On some iterative methods
for solving nonlinear equations

1. Introduction

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

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

where, $f,$ $g:X\to X,$ $X$ is a Banach space, $f$ is a differentiable operator and $g$ is continuous but nondifferentiable. For this reason the Newton’s method, i.e. the approximation of the solution $x^{\ast }$ of the equation (1) by the sequence ${(x_n)}_{n\ge 0}$ given by

(2) $$x_{n+1}=x_n-{\left(f^{\prime }(x_n)+g^{\prime }(x_n)\right)}^{-1}\left(f(x_n)+g(x_n)\right),n=1,2,\mathrm{\dots },x_0\in X,$$

cannot be applied.

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

(3) $$x_{n+1}=x_n-f^{\prime }{(x_n)}^{-1}\left(f(x_n)+g(x_n)\right),n=1,2,\mathrm{\dots },x_0\in X,$$

or

(3^′) $$x_{n+1}=x_n-A{(x_n)}^{-1}\left(f(x_n)+g(x_n)\right),n=1,2,\mathrm{\dots },x_0\in X,$$

where $A$ is a linear operator approximating $f^{\prime }$. 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:

(4) $x_{n+1}=$ $x_n-{\left(f^{\prime }(x_n)+[x_{n-1},x_n;g]\right)}^{-1}\left(f(x_n)+g(x_n)\right),n=1,2,\mathrm{\dots },$

$x_0,x_1\in X$

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

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 $p,$ is the same as for the method of chord (where $p=\frac{1+\sqrt{5}}{2}\approx 1.618)$, 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 $p=2$).

But, unlike the method of chord, the proposed method has a better rate of convergence, because the use of $f^{\prime }(x_n)$ instead of $[x_{n-1},x_n;f],$ as it is in the method of chord, does not affect the coefficient $c_2$ from the inequalities of the type:

$$\Vert x_{n+1}-x_n\Vert \le c_1{\Vert x_n-x_{n-1}\Vert }^2+c_2\Vert x_n-x_{n-1}\Vert \Vert x_{n-1}-x_{n-2}\Vert ,$$

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 $ℒ(X,X)$ (the Banach space of the linear and bounded operators from $X$ to $X$) is called the first order divided difference of the operator $g:X\to X$ at the points $x_0,$ $y_0\in X$ if the following properties hold:

• $[x_0,y_0;g](y_0-x_0)=g(y_0)-g(x_0),$for $x_0\ne y_0;$

• if $g$ is Fréchet differentiable at $x_0\in X,$ then

  $$[x_0,x_0;g]=g^{\prime }(x_0).$$

Definition 2.

An operator belonging to the space $ℒ(X,ℒ(X,X))$, denoted by $[x_0,y_0,z_0;g]$ is called the second order divided difference of the operator $g:X\to X$ at the points $x_0,y_0,z_0\in X$ if the following properties hold:

• $[x_0,y_0,z_0;g]$ $(z_0-x_0)=[y_0,z_0;g]-[x_0,y_0;g]$ for the distinct points $x_0,y_0,z_0\in X;$

• if $g$ is two times differentiable at $x_0\in X,$ then

  $$[x_0,x_0,x_0;g]=\frac{1}{2}{g}^{\prime \prime }(x_0).$$

We shall denote by the ball having the center at $x_1\in X$ and the radius $r>0$ .

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

Theorem 3.

If there exist the elements $x_0,x_1\in X$ and the positive real numbers $r,$ $l,$ $M,$ $K$ and $\epsilon$ such that the conditions

• the operator $f$ is Fréchet differentiable on $B_r(x_1)$ and $f^{\prime }$ satisfies

  $$\Vert f^{\prime }(x)-f^{\prime }(y)\Vert \le l\Vert x-y\Vert ,\forall x,y\in B_r(x_1);$$

• the operator $g$ is continuous on $B_r(x_1),$

• for any distinct points $x,y\in B_r(x_1)$ there exists the application ${\left(f^{\prime }(y)+[x,y;g]\right)}^{-1}$ and the inequality

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

  is true;

• for any distinct points $x,$ $y,$ $z\in B_r(x_1)$ we have the inequality

  $$\Vert [x,y,z;g]\Vert \le K;$$

• the elements $x_0,$ $x_1$ satisfy

  $$\Vert x_1-x_0\Vert \le M\epsilon ,\text{where }\epsilon =\Vert f(x_1)+g(x_1)\Vert ;$$

