Say I have a simple function

F[x_, y_, z_, r_, a_, b_, c_, d_] := (1 - a + b - (b/3) (c x + d (y + z + r))) x

And I want to solve the set of equations given by:

x == F[x,y,z,r,a,b,c,d]
y == F[y,x,z,r,a,b,c,d]
z == F[z,x,y,r,a,b,c,d]
r == F[r,x,y,z,a,b,c,d]

The variables are x,y,z,r; a,b,c,d are parameters; and the only important position is the first of the 4 variables in each equation of the type x==F[x,y,z,r,...]

I figured I should solve this system by the command


However, Mathematica doesn't evaluate the Solve command, it simply prints back the command without giving it a try...

  • What am I missing?

  • Is there a simpler way to build the system of equations?

  • Is it possible to do for any number of variables? (here, I have 4 variables, ideally I'd want it working for n variables)

In math, the general system of equations would look like this:


and I want to solve for each component of $\vec{x}$


1 Answer 1


Are you sure you had defined F before trying Solve? Is the F in your notebook blue (undefined) or black (properly defined)? Does Solve return an error of any kind?

This is what I get:

F[x_, y_, z_, r_, a_, b_, c_, d_] := (1 - a + b - (b/3) (c x + d (y + z + r))) x

  {x == F[x, y, z, r, a, b, c, d], 
   y == F[y, x, z, r, a, b, c, d],
   z == F[z, y, x, r, a, b, c, d], 
   r == F[r, y, z, x, a, b, c, d]},
  {x, y, z, r}

enter image description here

  • $\begingroup$ I suspect F was defined as F[x,y,z,a,b,c,d] without the r $\endgroup$
    – Girardi
    Jul 8, 2020 at 17:03
  • 1
    $\begingroup$ @Girardi Yep that could be a reason as well. I tend to add a Clear before any function definition to make sure that old definitions don't linger etc. $\endgroup$
    – MarcoB
    Jul 8, 2020 at 17:04

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