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!
2
2
≤
0LUO
d
w (
t
)
dt
−
w (
t
)
−
4
2
2
2
≤
d
w (
t
)
)
−
2
2
=
(2.83)
−
w (
t
0OJAn
"
dt
2
2
≤
0
d
w (
t
)
dt
−
EXIN
and for all three neurons
dV
(
t
) /
dt
=
0for
d
w (
t
) /
dt
=
0(i.e.,
dV
(
t
) /
dt
is
negative semidefinite (i.e., the Lyapunov function is
weak
). For completeness
the Lyapunov function time derivative is given for OJA [83]:
2
2
+
(
1
dV
(w (
t
))
dt
d
w (
t
)
dt
=−
w (
t
)
−
2
2
2
w
(
t
)
R
w
(
t
)
2
2
T
2
−
w
(
t
)
−
R
w
(
t
)
Then, for the Lyapunov direct method (see, e.g., [132]) the ODE has the points
for
d
0 as asymptotic stable points in the limit of the ODE approxi-
mation, but the global asymptotic stability theorem does not hold because
V
(w)
is
radially bounded
. According to Ljung [118, Cor. 2], the ODEs have the point
{∞}
as stationary, too. For all three neurons, at the stable state
w (
t
st
)
,
d
w (
w (
t
) /
dt
=
t
st
)
T
=
0
⇒
w
(
t
st
) w (
t
st
)
R
w (
t
st
)
dt
T
=
w
(
t
st
)
R
w (
t
st
) w (
t
st
)
Then
T
R
w (
t
st
)
=
w
(
t
st
)
R
w (
t
st
)
w (
t
st
)
(2.84)
w
T
(
t
st
) w (
t
st
)
That is,
w (
t
st
)
is one eigenvector of
R
, and its corresponding eigenvalue
λ (
t
st
)
is
the Rayleigh quotient of
R
. It still has to be proven that
λ (
t
st
)
=
λ
n
. The weight
w (
t
)
can be expressed as a function of the orthogonal vectors:
n
w (
t
)
=
f
i
(
t
)
z
i
(2.85)
i
=
1
Replacing it into eq. (2.33), recalling the orthogonality of the eigenvectors, yields
df
i
(
t
)
dt
)
−
4
2
T
=
w (
t
[
w
(
t
)
R
w (
t
)
f
i
(
t
)
T
−
w
(
t
) w (
t
) λ
i
f
i
(
t
)
]
∀
i
=
1, 2,
...
,
n
(2.86)
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