A SIXTH-ORDER DERIVATIVE-FREE ITERATIVE METHOD FOR SOLVING NONLINEAR EQUATIONS

  Bulletin of Mathematics

  ISSN Printed: 2087-5126; Online: 2355-8202 Vol. 08, No. 01 (2016), pp. 1–8 http://jurnal.bull-math.org

  

A SIXTH-ORDER DERIVATIVE-FREE

  

ITERATIVE METHOD FOR SOLVING

NONLINEAR EQUATIONS

Iin Purnama Edwar, M. Imran and Leli Deswita

Abstract.

  This article discusses a derivative-free iterative method to solve a non- linear equation. The method derived by approximating derivatives on the method proposed by Rafiullah [International Journal of Computer Mathematics, 4: 2459– 2463, 2010] using a central difference formula. We show analytically that the method has sixth-order of convergence. Numerical experiments show that the new method is comparable with other methods in terms of the speed in obtaining a root.

  

1. INTRODUCTION

One of the well known topics discussed in the mathematical sciences is a

technique to obtain the solution of nonlinear equation of the form f (x) = 0. (1)

The numerical method that can be used to solve (1) is based on iterative

method. The famous iterative method appears in the literatures is Newton’s

method, given by f (x )

  n ′

  x , f n

  n = x n − (x n ) 6= 0, = 0, 1, 2, . . . . (2)

  • 1

  ′

  f (x n ) Received 19-12-2015, Accepted 27-01-2016.

  2016 Mathematics Subject Classification : 65Hxx, 40A05 Key words and Phrases

  : Central difference, derivative-free method, iterative method, order of convergence.

  

Iin Purnama Edwar, M. Imran & Leli Deswita – Iterative Method Derivative-Free

The method has quadratically convergence [4, p. 55]. Using Newton’s

method and Adomian’s decomposition method [1, p. 10], Basto et al. [3]

obtain a new iterative method of the form

  2 ′′

  

f f

(x n ) (x n )f (x n )

x ,

  n = x n − − (3)

  • 1

  ′ ′

  3

  f (x n ) 2f (x n ) which converges cubically [3].

  ′′

  Rafiullah [10] modifies (3) by estimating f (x n ) using a forward dif- ference [6],

  ′ ′

  f (y n ) − f (x n )

  ′′

  

f ,

(x n ) ≈ (4) y

  n − x n

  

and combines the resulting equation with another Newton’s method. He

ends up with f (x n ) y = x − ,

  n n ′

  f

(x n )

  ′ ′

  f (x n )(f (x n ) − f (y n )) z = y − ,

  n n ′

  2

  2f (x n ) f (z n ) x n = z n − . (5)

  • 1

  ′

  f (z n )

  ′

  

Then, by estimating f (z n ) in (5) by a linear interpolation [8] based on

  ′ ′

  

two known points, namely (x n , f (x n )) and (y n , f (y n )), and he obtains the

following a sixth-order iterative method f (x n ) y = x − , (6)

  n n ′

  f (x n )

  ′ ′

  f (x n )(f (x n ) − f (y n )) z = y − , (7)

  n n ′

  2

  

2f (x n )

  ′

  2f (z n )f (x n ) x = z − . (8)

  n +1 n ′ ′ ′ 2 ′

  2

  4f (x n )f (y n ) − f (x n ) − f (y n ) The rest of this paper is organized as follows. In section two, a new

sixth order derivative-free iterative method is obtained by approximating

the first derivative appear in (6)-(8) using central difference formulas [7,

p. 313]. Then in section three the proposed method is tested on four test

functions and compared with some known iterative methods.

  

Iin Purnama Edwar, M. Imran & Leli Deswita – Iterative Method Derivative-Free

2. A DERIVATIVE-FREE ITERATIVE METHOD

  ′ ′

  

If we approximate f (x n ) and f (y n ) in (6)-(8) using a central difference,

that is f (x n + f (x n )) − f (x n − f (x n ))

  ′

  f (x n ) ≈ =: T , (9)

  1x

  

2f (x n )

and

f (y + f (y )) − f (y − f (y ))

  n n n n ′

  f , (y n ) ≈ =: T (10)

  1y

  

2f (y n )

and substitute them into (6)-(8), we obtained the following new iterative

method f

  (x n ) y ,

  n = x n − (11)

  T

  1x

  f (x )(T − T )

  n 1x 1y

  z ,

  n = y n − (12)

  2

  2(T )

  1x

  2f (z n )T

  1x

x .

n +1 = z n − (13)

  2

  2 T

  4T − (T ) − (T )

  1x 1y 1x 1y

  In the following we show that the proposed method (11)–(13) is of order six.

