Re: problem with Quaternion polynomial root solver
- To: mathgroup at smc.vnet.net
- Subject: [mg68265] Re: problem with Quaternion polynomial root solver
- From: Roger Bagula <rlbagula at sbcglobal.net>
- Date: Mon, 31 Jul 2006 03:45:21 -0400 (EDT)
- References: <eahtq5$p0b$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
I tried using the built in to do the multiplication steps.
I've come to the conclusion that
for odd quaternions ,cubes =>
q.(q.q) and (q.q).q
there are two distinct polynomials,
but for even powers of q there is only one result.
Holding t constant I looked at a Pisot like root structure as a 3d implict
and there are multi-surface intersections that produce distinct sets of
roots
in these equations.
I'm no closer to finding a definitive Quaternion Pisot.
Mathematica:
\!\(<< Algebra`Quaternions`\n
q = Quaternion[t, x, y, z]\n
q2 = ExpandAll[q ** q]\n
q3 = \((q ** q)\) ** q\n
FullSimplify[q3 - q2 - q - Quaternion[1, 0, 0, 0]]\n
NSolve[{\(-1\) +
x\^2 + y\^2 + z\^2 + t\ \((\(-1\) + \((\(-1\) + t)\)\ t - 3\ x\^2 - 3\
y\^2 - 3\ z\^2)\) == 0, \(-
x\)\ \((1 + \((2 - 3\ t)\)\ t + x\^2 + y\^2 + z\^2)\) ==
0, \(-y\)\ \((
1 + \((2 - 3\ t)\)\ t + x\^2 + y\^2 + z\^2)\) ==
0, \(-z\)\ \((1 + \((
2 - 3\ t)\)\ t + x\^2 + y\^2 + z\^2)\) == 0}, {t, x, y, z}]\n
a = Table[{Re[t] - Im[x], Re[x] + Im[t], Re[y] - Im[z], Re[z] + Im[y]}
/. \
NSolve[{\(-1\) +
x\^2 + y\^2 + z\^2 + t\ \((\(-1\) + \((\(-1\) + t)\)\
t - 3\ x\^2 - 3\ y\^2 - 3\ z\^2)\) == 0, \(-x\)\ \((1 + \((
2 - 3\ t)\)\ t +
x\^2 + y\^2 + z\^2)\) == 0, \(-y\)\ \((1 + \((2 - 3\ t)\)\ t +
x\^2 + y\^2 + z\^2)\) == 0, \(-z\)\ \((1 + \((2 - 3\ t)\)\ t +
x\^2 + y\^2 + z\^2)\) == 0}, {t, x, y, z}], {n, 1, 6}]\n
b = Table[Sqrt[\((Re[t] - Im[
x])\)^2 + \((Re[x] + Im[t])\)^2 + \((Re[y] - Im[z])\)^2 + \
\((Re[z] + Im[y])\)^2] /. \ NSolve[{\(-1\) + x\^2 + y\^2 + z\^2 +
t\ \((\(-1\) + \((\(-1\) + t)\)\ t - 3\
x\^2 - 3\ y\^2 - 3\
z\^2)\) == 0, \(-x\)\ \((1 + \((2 - 3\
t)\)\ t + x\^2 + y\^2 +
z\^2)\) == 0, \(-y\)\ \((1 + \((2 - 3\ t)\)\ t +
x\^2 + y\^2 + z\^2)\) ==
0, \(-z\)\ \((1 + \((2 - 3\ t)\)\ t + x\^2 +
y\^2 + z\^2)\) == 0}, {t, x, y, z}], {n, 1, 6}]\)
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