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Re: How can I solve these simultaneous quadratic equations in Mathemetica?
*To*: mathgroup at smc.vnet.net
*Subject*: [mg65775] Re: How can I solve these simultaneous quadratic equations in Mathemetica?
*From*: "Chris Chiasson" <chris.chiasson at gmail.com>
*Date*: Mon, 17 Apr 2006 02:28:16 -0400 (EDT)
*References*: <e1stfp$bf5$1@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
Thoihen (cool name),
There are some ways to get some solutions to your system. I am not
sophistocated at math (or spelling), but here are two solutions, one at
c=14/10 and the other at c=1/2:
eqns={1+3c/2+x(1+c+y+z-x)+c*y+(y+z+x)/2-w(2y+2x-3w+3+2c)\[Equal]0,
c/2+w(y+z-w+3/2+c)-x(2w+2z-3x+2)\[Equal]0,
1+2c+w(1+c)+z(w+x-z+3/2+c)-y(2w+2z-3y+4+2c)\[Equal]0,
y(x+w-y+1+c)-c(1+w)+(y+w-x)/2-z(2y+2x-3z)\[Equal]0}
FindRoot[eqns/.c\[Rule]1/2,{{w,2},{x,3},{y,4},{z,5}}]
FindRoot[eqns/.c\[Rule]14/10,{{w,2},{x,3},{y,4},{z,5}}]
One way to search for solutions (assuming the starting guesses are
sufficient) is to change the value of c and look at the residuals. In
this problem, it seems as if FindRoot will warn when its results won't
be correct. At first I just made a table and changed the step size for
c until I got some solutions. Eventually, I changed it to a graphing
routine.
If you could provide more information about what these parameters
represent and how you came to these equations, I am sure someone could
suggest a better method to go about finding solutions.
(for the code below: where the residuals go to zero, you may find a
good solution)
gendat=Table[{c,
Evaluate[eqns]/.Equal\[Rule]Subtract/.FindRoot[
Evaluate[eqns],{{w,2},{x,3},{y,4},{z,5}}]},{c,0,2,0.001}];
gendatplotdat=
Transpose[Map[Function[{datrow},{datrow[[1]],#}&/@datrow[[2]]],gendat]];
Show[Block[{$DisplayFunction=Identity},
MapIndexed[
ListPlot[#1,PlotJoined\[Rule]True,
PlotStyle\[Rule]Hue[#2[[1]]/Length[gendatplotdat]]]&,
gendatplotdat]],PlotRange\[Rule]All,ImageSize\[Rule]72*8]
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