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