Let $f:[a,b]\to ℝ,$ where $a,b\in ℝ,$ , and suppose that $f$ has the first order derivative, which is positive: $f^{\prime }\left(x\right)>0,\forall x\in [a,b]$. Consider the function $h:[a,b]\to ℝ$

In [2] there is shown that the Halley method for solving

is in fact the Newton method for solving (1.1). This method consists therefore in generating the sequence ${\left(x_n\right)}_{n\ge 0}$ by

The first and second order derivatives of $h$ are given by

and

These relations imply

and

where $\overline{x}\in [a,b]$ denotes the solution of (1.2). As shown in [1], equality (1.7) characterizes the Halley method, in the sense that ensures its convergence order $3.$ The authors of [4], analyzing an algorithm of Heron for approximating $\sqrt[3]{100}$, give a general algorithm which can be used for approximating the cubic root of any real positive number.

In [7] it is shown that the algorithm from [4] is nothing else than the chord method applied to equation $h\left(x\right)=0$ where $h\left(x\right)=\frac{f\left(x\right)}{\sqrt{f^{\prime }\left(x\right)}}$, with $f\left(x\right)=x^3-N$. In this case the equation $h\left(x\right)=0$ has the form $x^2-\frac{N}{x}=0$, when $N>0$, $N\in ℝ.$

It is clear that between the Heron algorithm and the Halley method there exists a connection, in the sense that the transformed equation to which we apply the Newton or the chord method is the same. In [7] and [10], the authors study the convergence and error bounds for the Steffensen and Aitken-Steffensen methods applied to (1.1). In this note we shall study a variant of the Aitken-Steffensen method, which differs from those presented in [10] and [11].

We shall consider other two equations, equivalent to (1.2), having the form

and

where ${\phi }_1,{\phi }_2:[a,b]\to [a,b]$ will be convenably chosen.

We shall study the sequence ${\left(x_n\right)}_{n\ge 0}$ given by

We shall consider the following assumptions on $f,{\phi }_1$ and ${\phi }_2$:

i. $f\in C^4[a,b];$

ii. equation (1.2) has a solution $\overline{x}\in (a,b);$

iii. $f^{\prime }\left(x\right)>0,$ $\forall x\in [a,b];$

iv. ${\phi }_1$ obeys where $[x,y;{\phi }_1]$ denotes the first order divided difference of ${\phi }_1$ on $x$ and $y$:

v. ${\phi }_2$ obeys

We shall use the following identities:

and also the Newton identity

where $[x,y,z;h]$ is the second order divided difference of $h$ on $x,y,z.$

We notice that equality (1.6) and hypothesis i. ensure the existence of $\alpha ,\beta \in ℝ,$ such that $h^{\prime }\left(x\right)>0$ $\forall x\in [\alpha ,\beta ]$.

The following theorem holds:

Proof. We shall analyses two cases.

I. Then i.e., . Denote $\mathrm{\Psi }\left(x\right)=x-{\phi }_1\left(x\right),$ and so , i.e. . Now we show that ${\phi }_2\left({\phi }_1\left(x_0\right)\right)>\overline{x}.$ From it follows ${\phi }_2\left({\phi }_1\left(x_0\right)\right)-\overline{x}={\phi }_2\left({\phi }_1\left(x_0\right)\right)-{\phi }_2\left(\overline{x}\right)=[\overline{x},{\phi }_1\left(x_0\right);{\phi }_2]\left({\phi }_1\left(x_0\right)-\overline{x}\right)>0,$ i.e. ${\phi }_2\left({\phi }_1\left(x_0\right)\right)>\overline{x}.$ Next, we show that $x_1\in [{\phi }_1\left(x_0\right),{\phi }_2\left({\phi }_1\left(x_0\right)\right)]$, where $x_1$ is obtained from (1.10) for $n=0$. Since $h^{\prime }\left(x\right)>0,\forall x\in [\alpha ,\beta ]$, and ${\phi }_1\left(x_0\right)\in [\alpha ,\beta ]$, we get that (we know that ) and so $x_1$satisfies $x_1>{\phi }_1\left(x_0\right)$. We have used the fact that $h^{\prime }\left(x\right)>0\forall x\in [\alpha ,\beta ]$ implies $[{\phi }_1\left(x_0\right),{\phi }_2\left({\phi }_1\left(x_0\right)\right);h]>0.$ Now we show that . This inequality follows from ${\phi }_2\left({\phi }_1\left(x_0\right)\right)>\overline{x}$, $h\left({\phi }_2\left({\phi }_1\left(x_0\right)\right)\right)>0$ and from (2.1) for $n=0$. we have shown that

