Proving ∑ r = 0 n − 1 ( − 1 ) r cos n...

Jamison Rios

Jamison Rios

Answered

2022-07-07

Proving r = 0 n 1 ( 1 ) r cos n ( r π n ) = n 2 n 1

Answer & Explanation

Jayvion Mclaughlin

Jayvion Mclaughlin

Expert

2022-07-08Added 14 answers

The statement above seems to be erroneous. It can be shown that
S = r = 0 n 1 ( 1 ) r cos n ( r π n ) = n 2 n 1
On the LHS, rewrite cos x = e i x + e i x 2 and expand using the binomial theorem to get
S = 1 2 n k = 0 n ( n k ) r = 0 n 1 ( 1 ) r e i π r n ( 2 k n )
However,
r = 0 n 1 ( 1 ) r e i π r n ( 2 k n ) = r = 0 n 1 ( e i 2 k π / n ) r = ( e i 2 k π / n ) n 1 e i 2 k π / n 1 = n m = δ k , m n
and thus
S = n 2 n ( ( n 0 ) + ( n n ) ) = n 2 n 1

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