bio | website | maths.ox.ac.uk/~greenbj |
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location | Auckland, New Zealand | |
age | 38 | |
visits | member for | 5 years, 1 month |
seen | 2 days ago | |
stats | profile views | 12,505 |
I'm a professor at Oxford University, on sabbatical in New Zealand until April 2014
Nov 11 |
comment |
A question about groups of intermediate growth
When making my comment above I was addressing the question for "supremum" instead of "infimum". I feel one does need Gromov/Pansu to do that. |
Nov 10 |
comment |
A question about groups of intermediate growth
Nice question. My impression is that there is no simpler proof of 2. You can get some way beyond polynomial growth by quoting Shalom and Tao. |
Oct 25 |
comment |
Arithmetic Progressions of Squares
Kevin - I'd suggest asking one of them. I think there is a preprint. |
Oct 25 |
awarded | Enlightened |
Oct 24 |
awarded | Nice Answer |
Oct 24 |
revised |
Arithmetic Progressions of Squares
added 270 characters in body; added 310 characters in body |
Oct 24 |
answered | Arithmetic Progressions of Squares |
Oct 21 |
awarded | Enlightened |
Oct 20 |
awarded | Nice Answer |
Oct 20 |
comment |
Infimums of exponential sums involving primes
For a long time it was a conjecture of Littlewood that the infimum is $o(\sqrt{x})$. Actually it was Konyagin who got down to $x^{-1/2 + o(1)}$. |
Oct 20 |
revised |
Infimums of exponential sums involving primes
added 385 characters in body; deleted 2 characters in body |
Oct 20 |
answered | Infimums of exponential sums involving primes |
Oct 12 |
answered | Cliques, Paley graphs and quadratic residues |
Oct 4 |
comment |
density of a set
Why aren't they all dense? It's just a question of whether $(n^k/2\pi)$ is equidistributed modulo 1, and that will be so for any fixed k since $\pi$ is irrational. You'd use Weyl's inequality for a rigorous proof. |
Sep 27 |
comment |
Extension of Tao-Green Theorem
I'm certain $A_1$ does contain long progressions, but proving it using the Tao-Green technique might not be so easy. In fact this set has density roughly $1/\log^2 N$ in the integers, and any such set is conjectured to have arbitrarily long progressions. |
Sep 19 |
comment |
Elementary proof of the equidistribution theorem
This would basically be Weyl's proof, I think - the key idea there is to smooth the characteristic function of the interval [a,b] on which one wishes to count points n\alpha. The Fejer kernel (or de la Vallee Poussin kernel) is one way to do that. (I'd have to admit that I haven't bothered actually going to look at the book before making this comment.) |
May 16 |
comment |
Size of Sum Sets
It doesn't prove anything, but the answer is indeed no, for essentially the reason you say. It can be proven rigorously using the circle method a la Vinogradov. |
May 11 |
comment |
Lower bounds on the easier Waring problem
Boris, One would think so, because your set is contained in the difference set S of the set of things which are the sum of at most 5 positive kth powers, and there seems little reason to suspect that $S$ behaves so much unlike a random set with $n^{5/k}$ elements up to $n$. Proving it would be quite a different matter. Maybe some papers of Browning and Heath-Brown are relevant.... |
May 11 |
awarded | Nice Answer |
May 7 |
answered | Roth's theorem and Behrend's lower bound |