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Consider the next sum \begin{equation} \sum_{k = 0}^{N - 1}e^{-j\frac{2\pi}{N}k} \end{equation}

Its geometric meaning is the sum of uniformly distributed vectors on the unit circle. Thus, we can say that the sum is zero. If we try to be more precise, then for the case of even N, we can single out pairs of opposite vectors, each of which in the sum 0. What to do in the case of an odd N? Is there a way to calculate the sum analytically, i.e. to prove this in general?

Marcus Müller
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Alexey K
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  • don't understand what you mean with "analytically": you've got the formula right there; that's pretty analytical, isn't it? – Marcus Müller Nov 17 '20 at 23:18
  • I would like to prove that this sum is equal to 0. For me, only the case of an even number of members of the sum is clear. How to prove this in general terms? – Alexey K Nov 17 '20 at 23:22
  • All the proofs in the answers by Marcus Muller, Hilmar, and Laurent Duval implicitly assume that $k$ is not an integer multiple of $N$, which is fine since the wording of the question implies that this is indeed what the OP had in mind. It is worth noting that the sum in question equals $N$, not $0$, when $k$ is an integer multiple of $N$. And yes, we can get this by applying L'Hopital's rule to $\frac{1-z^N}{1-z}$ but that is putting the cart before the horse because the geometric series sum is explicitly predicated on $z\neq 1$ so that $\frac{1-z^N}{1-z}$ is not valid as the sum. – Dilip Sarwate Nov 28 '20 at 15:40

3 Answers3

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For me, only the case of an even number of members of the sum is clear. How to prove this in general terms?

ah, "sum orthogonality proof" is something you'll want to google. You can do it (under a relaxed meaning of "proof") very geometrically. Understand each part in your sum as vector in a plane:

  1. take the first vector: $e^{0}\equiv 1$. It goes from point 0+0j to 1+0j.
  2. attach the second vector to the tip of that vector. Note the inner angle between these two. Also note how the position of the tip is the partial sum of the first two elements from your sum.
  3. attach the third to the tip of the second. Note that the angle between these two is the same.
  4. Repeat until you're out of vectors.

You clearly followed a polygonal chain, where each of the $N$ edges had length 1. Now, what is the condition on the angles for such a polygonal chain to be a closed polygon?

Marcus Müller
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    Yes, now it's more or less clear :) Angle step $\delta \phi = (\pi - \frac{2\pi}{N})$, so this is a regular polygon with the sum of interior angles $\pi(N-2)$ – Alexey K Nov 17 '20 at 23:42
  • yep, now you only need to proof that having this angle sum is a sufficient condition, and you're all set :) – Marcus Müller Nov 17 '20 at 23:51
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Geometric sum: (https://mathworld.wolfram.com/GeometricSeries.html)

$$\sum_{n=0}^{N-1} z^n = \frac{1-z^N}{1-z}$$

Just poke in $z = e^{-j2\pi/N}$ and it's obvious that $z^N =1$ and the whole term becomes zeros, at east for $N > 1$.

Hilmar
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Since it is often useful in DSP (filters, Fourier transforms of flat windows), let us derive the finite sum of geometric series again: $$s_{p,q} = \sum_{k=p}^q a^k\,. $$

with $a\neq1$ and $p\le q $. The trick here is that the sum is almost the same if you factorize some integer power of $a$ (a kind of scale invariance), only the first and the last terms change:

$$s_{p,q} = a^p + \sum_{k=p+1}^{q+1} a^k - a^{q+1} = a^p - a^{q+1} + a\sum_{k=p}^{q} a^k \,, $$ hence $$s_{p,q} -a s_{p,q} = a^p - a^{q+1}\,. $$ or $$s_{p,q} = \frac{a^p - a^{q+1}}{1-a} = a^p\frac{1 - a^{q+1-p}}{1-a} \,. $$

So whenever $p-q+1$ is a multiple of $N$, and $a=e^{-j2\pi/N}$, the sum $s_{p,q}$ vanishes to zero. However, the visual proof of Markus is way better. I will thus add another geometrical proof, also based on the concept of invariance: the complex roots of unity define a regular polygon. This polygon is invariant by a rotation of angle $2\pi/N$, so the sum $s$ of the original vertices is the same as the sum of the vertices rotated, which is $s\times e^{-j2\pi/N}$. And therefore

$$s = s\times e^{-j2\pi/N}$$ which has only one solution: $s=0$.

Laurent Duval
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