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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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