bio | website | |
---|---|---|
location | University of California, Berkeley | |
age | 68 | |
visits | member for | 5 years, 5 months |
seen | Mar 24 '14 at 1:18 | |
stats | profile views | 1,095 |
Aug
16 |
comment |
How well can we localize the “exoticness” in exotic R^4?
Removing a standard $D^4$ from a potentially-exotic smooth $S^4$ yields a potentially-exotic $\mathbb{R}^4$ that's standard at infinity. So if it were known that the latter must be globally standard, then replacing the $D^4$ would imply the original $S^4$ is standard, and hence the 4-dimensional smooth Poincaré conjecture. And there's essentially only one way to replace the $D^4$, since Gamma_4 = 0 (Cerf) implies oriented diffeos of $S^3$ are smoothly isotopic. |
Aug
15 |
comment |
Measure on real Grassmannians
For one application, an explicit curve {C(t) : t in [0,oo)}, dense in a Grassmannian of 2-planes in n-space, is the basis for the animation technique in statistical computer graphics known as the Grand Tour. It's important to ensure that as t -> oo, the curve C spends time in any open set U proportional to the invariant measure* of U. * Though the invariant measure on a Grassmannian is unique up to a scalar multiple, the invariant metric is not in the sole case of 2-planes in 4-space. This oriented Grassmannian's metric is the product of two round 2-spheres whose radii may be in any ratio. |
Aug
15 |
revised |
Widely accepted mathematical results that were later shown wrong?
deleted redundant word |
Aug
14 |
answered | Widely accepted mathematical results that were later shown wrong? |
Aug
10 |
comment |
Topological spaces that resemble the space of irrationals
Hello, Ethan. Yes, indeed -- as you may recall, I proved that (as well as an n-dimensional version) in a class of yours on PL topology around 1970. |
Aug
7 |
comment |
Topological spaces that resemble the space of irrationals
It's also amusing that given n, the complement of any countable dense subset of R^n is homeomorphic to the complement of any other such. |
Jul
30 |
asked | Topological spaces that resemble the space of irrationals |
Jul
16 |
comment |
Infinite games: are they well defined?
Oops, that should have read, "There must be a first countable ordinal beta at which there are no unclaimed members of J; the player whose turn it is at beta is defined as the winner. |
Jul
16 |
comment |
Infinite games: are they well defined?
(Cont'd.) Assume "whose turn it is" has been defined for all countable limit ordinals, and hence by alternation for all countable ordinals. Then it's not even necessary to predefine the length of the game. A simple example is this: let J be any countably infinite set. Each player on their turn claims some member of J. There must be a first countable ordinal at which the last member of J is claimed: the player who did that is defined as the winner. (I hope to expand on this soon in a short article.) |
Jul
16 |
comment |
Infinite games: are they well defined?
Incidentally, there's no reason that the plays of an infinite ordinal game must be indexed by omega. Any infinite ordinal lambda will do, though let's just be concerned about countable ordinals here. Special care must be taken to know whose turn it is at each play (whose index is some ordinal < lambda). This is easily accomplished in a "fair" way as long as whose turn it is is defined for limit ordinals; this is not difficult. (Successor ordinals just follow the alternating turn rule, as for the case lambda = omega.) |
Jul
15 |
comment |
Google question: In a country in which people only want boys
"... larger populations have more girls ..." No, since the population size is meaningful only if something is known about the number of families. In order to use the Strong Law of Large Numbers we assume infinitely many families #1,#2,#3,... with each one's sequence of births (G^n)B being concatenated in order of family #. Then the stochastic process generating this infinite sequence of G's and B's is isomorphic (up to measure 0) with repeated flips of a fair coin. Hence the SLLN implies the asymptotic fraction of B's or G's is 1/2, each with probability = 1. This is, to me, conclusive. |
Jul
8 |
comment |
Google question: In a country in which people only want boys
1. The words "a social convention cannot override biology" (not mine) mean just that the ultimate proportions of boys and girls are the same as the proportions in which boys and girls are born. 2. The stopping rule (in question) is a red herring because any stopping rule of the form "Stop as soon as a certain consecutive string of B's and G's occurs" will result in the same ratio of 1:1 (or more generally p:q) as the probabilities of B vs. G (or H vs. T) are in. 3. You are right that the 2:1 stopping rule is not almost certain to occur. Oops. 4. What modification are you thinking of? |
Jul
8 |
revised |
Google question: In a country in which people only want boys
Removed false statement. |
Jul
8 |
revised |
Google question: In a country in which people only want boys
Mentioned stopping condition for which the previous reasoning fails |
Jul
8 |
answered | Google question: In a country in which people only want boys |
Jul
6 |
comment |
Google question: In a country in which people only want boys
Zare falls into exactly the trap I mention in my first paragraph. |
Jul
5 |
revised |
Google question: In a country in which people only want boys
added comment about Google's phrasing of the problem |
Jul
5 |
revised |
Google question: In a country in which people only want boys
corrected grammar |
Jul
5 |
answered | Google question: In a country in which people only want boys |
Jul
5 |
answered | Which mathematicians have influenced you the most? |