Re: sum of roots of unity
- From: Narcoleptic Insomniac <i_have_narcoleptic_insomnia@xxxxxxxxx>
- Date: Wed, 17 Jun 2009 23:28:26 EDT
On Jun 17, 2009 at 9:32 PM CT, Rob Johnson wrote:
In article
<3279063.10397.1245290228761.JavaMail.jakarta@xxxxxxxxxxxxxxxxxxxxxx>,
Narcoleptic Insomniac
<i_have_narcoleptic_insomnia@xxxxxxxxx> wrote:
On Jun 17, 2009 at 6:22 PM CT, Rob Johnson wrote:
In article
<0ac4b3e9-7b7b-409e-99f5-1d4d9526a9e4@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
"oferlock@xxxxxxxxx" <oferlock@xxxxxxxxx> wrote:
this might be trivial, but I am a bit confused:
let N be some composite integer. Let t be some
integer from 1 to N-1. then, is the following
true for all such t?
sum_{k in {0..N-1}} e^{i pi t k/N} = 0
(note we do not necessarily have powers of a
primitive Nth roots of unity here)
There have been a few approaches posted here, but I
am surprised that what I think is the simplest
explanation has not appeared yet:
The sum of the roots of a monic polynomial is
negative of the coefficient of the penultimate term.
Thus, the sum of the roots of x^N - 1 is the
negative of the coefficient of x^{N-1}, namely 0.
Actually, this is exactly what I said a few days ago...
just in a more general sense using elementary symmetric
polynomials.
I'm guessing my posts are being filtered? Rob?
No, they are there, I just thought that you were doing
something more. I see that your post just looked more
complex because you pretty much rederived the fact that
I simply stated.
Ahhhh, I see ^_^ Well, assuming that the OP missed a
factor of 2 (as Achava Nakhash pointed out) and actually
meant to ask if the summation
p_t = sum_{k = 0}^{N - 1} e^{2 pi i k t / N} = 0
...for all 1 <= t <= N - 1, then equating the coefficients
of the "penultimate" term only shows that p_1 = 0. To show
that p_t = 0 for the remaining values -- using this method
of comparing coefficients -- I believe that the values of
all elementary symmetric polynomials must be found first
(an easy task, as they are just the other coefficients).
Given all of the ESPs, the values of all of the power
sums, p_t, can be found using an identity from Newton...
It seems a lot less simple, though no less valid,.
when you include the derivation.
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