12,692 reputation
1234
bio website math.ubc.ca/~israel
location Vancouver BC
age 63
visits member for 3 years, 1 months
seen 2 days ago
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.