MSC 2000. 65H05.
Keywords. Aitken-Steffensen iterations.

In this paper we consider the Steffensen method for approximating the solutions of the equations of the form

with $f:[a,b]\to \mathbb{R},a,b\in \mathbb{R},a<b$$f:[a,b]\to \mathbb{R},a,b\in \mathbb{R},a<b$f:[a,b]rarrR,a,b inR,a < bf:[a, b] \rightarrow \mathbb{R}, a, b \in \mathbb{R}, a<b$f:[a,b]\to \mathbb{R},a,b\in \mathbb{R},a<b$. Let $g:[a,b]\to \mathbb{R}$$g:[a,b]\to \mathbb{R}$g:[a,b]rarrRg:[a, b] \rightarrow \mathbb{R}$g:[a,b]\to \mathbb{R}$ be such that the equation

is equivalent to (1).
As it is well known, the Steffensen method consists in approximating the solution $x^{\ast }$$x^{\ast }$x^(**)x^{*}$x^{\ast }$ of (1) by the sequence ${\left(x_n\right)}_{n\ge 1}$${\left(x_n\right)}_{n\ge 1}$(x_(n))_(n >= 1)\left(x_{n}\right)_{n \geq 1}${\left(x_n\right)}_{n\ge 1}$ given by

We are interested in the following in the conditions under which the sequences ${\left(x_n\right)}_{n\ge 1}$${\left(x_n\right)}_{n\ge 1}$(x_(n))_(n >= 1)\left(x_{n}\right)_{n \geq 1}${\left(x_n\right)}_{n\ge 1}$ and ${\left(g\left(x_n\right)\right)}_{n\ge 1}$${\left(g\left(x_n\right)\right)}_{n\ge 1}$(g(x_(n)))_(n >= 1)\left(g\left(x_{n}\right)\right)_{n \geq 1}${\left(g\left(x_n\right)\right)}_{n\ge 1}$ are monotone, and offer bilateral approximations to $x^{\ast }$$x^{\ast }$x^(**)x^{*}$x^{\ast }$. The importance of such sequences resides in the fact that at each iteration step we obtain a rigorous error bound. We shall construct the function $g$$g$gg$g$ without assuming that $f$$f$ff$f$ is differentiable on the whole interval $[a,b]$$[a,b]$[a,b][a, b]$[a,b]$. In this sense, we shall use the divided differences of $f$$f$ff$f$.

Regarding the monotony and convexity of the function $f$$f$ff$f$ we shall adopt the following definitions.

Definition 1. The function $f$$f$ff$f$ is nondecreasing (increasing) on $[a,b]$$[a,b]$[a,b][a, b]$[a,b]$ if $[u,v;f]\ge 0(>0)\mathrm{\forall }u,v\in [a,b]$$[u,v;f]\ge 0(>0)\mathrm{\forall }u,v\in [a,b]$[u,v;f] >= 0( > 0)AA u,v in[a,b][u, v ; f] \geq 0(>0) \forall u, v \in[a, b]$[u,v;f]\ge 0(>0)\mathrm{\forall }u,v\in [a,b]$, while $f$$f$ff$f$ is nonincreasing (decreasing) if $[u,v;f]\le 0(<0)\mathrm{\forall }u,v\in [a,b]$$[u,v;f]\le 0(<0)\mathrm{\forall }u,v\in [a,b]$[u,v;f] <= 0( < 0)AA u,v in[a,b][u, v ; f] \leq 0(<0) \forall u, v \in[a, b]$[u,v;f]\le 0(<0)\mathrm{\forall }u,v\in [a,b]$.

