Another Sixth-Order Iterative Method Free from Derivative for Solving Multiple Roots of a Nonlinear Equation

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1 Applied Mathematical Sciences, Vol. 11, 2017, no. 43, HIKARI Ltd, Another Sixth-Order Iterative Method Free from Derivative for Solving Multiple Roots of a Nonlinear Equation Rahma Qudsi Pendidikan Matematika, Fakultas Keguruan dan Ilmu Pendidikan Universitas Islam Riau, Pekanbaru, Riau, Indonesia M. Imran, Syamsudhuha Numerical Computing Group, Department of Mathematics University of Riau, Pekanbaru 28293, Indonesia Copyright c 2017 Rahma Qudsi, M. Imran and Syamsudhuha. This article is 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 A three-step iterative method with sixth-order convergence as another modification of Newton Method was showed. We approximate them with a central difference and divided difference to avoid appearing the derivatives in the method. We show analytically that the method is of order six. We verify the theoretical result on relevant numerical problems and compare the behavior of the proposed method with some existing methods. Keywords: central difference, free derivative method, iterative method, multiple roots 1 Introduction Finding the roots of a nonlinear equation of the form gx) = 0 1)

2 2122 Rahma Qudsi, M. Imran and Syamsudhuha is one of foremost problem in numerical analysis. Some solutions of this problem are of the form multiple roots. In several times, there are many methods that can be used to find a multiple root of the nonlinear equation 1). Some of them requires the multiplicity of the root to be known in advance to apply the methods, and others transform the nonlinear equation 1) into a simple root problem. Schröder [6] was suggested a modified Newton s method by adding multiplicity, m, into Newton s method for a simple root, he ends up with the following formula x n+1 = x n m gx g x n ), for given x 0 and m. 2) This method converges quadratically for multiple roots[9, p. 135]. Neta [4] proposed a new method by modifying Chebyshev s method to have a third order iterative method for m 3, that is x n+1 = x n + mm 3) 2 with α = mm 3) and β = m 2 a fourth-order method y n = x n x n+1 = x n m 3 2m m + 2 u n α gx g x n ) 1 mm 2) m 2 m β gx g x n ) ) g x n ) g x n ). Shengguo, Xiangke and Lizhi [7] developed ) m g y n ) m2 2 g x n ) g x n ) m m+2) m g y n ) where u n. g x Li et al. [3] suggested a fifth-order method, using the transformation [8, p. 5] = gx u n fx) = { gx) g x) if gx) 0, 0 if gx) = 0, 3) and f 2 x n ) y n = x n fx n + fx n )) fx n )), fx n )fy n ) z n = y n fx n + fx n )) fx n )), x n+1 = z n fz B A,

3 Another sixth-order iterative method for solving multiple roots 2123 where A = z n x n ) 2 fx n )fz n ) fy n )) + fx n )fz n ) fx n )) z n x n )fx n 0 + fx n )) fx n ))) z n y n ) 2 B = z n y n )z n x n ) 2 fx n ). Qudsi et al. [5] suggested sixth order iterative method given as where 2f 2 x n ) y n = x n fx n + fx n )) fx n fx n )), 2fx n )fy n ) z n = y n fx n + fx n )) fx n fx n )), x n+1 = z n fz D C, C = 2z n x n ) 2 fx n )fz n ) fy n )) + 2fx n )fz n ) fx n )) z n x n ) fx n + fx n )) fx n fx n )) )) z n y n ) 2, D = 2z n y n )z n x n ) 2 fx n ). Khattri that is [2] also proposed another transformation for finding multiple roots, y n = x n fx f x n ), 4) z n = x n fx 1 + fy fy )), 5) f x n ) fx n ) fx n ) x n+1 = x n fx 1 + fy fy ) f x n ) fx n ) fx n ) + fz fy )). 6) fx n ) fx n ) This form is sixth order convergent iterative method.the purpose of this paper is to construct a new iterative method free from derivative for finding multiple roots of a nonlinear equation. We present this new method and its convergence analysis in Section 2. In Section 3, we present some numerical examples showing the performance of our proposed method compared to some mention methods.

