DERIVATIVE-FREE TWO STEP ITERATIVE METHOD USING CENTRAL DIFFERENCE

  Bulletin of Mathematics

  ISSN Printed: 2087-5126; Online: 2355-8202 Vol. 08, No. 01 (2016), pp. 9–17 http://jurnal.bull-math.org

  

DERIVATIVE-FREE TWO STEP ITERATIVE

METHOD USING CENTRAL DIFFERENCE

Riski Amelia, M. Imran and Syamsudhuha

Abstract.

  This article discusses an iterative method to solve a nonlinear equation, which is free from derivatives by approximating a derivative in the two-step of the method of Xiaojian [Applied Mathematics and Computation, 203: 824-827, 2008] by the method of central difference with one parameter θ. We show analytically that the method is of order four. Numerical experiments show that the new method is comparable with other discussed method.

  1. INTRODUCTION Many problems in the chemistry, economics, marketing and electricity often end in the form of nonlinear equation models [9, p. 4-10], f (x) = 0. (1)

  Many of the numerical methods can be used to solve nonlinear equations, like Newton’s method given by [1, p. 79-80], [3, p. 67] x n

  = x n − f (x n ) f

• 1

  ′ (x n

  ) , f

  ′ (x n ) 6= 0 dan n = 0, 1, 2, · · · , (2) Received 12-01-2016, Accepted 23-03-2016.

  2016 Mathematics Subject Classification: 49Mxx, 41A25 Key words and Phrases: Central difference, efficiency index, derivative-free method, order of convergence.

  Riski Amelia, M. Imran & Syamsudhuha – Iterative Method Derivative-Free

  which has quadratically convergence. Using equation (2) and Taylor expan- sion [2, p. 189], Gutierrez and Hernandez [4] obtain the classical Chebyshev- Halley’s method whose form is

  L 1 f (x n ) f (x n ) x , n +1 = x n − 1 + (3)

  ′ 2 1 − βL f (x n ) f (x n )

  ′′ f

  (x n )f (x n ) L f (x n ) = , β (4)

  ∈ R, n = 0, 1, 2, · · · , ′

  2 (f (x n )) which has third-order convergent [4].

  ′′ Xiaojian [12] develops the Ostrowski’s method [8] by estimating f (x ) n

  , f into (4) using hyperbola one point (x n (y n )), which produces

  2 ′ f

  2(f (x n )) (y n ) ′′ f (x n ) ≈ . (5)

  2 f

  (x n ) − f (x n )f (y n ) Substituting (5) into (4), we obtain

  2f (y ) n L . f (x n ) ≈ (6) f (x n ) − f (y n ) and then substituting (6) into (3), we obtain

   f (x n )  y ,  n = x n −

   ′ f (x n )

  (7) f (x n ) − f (y n ) f (x n )  x ,  n +1 = x n −

   ′ f (x n ) − 2f (y n ) f (x n ) which has the fourth-order convergence.

  Besides Xiaojian’s method, there are several other methods to solve nonlinear equations like Liu’s method [5] and Ren’s method [7] which have the fourth-order convergence.

  This article discusses a derivative-free iterative method form to deter- mine the root of (1), by approximating the derivative of (7) using central difference with one parameter θ [10, p. 171]. By definition [6, p. 75-76], we showed the order of convergence of the proposed methods and using numerical computation in section three for five test functions.

  Riski Amelia, M. Imran & Syamsudhuha – Iterative Method Derivative-Free

  2. DERIVATIVE-FREE TWO STEP ITERATIVE METHOD USING CENTRAL DIFFERENCE

  Derivative-free iterative method is a method to determine the roots of (1) ′ by estimating f (x ) to (7) with central difference one parameter θ n f

  (x n + θf (x n )) − f (x n − θf (x n )) ′ f (x ) ≈

  (8) n

  2θf (x n ) If we substitute (8) into (7) then we have, the new iterative method given by

  

  2 2θf (x n )

    y n = x n − ,  f

  (x n + θf (x n )) − f (x n − θf (x n )) (9)

  2 f (x ) − f (y ) 2θf (x ) n n n

    x

  . n = x n −

• 1

   f (x ) − 2f (y ) f (x + θf (x )) − f (x − θf (x )) n n n n n n

  In the following theorem, we show the order of convergence of the iterative method (9). Theorema 1.1.

