Biomedical Engineering Reference
In-Depth Information
Hence.we.can.get
L L
−
L L
−
M
M
M
M
g
g
∑
∑
∑
∑
( )
i
a
=
K
∆ ξ
(
,
V
;
ξ
,
G
)
× ∆ ξ
(
,
V
′ ξ
;
,
V
′
)
mn
k
m
c k
,
i
k
n
k
i
m
=
1
n
=
1
m
=
1
n
=
1
c
=
0
k
=
0
L L
−
L L
−
g
g
M
M
∑
∑
∑
∑
×
=
K
∆ ξ
(
,
V
;
ξ
,
G
)
∆ ξ
(
,
V
′ ξ
;
,
V
′
)
k
m
c k
,
i
k
n
k
i
c
=
0
k
=
0
m
=
1
n
=
1
L L
−
L L
−
g
g
∑
∑
(
)
(
)
N u
−
−
v
N w
−
=
K
2
×
2
×
2
×
1
c k
,
c k
,
k
c
=
0
k
=
0
L L
−
L L
−
g
g
∑
∑
2
N u
v
w
−
−
−
=
K
2
c k
,
c k
,
k
c
=
0
k
=
0
L L
−
L L
−
g
g
∑
∑
N
=
K
2
c
=
0
k
=
0
=
M
.
This.is.actually.another.way.to.prove.Lemma.4.2.
Apparently,
L L
−
L L
−
g
g
∑
∑
( )
i
a
=
K
∆ ξ
(
,
V
;
ξ
,
G
)
× ∆ ξ
(
,
V
′ ξ
;
,
V
′
)
i
k
m
c k
,
i
k
i
k
mi
c
=
0
k
=
0
L L
−
L L
−
g
g
∑
∑
=
K
∆ ξ
(
,
V
;
ξ
,
G
) 1
×
i
k
m
c k
,
.
(A.14)
c
=
0
k
=
0
L L
−
L L
−
g
g
∑
∑
≥
K
∆ ξ
(
,
V
;
ξ
,
G
)
× ∆ ξ
(
,
V
′ ξ
;
,
V
′
)
i
k
m
c k
,
i
k
n
k
c
=
0
k
=
0
( )
i
=
a
.
mn
A
i
,. the. element. in. the.
i
th.
Hence,. for. each. row. of. the. distribution. matrix.
( )
column.has.the.largest.value.
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