Definition 2. The function $f$$f$ff$f$ is nonconcave (convex) on $[a,b]$$[a,b]$[a,b][a, b]$[a,b]$ if

and is nonconvex (concave) if

Consider the function $p_{x_0}:[a,b]\mathrm{\setminus }\left\{x_0\right\}\to \mathbb{R}$$p_{x_0}:[a,b]\mathrm{\setminus }\left\{x_0\right\}\to \mathbb{R}$p_(x_(0)):[a,b]\\{x_(0)}rarrRp_{x_{0}}:[a, b] \backslash\left\{x_{0}\right\} \rightarrow \mathbb{R}$p_{x_0}:[a,b]\mathrm{\setminus }\left\{x_0\right\}\to \mathbb{R}$ given by

Recall the following result:
Theorem 3. [3, p. 290].
a) If $f$$f$ff$f$ is nonconcave on $[a,b]$$[a,b]$[a,b][a, b]$[a,b]$ then $p_{x_0}$$p_{x_0}$p_(x_(0))p_{x_{0}}$p_{x_0}$ is nondecreasing on $[a,b]$$[a,b]$[a,b][a, b]$[a,b]$;
b) If $f$$f$ff$f$ is convex on $[a,b]$$[a,b]$[a,b][a, b]$[a,b]$ then $p_{x_0}$$p_{x_0}$p_(x_(0))p_{x_{0}}$p_{x_0}$ is increasing on $[a,b]$$[a,b]$[a,b][a, b]$[a,b]$;
c) If $f$$f$ff$f$ is nonconvex on $[a,b]$$[a,b]$[a,b][a, b]$[a,b]$ then $p_{x_0}$$p_{x_0}$p_(x_(0))p_{x_{0}}$p_{x_0}$ is nonincreasing on $[a,b]$$[a,b]$[a,b][a, b]$[a,b]$;
d) If $f$$f$ff$f$ is concave on $[a,b]$$[a,b]$[a,b][a, b]$[a,b]$ then $p_{x_0}$$p_{x_0}$p_(x_(0))p_{x_{0}}$p_{x_0}$ is decreasing on $[a,b]$$[a,b]$[a,b][a, b]$[a,b]$.

Consider now $u,v,w,t\in [a,b]$$u,v,w,t\in [a,b]$u,v,w,t in[a,b]u, v, w, t \in[a, b]$u,v,w,t\in [a,b]$ such that $u\le min\{v,w,t\}$$u\le min\{v,w,t\}$u <= min{v,w,t}u \leq \min \{v, w, t\}$u\le min\{v,w,t\}$ and $t\ge max\{u,v,w\}$$t\ge max\{u,v,w\}$t >= max{u,v,w}t \geq \max \{u, v, w\}$t\ge max\{u,v,w\}$. The following result is known:

Lemma 4. [8]. If $f$$f$ff$f$ is nonconcave (convex) on $[a,b]$$[a,b]$[a,b][a, b]$[a,b]$ then the following relation holds:

An inequality analogous to (5) holds when $f$$f$ff$f$ is nonconvex (concave) on $[a,b]$$[a,b]$[a,b][a, b]$[a,b]$.

We shall consider that $f$$f$ff$f$ obeys the following hypotheses:
i. $f$$f$ff$f$ is continuous at $a$$a$aa$a$ and $b$$b$bb$b$;
ii. $f(a)\cdot f(b)<0$$f(a)\cdot f(b)<0$f(a)*f(b) < 0f(a) \cdot f(b)<0$f(a)\cdot f(b)<0$;
iii. $f$$f$ff$f$ is increasing on $[a,b]$$[a,b]$[a,b][a, b]$[a,b]$;
iv. $f$$f$ff$f$ is convex on $[a,b]$$[a,b]$[a,b][a, b]$[a,b]$ and $f$$f$ff$f$ is continuous at $a$$a$aa$a$ and $b$$b$bb$b$;
v. $f$$f$ff$f$ is differentiable at $x^{\ast }$$x^{\ast }$x^(**)x^{*}$x^{\ast }$, the solution of (1), and $x^{\ast }\in (a,b)$$x^{\ast }\in (a,b)$x^(**)in(a,b)x^{*} \in(a, b)$x^{\ast }\in (a,b)$.

Remark 1. Hypotheses iv. ensures the continuity of $f$$f$ff$f$ on $(a,b)$$(a,b)$(a,b)(a, b)$(a,b)$ (see, e.g. [3, p. 295]).