  Let α ∈ I be a simple root of the sufficiently differ- entiable function f is suffeciently

  : I ⊂ R → R for an open interval I. If x close to α, then the method defined by equation (9) is fourth-order conver- gence.

  ′ Proof.

  Let α is a simple root of f (x) = 0 so f (α) = 0, and f (α) 6= 0. The Taylor expansion of f (x ) arround x = α until the fourth order is n n

  3

  2 (x n − α) (x n − α) (x n − α)

  ′ (2) (3) f (x n ) = f (α) + f (α) + f (α) + f (α)

  1! 2! 3!

  2 (x n − α)

  (4)

  5

• f (α) + O(x n − α) . (10)

  4! Because f (α) = 0 and e = x − α, so (10) can be written as n n

  2

  3

  4 e e e n n n

  ′ (2) (3) (4)

  5 f (x n ) = f (α)e n + f (α) + f (α) + f (α) + O(e ), n

  2! 3! 4! !

  (2) (3) (4) f (α) f (α) f (α)

  2

  3

  4

  5 ′

  e e e e , (x + n ) = f (α) n + O(e ) (11) + n n n n

• f

  ′ ′ ′ 2!f (α) 3!f (α) 4!f (α)

  (k) f

  (α) and if c = , k = 2, 3, ..., , then equation (11) is given by k

  ′ k

  !f (α) ′

  2

  3

  4

  5 f e e e

  (x n ) = f (α) e n + c + c + c + O(e ) . (12) 2 n 3 n 4 n n Riski Amelia, M. Imran & Syamsudhuha – Iterative Method Derivative-Free

2 To obtain 2θf (x n ), squaring the equation (12) and then multiplied it by

  2θ, so we have 2 ′

  2 2 ′

  2 3 ′

  2 2 ′

  2

  4 e c e f f θ .

  2θf (x n ) = 2θf (α) + 4θf (α) + (4θc (α) + 2c (α) )e n 2 n 3 n

  2 (13)

  Then calculating x + θf (x ), denoted by w , where x = e + α, we n n n n n obtain

  ′ ′ 2 ′ 3 ′

  4 w f f f f . n = α + θ (α)e n + c (α)e + c (α)e + c (α)e (14) 2 n

  3 n 4 n Similarly, the Taylor expansion f (w n ) arround w n = α is

  2

  2

  3

  2

  2 ′ ′ ′ ′ ′

  (w n ) = (α) + f (α) n 3θc 2 (α) + c 2 (α) + c 2 (α) n

• f θf e f f θ f e

  4

• · · · + O(e ). (15)

  n Computing x n − θf (x n ), denoted by u n , by substituting x n = e n + α, we have

  ′ ′ 2 ′ 3 ′

  4 u f f f f . n = α − θ (α)e n + c (α)e + c (α)e + c (α)e (16) 2 n

  3 n 4 n Using Taylor expansion f (u n ) about u n = α, and ignore the higher order after simplifying process, we obtain

  2

  2

  3

  2

  2

  f (u ) = (θf (α) − f (α))e − 3θc f (α) + c f (α) θ + c f (α) e n n

• ′ ′ ′ ′ ′

  2

  2

  2 n

  4

• · · · + O(e ). (17)

  n Calculating f (w n ) − f (u n ), we have

  2

  2

  4 f c

  ′ 2 ′

  (w n ) − f (u n ) = (2θf (α) )e n + (6θf (α) )e + · · · + O(e ). (18) 2 n n

  2 _n

  2θf (x ) Calculating using the equation (13) and (18), provides f (w n n )

  )−f (u

  2

  2

  2

  2 2θf (x ) e (1 + 2c e + c e + 2c e ) n n

  3 2 n n n 2 .

