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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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