Biomedical Engineering Reference
In-Depth Information
G
c,k
G
c,k
~
ξ
m
ξ
m
c
c
~
ξ
n
ξ
n
k
k
~
ξ
n
ξ
n
V
k
V
k
V
k
V
k
ξ
i
ξ
j
p
1
p
2
p
1
p
2
Figure a.1
Visualization.of.finding.
ξ
.and.
ξ
.(logical.XOR.is.performed.in.striped.areas).
m
n
q
(
c k
)
q
(
,
,
c k
,
)
Therefore,.for.
= ξ
,
ξ
, ,
,.there.exists.a.unique.
= ξ
ξ
.such.that
m
n
m
n
( )
i
f
( )
q
= δ
(
f
(
ξ
,
V f
),
(
ξ
,
G
))
× δ
(
f
(
ξ
,
V f
′
),
(
ξ
,
V
′
))
T
i
k
T
m
c k
,
T
i
k
T
n
k
(
f
(
,
V f
),
(
,
G
))
(
f
(
,
V f
),
(
,
V
))
= δ
ξ
ξ
× δ
ξ
′
ξ
′
T
j
k
T
m
c k
,
T
j
k
T
n
k
j
=
f q
( )
.
( )
j
f
( )
i
q
)
f
( )
i
(
q
)
f
( )
q
=
=
,.then.
1
q
= ..This.completes.the.proof.
1
2
2
ProofofLemma4.2
Obviously,.
(
)
(
)
P
ξ
,
t
Σ
P
ξ
,
t
=
Σ
1
=
1
.satisfies.the.constraint.of.
..Assuming.
i
i
S
M
ξ ∈
S
ξ ∈
i
ξ
i
ξ
(
)
∀ξ ∈
S
that.
P
ξ
,
t
=
1
,.
,.one.has
i
i
M
L L
−
L L
−
M
M
g
g
∑
∑
∑
1
(
)
(
)
×
(
)
(
)
(
)
(
)
(
)
P
ξ
,
t
+
1
=
K
δ
f
ξ
,
V f
,
ξ
,
G
× δ
f
ξ
,
V f
′
,
ξ
,
V
′
i
T
i
k
T
m
c k
,
T
i
k
T
n
k
2
M
m
=
1
n
=
1
c
=
0
k
=
0
L L
−
L L
−
M
M
g
g
K
M
∑
∑
∑
(
)
(
)
(
)
(
)
(
)
(
)
=
δ
f
ξ
,
V f
,
ξ
,
G
× δ
f
ξ
,
V f
′
,
ξ
,
V
′
T
i
k
T
m
c k
,
T
i
k
T
n
k
2
m
=
1
n
=
1
c
=
0
k
=
0
K
M
∑
( )
i
=
f
( )
q
.
2
.
(A.5)
q Q
∈
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