How to prove <munderover> &#x2211;<!-- ∑ --> <mrow class="MJX-TeXAtom-ORD"> k

Cierra Castillo 2022-07-09 Answered
How to prove k = 1 n cos ( 2 π k n ) = 0 for any n>1?
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Answers (2)

cefflid6y
Answered 2022-07-10 Author has 13 answers
We have
k = 1 n cos ( 2 π k n ) = Re ( k = 1 n e 2 π i k / n )
and so
k = 1 n e 2 π i k / n = e 2 π i / n ( 1 e 2 π i ) 1 e 2 π i / n
and notice that
e 2 π i = cos ( 2 π ) + i sin ( 2 π ) = 1
so the claim follows.
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fythynwyrk0
Answered 2022-07-11 Author has 7 answers
One approach is to write
cos ( 2 π k n ) = 1 2 sin ( 2 π n ) ( sin ( 2 π ( k + 1 ) n ) sin ( 2 π ( k 1 ) n ) )
Now, we have converted the sum into a telescoping sum, which we can evaluate directly as
k = 1 n cos ( 2 π k n ) = 1 2 sin ( 2 π n ) ( sin ( 2 π ( n + 1 ) n ) + sin ( 2 π ( n ) n ) ) 1 2 sin ( 2 π n ) ( sin ( 2 π ( 2 1 ) n ) + sin ( 2 π ( 1 1 ) n ) ) = 1 2 sin ( 2 π n ) ( sin ( 2 π ( n + 1 ) n ) sin ( 2 π ( 2 1 ) n ) ) = 1 2 sin ( 2 π n ) ( sin ( 2 π n ) sin ( 2 π n ) ) = 0
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