  6 Theorem 1.1

  . Let f (D). If f (x) = 0 : D ⊂ R → R. Assume that f ∈ C

has a simple root at α is given and sufficiently close to α,

  ∈ D and x then the new method defined by (11)–(13) is of order six and satisfies the following error equation

  6

  2

  4

  2

  2

  2

  2

  4

  6

  c c c + 6c c + 5c c + 16c c c − 16c e

  1

  2

  1 3 n

  1

  3

  1

  3

  3

  1

  2

  2

  e n = − ,

  • 1

  5

  c

  4

  1

  (j)

  f (α) where c j = , j = 1, 2, 3, . . ., and e n = x n − α. j !

  Proof . Substituting (9) into (11), we obtain

  2

  2f (x n )

y .

  n = x n −

  (14) f (x n + f (x n )) − f (x n − f (x n ))

  ′ Let α be a simple root of f (x) = 0, then f (α) = 0 and f (α) 6= 0.

  Taylor’s expansion of f (x n ) about x n = α is given by [2, p. 189]

  2

  3

  4

  5

  6

  7

  f e e e e e e (x n ) = c n + c + c + c + c + c + O(e ), (15)

  1 2 n 3 n 4 n 5 n 6 n n Iin Purnama Edwar, M. Imran & Leli Deswita – Iterative Method Derivative-Free (j)

  f (α)

  

where c j = , j = 1, 2, 3, . . . and x n − α = e n . Using (15), we obtain

  j !

  respectively

  2

  2

  2

  3

  2

  4

  5

  e c e c c c

2f (x n ) = 2c + 4c + 4c + 2c e + (4c + 4c ) e

  n

  1 2 n

  1 3 n

  1

  4

  2 3 n

  1

  2

  2

  6

  7

  c c

  • (4c

  1 5 + 4c

  2 4 + 2c )e + O(e ), (16) n n

  3

  and

  2

  4

  5

  6

  7

  x e e e e e e n + f (x n ) = e n + α + c n + c + c n + c + c + c + O(e ).

  1 2 n

  3 4 n 5 n 6 n n

  

Taylor’s expansion of f (x n +f (x n )) about x n +f (x n ) = α, and f (x n −f (x n ))

about x − f (x ) = α are given respectively by

  n n

  2

  2

  2

  f e c c e (x n + f (x n )) = c

  1 + c n + 3c

  1 2 + c 2 + c

  2

  1 1 n

  2

  2

  2

  3

  3

  • 2c + 3c c + 4c c + 2c c + c + c c e

  3

  1

  3

  1

  3

  3

  2

  1

  2 1 n

  7

  • · · · + O(e ), (17)

  n

  and

  2

  2

  2

  f (x n − f (x n )) = −c + c e n + −3c c + c c + c e

  1

  1

  2

  2 2 n

  1

  1

  2

  2

  2

  3

  3

  c c c c

  • −2c + 3c − 4c + 2c + c − c e +

  3

  1

  3

  1

  3 3 n

  2

  1

  2

  1

  7

  • · · · + O(e ). (18)

  n

  Subtracting (18) from (17) gives

  2

  2

  3

  2

  3

  f e c e c c

(x n + f (x n )) − f (x n − f (x n )) = 2c n + 6c + 2c + 8c + 4c e

  1 2 n

  3

  1 3 n

  1

  1

  2

  3

  2

  4

  • 8c c + 6c c c + 10c c + 10c c e

  4

  2

  3

  1

  4

  2

  3

  1 1 n

  7

  • · · · + O(e ). (19)

  n

  Dividing (16) by (19) yields

  2

  2

  2

  3

  4

  5

  6

  e c e 2f (x n ) 2c + 4c

  1 2 + Ae + Be + Ce n n n n n

  1

  , =

  2

  2

  3

  4

  5

  6

  f (x + f (x )) − f (x − f (x )) e

  n n n n 2c n + De + Ee + F e + Ge + He n n n n n

  1

  (20) where

  2 A c ,

  = 4c + 2c

  1

  3

  2 B c c ,

  = 4c

  1 4 + 4c

  2

  3

  2 C = 4c c + 4c c + 2c ,

  1

  5

  2

  4

  3 D c ,

  = 6c

  1

  2

  

