A New Derivative-free Iterative Method for Solving Nonlinear Equations with Third Order Convergence

Abstract

In this paper a new iterative method is proposed to find a root of a nonlinear equation. The new method does not use derivatives. When the starting value is selected to be close to a root, the proposed method has a cubic convergence order. To show the efficiency of the method, we give some numerical examples using the test functions in the references and compare the results obtained with the classical Newton method of second order.

Full text

ISSN 1749-3889 (print), 1749-3897 (online)
International Journal of Nonlinear Science
Vol.13(2012) No.4,pp.505-512
A New Derivative-free Iterative Method for Solving Nonlinear Equations with
Third Order Convergence
Gustavo Fern´
andez-Torres∗
Department of Oil Engineering, Universidad Del Istmo Tehuantepec, Oaxaca, 70760, M´
exico
(Received 7 October 2011, accepted 4 May 2012)
Abstract: In this paper a new iterative method is proposed to find a root of a nonlinear equation. The new
method does not use derivatives. When the starting value is selected to be close to a root, the proposed method
has a cubic convergence order. To show the efficiency of the method, we give some numerical examples us-
ing the test functions in the references and compare the results obtained with the classical Newton method of
second order.
Keywords: Cubic approximation; iterative method; nonlinear equations; root finding method
1 Introduction
Until recently many iterative methods for solving the non-linear equation f(x) = 0 have been proposed [1–8, 12]. Most
of these methods are based on extensions of Newton’s method and use differential derivatives of higher orders. The use
of these derivatives is a serious disadvantage for these methods because of the cost of computational calculates and the
difficulty in finding the derivative.
To overcome the difficulties many methods have been proposed without the use of derivatives as in [2], but the order
of convergence of these methods is small. In [5] the author has proposed an iterative method with a convergence order
near 2 and without the use of derivatives of the function f(x). The motivation to use this method is Muller’s method but
using a cubic polynomial.
In [5], we take an initial point x0sufficiently close to the desired root and the iterative method is defined as
xn+1 =xn+6(xn−xn−1)xnK
(K−L)(2xn+xn−1)+(M−K)(4xn−xn−1),
where
M=f(xn−1), K =f(xn), L =f(2xn−xn−1).
The error in this method depends on the choice of the Newton method to approximate the root of a cubic polynomial
and in the choice of xn−xn−1to maintain the Newton method convergence order.
The purpose of this work is to develop some new simple iterative methods to improve in a significative way, the
error of the previous method. As a result, the proposed methods will have several convergence orders without requiring
derivatives of the function f(x). The method maintains the initial considerations of Muller’s method.
2 New Iterative Method
Under the assumption that a sufficiently derivable function f(x)has a unique zero on an interval (a, b)with f(a)f(b)<0,
let x0= 0 be an approximation sufficiently close to the root rof f(x). Take −1< β < 1sufficiently small, β= 0, and
consider the following particular interpolating polynomial that passes through the three points x0, x0−β, x0+β,
p(x) = A(x−x0)(x−x0−β)(x−x0+β) + 3Ax0(x−x0)(x−x0−β) + B(x−x0) + C. (1)
∗Corresponding author.E-mail address: gusfer@sandunga.unistmo.edu.mx
Copyright c
World Academic Press, World Academic Union
IJNS.2012.06.30/631

506 International Journal of Nonlinear Science, Vol.13(2012), No.4, pp. 505-512
Consider the following system of equations that the polynomial p(x)must verify
f(x0) = K=C=p(x0),(2)
f(x0+β) = L=βB +C=p(x0+β),(3)
f(x0−β) = M= 6β2x0A−βB +C=p(x0−β).(4)
The solutions of the system are very clear
C=K, B =L−K
β, A =M+L−2K
6β2x0
.(5)
Now, consider the derivatives of polynomial p(x)
p′(x) = 3A(x−x0)2−Aβ2+ 6Ax0(x−x0)−3Aβx0+B= 3A(x−x0)(x+x0)−Aβ2−3Aβx0+B,
p′′(x) = 6Ax, p′′′ (x) = 6A.
The function f(x)can be approximated with the Taylor’s polynomial of second order about x0and since
p(x0) = C, p′(x0) = −Aβ2−3Aβ x0+B, p′′ (x0) = 6Ax0, p′′′ (x0)=6A,
we have
f(x)≈C+ (−Aβ2−3Aβx0+B)(x−x0) + 3Ax0(x−x0)2.(6)
Observe that the coefficient of x−x0in the above equation can be written as
−Aβ2−3Aβx0+B=−Aβ(3x0+β) + B=
=3x0(L−M) + β(2K−L−M)
6βx0
=L−M
2β+2K−L−M
6x0
.
Define Sas
S=L−M+β(2K−L−M)
3x0
.