4 2124 Rahma Qudsi, M. Imran and Syamsudhuha 2 Proposed Method We consider an iterative method of the form 3) and applying three step iterative method of the form 4), 5) and 6). To avoid computing f x n ) in 4), 5) and 6), we approximate it by central difference, as follows f x n ) = fx n + fx n )) fx n fx n )). 7) 2fx n ) Replacing approximation of f x n ) given by 7), we get a new iterative method y n = x n t, 8) z n = x n t 1 + fy fy )), 9) fx n ) fx n ) x n+1 = x n t 1 + fy fy ) fx n ) fx n ) + fz fy )). 10) fx n ) fx n ) 2f where t = 2 x fx n+fx ) fx n fx ). For the proposed method defined by 3), 8), 9) and 10), we have the following analysis of convergence. Theorem 1. Let x D be a multiple roots of multiplicity m of a sufficiently differentiable function g : D R on open interval D. If x 0 is sufficiently close to x, then iterative method defined by 3), 8), 9) and 10) has a sixth-order convergence. Proof. From the assumption, 1) can be write down as gx) = x x ) m hx), 11) where x is a root of 1) with multiplicity m, and hx) is a continuous function with hx ) 0. Taking the derivative of g with respect to x, we obtain g x) = mx x ) m 1 hx) + x x ) m h x). 12) Substituting 11) and 12) into 3), we get fx) = gx) g x) = x x )hx) mhx) + x x )h x). 13)

5 Another sixth-order iterative method for solving multiple roots 2125 Here we have transformed a multiple root problem into a simple root problem. Let e n = x n x. Using Taylor s expansion about x n = x, we have respectively and hx n ) = hx )1 + b 1 e n + b 2 e 2 n + b 3 e 3 n + b 4 e 4 n + b 5 e 5 n + b 6 e 6 n + Oe 7 ), 14) h x n ) = hx )b 1 + 2b 2 e n + 3b 3 e 2 n + 4b 4 e 3 n + 5b 5 e 4 n + 6b 6 e 5 n + 7b 7 e 6 n + Oe 7 ), 15) where b k = hk) x ). Substituting 14)and 15) to 13), and using geometric k!hx ) series, we obtain where fx n ) = c 1 e n + c 2 e 2 n + c 3 e 3 n + c 4 e 4 n + c 5 e 5 n + c 6 e 6 n + Oe 7 16) c 1 = 1 m, c 2 = b 1 m 2, c 3 = 1 m 3 b2 1 2b 2 m + b 2 1m), c 4 = 1 m 4 b3 1m 2 3b 1 b 2 m 2 + 2b 3 1m 4b 1 b 2 m + 3b 3 m 2 + b 3 1m 4 ), c 5 = 1 m 5 10m2 b 2 1b 2 + 3mb m 2 b 1 b 3 + 4b 1 b 3 m 3 + 2b 2 2m 3 6mb 2 1b 2 + b m 2 b 2 2 4m 3 b 2 1b 2 4b 4 m 3 + 3m 2 b m 3 b 4 1), c 6 = 1 m 6 16m3 b 1 b 2 2 8b 4 b 1 m 3 8mb 3 1b 2 5m 4 b 3 1b m 2 b 1 b m 2 b 3 1b 2 5b 2 b 3 m 4 + 4mb b 2 2b 1 m 4 + 6m 2 b b 5 m 4 + 4m 3 b b 2 b 3 m 3 + m 4 b m 3 b 3 1b 2 5b 4 b 1 m 4 + b m 3 b 2 1b 3 + 9m 2 b 2 1b 3 + 5b 2 1b 3 m 4 ), c 7 = 1 m 7 36m3 b 1 b 2 b 3 12m 3 b 2 1b m 2 b 3 1b m 2 b 2 1b m 3 b 2 1b m 4 b 2 1b m 4 b 4 1b 2 6m 5 b 4 1b 2 48m 3 b 4 1b 2 36m 2 b 4 1b 2 10mb 4 1b 2 + b m 5 b m 4 b m 3 b m 2 b mb b 1 b 2 b 3 m 5 44b 1 b 2 b 3 m 4 6b 4 b 2 1m 5 + 6b 5 b 1 m 5 + 6b 4 b 2 m 5 18b 4 b 2 1m 4 + 6b 3 1b 3 m 5 + 9b 2 1b 2 2m b 3 1b 3 m b 3 1b 3 m b 5 b 1 m b 4 b 2 m 4 8b 3 2m 4 2b 3 2m 5 + 3b 2 3m 5 + 9b 2 3m 4 6b 6 m 5 8b 3 2m 3 ).

