39
THREE-STEP ITERATIVE METHOD WITH FIFTH-ORDER
CONVERGENCE FOR SOLVING MULTIPLE ROOTS OF NONLINEAR EQUATIONS
Zhinan Wu
School of Mathematics and Computer Science, Yichun University, Yichun 336000,PR China
ABSTRACT
This paper presents a three-step iterative method with fifth-order iterative convergence as a new modification of Newton’s method for finding multiple roots of nonlinear equations with unknown multiplicity
m
and for finding system of nonlinear equations. Its order of convergence is analyzed and proved. Results for some numerical examples show the efficiency of the new method.Keywords: Nonlinear equation; Multiple roots; Newton-like method; High-order convergence; Iterative methods.
1. Introduction
This paper addresses the problem of finding multiple roots
x
* of nonlinear equationf (x ) = 0
, whereR R b a
f : [ , ]
is a nonlinear differential function on[ b a , ]
. In case the multiplicitym
is given explicitly, there are many iterative methods established via various techniques(see[1-15] for more details). If the multiplicitym
is not known explicitly, Traub [16] utilized a simple transformation
F ( x ) = f ( x )/ f ' ( x )
instead off (x )
forcomputing a multiple root of
f (x ) = 0
. In this case, the aim of solving a multiple root is reduced to that of solving a simple root of the transformed equationf (x ) = 0
, and thus any iterative method can be used to preserve the original convergence order. However, Newton’s method for this transformed equation requires evaluations of the derivativesf ' x ( )
andf ' x ( )
. In order to avoid the evaluations of these derivatives with the multiplicitym
unknown, for multiple roots, King[17] proposed the secant method which does not use the function
F = f / f '
, butrather use
) ( )) ( (
)
= (
)) ( (
) ( )) ( (
)
= (
2
x f x f x f
x f x
x f x
x f x f x f
x F f
. Wu and Fu [18] further used)) ( ( ) (
)
= ( ) (
2
x f x f x f
x x f
F
and transformed the problem of solving multiple roots off (x ) = 0
into that of solving simple root off (x ) = 0
. Actually, they established the following iteration formulae:)) , ( (
) ( ) (
)
=
2(
2
1
n n
n n
n n
n
p F x F x F x F x
x x F
x
(1)where
p R
,p <
. So, the sequence x
n produced by the iteration formulae (1) is at least quadratically convergent for multiple roots. Moreover, Wu et al.[19] defined a function, ) ( ) ) ( )) ( ( (
) ( ) ( )) ( ) ) ( )) ( ( (
(
) ( ) ( )) (
= ( )
(
1/ 1/ 1/1/
x f x
f x f sign x f x
f x f x f x
f x f sign x f sign
x f x f x f x sign
F
m m mm
(2)
where
m
is the multiplicity, and employed the modified Steffensen’s method (see[20],[21])40
)) ( ))/(
( )) ( (
( ) (
)
= (
) ( )) ( (
) (
)
=
2(
2
1
n n
n n
n
n n
n
n n
n n
n n
n n
x F x F x F x F x F t
x h F
x
x F x F x F x F t
x h F
x x
(3)
to compute the approximate solution of the equation
f (x ) = 0
, whereh
n(> 0)
is the step size of iteration and
<
t
. Parida and Gupta [22] suggested another transformation
. 0
= iff(x) 0
, 0 iff(x) )
( )) ( (
) (
= ) (
2
x f x f x f
x f x
F
(4)where
= sign ( f ( x f ( x )) f ( x )) f
2( x )
, and transform the task of solving multiple zeros off
into that of solving simple zero ofF
. In this case, they utilized a quadratically convergent derivative free Newton-like iterative method:)) ( (
) ( ) (
)
=
2(
2
1
n n
n n
n n
n
p F x F x F x F x
x x F
x
(5)where the parameter
p
should be chosen such that the denominator is the largest in magnitude. Yun [23] suggested a new transformation off (x )
as) , ( )) ( (
)
= ( ) (
2
x f x f x f
x x f
H
(6)
took
such thatmax f ( x ) =
b x
a , that is
( ) , ( ) ,
= max ) max (
= f x f a f b
b x a
and proposed an iterative method as follows:
) . ( ) (2
) ( )
= 2(
1 1
1 1
n n
n
n n
n n
n
H x x H x
x H x x x
x
(7)
Recently, for the transformed equation
K (x ) = 0
with a simple root, Yun[24] proposed a Steffensen-type iterative formula0, ) ,
( )) ( (
)
= (
2
1
k
p K p K p K
p p K
p
k k
k
k k
k
(8)where
K ( x ) = K ( ; x ) = {
f(x) 0,0, if
f(x) = 0..
