RE: Re: Solving simultaneous eqns

*To*: mathgroup at smc.vnet.net*Subject*: [mg14358] RE: [mg14287] Re: Solving simultaneous eqns*From*: "Ersek, Ted R" <ErsekTR at navair.navy.mil>*Date*: Thu, 15 Oct 1998 00:28:59 -0400*Sender*: owner-wri-mathgroup at wolfram.com

Yeoung-Sang Yun wrote: > > I want to know how to solve the following simulaneous equations in > Mathematica: > > x=x^1.2/(2*x^3+y^0.7+4*z^2.5) > y=y^0.7/(2*x^0.6+y^2+z^2.2) > z=0.9*z^1.5/(x^0.7+2*y^0.2+z^1.1) > Bill Bertram replied with: > >Use: > >FindRoot[{x == x^1.2/(2*x^3+y^0.7+4*z^2.5), > y == y^0.7/(2*x^0.6+y^2+z^2.2), > z == 0.9*z^1.5/(x^0.7+2*y^0.2+z^1.1)}, > {x,1},{y,1},{z,1}] > >to get: > >{x -> 0.2071155206958545 - 4.456053142803247*^-10*I, > y -> 0.6156594604688913 + 1.237663943046694*^-9*I, > z -> 2.09226461646654*^-15 + 2.130084871188269*^-16*I} > > It looks like this is converging on a solution where z=0, Im[x]=0, Im[y]=0. So I set z=0 in the original equations. Then I am left with only the first two equations since the third equation is reduced to (0==0). Next I convert the machine numbers to rationals and give the two equations to FindRoot. The approximate root returned is very close to an exact root. Cheers, Ted ___________________________ In[1]:= (soln= FindRoot[{ x==x^(12/10)/(2*x^3+y^(7/10)), y==y^(7/10)/(2*x^(6/10)+y^2) },{x,0.207},{y,0.616}, AccuracyGoal->Infinity, WorkingPrecision->17])//InputForm Out[1]//InputForm= {x -> 0.2071155092220912707736701893`17, y -> 0.6156595017877353994690851319`17} In[6]:= test=eqs/.Equal[lhs_,rhs_]:>lhs-rhs; test/.soln//InputForm Out[7]//InputForm= {-2.775557561562891*^-17, 0.}