• the following relations hold:

  $\Vert x_2-x_1\Vert \le$ $\Vert x_1-x_0\Vert ,\text{with }{x}_2\text{ given by }\text{(}\text{<a href="\#S1.E4" title="In 1. Introduction ‣ On some iterative methods for solving nonlinear equations" class="ltx\_ref"><span class="ltx\_text ltx\_ref\_tag">4</span></a>}\text{)}\text{ for }n=1,$

  $q=$

  $r=$ $\frac{M\epsilon }{q}{\displaystyle \sum _{k=1}^{\mathrm{\infty }}}{q}^{u_k},$

  where ${(u_k)}_{k\ge 0}$ is the Fibonacci’s sequence $u_{k+1}=u_k+u_{k-1},k\ge 1,$ $u_0=u_1=1;$

are fulfilled, then the sequence ${(x_n)}_{n\ge 0}$ generated by (4) is well defined, all its terms belonging to $B_r(x_1).$

Moreover, the following properties are true:

• the sequence ${(x_n)}_{n\ge 0}$ is convergent;

• let $x^{\ast }={lim}_{n\to \mathrm{\infty }}{x}_n$. Then $x^{\ast }$ is a solution of the equation (1);

• we have the a priori error estimates:

  $$\Vert x^{\ast }-x_n\Vert \le \frac{M\epsilon }{q\left(1-q^{\frac{p^n(p-1)}{\sqrt{5}}}\right)}{\left(q^{\frac{1}{\sqrt{5}}}\right)}^{p^n},n\ge 1,p=\frac{1+\sqrt{5}}{2}.$$

Proof.

We shall prove first by induction that, for any $n\ge 2,$

(5) $x_n\in$ $B_r(x_1),$

(6) $\Vert x_n-x_{n-1}\Vert \le$ $\Vert x_{n-1}-x_{n-2}\Vert ,\text{ and}$

(7) $\Vert x_n-x_{n-1}\Vert \le$ $q^{u_{n-1}-1}M\epsilon .$

For $n=2,$ from $v)$ and $vi)$ we infer the above relations.

Let us suppose now that relations (5), (6) and (7) hold for $n=2,3,\mathrm{\dots },k,$ where $k\ge 2.$ Since $x_k,$ $x_{k-1}\in B_r(x_1),$ we can construct $x_{k+1}$ from (4), whence, using $iii),$ we have

$$\Vert x_{k+1}-x_k\Vert =\Vert {\left(f^{\prime }(x_k)+[x_{k-1},x_k;g]\right)}^{-1}\left(f(x_k)+g(x_k)\right)\Vert \le M\Vert f(x_k)+g(x_k)\Vert .$$

For the estimation of $\Vert f(x_k)+g(x_k)\Vert$ we shall rely on the equality

$g(x_k)-g(x_{k-1})-[x_{k-2},x_{k-1};g](x_k-x_{k-1})=$

$=[x_{k-2},x_{k-1},x_k;g](x_k-x_{k-1})(x_k-x_{k-2})$

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

(8) $\Vert g(x_k)-g(x_{k-1})-[x_{k-2},x_{k-1};g](x_k-x_{k-1})\Vert \le$

$\le K\Vert x_k-x_{k-1}\Vert \left(\Vert x_k-x_{k-1}\Vert +\Vert x_{k-1}-x_{k-2}\Vert \right)$

and on the inequality

(9) $$\Vert f(x_k)-f(x_{k-1})-f^{\prime }(x_{k-1})(x_k-x_{k-1})\Vert \le \frac{l}{2}{\Vert x_k-x_{k-1}\Vert }^2,$$

valid because of the assumptions $i)$ concerning $f.$

For $n=k-1,$ by (4), we get

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

whence

$f(x_k)+g(x_k)=$ $f(x_k)-f(x_{k-1})-f^{\prime }(x_{k-1})(x_k-x_{k-1})+g(x_k)-g(x_{k-1})-$

