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As part of a larger proof, I must show that:

$$\sum_{n=1}^{N-1}\cos(2\pi\frac{n}{N})=-1$$

I have thought about this but can't figure out how to get my hands on the value since I don't know any relations between sums of sinusoids and other stuff.

My gut feeling is that this is fairly obvious.

I can't use exponents because the proof this is a part of is showing that $1+\sum_{n=1}^{N-1}z^n=0$ for $z$ on the unit circle, and this argument would require the assumption.

I was thinking of using an argument similar to this:

Consider $\sum_{n=1}^{N}\cos(2\pi\frac{n}{N})$. Notice that $\cos(2\pi\frac{n}{N})=-\cos(\pi-2\pi\frac{n}{N})=-\cos(2\pi\frac{N/2-n}{N})=-\cos(2\pi\frac{3N/2-n}{N})$ so all the terms on the first half of the sum cancel (due to the 2nd-to-last identity), and all the terms on the second half cancel (due to the last identity).

However, formalizing this gets hairy when considering $\floor{\frac{N}{2}}$ that are odd. Is there a simple way to fix my argument?

Christian Chapman
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2 Answers2

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How naive? We show why the sum starting from $n=0$ is $0$, using an argument from Physics.

Place weights of $\frac{1}{N}$ at the vertices of a regular $N$-gon with centre at the origin. Then our sum is the $x$-coordinate of the centre of mass. But by symmetry the centre of mass is at the origin.

André Nicolas
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$\sum_{n=1}^{N} cos(\frac{2n\pi}{N})= \sum_{n=1}^{N} \Re(exp(i*\frac{2n\pi}{N})) = \Re (\sum_{n=1}^{N} exp(i*\frac{2n\pi}{N}))$

This is a geometric sum.

So, it is : $\Re( \frac{1-exp(i.\frac{2\pi}{N})^N}{1-exp(i.\frac{2\pi}{N})})= \Re(\frac{1-1}{1-exp(i.\frac{2\pi}{N})})=0$

So $\sum_{n=1}^N \cos(\frac{2n\pi}{N})=0= 1+ \sum_{n=1}^{N-1} \cos(\frac{2n\pi}{N})$.

And you have your result.

D.L.
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