one hand that

xk+1xk\displaystyle\left\|x_{k+1}-x_{k}\right\| M(l2+2K)quk21Mεxkxk1\displaystyle\leq M\left(\frac{l}{2}+2K\right)q^{u_{k-2}-1}M\varepsilon\left\|x_{k}-x_{k-1}\right\|
=quk2xkxk1\displaystyle=q^{u_{k-2}}\left\|x_{k}-x_{k-1}\right\|
<xkxk1\displaystyle<\left\|x_{k}-x_{k-1}\right\|

that is, (6) for n=k+1n=k+1, and, on the other hand

xk+1xkquk2xkxk1quk2quk1Mε=qukMε\left\|x_{k+1}-x_{k}\right\|\leq q^{u_{k-2}}\left\|x_{k}-x_{k-1}\right\|\leq q^{u_{k-2}}q^{u_{k-1}}M\varepsilon=q^{u_{k}}M\varepsilon

that is, (7) for n=k+1n=k+1.
The fact that xk+1Br(x1)x_{k+1}\in B_{r}\left(x_{1}\right) results from:

xk+1x1\displaystyle\left\|x_{k+1}-x_{1}\right\| x2x1+x3x2++xk+1xk\displaystyle\leq\left\|x_{2}-x_{1}\right\|+\left\|x_{3}-x_{2}\right\|+\cdots+\left\|x_{k+1}-x_{k}\right\|
Mεq(qu1+qu2++quk)<r\displaystyle\leq\frac{M\varepsilon}{q}\left(q^{u_{1}}+q^{u_{2}}+\cdots+q^{u_{k}}\right)<r

Now we shall prove that (xn)n0\left(x_{n}\right)_{n\geq 0} is a Cauchy sequence, whence jj ) follows.
It is obvious that

uk=15((1+52)k+1(152)k+1)15(1+52)k=pk5u_{k}=\frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2}\right)^{k+1}-\left(\frac{1-\sqrt{5}}{2}\right)^{k+1}\right)\geq\frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^{k}=\frac{p^{k}}{\sqrt{5}}

for k1k\geq 1.

So, for any k1,m1k\geq 1,m\geq 1 we have

xk+mxk\displaystyle\left\|x_{k+m}-x_{k}\right\| xk+1xk+xk+2xk+1++xk+mxk+m1\displaystyle\leq\left\|x_{k+1}-x_{k}\right\|+\left\|x_{k+2}-x_{k+1}\right\|+\cdots+\left\|x_{k+m}-x_{k+m-1}\right\|
Mεq(quk+quk+1++quk+m1)\displaystyle\leq\frac{M\varepsilon}{q}\left(q^{u_{k}}+q^{u_{k+1}}+\cdots+q^{u_{k+m-1}}\right)
Mεq(qpk5+qpk+15++qpk+m15)\displaystyle\leq\frac{M\varepsilon}{q}\left(q^{\frac{p^{k}}{\sqrt{5}}}+q^{\frac{p^{k+1}}{\sqrt{5}}}+\cdots+q^{\frac{p^{k+m-1}}{\sqrt{5}}}\right)

Using Bernoulli’s inequality, it follows

xk+mxk\displaystyle\left\|x_{k+m}-x_{k}\right\| Mεqqpk5(1+qpk+1pk5+qpk+2pk5++qpk+m1pk5)\displaystyle\leq\frac{M\varepsilon}{q}q^{\frac{p^{k}}{\sqrt{5}}}\left(1+q^{\frac{p^{k+1}-p^{k}}{\sqrt{5}}}+q^{\frac{p^{k+2}-p^{k}}{\sqrt{5}}}+\cdots+q^{\frac{p^{k+m-1}-p^{k}}{\sqrt{5}}}\right)
=Mεqqpk5(1+qpk(p1)5+qpk(p21)5++qpk(pm11)5)\displaystyle=\frac{M\varepsilon}{q}q^{\frac{p^{k}}{\sqrt{5}}}\left(1+q^{\frac{p^{k}(p-1)}{\sqrt{5}}}+q^{\frac{p^{k}\left(p^{2}-1\right)}{\sqrt{5}}}+\cdots+q^{\frac{p^{k}\left(p^{m-1}-1\right)}{\sqrt{5}}}\right)
Mεqqpk5(1+qpk(p1)5+qpk(1+2(p1)1)5++qpk(1+(m1)(p1)1)5)\displaystyle\leq\frac{M\varepsilon}{q}q^{\frac{p^{k}}{\sqrt{5}}}\left(1+q^{\frac{p^{k}(p-1)}{\sqrt{5}}}+q^{\frac{p^{k}(1+2(p-1)-1)}{\sqrt{5}}}+\cdots+q^{\frac{p^{k}(1+(m-1)(p-1)-1)}{\sqrt{5}}}\right)
=Mεqqpk5[1+qpk(p1)5+(qpk(p1)5)2++(qpk(p1)5)m1]\displaystyle=\frac{M\varepsilon}{q}q^{\frac{p^{k}}{\sqrt{5}}}\left[1+q^{\frac{p^{k}(p-1)}{\sqrt{5}}}+\left(q^{\frac{p^{k}(p-1)}{\sqrt{5}}}\right)^{2}+\cdots+\left(q^{\frac{p^{k}(p-1)}{\sqrt{5}}}\right)^{m-1}\right]
=Mεqqpk51qpk(p1)5m1qpk(p1)5.\displaystyle=\frac{M\varepsilon}{q}q^{\frac{p^{k}}{\sqrt{5}}}\frac{1-q^{\frac{p^{k}(p-1)}{\sqrt{5}}m}}{1-q^{\frac{p^{k}(p-1)}{\sqrt{5}}}}.