6 2126 Rahma Qudsi, M. Imran and Syamsudhuha Let w n = x n + fx n ) and v n = x n fx n ). On computing fw n ) and fv n ) using Taylor series, we end up respectively with fw n ) = 1 m 1 + m) e 2 n + 1 ) b1 1 + m 2 + 3m) ) e 2 m 4 n + and 1 m 6 ) b b 2 + 6b 2 1)m + 6b 2 + 9b 2 1)m 2 + 5b 2 1 8b 2 )m 3 + 2b 2 + b 2 1)m 4) e 3 n 1 ) b 3 + m b 3 1 4b 1 b 2 )m + 27b b 3 25b 1 b 2 )m b 1 b b b 3 )m b 3 48b 1 b b 3 1)m b 1 b b 3 + 7b 3 1)m 5 + b 3 1 3b 1 b 2 + 3b 3 )m 6) e 4 n + + Oe 7, 17) fv n ) = 1 m 2 m 1) e n + 1 ) 1 ) b1 3m + m 2 + 1))e 2 m 4 n + b 2 m b b 2 )m + 6b 2 3b 2 1)m 2 + 3b b 2 )m 3 + 2b 2 + b 2 1)m )e 4 3 n + 1 ) b 3 m b 3 1 4b 1 b 2 )m + 19b 1 b 2 + 3b 3 + 5b 3 1)m b 3 28b 1 b 2 + 6b 3 1)m b 1 b 2 8b b 3 )m 4 + 3b b b 1 b 2 )m 5 + b 3 1 3b 1 b 2 + 3b 3 )m 6 )e 4 n + + Oe 7. Substituting 17), 18) and 16) into y n in 8) and simplifying, we obtain 18) y n = x + b ) 1 1 ) b e 2 2 n + m m mb 2 1 2b 2 ) + m 3 2b 2 1 4b 2 ) ) e 3 n + 1 ) 3b 3 m m7b b 1 b 2 ) + m 2 4b b 1 b b 3 ) + 9m 4 b 3 + m 3 2b 1 b 2 b 3 1) + m 4 3b 3 1 9b 1 b 2 ) ) e 4 n + + Oe 7. 19) Expanding fy n ) about y n = x to get fy n ) = b ) 1 1 ) 2m e 2 m 2 n + 3 b 2 m 5 1 4m 3 b 2 + b 2 1 2mb 2 + mb1) 2 e 3 n + 1 ) 3b 3 m b b 1 b 2 )m + 12b 1 b 2 + 5b b 3 )m 2 + 2b 1 b 2 b 3 1)m 3 + 9b 3 + 3b 3 1 9b 1 b 2 )m 4) e 4 n + + Oe 7. 20)

7 Another sixth-order iterative method for solving multiple roots 2127 and dividing by fx n ), we get fy n ) fx n ) = b ) ) 1 1 e n + m m 4 b2 1 + b 2 1 2b 2 )m b 2 1m 2 + 2b 2 1 4b 2 )m 3 ) + + Oe 8. 21) Substituting 16) 21) into z n in 9), we get after simplifying, ) 1 z n = m b 1b b 2 1 2b 2 )m 4b 2 1m 2 2b 2 + b 2 1)m 3 ) e 4 n + 1 m 8 b4 1 +2b 4 1 4b 2 1b 2 )m + 4b 2 2 4b 2 1b 2 6b 4 1)m b b b 2 1b 2 +12b 1 b 3 )m b 2 1b 2 24b 4 1)m 5 14b 2 1b 2 + 6b 1 b 3 + 8b b 4 1)m 6 ) ) e 5 n + + Oe 8. Using tranformation 3) we get ) 1 fz n ) = m b 1b b 2 1 2b 2 )m 4b 2 1m 2 2b 2 + b 2 1)m 3 )) + + Oe 8. 22) Based on 10), 22) divided by 16) to get ) fz n ) 1 fx n ) = m b 1b b 2 1 2b 2 )m 4b 2 1m 2 2b 2 + b 2 1)m 3 )) e 3 n + 1 m 8 b b 4 1 4b 2 1b 2 )m 7b b 2 2 4b 2 1b 2 )m 2 b b 2 1b 2 )m b b b 2 1b b 1 b 3 )m 4 +50b 2 1b 2 25b 4 1)m 5 + 6b 1 b 3 14b 2 1b 2 + 8b b 4 1)m 6 ) ) e 4 n + + Oe 8. 23) Substituting 16) 23) into 10), noting e n = x n x, we get e 4 n e 2 n e n+1 = 1 m 9 b 1b 4 1 4b 2 1b 2 + 2b 4 1)m + 4b 2 2 4b 2 1b 2 8b 4 1)m b 2 1b 2 7b 4 1)m 3 8b 2 1b b b 2 2)m 4 9b b 2 1b 2 )m 5 + b b 2 2 4b 2 1b 2 )m 6 )e 6 n + + Oe 8 From the definition of the order of convergence [1, p. 56], this method has a sixth-order convergence, then the theorem proved.

