2
$\begingroup$

I try to plot such a thing:

 f[x_, y_, z_] := Resolve[
                    Exists[i, i ∈ Integers && i > 0 && i < 10 && i^2 + x + y + z < 2]
                  ]

RegionPlot3D[f[x, y, z], {x, -1, 1}, {y, -1, 1}, {z, -1, 1}]

But I get an error: "...must be a Boolean function". (I know that this function is very stupid, it should only acts as an example).

So even if the output of the function f[x,y,z] is True of False, Mathematica doesn't seem to recognize it as a Boolean function.

If I change the function to something which doesn't depend on x,y,z (only on i), then everything is fine.

Improve this question
$\endgroup$
  • 3
    $\begingroup$ It will work if you define your function with: f[x__?NumericQ] := Resolve[Exists[i, i \[Element] Integers && i > 0 && i < 10 && i^2 + Plus[x] < 2]] $\endgroup$ – Kuba♦ Apr 21 '16 at 10:23
  • $\begingroup$ Yes, thanks, now it works! $\endgroup$ – Agnieszka Apr 21 '16 at 10:52
  • $\begingroup$ I think this answer covers that problem, what do you think? mathematica.stackexchange.com/a/26037/5478 $\endgroup$ – Kuba♦ Apr 21 '16 at 11:37
  • 1
    $\begingroup$ @Kuba why not answer anyway? The question title is quite clear and much simpler to find than the pitfalls thread. Personally, I'd not vote to close. $\endgroup$ – Yves Klett Apr 21 '16 at 11:57
3
$\begingroup$

As mentioned in a liked topic and topics linked to it, plotting functions often do symbolic preprocessing while your function gives True/False only for numerical arguments.

Let's assure that only such will be provided:

ClearAll[f];
f[x__?NumericQ] := Resolve[
  Exists[i, i ∈ Integers && i > 0 && i < 10 && i^2 + Plus[x] < 2]
]

enter image description here

Improve this answer
$\endgroup$
  • $\begingroup$ Do you also get a bunch of ratnz errors with this? I know it produces the requested plot, but wondering how to do it without errors (and without Quiet) $\endgroup$ – Jason B. Apr 22 '16 at 9:51
  • $\begingroup$ @JasonB Yes I do, I wasn't paying attention but that's probably a different question. And I don't know how to easily fix that :) $\endgroup$ – Kuba♦ Apr 22 '16 at 10:20
2
$\begingroup$

This particular set of inequalities can also be done as follows (note the regions are slightly different to @RunnyKine. I think this is just the differences in limits):

reg = And @@ (# @@ +x + y + z < 2 & /@ Range[1, 9]);
rp1 = RegionPlot3D[reg, {x, -1, 1}, {y, -1, 1}, {z, -1, 1}];
ir = ImplicitRegion[reg, {{x, -1, 1}, {y, -1, 1}, {z, -1, 1}}];
rp2 = RegionPlot3D[ir];
vol = Volume[ir];
Row[{rp1, rp2, Row[{"Volume= ", vol}]}]

enter image description here

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.