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MathGroup Archive 1998

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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



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