Solving Nonlinear Algebraic Equation By Homotopy Analysis Method

Proceedings of the 6th IMT-GT Conference on Mathematics, Statistics and its Applications (ICMSA2010)
Universiti Tunku Abdul Rahman, Kuala Lumpur, Malaysia 712
Solving Nonlinear Algebraic Equation
By Homotopy Analysis Method
Chin Fung Yuen (1) , Lem Kong Hoong (2) , Chong Fook Seng (3)
Department of Physical and Mathematical Sciences, Faculty Science,
University Tunku Abdul Rahman (UTAR), Perak Campus,
Jalan Universiti, Bandar Barat
31900 Kampar, Perak, Malaysia.
chinfy@utar.edu.my (1) , lemkh@utar.edu.my (2) , chongfs@utar.edu.my (3)
Abstract. This paper presents an efficient method – Homotopy Analysis Method
(HAM) that based on Newton Raphson's method to solve the nonlinear algebraic
equation. This method is more powerful than the Newton Raphson's method in
the sense that this method allows more freedom in choosing the initial
approximation x0
instead of a good x0
that is closed to the root must be
chosen. By applying the HAM, it can speed up the iteration and we can control
the convergence rate by the h- curve. Some examples are given to show the
efficiency of the algorithm compare to the Newton- Raphson's method.
1. Introduction
Solving nonlinear algebraic equation is a subject of considerable interest and
many developments of the numerical technique have been done, for example
solving the nonlinear algebraic equation by homotopy perturbation method
[1], modified homotopy perturbation method [2], Newton-homotopy
continuation method [3], modified Adomian decomposition method [4],
Newton-homotopy analysis method [5]. Newton Raphson's method is a wellknown
and popular method for solving nonlinear algebraic equation. It has
high efficiency in the convergence speed by choosing a good initial
approximation x 0
. However bad structure of equation or bad initial
approximation x0
will impel some uncontrollable situation, such as overflow,
divergence and etc. In the recent paper [5], an efficient method – Homotopy
Analysis Method (HAM) that based on Newton Raphson's method is proposed
for solving nonlinear algebraic equation. This proposed method is applied to
solve test problems in order to assess its validity and accuracy.

Proceedings of the 6th IMT-GT Conference on Mathematics, Statistics and its Applications (ICMSA2010)
Universiti Tunku Abdul Rahman, Kuala Lumpur, Malaysia 713
2. Homotopy Analysis Method (HAM)
Consider the nonlinear algebraic equation
( x) = x ∈ R
f 0 ,
(1)
We assume that α is a zero of the function f ( x)
in (1).
By using Taylor’s expansion near x
2
3
δ δ
4
( x δ ) = f ( x) + δf
′( x) + f ′′( x) + f ′′′ ( x) O( δ )
f + + .
2 6
Neglecting terms of higher order, we have
2
3
δ δ
f ( x + δ ) = 0 ≈ f ( x) + δf
′( x) + f ′′( x) + f ′′′
( x)
.
2 6
Solving for δ
yields
( x)
′( x)
( x)
( x)
( x)
( x)
2
3
f δ f ′′ δ f ′′′
δ = − − − .
f 2 f ′ 6 f ′
We define a function
( x)
( x)
( x)
( x)
2
3
δ f ′′ δ f ′′′
g ( δ ) = −δ
− − = c ,
2 f ′ 6 f ′
where
f ( x)
c = .
f ′ x
( )
Now homotopy analysis method (HAM) is proposed to solve equation (1) by
using Newton- Raphson method. Let q ∈ [ 0,1]
denotes the so called
embedding parameter, h ≠ 0 an auxiliary parameter, H ( δ ) ≠ 0 an auxiliary
function, and L an auxiliary linear operator. We construct the zero- order

Proceedings of the 6th IMT-GT Conference on Mathematics, Statistics and its Applications (ICMSA2010)
Universiti Tunku Abdul Rahman, Kuala Lumpur, Malaysia 714
deformation equation [6]
( 1 q) L[ v( q)
− δ
0
] = qhH
( δ ){
g v( q)
− [ ]} , (2)
δ is the initial approximation of δ , and ( )
where 0
v q is an unknown
function. It should be emphasized that we have great freedom to choose the
auxiliary parameter h , the auxiliary function H ( δ ), the initial approximation
δ
0 , and the auxiliary linear operator L .
When q = 0 , equation (2) becomes
[ v( 0)
− δ
0
] = 0
L ,
when q = 1, equation (2) becomes
[ v( q)
] = 0
g ,
since ≠ 0
h and ( δ ) ≠ 0
H .
So, we see that, when q increases from 0 to 1, v( q)
δ to the solution δ . Expanding v( q)
guess 0
q , we have
varies from the initial
in Taylor series with respect to
v
( ) ∑ ∞ q =
0
+
m=
where
m
δ δ mq , (3)
1
( q)
m
1 d v
δ
m
=
m q= 0 .
m!
dq
If the auxiliary linear operator, the initial guess, the auxiliary parameter and the
auxiliary function are so properly chosen, the series (3) converges at q = 1
such that
+ ∑ ∞ m
δ = δ
0
δ m
q
(4)
m=
1

