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ITERATIVE METHOD FOR SOLVING NONLINEAR EQUATIONS USING DECOMPOSITION METHOD
F. Ahmad1, a, N. A. Mir2, M. Z. Sarikaya3, S. Hussain4, A. Rafiq5
1[email protected], 2[email protected], 3[email protected], 4[email protected],
2Mathematics Department, Rapha University, Islamabad, Pakkistan
3Mathematics Department, Faculty of Science and Arts, Düzce University, Düzce, Turkey
1,4 Mathematics Department, Majmaah University, College of Science, Alzulfi, Saudi Arabia
5Mathematics Department, COMSATS, Institute of Information Tech. Islamabad, Pakkistan a: Correspondence author: [email protected]
Received: November 30, 2014 Accepted: February 13, 2015
ABSTRACT
In this paper, we suggest and analyze a new three-step method for solving nonlinear equations using the decomposition technique which is mainly due to Noor et al. [5]. We show that this new iterative method has fourth- order of convergence. To demonstrate the efficiency and performance of the new method, we tested several numerical examples and results are shown in the table-1.
KEYWORDS: Iterative methods, Two-step methods, Three-step methods, Nonlinear equations, Order of convergence
1. INTRODUCTION
In recent years, much attention has been given to develop several iterative methods for solving nonlinear equations, see [2-6] and the references therein. These methods can be classified as one-step, two-step and three-step methods. Two-step methods have been suggested by combining the well-known Newton method with other one-step implicit methods. In [2] Chun has proposed and studied several one-step and two-step iterative methods with higher- order convergence by using the decomposition technique of Adomian [1]. In the methods of Chun [2], higher-order differential derivatives are involved, which is a serious drawback. To overcome this draw back, we suggest and analyze a family of multi-step methods for solving nonlinear equations using a different type of decomposition, which does not involve the high-order differentials of the function. This new decomposition is mainly due to Noor et al. [5].
2. ITERATIVE METHODS Consider the nonlinear equation
. 0 ) ( x =
f
(2.1)We assume that
α
is a simple root of (2.1) and is an initial guess sufficiently close toα
. We can rewrite the nonlinear equation (2.1) as a coupled system:, 0 ) 2 (
) ( ) ) ( ( )
( ′ + ′ + =
−
+ f f x g x
x
f γ
γ
γ
(2.2)2 , ) ( ) ) ( ( ) ( ) ( )
( f f x
x f
x f x
g ′ + ′
−
−
−
= γ
γ
γ
(2.3)where
γ
is the initial approximation for a zero of( 2 . 1 )
. We can rewrite( 2 . 2 )
in the following form:). ( ) (
) ( 2 )
( ) (
) ( 2
x f f
x g x
f f
x f
+ ′
− ′ + ′
− ′
= γ γ
γ γ (2.4)
We can also rewrite
( 2 . 4 )
as:), (x N c
x = +
(2.5)where
Farooq Ahmad et al., 2015
) , ( ) (
) ( 2
x f f
c f
+ ′
− ′
=
γ
γ γ
(2.6)and
) . ( ) (
) ( ) 2
( f f x
x x g
N = − ′ + ′
γ
(2.7)
Here
N (x )
is a linear operator.We note that if
x
0 is the initial guess, then from( 2 . 2 )
and( 2 . 3 ) ,
we have).
( ) ( x
0g x
0f =
(2.8)As in [6], the solution of (2.5) has the series form,
. 0
i i
x x ∑
∞=
=
(2.9)The nonlinear operator
N (x )
can be decomposed as. )
( ) (
1
0 0
1 0
0
−
+
= ∑ ∑ ∑
∑
−=
=
∞
=
∞
=
j i
j j
i
j i
i i
x N x N x
N x
N
(2.10)Combining (2.4), (2.7) and (2.10), we have
. )
(
1 0 0
1 0
0
−
+
+
=
= ∑ ∑ ∑ ∑
−=
=
∞
=
∞
=
j i
j j
i
j i
i i
x N x N x
N c x
x
(2.11)Thus we have the following iterative scheme:
. , 2 , 1 ), (
) (
), ( ) (
), ( ,
1 1
0 1
0 1
0 1
0 2
0 1
0
K K
K M
= +
+ +
− + + +
=
− +
=
=
=
−
+
N x x x N x x x n
x
x N x x N x
x N x
c x
n n
n
(2.12)
Then
, , 2 , 1 ), (
0 11 2
1
+ + K + = + + K + = K
+
N x x x n
x x
x
n nand
.