For x=r, and the above equations in (6), we have
6Ax0β(r−x0)2+S(r−x0) + 2βK ≈0.
We can resolve the above equation to give
r≈x0+−S±S2−48AKβ 2x0
12Aβx0
that can be written as
r≈x0+β
2(2K−L−M)[S∓S2+ 8K(2K−L−M)].
As in Muller’s method, to avoid the problems in the calculation of the error caused by the subtraction of nearly equal
numbers, we apply the formula
r≈x0−4Kβ
S±S2+ 8K(2K−L−M)
This formula provides two possibilities, depending on the sign preceding the radical term. The sign is chosen in such a
way that the denominator will be the largest in magnitude and will result in being selected as the closest point to x0. With
this we have the following iterative method.
Let x0= 0 be an approximation sufficiently close to a root rof f(x). Take −1< β < 1sufficiently small, β= 0,
then for n≥0we have
xn+1 =xn−4Kβ
S+sign(S)S2+ 8K(2K−L−M),(7)
with
S=L−M+β(2K−L−M)
3xn
,(8)
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G. Fern´
andez-Torres: A New Derivative-free Iterative Method for Solving Nonlinear Equations with Third Order Convergence 507
and
f(xn) = K, f (xn+β) = L=βB +C, f (xn−β) = M. (9)
These methods depend on the selection of β. Now we analyse four different values for βto define the new iterative
methods.
a) β= (xn−xn−1)2,
b) β=xn|xn−xn−1|,
c) β=xn(xn−xn−1),
d) β= [xn(xn−xn−1)]9/8.
3 Convergence Analysis
Theorem 1 Let r∈Ibe a zero of a sufficiently differentiable function f:I→ ℜ for an interval I. If x0= 0 is
sufficiently close to r, then the iterative method (7) with conditions (8) and (9) has orders of convergence 2, 2.4, 2.7 and 3
for the selection of β, a), b), c) and d), respectively. Moreover, for any choice for βthe maximum convergence order is 3.
Proof. Suppose that the point xnhas been calculated starting the process with x0. Using (1), we know that
f(x)−p(x) = E(x),
where E(x)is the error of an interpolating polynomial that can be calculated for the common process and is given by
E(x) = (x−xn)(x−xn+β)(x−xn−β)f′′′(ξ1)−6A
6
for some ξ1between xn−βand xn+β. Thus, f(x)can be written as
C+ (−Aβ2−3Aβxn+B)(x−xn)+3Axn(x−xn)2+ (x−xn)3f′′′(ξ1)
6−β2(x−xn)f′′′(ξ1)
6+β2(x−xn)A.
If xn+1 is a root of polynomial (6), C+ (−Aβ2−3Aβxn+B)(x−xn)+3Axn(x−xn)2then
f(xn+1) = (xn+1 −xn)3f′′′ (ξ1)
6−β2(xn+1 −xn)f′′′(ξ1)
6+β2(xn+1 −xn)A.
We consider that xn+1 = 0, this is always possible since it represents a shift with respect to origin. Thus the above
expression can be written as
f(0) = −f′′′(ξ1)
6x3
n+β2xn
f′′′(ξ1)
6−Axn
We know that we can write (5) using divided differences and for ξ2∈(xn−β, xn+β),Acan be expressed as
A=−f′′(ξ2)
3xn
.
Therefore
f(0) = xn(β2−x2
n)f′′′(ξ1)
6+β2f′′(ξ2)
3.
We now assume that the points xn, xn−1lie in a neighborhood of the root r. Thus if we let εn+1 =xn+1 −r=−r, εn=
xn−r, we have xn−xn−1=εn−εn−1. Consider also the cases for β, then we have
a)β= (xn−xn−1)2
f(0) = xn(xn−xn−1)4−x2
nf(ξ1)
6+ (xn−xn−1)4f′′(ξ2)
3.
b)β=xn|xn−xn−1|
f(0) = x3
n[xn−xn−1−1] f(ξ1)
6+x2
n(xn−xn−1)f′′(ξ2)
3.
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508 International Journal of Nonlinear Science, Vol.13(2012), No.4, pp. 505-512
c)β=xn(xn−xn−1)
f(0) = x3
n(xn−xn−1)2−1f(ξ1)
6+x2
n(xn−xn−1)2f′′(ξ2)
3.
d)β= [xn(xn−xn−1)]9/8
f(0) = xn[xn(xn−xn−1)]9/4−x2
nf(ξ1)
6+ [xn(xn−xn−1)]9/4f′′(ξ2)
3.
