Re: a Quaternion quadratic level Pisot polynomial

*To*: mathgroup at smc.vnet.net*Subject*: [mg68621] Re: a Quaternion quadratic level Pisot polynomial*From*: "jdturner" <jamesdanielturner at hotmail.com>*Date*: Sun, 13 Aug 2006 05:52:29 -0400 (EDT)*Sender*: owner-wri-mathgroup at wolfram.com

Roger Bagula wrote: > I tried all kinds of combinations. > q^4-4*q^2-1 > is the opposite of a Pisot ( one root inside the Uniot Quaternion and > the rest outside). > > (-1+I+J+K)*q^2-2*q-1==0 > solves to give a set of Pisot like roots. > {{-0.34833, 0.385615, -0.34833, -0.15167}, > {-0.15167, -0.885615, -0.15167, -0.34833}, > {-0.420313, 0.116972, 0.116972, 0.116972}, > {-0.0796875, -0.616972, -0.616972, -0.616972}} > I have a question for you. Why four roots, since the equation is easily solved by completing the square. First rewrite the equation as q^2+c1*q+c0=0, c1=-2/(-1,I+J+k), c0 = -1/(-1+I+J+K) yielding the roots qp = -c1/2+qsqrt( -c0+c1~c1/4 ) = [ - 0.0796874903417d0, [ - 0.61697245625346d0, - 0.61697245625346d0, - 0.61697245625346d0]] qm= -c1/2-qsqrt( -c0+c1~c1/4 ) = [ - 0.4203125096583d0, [0.11697245625346d0, 0.11697245625346d0, 0.11697245625346d0]] which coorespond to your last two rts. Jim Turner