Re: Solving simultaneous eqns

*To*: mathgroup at smc.vnet.net*Subject*: [mg14278] Re: Solving simultaneous eqns*From*: Alan Lewis <alan at enfs.com>*Date*: Mon, 12 Oct 1998 13:51:57 -0400*Organization*: @Home Network*References*: <6vf3ta$dde@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Hi, FindRoot works quite well; of course, there are lots of sols and it depends on where you start. Some examples: 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,0}, {y,1},{z,0}] yields x= 0, y=1, z=0 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,0}] yields x = 0.666841,y = 0.204393, z=0 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,.2}, {y,.7},{z,0}] yields x=0.207116, y=0.615659,z=0 Alan Yeoung-Sang Yun wrote: > > Hello! > > 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) > > Thanks for helping. > Y.-S. Yun > Department of Chemical Engineering > Pohang University of Science and Technology San 31, Hyoja-dong, Pohang > 790-784, Republic of Korea