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Theorem 1.
The cardinality of the set
F
n
,
m
is given by the expression
F
m
=
m
k
=
1
(
1
−
2cosβ
k
)
m
+
k
1
1
+(
−
1
)
tg
2
β
k
0
n
n
−
2
,
(7.16)
,
(
m
+
1
)
where
β
k
is given by (
7.13
).
n
,
m
as
Proof.
We can express the cardinality of
F
m
h
=
1
c
hk
2
F
m
=
h
,
r
a
(
n
)
m
k
=
1
λ
m
k
=
1
λ
0
n
k
h
,
r
c
hk
c
rk
=
k
hr
=
.
(7.17)
,
First we will compute the sum
k
2
m
k
2
m
m
h
=
1
c
hk
=(
−
1
)
m
h
=
1
(
−
1
)
h
sin
(
h
β
k
)=(
−
1
)
1
S
.
(7.18)
+
1
+
The value of
S
is obtained from the imaginary part of the expression
m
h
=
0
(
−
1
)
h
exp
(
ih
β
k
)
m
e
i
(
m
+
1
)
β
k
+(
−
)
1
1
=
1
+
e
i
β
k
m
e
ik
π
m
cos
1
+(
−
1
)
1
+(
−
1
)
(
k
π
)
=
=
e
i
β
k
e
i
β
k
1
+
1
+
e
−
i
β
2
m
+
k
m
+
k
1
+(
−
1
)
(
1
+(
−
1
)
)
=
,
2cos
β
k
2
e
i
β
2
e
−
i
β
k
/
2
e
i
β
k
/
2
(
+
)
and the imaginary part of this value is equal to
(
−
β
2
)(
m
+
k
m
+
k
sin
1
+(
−
1
)
)
tg
β
k
2
1
+(
−
1
)
=
−
=
S
.
(7.19)
2cos
β
2
2
Therefore,
k
2
m
1
2
m
m
h
=
1
c
hk
=(
−
1
)
m
+
k
1
tg
β
k
1
+(
−
1
)
k
+
1
S
=(
−
1
)
+
+
2
2
and substituting the last value into (
7.17
) we conclude that
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