Hence

xk+mxkMεqpk5(1qpk(p1)5m)q(1qpk(p1)5),k1\left\|x_{k+m}-x_{k}\right\|\leq\frac{M\varepsilon q^{\frac{p^{k}}{\sqrt{5}}}\left(1-q^{\frac{p^{k}(p-1)}{\sqrt{5}}m}\right)}{q\left(1-q^{\frac{p^{k}(p-1)}{\sqrt{5}}}\right)},\quad k\geq 1

and (xn)n0\left(x_{n}\right)_{n\geq 0} is a Cauchy sequence.
It follows that (xn)n0\left(x_{n}\right)_{n\geq 0} is convergent, and let x=limnxnx^{*}=\lim_{n\rightarrow\infty}x_{n}. For nn\rightarrow\infty in (4) we get that xx^{*} is a solution of (1). For mm\rightarrow\infty in the above nn equality we obtain the very relation jjjjjj ).

The theorem is proved.

3. NUMERICAL EXAMPLE

Given the system

{3x2y+y21+|x1|=0x4+xy31+|y|=0\left\{\begin{array}[]{c}3x^{2}y+y^{2}-1+|x-1|=0\\ x^{4}+xy^{3}-1+|y|=0\end{array}\right.

we shall consider X+(2,),x=(x,x′′)=max{|x|,|x′′|},f=(f1,f2),g=(g1,g2)X+\left(\mathbb{R}^{2},\|\cdot\|_{\infty}\right),\|x\|_{\infty}=\left\|\left(x^{\prime},x^{\prime\prime}\right)\right\|_{\infty}=\max\left\{\left|x^{\prime}\right|,\left|x^{\prime\prime}\right|\right\},f=\left(f_{1},f_{2}\right),g=\left(g_{1},g_{2}\right). For x=(x,x′′)2x=\left(x^{\prime},x^{\prime\prime}\right)\in\mathbb{R}^{2} we take f1(x,x′′)=3(x)2x′′+(x′′)21,f2(x,x′′)=(x)4+x(x′′)31,g1(x,x′′)=|x1|,g2(x,x′′)=|x′′|f_{1}\left(x^{\prime},x^{\prime\prime}\right)=3\left(x^{\prime}\right)^{2}x^{\prime\prime}+\left(x^{\prime\prime}\right)^{2}-1,f_{2}\left(x^{\prime},x^{\prime\prime}\right)=\left(x^{\prime}\right)^{4}+x^{\prime}\left(x^{\prime\prime}\right)^{3}-1,g_{1}\left(x^{\prime},x^{\prime\prime}\right)=\left|x^{\prime}-1\right|,g_{2}\left(x^{\prime},x^{\prime\prime}\right)=\left|x^{\prime\prime}\right|.

We shall take [x,y;g]M2×2()[x,y;g]\in M_{2\times 2}(\mathbb{R}) as

[x,y;g]i,1\displaystyle{[x,y;g]_{i,1}} =gi(y,y′′)gi(x,y′′)yx\displaystyle=\frac{g_{i}\left(y^{\prime},y^{\prime\prime}\right)-g_{i}\left(x^{\prime},y^{\prime\prime}\right)}{y^{\prime}-x^{\prime}}
[x,y;g]i,2\displaystyle{[x,y;g]_{i,2}} =gi(x,y′′)gi(x,x′′)y′′x′′,i=1,2.\displaystyle=\frac{g_{i}\left(x^{\prime},y^{\prime\prime}\right)-g_{i}\left(x^{\prime},x^{\prime\prime}\right)}{y^{\prime\prime}-x^{\prime\prime}},\quad i=1,2.

