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2.7.1 OJAn, LUO, and MCA EXIN
By using eq. (2.96), the following formulas hold:
1
u
p
T
2
2
w
(
t
+
1
)
x
(
t
)
=
y
(
t
)
−
α
s
(
p
)
x
(
t
)
−
(2.144)
2
2
2
s
2
w (
t
+
1
)
=
p
+
α
(
p
)
uq
(2.145)
where
2
y
(
t
) w (
t
)
p
q
=
x
(
t
)
−
(2.146)
2
Notice that
12
u
p
2
2
2
2
sin
2
q
=
x
(
t
)
−
=
x
(
t
)
ϑ
(2.147)
x
w
where
ϑ
x
w
is the angle between the directions of
x
(
t
)
and
w (
t
)
. So eq. (2.144)
can be rewritten as
)
1
q
T
w
(
t
+
1
)
x
(
t
)
=
y
(
t
−
α
s
(
p
)
(2.148)
Hence,
p
1
−
α
s
(
p
)
q
2
p
+
α
ρ (α)
=
(2.149)
2
s
2
(
p
)
uq
It follows that
2
2
s
2
ρ (α) >
⇔
p
(
−
α
s
(
p
)
q
)
>
p
+
α
(
p
)
uq
1
1
(2.150)
which yields
α (α
−
α
b
)
>
0
(2.151)
being
2
ps
−
1
(
p
)
α
b
=
(2.152)
pq
−
u
In particular,
α
b
EXIN
=
p
α
b
OJAn
=
p
2
α
b
LUO
(2.153)
12
From this point on, this analysis diverges from the analysis in [195, proof of Prop. 2], because
of a mistake in the unnumbered equation about
2
proceding eq. (A13). Then all the consequent
theory in [195, proof of Prop. 2] is incorrect.
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