Reputation
1,102
Next privilege 2,000 Rep.
Edit questions and answers
Badges
7 21
Newest
 Yearling
Impact
~20k people reached

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