David Čepelík
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 Feb 6 comment Solving $x^2 \equiv -x\pmod{2015}$ Sure, I am doing that :). It's easy to find solutions to these individual subproblems, but the question is, how do I find the intersection of those solutions? Feb 6 comment Solving $x^2 \equiv -x\pmod{2015}$ @DietrichBurde Great, Chinese remainder theorem will give me range in which I can found the unique $x$ for which these properties hold; but how do I actually find it? (Sorry to be a nob.) For each case (modulo $5$, $13$, $31$), I still have two options: either it divides $x$ or $x + 1$. Feb 6 comment Length of the Union of Intervals is less than the Sum of Each Length of Intervals? Sure, it does, except your previous version of the comment read $\dots \le \dots$, so I assumed it was a syntax error. Feb 6 comment Length of the Union of Intervals is less than the Sum of Each Length of Intervals? Yes, but the end should read "Since each ${\mathcal l}(E_n) \le {\mathcal l}(I_n)$", right? Feb 6 comment Length of the Union of Intervals is less than the Sum of Each Length of Intervals? Yes, but for two open intervals $A$ and $B$ such that $A \subset B$, this should be simple, right? Feb 6 asked Solving $x^2 \equiv -x\pmod{2015}$ Feb 6 revised Length of the Union of Intervals is less than the Sum of Each Length of Intervals? added 291 characters in body Feb 6 answered Length of the Union of Intervals is less than the Sum of Each Length of Intervals? Jan 26 accepted Global extrema of $(x^2 + y^2)e^{-(x^2 + y^2)}$ Jan 26 comment Global extrema of $(x^2 + y^2)e^{-(x^2 + y^2)}$ Great, thanks again! As I was looking for (a) confirmation of my argument and (b) a notion of general approach I should take when dealing with such problems, I am accepting your answer. Thanks for everyone's help! Jan 26 comment Global extrema of $(x^2 + y^2)e^{-(x^2 + y^2)}$ Thanks for reply. Is my approach a standard one, or is there some other approach that may be applied to such problems (proving local extrema points are global) in general, such as the one you propose? Jan 26 awarded Yearling Jan 26 comment Global extrema of $(x^2 + y^2)e^{-(x^2 + y^2)}$ Now this is clever! I always enjoy solutions that build the problem from known facts. Thanks a lot. Jan 26 asked Global extrema of $(x^2 + y^2)e^{-(x^2 + y^2)}$ Jan 24 accepted Is the set $M=\{[x,y,z]\in{\mathbf R}^3 :\ x^2 + y^2 +z^2 + xy + yz + xz = 1,\ x \ge 0,\ y \ge 0\}$ compact? Jan 24 comment Is the set $M=\{[x,y,z]\in{\mathbf R}^3 :\ x^2 + y^2 +z^2 + xy + yz + xz = 1,\ x \ge 0,\ y \ge 0\}$ compact? I like this approach best. Thanks! Jan 24 answered Is the set $M=\{[x,y,z]\in{\mathbf R}^3 :\ x^2 + y^2 +z^2 + xy + yz + xz = 1,\ x \ge 0,\ y \ge 0\}$ compact? Jan 24 comment Is the set $M=\{[x,y,z]\in{\mathbf R}^3 :\ x^2 + y^2 +z^2 + xy + yz + xz = 1,\ x \ge 0,\ y \ge 0\}$ compact? Thanks for insight! You do algebra, right? I could tell! Jan 24 comment Is the set $M=\{[x,y,z]\in{\mathbf R}^3 :\ x^2 + y^2 +z^2 + xy + yz + xz = 1,\ x \ge 0,\ y \ge 0\}$ compact? @Piquito Thanks, but I think I found a simple answer; the part of the set for $z \le 0$ is point reflection of the part for $z \ge 0$, which fits into the unit ball. Therefore, the set as a whole is bounded and fits a unit ball. Jan 24 revised Is the set $M=\{[x,y,z]\in{\mathbf R}^3 :\ x^2 + y^2 +z^2 + xy + yz + xz = 1,\ x \ge 0,\ y \ge 0\}$ compact? Fixed the set