A Three-Step Iterative Method to Solve A Nonlinear Equation via an Undetermined Coefficient Method

Transcription

1 Global Journal of Pure and Applied Mathematics. ISSN Volume 14, Number 11 (018), pp Research India Publications A Three-Step Iterative Method to Solve A Nonlinear Equation via an Undetermined Coefficient Method Nora Fitriyani, M. Imran, Syamsudhuha Numerical Computing Group, Department of Mathematics University of Riau, Pekanbaru 893, Indonesia. Abstract This article discusses a three-step iterative method which is a combination of the iterative method that contains the function and first derivative with Newton s method. The process of the combination uses the principle of an undetermined coefficient method by allowing only one additional function evaluation that may occur in the process. By combining the methods proposed by Khattri and Abbasbandy [Math. Vesn., 63 (011), 67-7] with Newton s method, then through the convergence analysis it was shown that the proposed method has a sixth order convergence and requires two function and two first derivative evaluations in each iteration. Hence its efficiency index is Some examples of nonlinear equations are solved using the proposed method. Then the obtained solutions are compared with those of other mention methods to see the effectiveness of proposed methods. AMS Subject classification: 65H05 Keywords: Efficiency index, three-step iterative method, order of convergence 1. INTRODUCTION Finding the root of the nonlinear equation in the form of f(x) = 0 (1) is one of the important problems in numerical analysis. To find the simple root of the equation (1) many methods can be used and the most famous classical method is Newton s method with the following iteration: x n+1 = x n f(x n), n = 0,1,,, f (x n ) where x 0 is the given initial approximation. This method has second order convergence [1, h.58].

2 146 Nora Fitriyani, M. Imran, Syamsudhuha Many researchers have modified Newton s method which produces high orders convergence such as [3] - [6], [8], [10], [11], [13] dan [15]. Fang et al. [8] modified Newton s method with five evaluation functions and produced a sixth order convergence method having the following iterations: f(x n ) y n = x n α n f(x n ) + f (x n ), f(y n ) z n+1 = y n β n f(y n ) + f (y n ), x n+1 = z n f(z n ) γ n f(z n ) + f (y n ), where is α n, β n, γ n real numbers chosen in such a way that 0 α n, β n, γ n 1, and sign(α n f(x n )) = sign(f (x n )), sign(β n f(y n )) = sign(f (y n )), sign(γ n f(z n )) = sign(f (y n )), where n = 1,,..., and sign(x) is a sign function. Sharma and Guha [16] modified Ostrowski s method having four function evaluations and a sixth order convergence. Their formula is given as follows: y n = x n f(x n) f (x n ), z n = y n f(y n) f(x n ) f (x n ) f(x n ) f(y n ), x n+1 = z n f(z n) f(x n ) + af(y n ) f (x n ) f(x n ) + bf(y n ), where a and b are parameters, with b = a. Grau and Diaz-Barero [10] introduced a new method by correcting Ostrowski s method. The method has a sixth order convergence with the following iterations: y n = x n f(x n) f (x n ), f(x n ) f(y n ) z n = y n f(x n ) f(y n ) f (x n ), f(x n ) f(z n ) x n+1 = z n f(x n ) f(y n ) f (x n ).

