Biomedical Engineering Reference
In-Depth Information
X
X
[
(i)
F
n
(U
(i)
)]
2
[
(i)
F
n
(U
(i)
)]
2
RSS(
0
)
=
i=1
i=1
X
[(
(i)
0
) ( F
n
(U
(i)
)
0
)]
2
=
i=1
X
[(
(i)
0
) ( F
n
(U
(i)
)
0
)]
2
i=1
X
( F
n
(U
(i)
)
0
)
2
X
i2I
n
( F
n
(U
(i)
)
0
)
2
=
i2I
n
+2
X
i2I
n
(
(i)
0
) ( F
n
(U
(i)
)
0
)
2
X
i2I
n
(
(i)
0
) ( F
n
(U
(i)
)
0
)
X
( F
n
(U
(i)
)
0
)
2
X
i2I
n
( F
n
(U
(i)
)
0
)
2
;
=
i2I
n
where this last step uses the facts that
X
(
(i)
0
) ( F
n
(U
(i)
)
0
) =
X
i2I
n
( F
n
(U
(i)
)
0
)
2
i2I
n
and
X
(
(i)
0
) ( F
n
(U
(i)
)
0
) =
X
i2I
n
( F
n
(U
(i)
)
0
)
2
:
i2I
n
For the case of F
n
, this equality is an outcome of the fact that I
n
can be
decomposed into a number of consecutive blocks of indices, say B
1
;B
2
;:::;B
r
on each of which F
n
is constant (denote the constant value of B
j
by w
j
) and,
furthermore, on each block B
j
such that w
j
6=
0
, we have for each k 2 B
j
,
P
i2B
j
(i)
n
j
F
n
(U
(k)
) =
w
j
=
;
where n
j
is the size of B
j
. A similar phenomenon holds for F
n
. The equalities
in the two displays preceding the above one then follow by writing the sum
over I
n
as a double sum where the outer sum is over the blocks and the inner
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