Re: problem solving polynomial equations

*To*: mathgroup at smc.vnet.net*Subject*: [mg61342] Re: [mg61306] problem solving polynomial equations*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Sun, 16 Oct 2005 00:17:57 -0400 (EDT)*References*: <200510140953.FAA28482@smc.vnet.net> <200510150222.WAA17287@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

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 > > Basically what happens is this. Solve can't find the general solution to parametric equations of this kind. (It is not the fault of Solve; there is really no way to do it). So what it does is returns basically any solutions it can find, by using certain "heuristic" methods. One method is simply to set some variables to 0 and see if the equations can be solved for the remaining ones. This is what seems to happen here: Solve[{eqy == 0, eqz == 0}, {y, z}] /. Sign[s]^2 -> 1 {{y -> (1/4)*((-r)*Sign[s] - Sqrt[r^2 - 8*b]), z -> 0}, {y -> (1/4)*(r*Sign[s] - Sqrt[r^2 - 8*b]), z -> 0}, {y -> (1/4)*(Sqrt[r^2 - 8*b] - r*Sign[s]), z -> 0}, {y -> (1/4)*(r*Sign[s] + Sqrt[r^2 - 8*b]), z -> 0}, {y -> 0}} These are the kind of solutions you could find yourself by hand! Now for the other case. Solve[{Numerator[eqy] == 0, Numerator[eqz] == 0}, {y, z}] /. Sign[s] ^2 -> 1 This is pure speculation, but I would guess Solve has for some reason started to apply some algorithm that might give additional solutions. Unfortunately there is no reason to believe that the computation will finish in reasonable time. This could in fact be one of the cases where "better is worse", meaning that the fact that the equations are "better" (so Mathematica makes a serious attempt to find solutions) cases no solutions to be returned. To see this phenomenon in an extreme form on a trivial example, compare these two equations: Solve[x (Exp[Sin[x]-Cos[x]])\[Equal]0,x] Inverse functions are being used by Solve, so some solutions may not be \ found; use Reduce for complete solution information. {{x->0}} Solve can do nothing with this except find the trivial solution so it quickly returns the trivial solution. Now consider the following equation that I made up at random: Solve[(x^32 - x^31)^(1/12) + (x^7 - x^11/3)^(1/5) + 17x + 5 == 0, x] This is "better" than the other one, since Mathematica, in principle knows algorithms for solving equations of this type. But it is "worse" since the computation is unlikely to finish (I did not wait long but even if this ever finishes one can produce a more complicated example that will not). So, I think what happened in your case is that your second set of equations is sufficiently "better" than the first to make it actually "worse". Well, of course, this is only speculation. Daniel Lichtblau can almost certainly tell if it is true, if he chooses to. Andrzej Kozlowski Andrzej Kozlowski Tokyo, Japan

**Follow-Ups**:**Re: Re: problem solving polynomial equations***From:*Daniel Lichtblau <danl@wolfram.com>

**References**:**Re: Solving Diophantine Equations***From:*Andrzej Kozlowski <andrzej@yhc.att.ne.jp>

**problem solving polynomial equations***From:*wtplasar@ehu.es