x == x^1.2/(2*x^3+y^0.7+4*z^2.5)

*To*: mathgroup at smc.vnet.net*Subject*: [mg14272] x == x^1.2/(2*x^3+y^0.7+4*z^2.5)*From*: wself at viking.emcmt.edu (Will Self)*Date*: Mon, 12 Oct 1998 13:51:51 -0400*Sender*: owner-wri-mathgroup at wolfram.com

Sorry, I lost the original message so I am sending this to the whole group-- Someone wanted to solve the equations 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) Here is my reply: The form of your equations suggests trying something like this: f[{x_,y_,z_}]:={x^1.2/(2*x^3+y^0.7+4*z^2.5), y^0.7/(2*x^0.6+y^2+z^2.2), 0.9*z^1.5/(x^0.7+2*y^0.2+z^1.1)} FixedPoint[f,{1,1,1}, 50] You could raise that 50 on a faster machine than the one I have. I get a solution of approximately {0.666839, 0.204396, 0} You could try other starting points and do the same thing. FixedPoint[f,{0,0,1}, 50] gives {0, 0, 0.838953}. FixedPoint[f,{0, 1, 0}, 50] gives {0, 1, 0}. FixedPoint[f,{1, 0, 0}, 50] gives a computational underflow. If this is from a real situation then your intuition should suggest where to look for a solution. There is also a three-dimensional version of Newton's method that you could try if this doesn't seem to be working. Let me know if you want anything on that. Best wishes, Will Self