Applied Mathematical Sciences, Vol. 8, 2014, no. 161, 8021-8027 HIKARI Ltd, www.m-hikari.com http://dx.doi.org/10.12988/ams.2014.49710 A Two-step Iterative Method Free from Derivative for Solving Nonlinear Equations Alyauma Hajjah 1, M. Imran and M. D. H. Gamal Numerical Computing Group, Department of Mathematics University of Riau, Pekanbaru 28293, Indonesia 1 Corresponding author Copyright c 2014 Alyauma Hajjah, M. Imran and M. D. H. Gamal. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Abstract We address a two-step iterative method to solve a nonlinear equation, which is free from derivative by approximating a derivative in the two-step King method by the method of forward difference with one parameter θ. We show analytically that the method is of an order three for a simple root. Numerical experiments show that the new method gives the encouraging results. Although the efficiency index of our method is worse than that of King s methods, numerical experiments show that the proposed method is comparable to the existing methods in terms of the number of iterations. Mathematics Subject Classification: 65H05, 65D99 Keywords: Forward difference, free derivative method, iterative method, order of convergence 1 Introduction Having a particular technique to solve a nonlinear equation f(x) = 0 (1)
8022 Alyauma Hajjah, M. Imran and M. D. H. Gamal is still an active research in numerical analysis. Not all cases of equation (1) can be analytically solved, so solving numerically becomes an alternative. A numerical method used for finding the solution of the nonlinear equation (1) is a two-step iterative method, that is a combination of Newton method [1, h.68] and Halley method [10, h.86] with order of convergence three. We express their formula as y n = x n f(x n) f (x n ), 2f(y n )f (y n ) 2f (y n ) 2 (f(y n )f (y n )). The other two-step iterative methods are the iterative method proposed by Jisheng [6] in the form y n = x n f(x n) f (x n ), f 2 (x n ) + f 2 (y n ) f (x n )(f(x n ) f(y n )), and the one proposed by Weerakon and Fernando [11] in the form y n = x n f(x n) f (x n ), 2f(x n ) f (x n ) + f (y n ). Two-step iterative method free from derivative is a method proposed by Dehghan-Hajarian [3], with order of convergence three (Dehghan Method 1). We can express their formula as 2f(x n ) 2 y n =x n f(x n + f(x n )) f(x n f(x n )), (2) x n+1 =x n and (Dehghan Method 2) 2f(x n ) (f(x n ) + f(y n )) f(x n + f(x n )) f(x n f(x n )). (3) 2f(x n ) 2 y n =x n + f(x n + f(x n )) f(x n f(x n )), (4) x n+1 =x n 2f(x n ) (f(y n ) f(x n )) f(x n + f(x n )) f(x n f(x n )). (5)
A two-step iterative method free from derivative 8023 Some other papers discuss the two-step type iterative methods and their applications, these include [2]. The purpose of this paper is to develop a new two-step iterative method free from derivative and provide the convergence analysis. We present this new method in section 2 in this paper. In the new method we approximate the derivative of the function by forward difference with one parameter θ. Numerical examples show better performance of our method in section 3. We conclude our discussion in section 4. 2 Proposed Iterative Method and Analysis of Convergence The iterative method discussed in this paper is that of proposed by King method with order of convergence four [7] in the form y n =x n f(x n) f (x n ), (6) f(x n ) + βf(y n ) f(y n ) x n+1 =y n f(x n ) + (β 2)f(y n ) f (x n ). (7) The derivative f (x n ) in equation (6) and equation (7) are approximated by the forward difference with one parameter θ [1], f (x n ) f(x n + θf(x n )) f(x n ). (8) θf(x n ) Substituting equation (8) into equation (6) and equation (7), we have θf 2 (x n ) y n = x n f(x n + θf(x n )) f(x n ), (9) f(x n ) + βf(y n ) θf(x n )f(y n ) f(x n ) + (β 2)f(y n ) f(x n + θf(x n )) f(x n ). (10) Equations (9) and (10) are called Two-Step Iterative Method Free Derivative. Now we are going to prove the order of convergence of the method. Theorem 1 Let f : D R for an open interval D. Assume that f has sufficiently continuous derivatives in the interval D. If x has a simple root at f(x), and if x 0 is sufficiently close to x, then the new iterative method defined by equation (9) and (10) satisfies the following error equation: e n+1 = ( c 2 2θ 2 F 2 1 c 2 2θF 1 )e 3 n + O(e 4 n), (11)