Iin Purnama Edwar, M. Imran & Leli Deswita – Iterative Method Derivative-Free

  3

  2 E = 2c c + 8c c + 4c ,

  3

  1

  3

  1

  2

  3

  2 F c c c c c ,

  = 8c + 6c + 10c + 10c

  4

  2

  3

  1

  4

  2

  3

  1

  1

  5

  3

  2

  2

  2

  2

  2 G c c c c c c c c c ,

  = 2c

  5 + 20c 5 + 24c

  2 4 + 6c + 6c

  1 3 + 12c

  1 5 + 12c

  2 4 + 6c

  1

  1

  1

  3

  1

  2

  3

  5

  4

  3

  2

  2

  2 H = 12c c + 10c c c + 40c c + 60c c c + 30c c c + 24c c c

  6

  2

  5

  6

  5

  2

  3

  4

  1

  4

  1

  1

  1

  1

  1

  2

  2

  3 c c c c c c .

  • 12c + 2c + 14c + 14c + 14c

  1

  2

  3

  1

  6

  2

  5

  3

  4

  3

2 Simplifying (20) and applying a geometry series to the resulting equation,

  we end up with

  2

  2f (x ) c

  n

  2

  2

  7

  e

= e − + · · · + O(e ). (21) + n

  n n

  f (x + f (x )) − f (x − f (x )) c

  n n n n

  1 Substituting (21) into (14), and noting x = e + α, we obtain n n

  c

  2

  2

  7

  y e

  n = α + − + · · · + O(e ). (22) n n

  c

1 Expanding f (y n ) using Taylor’s series about y n = α and using (22)

  yield

  2

  −2c

  2

  2

  2

  3

  7

  • f (y ) = c e + c c + 2c e + · · · + O(e ). (23)

  n

  2

  3

  3 n 1 n n

  c

1 Computing (9) using (15) and (19), after simplifying, we have

  2

  2

  7 T e c

  = c + 2c n + (c + 3c )e + · · · + O(e ). (24)

  1x

  1

  2

  3 3 n n

1 Similarly, we obtain

  2

  2c

  2

  2

  7

  • T e = c + · · · + O(e ). (25)

  1y 1 n n

  c

1 Substituting (15), (23), (24), and (25) into (12), and after some algebra, we

  obtain

  2

  1 c 2c

  

1

  3

  2

  3

  7

  z c c e

  • 1
  • n = α + + · · · + O(e ). (26)

  3 n n

  2

  2 c c

  

2

  1

1 By the same strategy to obtain f (y n ), using (26) we obtain

  2

  2c

  1

  1

  2

  2

  3

  7

  • f c c
  • +

    c e (z n ) = + · · · + O(e ). (27)

  3 3 n n

  1

  c

  2

  2

  

Iin Purnama Edwar, M. Imran & Leli Deswita – Iterative Method Derivative-Free

Substituting (24), (25), (26), and (27) into (13), using geometric series and

simplifying the resulting equations we ends up with

  6

  2

  4

  2

  2

  2

  2

  4

  6

  c c c + 6c c + 5c c + 16c c c − 16c e

  1

  2

  1

  3 n

  1

  3

  1

  3

  3

  1

  2

  2

  x .

  n +1 = α −

  (28)

  5

  4 c

  1 Since e = x − α then (28) becomes n +1 n +1

  6

  2

  4

  2

  2

  2

  2

  4

  6

  c c c + 6c c + 5c c + 16c c c − 16c e

  1

  2

  1

  3 n

  1

  3

  1

  3

  3

  1

  2

  2

  e .

  n +1 = −

  5

  4 c

  1 From the definition of the order of convergence, we see that (11)–(13) is of order six .

  

3. NUMERICAL SIMULATION

In this section, we compare the number of iteration to obtain an approxi-

mated root for Rafiullah’s Method (MR), equation (8), Second Derivative-

Free Variant of Halley’s Method (MH) [5], A Sixth-Order Iterative Method

Free from Derivative (MP) [9] and the Derivative-Free Iterative Method

(MT) given by (11)-(13) using four test functions. The computation was

carried out using Maple 17. The criteria to stop the iteration are |f (x n +1 )| ≤

  −50

  

T ol , and |x −x | ≤ T ol, where T ol ≤ 1.0×10 . The maximum iteration

  k k−1

  allowed is 100.

