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Re: problem solving polynomial equations


ezsubs is there my mistake, replace 

Solve[{eqy == 0, eqzsubs == 0}, {y, z}] /. Sign[s]^2 -> 1

with 

Solve[{eqy == 0, eqz == 0}, {y, z}] /. Sign[s]^2 -> 1



> 
> On 15 Oct 2005, at 11:22, wtplasar at ehu.es wrote:
> 
> >
> > Hi,
> >
> >
> > I have  two equations. The first one is
> >
> > eqy=(y*(3*b^2 + 12*b*y^2 + 12*y^4 + 3*b^2*z^4 + 4*b*y^2*z^4 +
> > 3*r*y*Sqrt[4*y^4 + 4*b*y^2*(1 + z^4) + b^2*(1 + 3*z^4) + z^2*
(8*b*y^2
> > + 4*y^4 + b^2*(3 + z^4))*Sign[k]]*
> >     Sign[s] + z^2*Sign[k]*(6*b^2 + 16*b*y^2 + 8*y^4 + r*y*Sqrt
[4*y^4 +
> > 4*b*y^2*(1 + z^4) + b^2*(1 + 3*z^4) + z^2*(8*b*y^2 + 4*y^4 + b^2*
(3 +
> > z^4))*Sign[k]]*
> >       Sign[s])))/(3*(3*b^2 + 8*b*y^2 + 4*y^4 + 3*b^2*z^4 + 2*b*
(3*b +
> > 4*y^2)*z^2*Sign[k] +
> >    2*r*y*Sqrt[4*y^4 + 4*b*y^2*(1 + z^4) + b^2*(1 + 3*z^4) + z^2*
> > (8*b*y^2 + 4*y^4 + b^2*(3 + z^4))*Sign[k]]*Sign[s]));
> >
> >
> > and the second one is
> >
> > eqz=(y*z*(4*b*y + 8*y^3 + 4*b*y*z^4 + r*Sqrt[4*y^4 + 4*b*y^2*(1 + 
z^4)
> > + b^2*(1 + 3*z^4) + z^2*(8*b*y^2 + 4*y^4 + b^2*(3 + z^4))*Sign[k]] 
> > *Sign
> > [s] +
> >    z^2*Sign[k]*(8*b*y + 8*y^3 + r*Sqrt[4*y^4 + 4*b*y^2*(1 + z^4) 
+  
> > b^2*
> > (1 + 3*z^4) + z^2*(8*b*y^2 + 4*y^4 + b^2*(3 + z^4))*Sign[k]]*Sign
> > [s])))/
> >  (3*(3*b^2 + 8*b*y^2 + 4*y^4 + 3*b^2*z^4 + 2*b*(3*b + 4*y^2)
*z^2*Sign
> > [k] +
> >    2*r*y*Sqrt[4*y^4 + 4*b*y^2*(1 + z^4) + b^2*(1 + 3*z^4) + z^2*
> > (8*b*y^2 + 4*y^4 + b^2*(3 + z^4))*Sign[k]]*Sign[s]));
> >
> > Now when I do
> >
> > Solve[{eqy == 0, eqzsubs == 0}, {y, z}] /. Sign[s]^2 -> 1
> >
> > I get the result fairly quickly, but if I do
> >
> > Solve[{Numerator[eqy]== 0, Numerator[eqz] == 0}, {y, z}]/. Sign[s]
^2 -
> >
> >> 1
> >>
> >
> > it seems to get stuck. I may get an answer eventually, but it 
seemed
> > to be taking too long and aborted it.
> >
> > Any clues? Thanks.
> >
> > Ruth
> >
> >
> 
> 
> What is eqzsubs?
> 
> 
> Andrzej Kozlowski
> Tokyo, Japan
> 
> 
> 



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