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5.3.2.4 The λ
1
/
2
ζ
≤
γ
≤
γ
max
Case
g
(γ
,
ζ)
is strictly increasing from
−∞
for
γ
=
λ
1
/
2
ζ
+
to
+∞
for
γ
=+∞
; this implies a zero for
g
at
γ
=
γ
∗∗∗
.This
zero necessarily represents the maximum level of the cost function (
γ
∗∗∗
=
γ
max
)
because all
k
i
s have the same sign and only for
g
(γ
,
ζ)
≤
0 are the hyperconics
real hyperellipsoids [(
n
−
1)-dimensional ellipsoids] with axes parallel to the
A
∀
γ
∈
(γ
∗∗∗
,
the
k
i
s are all negative and the family is composed
eigenvectors [
∞
)
ζ
+
, the hyperellipsoids tend to stretch in
of imaginary ellipsoids]. As
γ
→
λ
/
2
1
v
1
direction. As
the
max
, the hyperellipsoids tend to collapse into a point,
but their shape spreads out. The density of these hypersurfaces is proportional to
the gap between
γ
→
γ
γ
∗∗∗
=
σ
2
λ
/
2
ζ
and
γ
max
.For
ζ
=
0
.
5 (TLS),
1
.
1
5.3.3 Considerations and the Two-Dimensional Case
Theorem 99 (GTLS EXIN Critical Points)
The GeTLS EXIN cost function of
an n-dimensional unknown has n
+
1
critical points: a minimum, n
−
1
saddles,
and a maximum.
Remark 100 (Link with the MCA EXIN Stability Analysis)
All the previous
considerations about the TLS EXIN stability analysis and cost landscape are very
similar to the corresponding considerations for MCA EXIN stability, in particular
Section 2.5.1.2. This was expected because, in the MCA space, the only difference
lies in the fact that the TLS EXIN problem takes into consideration only inter-
sections with a hyperplane perpendicular to the last Cartesian coordinate
(
TLS
hyperplane
)
instead of hyperplanes perpendicular to the
[
A
;
b
]
eigenvectors.
In the following, much insight will be dedicated to the two-dimensional case
(i.e.,
n
2) because it is easy to visualize. Figure 4.1 shows this general case
in the TLS hyperplane. The following overdetermined system of linear equations
in two variables, taken from [22, Ex. 3], will be the main benchmark for the
GeTLS EXIN neuron from now on.
=
11
12
13
14
15
1
1
2
2
2
x
1
x
2
=
(5.57)
The exact results, obtained by using MATLAB, are
x
=
[0
.
7, 0
.
3]
T
(5.58a)
=
[0
.
8544, 0
.
2587]
T
x
(5.58b)
x
=
[1
.
1200, 0
.
1867]
T
(5.58c)
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