bio | website | maths.ox.ac.uk/~greenbj |
---|---|---|
location | Auckland, New Zealand | |
age | 38 | |
visits | member for | 5 years, 2 months |
seen | 2 days ago | |
stats | profile views | 12,563 |
I'm a professor at Oxford University, on sabbatical in New Zealand until April 2014
Mar 7 |
comment |
On the $L^1$-norm of certain exponential sums.
I chatted with Tom Sanders about this today in Oxford, and I think we can more-or-less solve it, at least the second case. The key idea is to write |1 + e(t)| as 2^{1/2} (1 + cos(2 pi t))^{1/2}, then use the inequality (1 + x)^{1/2} \leq 1 + x/2 - cx^2, valid for some $c > 0 (in fact for c = 3/2 - 2^{1/2}). Now expand everything out, and you get a bound for your integral of 2^{n/2}(1 - c/4)^n if S is "good": has no relations with coefficients <= 2. A bit of fiddling should give exactly what you want, with your weaker assumption on S; in the powers of two case you can split S into two good sets |
Mar 6 |
comment |
On the $L^1$-norm of certain exponential sums.
Joel, this is a nice question. I haven't thought about it yet, but the first thing that comes to my mind (especially in connection with your second condition) is the paper of Mauduit and Rivat on binary digits of primes. They have to estimate the L^1 norm of the exponential sum of some Riesz products quite similar to yours, and they do beat the trivial Cauchy-Schwarz bound by an expontial factor. |
Jan 22 |
awarded | Good Answer |
Dec 16 |
awarded | Popular Question |
Oct 19 |
comment |
Showing non-expansion for $x\rightarrow x+1, x\rightarrow 2x.$
Harald: I think his name is Gonzalo Fiz-Pontiveros. |
Sep 24 |
comment |
Möbius Randomness of the Rudin-Shapiro Sequence
Terry is right. To get more than $o(n)$ cancellation one would need to show that more general bilinear sums involving the Rudin-Shapiro sequence are small, not a tempting task, but probably fairly necessary. I note that it would also be enough to show that a weird variant of the Gowers $U^3$-norm of $\mu$ is small -- not the usual Gowers norm $U^3[N]$, but the norm $U^3[F_2^n]$ in which M\"obius is considered as a function on the binary cube via its binary digits. Not a very natural thing to try and compute. |
Sep 16 |
comment |
There exists B subset A, |B| = log n, A \cap 2*B = \emptyset
A masters student from Lyon, Jehanne Dousse, recently improved [SSV] to log n logloglog n or thereabouts, by replacing the use of Szemeredi's theorem by a density increment argument generalising that of Roth, allowing one to locate inside a dense set A some elements x_1,...,x_k, all of whose midpoints also lie in A. The key is that these configurations can still be detected by Gowers' U2-norm. I have no clue what the correct bound is.... |
Jul 28 |
comment |
Growth rate of the infinity norm of Discrete Fourier Transform of +1,-1 vectors
The answer is basically $\sqrt{n}$ for large $n$.When $n$ is prime, an example should be given by the Legendre symbol $f_i = (i | n)$. |
Jun 18 |
awarded | Enlightened |
Jun 18 |
awarded | Nice Answer |
Jun 18 |
answered | Can you cover the Boolean cube {0,1}^n with O(1) Hamming-balls each of radius n/2-c*sqrt(n)? |
Jun 18 |
comment |
Can you cover the Boolean cube {0,1}^n with O(1) Hamming-balls each of radius n/2-c*sqrt(n)?
I agree it should probably be linear in n, but this might be rather hard to prove. Benny Sudakov was telling me that it is known (but difficult) that Hamming Balls of radius 1 cover with efficiency 1 + o(1), but this is an open problem for balls of radius 2. He may known about your question too; I'll ask him this afternoon. |
May 22 |
comment |
Nonlinear equations in integers
In fact there is no set of size $o(N)$ which misses all triples $(3k, 4k, 5k)$ by the following type of argument: consider such triples with $|k - N/10| < N/100$ (say). These are all disjoint. Therefore any set consisting of $98\%$ of ${1,..,N}$ contains a pythagorian triple. Computation of the exact density might be tricky. |
May 22 |
comment |
Nonlinear equations in integers
One of the references here, I bet: algo.inria.fr/csolve/triple |
May 22 |
comment |
Nonlinear equations in integers
I'm pretty sure, actually, that all sets of $99\%$ of $[1,...,N]$ contain a triple $(3k, 4k, 5k)$. I'm also pretty sure I can find a reference for this fact given a minute or two. |
May 22 |
comment |
Nonlinear equations in integers
Mark, for the Pell equation there is a huge set $A$, of size $N - O(\log N)$, just by deleting all $x$ and $y$ that are solutions to the Pell equation? |
May 22 |
comment |
Nonlinear equations in integers
Quid: what about $3^2 + 4^2 = 5^2$? |
May 22 |
comment |
Nonlinear equations in integers
Siming, I think that taking the squares $x^2$ with $x$ either odd or $\equiv 2 \mod{4}$ gives you a set consisting of $\frac{3}{4}$ of the squares with no solution to $x^2 + y^2 = z^2$ (look mod $8$). Probably it's true that if you take $99 \%$ of the squares then there's a solution to $x^2 + y^2 = z^2$. I don't immediately have a feel for whether this is doable or not; I'll get back to you. Certainly use of the circle method will be problematic if one proceeds naively. |
May 22 |
answered | Nonlinear equations in integers |
May 17 |
comment |
Gauss sums over multiplicative subgroups
Igor, Thanks for this mention. I did try pretty hard with that particular set of notes, though I might change one or two things if I were doing them again. I had some help from Bourgain and Lindenstrauss. |