I have a function $f(m,n)$ where m and n are positive integers. Suppose $f(m,n)$ is like a black box, I can get values of it when I enter values of m and n. I want to scan over a region for example $4<m<100, 4<n<1000$ to find where $f(m,n)<0$. Then make a plot of the region where it satisfies the requirement.

For the plot part, I think I can directly use the Listplot if I can successfully generate a parameter list for m and n where $f(m,n)<0$. For the first part, I could use the command For to generate the list satisfying the requirement. I guess this is not the most ideal command I want to use in Mathematica.

  • $\begingroup$ So if you have something like this with 9025 randomly positive and negative points would you be interested in defining positive/negative parametric regions, shown in two colors? $\endgroup$
    – Syed
    Commented May 10, 2022 at 15:57

1 Answer 1


Does RegionPlot work for what you need? It takes an inequality and a range as input and shows the regions in which that function is true. Below, I have included code that gives an example.

RegionPlot[Sin[x]^y - 10 < 0, {x, 0, 1000}, {y, 0, 1000}] displays plot goes here

  • $\begingroup$ Thanks for the help. I should be more precise with my question. My function is sort of like a black box. When I enter values of m and n, it returns me some value. $\endgroup$
    – Vayne
    Commented May 10, 2022 at 14:58
  • 1
    $\begingroup$ It's not perfect, but you could do something like this: f[x_, y_] := RandomReal[{0, x}] - RandomReal[{0, y}]*2 ListPlot@ Table[If[f[x, y] < 0, {x, y}, Nothing], {x, 0, 10, .1}, {y, 0, 10, .1}] $\endgroup$
    – Romanp
    Commented May 10, 2022 at 15:08
  • $\begingroup$ @Vayne You only have to substitude Sin[x]^y - 10 by your mystical f[x,y] in @Romanp 's nice answer and it should work! $\endgroup$ Commented May 10, 2022 at 15:26
  • $\begingroup$ That won't work if there's random number generation or something involved in f, but otherwise, should be fine. $\endgroup$
    – Romanp
    Commented May 10, 2022 at 15:26
  • $\begingroup$ @Romanp Clearly , but OP speaks of a function which returns a hopefully unique value! $\endgroup$ Commented May 10, 2022 at 15:30

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