Comments on: EDP27 — the modular version of Roth’s AP-discrepancy theorem http://gowers.wordpress.com/2012/09/19/edp27-the-modular-version-of-roths-ap-discrepancy-theorem/ Mathematics related discussions Thu, 20 Jun 2013 01:03:30 +0000 hourly 1 http://wordpress.com/ By: Vedic Mathematics http://gowers.wordpress.com/2012/09/19/edp27-the-modular-version-of-roths-ap-discrepancy-theorem/#comment-31877 Tue, 25 Dec 2012 06:28:57 +0000 http://gowers.wordpress.com/?p=4528#comment-31877 Nice solutions is given by author.

]]> By: A.Czuron http://gowers.wordpress.com/2012/09/19/edp27-the-modular-version-of-roths-ap-discrepancy-theorem/#comment-31689 Thu, 06 Dec 2012 15:47:32 +0000 http://gowers.wordpress.com/?p=4528#comment-31689 Does HAP of length 1 are satisfactory?
For example if we have sequence x=(1,2,3,4) and we want to find r=4(mod 5), dose the answer is HAP= {x_4}?

]]> By: Liz http://gowers.wordpress.com/2012/09/19/edp27-the-modular-version-of-roths-ap-discrepancy-theorem/#comment-30057 Tue, 06 Nov 2012 19:07:10 +0000 http://gowers.wordpress.com/?p=4528#comment-30057 Hi there, just checking in with you, I don’t know if you were hit by super storm Sandy but I hope you are doing well. If you have a minute or so, I hope you check out our site, thefactortree.com, and provide us with some feedback, that would be greatly appreciated. Thanks. -L

]]> By: Liz http://gowers.wordpress.com/2012/09/19/edp27-the-modular-version-of-roths-ap-discrepancy-theorem/#comment-29134 Thu, 18 Oct 2012 12:46:22 +0000 http://gowers.wordpress.com/?p=4528#comment-29134 Hi there,

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]]> By: The Quantum Debate is Over! (and other Updates) | Combinatorics and more http://gowers.wordpress.com/2012/09/19/edp27-the-modular-version-of-roths-ap-discrepancy-theorem/#comment-28977 Mon, 15 Oct 2012 03:56:46 +0000 http://gowers.wordpress.com/?p=4528#comment-28977 [...] posts on the Erdos Discrepancy Problem over Gowers’s blog (Here is the link to the last one EDP27). Tim contributed a large number of comments and it was interesting to follow his line of thought. [...]

]]> By: Anthony Wang http://gowers.wordpress.com/2012/09/19/edp27-the-modular-version-of-roths-ap-discrepancy-theorem/#comment-28965 Mon, 15 Oct 2012 01:37:50 +0000 http://gowers.wordpress.com/?p=4528#comment-28965 This seems like something someone might have tried already, but I don’t see it on the general strategies list, so here goes.

The strategy is to rephrase the question in terms of power series. Given a \pm 1 valued sequence a_1,a_2,\cdots, we can associate a power series

p(z) = a_1z+a_2z^2+\cdots

which has radius of convergence 1.

The benefit of this approach is that we can use the following trick to remove the terms not divisible by some integer n: let \zeta=e^{2\pi i/n} be a primitive nth root of unity. Then the function

p_n(z)= \frac{1}{n} \sum_{i=0}^{n-1} p(\zeta^i z)

removes all a_i with i not divisible by n.

So the plan is to get a good understanding of which p(z) will have bounded partial sums when z=1, and then apply it to each of the functions p_n(z) to get a better understanding of the problem.

As an example, if we are going to restrict to complex numbers, we should probably expect that p_n(z) do not have any poles at 1. Then, we can check this for the sequence 1,-1,0,1,-1,0,… Its associated power series is

p(z)=z-z^2+z^4-z^5+\cdots = \frac{z(1-z)}{1-z^3} = \frac{z}{(z-\omega)(z-\omega^2)},

where \omega=e^{2\pi i/3}. Note that p_n(z) will clearly not have a pole when n is not divisible by 3. In the case when 3|n, it turns out that the poles ‘cancel’:

p(\omega z)+p(\omega^2 z)=\frac{z(\omega+1)}{\omega(z-\omega)(\omega z-1)},

where we ended up taking out a z-1 from both the top and the bottom. We can then conclude that p_n(z) will also not have any poles when n is divisible by 3.

(Note: I am actually an undergrad at MIT taking complex analysis right now, so I don’t understand much of what you guys say (yet! hopefully I will someday). Sorry if you already knew this or it doesn’t help).

]]> By: Gil Kalai http://gowers.wordpress.com/2012/09/19/edp27-the-modular-version-of-roths-ap-discrepancy-theorem/#comment-26120 Thu, 20 Sep 2012 01:15:07 +0000 http://gowers.wordpress.com/?p=4528#comment-26120 Maybe, if f is not quasirandom then f is even less quasirandom on an AP?

]]> By: gowers http://gowers.wordpress.com/2012/09/19/edp27-the-modular-version-of-roths-ap-discrepancy-theorem/#comment-26090 Wed, 19 Sep 2012 22:23:16 +0000 http://gowers.wordpress.com/?p=4528#comment-26090 A thought about the last example (in point 10 in the post) that might perhaps be the basis for an appropriate definition of quasirandomness is that the example is set up in such a way that the sum \sum_{x\in Q}f(x) is zero for many short progressions Q. This helps to ensure that many longer progressions give rise to the same sums. For different reasons, many sums over short APs are zero in the example with lots of 1s and lots of -1s.

So perhaps what we want is not a quasirandomness property, but simply a property that says in some way or other that there are not too many short APs along which the sum is zero. It would still be helpful for this property to be as weak as possible, so that if it fails we can get as much structural information as possible about the function f.

Note that the proportion of APs (of all lengths) along which the sum is zero is at least p^{-1}. To see this, consider APs of common difference 1. For each a let n(a) be the number of m such that \sum_{x=0}^{m-1}f(x)=a. Then the number of pairs (s,t) such that the sums \sum_{x=0}^{s-1}f(x) and \sum_{x=0}^{t-1}f(x) are both equal to a is n(a)^2. Therefore, the number of sums \sum_{x=s}^{t-1}f(x) equal to zero is at least \sum_an(a)^2. Since \sum_an(a)=n and there are p different values of a, \sum_an(a)^2\geq n^2/p, which is a proportion 1/p of all APs mod n of common difference 1. Then one can do the same for the other common differences.

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