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 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".)