1
$\begingroup$

I has implicit function and square region. I need to check this region for presence of part of curve, with interval arithmetic only.

My attempt:

ClearAll["Global`*"];
f[a_Interval, b_Interval] := Sin[a + b] - Cos[a*b] + 1;
s = 1.1;(*Square size*)
cx = -3.9; (*Square center x*)
cy = -1.2; (*Square center y*)

x = Interval[{cx - s/2, cx + s/2}];
y = Interval[{cy - s/2, cy + s/2}];

IntervalMemberQ[f[x, y], 0]

Does this code are correctly solve the problem?

Update: I need to get a false if the curve is guaranteed not in the specified square(include inner area). But true value cannot guaranteed the opposite

Improve this question
$\endgroup$
5
  • $\begingroup$ BTW, AroundReplace[ Sin[a + b] - Cos[a*b] + 1, {a -> Around[-3.9, 0.55], b -> Around[-1.2, 0.55]}] performs a worser result Around[2,2.5]. $\endgroup$
    – user64494
    Nov 22, 2019 at 20:19
  • $\begingroup$ The implicit curve is given by f[a,b]==0? And you want to determine whether f[square1,square2] hits zero? $\endgroup$
    – Daniel Lichtblau
    Nov 23, 2019 at 16:17
  • $\begingroup$ The above code giving False is sufficient but not necessary condition for the square in question to contain no zero. Changing s to 2.0, for example, will give a result of True when in fact this larger square still contains no zero. $\endgroup$
    – Daniel Lichtblau
    Nov 24, 2019 at 16:55
  • $\begingroup$ @DanielLichtblau I need to get a false if the curve is guaranteed not in the specified square(include inner area). But true value cannot guaranteed the opposite. $\endgroup$
    – PavelDev
    Nov 24, 2019 at 17:02
  • 1
    $\begingroup$ A False is a guarantee of no intersection. So if that is all you require, the code is fine for that purpose. $\endgroup$
    – Daniel Lichtblau
    Nov 24, 2019 at 17:04

2 Answers 2

Reset to default
1
$\begingroup$

Likely correctly in view of

NMinimize[{Sin[a + b] - Cos[a*b] + 1, a >= -3.9 - 0.55&& a <= -3.9 + 0.55 && b >= -1.2 - 0.55 &&   
 b <= -1.2 + 0.55}, {a, b}]

{0.38735,{a->-4.45,b->-1.4593}}

Improve this answer
$\endgroup$
1
$\begingroup$

The function does not achieve a zero within the square (as is supported byNMinimize. 

reg = ImplicitRegion[Abs[x + 3.9] < 0.55 &&Abs[y + 1.2] < 0.55, {{x, -5, 0}, {y, -2, 0}}];
Show[ContourPlot[f[x, y] == 0, {x, -5, 0}, {y, -5, 0}],RegionPlot[reg]]

enter image description here

Or it can be seen by plotting region and plane z=0:

{m, w} = NMinimize[f[x, y], {x, y} \[Element] reg];
pnt = {Sequence[x, y], m} /. w;
Show[Plot3D[{f[x, y], 0}, {x, y} \[Element] reg, Mesh -> None], 
 Graphics3D[{Red, PointSize[0.02], Point[{-3.9, -1.2, f[-3.9, -1.2]}],
    Green, Point[pnt]}]]

enter image description here

Improve this answer
$\endgroup$
1
  • $\begingroup$ I updated my question, please take a look $\endgroup$
    – PavelDev
    Nov 24, 2019 at 17:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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