bio | website | maths.ox.ac.uk/~greenbj |
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location | Auckland, New Zealand | |
age | 38 | |
visits | member for | 4 years, 11 months |
seen | 19 hours ago | |
stats | profile views | 12,384 |
I'm a professor at Oxford University, on sabbatical in New Zealand until April 2014
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 |
Apr 24 |
awarded | Yearling |
Apr 6 |
awarded | Enlightened |
Apr 2 |
awarded | Nice Question |