  =

  2 ′

  2

  2

  2

  2

  3 f (w ) − f (u ) e f θ e n n 1 + (3c 2 n ) + (c 3 (α) + 2c + 4c 3 )e + · · · + O(e ) n n n

  2 (19)

  Using Geometric series and let, 2 ′

  2

  2

  2

  3 r e f θ c ,

  = 3c n + 2c + c (α) + 4c e + 5c + · · · + 5c e (20)

  2

  3 3 n

  4

  2 3 n Riski Amelia, M. Imran & Syamsudhuha – Iterative Method Derivative-Free

  then

  2 2θf (x n )

  2

  2

  3

  2

  2

  4 ′ = e − c e − 2c − · · · − 2c e + c f (α) θ c + · · · − 5c e . n

  2

  3

  3

  2

  4 n 2 n n f

  (w n ) − f (u n ) (21)

  Substituting equation (21) into (9) and if (x n = α + e n ), then

  2

  2

  2

  2

  3

  2

  2

  4 ′ ′ y e α θ e α θ c e . n = α + c 2 + − 2c + c 3 (f ) + 2c 3 + − c 3 (f ) 2 + · · · + 5c

  4 n 2 n n

  (22) Expanding f (y n ) using Taylor expansion about y n = α until the fourth order and ignore the higher order, we have

  2 ′′ f

  (α)(y n − α) ′ f (y n ) = f (α) + f (α)(y n − α) +

  2 ′′′ 3 (4)

  4 f f

  (α)(y n − α) (α)(y n − α) .

  (23) + +

  6

  24 Substituting equation (22) into (23), after simplifying, we obtain ′ 2 ′

  3 2 ′

  2

  3 f α c α θ α e

  (y n ) = c (f )e (f ) + · · · − 2(f )c + 2 n 3 n

  2 ′ ′

  4 − 5(f + α )c c + · · · + 5(f α )c e . (24)

  2

  3 4 n Computing f (x n ) − f (y n ), we get

  ′ ′

  3 2 ′

  3

  (x n ) − f (y n ) = f (α)e n − c (α) + · · · − c (α)

• f f θ f e

  3 3 n ′

  3 2 ′

  4

• c f θ c f e .

  (α) + · · · − 4c (α) (25)

  2

  3 4 n Computing f (x n ) − 2f (y n ), we have

  2

  3

  2

  3 ′ ′ ′ ′

  (x n ) − 2f (y n ) = f (α)e n − (c (α))e − 2c (α) + · · · − 3c (α) 2 n

• f f f θ f e

  3 3 n ′

  3 2 ′

  4 f θ c f e .

• 2c (α) + · · · − 9c (α) (26)

  2

  3 4 n f

  (x n ) − f (y n ) Computing , denoted by S, we get f

  (x n ) − 2f (y n ′

  2

  2 2 ′

  2

  2

  3 1 + − c f (α) θ + · · · − c e + c f (α) θ c + · · · − 4c e

  3

  3

  3

  2

  4 n n S = .

  3 ′

  2

  2

  2

  3 e f θ e e 1 + (−c

  2 n ) − − 3c 3 − · · · − 2c 3 (α) + 10c − · · · − 9c

  4 n n

  2 (27) Riski Amelia, M. Imran & Syamsudhuha – Iterative Method Derivative-Free

  Applying geometric series to avoid the division of two polynomials, denoted by R, we obtain ′

  2

  2

  2

  3

  3

  e f θ e e ,

• R
• = −c n − 3c − · · · − 2c (α) − 10c − · · · − 9c

  2

  3 3 n 4 n

  2 (28)

  S so we get as R S

  ′

  2

  2

  2 2 ′

  2

  2

  3 e c f θ e f θ e

• = 1+c + n (α) + 2c − c 5c + · · · + 4c (α)

  2

  3 3 n

  4 4 n

  2 R ′

  2

  2

  2

  4

• 14c c + · · · − 6c f (α) θ c e . (29)

  4

  2 3 n

  2 Substituting equations (22) and (29) into (9), ′

  2

  2

  3

  4 x n = α + − c f (α) θ c + c − c c e . (30)

• 1

  3

  2

  2 3 n

  2 If x n +1 − α = e n +1 , then subtracting both sides of the equation (30) from α , we obtain

  ′

  2

  2

  3

  4 e f θ c c e . n = − c (α) + c − c (31)

• 1

  3

  2

  2 3 n

  2 From the definition of the order of convergence, we see that equation (9) is fourth order. ✷

  3. NUMERICAL SIMULATION In this section, we compare the number of iterations to obtain an approxi- mated root for Newton’s method (MN), Ostrowski’s method (MO) [8], Liu’s method (MLI) [5], Ren’s method (MR) [7], and derivative-free iterative method (MIBT) given by (9) using five test functions. The computation is carried out using Maple 13. The stopping criteria of the iteration are

  −15 |x n − α| ≤ tol and |f (x n )| ≤ tol , where tol ≤ 1.0 × 10 . The maxi-

• 1 +1 mum iteration allowed is 300.

