Download Numerical Solution of Fuzzy Polynomials by Newton

Journal of Applied Mathematics, Islamic Azad University of Lahijan, Vol.7, No.4(27), Winter 2011, pp 17-24
ISSN: 2008-594x
Numerical Solution of Fuzzy Polynomials by
Newton-Raphson Method
T. Allahviranloo*, S. Asari
Department of Mathematics, Science and Research Branch, Islamic Azad University, Tehran, Iran
Received: 3 June 2009
Accepted: 19 November 2009
Abstract
The main purpose of this paper is to find fuzzy root of fuzzy polynomials (if exists) by using
Newton-Raphson method. The proposed numerical method has capability to solve fuzzy
polynomials as well as algebric ones. For this purpose, by using parametric form of fuzzy
coefficients of fuzzy polynomial and Newton-Rphson method we can find its fuzzy roots.
Finally, we illustrate our approach by numerical examples.
Keywords: Fuzzy Numbers, Fuzzy Polynomials, Newton-Raphson Method.
1 Introduction
Polynomials play a major role in various areas such as pure and applied
mathematics, engineering and social sciences. In this paper we propose to find fuzzy
roots of a fuzzy polynomial like A1 x  A2 x 2    An x n  A0 where x i , A j  E 1 for
(if exists). The set of all the fuzzy numbers is denoted by E 1 . The
applications of fuzzy polynomials are considered by [1].
The concept of fuzzy numbers and arithmetic operation with this numbers were first
introduced by Zadeh, [2]. Many researchers have studied on solution methods of fuzzy
polynomials. Buckley and Eslami, [3] considered neural net solutions to fuzzy
problems. Otadi [4, 5] proposed architecture of fuzzy neural networks with crisp
weights for fuzzy input vector and fuzzy target. Abasbandy and Asady, [6] considered
Newton’s method for solving fuzzy nonlinear equations. Linear and nonlinear fuzzy
equations are solved by [5, 6, 7, 8, 9, 10]. In this paper we want to solve fuzzy
polynomials with fuzzy coefficients and fuzzy variable, numerically. The fuzzy
quantities are presented in parametric form. We first convert the polynomial fuzzy
coefficients into parametric form then apply Newton method on each limit. Finally, in
order to finding root, which is also fuzzy number/point, we numerically calculate level
sets (i.e.,  -cuts) of fuzzy coefficients on each limits. (Note that fuzzy polynomials may
have no root).
i  1,2,, n j  0,1,, n
In this paper the algorithm is illustrated by solving several numerical examples in last
*Corresponding
section. author
E-mail address: [email protected]
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A. Allahviranloo, et al, Numerical Solution of Fuzzy Polynomials by Newton-Raphson Method
2 Preliminaries
A popular fuzzy number is the triangular fuzzy number U  (m,  ,  ) for
  0,   0 with membership function
x  m
   1, m    x  m,

m  x
 u ( x)  
 1, m  x  m   ,
 
0,
otherwise.


