Image Processing Reference
In-Depth Information
Taking the derivative of
J
with respect to
u
and setting the first derivative to
zero, we get
δ
δ
J
u
m
−
(
1
=
2
mu
1
−
Kx v
( , )
−=
λ
0
11
11 1
11
1
/(
m
−
1
)
⎛
⎜
⎞
⎟
λ
1
u
=
( 7.11)
11
21
mKxv
(
−
( , )
11
Likewise, for
1
/(
m
−
1
)
⎛
⎜
⎞
⎟
λ
1
u
=
12
21
mKxv
(
−
(
, )
12
1
/(
m
−
1
)
⎛
⎜
λ
⎞
1
u
=
⎟
1
c
21
mKxv
(
−
( , )
1
c
Summing up all
u
's we get
1
/(
m
−
1
)
1
/(
m
−
1
)
⎛
⎜
λ
⎞
⎟
⎛
⎜
λ
⎞
⎟
1
1
uu u
+++=
−
+
11
12
1
c
21
mKxv
(
( , )
21
mKxv
( (,
−
))
11
12
1
/(
m
−
1
)
⎛
⎜
⎞
⎟
λ
1
++
21
mKxv
(
−
( , )
1
c
1
/(
m
−
1
)
⎡
⎤
1
/(
m
−
1
)
c
λ
1
=
⎛
⎞
⎠
∑
1
⎢
⎢
⎥
⎥
( 7.1 2)
⎜
⎟
(
)
2
m
1
−
Kx v
(
−
)
1
j
⎣
⎦
j
=
1
∑
c
As per the objective criterion,
μ
ik
=
1
.
k
=
1
So,
1
/(
m
−
1
)
⎡
⎤
1
/(
m
−
1
)
c
λ
1
1
⎛
⎜
⎞
⎟
∑
⎢
⎢
⎥
⎥
1
=
(
)
2
m
1
−
K xv
(
−
)
1
j
⎣
⎦
j
=
1
1
/(
m
−
1
)
λ
1
1
⎛
⎜
⎞
⎟
( 7.1 3)
=
∑
c
2
m
(
)
1
/(
m
−
1
)
11
/
(
−
Kx v
(
−
))
1
j
j
=
1
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