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m
h
=
1
c
hk
2
F
m
=
m
k
=
1
λ
m
k
=
1
λ
m
+
k
2
2
1
(
1
+(
−
1
)
)
tg
2
β
k
2
n
n
−
1
k
n
−
1
k
=
,
m
+
4
m
k
=
1
(
1
−
2cosβ
k
)
m
+
k
1
1
+(
−
1
)
tg
2
β
k
n
−
=
2
,
(
m
+
1
)
which proves the theorem.
r
=
1
sin
2
nr
π
p
Lemma 2.
The sum
is equal to 0 if n is a multiple of
(
p
+
1
)
, and to
∑
p
+
1
p
+
1
otherwise.
2
Proof.
It is clear that, if
n
is a multiple of
(
p
+
1
)
the equality
p
r
=
1
sin
2
nr
π
1
=
0
p
+
is satisfied. Let us suppose that
n
is not a multiple of
(
p
+
1
)
,then
p
r
=
0
sin
2
nr
π
p
r
=
0
p
r
=
0
cos
2
nr
π
+
1
2
(
cos
2
nr
π
p
1
2
−
1
2
1
=
1
−
1
)=
1
.
(7.20)
p
+
p
+
p
+
0
exp
i
2
nr
π
p
r
The last sum of the above expression is the real part of
1
, and with the
∑
=
p
+
notation
i
2
n
π
z
=
exp
1
=
1
p
+
we have
p
r
=
0
exp
i
2
nr
π
p
r
=
0
z
r
z
p
+
1
1
−
1
=
=
z
=
0
p
+
1
−
and (
7.20
) gives the desired conclusion, which finishes the proof.
n
Theorem 2.
The cardinality of
F
m
is equal to
,
F
n
,
m
=
m
k
=
1
(
1
−
2cosβ
k
)
2
n
where
β
k
is given by (
7.13
).
n
F
Proof.
We can express the cardinality of
m
in the way
,
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