bio | website | math.ubc.ca/~israel |
---|---|---|
location | Vancouver BC | |
age | 63 | |
visits | member for | 3 years, 1 months |
seen | 2 days ago | |
stats | profile views | 4,303 |
Associate Professor Emeritus, University of British Columbia, and Optimization Algorithms Researcher, D-Wave Systems, Burnaby BC
Apr 20 |
revised |
Distance between poisson points in two disjoint unit discs
added 726 characters in body |
Apr 20 |
revised |
Distance between poisson points in two disjoint unit discs
added 726 characters in body |
Apr 18 |
revised |
Distance between poisson points in two disjoint unit discs
added 50 characters in body |
Apr 18 |
comment |
Distance between poisson points in two disjoint unit discs
Oops, it looks like the anon user is correct. The product of two independent Poisson processes is not a Poisson process. |
Apr 10 |
answered | Random walk on a Penrose tiling |
Apr 9 |
comment |
$\aleph$ looks like $\mathbb N$?
Cantor may not have had much contact with Judaism, but I suspect that many of the Protestant theologians with whom he did have contact would have had a good working knowledge of Hebrew: certainly biblical Hebrew, but maybe also some of those esoteric texts. |
Apr 8 |
revised |
Homeomorphisms that admit a decomposition
deleted 39 characters in body |
Apr 7 |
comment |
Homeomorphisms that admit a decomposition
Homeomorphisms don't preserve null sets. For example, there is a homeomorphism of $[0,1]$ to itself that maps the usual Cantor set to a "fat" Cantor set of positive Lebesgue measure. Homeomorphisms that are bi-Lipschitz preserve null sets. |
Apr 7 |
revised |
Homeomorphisms that admit a decomposition
added 345 characters in body |
Apr 7 |
answered | Homeomorphisms that admit a decomposition |
Apr 6 |
comment |
Homeomorphisms that admit a decomposition
@plusepsilon.de: Functions with different numbers of fixed points can't be conjugates, and the number of fixed points could be any integer $\ge 1$. |
Apr 3 |
awarded | Nice Answer |
Mar 19 |
awarded | Nice Answer |
Mar 14 |
awarded | Yearling |
Mar 10 |
answered | Numerical calculation of Fourier transform with a nice error bound |
Feb 20 |
comment |
Integer Solutions of $x+y^n = y + x^m$ for $n < m$
These are all the solutions with $2 \le n < m \le 200$ and $2 \ge x,y \le 20000$. |
Feb 5 |
comment |
Exponential of a specific hypergeometric series
If $f$ has only positive coefficients, then so does $\exp(f)$. |
Jan 20 |
answered | Perturbations of positive-definite self-adjoint operators |
Jan 15 |
answered | Estimating the vector potential |
Dec 26 |
comment |
Relating the roots of polynomials to the solution sets of certain functional equations
@Suvrit: $f(x) = 1/x$ is not a solution of this functional equation for $n=1$. It is a solution for $n=2$ of the functional equation $f(f(x)) - x = 0$, for which the polynomial $z^2 - 1$ does have real roots. |