A New Derivative-free Iterative Method for Solving Nonlinear ...

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 Fernández-Torres∗ Department of Oil Engineering, Universidad Del Istmo Tehuantepec, Oaxaca, 70760, México (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 using 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 x0 sufficiently close to the desired root and the iterative method is defined as xn+1 = xn +

6(xn − xn−1 )xn K , (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−1 to 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 r of f (x). Take −1 < β < 1 sufficiently 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. ∗ Corresponding

author.

E-mail address: [email protected] c Copyright⃝World Academic Press, World Academic Union IJNS.2012.06.30/631

(1)

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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 ) = f (x0 + β) = f (x0 − β) =

K = C = p(x0 ), L = βB + C = p(x0 + β), M = 6β 2 x0 A − βB + C = p(x0 − β).

(2) (3) (4)

The solutions of the system are very clear C = K, B =

L−K M + L − 2K . ,A = β 6β 2 x0

(5)

Now, consider the derivatives of polynomial p(x) p′ (x) = 3A(x − x0 )2 − Aβ 2 + 6Ax0 (x − x0 ) − 3Aβx0 + B = 3A(x − x0 )(x + x0 ) − Aβ 2 − 3Aβx0 + B, p′′ (x) = 6Ax, p′′′ (x) = 6A. The function f (x) can be approximated with the Taylor’s polynomial of second order about x0 and since p(x0 ) = C, p′ (x0 ) = −Aβ 2 − 3Aβx0 + B, p′′ (x0 ) = 6Ax0 , p′′′ (x0 ) = 6A, we have f (x) ≈ C + (−Aβ 2 − 3Aβx0 + B)(x − x0 ) + 3Ax0 (x − x0 )2 .

(6)

Observe that the coefficient of x − x0 in the above equation can be written as −Aβ 2 − 3Aβx0 + B = −Aβ(3x0 + β) + B = =

L−M 2K − L − M 3x0 (L − M ) + β(2K − L − M ) = + . 6βx0 2β 6x0

Define S as S =L−M +

β(2K − L − M ) . 3x0

For x = r, and the above equations in (6), we have 6Ax0 β(r − x0 )2 + S(r − x0 ) + 2βK ≈ 0. We can resolve the above equation to give r ≈ x0 + that can be written as r ≈ x0 +

−S ±

√

S 2 − 48AKβ 2 x0 12Aβx0

√ β [S ∓ S 2 + 8K(2K − L − M )]. 2(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 4Kβ √ r ≈ x0 − 2 S ± S + 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 r of f (x). Take −1 < β < 1 sufficiently small, β ̸= 0, then for n ≥ 0 we have 4Kβ √ , (7) xn+1 = xn − 2 S + sign(S) S + 8K(2K − L − M ) with S =L−M +

β(2K − L − M ) , 3xn

IJNS email for contribution: [email protected]

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G. Fernández-Torres: A New Derivative-free Iterative Method for Solving Nonlinear Equations with Third Order Convergence

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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 β to 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 ∈ I be 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 β the maximum convergence order is 3. Proof. Suppose that the point xn has 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 ξ1 between xn − β and xn + β. Thus, f (x) can be written as C + (−Aβ 2 − 3Aβxn + B)(x − xn ) + 3Axn (x − xn )2 + (x − xn )3

f ′′′ (ξ1 ) f ′′′ (ξ1 ) − β 2 (x − xn ) + β 2 (x − xn )A. 6 6

If xn+1 is a root of polynomial (6), C + (−Aβ 2 − 3Aβxn + B)(x − xn ) + 3Axn (x − xn )2 then f (xn+1 ) = (xn+1 − xn )3

f ′′′ (ξ1 ) f ′′′ (ξ1 ) − β 2 (xn+1 − xn ) + β 2 (xn+1 − xn )A. 6 6

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 ′′′ (ξ1 ) 3 f ′′′ (ξ1 ) f (0) = − xn + β 2 xn − Axn 6 6 We know that we can write (5) using divided differences and for ξ2 ∈ (xn − β, xn + β), A can be expressed as A=−

f ′′ (ξ2 ) . 3xn

Therefore

f ′′′ (ξ1 ) f ′′ (ξ2 ) + β2 . 6 3 We now assume that the points xn , xn−1 lie 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 (β 2 − x2n )

[ ] f (ξ1 ) f ′′ (ξ2 ) + (xn − xn−1 )4 . f (0) = xn (xn − xn−1 )4 − x2n 6 3 b)β = xn

√ |xn − xn−1 | f (0) = x3n [xn − xn−1 − 1]

f (ξ1 ) f ′′ (ξ2 ) + x2n (xn − xn−1 ) . 6 3

IJNS homepage: http://www.nonlinearscience.org.uk/

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c)β = xn (xn − xn−1 ) [ ] f (ξ1 ) f ′′ (ξ2 ) f (0) = x3n (xn − xn−1 )2 − 1 + x2n (xn − xn−1 )2 . 6 3 d)β = [xn (xn − xn−1 )]9/8 [ ] f (ξ ) ′′ 1 9/4 f (ξ2 ) f (0) = xn [xn (xn − xn−1 )]9/4 − x2n + [xn (xn − xn−1 )] . 6 3 We assume that the magnitude of the quantity εn is less than some upper bound εm . With this we make the assumption εn+1 < εm that is clear by the equality xn+1 − r = (xn+1 − y) + (y − r) with y the root of polynomial p(x). Note that εn+1 = O(ε3m ), thus if we expand all functions around r, we obtain [ ′′ ] f (r) a)εn+1 f ′ (r) = ε4n−1 + O(ε3m ). 3 [