Remark 2. Hypotheses i.-iv. ensure the existence and the unicity of the solution $x^{\ast }\in (a,b)$$x^{\ast }\in (a,b)$x^(**)in(a,b)x^{*} \in(a, b)$x^{\ast }\in (a,b)$ of equation (1).

Let $\alpha ,\beta \in (a,b)$$\alpha ,\beta \in (a,b)$alpha,beta in(a,b)\alpha, \beta \in(a, b)$\alpha ,\beta \in (a,b)$ be such that $f(\alpha )<0$$f(\alpha )<0$f(alpha) < 0f(\alpha)<0$f(\alpha )<0$ and $f(\beta )>0$$f(\beta )>0$f(beta) > 0f(\beta)>0$f(\beta )>0$ (their existence is ensured by hypotheses i.-iv.).

Consider the function $g:[\alpha ,\beta ]\to \mathbb{R}$$g:[\alpha ,\beta ]\to \mathbb{R}$g:[alpha,beta]rarrRg:[\alpha, \beta] \rightarrow \mathbb{R}$g:[\alpha ,\beta ]\to \mathbb{R}$ given by

By iii. and iv. and Lemma 4 it follows that

i.e., $g$$g$gg$g$ is decreasing.
We shall make the following hypotheses regarding the initial approximation $x_1$$x_1$x_(1)x_{1}$x_1$ in (3):
а) $f\left(x_1\right)<0$$f\left(x_1\right)<0$f(x_(1)) < 0f\left(x_{1}\right)<0$f\left(x_1\right)<0$;
b) $g\left(x_1\right)<\beta$$g\left(x_1\right)<\beta$g(x_(1)) < betag\left(x_{1}\right)<\beta$g\left(x_1\right)<\beta$.

Regarding the convergence of the Steffensen method (3) we prove the following result:

Theorem 5. Assume that $f$$f$ff$f$ obeys assumptions i.-v., that the function $g$$g$gg$g$ is given by (6) and $x_1$$x_1$x_(1)x_{1}$x_1$ obeys a) and b). Then the sequence ${\left(x_n\right)}_{n\ge 1}$${\left(x_n\right)}_{n\ge 1}$(x_(n))_(n >= 1)\left(x_{n}\right)_{n \geq 1}${\left(x_n\right)}_{n\ge 1}$ and ${\left(g\left(x_n\right)\right)}_{n\ge 1}$${\left(g\left(x_n\right)\right)}_{n\ge 1}$(g(x_(n)))_(n >= 1)\left(g\left(x_{n}\right)\right)_{n \geq 1}${\left(g\left(x_n\right)\right)}_{n\ge 1}$ generated by (3) satisfy the following properties:
j. the sequence ${\left(x_n\right)}_{n\ge 1}$${\left(x_n\right)}_{n\ge 1}$(x_(n))_(n >= 1)\left(x_{n}\right)_{n \geq 1}${\left(x_n\right)}_{n\ge 1}$ is increasing and bounded;
jj. the sequence ${\left(g\left(x_n\right)\right)}_{n\ge 1}$${\left(g\left(x_n\right)\right)}_{n\ge 1}$(g(x_(n)))_(n >= 1)\left(g\left(x_{n}\right)\right)_{n \geq 1}${\left(g\left(x_n\right)\right)}_{n\ge 1}$ is decreasing and bounded;
jjj. the following is true:

Proof. By (6) we get that $x^{\ast }=g\left(x^{\ast }\right)$$x^{\ast }=g\left(x^{\ast }\right)$x^(**)=g(x^(**))x^{*}=g\left(x^{*}\right)$x^{\ast }=g\left(x^{\ast }\right)$. Since $x_1<x^{\ast }$$x_1<x^{\ast }$x_(1) < x^(**)x_{1}<x^{*}$x_1<x^{\ast }$, and $g$$g$gg$g$ is decreasing, it follows $g\left(x_1\right)>g\left(x^{\ast }\right)=x^{\ast }$$g\left(x_1\right)>g\left(x^{\ast }\right)=x^{\ast }$g(x_(1)) > g(x^(**))=x^(**)g\left(x_{1}\right)>g\left(x^{*}\right)=x^{*}$g\left(x_1\right)>g\left(x^{\ast }\right)=x^{\ast }$ and so $x_1<x^{\ast }<g\left(x_1\right)$$x_1<x^{\ast }<g\left(x_1\right)$x_(1) < x^(**) < g(x_(1))x_{1}<x^{*}<g\left(x_{1}\right)$x_1<x^{\ast }<g\left(x_1\right)$.