  3

  2 1. f 1 (x) = x + 4x − 10 2. f (x) = cos(x) − x

  2

  2

  2

• x

  4 −0.1 *

  3

  4

  4

  3

  6

  1.3

  3 0.6586048471181404

  5

  4

  4

  3 f

  6

  4

  4

  3

  6

  2.1

  3

  5

  4

  3

  5

  2.0

  1.5

  4

  3

  4

  4 to x = −0.1, MIBT has a number of iterations less than MO, MLi, and MR, while MN converges to the other roots. Therefore this method can be used as an alternative method for solving nonlinear equations.

  is sightly better than other methods. Based on the number of iterations, f

  4 The simulation shows that the derivative-free iterative method (MIBT)

  4

  5

  4

  7

  4.0

  3

  4

  3

  4

  6

  2.5

  3 0.2575302854398608

  3

  4

  3

  4

  1.0

  5

  4 f

  6

  3 1.4044916482153412

  3

  Riski Amelia, M. Imran & Syamsudhuha – Iterative Method Derivative-Free

  0.5

  5

  5

  2

  4

  1.2

  3 1.3652300134140968

  5

  5

  3

  7

  1

  4.0

  α MN MO MLi MR MIBT f

  Number function x The number of iterations

  Table 1: Comparison the number of iteration of discussed iterative methods

  − 3x + 2 Table 1 presents, the number of iteration needed to obtain the approximated root for several mention methods by varying an initial guess x . The star mark ∗ indicates that the method converges to a different root.

  2 − e x

  5 (x) = x

  5. f

  2 − x + 3

  3

  6 − 10x

  4. f 4 (x) = x

  2

  7

  3

  3

  6

  1.0

  3

  3 f

  3

  3

  3

  4

  2.0

  3

  3

  3

  4

  5

  0.0

  3 0.7390851332151606

  3

  3

  3

  6

  2 −0.5

  4 f

  5

  5

  5. CONCLUSIONS The simulation shows that MO and MIBT are comparable with the other methods. Because MO is not a derivative-free, then MIBT is slightly better than MO. In general MIBT is comparable with the other methods, therefore this method can be used as an alternative method for solving nonlinear

  

Riski Amelia, M. Imran & Syamsudhuha – Iterative Method Derivative-Free

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  4. Gutierrez, J. M. & Hernandez, M. A. 1997. A Family of Chebyshev-Halley

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5. Z. Liu & Q. Zheng, A variant of Steffensen’s method fourth-order conver-

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  with fourth-order convergence , Applied Mathematics and Computation, 209 (2009), 206–210.

  

8. J. F. Traub, Iterative Methods for the Solution of Equations, Prentice-Hall

Inc. Englewood Cliffs, 1964.

  9. R. Wait, The Numerical Solution of Algebric Equation, Jhon Wiley and Sons, New York, 1979.

  

10. Cheney, W. & Kincaid, D. 2004. Numerical Methods for Mathematics

  and Computing , Sixth Edition. Brook/Cole Publishing Company,

  California 11. S. Weerakoon & T. G. I. Fernando, A variant of Newton’s method with accelerated third order convergence , Applied Mathematics Letters, 13 (2000), 87–93.

  

12. Z. Xiaojian, Modified Chebyshev-Halley methods free from second deriva-

  tive , Applied Mathematics and Computation, 203 (2008), 824-827.

RISKI AMELIA

  : Magister Student, Department of Mathematics, Faculty of Math-

ematics and Natural Sciences University of Riau, Bina Widya Campus, Pekanbaru

28293, Indonesia.

  E-mail: riskiameliabintikhaidir@gmail.com

  Riski Amelia, M. Imran & Syamsudhuha – Iterative Method Derivative-Free M. IMRAN : Numerical Computing Group, Department of Mathematics, Faculty

of Mathematics and Natural Sciences University of Riau, Bina Widya Campus,

  Pekanbaru 28293, Indonesia.

  E-mail: mimran@unri.ac.id

  Syamsudhuha : Numerical Computing Group, Department of Mathematics, Faculty

of Mathematics and Natural Sciences University of Riau, Bina Widya Campus,

  Pekanbaru 28293, Indonesia.

  E-mail: syamsudhuha@lecturer.unri.ac.id