Its parametric form is
u(h)  m   (h  1),
u (h)  m   (h  1) .
An arbitrary fuzzy number is represented by an ordered pair of function (u (h) , u (h))
0  h  1, which satisfies the following requirements.
u(h) is bounded left continuous no decreasing function over [0,1].
u (h) is bounded left continuous no increasing function over [0,1].
u (h)  u (h), 0  h  1.
Let E be the set of all upper semi continuous normal convex fuzzy numbers with
bounded h -level intervals. Meaning if u  E then the h -level set
U h  {u U ( x)  h , x  R} For 0  h  1 , and U 0  {u U ( x) , x  R}   h(0 , 1][U ]h . Since
level sets of fuzzy numbers become closed intervals, we denote
U h as
[U ]  [u (h) , u (h)] ,
l
u
Where u l (h) and u u (h) are the lower limit and the upper limit of the h -level set U h ,
respectively.
2.1 Interval arithmetic
Let A and B be fuzzy numbers with Ah  [a l (h), au (h)] and Bh  [bl (h), bu (h)] .
Then interval operations are as follow
 Ah  [au
(h),al (h)]
Ah  [al (h), au (h)]
if   0
Ah  [au (h), al (h)]
if
  0.
Ah (+) Bh = al (h)  bl (h) , au (h)  bu (h)
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Journal of Applied Mathematics, Islamic Azad University of Lahijan, Vol.7, No.4(27), Winter 2011, pp 17-24
Ah (-) Bh = al (h)  bu (h) , al (h)  bu (h)
In general case, we obtain a very complicated expression for h -level sets of the product
and division.
Ah (.) Bh = [mh , M h ]
m h  min{ a1 (h)b1 (h), a1 (h)b3 (h), a3 (h)b1 (h), a3 (h)b3 (h)}
M h  max{ a1 (h)b1 (h), a1 (h)b3 (h), a3 (h)b1 (h), a3 (h)b3 (h)}
Ah (/) Bh =[ mh , M h ]
mh  min{ a1 (h) / b1 (h), a1 (h) / b3 (h), a3 (h) / b1 (h), a3 (h) / b3 (h)}
M h  max{ a1 (h) / b1 (h), a1 (h) / b3 (h), a3 (h) / b1 (h), a3 (h) / b3 (h)}
Definition 2.1 (see[11] ). Consider x, y  E. If there exists z  E such that x  z  y ,
then z is called the H-difference of x and y and is denoted by x  H y .
Definition 2.2 We say that f (x) is fuzzy-valued function if f :   E .
Definition 2.3
The supremum metric d on E is defined by
d (U ,V )  sup{ d h ([U ]h ,[V ]h ) : h  I }
And ( E, d ) is a complete metric space. I is a real interval.
Definition 2.4 (see [12]). Let f : (a, b)  E and x0  (a, b) . We say that f if strongly
generalized differentiable x 0 at (Bede-Gal differentiability), if there exists an
element f ( x0 )  E , such that
i) for all h  0 sufficiently small, f ( x0  h)  H f ( x0 ) , f ( x0 )  H f ( x0  h) and the
limits (in the metric d  ):
f ( x0  h)  H f ( x0 )
f ( x0 )  H f ( x0  h)
lim h0
 lim h0
 f ( x0 )
h
h
Or
ii) for all h  0 sufficiently small, f ( x0  h)  H f ( x0 ) , f ( x0 )  H f ( x0  h) and the
following limits hold (in the metric d  ):
f ( x0 )  H f ( x0  h)
f ( x0  h)  H f ( x0 )
lim h0
 lim h0
 f ( x0 )
h
h
or
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A. Allahviranloo, et al, Numerical Solution of Fuzzy Polynomials by Newton-Raphson Method
iii) for all h  0 sufficiently small, f ( x0  h)  H f ( x0 ) , f ( x0  h)  H f ( x0 ) and the
following limits hold (in the metric d  ):
f ( x0  h)  H f ( x0 )
f ( x0  h)  H f ( x0 )
lim h0
 lim h0
 f ( x0 )
h
h
or
iv) for all h  0 sufficiently small, f ( x0 )  H f ( x0  h) , f ( x0 )  H f ( x0  h) and the
following limits hold (in the metric d  ):
f ( x0 )  H f ( x0  h)
f ( x0 )  H f ( x0  h)
lim h0
 lim h0
 f ( x0 )
h
h
Theorem 2.1 (see [11]) let f :R  E be a function and set f (t )  ( f ( x, h) , f ( x, h)), for
each h  [0,1]. Then
1) If is differentiable in the first form (i), then
f ( x, h) and f ( x, h) are differentiable
functions and f (h)  ( f ( x, h) , f ( x, h)).
2) If is differentiable in the second form (ii), then f ( x, h) and f ( x, h) are
differentiable functions and f (h)  ( f ( x, h) , f ( x, h)).
3 Fuzzy polynomials
As mentioned before we are interested in finding solution of
A1 x  A2 x 2    An x n  A0
where Ai , x  E for i=1,2,…,n
j
1
(1)
j  0,1, n.
n
let f ( x)   Aj x j  A0 . f (x) is called fuzzy polynomial of degree at most n.
j 1
Therefore, Eq. (1) can be written such as f ( x)  A0 . We use of parametric form of the
f (x) , So, Eq. (1) can be in the form of following formulas
 A ( h) x
f l ( x, h ) 
l
j
j

x j 0
f u ( x, h ) 
 A ( h) x
u
j
j
 A ( h) x
u
j
j
 A0l (h)
(2)
x j 0

x j 0
 A ( h) x
l
j
j
 A0u (h)
x j 0
Assume f l ( x, h) and f u ( x, h) has at least two continuous derivatives for all x in some
intervals about the roots  l (h) and  u (h) respectively.
By using (2),  f ( x)h has such form as follows
f l ( x, h) 
 A ( h) x
l
j
x 0
j
j