3 A Three-Step Iterative Method to Solve a Nonlinear Equation 147 Khattri and Abbasbandy [1] introduced an iterative method having a fourth order convergence with the following formula: f(x n ) y n = x n, 3 f (x n ) x n+1 = y n [1 + ( 1 8 )(f (y n) ) + f (x n ) ( 9 )(f (y n) f (x n ) ) + ( 15 8 )(f (y n) f (x n ) )3 ] f(x n). f (x n ) In this article, the authors present a combination of the Khattri and Abbasbandy method [1] with Newton s method using the principle of an undetermined coefficient method. The derivation and analysis of convergence of the proposed method are presented in section two. In section three numerical computations are carried out on the four test functions. Then the results are compared to the obtained results from the some known methods to see the efficiency of the proposed method.. DEVELOPMENT OF METHOD AND CONVERGENCE ANALYSIS Suppose that φ(x; f(x), f (x), f (y)) is a function from R to R. Consider the modification of Newton s method given as follows: z n = φ(x n ; f(x n ), f (x n ), f (y n )), x n+1 = z n f(z n) f (z n ). } () to eliminate f (z n ) in (), first we consider the following equation: f (z n ) = Af(x n ) + Bf (x n ) + Cf (y n ) + Df(z n ) + Ef ( x n+y n ). (3) Next we expand f(x), f (x) about x n to the fourth derivative, and evaluate f (y n ), f(z n ), f ( x n+y n ), and f (z n ), then substitute the resulting equations into equation (3). On comparing the coefficient of derivative of f at x n, we obtain A + D = 0, (4) B + C + D(z n x n ) + E = 1, (5) C(y n x n ) + D (z n x n ) + E (y n x n ) = (z n x n ), (6) C (y n x n ) + D (z 6 n x n ) 3 + E (y 8 n x n ) = (z n x n ), (7) C (y 6 n x n ) 3 + D (z 4 n x n ) 4 + E (y 48 n x n ) 3 = (z n x n ) 3. (8) 6

4 148 Nora Fitriyani, M. Imran, Syamsudhuha Then on solving the system equation (4)-(8), we have A = (α β) ( β+α)α, (9) B = (3β3 9β α α 3 +7βα ), (10) 3β ( β+α) C = α (α β) 3β ( β+α), (11) D = (α β) ( β+α)α, (1) E = 4α (α β) 3β ( β+α), (13) where α = z n x n and β = y n x n. On substituting equations (9)-(13) into equation (3), we end up with f (z n ) = (α β) ( β + α)α f(x η n) + 3β ( β + α) f (x n) + α (α β) 3β ( β + α) f (y n) + (α β) f(z ( β+α)α n) + 4α (α β) 3β ( β+α) f (x n+y n ), (14) with η = 3β 3 9β α α 3 + 7βα. Replacing the value f ( x n+y n ) by the arithmetic mean of f (x n ) and f (y n ), we can rewrite (14) as (α β) f (z n ) = ( β + α)α f(x n) + ( β 3βα + α α )f (x β( β + α) n ) + ( β( β + α) )f (y n) +( (α β) )f(z ( β+α)α n), (15) with η = 3β 3 9β α α 3 + 7βα. Then on substituting equation (15) into the second step of equation (), we obtain x n+1 = z n with μ = α(β 3βα + α ). αβ( β+α)f(z n ) μf (x n )+( α 3 )f (y n )+β(α β)(f(z n ) f(x n )), (16) From equation (16) we suggest a new family of iterative method of the form z n = φ(x n ; f(x n ), f (x n ), f (y n )), x n+1 = z n αβ( β+α)f(z n ) μf (x n )+( α 3 )f (y n )+β(α β)(f(z n ) f(x n )),} (17)