and

II. $x_0>\overline{x}$. Similarly to the above reason, we get that

and

Denoting by $I_0$ the open interval determined by ${\phi }_1\left(x_0\right)$ and ${\phi }_2\left({\phi }_1\left(x_0\right)\right)$, then obviously relations (2.3) - (2.6) may be sinthetized as

It can be easily seen that if we denote by $I_1$ the open interval determined by ${\phi }_1\left(x_1\right)$ and ${\phi }_2\left({\phi }_1\left(x_1\right)\right)$ then get

and

where $x_2$ is obtained from (1.10) for $n=1$.

Let $I_n$ be the open interval determined by ${\phi }_1\left(x_n\right)$ and ${\phi }_2\left({\phi }_1\left(x_n\right)\right)$ for some $n\in \mathbb{N}$. Then repeating the above reason, we may show that

and

where $I_{n+1}$ is determined by ${\phi }_1\left(x_{n+1}\right)$ and ${\phi }_2\left({\phi }_1\left(x_{n+1}\right)\right)$. From the above reason and from (2.7) it follows j., which yields an error bound bound for each iteration step.

For jj. using identity (2.2) we get

whence, taking into account (1.10) and $h\left(\overline{x}\right)=0,$ we get

The mean value formulae for the divided differences lead us to

and

From i. and using the Lagrange formula it follows

Denoting

and

from (2.8) and taking into account (2.9) and (2.10) we get

The property jj. follows easily by denoting $k=\frac{M_1}{2m_1}$ and taking into account iv, v, and the fact that ${\eta }_n\in I_n.$

Property jjj is an immediate consequence of j and jj.  

we shall present a modality of choosing ${\phi }_1$ and ${\phi }_2$ in order to obey the assumptions of Theorem 2.1

Suppose that $f$ is strictly convex on $[a,b]$, i.e. $f^{\prime \prime }\left(x\right)>0,\forall x\in [a,b]$. This assumption, together with $f^{\prime }\left(x\right)>0$ $\forall x\in [a,b]$, lead, by (1.4) to $h^{\prime }\left(x\right)>0\forall x\in [a,\overline{x}]$. Relation (1.4) again and $f^{\prime }\left(x\right)>0$ and $f\left(\overline{x}\right)=0$ imply the existence of $\beta$, such that $h^{\prime }\left(x\right)>0,\forall x\in [\overline{x},\beta ].$These hypotheses ensure the existence of an interval $[\alpha ,\beta ]$ for which $h^{\prime }\left(x\right)>0,\forall x\in [\alpha ,\beta ]$. Since $f^{\prime \prime }\left(x\right)>0$ it follows that $f^{\prime }\left(x\right)$ is increasing on $[a,b]$.

Taking

and

with $\mu \ge f_s^{\prime }\left(b\right)$ and and assuming that , then the functions ${\phi }_1$ and ${\phi }_2$ defined above obey hypotheses iv and v of Theorem 2.1. For $a\le x_0\le \overline{x}$ in Theorem 2.1, hypothesis ${\phi }_1\left(x_0\right)\in [\alpha ,\beta ]$ is automatically verified, but the assumption ${\phi }_2\left({\phi }_1\left(x_0\right)\right)\in [\alpha ,\beta ]$ must be kept.