$-[x_{k-2},x_{k-1};g](x_k-x_{k-1}).$

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

$\Vert x_{k+1}-x_k\Vert \le$

$\le M\Vert f(x_k)+g(x_k)\Vert$

$\le \frac{Ml}{2}{\Vert x_k-x_{k-1}\Vert }^2+MK\Vert x_k-x_{k-1}\Vert \left(\Vert x_k-x_{k-1}\Vert +\Vert x_{k-1}-x_{k-2}\Vert \right)$

$\le M\Vert x_k-x_{k-1}\Vert \left(\frac{l}{2}\Vert x_{k-1}-x_{k-2}\Vert +2K\Vert x_{k-1}-x_{k-2}\Vert \right)$

$=M\left(\frac{l}{2}+2K\right)\Vert x_k-x_{k-1}\Vert \Vert x_{k-1}-x_{k-2}\Vert .$

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

$\Vert x_{k+1}-x_k\Vert \le$ $M\left(\frac{l}{2}+2K\right)q^{u_{k-2^{-1}}}M\epsilon \Vert x_k-x_{k-1}\Vert$

$=$ $q^{u_{k-2}}\Vert x_k-x_{k-1}\Vert$

$\Vert x_k-x_{k-1}\Vert ,$

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

$\Vert x_{k+1}-x_k\Vert \le q^{u_{k-2}}\Vert x_k-x_{k-1}\Vert \le q^{u_{k-2}}{q}^{u_{k-1}}M\epsilon =q^{u_k}M\epsilon ,$

that is, (7) for $n=k+1.$

The fact that $x_{k+1}\in B_r(x_1)$ results from:

$\Vert x_{k+1}-x_1\Vert$ $\le \Vert x_2-x_1\Vert +\Vert x_3-x_2\Vert +\mathrm{\cdots }+\Vert x_{k+1}-x_k\Vert$

Now we shall prove that ${(x_n)}_{n\ge 0}$ is a Cauchy sequence, whence $j)$ follows.

It is obvious that

$$u_k=\frac{1}{\sqrt{5}}\left({\left(\frac{1+\sqrt{5}}{2}\right)}^{k+1}-{\left(\frac{1-\sqrt{5}}{2}\right)}^{k+1}\right)\ge \frac{1}{\sqrt{5}}{\left(\frac{1+\sqrt{5}}{2}\right)}^k=\frac{p^k}{\sqrt{5}},$$

for $k\ge 1.$

So, for any $k\ge 1,$ $m\ge 1$ we have

$\Vert x_{k+m}-x_k\Vert$ $\le \Vert x_{k+1}-x_k\Vert +\Vert x_{k+2}-x_{k+1}\Vert +\mathrm{\cdots }+\Vert x_{k+m}-x_{k+m-1}\Vert$

$\le \frac{M\epsilon }{q}(q^{u_k}+q^{u_{k+1}}+\mathrm{\cdots }+q^{u_{k+m-1}})$

$\le \frac{M\epsilon }{q}\left(q^{\frac{p^k}{\sqrt{5}}}+q^{\frac{p^{k+1}}{\sqrt{5}}}+\mathrm{\cdots }+q^{\frac{p^{k+m-1}}{\sqrt{5}}}\right).$

Using Bernoulli’s inequality, it follows

$\Vert x_{k+m}-x_k\Vert$ $\le \frac{M\epsilon }{q}{q}^{\frac{p^k}{\sqrt{5}}}\left(1+q^{\frac{p^{k+1}-p^k}{\sqrt{5}}}+q^{\frac{p^{k+2}-p^k}{\sqrt{5}}}+\mathrm{\cdots }+q^{\frac{p^{k+m-1}-p^k}{\sqrt{5}}}\right)$

$=\frac{M\epsilon }{q}{q}^{\frac{p^k}{\sqrt{5}}}\left(1+q^{\frac{p^k(p-1)}{\sqrt{5}}}+q^{\frac{p^k(p^2-1)}{\sqrt{5}}}+\mathrm{\cdots }+q^{\frac{p^k(p^{m-1}-1)}{\sqrt{5}}}\right)$

$\le \frac{M\epsilon }{q}{q}^{\frac{p^k}{\sqrt{5}}}\left(1+q^{\frac{p^k(p-1)}{\sqrt{5}}}+q^{\frac{p^k(1+2(p-1)-1)}{\sqrt{5}}}+\mathrm{\cdots }+q^{\frac{p^k(1+(m-1)(p-1)-1)}{\sqrt{5}}}\right)$