5 A Three-Step Iterative Method to Solve a Nonlinear Equation 149 where μ = α(β 3βα + α ), α = z n x n, β = y n x n and φ(x n ; f(x n ), f (x n ), f (y n )) is any fourth order method. Furthermore, using the fourth order method derived by Khattri and Abbasbandy [1], we propose the following iterative method. f(x n ) y n = x n 3 f (x n ) (18) z n = x n [1 + ( 1 8 )(f (y n) f (x n ) ( 9 )(f (y n) f (x n ) ) + ( 15 8 )(f (y n) f (x n ) )3 ] f(x n) f (x n ) (19) αβ( β+α)f(z x n+1 = z n n ) μf (x n )+( α 3 )f (y n )+β(α β)(f(z n ) f(x n )) (0) with μ = α(β 3βα + α ), α = z n x n, β = y n x n. The method defined in equation (18)-(0) is called the method MKA whose convergence analysis is given in Theorema 1. Teorema 1 (Order of Convergence) Suppose that λ I is a simple zero of a function f: I R R sufficiently differentiable at open interval I. If x 0 is close enough to λ, the iterative method defined in equation (18)-(0) has a sixth order convergence and satisfies the error equation: e n+1 = ( 85 9 A A 4 + A )e n 6 + O(e n 7 ), with and e n = x n λ. A j = f(j) (λ), j =,3,,8, j!f (λ) Proof. Suppose that λ is a simple root of f(x) = 0 then f(λ) = 0, and f (λ) 0. Then by Taylor s expansion [, h.16] of f(x) about λ until the sixth order derivative, we get f(x) = f(λ) + f (λ)(x n λ) + f () (x λ) (λ)! +f 5 (λ) (x λ)5 5! + f 4 (x λ)4 (λ) 4! + f 6) (λ) (x λ)6 6! + f (3) (x λ)3 (λ) 3! + O((x λ) 7 ). (1) By evaluating equation (1) at x = x n, and recalling e n = x n λ and f(λ) = 0, equation (1) can be written as f(x n ) = f (λ)(e n + A e n + e n 3 + A 4 e n 4 + A 5 e n 5 + A 6 e n 6 ) + O(e n 7 ), ()

6 1430 Nora Fitriyani, M. Imran, Syamsudhuha with Furthermore by the same way, we obtain A j = f(j) (λ), j =,3,,8. j!f (λ) f (x n ) = f (λ)(1 + A e n + 3 e n + 4A 4 e n 3 + 5A 5 e n 4 + 6A 6 e n 5 + 7A 7 e n 6 ) + O(e n 7 ). Then using equation (), equation (3) and the aid of geometric series [17], we have f(x n ) f (x n ) = e n A e n + (A )e n 3 + ( 4A 3 3A 4 + 7A )e n 4 + (6 4A 5 0A + 10A A 4 + 8A 4 )e n 5 + ( 16A A A A 4 +5A 3 5A 6 8A A 4 33A )e n 6 + O(e n 7 ). (4) (3) Next equation (4) is substituted to equation (18) and simplifying the resulting equation we get y n = λ e n + 3 A e n 3 (A )e n 3 3 ( 4A 3 3A 4 + 7A )e n 4 3 (6 4A 5 0A + 10A A 4 + 8A 4 )e n 5 3 ( 16A A A A 4 +5A 3 5A 6 8A A 4 33A )e n 6 + O(e n 7 ). (5) Following the idea how to obtain f (x n ), we have f (y n ) by considering equation (5) that is f (y n ) = f (λ)(1 + 3 A e n + ( 4 3 A )e n + (4A A A3 3 )e n +( 44 A 9 A A )e 4 n + ( 5 A 9 3A 4 1A A A A A 4 3 A5 5 3 )e n +( 64A A 3 8 A A 5 44 A 9 A A A A A3 7 A A )e 6 n ) + O(e 7 n ). (6)

7 A Three-Step Iterative Method to Solve a Nonlinear Equation 1431 From equation (3) and equation (6) and with the aid of the geometry series [17], we obtain f (y n ) f (x n ) = A e n + ( A )e n + ( A A )e n +( A A A A A 4 )e n 4 + ( 0 3 A +8 A A A A 6 64A A A A 3 )e n 5 +( 1376 A A A A A 9 A A A A A A A A 5 )e 6 n + O(e 7 n ). (7) Next equation (4) and equation (7) are substituted to equation (19) and simplifying the resulting equation, we end up with z n = λ + ( A A A 4)e 4 n + ( A A A A 5 0 A 9 A 4 )e 5 n + ( 306 A 7 A A A 3 10 A A A 6)e n 6 + O(e n 7 ). (8) Then by evaluating equation (1) at x = z n as in equation (8), we get f(z n ) = f (λ)(( A A A 4)e 4 n + ( A A A + 8 A A 9 A 4 )e 5 n + ( 306 A 7 A A A 3 10 A A A 7 6)e 6 n + O(e 7 n )). (9) Since α = z n x n, β = y n x n, from equation (8) and equation (5), we have respectively α = e n + ( 1 9 A A3 A )e 4 n + ( 0 9 A A A A 5 74 A4 9 )e 5 n + ( A A A 5 9 3A A A 7 A 4 )e 6 n + O(e 7 n ), (30) and β = 3 e n + 3 A e n 3 ( + A )e n 3 3 ( 4A 3 + 7A 3A 4 )e n 4 3 ( 4A A A A 4 0A )e n 5 3 (5A 3 16A 5 5A A A 5 33A 8A A A 4 )e n 6 + O(e n 7 ). (31)

