Question

Hey folks!

Looking to prove to myself that a consequence of Euler's formula is true:
\[\sum_{k=1}^{n}e^{i\frac{k}{n}\tau} = 0 \text{ for all } n \ge 2 \text{ ; } \tau=2\pi\]
It certainly works for all values of n I've tested, and wolfram alpha reckons it's true (http://bit.ly/vyruMb). Using De Moivre's Theorem:
\[\sum_{k=1}^{n}e^{i\frac{k}{n}\tau}=\sum_{k=1}^{n}cos(k\frac{\tau}{n} + isin(k\frac{\tau}{n}\] and plotting the values for some n, I can see visually how the imaginary parts at least will always cancel out.

Anybody got some thoughts?

Answer 1

Sure. If n is even, then each term in your original sum can be paired off with its negative. E.g., if n = 6, then the terms k = 1 and k = 4 are additive opposites; as are k = 2 and k = 5, k = 3 and k = 6.

If n is odd, it's a little more subtle.

Answer 2

If n is odd, let z = e^(2.pi.i/n) and let w = e^(2.pi.i/2n) = e^(pi.i/n)

Then your sum is the sum of z, z^2, .... z^n; call that sum S.

Then wS = wz + wz^2 + ... + wz^n.

Now \( wS + S = \sum_{i=1}^{2n} w^i \).

Using the argument for n even, this sum is zero, i.e.,
\[ (w+1)S = 0 \]
But as \( w + 1 \neq 0 \), it must be that S is zero.

Answer 3

Nice, The Approach is really nice

Answer 4

Umm \(wS + S = \sum_{i=1}^{2n} w^i\) << How did you get it?

Answer 5

Ah, I got stuck on \(wS + S = \sum_{i=1}^{2n} w^i\) for a while. The wS will be the odd*2pi/2n, and the S are the even*2pi/2n. I don't see where \(w^1\) is though, since \(wz = e^{i\frac{3\pi}{n}}\), and \(z=e^{i\frac{2\pi}{n}}\). Should the sum begin at i=2, or am I missing something?

Answer 6

w^2 = z and z^n = 1 hence
z + wz + z^2 + .... + wz^(n-1) + z^n + wz^n
= w^2 + w^3 + w^4 + w^(2n-1) + w^(2n) + w

Answer 7

there should be an ellipsis ... in there

Answer 8

Ah, quite so, quite so. Thanks JamesJ!

Answer 9

ah I see

Answer 10

I'm writing the algebra but I'm seeing it geometrically. Multiplication by w rotates all of those numbers on the unit circle by half the angle of z.

Answer 11

I suppose the proof for even n would be easier for \(-\frac{\tau}{2}\) to \(+\frac{\tau}{2}\) than from 0 to \(\tau\)

Answer 12

it would be easier slightly if it were i = 0 to n-1

Answer 13

at least easier for me.

Answer 14

Well I shalln't need the proof for another year or so, if at all. For the time being, I'm satisfied that it's true.