We show now that $x_2$$x_2$x_(2)x_{2}$x_2$ given by (3) verifies $x_1<x_2<x^{\ast }$$x_1<x_2<x^{\ast }$x_(1) < x_(2) < x^(**)x_{1}<x_{2}<x^{*}$x_1<x_2<x^{\ast }$. Since $f\left(x_1\right)<0$$f\left(x_1\right)<0$f(x_(1)) < 0f\left(x_{1}\right)<0$f\left(x_1\right)<0$ and $f$$f$ff$f$ is increasing, it follows $x_2=x_1-\frac{f\left(x_1\right)}{[x_1,g\left(x_1\right);f]}>x_1$$x_2=x_1-\frac{f\left(x_1\right)}{\left[x_1,g\left(x_1\right);f\right]}>x_1$x_(2)=x_(1)-(f(x_(1)))/([x_(1),g(x_(1));f]) > x_(1)x_{2}=x_{1}-\frac{f\left(x_{1}\right)}{\left[x_{1}, g\left(x_{1}\right) ; f\right]}>x_{1}$x_2=x_1-\frac{f\left(x_1\right)}{[x_1,g\left(x_1\right);f]}>x_1$. Further, it can be easily seen that the following identity holds:

whence, by (3) for $n=1$$n=1$n=1n=1$n=1$ it follows $x_2<g\left(x_1\right)$$x_2<g\left(x_1\right)$x_(2) < g(x_(1))x_{2}<g\left(x_{1}\right)$x_2<g\left(x_1\right)$, since $f\left(g\left(x_1\right)\right)>0$$f\left(g\left(x_1\right)\right)>0$f(g(x_(1))) > 0f\left(g\left(x_{1}\right)\right)>0$f\left(g\left(x_1\right)\right)>0$ and $[x_1,g\left(x_1\right);f]>0$$\left[x_1,g\left(x_1\right);f\right]>0$[x_(1),g(x_(1));f] > 0\left[x_{1}, g\left(x_{1}\right) ; f\right]>0$[x_1,g\left(x_1\right);f]>0$.

From the identity

taking into account (3) for $n=1$$n=1$n=1n=1$n=1$ and the fact that $f$$f$ff$f$ is convex, we get $f\left(x_2\right)<$$f\left(x_2\right)<$f(x_(2)) <f\left(x_{2}\right)<$f\left(x_2\right)<$ 0 and so $x_2<x^{\ast }$$x_2<x^{\ast }$x_(2) < x^(**)x_{2}<x^{*}$x_2<x^{\ast }$.

By $x_2>x_1$$x_2>x_1$x_(2) > x_(1)x_{2}>x_{1}$x_2>x_1$ it results $g\left(x_2\right)<g\left(x_1\right)$$g\left(x_2\right)<g\left(x_1\right)$g(x_(2)) < g(x_(1))g\left(x_{2}\right)<g\left(x_{1}\right)$g\left(x_2\right)<g\left(x_1\right)$. We prove that $g\left(x_2\right)>x^{\ast }$$g\left(x_2\right)>x^{\ast }$g(x_(2)) > x^(**)g\left(x_{2}\right)>x^{*}$g\left(x_2\right)>x^{\ast }$. Since $x_2<x^{\ast }$$x_2<x^{\ast }$x_(2) < x^(**)x_{2}<x^{*}$x_2<x^{\ast }$, from the monotony of $g$$g$gg$g$ it follows $g\left(x_2\right)>g\left(x^{\ast }\right)=x^{\ast }$$g\left(x_2\right)>g\left(x^{\ast }\right)=x^{\ast }$g(x_(2)) > g(x^(**))=x^(**)g\left(x_{2}\right)>g\left(x^{*}\right)=x^{*}$g\left(x_2\right)>g\left(x^{\ast }\right)=x^{\ast }$. In conclusion, we get