  3 f 3 = √x

  3

  

3

  3

  0.5

  − x

  

Iin Purnama Edwar, M. Imran & Leli Deswita – Iterative Method Derivative-Free

Table 1: Comparison the number of iterations of the discussed iterative

methods f (x) x The number of iterations

  3

  

3

  3.2

  3 1.000000000000000

  3

  3

  

3

  3

  3.0

  3

  3

  3

  0.7

  3

  1.9

  

3

  3

  2.2

  3

  3

  

3

  3

  3

  3

  3

  

3

  3

  1.2

  3

  3

  

3

  

3

  1.5

  3 f 4 = xe −x

  3

  2

  

2

  2

  1.8

  3

  3

  

3

  1.5

  2.0

  3 1.7461395304080124

  3

  

3

  3

  1.2

  1 = e −x + cos(x)

  α MR MH MP MT f

  2

  3

  3

  3

  3

  

3

  3

  1.1

  3 2.120028238987641

  3

  

2

  0.5

  

3

  2 = x − 2 − e −x

  3 f

  3

  

3

  3

  2.3

  3

  3

  3

  • 1.2
  • 0.6
  • 0.1

  − 0.1

  3

  Ed , John Wiley & Sons, Inc., New York, 1999.

  st

  

REFERENCES

1.

  The simulation shows that MH is sightly better than other methods,

however this method is not derivative free. In general MT is comparable to

other methods, therefore this method can be used as an alternative method

for solving nonlinear equation.

  3 Table 1 shows the number of iterations needed to obtain the approxi-

mated root for several mention methods by varying an initial guess x . The

star mark (*) indicates that the method converges to a different root. How-

ever, there is no significant difference among mention methods in terms of

the number of iteration needed to obtain an approximate root.

  3

  

3

  3

  0.2

  3

  

3

  7 4 *

  3

  0.0

  4

  3

  

3

  3

  4

  4

  

3

  4

  4 0.111832559158963

G. Adomian, Nonlinear Stochastic Operator Equations, Academic Press, New York, 1986.

2. R. G. Bartle dan D. R. Shebert. Introduction to Real Analysis, 4

  

Iin Purnama Edwar, M. Imran & Leli Deswita – Iterative Method Derivative-Free

3.

  M. Basto, V. Semiao dan F. L. Calheiros, A new iterative method to compute nonlinear equation , Applied Mathematics and Computation, 173 (2006), 468–483.

  rd 4.

  Ed J. D. Faires and R. L. Burden, Numerical Methods, 3 , Brooks Cole, Belmont, 2002.

  

5. J. Han, H. He, A. XU and Z. Cen, A second derivative free variant of

Halley’s method with sixth orde convergence , Journal of Pure and Applied Mathematics: Advances and Applications, 3 (2011), 195–203.

  

6. J. Kou, Y. Li, and X. Wang, Modied Halley’s method free from second

derivative , Applied Mathematics and Computation, 183 (2006), 704–708.

  rd 7.

  J. H. Mathews and K. D. Fink, Numerical Methods Using MATLAB, 3 Ed . Prentice Hall, New Jersey, 1999.

  

8. S. K. Parhi and D. K. Gupta, A six order method for nonlinear equation,

Applied Mathematics and Computation, 203 (2008), 50–55.

  

9. 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, 8 (2014), 5721–5730.

10. M. Rafiullah, Three-step iterative method with sixth order convergence for

  solving nonlinear equations , International Journal of Computer Mathe- matics, 4 (2010), 2459–2463.

  Iin Purnama Edwar : Magister Student, Department of Mathematics, Faculty of

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

baru 28293, Indonesia.

  E-mail: edwar.iinpurnama@gmail.com M. Imran

  : Numerical Computing Group, Department of Mathematics, Faculty of

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

baru 28293, Indonesia.

  E-mail: mimran@unri.ac.id Leli Deswita

  : Applied Mathematics Laboratory, Department of Mathematics, Fac-

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

Pekanbaru 28293, Indonesia.

  E-mail: deswital@yahoo.com