Information Technology Reference
In-Depth Information
For a fixed
µ
0
we write
β
k
(µ
0
)
=
α
1
α
2
and, integrating (5.17), we have
µ
d
d
t
(µ)
l
2
(µ)
d
µ
+
k
.
b
1
(µ)
l
2
(µ)
=
l
1
(µ)
(5.18)
µ
µ
0
To solve for the constant
k
,wehave
µ
0
d
µ
t
(µ)
l
2
(µ)
d
µ
+
k
,
α
1
=
b
1
(µ
0
)
=
l
2
(µ
0
)
l
1
(µ)
µ
0
which simplifies to
α
1
=
l
2
(µ
0
)(
0
+
k
)
, implying
α
1
k
=
l
2
(µ
0
)
.
Similarly,
d
µ
t
(µ)
l
1
(µ)
d
µ
+
µ
α
2
l
1
(µ
0
)
.
b
2
(µ)
=
l
1
(µ)
l
2
(µ)
µ
0
For convenience we choose
(α
1
α
2
)
as the origin of
L
, which implies that
t
(µ
0
)
=
0.
Then the coordinates in
L
for the marker
µ
on the
k
th biplot trajectory are given by
µ
d
d
µ
t
(µ)
l
2
(µ)
d
l
2
(µ)
l
1
(µ)
µ
b
1
(µ)
b
2
(µ)
.
µ
0
d
d
µ
t
(µ)
l
1
(µ)
d
µ
β
k
=
=
µ
(5.19)
l
1
(µ)
l
2
(µ)
µ
0
To construct the biplot with our R function
Nonlinbipl
,aseriesof
µ
values are
selected and the trajectory is computed by substituting the symbolic differentiation into
the integrand which is then solved by numerical integration (see Section 5.6 for details
of the method employed). The symbolic differentiation of
t
(µ)/
l
i
(µ)
contains the first-
and second-order derivatives of
d
n
+
1
(µ)
. This follows since
t
∗
(µ)
a
1
(µ)
+
a
2
(µ)
a
i
(µ)
a
1
(µ)
+
a
2
(µ)
l
i
(µ)
=
and
t
(µ)
=
,
where
1
λ
1
y
1
1
λ
2
y
2
)
d
d
d
n
+
1
(µ)
}
(
t
∗
(µ)
=−
1
n
1
{
d
d
a
(µ)
={
and
d
n
+
1
(µ)
}
.
µ
µ
This means that for normal projection the distance function must not allow second-order
derivatives to vanish.
