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ω
If eigenvector co-ordinates do not belong to a segment [0. 1], we'll determine
>
0
.
m
,
i
j
~
∑
=
ω
=
ω
m
m
,
i
i
1
and we obtain
ω
~
m
,
i
ω
=
.
~
m
,
j
ω
m
~
ω
Y
We consider
as grade of membership of the object
to term
. As for
X
()
m
,
i
i
m
()
[]
each object
evaluations
are determined, let us consider that they
Y
y
i
∈
0
,
()
i
~
ω
X
belong to term
, accordingly.
In all further buildings the assumption is made that co-ordinates of the
eigenvector corresponding to a maximum eigenvalue belong to the segment [0, 1],
otherwise normalization is used.
In order to obtain membership function
with membership degrees
m
,
i
m
()
μ
x
of the term
the left
X
m
m
y
=
a
x
+
b
x
=
1
boundary of which looks like
, and right, in turn, is
, let us use
m
m
a method of least squares
j
(
)
∑
=
2
F
=
a
y
+
b
−
ω
→
min
.
()
m
m
i
m
m
,
i
i
1
From the system of normal equations
∂
F
⎧
⎧
j
j
j
∑
∑
∑
m
=
0
2
a
y
+
b
y
=
y
ω
;
⎪
⎪
⎨
⎪
⎪
⎨
()
()
()
m
i
m
i
i
m
,
i
∂
a
m
⇔
i
=
1
i
=
1
i
=
1
∂
F
j
j
⎪
⎪
⎪
⎪
∑
∑
m
=
0
a
y
+
jb
=
ω
.
()
m
i
m
m
,
i
∂
b
⎩
⎩
m
i
=
1
i
=
1
we obtain unknown coefficients:
j
⎛
j
⎞
⎛
j
⎞
∑
∑
∑
j
y
ω
−
⎜
⎜
⎝
y
⎟
⎟
⎠
⎜
⎜
⎝
ω
⎟
⎟
⎠
()
()
i
m
,
i
i
m
,
i
i
=
1
i
=
1
i
=
1
b
=
;
m
2
j
⎛
j
⎞
∑
∑
2
j
y
−
⎜
⎜
⎝
y
⎟
⎟
⎠
()
()
i
i
i
=
1
i
=
1
j
⎛
j
⎞
⎛
j
⎞
∑
∑
∑
j
y
ω
−
⎜
⎜
⎝
y
⎟
⎟
⎠
⎜
⎜
⎝
ω
⎟
⎟
⎠
()
()
i
m
,
i
i
m
,
i
1
j
1
j
∑
∑
i
=
1
i
=
1
i
=
1
a
=
ω
−
y
.
()
m
m
,
i
i
j
j
2
j
⎛
j
⎞
=
1
=
1
i
i
∑
∑
2
j
y
−
⎜
⎜
⎝
y
⎟
⎟
⎠
()
()
i
i
i
=
1
i
=
1