$=\frac{M\epsilon }{q}{q}^{\frac{p^k}{\sqrt{5}}}\left[1+q^{\frac{p^k(p-1)}{\sqrt{5}}}+{\left(q^{\frac{p^k(p-1)}{\sqrt{5}}}\right)}^2+\mathrm{\cdots }+{\left(q^{\frac{p^k(p-1)}{\sqrt{5}}}\right)}^{m-1}\right]$

$=\frac{M\epsilon }{q}{q}^{\frac{p^k}{\sqrt{5}}}{\displaystyle \frac{1-q^{\frac{p^k(p-1)}{\sqrt{5}}m}}{1-q^{\frac{p^k(p-1)}{\sqrt{5}}}}}.$

Hence

$$\Vert x_{k+m}-x_k\Vert \le \frac{M\epsilon q^{\frac{p^k}{\sqrt{5}}}\left(1-q^{\frac{p^k(p-1)}{\sqrt{5}}m}\right)}{q\left(1-q^{\frac{p^k(p-1)}{\sqrt{5}}}\right)},k\ge 1,$$

and ${(x_n)}_{n\ge 0}$ is a Cauchy sequence.

It follows that ${(x_n)}_{n\ge 0}$ is convergent, and let $x^{\ast }={lim}_{n\to \mathrm{\infty }}{x}_n.$ For $n\to \mathrm{\infty }$ in (4) we get that $x^{\ast }$ is a solution of (1). For $m\to \mathrm{\infty }$ in the above $n$ equality we obtain the very relation $jjj).$

The theorem is proved. ∎

3. Numerical example

Given the system

$$\{\begin{array}{c}3x^{2}y+y^2-1+\left|x-1\right|=0\\ x^4+x{y}^3-1+\left|y\right|=0,\end{array}$$

we shall consider $X+({ℝ}^2,\parallel \cdot {\parallel }_{\mathrm{\infty }}),\parallel x{\parallel }_{\mathrm{\infty }}=\parallel (x^{\prime },x^{\prime \prime }){\parallel }_{\mathrm{\infty }}=\max \{|x^{\prime }|,|x^{\prime \prime }|\},f=(f_1,f_2),g=(g_1,g_2).$ For $x=(x^{\prime },x^{\prime \prime })\in {ℝ}^2$ we take $f_1(x^{\prime },x^{\prime \prime })=3{(x^{\prime })}^2{x}^{\prime \prime }+{(x^{\prime \prime })}^2$ $-1,f_2(x^{\prime },x^{\prime \prime })={(x^{\prime })}^4+x^{\prime }{(x^{\prime \prime })}^3-1,$ $g_1(x^{\prime },x^{\prime \prime })=\left|x^{\prime }-1\right|,g_2(x^{\prime },x^{\prime \prime })=\left|x^{\prime \prime }\right|.$

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

${[x,y;g]}_{i,1}=$ $\frac{g_i(y^{\prime },y^{\prime \prime })-g_i(x^{\prime },y^{\prime \prime })}{y^{\prime }-x^{\prime }},$

${[x,y;g]}_{i,2}=$ $\frac{g_i(x^{\prime },y^{\prime \prime })-g_i(x^{\prime },x^{\prime \prime })}{y^{\prime \prime }-x^{\prime \prime }},i=1,2.$