Proceedings of the 6th IMT-GT Conference on Mathematics, Statistics and its Applications (ICMSA2010)
Universiti Tunku Abdul Rahman, Kuala Lumpur, Malaysia 715
which must be one of solutions of the original nonlinear equation, as proved by
Liao [7]. Choosing h = −1
and H ( δ ) = 1, equation (2) becomes
( 1−
q) L[ v( q)
− δ
0
] + q{
A[ v( q)
] − c}
= 0
which is used mostly in the homotopy perturbation method (HPM), whereby
the solution is obtained directly, without using Taylor series. The comparison
between HAM and HPM can be found in [8].
Setting H ( δ ) = 1 and differentiating equation (2) m times with respect to the
embedding parameter q and then setting q = 0 and finally dividing them by
m!, we have the so- called mth- order deformation equation
L
r
[ δ − χ δ ] R ( δ )
m
where
( )
m
h , (5)
m− 1
=
m m−1
⎧0,
m ≤ 1
χ
m
= ⎨ ,
⎩1,
m > 1
1 d
1 !
( m − )
[ A[ v( q)
] − c]
m−1
R r m
δ
m−1
=
m−1
q=
0 , (6)
and
r
δ =
n
δ
dq
( t) , δ ( t) , δ ( t) ,..., δ ( )}
{
0 1 2 m
t
.
The δ from HAM is then incorporated into the follow Newton Raphson’s
method like iteration formula,
x
δ
( x )
n+1 = xn
+
n
.
The numerical iteration formula of Newton Raphson’s method for solving
equation is
x
= x
−
f
( xn
)
′( x )
n+1 n
. (7)
f
n

Proceedings of the 6th IMT-GT Conference on Mathematics, Statistics and its Applications (ICMSA2010)
Universiti Tunku Abdul Rahman, Kuala Lumpur, Malaysia 716
If a good approximation x0
is chosen, iteration (7) may converge to the root
of the equation (up to precision required) in just a few iterations. When a bad
approximation x0
is chosen, the iteration (7) may converge slowly such that
many iterations are needed for precision required or it may even lead to
divergent results. By means of HAM, the initial approximation x0
may be
chosen freely instead of a good x0
closed to the exact root. Moreover, the
iteration converges faster and we can control the convergence rate by altering
the auxiliary parameter h . The value of h can be determined by potting the
so- called h - curve, as suggested by Liao.
3. Test problem
We present some examples to illustrate the efficiency of the algorithm compare
to the Newton- Raphson's method.
Example 1: f ( x) = x − cos x = 0
The table below lists the results obtained by HAM and the Newton Raphson’s
method. The initial value and the number of iteration to obtain the root up to 6
decimal places are shown in the table. As we see from the table below, it is
clear that the result obtained by the HAM is more efficient in solving the
equation. This example shows that even a bad initial value is chosen, which is
far from the root, the HAM still convergences in a few iterations. Moreover,
even though the Newton- Raphson’s method is convergent for certain initial
value, HAM converges faster to the root.
Method
Initial value,
x 0
Number
iterations, n
f
( ) x n
HAM
1 2 −7
− 2.229505×
10
( m = 3,
h = −0.5)
50 6 −7
− 2.229505×
10
Newton- Raphson's 1 3 −7
− 2.229505×
10
50 Divergent -
Here are some iterations show by HAM and Newton Raphson’ method for
initial value x = 0
50 ,