1 i i
x c
x ∑
∞=
+
=
(2.13)It can be shown that the series i
i
x
∞
=
∑
0
converges absolutely and uniformly to a unique solution of equation (2.5).
From (2.6), (2.9), (2.10), (2.11) and (2.12), we have
) , ( ) (
) ( 2
0
f f x
c f
x = = − ′ + ′ γ
γ γ
(2.14)and
). ( ) (
) ( 2 )
( ) (
) ( ) 2
(
0 0 0
0 0
1 f f x
x f x
f f
x x g
N
x =− ′ + ′
+ ′
− ′
=
= γ γ
(2.15)
This enables us to suggest the following method for solving the nonlinear equation (2.1).
Algorithm For the give
x
o compute the approximate solutionx
n+1 by the iterative schemes:), ( ) (
) ( 2
, , 2 , 1 , 0 , 0 ) ( ), (
) (
1
n n
n n
n
n n
n n n
z f x f
x x f
x
n x
x f f
x x f z
′ +
− ′
=
⋅
⋅
⋅
=
′ ≠
− ′
=
+
(2.16)
which is known as Weerakoon and Fernando method [7] and is cubically convergent.
Again by using
( 2 . 12 ) ( , 2 . 13 ) ( , 2 . 14 )
and( 2 . 15 ) ,
we conclude that) . ( ) (
) ( 2 )
( ) (
) ( 2
) (
0 0 0
0 1
x f f
x f x
f f
f x N x
x c x
+ ′
− ′ + ′
− ′
= +
= +
=
γ γ
γ γ
(2.17)
Using (2.17), we can suggest the following three-step iterative method for solving nonlinear equation (2.1) as:
Algorithm For the given
o
x
compute the approximate solution +1n
x
by the iterative schemes (AAU):( )
) . ( ) (
) ( 2
) , ( ) (
) ( 2
, 0 ) ( ) , (
1
n n
n n
n
n n
n n
n
n n
n n
n
y f x f
y y f
x
z f x f
x x f
y
x x f
f x x f
z
+ ′
− ′
=
+ ′
− ′
=
′ ≠
− ′
=
+
Again, using (2.8) and (2.12), we can calculate
) . (
) (
) (
) 2 (
1 0
1 0 1
0
2
f f x x
x x x f
x N
x ′ + ′ +
− +
= +
=
γ
(2.18) From (2.11), (2.12), (2.13), (2.14), (2.15) and (2.18), we get
) . (
) (
) (
2 )
( ) (
) ( 2
) ( ) (
) (
1 0
1 0
0 1
0 0
0
2 1
x x f f
x x f x
f f
f
x N x x N x N x
x x c x
′ +
′ +
− + + ′
− ′
=
− + +
+
=
+ +
=
γ γ
γ γ
Using this, we can suggest and analyze the following four-step iterative method for solving nonlinear equation (2.1).
Algorithm: For a given
x
0,compute the approximate solution by the iterative schemes:
( ) , ( ) 0 , )
( ′ ≠
− ′
=
nn n n
n
f x
x f
x x f
z
) . (
) (
) (
2 ) , ( ) (
) ( 2
) , ( ) (
) ( 2
1
n n n
n n n
n
n n
n n
n n
n n
n
s y f x f
s y y f
x
y f x f
y s f
z f x f
x x f
y
′ +
′ +
− +
=
+ ′
− ′
=
+ ′
− ′
=
+
Farooq Ahmad et al., 2015
3. ANALYSIS OF CONVERGENCE
Theorem: Assume that the function
f : D ⊂ R → R
for an open intervalD
has a simple rootα ∈ D
.Let
f (x )
be sufficiently smooth in the neighborhood of the rootα
, then the order of convergence of the method defined by Algorithm 2 is four.Proof: The iterative scheme is given by
( ) , ( ) 0 , )
( ′ ≠
− ′
=
nn n n
n
f x
x f
x x f
z
(3.1)) , ( ) (
) ( 2
n n
n n
n
f x f z
x x f
y = − ′ + ′
(3.2)) . ( ) (
) ( 2
1
n n
n n
n
f x f y
y y f
x
+= − ′ + ′
(3.3)Let
☺
be a simple zero off
. By Taylor's expansion, we have,[ ( ) ] ,
) ( )
( x
nf e
nc
2e
n2c
3e
n3c
4e
n4c
5e
5nc
6e
n6O e
n7f = ′ α + + + + + +
(3.4)[ 1 2 3 4 5 6 ( ) ] ,
) ( )
( x
nf c
2e
nc
3e
n2c
4e
n3c
5e
n4c
6e
n5O e
n6f ′ = ′ α + + + + + +
(3.5)where
. and
..., , 3 , 2 ) ,
( ) (
! 1
( )α α
α = = −
= ′
n nk
k
k e x
f f
c k
(3.6)From
3. 4
and3. 5,
we have).