Assume now that for some $n\ge 2$$n\ge 2$n >= 2n \geq 2$n\ge 2$, the elements obtained by (3) verify:

Repeating the above reason for $x_1=x_n$$x_1=x_n$x_(1)=x_(n)x_{1}=x_{n}$x_1=x_n$ we get

From (10) and (11) one obtains the monotony of the sequences ${\left(x_n\right)}_{n\ge 1}$${\left(x_n\right)}_{n\ge 1}$(x_(n))_(n >= 1)\left(x_{n}\right)_{n \geq 1}${\left(x_n\right)}_{n\ge 1}$ and ${\left(g\left(x_n\right)\right)}_{n\ge 1}$${\left(g\left(x_n\right)\right)}_{n\ge 1}$(g(x_(n)))_(n >= 1)\left(g\left(x_{n}\right)\right)_{n \geq 1}${\left(g\left(x_n\right)\right)}_{n\ge 1}$. Obviously, these sequence are bounded, so there exists $\overline{x}=\underset{n\to \mathrm{\infty }}{lim}{x}_n$$\overline{x}=\underset{n\to \mathrm{\infty }}{lim} x_n$bar(x)=lim_(n rarr oo)x_(n)\bar{x}= \lim _{n \rightarrow \infty} x_{n}$\overline{x}=\underset{n\to \mathrm{\infty }}{lim}{x}_n$, and $\underset{n\to \mathrm{\infty }}{lim}g\left(x_n\right)=g(\overline{x})$$\underset{n\to \mathrm{\infty }}{lim} g\left(x_n\right)=g(\overline{x})$lim_(n rarr oo)g(x_(n))=g( bar(x))\lim _{n \rightarrow \infty} g\left(x_{n}\right)=g(\bar{x})$\underset{n\to \mathrm{\infty }}{lim}g\left(x_n\right)=g(\overline{x})$, since $g$$g$gg$g$ is continuous.

Passing to limit in (3) implies $\overline{x}=\overline{x}-\frac{f(\overline{x})}{[\overline{x},g(\overline{x});f]}$$\overline{x}=\overline{x}-\frac{f(\overline{x})}{[\overline{x},g(\overline{x});f]}$bar(x)= bar(x)-(f(( bar(x))))/([( bar(x)),g(( bar(x)));f])\bar{x}=\bar{x}-\frac{f(\bar{x})}{[\bar{x}, g(\bar{x}) ; f]}$\overline{x}=\overline{x}-\frac{f(\overline{x})}{[\overline{x},g(\overline{x});f]}$ i.e. $f(\overline{x})=0$$f(\overline{x})=0$f( bar(x))=0f(\bar{x})=0$f(\overline{x})=0$, and so $\overline{x}=x^{\ast }$$\overline{x}=x^{\ast }$bar(x)=x^(**)\bar{x}=x^{*}$\overline{x}=x^{\ast }$.

Relations (11) imply the following a posteriori errors

Remark 3. Consider in (3) the function $g:[\alpha ,\beta ]\to \mathbb{R}$$g:[\alpha ,\beta ]\to \mathbb{R}$g:[alpha,beta]rarrRg:[\alpha, \beta] \rightarrow \mathbb{R}$g:[\alpha ,\beta ]\to \mathbb{R}$,