We assume that the magnitude of the quantity εnis less than some upper bound εm. With this we make the assumption
εn+1 < εmthat is clear by the equality xn+1 −r= (xn+1 −y) + (y−r)with ythe root of polynomial p(x). Note that
εn+1 =O(ε3
m), thus if we expand all functions around r, we obtain
a)εn+1f′(r) = ε4
n−1f′′(r)
3+O(ε3
m).
b)εn+1f′(r) = ε2
nεn−1f′′(r)
3+O(ε3
m).
c)εn+1f′(r) = ε2
nε2
n−1f′′(r)
3+O(ε3
m).
d)εn+1f′(r) = ε9/4
nε9/4
n−1f′′(r)
3+O(ε3
m).
We suppose that εn+1 is asymptotic to cεp
nwith p > 1. Consequently, by expressing the above equations in terms of
εn−1and neglecting the terms of O(εm), we obtain
a)cp+1εp2
n−1∼ε4
n−1f′′(r)
3f′(r).
b)cp+1εp2
n−1∼ε2p+1
n−1f′′(r)
3f′(r).
c)cp+1εp2
n−1∼ε2p+2
n−1f′′(r)
3f′(r).
d)cp+1εp2
n−1∼ε
9
4(p+1)
n−1f′′(r)
3f′(r).
In order to satisfy the previous asymptotic equations, it is evident that in each case we have
a) p2= 4 therefore the maximun value is p= 2.
a) p2= 2p+ 1 therefore the maximun value is p= 1 + √2≈2.41.
a) p2= 2p+ 2 therefore the maximun value is p= 1 + √3≈2.73.
a) p2=9
4(p+ 1) therefore the maximun value is p= 3.
For any other value of β, such as β=xn(xn−xn−1)2or β= [xn(xn−xn−1)]3/2), it signifies that the term ϵ2
nε4
n−1
or ϵ3
nε3
n−1gives the equations p2= 2p+ 4 or p2= 3p+ 3 and the solutions p= 3.23 or p= 3.79 respectively. These are
greater than p= 3. In these cases, we have
εn+1f′(r) = −ε3
nf′′′(r)
6+O(ε3
m)
Thus the associated equation will be
εn+1f′(r)
ε3
n
=−f′′′(r)
6+O(ε3
m)
therefore the convergence order will also be 3.
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G. Fern´
andez-Torres: A New Derivative-free Iterative Method for Solving Nonlinear Equations with Third Order Convergence 509
4 Numerical experiments
We present some numerical test functions taking into account the proposed method. The test functions are the following
1) f(x) = sin2x−x2+ 1. 6) f(x) = 1
x−sin x+ 1.
2) f(x) = cos x−x. 7) f(x) = x3+ 4x2−10.
3) f(x) = (x−1)3−1. 8) f(x) = ex−3x2.
4) f(x) = x3−10. 9) f(x) = x2sin x−cos x.
5) f(x) = x3−2x2−5. 10) f(x) = sin x−x
2.
The next tables present the method with cubic order convergence (New) and is compared with other methods proposed
in [1–4, 6–8, 12] and with the classical Newton method of second order. In the tables below the initial value (x0), the
number of iterations (N) to obtain the approximation, the number of evaluations of the function (NOFE) and the error for
f(xn)are shown. The abbreviation (ND) indicates that the result is not defined in the paper referred above.
Table 1: Number of iterations and solution obtained for different methods
f(x) = sin2x−x2+ 1
Method x0N Obtained solution f(xn)NOFE
Newton 1 6 1.40449164821621534112 2.75E-16 12
3 6 1.40449164821621534112 2.75E-16 12
[12] 1 4 1.40449164821621 1E-14 12
3 3 1.40449164821621 1E-14 9
[8] 1 3 ND 1E-14 9
3 3 ND 1E-14 9
[6] 1 4 ND 1E-14 12
[3] 1 4 ND 1E-15 9
New 1 3 1.40449164821534134 2.75E-16 9
3 4 1.40449164821621534112 2.75E-16 12
The exact solution expected is r= 1.40449164821534122603509
Table 2: Number of iterations and solution obtained for different methods
f(x) = cos x−x
Method x0N Obtained solution f(xn)NOFE
Newton -0.3 6 0.73908513321416067 5.12E-16 12
1 4 0.73908513321416067 5.12E-16 8
[12] -0.3 3 0.739085133214758 1E-14 9
1 2 0.739085133214758 1E-14 6
1.7 3 0.739085133214758 1E-14 9
[2] 0 3 ND 1E-10 9
New -0.3 3 0.73908513321416067 5.12E-16 9
1 3 0.73908513321416067 5.12E-16 9
1.7 3 0.73908513321416089 4.22E-16 9
The exact solution expected is r= 0.739085133215160641655312
Table 3: Number of iterations and solution obtained for different methods
f(x) = (x−1)3−1
Method x0N Obtained solution f(xn)NOFE
Newton 2.5 6 2 0 12
[12] 2.5 4 2 0 10
[8] 2.5 3 ND ND 9
New 2.5 4 2 0 12
The exact solution expected is r= 2
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510 International Journal of Nonlinear Science, Vol.13(2012), No.4, pp. 505-512