In this paper we construct a new modification of Newton’s method based on Newton’s method. We will present the proof that the method is three-step iterative method with fifth-order convergence for nonlinear equations of multiple roots with unknown multiplicity
m
and without requiring the use of the second derivative.2. Iterative method with fifth-order convergence for solving multiple roots We consider the simple transformation (see[16],[25]):
0,
= iff(x) 0,
, 0 iff(x) ) ,
( '
) (
= )
( f x
x f x
F
(9)and use a Newton-like iterative method:
41
.
) ( '
)
= (
) , ( '
)
= (
) , ( '
)
= (
1
n n n
n
n n n
n
n n n
n
z F
z z F
x
y F
y y F
z
x F
x x F
y
(10)
In order to avoid computing the first derivatives of function
F ( x
n)
,F ( y
n)
andF ( z
n)
, we approximate them as follows:), ( ) =
(
) ( )) ( ) (
(
'
1 nn
n n
n
n
g x
x F
x F x F x x F
F
(11)
) (
= ) )) (
( ) ( ) 2(
(
'
1 n 2 nn n
n n
n
g x g x
x y
x F y y F
F
(12)
) (
=
) (
) ) (
( ) ( ) ( )
= (
) ](
, , [ ] , [ ) ( '
3
1
n
n n n
n
n n
n
n n
n n
n n
n n n n n n
n n
x g
y x z
z
x x g
z
x F z F y
z
y F z F
y z x x z F y z F z
F
(13)
(See [25-28] for the detail discussions of (11-13) respectively.) Substituting the approximations of
F ' ( x
n)
,
F ' ( y
n)
andF ' ( z
n)
given by (11)-(13) in (10), we establish the following new iterative method:
.
) (
)
= (
) , (
)
= (
) , (
)
= (
3 1
2 1
n n n
n
n n n
n
n n n
n
x g
z z F
x
x g
y y F
z
x g
x x F
y
(14)
We give the following convergence theorem for the proposed method (14) as follows.
Theorem 1 Suppose that
F C
1( D )
(D R R
) has a single rootx
* D
, whereD
is an open interval.If the initial point
x
0 is sufficiently close tox
*, the iterative method defined by (14) has fifth-order convergence.Proof: We assume that
f (x )
can be written as), ( ) (
= )
( x x x
*h x
f
m (15)where
x
* is a multiple root of (15) with multiplicitym
,h (x )
is a continuous function withh ( x
*) 0
. According to (15), we have).
( ' ) ( ) ( ) (
= )
( x m x x
* 1h x x x
*h x
f
'
m
m (16)Dividing (15) by (16), we get
42
) . ( ' ) ( ) (
) ( )
= ( ) (
)
= ( )
(
**
x h x x x mh
x h x x x
f x x f
F
'
(17)From (17), we can see that the problem of computing multiple roots of
f (x ) = 0
can be reduced to the equivalent problem of computing simple rootx
* ofF (x ) = 0
.Using Taylor’s expansion, we have
)]
( )[1
(
= )
( x
nh x
*b
1e
nb
2e
n2b
3e
3nb
4e
n4b
5e
n5b
6e
n6o e
n7h
(18)where
, = 1,2,...,
) (
! )
= (
*
* ) (
x k h k
x b h
k
k and
e
n= x
n x
*.By (18), we obtain
)].
