Re: problem with Quaternion polynomial root solver
- To: mathgroup at smc.vnet.net
- Subject: [mg68352] Re: problem with Quaternion polynomial root solver
- From: Roger Bagula <rlbagula at sbcglobal.net>
- Date: Wed, 2 Aug 2006 05:24:08 -0400 (EDT)
- References: <eahtq5$p0b$1@smc.vnet.net> <eakd82$qhf$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
In an attempt to find a Quaternion Pisot like solution
I attempt to find fixed point solutions to a 4d Quaternion Migdal-Kadanov
under the impression that the 4d lattice involved would be Ising like
and have a Pisot related solution as happens in two dimensions.
( well known 2d fixed point solution is b1=Log[1+Sqrt[2]]/2
which is related to the Pisot polynomial x^2-2*x-1)
I get a solution in Mathematica of the form:
b0=F[b1,b2,b3]
but not a true fixed point numerical solution.
Can anyone suggest a better way to find the fixed point solution in
Mathematica.
I also tried an SU(3) Migdal -Kadanov and got a recursion solution,
but Mathematica refused to solve for the fixed point.
\!\(Clear[t, x, y, z, ta, xa, ya, za, b0, b1, b2, b3, b0a, b1a, b2a,
b3a]\n (*\ since\ the\
fixed\ point\ of\ the\ Migdal\ - Kadanov\ recursion\ is\ connected\
to\ a\ quadratic\ Pisot\ with\ root\ 1 + Sqrt[2]*) \n (*I' m\
trying\ \
to\ get\ a\ fixed\ point\ for\ a\ Quaternion\ Migdal - Kadanov\ to\ see\
what\
\ kind\ of\ root\ it\ involves*) \[IndentingNewLine]
<< Algebra`Quaternions`\n (*\ four\ variable\ spin/
magnetic\ exponential\ definition\ of\ the\ Migdal\ - Kadanov\ \
form*) \n
t = Exp[b0 + b1]; ta = Exp[b0a + b1a];\n
x = Exp[b0 - b1 + b2]; xa = Exp[b0a - b1a + b2a];\n
y = Exp[b0 - b1 - b2 + b3]; ya = Exp[b0a - b1a - b2a + b3a];\n
z = Exp[b0 - b1 - b2 - b3]; za = Exp[b0a - b1a - b2a - b3a];\n
\(q = Quaternion[t, x, y, z];\)\n
\(qa = Quaternion[ta, xa, ya, za];\)\n
\(q2 = ExpandAll[q ** q];\)\n
FullSimplify[q2 - qa]\n (*Solve\ for\ Migdal_Kadanov\ recursion*) \n
Solve[{\(-\[ExponentialE]\^\(b0a + b1a\)\) - \[ExponentialE]\^\(2\ b0
- 2\ \
b1 - 2\ b2 - 2\ b3\)\ \((1 + \[ExponentialE]\^\(4\ b3\) - \
\[ExponentialE]\^\(2\ \((2\ b1 + b2 + b3)\)\) + \[ExponentialE]\^\(4\ b2
+ 2\ \
b3\))\) == 0, 2\ \[ExponentialE]\^\(2\ b0 + b2\) -
\[ExponentialE]\^\(b0a - \
b1a + b2a\) == 0, 2\ \[ExponentialE]\^\(2\ b0 - b2 + b3\) -
\[ExponentialE]\^\
\(b0a - b1a - b2a + b3a\) == 0, 2\ \[ExponentialE]\^\(2\ b0 -
b2 - b3\) - \[ExponentialE]\^\(b0a - b1a - b2a - b3a\) ==
0}, {b0a, b1a, b2a, b3a}]\n
FullSimplify[{b0 - \((1\/8\ \((2\ Log[2\ \[ExponentialE]\^\(2\ b0 +
b2\)] + \
Log[2\ \[ExponentialE]\^\(2\ b0 - b2 - b3\)] + Log[2\ \[ExponentialE]\^\(2\
b0 - b2 + b3\)] + 4\ Log[\(-\[ExponentialE]\^\(2\ b0 - 2\ b1
- 2\ \
