ON SOME AITKEN-STEFFENSEN-HALLEY-TYPE METHODS FOR APPROXIMATING THE ROOTS OF SCALAR EQUATIONS

. In this note we extend the Aitken-Steﬀensen method to the Halley transformation. Under some rather simple assumptions we obtain error bounds for each iteration step; moreover, the convergence order of the iterates is 3, i

As it was shown in [2], the Halley method for solving: (1.1) f (x) = 0, is given by (1.2) This sequence is in fact generated by the Newton method for solving h (x) = 0.
The first and second order derivatives of h are given by (1.3) which yield the following equalities for a solution x of (1.1): 1/2 , and (1.5) Relation (1.6) ensures the convergence order 3 for the sequence (x k ) k≥0 .In the papers [2]- [8] and [12] there are studied the convergence and the convergence order of some sequences generated by some interpolatory methods applied to equation h (x) = 0.
We shall consider other two equations equivalent to (1.1) of the form: x − ϕ 1 (x) = 0, and (1.7) The Aitken method for solving h (x) = 0 is given by the iteration (1.9) In this note we shall study the convergence of these iterates.We shall show that the functions ϕ 1 and ϕ 2 may be chosen in order to obtain bilateral approximations at each iteration step; this fact allows the control of the errors.On the other hand, the convergence order of (x n ) n≥0 given by (1.9) is at least equal to 3. Hypotheses

ERROR EVALUATION AND LOCAL CONVERGENCE
Consider the interval [α, β] given above.We shall make the following assumptions on ϕ 1 and ϕ 2 : where [x, y; ϕ] denotes the first order divided difference of ϕ on x and y; v. the function ϕ 2 verifies the relations − We can state the following result: Theorem 2.1.Assume that i-v hold, and for some Then the following relations hold: Denote by I 0 the interval having the extremities ϕ 1 (x 0 ) and ϕ 2 (x 0 ).We notice, taking into account the mean formula, that [ϕ 1 (x 0 ), ϕ 2 (x 0 ); h] > 0. When x 0 < x, by iv. and x = ϕ 1 (x) it follows ϕ 1 (x 0 ) < x.Analogously, ϕ 1 (x 0 ) > x for x 0 > x.Taking into account v. and x = ϕ 2 (x) we get ϕ 2 (x 0 ) > x for x 0 < x and ϕ 2 (x 0 ) < x for x 0 > x.It is obvious that in both situations x ∈ I.It can be easily seen that for all n = 0, 1, . . .we have which, for n = 0 imply x 1 > ϕ 1 (x 0 ) and Denoting by I 1 the interval determined by ϕ 1 (x 1 ) and ϕ 2 (x 1 ) then and the element x 2 constructed by (1.9) satisfies x 2 , x ∈ I 1 .
Repeating the above reason we get x n+1 , x ∈ I n , the interval being determined by ϕ 1 (x n ), ϕ 2 (x n ), and also that and x ∈ I n+1 .It is clear now that jj.holds.In order to obtain jjj.we shall use the identity which, together with (1.9) and h(x) = 0, imply For the difference x − ϕ 1 (x n ), by iv.one gets Analogously, by v. we get The mean formula for divided differences implies For h ′′ (ξ n ) we have Since the initial approximation x 0 was supposed sufficiently close to the solution x, then implies, together with jjj., statement j.
Remark 2.1.Supposing that f ′′ (x) ≥ 0 for all x ∈ [a, b], and if instead of iii.we assume that f ′ (x) > 0, then obviously f (x) < 0 for a ≤ x < x, and so in Theorem 2.1 we may take α = a.From the above conditions it follows that h ′ (x) > 0 for x ∈ [a, x], and since f (x) = 0, one gets β > x.
3. DETERMINING THE FUNCTIONS ϕ 1 AND ϕ 2 Under reasonable hypotheses on f , we shall show that there exist two classes of functions among we can choose the functions ϕ 1 and ϕ 2 such that hypotheses iv. and v. to be satisfied.
Let f : [a, b] → R, a, b ∈ R, a < b and suppose that f ∈ C 4 [a, b], and f ′ (x) > 0, ∀x ∈ [a, b].Consider the function h : [a, b] → R given by