( )
6 8 20
10 4 (
) 3 4 7
( ) 2 2 ) (
( ) (
6 5
2 3 4 2 2 2 3 4
2 5
4 4 3 2 3 2 3 3 2 2 2 2
n n
n n
n n n n
e O e c c c c c
c c
e c c c c e c c e c x e
f x f
+ +
+
− +
− +
−
− +
− +
−
′ =
(3.7)Using
3. 1
and3. 7,
we get).
( )
4 7
3 ( ) 2 2
(
3 22 3 4 2 3 23 4 52
2 n n n n
n
c e c c e c c c c e O e
z = α + + − + − + +
(3.8)By Taylor's series, we have
( ) z
nf ( ) [ c
2e
n2( 2 c
32 c
22) e
n3( 3 c
47 c
2c
35 c
32) e
n4O ( e
n5) ] ,
f = ′ α + − + − + +
(3.9)and
)].
( ) 16 28
20 8
(
) 8 11
6 ( ) 4 4
( 2
1 )[
( ) (
6 5
5 2 3 2 3 4 2 2 5 2
4 4 2 2 2 3 4 2 3 3 2 3 2 2 2 2
n n
n n
n n
e O e c c c c c c c
e c c c c c e c c c e c f
z f
+
− +
− +
+
− +
− +
′ +
′ = α
(3.10)
From (3.2), (3.5) and (3.10) we get
), ( 4 )
6 3
9 2 2
( 3 ) 2 3
( 3 2 )
( 1
6 5
2 3 4 2
2 2 3 4 2 5 4
3 2 3 2 4 3 2 2 3
n n
n n
n
e O e c c
c c c c c e
c c c c e c c y
+
− +
− +
+
− +
+ +
+
= α
(3.11)
implies by Taylor's series,
)], ( 3 )
6 9
2 2 ( 3 ) 2 3
( 3 2 )
)[( 1 ( ) (
6 5
2 4 2
4 2 5 4
3 2 3 2 4 3 2 2
3 n n
n
e O e c c c c
c c c e
c c c c e c c f
y f
+
− +
−
+ +
− +
+
′ +
= α
(3.12)
)].
( 2 )
12 3 18
4 3
(
) 6 3
2 ( ) 2 (
1 )[
( ) (
6 5
2 3 2 5 2 3 2 3 4 2 2 5 2
4 4 2 2 2 3 4 2 3 3 2 3 2
n n
n n
n
e O e c c c c c c c c c
e c c c c c e c c c f
y f
+
− +
− +
+
− +
+ +
′ +
′ = α
(3.13)
Finally using (3.3), (3.5), (3.12) and (3.13), we obtain
), ( )
2 4 5 4 ( 3 2 )
( 1
2 3 23 4 32 3 22 2 4 24 5 61 n n n
n
c c c e c c c c c c e O e
x
+= α + + + + + − +
(3.14)implies
).
( )
2 4 5 4 ( 3 2 )
( 1
2 3 23 4 32 3 22 2 4 24 5 61 n n n
n
c c c e c c c c c c e O e
e
+= + + + + − +
Hence proved.
4. Numerical Examples
In this section we consider some numerical examples to demonstrate the performance of the new developed iterative method. We compare Chun method [2] (CHU), Noor and Noor method [4] (NR3), Noor and Noor method [3] (NR1) and Noor and Noor method [5] (NR2) with the new developed method (AAU). All the computations for above mentioned methods, are performed using software Maple
9
, precision60
digits andε = 10
−18as tolerance and also the following criteria is used for estimating the zero:(i)
= x
n+− x
n<
δ
1ε .
(ii)
f ( ) x
n< ε .
(iii) Maximum numbers of iterations
= 500 .
We use the following examples for numerical testing and results are given in the Tables of Results.
