Modified Iterative Method for Solving Nonlinear Equation

In this paper, we present new oneand two-steps iterative methods for solving nonlinear equation f(x)=0. It is proved here that the iterative methods converge of order three and six respectively. Several numerical examples are given to illustrate the performance and to show that the iterative methods in this paper give better result than the compared methods. Introduction Solving non-linear equations is one of the most important problems in numerical analysis. Multi-step iterative methods for finding solutions of nonlinear equations have been a constant interesting field of study in numerical analysis. Important research on these methods was carried out during the last ten years (Babajee, 2008). Several authors (Ostrowski 1973, Abbasba ndy 2003, Noor 2007, Chun 2005, Kanwar et. al 2005, Noor & Ahmed 2006 and Noor et. al 2006) attempted to develop new higher order methods of higher efficiency index by adding finite evaluations of the function in the multi-step methods to obtain less iteration than the classical Newton method. Further, Noor and Ahmed (2006) derived a new two step iterative method of order four depending on an auxiliary equation, Newton’s method and by considering terms up to first-order in Taylor series. The purpose of this paper is to present oneand twosteps iterative methods having third and sixth order of convergence respectively in a similar manner of (Noor & Ahmed, 2006) by considering terms up to second-order in Taylor series. Iterative methods Consider the nonlinear equation . 0 ) (  x f ...(1) Let  be a simple root and 0 x be the initial guess known for the required root. To derive our iterative methods in a similar manner of (Noor & Ahmed, 2006), assume Journal of Kirkuk University –Scientific Studies, vol.7, No.1, 2012 147 h x x   0 1 , 1  h , ... (2) be the first approximation to the root. Consider the following auxiliary equation with a parameter p : 0 ) ( ) ( ) ( ) ( 2 2 0 3     x f x f x x p x g , ... (3) where R p . It is clear that the root of (1) is also the root of (3) and vice versa. If h x x   0 1 is the better approximation for the required root, then (3) gives . 0 ) ( ) ( 0 2 0 2 3     h x f h x f h p ...(4) Expanding ) ( 0 h x f  by the Taylor’s theorem and simplifying, we get                2 ) ( ) ( ) ( 4 ) ( ) ( ) ( 2


Introduction
Solving non-linear equations is one of the most important problems in numerical analysis.Multi-step iterative methods for finding solutions of nonlinear equations have been a constant interesting field of study in numerical analysis.Important research on these methods was carried out during the last ten years (Babajee, 2008).Several authors (Ostrowski 1973, Abbasba ndy 2003, Noor 2007, Chun 2005, Kanwar et. al 2005, Noor & Ahmed 2006and Noor et. al 2006) attempted to develop new higher order methods of higher efficiency index by adding finite evaluations of the function in the multi-step methods to obtain less iteration than the classical Newton method.Further, Noor and Ahmed (2006) derived a new two step iterative method of order four depending on an auxiliary equation, Newton's method and by considering terms up to first-order in Taylor series.The purpose of this paper is to present one-and two-steps iterative methods having third and sixth order of convergence respectively in a similar manner of (Noor & Ahmed, 2006) by considering terms up to second-order in Taylor series.

Iterative methods
Consider the nonlinear equation Let  be a simple root and 0 x be the initial guess known for the required root.To derive our iterative methods in a similar manner of (Noor & Ahmed, 2006), assume ) be the first approximation to the root.Consider the following auxiliary equation with a parameter p : It is clear that the root of ( 1) is also the root of (3) and vice versa.If is the better approximation for the required root, then (3) gives by the Taylor's theorem and simplifying, we get in which sign should be chosen to make the denominator largest in magnitude(see Noor and Ahmed, 2006).
We suggest the following one step iteration method for solving nonlinear equation Algorithm 1: Here p is chosen so that and p have the same sign.For a given 0 x , calculate , where sign is chosen such as to make the denominator largest in magnitude.Now, by combining Algorithm 2.1and the Newton's method we obtain the following two step iterative method: Algorithm 2: Here p is chosen so that and p have the same sign.For a given 0 x , calculate where sign is chosen such as to make the denominator largest in magnitude.

Convergence analysis
In this section, we consider the convergence analysis of iterative technique given by Algorithm 2.1 and Algorithm 2.2 by the following theorems respectively: for an open interval I is sufficiently differentiable near a simple root x is sufficiently closed to  , then the one step iterative method defined by Algorithm 2.1 has third-order convergence.Proof.The technique is given by ...( 6) From ( 6) (see [6] and [7]), we get then by Taylor's expansion, we have for j=2, 3… .
From (8), ( 9) and (10) we have From ( 11), ( 12) and ( 13) we have …( 14) From ( 7) and ( 14) we get This shows that the method given in Algorithm 2.1 has third-order convergences.□ Theorem 2: Assume that the function for an open interval I is sufficiently differentiable near a simple root x is sufficiently closed to  , then the one step iterative method defined by Algorithm 2.2 has sixth-order convergence.

Numerical examples
We now present some examples to illustrate the efficiency of our developed methods in this paper which are given by the Algorithm 2.1 and Algorithm 2.2.We compare the Newton's method (NM) Ostrowski (1973), the method of Noor and Ahmed (2006) (NA) and the methods of Noor et al. (2006) (NRM), with the iterative methods introduced in this paper.We used 15 10    .The following stopping criteria are used for computer programs: The following examples are used for numerical testing, see( (Noor and Ahmed (2006)) and (Noor et. al (2006)

Conclusion
From Table 1, we conclude that the one-step iterative method (Algorithm 1) performs better than the Newton's method and two-steps iterative method (Algorithm 2) performs better than the Newton's method and the methods presented by (Noor &Ahmed, 2006 andNoor et.al, 2006).


This shows that the two step iterative method given in Algorithm 2.2 has sixth-order convergences.□