Jamison Rios

2022-07-07

Proving $\sum _{r=0}^{n-1}\left(-1{\right)}^{r}{\mathrm{cos}}^{n}\left(\frac{r\pi }{n}\right)=\frac{n}{{2}^{n-1}}$

Jayvion Mclaughlin

Expert

The statement above seems to be erroneous. It can be shown that
$S=\sum _{r=0}^{n-1}\left(-1{\right)}^{r}{\mathrm{cos}}^{n}\left(\frac{r\pi }{n}\right)=\frac{n}{{2}^{n-1}}$
On the LHS, rewrite $\mathrm{cos}x=\frac{{e}^{ix}+{e}^{-ix}}{2}$ and expand using the binomial theorem to get
$S=\frac{1}{{2}^{n}}\sum _{k=0}^{n}\left(\genfrac{}{}{0}{}{n}{k}\right)\sum _{r=0}^{n-1}\left(-1{\right)}^{r}{e}^{\frac{i\pi r}{n}\left(2k-n\right)}$
However,
$\sum _{r=0}^{n-1}\left(-1{\right)}^{r}{e}^{\frac{i\pi r}{n}\left(2k-n\right)}=\sum _{r=0}^{n-1}\left({e}^{i2k\pi /n}{\right)}^{r}=\frac{\left({e}^{i2k\pi /n}{\right)}^{n}-1}{{e}^{i2k\pi /n}-1}=n\sum _{m=-\mathrm{\infty }}^{\mathrm{\infty }}{\delta }_{k,mn}$
and thus
$S=\frac{n}{{2}^{n}}\left(\left(\genfrac{}{}{0}{}{n}{0}\right)+\left(\genfrac{}{}{0}{}{n}{n}\right)\right)=\frac{n}{{2}^{n-1}}$

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