] f ′′ (r) b)εn+1 f (r) = + O(ε3m ). 3 [ ′′ ] f (r) ′ 2 2 c)εn+1 f (r) = εn εn−1 + O(ε3m ). 3 [ ′′ ] f (r) 9/4 d)εn+1 f ′ (r) = ε9/4 ε + O(ε3m ). n n−1 3 ′

ε2n εn−1

We suppose that εn+1 is asymptotic to cεpn with p > 1. Consequently, by expressing the above equations in terms of εn−1 and neglecting the terms of O(εm ), we obtain [ ′′ ] 2 f (r) p+1 p 4 a)c εn−1 ∼ εn−1 . 3f ′ (r) ] f ′′ (r) . 3f ′ (r) ] [ ′′ 2 2p+2 f (r) p+1 p c)c εn−1 ∼ εn−1 . 3f ′ (r) [ ′′ ] 9 2 f (r) 4 (p+1) d)cp+1 εpn−1 ∼ εn−1 . 3f ′ (r) 2

[

p b)cp+1 εn−1 ∼ ε2p+1 n−1

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 = 94 (p + 1) therefore the maximun value is p = 3. 3/2

For any other value of β, such as β = xn (xn − xn−1 )2 or β = [xn (xn − xn−1 )] ), it signifies that the term ϵ2n ε4n−1 or ϵ3n ε3n−1 gives 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 [ ′′′ ] ′ 3 f (r) + O(ε3m ) εn+1 f (r) = −εn 6 Thus the associated equation will be

εn+1 f ′ (r) f ′′′ (r) + O(ε3m ) =− 3 εn 6

therefore the convergence order will also be 3.



4

509

Numerical experiments

We present some numerical test functions taking into account the proposed method. The test functions are the following 1) f (x) = sin2 x − x2 + 1. 6) f (x) = x1 − sin x + 1. 2) f (x) = cos x − x. 7) f (x) = x3 + 4x2 − 10. 3 3) f (x) = (x − 1) − 1. 8) f (x) = ex − 3x2 . 3 4) f (x) = x − 10. 9) f (x) = x2 sin x − cos x. 3 2 5) f (x) = x − 2x − 5. 10) f (x) = sin x − x2 . 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) = sin2 x − x2 + 1 x0 N Obtained solution f (xn ) NOFE 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

Method Newton

Table 2: Number of iterations and solution obtained for different methods f (x) = cos x − x x0 N Obtained solution f (xn ) NOFE -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

Method Newton

Table 3: Number of iterations and solution obtained for different methods f (x) = (x − 1)3 − 1 Method Newton [12] [8] New

x0 2.5 2.5 2.5 2.5

N Obtained solution f (xn ) 6 2 0 4 2 0 3 ND ND 4 2 0 The exact solution expected is r = 2


NOFE 12 10 9 12

510


Table 4: Number of iterations and solution obtained for different methods f (x) = x3 − 10 Method x0 N 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 x0 N 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) = x1 − sin x + 1 Method x0 N 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 x0 N Obtained solution f (xn ) NOFE 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

Method Newton



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Table 8: Number of iterations and solution obtained for different methods f (x) = ex − 3x2 Method x0 N 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) = x2 sin x − cos x Method x0 N 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 − x2 x0 N Obtained solution f (xn ) NOFE 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

Method Newton

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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References [1] I. Abu-Alshaikh. A New Iterative Method for Solving Nonlinear Equations. World Academic Science, Engineering and Technology, 5(2005):190-193. [2] G. Calderón. Métodos Iterativos para Resolver Ecuaciones no Lineales. Revista Notas de Matemática, 3(1) (2007):33-44. [3] C. Chun. A Family of Composite Fourth-Order Iterative Methods for Solving Nonlinear Equations. Applied Mathematics and Computation, 2 (2007):951-956. [4] J. Feng. A New Two-step Method for Solving Nonlinear Equations. International Journal of Nonlinear Science, 8 (2009):40-44. [5] G. Fernández-Torres. A New Derivative-free Iterative Method for Solving Nonlinear Equations, preprint. [6] K. Jisheng, L. Yitian, W. Xiuhua. A Modification of Newton Method with Third Order Convergence. Applied mathematics and computation, 181 (2006):1106-1111. [7] K. Jisheng, L. Yitian, W. Xiuhua. Composite Fourth-Order Iterative Method for Solving non-linear Equations. Applied Mathematics and Computation, 2 (2007):951-956. [8] T. Lukić, N. M. Ralević. Newton’s Method with Accelerated Convergence Modified by an Aggregation Operator. Proceedings of 3rd Serbian-Hungarian Joint Symposium on Intelligent Systems, SCG, Subotica, (2005):121-128. [9] J. M. Ortega and W. C. Reinboldt. Iterative Solution of Nonlinear Equations in Several Variables. Academic Press, New York. 1970. [10] A. M. Ostrowski. Solution of Equations and Systems of Equations. Academic Press, New York. 1960. [11] J. F. Traub. Iterative Methods for the Solutions of Equations. Chelsea publishing, New York. 1977. [12] S. Weerakoon, and T. G. I. Fernando. A Variant of Newtons Method with Accelerated Third Order Convergence. Applied Mathematics Letters, 13 (2000):87-93.