8 2128 Rahma Qudsi, M. Imran and Syamsudhuha 3 Numerical Experiments In this section, we give some numerical results to demonstrate the efficiency of the proposed method. We compare this method MITS) with Li s method MLi) defined by [3], Khattri s method MK)defined by [2] and Qudsi s method MQud) defined by [5]. All computation were done using MAPLE. We use the following functions which also have been consider in [3] 1. f 1 x) = 8xe x2 2x 3) 8 with m = 8, and x = f 2 x) = e x2 +x+3 x + 2) 9 with m = 9, and x = f 3 x) = 2x cosx)+x2 3) 10 x 2 +1 with m = 10, and x = f 4 x) = e x + 2 sinx)) 4 with m = 4, and x = f 5 x) = x 3)e x ) 5 with m = 5, and x = Table 1: Comparison the number of iteration and error of the discussed iterative methods MITS MQud MLi gx) x 0 n error COC n error COC n error COC g e-24 6 * e e e e e e e-68 5 g e e e e e e e e e-25 5 g e-24 6 * e e e e e-14 6 div e-14 5 g e-15 6 div * e e e e e e-28 5 g div e e div e e div e e-22 5 We set a tolerance, T oll = , and we use x n+1 x n < T oll and fx n+1 ) < T oll as stopping criteria. Table 1 shows the number of iteration, error of each method and the Computational Order of Convergence COC). In Tabel 1, we denote n as the number of maximum iteration, err as computational error of the iterative methods, NC as the method does not convergence and F a as the method is fail to obtain the root. Based on numerical computation in Table 1,

9 Another sixth-order iterative method for solving multiple roots 2129 References [1] K. E. Atkinson, Elementary Numerical Analysis, 2 nd Ed., John Wiley & Sons, Inc., New York, [2] S. K. Khattri, How to Increase Convergence Order of The Newton Method to 2 m, Applications of Mathematics, ), [3] X. Li, C. Mu, J. Ma and L. Hou, Fifth-Order Iterative Method for Finding Multiple Roots of Nonlinear Equations, Numer Algor., ), [4] B. Neta, New Third Order Nonlinear Solvers for Multiple Roots, Applied Mathematics and Computation, ), [5] R. Qudsi, M. Imran and Syamsudhuha, A Sixth-Order Iterative Method Free from Derivative for Solving Multiple Roots of a Nonlinear Equation, Applied Mathematical Sciences, ), [6] E. Schröder, On Infinitely Many Algorithms for Solving Equation, TR- 2990, University of Maryland, [7] L. Shengguo, L. Xiangke and C. Lizhi, A New Fourth-Order Iterative Method for Finding Multiple Roots of Nonlinear Equations, Applied Mathematics and Computation, ), [8] J. F. Traub, Iterative Methods For the Solution of Equations, Prentice- Hall, Inc., New York, [9] R. Wait, The Numerical Solution of Algebraic Equations, John Wiley, Ltd., New York, [10] S. Wu and X. Li, A Fourth-Order Modification of Newton Method For Multiple Roots, International Journal of Research and Reviews in Applied Sciences IJRRAS, ), Received: July 11, 2017; Published: August 23, 2017