Proving ∑ r = 0 n − 1 ( − 1 ) r cos n ⁡ ( r π n ) = n 2 n − 1

Question

Proving       ∑          r      =      0              n      −      1        (  −  1      )    r        cos    n    ⁡            (                        r          π                n            )        =      n          2              n        −        1

Jayvion Mclaughlin · Accepted Answer

The statement above seems to be erroneous. It can be shown that

S = \sum_{r = 0}^{n - 1} (- 1)^{r} \cos^{n} (\frac{r π}{n}) = \frac{n}{2^{n - 1}}

On the LHS, rewrite

\cos x = \frac{e^{i x} + e^{- i x}}{2}

and expand using the binomial theorem to get

S = \frac{1}{2^{n}} \sum_{k = 0}^{n} (\binom{n}{k}) \sum_{r = 0}^{n - 1} (- 1)^{r} e^{\frac{i π r}{n} (2 k - n)}

However,

\sum_{r = 0}^{n - 1} (- 1)^{r} e^{\frac{i π r}{n} (2 k - n)} = \sum_{r = 0}^{n - 1} (e^{i 2 k π / n})^{r} = \frac{(e^{i 2 k π / n})^{n} - 1}{e^{i 2 k π / n} - 1} = n \sum_{m = - \infty}^{\infty} δ_{k, m n}

and thus

S = \frac{n}{2^{n}} ((\binom{n}{0}) + (\binom{n}{n})) = \frac{n}{2^{n - 1}}

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