Using method (3) with $x_0=(1,0)$ we obtain

$n$	$x_n^{\left(1\right)}$	$x_n^{\left(2\right)}$	$\Vert x_n-x_{n-1}\Vert$
$0$	$1$	$0$
$1$	$1$	$\mathrm{0.333\hspace{0.17em}333\hspace{0.17em}333\hspace{0.17em}333\hspace{0.17em}333}$	$3.333\cdot {10}^{-1}$
$2$	$\mathrm{0.906\hspace{0.17em}550\hspace{0.17em}218\hspace{0.17em}340\hspace{0.17em}611}$	$\mathrm{0.354\hspace{0.17em}002\hspace{0.17em}911\hspace{0.17em}208\hspace{0.17em}151}$	$9.344\cdot {10}^{-2}$
$3$	$\mathrm{0.885\hspace{0.17em}328\hspace{0.17em}400\hspace{0.17em}663\hspace{0.17em}412}$	$\mathrm{0.338\hspace{0.17em}027\hspace{0.17em}276\hspace{0.17em}361\hspace{0.17em}332}$	$2.122\cdot {10}^{-2}$
$4$	$\mathrm{0.891\hspace{0.17em}329\hspace{0.17em}556\hspace{0.17em}832\hspace{0.17em}800}$	$\mathrm{0.326\hspace{0.17em}613\hspace{0.17em}976\hspace{0.17em}593\hspace{0.17em}566}$	$1.141\cdot {10}^{-2}$
$5$	$\mathrm{0.895\hspace{0.17em}238\hspace{0.17em}815\hspace{0.17em}463\hspace{0.17em}844}$	$\mathrm{0.326\hspace{0.17em}406\hspace{0.17em}852\hspace{0.17em}843\hspace{0.17em}625}$	$3.909\cdot {10}^{-3}$
$6$	$\mathrm{0.895\hspace{0.17em}154\hspace{0.17em}671\hspace{0.17em}372\hspace{0.17em}635}$	$\mathrm{0.327\hspace{0.17em}730\hspace{0.17em}334\hspace{0.17em}045\hspace{0.17em}043}$	$1.323\cdot {10}^{-3}$
$7$	$\mathrm{0.894\hspace{0.17em}673\hspace{0.17em}743\hspace{0.17em}471\hspace{0.17em}137}$	$\mathrm{0.327\hspace{0.17em}979\hspace{0.17em}154\hspace{0.17em}372\hspace{0.17em}032}$	$4.809\cdot {10}^{-4}$
$8$	$\mathrm{0.894\hspace{0.17em}598\hspace{0.17em}908\hspace{0.17em}977\hspace{0.17em}448}$	$\mathrm{0.327\hspace{0.17em}865\hspace{0.17em}059\hspace{0.17em}348\hspace{0.17em}755}$	$1.140\cdot {10}^{-4}$
$9$	$\mathrm{0.894\hspace{0.17em}643\hspace{0.17em}228\hspace{0.17em}355\hspace{0.17em}865}$	$\mathrm{0.327\hspace{0.17em}815\hspace{0.17em}039\hspace{0.17em}208\hspace{0.17em}286}$	$5.002\cdot {10}^{-5}$
$10$	$\mathrm{0.894\hspace{0.17em}659\hspace{0.17em}993\hspace{0.17em}615\hspace{0.17em}645}$	$\mathrm{0.327\hspace{0.17em}819\hspace{0.17em}889\hspace{0.17em}264\hspace{0.17em}891}$	$1.676\cdot {10}^{-5}$
$11$	$\mathrm{0.894\hspace{0.17em}657\hspace{0.17em}640\hspace{0.17em}195\hspace{0.17em}329}$	$\mathrm{0.327\hspace{0.17em}826\hspace{0.17em}728\hspace{0.17em}208\hspace{0.17em}560}$	$6.838\cdot {10}^{-6}$
$12$	$\mathrm{0.894\hspace{0.17em}655\hspace{0.17em}219\hspace{0.17em}565\hspace{0.17em}091}$	$\mathrm{0.327\hspace{0.17em}827\hspace{0.17em}351\hspace{0.17em}826\hspace{0.17em}856}$	$2.420\cdot {10}^{-6}$
$13$	$\mathrm{0.894\hspace{0.17em}655\hspace{0.17em}074\hspace{0.17em}977\hspace{0.17em}661}$	$\mathrm{0.327\hspace{0.17em}826\hspace{0.17em}643\hspace{0.17em}198\hspace{0.17em}819}$	$7.086\cdot {10}^{-7}$
$\mathrm{\dots }$
$39$	$\mathrm{0.894\hspace{0.17em}655\hspace{0.17em}373\hspace{0.17em}334\hspace{0.17em}687}$	$\mathrm{0.327\hspace{0.17em}826\hspace{0.17em}521\hspace{0.17em}746\hspace{0.17em}298}$	$5.149\cdot {10}^{-19}$

