1

I want to prove that the below equation can be held.

$$\sum_{ k=1 }^{ n } \cos\left(\frac{ 2 \pi k }{ n } \right) =0, \qquad n>1 $$

Firstly I tried to check the equation with small values of $n$

$$ \text{As } n=2 $$

$$ \cos\left(\frac{ 2 \pi \cdot 1 }{ 2 } \right) + \cos\left(\frac{ 2 \pi \cdot 2 }{ 2 } \right) $$

$$ = \cos\left(\pi\right) + \cos\left(2 \pi\right) $$

$$ = -1+ 1 =0 ~~ \leftarrow~~ \text{Obvious} $$

But

$$ \text{As}~~ n=3 $$

$$ \cos\left(\frac{ 2 \pi \cdot 1 }{ 3 } \right) +\cos\left(\frac{ 2 \pi \cdot 2 }{ 3 } \right) + \cos\left(\frac{ 2 \pi \cdot 3 }{ 3 } \right) $$

$$ = \cos\left(\frac{ 2 \pi }{ 3 } \right) + \cos\left(\frac{ 4 \pi }{ 3 } \right) + \cos\left( 2\pi \right) $$

$$ = \cos\left(\frac{ 2 \pi }{ 3 } \right) + \cos\left(\frac{ 4 \pi }{ 3 } \right) + 1 =?$$

What formula(s) or property(s) can be used to prove the equation?

Blue
  • 75,673
electrical apprentice
  • 1,519
  • This may be helpful: https://math.stackexchange.com/questions/4136552/limit-of-series-involving-sum-of-cosines-lim-n-to-infty-sum-k-0n-co – Vasili Jun 12 '21 at 03:04
  • 1
    Also https://math.stackexchange.com/questions/17966/how-can-we-sum-up-sin-and-cos-series-when-the-angles-are-in-arithmetic-pro : this question has come up many times going back to the start of the site... –  Jun 12 '21 at 03:57
  • 1
    And the last time it came up was 6 hours ago... https://math.stackexchange.com/questions/4170389/summation-from-0-to-n-of-cosk-n2-pi – Hans Lundmark Jun 12 '21 at 04:51

3 Answers3

5

Your sum is the sum of the real parts of the $n$th roots of the unity. That is, they are the roots of the polynomial $z^n-1$. The sum of the roots is equal to the coefficient of $z^{n-1}$ by Vieta's formulas, hence it is zero.

GReyes
  • 16,446
  • 11
  • 16
  • 2
    I originally thought of the geometric series approach, but this is very slick! – user580918 Jun 12 '21 at 03:13
  • @user580918 When I was writing the answer, I was going to use the symmetry of the roots, but all of a sudden I realized that I could use Vieta's formulas.. – GReyes Jun 12 '21 at 04:05
2

Observe that $$\sum_{k=1}^n\cos(\frac{2\pi k}{n})=\operatorname{Re}\left(\sum_{k=1}^ne^{\frac{2\pi ki}{n}}\right).$$ Consider the sum $$\sum_{k=1}^ne^{\frac{2\pi ki}{n}}=\sum_{k=1}^n(e^{\frac{2\pi i}{n}})^k$$ and use the fact that $$1+\sum_{k=1}^n r^k =\frac{1-r^{n+1}}{1-r}$$ to show it's zero.

Stuck
  • 1,712
0

The cosines are the $x$-coordinates of points on the unit circle equally spaced around the origin. Since the average $x$ value is zero, it follows that the sum of the $x$ values is also zero.

RobertTheTutor
  • 7,415
  • 2
  • 10
  • 34