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watch5826c 2022-06-23 Answered
I am trying to prove that
lim N 1 2 N k = 1 N [ cos ( ( x k L ) q ) + cos ( ( x + k L ) q ) ] = 0
for every x , L R + with x L and for every q 2 n π L , n Z .
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Answers (1)

America Barrera
Answered 2022-06-24 Author has 23 answers
cos ( α ± β ) = cos α cos β sin α sin β .
Plugging this into your expression yields
cos ( x q ) 1 N k = 1 N cos ( k L q ) .
Also,
sin ( α ± β ) = sin α cos β ± sin β cos α .
Therefore,
2 sin ( α 2 ) cos ( k α ) = sin ( ( k + 1 2 ) α ) sin ( ( k 1 2 ) α ) .
This allows for a telescopic sum argument.
1 N k = 1 N cos ( k L q ) = sin ( ( N + 1 2 ) L q ) sin ( 1 2 L q ) 2 N sin ( L q 2 ) N 0
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