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New Newton's Method with Third-order Convergence for Solving Nonlinear Equations

New Newton’s Method with Third-order Convergence for Solving Nonlinear Equations

Osama Yusuf Ababneh

Abstract—For the last years,the variants of the Newton’s method

with cubic convergence have become popular iterative methods to

find approximate solutions to the roots of non-linear equations.These

methods both enjoy cubic convergence at simple roots and do not

require the evaluation of second order derivatives.In this paper,we

present a new Newton’s method based on contra harmonic mean with

cubically convergent.Numerical examples show that the new method

can compete with the classical Newton’s method.

Keywords—Third-order convergence,Non-linear equations,Root-

finding,Iterative method.

I.I NTRODUCTION

S OLVING non-linear equations is one of the most impor-

tant problems in numerical analysis.In this paper,we

consider iterative methods tofind a simple root of a non-

linear equation f(x)=0,where f:D⊂R→R for an

open interval D is a scalar function.The classical Newton

method for a single non-linear equation is written as

x n+1=x n−f(x n)

(x n)

.(1)

This is an important and basic method[8],which converges quadratically.Recently,some modified Newton methods with cubic convergence have been developed in[1],[2],[3],[4],[5], [6]and[7].Here,we will obtain a new modification of New-tons method.Analysis of convergence shows the new method is cubically convergent.Its practical utility is demonstrated by numerical examples.

Letαbe a simple zero of a sufficiently differentiable function f and consider the numerical solution of the equation f(x)=0.It is clear that

f(x)=f(x n)+

x

x n

f (t)dt.(2)

Suppose we interpolate f on the interval[x n,x]by the con-stant f (x n),then(x−x n)f (x n)provides an approximation for the indefinite integral in(2)and by taking x=αwe obtain 0≈f(x n)+(α−x n)f (x n).

Thus,a new approximation x n+1toαis given by

x n+1=x n−f(x n) f (x n)

.

Dr.Osama Yusuf Ababneh is with the Department of Mathematics,Irbid National University,Irbid,Jordan e-mail:(ababnehukm@http://doc.xuehai.net).On the other hand,if we approximate the indefinite integral in(2)by the trapezoidal rule and take x=α,we obtain 0≈f(x n)+

1

2

(α−x n)(f (x n)+f (α)),

and therefore,a new approximation x n+1toαis given by

x n+1=x n−

2f(x n)

f (x n)+f (x n+1)

.

If the Newton’s method is used on the right-hand side of the above equation to overcome the implicity problem,then

x n+1=x n−

2f(x n)

f (x n)+f (z n+1)

,(3) where

z n+1=x n−

f(x n)

f (x n)

is obtained which is,for n=0,1,2,...,the trapezoidal Newton’s method of Fernando et al.[1].Let us rewrite equation(3)as

x n+1=x n−

f(x n)

(f (x n)+f (z n+1))/2

,n=0,1, (4)

So,this variant of Newton’s method can be viewed as obtained by using arithmetic mean of f (x n)and f (z n+1)instead of f (x n)in Newton’s method defined by(1).Therefore,we call it arithmetic mean Newton’s(AN)method.

In[3],the harmonic mean instead of the arithmetic mean is used to get a new formula

x n+1=x n−

f(x n)(f (x n)+f (z n+1))

2f (x n)f (z n+1)

,n=0,1, (5)

which is called harmonic mean Newton’s(HN)method and used the midpoint to get

x n+1=x n−

f(x n)

((x n+z n+1)/2)

,n=0,1, (6)

which is called midpoint Newton’s(MN)method.

II.N EW ITERATIVE METHOD AND CONVERGENCE

ANALYSIS

If we use the contra harmonic mean instead of the arithmetic mean in(4),we get new Newton method

x n+1=x n−

f(x n)(f (x n)+f (z n+1))

f 2(x n)+f 2(z n+1)

,n=0,1, (7)

World Academy of Science, Engineering and Technology 61 2012

1071

第1页

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