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