( 6
5 4
3 2
)[
(
= ) (
' x
nh x
*b
1b
2e
nb
3e
n2b
4e
n3b
5e
n4b
6e
n5o e
n6h
(19)Substituting (18)-(19) into (17), we get
) ( ) =
( ' ) (
)
= ( )
(
1 n 2 n2 3 n3 4 n4 5 5n 6 n6 n7n n n
n n
n
c e c e c e c e c e c e o e
x h e x mh
x h x e
F
(20)where
3
2 1 2 1 2 2 3
1 2 1
= 2 ,
= 1 ,
= m
b m b m c b
m c b
c m
(21)
4
3 1 3 1 2 3 1 2 3 2
1 2 2 1 4
2 3
4
= 3
m
b m b m b m b m b b m b
c b
(22)
) 3
3 6
10
4 4
2 4
4 1 (6
=
4 1 4 1 2 4 1 3 4 1 2 1 2 2 2 1 2
3 2 1 2 2 2 2 3 2 2 3 4 3 3 1 2 3 5 1 5
b m b m b m b m b b m b b
m b b m b m b m b m b b m b m b c
(23)
) 8 21
18 5
12 16
5 9
5 14
5 12
5 5
4 6
4 1 (8
=
2 3 1 2 2 3 1 3 2 3 1 4 2 3 1 2 2 2 1 3 2 2 1
4 2 2 1 2 3 2 1 4 3 2 1 3 3 2 1 4 3 2 3 3 2
4 5 4 4 1 5 1 5 1 2 5 1 3 5 1 4 5 1 3 4 6 1 6
m b b m b b m b b m b b m b b m b b
m b b m b b m b b m b b m b b m b b
m b m b b b m b m b m b m b m b m b c
(24)
Substituting (20) into (11), we obtain
) ( ) 8 3
3 7
2 10
5 10
(5 ) 4
6
4 2
(4 ) 3
(3 ) (2
= ) (
5 4 4 2 1 4 2 2 1 2 3 5 4 1 4 2
3 2 2 1 2 3 5 2 1 5 3 1 5 1 5 3 4 3 1 4 2 1 4 1
4 3 2 1 3 2 2 2 2 3 2 1 3 1 3 1
2 1 1
n n n
n n
n
e o e c c c c c c c c c c c
c c c c c c c c c c c e c c c c c c
c c c c c c e c c c c c c e c c c x
g
(25)
Substituting (20) and (25) into the first formula of (14), we have
43
) ( ) 2 7
7
10 4 3
5 4
6 1 (3
) 2
3 2
2 1 ( )
= (1
5 4 3 2 4 1 3 2 3 1 3 2 1
3 2 2 1 3 2 3 2 3 1 3 2 2 1 3 2 1 4 5 1 4 4 1 4 3 1 4 2 3 1 1
3 2 2 2 1 2 2 1 3 3 1 3 2 1 3 1 2 2 2 1 2
1 1
* 2
n n
n n
n
e o e c c c c c c c c c
c c c c c c c c c c c c c c c c c c c
e c c c c c c c c c c c c
c e c x c
y
(26)
With (26), we get
) ( ) 2 7
7
10 5 4
7 4
6 1 (3
) 2
3 2
2 1 ( ) (1
= ) (
5 4 3 2 4 1 3 2 3 1 3 2 1
3 2 2 1 3 2 3 2 3 1 3 2 2 1 3 2 1 4 5 1 4 4 1 4 3 1 4 2 2 1 1
3 2 2 2 1 2 2 1 3 3 1 3 2 1 3 1 2 2 1 2 1 2
n n
n n
n
e o e c c c c c c c c c
c c c c c c c c c c c c c c c c c c c
e c c c c c c c c c c c c
e c c y
F
(27)
Thereby, with (20), (25-27), we obtain
) ( ) 8 7
2 6
10 5
10 4
16
15 20
8 4
3 3
1 (10
) 6 4
4 2
4 4
6 1 (2
) 2 3
1 (
= ) (
5 4 2 1 4 2 3 1 4 2
3 1 4 2 4 2 2 1 4 2 1 7 1 5 6 1 5 5 1 5 3 2 2 4 1 3 2 2 1
2 2 3 1 3 3 2 1 2 2 5 1 4 2 4 2 2 3 2 1 5 3 1 2 3 3 1 5 4 3 1 1
3 3 2 1 3 2 2 1 3 2 3 2 2 1 3 2 1 4 5 1 4 4 1 4 3 1 4 2 2 1 1
2 2 2 2 2 1 3 3 1 3 2 1 3 1 1 2 1 1 2
n n
n n
n n
e o e c c c c c c
c c c c c c c c c c c c c c c c c c
c c c c c c c c c c c c c c c c c c c
e c c c c c c c c c c c c c c c c c c c c
e c c c c c c c c c c e c c c x
g
(28) From (26-28), it follows that
) ( ) 6
12 2
11 2
3 7
8 10
21 9
2 16
4 2
1 (9
) 2 7
6 2
3 ) (
= (1
6 5 4 2 4 1 2 3 6 1 2 3 5 1 2 3 4 1 2 1 4 2 3 1 4 2 4 2 3 1 4 2 2 1 4 2 1
3 2 2 1 3 2 2 3 1 3 2 2 2 1 5 1 4 2 4 2 6 1 4 2 4 1 2 2 2 1 2 3 3 1 2 4 3 1
4 3 4 1 3 1 3 1 3 2 1 3 2 2 2 1 2 2 1 2 3 2 1 3 2
1 1 2
* 2
n n n n
n
e o e c c c c c c c c c c c c c c c c c c c c
c c c c c c c c c c c c c c c c c c c c c c c c
e c c c c c c c c c c c c c c
e c c
c x c
z
(29)
It is similar to (27), we have
) (
) 2 7
6 3
( )
(1
= ) (
5
4 4 1 3 3 1 3 1 3 3 2 1 2 2 2 1 2 2 1 2 2 2 1 3 2 1 2 2
n
n n
n
e o
e c c c c c c c c c c c c c c
e c c c z
F
(30)
Moreover, substituting (20), (25), (26), (27), (29), (30) into (13), we get
) ( ) 2 2
7 7
(2 )
(1
= )
(
3 13 3 12 3 1 22 1 22 3 41 2 2 1 2 2 1
3 n
e n n
n
c c c c c c c c c c e o e
c e c c c c x
g
(31)Substituting (29-31) into the third formula of (14), we get
44
) ) (
2
= (1 ) (
)
= (
2 5 61 4 2 2 1
* 1
3
1 n n
n n n
n
e o e
c c c x c
x g
z z F
x
(32)which just means that the iterative method defined by (14) has fifth-order convergence. The proof is completed.