8 143 Nora Fitriyani, M. Imran, Syamsudhuha Using equation (30), equation (31) and equation (9) we get αβ( β + α)f(z n ) = f (λ)(( 9 A A3 81 A 4)e 7 n + ( A A4 346 A A A 9 3)e 8 n + ( 1964 A A 43 A A 4 8 A A 43 A A A 7 A 3 )e 9 n + O(e 10 n )). (3) Next using equation (), (3), (6), (9), (30), and (31), the denominator of the equation (0) becomes β(α β)f(x n ) + α(β 3βα + α )f (x n ) + ( α) 3 f (y n ) + β(α β)f(z n ) = ( 9 e n 3 9 A e n 4 + ( A )e n 5 +( A A + 81 A 4)e n 6 + O(e n 7 ))f (λ). (33) Next by dividing (3) with (33) and with the aid of geometry series, we have αβ( β + α)f(z n ) β(α β)f(x n ) + α(β 3βα + α )f (x n ) + ( α) 3 f (y n ) + β(α β)f(z n ) = ( A A A 4)e n 4 + ( 74 9 A A A A A 4 )e n 5 + ( A A A A A A 4 A A 7 A A 7 6)e 6 n + O(e 7 n ). (34) On substituting equation (8) and equation (34) into equation (0), we obtain x n+1 = λ + ( 85 9 A A 4 + A )e n 6 + O(e n 7 ). (35) Noting e n+1 = x n+1 λ, then the equation (35) becomes e n+1 = ( 85 9 A A 4 + A )e n 6 + O(e n 7 ). (36) From the definition of the order of convergence [14, h.75], we see that equation (18)-(0) is of order six and Theorem 1 is proven. From formula equation (18)-(0) we recognize that the proposed method required four function evaluations per iteration, Thus based on the definition of the efficiency index [9, h.61], the efficiency index of MKA iterative method is

9 A Three-Step Iterative Method to Solve a Nonlinear Equation NUMERICAL SIMULATIONS In this section numerical simulations are performed which aims to compare the number of iterations required by Fang s method (MF) defined in [8], Sharma s method (MSh) defined in [16], Grau s method (MG) which is defined in [10], and the MKA method. For this purpose we use the following nonlinear equations: 1. f 1 (x) = cos(x) x, [3],. f (x) = x 3 + 4x 10, [5], 3. f 3 (x) = (x 1) 3 1, [18], 4. f 4 (x) = sin(x) x,[6]. In doing comparisons, tolerance allowed is , and the stop criteria is x n+1 x n + f(x n+1 ) tol and a maximum iteration is 100. Whereas computational order of convergence [7] is estimated using a formula COC ln( x k+1 x k / x k x k 1 ) ln( x k x k 1 / x k 1 x k ). Tabel 1: Comparison of several iteration methods f n (x) x 0 Method n COC f(x n ) x n x n 1 MKA e e 147 f MF e e 309 MSh e e 14 MG e e 144 MKA e e 18 f 1.6 MF e e 13 MSh e e 193 MG e e 06 MKA e e 05 f MF e e 314 MSh e e 175 MG e e 01 MKA e e 61 f 4.0 MF e e 185 MSh e e 41 MG e e 51

