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T n (
) ψ 2
ψ 1 ψ 2
x
μ(
x
) :=
(5.9)
x n ,wehave f
Hence, as f
(
x
) =
f n (
x
) +
T n (
x
)
(
x
) =
f n (
x
) + (μ(
x
1 + (
1
x n so the polynomial p 1 in ( 5.5 ) is given by p 1 (
) + ψ 1 x n , and
μ(
x
))ψ 2 )
x
) =
f n (
x
) + ψ 2 x n .
p 2 (
x
) =
f n (
x
Note As in TS modeling, the representation ( 5.5 )is exact , i.e., there is no approx-
imation involved and there is no uncertainty in the membership functions, defined
in ( 5.9 ). As a conclusion, using the Taylor-based modeling any smooth nonlinear
system can be exactly expressed as a fuzzy polynomial one in a compact domain
.
Example 5.1 Let us model sin
(
x
)
, x
∈[−
1
,
1
]
, considering its Taylor series around
x
=
0:
x 3
3
x 5
5
x 7
7
sin
(
x
) =
x
+
+ ...
(5.10)
!
!
!
Let us define f 1 =
0, and T 1 (
x
) = (
sin
(
x
)
0
)/
x from the previously discussed
notation. The maximum and minimum of sin
(
x
)/
x in [
1
,
1] are 1 and 0
.
8415, so
we may express:
sin x
= μ 1 (
0
+
1
·
x
) + μ 2 (
0
+
0
.
8415 x
)
(5.11)
which is coincident with the standard sector-nonlinearity TS model. Now, using the
cubic term in the Taylor series, the fuzzy model needs considering f 3
=
0
+
x ,
x 3 ,so T 3 (
T 3 (
) = (
(
)
)/
)
.
.
x
sin
x
x
x
has maximum
0
1585 and minimum
0
1667
[−
,
]
in the interval
1
1
. Hence,
1585 x 3
1667 x 3
sin x
= μ 1 (
0
+
x
0
.
) + μ 2 (
0
+
x
0
.
)
(5.12)
x 3
x 3
x 5 ,
If we proceed to 5th order with f 5 =
0
+
x
/
6, T 5 (
x
) = (
sin
(
x
)
x
+
/
6
)/
then in
[−
1
,
1
]
, T 5 (
x
)
has maximum 0.0083336, minimum 0.0081376, so we get:
x 3
0083336 x 5
0081376 x 5
sin x
=
x
/
6
+ 1 ·
0
.
+ μ 2 ·
0
.
)
(5.13)
where, for instance,
μ 1 would have the usual interpolation expression:
μ 1 (
x
) = (
T 5 (
x
)
0
.
0081376
)/(
0
.
0083336
0
.
0081376
)
(5.14)
If the bounding polynomials are plotted, it can be seen that they are closer as their
degree increases, see Sala and Ariño ( 2009 ) for details. Both 3rd- and 5th-order
bounds are very accurate (maximum error 0.0082 and 0.0002, respectively) in the
chosen interval.
 
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