Using method (3) with x0=(1,0)x_{0}=(1,0) we obtain

nn xn(1)x_{n}^{(1)} xn(2)x_{n}^{(2)} xnxn1\left\|x_{n}-x_{n-1}\right\|
0 1 0
1 1 0.333333333333333 3.3331013.333\cdot 10^{-1}
2 0.906550218340611 0.354002911208151 9.3441029.344\cdot 10^{-2}
3 0.885328400663412 0.338027276361332 2.1221022.122\cdot 10^{-2}
4 0.891329556832800 0.326613976593566 1.1411021.141\cdot 10^{-2}
5 0.895238815463844 0.326406852843625 3.9091033.909\cdot 10^{-3}
6 0.895154671372635 0.327730334045043 1.3231031.323\cdot 10^{-3}
7 0.894673743471137 0.327979154372032 4.8091044.809\cdot 10^{-4}
8 0.894598908977448 0.327865059348755 1.1401041.140\cdot 10^{-4}
9 0.894643228355865 0.327815039208286 5.0021055.002\cdot 10^{-5}
10 0.894659993615645 0.327819889264891 1.6761051.676\cdot 10^{-5}
11 0.894657640195329 0.327826728208560 6.8381066.838\cdot 10^{-6}
12 0.894655219565091 0.327827351826856 2.4201062.420\cdot 10^{-6}
13 0.894655074977661 0.327826643198819 7.0861077.086\cdot 10^{-7}
39 0.894655373334687 0.327826521746298 5.14910195.149\cdot 10^{-19}

Using the method of chord with x0=(5,5),x1=(1,0)x_{0}=(5,5),x_{1}=(1,0), we obtain

nn xn(1)x_{n}^{(1)} xn(2)x_{n}^{(2)} xnxn1\left\|x_{n}-x_{n-1}\right\|
0 5 5
1 1 0 5.00010+005.000\cdot 10^{+00}
2 0.989800874210782 0.012627489072365 1.26210021.262\cdot 10^{-02}
3 0.921814765493287 0.307939916152262 2.95310012.953\cdot 10^{-01}
4 0.900073765669214 0.325927010697792 2.17410022.174\cdot 10^{-02}
5 0.894939851624105 0.327725437396226 5.13310035.133\cdot 10^{-03}
6 0.894658420586013 0.327825363500783 2.81410042.814\cdot 10^{-04}
7 0.894655375077418 0.327826521051833 3.04510063.045\cdot 10^{-06}
8 0.894655373334698 0.327826521746293 1.74210091.742\cdot 10^{-09}
9 0.894655373334687 0.327826521746298 1.07610141.076\cdot 10^{-14}
10 0.894655373334687 0.327826521746298 5.42110205.421\cdot 10^{-20}

Using method (4) with x0=(5,5),x1=(1,0)x_{0}=(5,5),x_{1}=(1,0), we obtain

nn xn(1)x_{n}^{(1)} xn(2)x_{n}^{(2)} xnxn1\left\|x_{n}-x_{n-1}\right\|
0 5 5
1 1 0 5.00010+005.000\cdot 10^{+00}
2 0.909090909090909 0.363636363636364 3.63610013.636\cdot 10^{-01}
3 0.894886945874111 0.329098638203090 3.45310023.453\cdot 10^{-02}
4 0.894655531991499 0.327827544745569 1.27110031.271\cdot 10^{-03}
5 0.894655373334793 0.327826521746906 1.02210061.022\cdot 10^{-06}
6 0.894655373334687 0.327826521746298 6.08910136.089\cdot 10^{-13}
7 0.894655373334687 0.327826521746298 2.71010202.710\cdot 10^{-20}

It can be easily seen that, given these data, method (4) is converging faster than (3) and than the method of chord.

REFERENCES

[1] G. Goldner, M. Balázs, On the method of the chord and on a modification of it for the solution of nonlinear operator equations, Stud. Cerc. Mat., 20 (1968), pp. 981-990 (in Romanian).
[2] G. Goldner, M. Balázs, Remarks on divided differences and method of chords, Rev. Anal. Numer. Teoria Aproximaţiei, 3 (1974) no. 1, pp. 19-30 (in Romanian). ©
[3] T. Yamamoto, A note on a posteriori error bound of Zabrejko and Nguen for Zincenko’s iteration, Numer. Funct. Anal. Optimiz., 9 (1987) 9&10, pp. 987-994. ©

Recevied: December 1, 1993 Institutul de Calcul (Academia Română)
Str. Republicii Nr. 37
P.O. Box 68

3400 Cluj-Napoca
Romania

1994

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