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Re: a Quaternion quadratic level Pisot polynomial
*To*: mathgroup at smc.vnet.net
*Subject*: [mg68637] Re: a Quaternion quadratic level Pisot polynomial
*From*: Roger Bagula <rlbagula at sbcglobal.net>
*Date*: Mon, 14 Aug 2006 06:44:08 -0400 (EDT)
*References*: <ebmtoj$6k1$1@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
jdturner wrote:
>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
>
>
>
Jim Turner,
That's a good question.
It appears Mathematica solves four equations in four unknowns and gives at least four answers
even with {t_Real, x_Real, y_Real, z_Real}
Peter Pleasants finds :
4x^4 + 4x^3 + 6x^2 + 4x + 1
Has Pisot roots too. Mathematica gives 10 roots... two of which are the Real root
that are the same as this polynomial.
You are welcome to try your method on that.
I was trying to get a method in Mathematica that gave
correct roots for any polynomial
and the specific polynomial that behaved as a Pisot:
1) Polynomial with integer coefficients
2) one real modulus/absolute values greater than one
3) other modulus/absolutue values less than one
Complex limited , Real limited ones ones are already known.
I can't say that I understand Quaternion roots all the well
even with my work.
Since you are the only one to respond,
I assume others are even less brave.
Roger Bagula
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