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