We further consider how to find the multiplicity of the root
x
* in the iterative method. Ifx
n is the nth iteration computed by an iterative method applied tof
, then from (9), we have) , ( ' ) (
)
= ( ) ( ' ) (
) (
) ( ) (
*
*
n n n
n n n
n n
n n
n
mh x h x
x h x
h x x x mh
x h x f x
where
f
n= f ( x
n)
. Because
n is small, we getf
n m
n. Similarly, we can compute that
f
n1 m
n1. Furthermore
n1
n= x
n1 x
n. Consequently, when the iteration becomes closer to the rootx
*, we can estimate its multiplicity by computing.
1 1
n n
n n
f f
x m x
In the practical computing root
x
* process, some iteration number is no more than two by using (14). According to this case, we compute the rootx
* by using (14), then we select the initial value near the rootx
*, and we use Newton iterative method to compute the multiplicity. From Table 1, we can know that Newton iterative method’s the iteration number is enough to ensure computing the correct multiplicity. Therefore,m
is approximately the reciprocal of the divided difference off
for successive iterationx
n andx
n1. It may be computed, and displayed at each step along with the current iteration(see[17],[22]).3. Numerical results
We employ the proposed modification of Newton’s method with three-step (14)(MNM) to solve some nonlinear equations and compare them with King’s method [17] (KM, (4),(13)), the high-order convergence iteration methods without employing derivatives given in [18](WFM,(6)), the improved method for finding multiple roots and it¡¯s multiplicity of nonlinear equations given in [22](PGM,((6),(11))), the derivative free iterative method for finding multiple roots of nonlinear equations given in [23](YM,((7),(10))), transformation methods for finding multiple roots of nonlinear equations [24](YM,((10),(12))), as well as Newton’s first order method(NM). Displayed in Table 1 are the number of iterations required such that
| f ( x
n) |< 1. E 17
,| x
n x
|< 1. E 17
. All the computations were done by using Visual C++ 6.0. Since King’s method and Yun’s method require two starting values, we have used0.1
=
01
x
x
. We used the following test functions and obtained the approximate zerosx
round up to the 17th decimal place:7499790 2.23606797
= 4,
= 1 , 1) (
) 5
= ( )
(
24
1
m x
x x x f
544138924 0.71483582
= 5,
= , 1) 2 ) sin ( (
= )
(
2 52
x m x
x x
f
91589544 1.79035317
= 8,
= , 3) 2 (8
= )
(
2 83
x xe
x m x
f
x94385368 2.98064527
= 10,
= 1) ,
(
3) )
( cos
= (2 )
(
210 2
4
m x
x x x x x
f
76083051 2.49053982
= 9,
= , 2) (
= )
(
2 3 95
x m x
e x
f
x x45
09263654 3.16274887
= 4,
= , )) ( sin 2 (
= )
(
46
x m x
e x
f
x59101421 5.46901233
= 8,
= , 7) 2 5) 3 ( ln (
= )
(
2 87
x x m x
x x
f
58839640 2.33196765
= 5,
= , 3) )
( sin 2 5 2 (
= )
(
2 2 58
x x x m x
x x
f
00000000 2.00000000
= 4,
= 1), 1) /((
2) (
= )
(
4 29
x m x
x x f
0000000 2.50000000
= 4 ,
= 15 ,
2.5) (
= )
(
4 *15
10
x x e m x
f
x5704787 2.14789903
= 7,
= , 1 1) (
= )
(
711
m x
x x x
f
4231572 8.30943269
= 3,
= , 5) )
( ln (
= )
(
312
x m x
x x
f
0687451 1.11707877
= 9,
= , 1) )
( cos ) ( sin (
= )
(
3 913
x x m x
x x
f
00000000 3.00000000
= 5,
