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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