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MathGroup Archive 1998

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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


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