= , )) ( exp 3) ((
= )
(
514
x m x
x x f
3628973 1.22281396
= 7,
= , 2) 1 )
( ln (
= )
(
4 7 *15
x x x m x
f
Note that we used NC in Table 1 to mean that the method does not convergence to the root. And these methods can converge to root by using closer initial values. The computational results in Table 1 demonstrate that our proposed iterative method (MNM) requires less number of iterations than those of KM, WFM, PGM, YM, and NM. Therefore, it is significant and applicable and can compete with other existing methods. Table 1 Comparison of various iterative methods
) (x
f
MNM KM[17] WFM[18] PGM[22] YM[23] YM[24] NMParameters
p = 1
p = 1
= 10
8 = 10
84.30
= ,
01
x
f
2 6 7 6 NC NC 1224.50
= ,
02
x
f
4 NC NC NC NC NC 1685.00
= ,
03
x
f
2 NC NC NC NC NC 4004.90
= ,
04
x
f
3 NC NC NC NC NC 3396.50
= ,
05
x
f
3 NC NC NC NC NC 2903.80
= ,
06
x
f
2 NC NC 3 NC NC 1198.30
= ,
07
x
f
2 NC NC NC NC NC 2575.50
= ,
08
x
f
2 NC NC NC NC NC 1643.50
= ,
09
x
f
2 5 6 5 4 4 1213.10
= ,
010
x
f
2 6 4 20 NC 3 864.50
= ,
011
x
f
2 4 NC NC NC 2 22412.0
= ,
012
x
f
2 4 9 8 NC NC 8746
2.20
= ,
013
x
f
3 NC NC NC NC NC 3007.00
= ,
014
x
f
3 NC NC NC NC NC 1826.00
= ,
015
x
f
3 NC NC NC NC NC 244
4. Efficiency of iterative methods
In the following we compare the efficiency of methods mentioned in Section 1. We consider the definition of efficiency index [29-31] as EFF =
1
r
, wherer
is the order of the method and
is number of function (and derivatives) evaluations per iteration required by the method. These results are presented in table 2. In table 2, we listed the methods according to decreasing order of efficiency index, and multiplicitym
of roots with all methods under the row of King’s method (including King’s method) are unknown. For unknown multiplicitym
, our new method comes second, next to King’s method. Table 2 Comparison the efficiency of various iterative methods
Method
Referencer
EFFNeta
[8](49)m 3
2.732 2 1.653Neta
[8](51) 2.732 2 1.653Neta and Johnson
[10]
m = 2
4 3 1.587Neta
[8](39) 3 3 1.442Neta
[8](29)m 3
3 3 1.442Chun and Neta
[7](22) 3 3 1.442
Chun
,Bae , and Neta
[9](14) 3 3 1.442
Victory and Neta
[4](3) 3 3 1.442
Hansen and Patrick
[2](8.1) 3 3 1.442
Halley
[12] 3 3 1.442Laguerre
[13] 3 3 1.442Dong
[14](7),(8) 3 3 1.442Dong
[5](9),(10) 3 3 1.442Osada
[6] 3 3 1.442E.Schröder [1] 2 2 1.414
Neta and Johnson
[10]
m 2
4 4 1.414Neta
[11] 4 4 1.414Neta
[8](32)m = 3
2 3 1.259Werner
[15](16)m = 2
1.5 3 1.145King
[17] 1.618 2 1.272Our method
This paper 5 8 1.223Xinyuan Wu
[18] 2 4 1.19047
Xinyuan Wu
[19] 2 4 1.190Parida
K
P . .
[22] 2 4 1.190Beong In Yun
[23] 2 4 1.190
Beong In Yun
[24] 2 4 1.190
5. Conclusion
A new iterative method with fifth-order convergence has been developed as a modification of Newton’s method for finding multiple roots of nonlinear equations with unknown multiplicity
m
and for finding system of nonlinear equations. Several numerical examples demonstrate that the proposed iterative method is more efficient and performs better than classical Newton’s method and many other existing methods.References
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