8024 Alyauma Hajjah, M. Imran and M. D. H. Gamal where F 1 = f (x ), c j = f (j) (x ) j!f (x ), j > 1, and e n = x n x. Proof. Let x is be a simple root of f(x) = 0, then f(x ) = 0. Let e n = x n x. With Taylor expansion of f(x n ) about x n = x, we obtain f(x n ) = F 1 (e n + c 2 e 2 n + c 3 e 3 n + O(e 4 n)). (12) Computing f 2 (x n ) using equation (12), then multiplied by θ, we get θf 2 (x n ) = F 2 1 ( θe 2 n + 2θc 2 e 3 n + O(e 4 n) ). (13) Then, computing f(x n + θf(x n )), we have f(x n + θf(x n )) = (F 1 + θf 2 1 )e n + (3c 2 θf 2 1 + c 2 θ 2 F 3 1 + c 2 F 1 )e 2 n + (2c 2 2θ 2 F 3 1 + 4c 3 θf 2 1 + c 3 θ 3 F 4 1 + 2c 2 2θF 2 1 + c 3 F 1 + 3c 3 θ 2 F 3 1 )e 3 n + O(e 4 n). (14) Using equation (12) and (14), we compute f((x n ) + θf(x n )) f(x n ), then we have f((x n ) + θf(x n )) f(x n ) = θf 2 1 e n + (3c 2 θf 2 1 + c 2 θ 2 F 3 1 )e 2 n + (2c 2 2θ 2 F 3 1 + 3c 3 θ 2 F 3 1 2c 2 2θF 2 1 + 4c 3 θf 2 1 + c 3 θ 3 F 4 1 )e 3 n + O(e 4 n). (15) Considering a geometric series and computing and (15), we obtain after simplifying θf(x n) 2 f((x n)+θf(x n)) f(x n) using (13) θf(x n ) 2 f((x n ) + θf(x n )) f(x n ) = e n + ( c 2 c 2 θf 1 )e 2 n + ( 2c 3 3c 3 θf 1 + c 2 2θ 2 F 2 1 + 2c 2 2 + 2c 2 2θF 1 c 3 θ 2 F 2 1 )e 3 n + O(e 4 n). (16) Substituting equation (16) into equation (9) and x n = x + e n, we get y n = x + (c 2 + c 2 θf 1 )e 2 n + (2c 3 2θc 2 2F 1 + c 3 F 2 1 θ 2 c 2 2F 2 1 θ 2 Then, Taylor expansion of f(y n ) about y n = x, gives 2c 2 2 + 3θF 1 c 3 )e 3 n + O(e 4 n). (17) f(y n ) =(c 2 θf 2 1 + c 2 F 1 )e 2 n + (c 3 θ 2 F 3 1 2c 2 2θF 2 1 2c 2 2F 1 c 2 2θ 2 F 3 1 + 3c 3 θf 2 1 + 2c 3 F 1 )e 3 n + O(e 4 n), (18) and computing f(x n)+βf(y n) f(x n)+(β 2)f(y n), we have
A two-step iterative method free from derivative 8025 f(x n ) + βf(y n ) f(x n ) + (β 2)f(y n ) = 1 + (2c 2 + 2c 2 θf 1 )e n + (2c 2 2θ 2 F 2 1 + 4c 3 2c 2 2β + 2c 2 2θF 1 + 2c 3 θ 2 F 2 1 4βc 2 2θF 1 2βc 2 2θ 2 F 2 1 2c 2 2 + 6c 3 θf 1 )e 2 n + (4βc 3 2 4βc 2 c 3 θ 3 F 3 1 20βc 2 c 3 θf 1 + 2c 4 θ 3 F 3 1 + 2c 3 2θ 2 F 2 1 16βc 2 c 3 θ 2 F 2 1 4c 3 2θF 1 Then, computing + 10c 4 + 8c 4 θ 2 F 2 1 + 2β 2 c 3 2 + 12c 4 θf 1 + 16c 2 c 3 θ 2 F 2 1 + 2c 3 2θ 3 F 3 1 + 12c 2 c 3 θf 1 + 4c 2 c 3 θ 3 F 3 1 4βc 3 2θ 3 F 3 1 8βc 3 c 2 + 6β 2 c 3 2θ 2 F 2 1 8βc 3 2θ 2 F 2 1 + 6β 2 c 3 2θF 1 + 2β 2 c 3 2θ 3 F 3 1 )e 3 n + O(e 4 n). (19) θf(x n)f(y n) f((x n)+θf(x n)) f(x n), we obtain θf(x n )f(y n ) f((x n ) + θf(x n )) f(x n ) =(c 2 + θc 2 F 1 )e 2 n + ( 5c 2 2θF 1 + c 3 θ 2 F 2 1 4c 2 2 + 2c 3 + 3c 3 θf 1 2c 2 2θ 2 F 2 1 )e 3 n + O(e 4 n). (20) Substituting equations (19),(20) dan (17) into equation (10), noting x n+1 x = e n+1, then after simplifying we have e n+1 = ( c 2 2θ 2 F 2 1 c 2 2θF 1 )e 3 n + O(e 4 n). This ends the proof. 3 Numerical Experiments In this section, we give the results of some numerical examples to compare Dehghan Method 1 (DM1) by equations (8)-(9), Dehghan Method 2 (DM2) by equations (10)-(11), King Method (KM) by equations (12)-(13) and Proposed Free Derivative Method (PFDM) by equations (15)(16) for the solution of the nonlinear equations. In this comparison we use the following test functions: f 1 = cos(x) x x = 0.7390851332151606 [5] f 2 = sin(x) 0.5 x = 0.5235987755982989 [8, 9] f 3 = sin 2 (x) x 2 + 1 x = 1.4044916482153412 [5] f 4 = x 3 x + 3 x = 1.6716998816571610 [8, 9] f 5 = tan(x) cos(x) 0.5 x = 0.8570567764718169 [8, 9] All the computation is done in Maple 13. We use tolerance, ɛ = 1.0 10 15, and the maximum number of iteration allowed is 100. We stop the iteration process by the following criteria 1. x n+1 x n < ɛ 2. f(x n+1 ) < ɛ.