Using the method of chord with $x_0=(5,5),$ $x_1=(1,0),$ we obtain

$n$	$x_n^{\left(1\right)}$	$x_n^{\left(2\right)}$	$\Vert x_n-x_{n-1}\Vert$
$0$	$5$	$5$
$1$	$1$	$0$	$5.000\cdot {10}^{+00}$
$2$	$\mathrm{0.989\hspace{0.17em}800\hspace{0.17em}874\hspace{0.17em}210\hspace{0.17em}782}$	$\mathrm{0.012\hspace{0.17em}627\hspace{0.17em}489\hspace{0.17em}072\hspace{0.17em}365}$	$1.262\cdot {10}^{-02}$
$3$	$\mathrm{0.921\hspace{0.17em}814\hspace{0.17em}765\hspace{0.17em}493\hspace{0.17em}287}$	$\mathrm{0.307\hspace{0.17em}939\hspace{0.17em}916\hspace{0.17em}152\hspace{0.17em}262}$	$2.953\cdot {10}^{-01}$
$4$	$\mathrm{0.900\hspace{0.17em}073\hspace{0.17em}765\hspace{0.17em}669\hspace{0.17em}214}$	$\mathrm{0.325\hspace{0.17em}927\hspace{0.17em}010\hspace{0.17em}697\hspace{0.17em}792}$	$2.174\cdot {10}^{-02}$
$5$	$\mathrm{0.894\hspace{0.17em}939\hspace{0.17em}851\hspace{0.17em}624\hspace{0.17em}105}$	$\mathrm{0.327\hspace{0.17em}725\hspace{0.17em}437\hspace{0.17em}396\hspace{0.17em}226}$	$5.133\cdot {10}^{-03}$
$6$	$\mathrm{0.894\hspace{0.17em}658\hspace{0.17em}420\hspace{0.17em}586\hspace{0.17em}013}$	$\mathrm{0.327\hspace{0.17em}825\hspace{0.17em}363\hspace{0.17em}500\hspace{0.17em}783}$	$2.814\cdot {10}^{-04}$
$7$	$\mathrm{0.894\hspace{0.17em}655\hspace{0.17em}375\hspace{0.17em}077\hspace{0.17em}418}$	$\mathrm{0.327\hspace{0.17em}826\hspace{0.17em}521\hspace{0.17em}051\hspace{0.17em}833}$	$3.045\cdot {10}^{-06}$
$8$	$\mathrm{0.894\hspace{0.17em}655\hspace{0.17em}373\hspace{0.17em}334\hspace{0.17em}698}$	$\mathrm{0.327\hspace{0.17em}826\hspace{0.17em}521\hspace{0.17em}746\hspace{0.17em}293}$	$1.742\cdot {10}^{-09}$
$9$	$\mathrm{0.894\hspace{0.17em}655\hspace{0.17em}373\hspace{0.17em}334\hspace{0.17em}687}$	$\mathrm{0.327\hspace{0.17em}826\hspace{0.17em}521\hspace{0.17em}746\hspace{0.17em}298}$	$1.076\cdot {10}^{-14}$
$10$	$\mathrm{0.894\hspace{0.17em}655\hspace{0.17em}373\hspace{0.17em}334\hspace{0.17em}687}$	$\mathrm{0.327\hspace{0.17em}826\hspace{0.17em}521\hspace{0.17em}746\hspace{0.17em}298}$	$5.421\cdot {10}^{-20}$

Using method (4) with $x_0=(5,5),$ $x_1=(1,0),$ we obtain

$n$	$x_n^{\left(1\right)}$	$x_n^{\left(2\right)}$	$\Vert x_n-x_{n-1}\Vert$
$0$	$5$	$5$
$1$	$1$	$0$	$5.000\cdot {10}^{+00}$
$2$	$0.909090909090909$	$0.363636363636364$	$3.636\cdot {10}^{-01}$
$3$	$0.894886945874111$	$0.329098638203090$	$3.453\cdot {10}^{-02}$
$4$	$0.894655531991499$	$0.327827544745569$	$1.271\cdot {10}^{-03}$
$5$	$0.894655373334793$	$0.327826521746906$	$1.022\cdot {10}^{-06}$
$6$	$0.894655373334687$	$0.327826521746298$	$6.089\cdot {10}^{-13}$
$7$	$0.894655373334687$	$0.327826521746298$	$2.710\cdot {10}^{-20}$

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

Recevied: December 1, 1993   Institutul de Calcul (Academia Română)

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