8
$\begingroup$

I'd like to prove algorithmically the fact that the integral of $e^{-x^2 - y^2}$ evaluated over a circular disk, ${\cal D}$, is greater than over a square, ${\cal S}$, of the same area regardless of the (positive) area.

enter image description here

The basic comparative relation test would seem to be:

Integrate[Exp[-x^2 - y^2], {x, y} \[Element] Disk[{0,0}, Sqrt[area/ \[Pi]]]] > 
Integrate[Exp[-x^2 - y^2], {x, y} \[Element] Rectangle[{-Sqrt[area]/2,-Sqrt[area]/2}, {Sqrt[area]/2, Sqrt[area]/2}]]]

I want to get a True output, but the above does not give it.

I've tried all forms such as Assuming[area > 0, ....] and Assuming[area \[Element] PositiveReals] and Reduce and Resolve and Simplify and FullSimplify, Subscript[\[ForAll], area > 0] ... in every combination, without success. I've also tried GreaterEqualThan[][] and related forms, again without success.

I can get the desired result if I assign area to some specific value, but I'd like to prove the relation regardless of the particular value.

Improve this question
$\endgroup$
9
  • 6
    $\begingroup$ it comes down to showing that $\left(1-e^{-\frac{x}{\pi }}\right) -\text{erf}\left(\frac{\sqrt{x}}{2}\right)^2>0$ for all positive $x$. But Mathematica can't show this. Plotting this shows this is indeed the case. But Reduce can't do it. In the above $x$ represents the area. !Mathematica graphics the $erf$ is not easy to work with for reduce. may be you can ask in the math forum. $\endgroup$
    – Nasser
    Commented Feb 1, 2023 at 4:48
  • 1
    $\begingroup$ @Nasser there's a relation of the error function to hypergeometrics. Perhaps a help for Reduce? $\endgroup$
    – bmf
    Commented Feb 1, 2023 at 5:04
  • 1
    $\begingroup$ @Syed are you sure it was 7 minutes? because after ;// AbsoluteTiming I get {0.349316, Null} $\endgroup$
    – bmf
    Commented Feb 1, 2023 at 8:24
  • 1
    $\begingroup$ I am on v12.2 Win7-x64 and the screenshot shows that the timings are 0.66s for the first integral and 13.4min for the second one. Thanks for indicating @bmf. I made a mistake above as I was timing manually with a wall clock. $\endgroup$
    – Syed
    Commented Feb 1, 2023 at 9:23
  • 1
    $\begingroup$ Algorithmically it looks tricky. Geometrically, your picture already gives a strong hint (maybe you knew that though). Nice question (and upvoted) in any case. $\endgroup$
    – Daniel Lichtblau
    Commented Feb 1, 2023 at 16:54

0

Reset to default

Your Answer

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

Browse other questions tagged or ask your own question.