8026 Alyauma Hajjah, M. Imran and M. D. H. Gamal In Table 1, 100+ indicates that the method does not converge after the maximum iteration allowed is reached, star sign (*) in the number of iterations indicates that the method converges to a different root. From Table 1 one can see that the computational results obtained are not far different. In f 1 for initial guess -1.5 DM1 and DM2 require 5 iterations, KM requires 36 iterations, and PFDM requires 4 iterations. For initial guess -1.0 DM1 and DM2 require 5 iterations, KM and PFDM require 3 iterations. So the quickest method to hit the root is PFDM. In f 2 the method having the least iteration is PFDM. As far as the numerical results are concerned, for most of the functions we tested, the proposed method can be competitive with the methods we are comparing. f 2 f 3 f 4 f 5 Table 1: Comparisons of the discussed methods The number of iterations Functions x 0 DM1 DM2 KM PFDM x 1.0 5 5 4 4 f 1 0.5 6 5 3 3 0.7390851332151606 1.5 5 5 36 4 0.0 4 4 3 3 0.0 3 3 3 3 0.5 2 2 2 2 0.5235987755982989 0.2 3 3 3 3 1.0 4 4 3 3 0.5 5 9 6 5 1.3 3 3 2 3 1.4044916482153412 0.3 7 7 9 5 1.5 3 3 2 3 0.9 4 17 90 4 0.5 5 100+ 23 5 1.6716998816571610 1.7 3 3 2 3 0.0 6 100+ 55 8 0.7 3 4 3 3 1.1 7 100+ 3 7 0.8570567764718169 0.5 3 6 3 4 2.5 3 4 3 9 4 Conclusion We have put forward a new two-step iterative methods free derivative to solve nonlinear equation whose efficiency index is 3 1/3 = 1.44225. Although the efficiency index of our method is worse than that of King s methods, numerical
A two-step iterative method free from derivative 8027 experiment show that it is comparable to the existing method in terms of the number of iterations. References [1] K. E. Atkinson. Elementary Numerical Analysis, 2 nd Ed. John Wiley, New York, 1993. [2] A. Cordero, J. L. Hueso, E. Martinez and J. R. Torre-grosa, 2012. Steffensen type methods for solving non-linear equations, Journal of Computational and Applied Mathematics, 236, 3058-3064. http://dx.doi.org/10.1016/j.cam.2010.08.043 [3] M. Dehghan and M. Hajarian. Some Derivative Free Quatratic and Cubic Convergence Iterative Formulas for Solving Nonlinear Equation, J. Comput. Appl. Math, 29 (2010), 19-31. http://dx.doi.org/10.1590/s1807-03022010000100002 [4] W. Gautschi. Numerical Analysis, 2 nd Ed, Birkhauser, New York, 2012. http://dx.doi.org/10.1007/978-0-8176-8259-0 [5] J. P. Jaiswal. A New Third-Order Derivative Free Method for Solving Equations, Univ. J. Appl. Math. 1 (2013), 131-135. [6] K. Jisheng, L. Yetian and W. Xiuhui. A Composite Fourth-Order Iterative Method for Solving, Appl. Math. Comput, 184 (2006), 471-475. [7] R. F. King. A Family of Fourth Order Methods for Nonlinear Equations, SIAM J. Numer. Anal. 10 (1973), 876-879. http://dx.doi.org/10.1137/0710072 [8] D. Nerinckx and D. Haegenans. A Comparison of Non-linear Equation Solver, J. Comput. Appl. Math. 2 (1976), 145148. http://dx.doi.org/10.1016/0771-050x(76)90017-6 [9] J. R. Rice. A set of 74 test functions for non-linear equation solvers, Report Purdue University CSD TR 34 (1969) [10] R. Wait. The Numerical Solution of Algebraic Equation, Jhon Wiley dan Sons, New York, 1979. [11] S. Weerakon and T.G.I. Fernando. A Variant of Newton s Method with Accelarated Third Order Convergence, Appl. Math. letters, 13 ( 2000), 87-93. http://dx.doi.org/10.1016/s0893-9659(00)00100-2 Received: September 7, 2014; Published: November 19, 2014