If $f$$f$ff$f$ is concave on $[\alpha ,\beta ]$$[\alpha ,\beta ]$[alpha,beta][\alpha, \beta]$[\alpha ,\beta ]$, then $[u,v;f]>[\beta ,b;f],\mathrm{\forall }u,v\in [\alpha ,\beta ]$$[u,v;f]>[\beta ,b;f],\mathrm{\forall }u,v\in [\alpha ,\beta ]$[u,v;f] > [beta,b;f],AA u,v in[alpha,beta][u, v ; f]>[\beta, b ; f], \forall u, v \in[\alpha, \beta]$[u,v;f]>[\beta ,b;f],\mathrm{\forall }u,v\in [\alpha ,\beta ]$ and so $g$$g$gg$g$ is decreasing on $[\alpha ,\beta ]$$[\alpha ,\beta ]$[alpha,beta][\alpha, \beta]$[\alpha ,\beta ]$. Suppose now that hypotheses iv. and a) resp. b) imposed on $f$$f$ff$f$ and $g$$g$gg$g$ are replaced by
iv ${}^{\prime }$${}^{\prime }$^(')^{\prime}${}^{\prime }$. the function $f$$f$ff$f$ is concave on $[a,b]$$[a,b]$[a,b][a, b]$[a,b]$;
the initial value $x_1$$x_1$x_(1)x_{1}$x_1$ in (3) is such that
${\mathrm{a}}^{\prime }).f\left(x_1\right)>0$$\left.{\mathrm{a}}^{\prime }\right).f\left(x_1\right)>0${:a^(')).f(x_(1)) > 0\left.\mathrm{a}^{\prime}\right) . f\left(x_{1}\right)>0${\mathrm{a}}^{\prime }).f\left(x_1\right)>0$;
${\mathrm{b}}^{\prime }$${\mathrm{b}}^{\prime }$b^(')\mathrm{b}^{\prime}${\mathrm{b}}^{\prime }$ ). $g\left(x_1\right)>\alpha$$g\left(x_1\right)>\alpha$g(x_(1)) > alphag\left(x_{1}\right)>\alpha$g\left(x_1\right)>\alpha$, with $g$$g$gg$g$ given by (13).
Then the sequences ${\left(x_n\right)}_{n\ge 1}$${\left(x_n\right)}_{n\ge 1}$(x_(n))_(n >= 1)\left(x_{n}\right)_{n \geq 1}${\left(x_n\right)}_{n\ge 1}$ and ${\left(g\left(x_n\right)\right)}_{n\ge 1}$${\left(g\left(x_n\right)\right)}_{n\ge 1}$(g(x_(n)))_(n >= 1)\left(g\left(x_{n}\right)\right)_{n \geq 1}${\left(g\left(x_n\right)\right)}_{n\ge 1}$ have the following properties: ${\mathrm{j}}^{\prime }.{\left(x_n\right)}_{n\ge 1}$${\mathrm{j}}^{\prime }.{\left(x_n\right)}_{n\ge 1}$j^(').(x_(n))_(n >= 1)\mathrm{j}^{\prime} .\left(x_{n}\right)_{n \geq 1}${\mathrm{j}}^{\prime }.{\left(x_n\right)}_{n\ge 1}$ is decreasing;
${\mathrm{jj}}^{\prime }.{\left(g\left(x_n\right)\right)}_{n\ge 1}$${\mathrm{jj}}^{\prime }.{\left(g\left(x_n\right)\right)}_{n\ge 1}$jj^(').(g(x_(n)))_(n >= 1)\mathrm{jj}^{\prime} .\left(g\left(x_{n}\right)\right)_{n \geq 1}${\mathrm{jj}}^{\prime }.{\left(g\left(x_n\right)\right)}_{n\ge 1}$ is increasing;
${\mathrm{jjj}}^{\prime }.g\left(x_n\right)<x^{\ast }<x_n,n=1,2,\dots$${\mathrm{jjj}}^{\prime }.g\left(x_n\right)<x^{\ast }<x_n,n=1,2,\dots$jjj^(').g(x_(n)) < x^(**) < x_(n),quad n=1,2,dots\mathrm{jjj}^{\prime} . g\left(x_{n}\right)<x^{*}<x_{n}, \quad n=1,2, \ldots${\mathrm{jjj}}^{\prime }.g\left(x_n\right)<x^{\ast }<x_n,n=1,2,\dots$
The proof of these properties is similar to that given for Theorem 5.

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Received by the editors: March 12, 2003.