Re: sum of roots of unity
- From: Narcoleptic Insomniac <i_have_narcoleptic_insomnia@xxxxxxxxx>
- Date: Thu, 18 Jun 2009 21:20:04 EDT
On Jun 18, 2009 at 7:29 PM CT, Rob Johnson wrote:
In article
<1645093.10506.1245295737376.JavaMail.jakarta@nitrogen
.mathforum.org>,
Narcoleptic Insomniac
<i_have_narcoleptic_insomnia@xxxxxxxxx> wrote:
On Jun 17, 2009 at 9:32 PM CT, Rob Johnson wrote:
In article
<3279063.10397.1245290228761.JavaMail.jakarta@nitrogen
.mathforum.org>,
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@xxxxxxxxxxxxxxxx
legroups.com>,
"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
Since the title referred to roots of unity, I had
assumed this.
..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...
I was thinking of the sum as a permutation of a
subgroup of the roots of unity.
[snip ASCII math]
My mistake, I guess I just thought that your original post
was discussing an alternative method (via coefficients),
and not the primitive root/permutation (group theoretic)
method which had already been discussed earlier by Chip.
I must have misread or misinterpreted your posts or
something...
My Apologies,
Kyle Czarnecki
However, it seems that you are looking at these as.
the sum of the
t^{th} powers of the N^{th} roots of unity.
[snip ASCII math]
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