Triangular formula (1 Viewer)

TuePhuong · Nov 5, 2016

p is a prime number. Proof that $\sum\limits_{i = 1}^{p - 1} {cos\left( {\frac{{2\pi i}}{p}} \right)} = - 1$ .

He-Mann · Nov 5, 2016

TuePhuong · Nov 5, 2016

why lol?

InteGrand · Nov 5, 2016

TuePhuong said:
p is a prime number. Proof that $\sum\limits_{i = 1}^{p - 1} {cos\left( {\frac{{2\pi i}}{p}} \right)} = - 1$ .

$\noindent (We don't actually need $p$ to be prime, any positive integer at least $2$ will do.) If we sum from $1$ to $p$, then we are summing the real parts of all the $p$-th roots of unity, which gives 0, because the sum of roots of the polynomial $f(z) = z^{p} -1$ is 0. Since the final term in the sum from 1 to $p$ is $\cos\left(\frac{2\pi p}{p}\right) = \cos \left(2\pi\right) = 1$, the sum in the question is $-1$.$

TuePhuong · Nov 5, 2016

Thank a lot!

Triangular formula (1 Viewer)

TuePhuong

New Member

He-Mann

Vexed?

TuePhuong

New Member

InteGrand

Well-Known Member

TuePhuong

New Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)