Table 4: Number of iterations and solution obtained for different methods
f(x) = x3−10
Method x0N Obtained solution f(xn)NOFE
Newton 1.5 6 2.15443469003188381 1.29E-15 12
[12] 1.5 4 2.15443469003367 1E-14 12
[8] 1.5 4 ND 1E-14 12
New 1.5 3 2.15443469003188381 1.29E-15 9
The exact solution expected is r= 2.15443469003188372175929
Table 5: Number of iterations and solution obtained for different methods
f(x) = x3−2x2−5
Method x0N Obtained solution f(xn)NOFE
Newton 2 6 2.69064744802861355 1.29E-15 12
[2] 2 3 ND 1E-10 9
New 2 3 2.69064744802861355 1.29E-15 9
The exact solution expected is r= 2.69064744802861375035079
Table 6: Number of iterations and solution obtained for different methods
f(x) = 1
x−sin x+ 1
Method x0N Obtained solution f(xn)NOFE
Newton -1.3 25 -0.62944648407333337 1.25E-16 50
[2] -1.3 4 ND 1E-15 12
New -1.3 4 -0.62944648407333337 1.25E-16 12
The exact solution expected is r=−0.629446484073333329964537
Table 7: Number of iterations and solution obtained for different methods
f(x) = x3+ 4x2−10
Method x0N Obtained solution f(xn)NOFE
Newton 1 5 1.36523001341409689 6.95E-16 10
2 5 1.36523001341409689 6.95E-16 10
[12] 1 3 1.36523001341448 1E-14 9
2 3 1.36523001341448 1E-14 9
[8] 1 3 ND 1E-14 9
2 3 ND 1E-14 9
[2] 1 4 ND 1E-14 12
2 4 ND 1E-14 12
[7] 1 3 ND 1E-15 9
2 3 ND 1E-14 9
[6] 1 3 ND 1E-15 9
2 3 ND 1E-14 9
New 1 3 1.36523001341409689 6.95E-16 9
2 3 1.36523001341409689 6.95E-16 9
The exact solution expected is r= 1.36523001341409684576081
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G. Fern´
andez-Torres: A New Derivative-free Iterative Method for Solving Nonlinear Equations with Third Order Convergence 511
Table 8: Number of iterations and solution obtained for different methods
f(x) = ex−3x2
Method x0N Obtained solution f(xn)NOFE
Newton 0.5 6 0.91000757248870912 -1.63E-16 12
[4] 0.5 4 0.9100075727 ND 16
[1] 0.5 11 0.910007572489 0.0268E-12 33
New 0.5 3 0.91000757248870912 -1.63E-16 9
The exact solution expected is r= 0.910007572488709060657338
Table 9: Number of iterations and solution obtained for different methods
f(x) = x2sin x−cos x
Method x0N Obtained solution f(xn)NOFE
Newton 6 5 6.30830895523815105 -1.31E-14 10
[6] 6 3 6.3083089552381511 -1.32E-14 9
New 6 3 6.30830895523815105 -1.31E-14 9
The exact solution expected is r= 6.30830895523815137755327
Table 10: Number of iterations and solution obtained for different methods
f(x) = sin x−x
2
Method x0N Obtained solution f(xn)NOFE
Newton 1.6 5 1.89549426703398075 3.65E-16 10
2 4 1.89549426703398075 3.65E-16 8
[7] 1.6 3 ND 1E-14 9
2 2 ND 1E-14 6
New 1.6 3 1.89549426703398094 6.17E-18 9
2 3 1.89549426703398094 6.17E-18 9
The exact solution expected is r= 1.895494267033980947144036
5 Conclusions
The tables show that the new method can compete with the other methods which are presented in [1–8, 12]. The methods
presented in those papers use the derivatives of the function, except in [1, 2, 5], but the results of the new method are more
accurate.
In this paper, we have developed a new simple iterative method given by (7) with the following important properties:
i. Any derivatives of f(x)is not required.
ii. The convergence order of the method is 3. In practice, for many numerical examples the present method is better
than the Newton’s method and equal to or better than other methods that use the derivatives and have the same convergence
order or higher.
iii. As in Muller’s method, the use of square root in the solution allows this method to obtain complex roots of
polynomials.
Acknowledgments
The author wishes to acknowledge the valuable participation of Professor Nicole Mercier and Professor Ted Fondulas in
the proof-reading of this paper.
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512 International Journal of Nonlinear Science, Vol.13(2012), No.4, pp. 505-512
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