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Re: Methods for plotting

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  • Subject: [mg86309] Re: Methods for plotting
  • From: "Steve Luttrell" <steve at>
  • Date: Sat, 8 Mar 2008 05:40:40 -0500 (EST)
  • References: <fqqs4b$kgg$>

Your expression is not written in Mathematica. I assume the left hand side 
of your equation is meant to be

lhs=Exp[-x]/((x^2 + 1) (y^2 + 2) (z^2 + 1)) +
 Exp[-2 y]/((x^2 + 2) (y^2 + 2) (z^2 + 1)) +
 Exp[-3 z]/((x^2 + 1) (y^2 + 2) (z^2 + 3))

Assuming x,y,z are all real-valued, then each of the 3 terms in this 
expression is positive, so lhs can never be zero, so there is no contour 
surface lhs==0 for you to plot. An exception is when x,y,z all tend to 
+Infinity which makes lhs-->0.

Stephen Luttrell
West Malvern, UK

"Paolo" <paolomarracino at> wrote in message 
news:fqqs4b$kgg$1 at
> Hi, I'm new in this forum and I hope you'll help me to solve a little 
> problem with Mathematica(5.0), that I just started to use. I have an 
> equation (coming from the divergence of a vector) and it's form is similar 
> to:
> [exp(-x)]/[(x^2+1)(y^2+2)(z^2+1)]+
> [exp(-2y)]/[(x^2+2)(y^2+2)(z^2+1)]+
> [exp(-3z)/[(x^2+1)(y^2+2)(z^2+3)]=0
> I have difficulties to plot this implicit function, i used Plot3d and 
> ImplicitPlot with no success and the error message that I get involves the 
> non-algebraic way the programm uses to solve the equation.
> Is there any other Plot-type function that can possibly solve this 
> problem?

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