Information Technology Reference
In-Depth Information
For instance, if
μ
S
(
x
)=
1
−
x
/
10 and
−
√
2), it is
K
N
1
(
μ
S
)=
{
1
−
x
1.
N
1
(
x
)=
(with the fix point 2
x
;1
−
x
/
10
≥
√
2
1
+
x
x
;
√
2
,and
√
2
2
−
}
=
{
−
10
≥
x
}
−
10
≈
4
.
14, with which the Kernel is
[
,
.
)
0
4
14
.
(
)=
−
x
(fix point 1/2), is
K
N
2
(
μ
)=
{
−
/
≥
/
}
=
{
≥
}
2.
N
2
x
1
x
;1
x
10
1
2
x
;5
x
,
S
and
s
=
5, with which the kernel is
[
0
,
5
)
.
(with fix point
√
12
−
2
4
√
12
1
−
x
−
2
3.
N
3
(
s
)=
), is
K
N
3
(
μ
S
)=
{
x
;1
−
x
/
10
≥
}
=
+
√
12
1
2
x
4
{
x
;5
/
2
(
6
−
)
≥
x
}
,and
s
≈
6
.
34, with which the kernel is
[
0
,
6
.
34
)
.
In the figure 15.2 it is shown how these points can vary with different functions
μ
S
(
=
−
)
N
1
id
.
(a)
(b)
(c)
(d)
Fig. 15.2
Remarks 2.
1. Instead of taking notS, which is not a linguistic term, it can be
taken an opposite aS (for instance aS
=
big) of S, the antonym which is a
linguistic term.
Since
μ
aS
≤
μ
notS
, it will follow
μ
S
(
x
)
≥
μ
notS
≥
μ
aS
, and
K
N
(
μ
S
)
⊂{
x
;
μ
S
(
x
)
≥
μ
aS
(
x
)
}
=
K
a
(
μ
S
)
.
Thus, K
a
(
μ
S
)
is a larger set than
K
N
(
μ
S
)
, therefore if s
a
=
supK
a
(
μ
S
)
,itiss
≤
s
a
.
For instance, with
μ
aS
(
x
)=
μ
S
(
α
(
x
))
and the symmetry
α
(
x
)=
10
−
x, it is
10
−
x
x
10
=
μ
big
(
μ
aS
(
x
)=
1
−
=
x
)
, and
10
x
10
≥
x
10
⇔
μ
S
(
)
≥
μ
aS
(
)
⇔
−
≥
,
x
x
1
5
x
that gives K
a
(
μ
S
)=
K
N
2
(
μ
S
)=[
0
,
5
)
,sofor
α
(
x
)=
10
−
x and N
2
,s
=
s
a
=
5
.