A
u
j
(h) x j  A0l (h)
x 0
j
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Journal of Applied Mathematics, Islamic Azad University of Lahijan, Vol.7, No.4(27), Winter 2011, pp 17-24

f u ( x, h) 

Auj (h) x j 
x j 0
Alj (h) x j  A0u (h)
(3)
x j 0
From (3), we obtain
 j A ( h) x
f  l ( x, h) 
l
j
j 1

x 0

f  u ( x, h ) 
jA
u
j
( h) x j
x 0
j
j
jAuj (h) x j 1 
x j 0
 j A ( h) x
l
j
j 1
(4)
x j 0
Now, by applying Newton’s formula we have
x nl 1 ( h)  x nl (h) 
f l ( x n , h)
f  l ( x n , h)
= x nl (h) -[  A lj (h) x j 
x 0
j
u
u
xn
1 ( h ) = x n ( h) 
A
(5)
u
j
(h) x j  A0l (h) ] /
x 0
f l ( x n , h)
f  u ( x n , h)
= x nu (h) - [  Auj (h) x j 
x j 0

[  j A lj (h) x j 1 
x 0
j
j
jA
u
j
( h) x j ]
x 0
j
Alj (h) x j  A0u (h) ]
x j 0
By applying Newton's method for parametric Eqs (3) with initial points
u
l
x0l (h)  m   (h  1) and x0 (h)  m   (1  h) , we can obtain roots of f ( x, h) and
f u ( x, h) respectively.
l
u
l
u
Two sequences { x n
1 ( h) } and { xn1 ( h ) } must convergent to  (h) and  (h)
respectively.
Where  (h)  ( l (h), u (h)) is the root of the polynomial?
3.1 Convergence of method
We are computing a sequence of iteration x n , and we would like to estimate their
accuracy to know when to stop the iteration.
Assume r be the root of polynomial and x n be the computed solution of it. To
estimate r  xn , note that, since f (r )  0 , we have
f ( xn )  f ( n )( xn  r )
For some  n between x n and  . Therefore
  xn 
 f ( xn )  f ( xn )
, which f (x) is continuous and f (r )  0

f ( n )
f ( xn )
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(6)
A. Allahviranloo, et al, Numerical Solution of Fuzzy Polynomials by Newton-Raphson Method
By considering (6) and Taylor’s theorem we have
f ( xn )
e f ( xn )  f ( xn )
en1  xn1  r  xn 
r  n
f ( xn )
f ( xn )
Since 0  f (r )  f ( xn  en ) , then
en f ( x n )  f ( x n ) 
1
2
f ( n )en .
2
(7)
From (6), (7), we obtain following result
en1 
1 f ( n ) 2 1 f (r ) 2
en 
en  cen2 .
2 f ( xn )
2 f (r )
(8)
This formula says that the error in x n 1 is proportional to square of the error in x n and
method has at least second degree convergence. And if f (r )  0 , then it will be second
degree convergence.
That means, when the initial error is sufficiently small, the error in the succeeding
iterates will decrease very rapidly.
Proof of Newton’s method convergence is [13].
Since both upper limit and lower limit of fuzzy polynomial satisfies in Newton’s
method theorem we can say by starting from arbitrary initial point, sequence of iteration
x n will be convergent to the root of polynomial (if exists).
4 Numerical examples
Example 4.1
Consider the following fuzzy polynomial:
(2,1,1) x  (1,1,1) x 2  (4,2,1) x 3  (0,1,1) x 4  (7,5,4)
The exact solution is x  1. The initial point is x0  0.5 and n=5. we have fuzzy point 1.
with e  10 6
1
1
Fuzzy point
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Journal of Applied Mathematics, Islamic Azad University of Lahijan, Vol.7, No.4(27), Winter 2011, pp 17-24
Example 4.2
Consider the following fuzzy polynomial: (0,1,1) x  (0,2,2) x 2  (1,1,1) x 3  (1,4,4)
With the exact solution x=-1.
1
1
Fuzzy point
The initial point is x0  1.5 and n=5. We have fuzzy point 1 with e  10 6 .
4 Conclusions
In this paper, we considered to fined root of fuzzy polynomial and saw we can get
the solution in initial steps with almost high accuracy
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