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Apr
14 |
comment |
Is a one-dimensional unstable manifold of an ODE a union of the associated equilibrium point and two full orbits?
It's possible for the stable and unstable manifolds to intersect, in what's called a "homoclinic point". This would violate #2, if I understand what you mean. |
Jan
6 |
reviewed | Approve Applications of forcing in model theory |
Nov
2 |
awarded | Civic Duty |
Oct
27 |
awarded | Yearling |
Oct
18 |
reviewed | Approve Linear transformation that preserves the determinant |
Sep
27 |
reviewed | Approve A family of convex bodies in Banach-Mazur position |
Sep
18 |
reviewed | Approve A problem of Keisler and Tarski |
Sep
3 |
comment |
Which polynomial's roots are its coefficients?
Lubin's suggestion gives a continuous map, which is nice. The process outlined here can't be made continuous (e.g. to define it for $z^2 + e^{i\theta}$ you essentially have to choose branches of the square root function for the coefficients). On its own, that seems unnatural, but it makes sense as picking a branch of the inverse of Lubin's map. Igor's idea of looking at all permutations then corresponds to taking the full inverse. |
Sep
1 |
awarded | Guru |
Aug
30 |
comment |
Is $\mathbb{R}^3 \setminus \mathbb{Q}^3$ simply connected?
Is it obvious that transversality arguments work for weird non-differentiable curves? In a manifold you know any curve is homotopic to a differentiable one, but showing that fact in $\mathbb{R}^3 \setminus \mathbb{Q}^3$ doesn't seem any easier than the original question. |
Aug
30 |
awarded | Good Answer |
Aug
30 |
awarded | Enlightened |
Aug
29 |
comment |
Is $\mathbb{R}^3 \setminus \mathbb{Q}^3$ simply connected?
@Victor The argument takes for granted that the ambient space stays simply connected (even connected) when you remove just a finite set of points--fine in $\mathbb{R}^3$ but not $\mathbb{R}$ or $\mathbb{R}^2$. |
Aug
29 |
awarded | Nice Answer |
Aug
29 |
answered | Is $\mathbb{R}^3 \setminus \mathbb{Q}^3$ simply connected? |
May
26 |
comment |
John Nash's Mathematical Legacy
It's amazingly prescient, from a computer science perspective. Here's a link to the paper: rand.org/content/dam/rand/pubs/research_memoranda/2008/… |
Apr
15 |
awarded | Nice Answer |
Apr
9 |
answered | Deep Learning / Deep neural nets for mathematician |
Mar
11 |
comment |
understanding the average height of a unit hyper-semisphere
On a high-dimensional sphere most of the mass is concentrated around the equator, where the height is lowest. That's a heuristic explanation for why the average height goes to zero. |
Feb
22 |
reviewed | Approve Is the Ford disk packing optimal? |