b2 - 2\ b3\)\)\ \((1 + \[ExponentialE]\^\(4\ b3\) -
\[ExponentialE]\^\(4\ b1 \
+ 2\ b2 + 2\ b3\) + \[ExponentialE]\^\(4\ b2 + 2\ b3\))\)])\))\), b1 - \
\((1\/8\ \((\(-2\)\ Log[2\ \[ExponentialE]\^\(2\ b0 +
b2\)] -
Log[2\ \[ExponentialE]\^\(2\ b0 - b2 - b3\)] - Log[2\ \
\[ExponentialE]\^\(2\ b0 - b2 + b3\)] + 4\ Log[\(-\[ExponentialE]\^\(2\
b0 - \
2\ b1 - 2\ b2 - 2\ b3\)\)\ \((1 + \[ExponentialE]\^\(4\ b3\) - \
\[ExponentialE]\^\(4\ b1 + 2\ b2 + 2\ b3\) + \[ExponentialE]\^\(4\ b2 + 2\
b3\))\)])\))\), b2 - \((1\/4\ \((2\ Log[2\ \
\[ExponentialE]\^\(2\ b0 + b2\)] - Log[2\ \[ExponentialE]\^\(2\
b0 - b2 - b3\)] -
Log[2\ \[ExponentialE]\^\(2\ b0 - b2 + b3\)])\))\), b3 - \
\((1\/2\ \((\(-Log[2\ \[ExponentialE]\^\(2\ b0 - b2 - b3\)]\) + Log[2\ \
\[ExponentialE]\^\(2\ b0 - b2 + b3\)])\))\)}]\n
NSolve[{1\/8\ \((8\ b0 - 4\ Log[2] - 2\ Log[\[ExponentialE]\^\(2\ b0 + \
b2\)] - Log[\[ExponentialE]\^\(2\ b0 -
b2 - b3\)] -
Log[\[ExponentialE]\^\(2\
b0 - b2 +
b3\)] - 4\ Log[\(-\[ExponentialE]\^\(2\ b0 - 2\ b1 -
2\ \
b2 - 2\ b3\)\)\ \((1 + \[ExponentialE]\^\(4\ b3\) -
\[ExponentialE]\^\(2\ \((
2\ b1 + b2 + b3)\)\) + \[ExponentialE]\^\(4\ b2 + 2\ b3\))\)])\) == 0,
1\/8\ \((8\ b1 + Log[16] + 2\
Log[\[ExponentialE]\^\(2\ b0 + b2\)] +
Log[\[ExponentialE]\^\(2\ b0 - b2 - b3\)] +
Log[\[ExponentialE]\
\^\(2\ b0 - b2 + b3\)] -
4\ Log[\(-\[ExponentialE]\^\(2\ b0 - 2\ b1 - 2\ b2 - 2\
b3\)\)\ \
\((1 + \[ExponentialE]\^\(4\ b3\) - \[ExponentialE]\^\(2\ \((2\ b1 + b2
+ b3)\
\)\) + \[ExponentialE]\^\(4\ b2 + 2\ b3\))\)])\) == 0, 1\/4\ \((4\ b2 - 2\ \
Log[\[ExponentialE]\^\(2\ b0 + b2\)] + Log[\[ExponentialE]\^\(2\ b0 -
b2 - b3\)] +
Log[\[ExponentialE]\^\(2\ b0 - b2 + b3\)])\) == 0, b3 + \
1\/2\ \((Log[\[ExponentialE]\^\(2\ b0 - b2 - b3\)] -
Log[\[ExponentialE]\^\(2\
\ b0 - b2 + b3\)])\) == 0}, {b0, b1, b2, b3}]\)
Try for further solution:
\!\(b0 = Log[\(1.`\ \[ExponentialE]\^\(b1 + b2 + b3\)\)\/\@\(\(-2.`\) -
2.`\ \
\[ExponentialE]\^\(4\ b3\) + 2.`\ \[ExponentialE]\^\(4\ b1 + 2\ b2 + 2\
b3\) \
- 2.`\ \[ExponentialE]\^\(4\ b2 + 2\ b3\)\)]\n
NSolve[{1\/8\ \((8\ b1 + Log[16] + 2\ Log[\[ExponentialE]\^\(2\ b0 +
b2\)] \
+ Log[\[ExponentialE]\^\(2\ b0 - b2 - b3\)] + Log[\[ExponentialE]\^\(2\
b0 - \
b2 + b3\)] - 4\ Log[\(-\[ExponentialE]\^\(2\ b0 - 2\ b1 - 2\ b2 - 2\
b3\)\)\ \
\((1 + \[ExponentialE]\^\(4\ b3\) - \[ExponentialE]\^\(2\ \((2\ b1 + b2
+ b3)\
\)\) + \[ExponentialE]\^\(4\ b2 + 2\ b3\))\)])\) ==
0, 1\/4\ \((4\ b2 - 2\ Log[\[ExponentialE]\^\(2\ b0 +
b2\)] + \
Log[\[ExponentialE]\^\(2\
b0 - b2 - b3\)] + Log[\[ExponentialE]\^\(2\ b0 - b2 + \
b3\)])\) == 0, b3 + 1\/2\ \((
Log[\[ExponentialE]\^\(2\ b0 - b2 - b3\)] -
Log[\[ExponentialE]\
\^\(2\ b0 - b2 + b3\)])\) == 0}, {b1, b2, b3}]\)