From: rusin@washington.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: algebraic number question
Date: 28 Oct 1995 15:51:12 GMT

In article <46tc38$spk@newsreader.wustl.edu>,
Jim Buddenhagen <jbuddenh@artsci.wustl.edu> wrote:
>Let x_n = (2*cos(Pi/n)-1)/(3-2*cos(Pi/n) ,   n=1,2,3...
>The x_n are algebraic numbers with surprizingly simple minimal
>polynomials.  Call them p_n ,  n=1,2,3,...
>
>Question:  Is it true that all roots of all the p_n are real and
>lie between -1 and +1 ?  If so, is there an easy proof?

So  x_n=(w+w^(-1)-1)/(3-w-w^(-1)), where  w=exp(pi/n) is s primitive
(2n)-th root of  1. Its minimal polynomial is  p_n(X)=Prod(X-x_n^\sigma)
where the product runs over the Galois group of the field  Q(x_n).

This field is the maximal real subfield of  Q(w). The Galois group of
the latter is the multiplicative group of  Z_(2n)  in a natural way:
the action of  [k mod (2n)]  on  w  is to send it to  w^k. The Galois
group of the real subfield is then the homomorphic image of this, that is,
it's  (Z_(2n))^*/{+-1}, but with essentially the same action.

Just in case I lost anyone there, the conclusion is that the other roots
of the minimal polynomial  p_n  are precisely the distinct real numbers
of the form x_n^\sigma_k=(w^k+w^(-k)-1)/(3-w^k-w^(-k)), which can be written
in real form as  (2cos(k*pi/n) - 1 )/( 3 - 2cos(k*pi/n) ). Since
(2x-1)/(3-2x) is an increasing function, it can for  x  between  -1  and  1
take on values only between  -3/5  and  -1/2.

dave
(PS-Surprisingly, some words have no "z".)