International Journal of the Physical Sciences Vol. 5(7), pp. 935-939, July 2010 Available online at http://www.academicjournals.org/IJPS ISSN 1992 - 1950 ©2010 Academic Journals
Full Length Research Paper
A new derivative-free method for solving nonlinear equations P. A. Phiri1* and O. D. Makinde2 1
Department of Mathematics and Applied Mathematics, University of Pretoria, Pretoria 0002, South Africa. Faculty of Engineering, Cape Peninsula University of Technology, P. O. Box 1906, Bellville 7535, South Africa.
2
Accepted 19 May, 2010
This paper presents a new method for solving nonlinear equations. The method is free of derivatives and easy to implement. Although its convergence is linear, numerical experiments conducted indicate that its performance compares well not only with Newton-based methods of higher order but with other derivative-free methods as well. Key words: Derivative-free iterative method, nonlinear equations. INTRODUCTION Newton's method is one of the most popular techniques for solving the nonlinear equation (Abbasbandy et al., 2007).
φ (x) = 0.
(1)
As it is well known, the method is quadratic in convergence but its convergence is slow or it fails to converge at all when either f '( x ) is too small at any stage of the computation, or when the initial values are too far removed from the true solution. This paper presents a new method for solving nonlinear equations. The method is derivative-free and its convergence is linear. However, as numerical experiments conducted indicate, although its convergence is of order one, for certain problems, the method converges faster than some Newton-based methods of higher order; in particular, when Newton-based methods fail to converge, the method can be used to find the solution of the problem. Numerical experiments conducted also show that the method compares well with other iterative methods that do not depend on derivatives. DEVELOPMENT OF THE METHOD Consider the solution of the equation
x ψ 1 +ψ
2
= 1,
(2)
where ψ 1 + ψ 2 is an index, by using the iterative scheme
of steps 1 - 6 as thus shown. Step 1 x ( 0 ) := x 0 ,
x0
given;
Step 2 i = 0;
Step 3 Set
y ( i ) :=
1
(x )
Step 4 Set x ( i + 1) := y ( i ) 1ψ 1 ( )
(4)
Step 5: If x ( i + 1 ) − x ( i ) < ε where
*Corresponding author. E-mail:
[email protected].
(3)
ψ2
ε
is a predetermined accuracy
limit, go to Step 6, otherwise increase Step 3;
i
by 1 and go to
936
Int. J. Phys. Sci.
Proof
Table 1. Tableau to the solution of the equation xψ 1 +ψ 2 = 1.
y
x (i )
i
( x0 )
x0
0
( x0 )
1
−ψ 2
ψ 22
( x0 )
2
( x0 )
3
−ψ 23
ψ 24
( x0 )
4
( x0 ) ( x0 )
ψ 12
( x0 ) ( x0 )
ψ 14
r =
−ψ
ψ 12
ψ ∞
ln{Si}i=0
ψ 13
−ψ 25
{ x i }i∞= 0
=
= x0, x1, x2,
,
.
The result follows since, if we let
ψ1
−ψ 23
ψ 24
ψ 13
{S i }i∞= 0
−ψ 2
ψ 22
ψ1
Let
(i )
lnx0
ψ 14
2
, it follows that
1
{}
= ri
∞ i=0
so that
{}
lim r i → 0
i→∞
when
r