Proceedings of the 6th IMT-GT Conference on Mathematics, Statistics and its Applications (ICMSA2010)
Universiti Tunku Abdul Rahman, Kuala Lumpur, Malaysia 717
HAM
x
x
x
x
x
x
x
x
0
1
2
3
4
5
6
7
= 50.000000
= 1.176610×
10
= −2.156798×
10
= 0.000000
= 0.734375
= 0.739086
= 0.739085
= 0.739085
9
58
Newton- Raphson method
x
x
x
x
x
x
0
1
2
3
4
5
7
= 50.000000
= −16.476901
= −7.182040
= 28.719786
= −23.328018
= −11.620473
x6
= −4.880916
x = −2.338584
3
Example 2: f ( x) = x = 0
A real root is x = 0. In the table below, we list the results obtained by HAM and
the Newton Raphson’s method. The initial value and the number of iteration to
obtain the root up to 6 decimal places are shown in the table. It is clear that the
result obtained by the HAM is more efficient while the result by Newton-
Raphson's method is divergent. Newton Raphson’s method diverges even an
initial value that is closed to the root is chosen. On the other hand, HAM
converges for both the initial values.
Here are some iterations obtained by HAM and Newton Raphson’ method for
initial value x 1 and x 50 ,
0 =
0 =
( ) x n
Method Initial value Number
iterations
f
HAM
1 115 0.000000
( m = 3,
h = −0.1)
50 146 0.000000
Newton- Raphson's 1 Divergent -
50 Divergent -

Proceedings of the 6th IMT-GT Conference on Mathematics, Statistics and its Applications (ICMSA2010)
Universiti Tunku Abdul Rahman, Kuala Lumpur, Malaysia 718
HAM
x
x
x
.
.
.
x
x
x
0
1
2
114
115
116
= 1.000000
= −0.880000
= 0.774400
= −0.000001
= 0.000000
= −0.000000
Newton- Raphson method
x
x
x
x
x
x
x
0
1
2
3
4
5
7
8
= 1.000000
= −2.000000
= 4.000000
= −8.000000
= 16.000000
= −32.000000
x6
= 64.000000
x = −128.000000
= 256.000000
HAM
x
x
x
.
.
.
x
x
x
0
1
2
145
146
147
= 50.000000
= −44.000000
= 38.720000
= −0.000001
= 0.000000
= −0.000000
Newton- Raphson method
x
x
x
x
x
x
x
x
0
1
2
3
4
5
6
8
= 50.000000
= −100.000000
= 200.000000
= −400.000000
= 800.000000
= −1600.000000
= 3200.000000
= 12800.000000
Example 3: f ( x) = x
3 + 4x
2 + 8x
+ 8 = 0
A real root is x = −2
. The initial value and the number of iteration to obtain the
root up to 6 decimal places are shown in the table. It is clear that the result
obtained by the HAM is more efficient in the sense that it requires less
iterations then the Newton- Raphson’s method to obtain the root.

Proceedings of the 6th IMT-GT Conference on Mathematics, Statistics and its Applications (ICMSA2010)
Universiti Tunku Abdul Rahman, Kuala Lumpur, Malaysia 719
Method Initial value Number iterations f ( )
HAM
1 3 0.000000
( m = 3,
h = −0.9)
50 7 0.000000
Newton- Raphson's 1 6 0.000000
50 13 0.000000
x n
A few iterations by both methods are listed below for initial value x = 0
1
and x = 0
50 ,
HAM
Newton- Raphson method
x
x
x
x
x
x
0
1
2
3
4
5
= 1.000000
= −0.839882
= −2.001045
= −2.000000
= −2.000000
= −2.000000
x
x
x
x
x
x
0
1
2
3
4
5
7
= 1.000000
= −0.105263
= −1.106642
= −2.060344
= −2.001816
= −2.000001
x6
= −2.000000
x = −2.000000
HAM
x
x
x
x
x
x
x
0
1
2
3
4
5
7
8
= 50.000000
= 23.680395
= 10.820140
= 4.494731
= 1.282956
= −0.625917
x6
= −2.020002
x = −2.000000
= −2.000000
Newton- Raphson method
x
x
x
.
.
.
x
x
x
0
1
2
12
13
14
= 50.000000
= 32.877086
= 21.455714
= −2.000080
= −2.000000
= −2.000000

Proceedings of the 6th IMT-GT Conference on Mathematics, Statistics and its Applications (ICMSA2010)
Universiti Tunku Abdul Rahman, Kuala Lumpur, Malaysia 720
4. Conclusion
The Newton- Raphson’s method is a popular numerical method. However, it
still has some well- known critical defects such as requirement of a good initial
value and the problem of divergence. HAM in general converges faster and
allows more freedom in choosing of the initial value. This paper provides an
alternative way for solving nonlinear equation. Our results show that HAM is
able to manage problems that Newton Raphson’s method fails.
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