10 1434 Nora Fitriyani, M. Imran, Syamsudhuha In Table 1, f n (x) denotes the function of the nonlinear equation, n states the number of iterations obtained from all four methods, COC stand for the computational order of convergence, f(x n ) is the absolute value of the value of function at x n and x n x n 1 denotes the absolute value of the difference between two consecutive approximation of the root. Based on Table 1 it can be seen that for the function f 1, MKA requires the same number of iterations as MSh and MG and less than MF. Whereas for the functions f and f 4 MKA requires the same number of iterations as MF, Msh and MG. For the f 3 function MKA requires the same number of iterations as Msh and MG and less than MF. Furthermore, the COC given in Table 1 is inagreement with the analytic result obtain in the previous section. Overall based on the results of the numerical simulations as stated in Table 1 the new method (MKA) can compete with the existing methods and therefore it can be used as an alternative method to solve the nonlinear equation. REFERENCES [1] K. E. Atkinson, An Introduction to Numerical Analysis, Second Edition, John Wiley & Son, New York,1989. [] R. G. Bartle, dan D. R. Shebert, Introduction to Real Analysis, Second Edition. John Wiley & Sons, Inc., New York,199. [3] S. Chen dan Y. Qian, A Family of Combined Iterative Methods for Solving Nonlinear Equations, American Journal of Applied Mathematics and Statistics., 5(017), 3. [4] C. Chun, Iterative Methods Improving Newton s Method by the Decomposition Method, An Int. J. Comput. Math. with Appl., 50(005), [5] C. Chun and B. Neta, Some Modification of Newtons Method by The Method of Undetermined Coefficients, Comput. Math. with Appl., 56(008), [6] C. Chun dan B. Neta, Certain Improvements of Newton s Method with Fourth-order Convergence, Appl. Math. Comput., 15(009), [7] A. Cordero and J.R. Torregrosa,Variants of Newton s Method Using Fifth-order Quadrature Formulas, Appl. Math. Comput., 190(007), [8] L. Fang, T. Chen, L. Tian, L. Sun, dan B. Chen, A Modified Newton-type Method with Sixth-order Convergence for Solving Nonlinear equations, Procedia Engineering., 15(011), [9] W. Gautschi, Numerical Analysis,Second Edition, Birkhauser, New York, 01. [10] M.Grau dan J.L. Diaz-Barrero, An Improvement to Ostrowski Root-Finding Method, Appl. Math. Comput., 173(006), [11] Y. Ham, C. Chun, dan S. G. Lee, Some Higher-order Modifications of Newton s Method for Solving Nonlinear Equations, J. Comput. Appl. Math., (008),

11 A Three-Step Iterative Method to Solve a Nonlinear Equation 1435 [1] S. K. Khattri dan S. Abbasbandy, Optimal Fourth Order Family of Iterative Methods, Mat. Vesn., 63(011), [13] J. Kou, Y. Li, dan X. Wang, A Modification of Newton Method with Third-order Convergence, Appl. Math. Comput., 181(006), [14] J. H. Mathews dan K. D. Fink, Numerical Method Using MATLAB, Third Edition. Prentice-Hall. Upper Saddle River, New Jersey, [15] T. J. McDougall dan S. J. Wotherspoon, A Simple Modification of Newton s Method to Achieve Convergence of order 1 +, Appl. Math. Lett., 9(014), 0-5. [16] J. R. Sharma dan R. K. Guha, A Family of Modified Ostrowski Methods with Accelerated Sixth Order Convergence, Appl. Math. Comput., 190(007), [17] J. Stewart, Calculus, Seventh Edition. Brooks/Cole Publishing. Belmont, 01. [18] S. Weerakoon dan T. G. I. Fernando, A variant of Newton s Method with Accelerated Third-order Convergence, Appl. Math. Lett., 13(000),

12 1436 Nora Fitriyani, M. Imran, Syamsudhuha