Example Exact Zero
153411 4044916482
. 1 ,
1 )
( sin
2 21
= x − x + α =
f
398608 2575302854
. 0 ,
2
2
3
2
= x − e − x + α =
f
x156067 7390851332
. 0 ,
)
3
= cos( x − x α =
f
( 1 )
31 , 2 . 0
4
= x − − α =
f
3188372 1544346900
. 2 ,
3
10
5
= x − α =
f
309191 2076478271
. 1 ,
5 ) cos(
3 ) ( sin
26
2
−
= +
+
−
= xe x x α
f
x3 ,
30
1
7 7
2
=
−
= e
x + x−α
f
Farooq Ahmad et al., 2015
.
AAU 5 0 1.2e - 39
15 - 8.0e 28
- 1.9e 6
CHU
21 - 4.4e 41
- 6.0e - 6
NR3
27 - 1.1e 54
- 2.6e 8
NR2
15 - 2.4e 29
- 1.7e 7
NR1
22 - 8.2e 42
- 2.0e 8
NW
5 . 3 ,
54 - 1.0e 0
4 AAU
17 - 1.9e 34
- 1.3e - 5
CHU
27 - 1.2e 55
- 6.2e 6
NR3
16 - 3.5e 32
- 7.5e - 5
NR2
17 - 7.5e 33
- 2.1e - 5
NR1
16 - 2.3e 32
- 2.0e - 5
NW 7 . 1 ,
31 - 1.5e 126
- 7.4e 4
AAU
16 - 2.0e 32
- 1.5e 5
CHU
15 - 5.8e 29
- 1.2e - 6
NR3
18 - 1.0e 37
- 5.1e 6
NR2
25 - 4.5e 50
- 7.3e 6
NR1
28 - 9.1e 55
- 2.9e 6
NW 2 ,
24 - 1.0e 96
- 1.4e - 4
AAU
28 - 1.0e 9
- 1.6e - 15
CHU
27 - 4.2e 53
- 3.8e 8
NR3
16 - 3.6e 31
- 1.5e - 6
NR2
16 - 8.9e 10
- 8.0e - 7
NR1
26 - 7.3e 50
- 1.0e - 7
NW
1
,
) ( n
1 - Table
0 4
0 3
0 2
0 1
=
=
=
−
=
x f
x f
x f
x f
x
f
nδ
f
5, x
01. 5
NW 7 2.0e-54 5.4e-28
NR1 7 1.3e-31 1.4e-16
NR2 6 3.8e-28 8.0e-15
NR3 7 -1.3e-39 1.4e-20
CHU 27 2.3e-31 1.9e-16
2 ,
06
x = −
f
NW 9 -2.2e-40 2.7e-21
NR1 8 -3.0e-9 1.3e-17
NR2 9 -4.3e-42 3.8e-22
NR3 7 1.8e-37 7.7e-20
CHU 7 -6.0e-9 6.4e-18
AAU 5 -1.3e-61 1.9e-16
f
7, x
03. 5
NW 13 1.5e-47 4.2e-25
NR1 11 5.0e-33 7.6e-18
NR2 13 1.0e-50 1.1e-26
NR3 8 -4.6e-37 7.3e-20
CHU 9 4.8e-31 7.5e-17
AAU 7 1.3e-59 2.2e-16
5. CONCLUSION
In the Table-1, we observe that our iterative method (AAU) is comparable with all the cited methods and gives better results. The technique and idea of this paper can be developed to higher-order multi-step iterative methods for solving nonlinear equations, as well as a system of nonlinear equations.
REFERENCES
1) R. G. Adomian, Nonlinear stochastic systems and application to physics, Kluwer Academic publishers, Dodrecht, 1989.
2) C. Chun, Iterative methods improving Newton's method by the decomposition method, Comp. Math. Appl., 50 (2005), 1559-1568.
3) M. A. Noor and K. I. Noor, Three step iterative methods for nonlinear equations, Appl. Math. Comput., 183(1), ( 2006), 322-327.
4) M. A. Noor and K. I. Noor, Some iterative schemes for nonlinear equations, Appl. Math. Comput., 183(2), ( 2006), 774-779.
5) M. A. Noor, K. I. Noor, S. T. Mohyud-Din and A. Shabbir, An iterative method with cubic convergence for nonlinear equations, App. Math. Comput. 183(2), ( 2006), 1249-1255.
6) K. I. Noor and M. A. Noor, Iterative methods with fourth-order convergence for nonlinear equations, Appl. Math.
Comput., 189(1), ( 2007), 221-227.