Archive for November, 2010

Random update

November 23, 2010

The reason for the fact that is has been more than 2 weeks since my last update is simple; contrary to popular belief, I have been busy. With a few things actually. I have been a bit busy studying (who would’ve thought). I have been busy writing my first article (thank you very much), which will be finished end of the week, hopefully. I have been busy playing guitar (obviously, I suck). And, last but not least, I have been juggling like there was no tomorrow (again, I suck). When I’ll improve, I might put some videos of me juggling here. That ought to be fun ūüôā My personal bests are, as of this moment:

3 balls: 12 minutes (close to 2500 catches)
4 balls: 57 seconds (over 200 catches)
5 balls: 15 catches
6 balls: 6 catches
7+ balls: err, no

I’ll update you on all of these things as soon as I’ve got some news.

EDIT 25 november: Woohoo! I flashed 6 balls ūüėÄ
EDIT 30 november: I updated a few personal bests. As of tomorrow I won’t do that anymore to be able to track my progress. But right now I’m too happy with my clean run of 15 ūüėÄ



November 5, 2010

A variant on a well-known poem;

The title (basically) said it and I again will say
I’m in my prime; today it’s my 21st¬†birthday!

Erd√∂sproblem 3, part III

November 4, 2010

So far I found about 10 more, what I like to call, NP-primitive counterexamples to the question¬†asked here. When do I call a counterexample¬†NP-primitive? When it’s easy to show (by the reasoning from the previous post) that¬†it’s indeed a¬†counterexample and if¬†no (strict) subset of¬†it is a NP-primitive¬†counterexample.¬†And given a NP-primitive counterexample, it’s easy to construct other easy to check counterexamples. The question now is of course: are there infinitely many NP-primitive counterexamples? Conceivably, this is not too hard. The¬†ones I found so far:


And you can even make more by e.g. replacing 60 in the fifth to last set with 120, 140 and 840. A more general, and to me more sensible, conjecture of Erdös, Graham and Spencer is the following:

If \displaystyle \sum_{i=1}^k \dfrac{1}{x_i} < N - \dfrac{1}{m}, then the x_i can be split into N subsets
A_1, A_2, .., A_n such that for all j with 1 \le j \le N we have: \displaystyle \sum_{a_i \in A_j} \dfrac{1}{a_i} \le 1, as long as m \le 30.

This is known for m = \dfrac{7}{2} (see here) and \{2,3,3,5,5,5,5\} shows that m = 30 is tight. Very probably this conjecture is cute enough to deserve a closer look at sometime.

EDIT: A counterexample to this last conjecture was found by Song Guo (Electronic J. of Combinatorics 15 (2008), #N43). You can find it online, if you’re interested. That paper also refers to other articles about these conjectures. The main question of finding the maximal m is still open, in any case

Erd√∂sproblem 3, part II

November 3, 2010

I checked my dictionary, but apparently there is no english word for the dutch word ‘tangconstructie’. Anyway, remember how I, rhetorically, asked ‘how else could it be?’ yesterday? Well,¬†I am more than delighted to say that I am able¬†to give a counterexample to the metaconjecture that every proof of a counterexample to the question of Erd√∂s and Graham I talked about in the previous post, works¬†via a case-by-case-argument.

Consider the set X = \{2,3,4,5,6,8,9,10,12,15,16\} and let lcm\{S\} be the least common multiple of all members of S, for a set S of positive integers. Then \displaystyle \sum_{x_i \in X} \dfrac{1}{x_i} = 2 - \dfrac{1}{lcm\{X\}}. Now, assume we have a partition of X in two sets A and B with \displaystyle \sum_{a_i \in A} \dfrac{1}{a_i} < 1 and \displaystyle \sum_{b_i \in B} \dfrac{1}{b_i} < 1. Then in particular we have:

2 - \dfrac{1}{lcm\{X\}}
= \displaystyle \sum_{x_i \in X} \dfrac{1}{x_i}
= \displaystyle \sum_{a_i \in A} \dfrac{1}{a_i} + \displaystyle \sum_{b_i \in B} \dfrac{1}{b_i}
\le 1 - \dfrac{1}{lcm\{A\}} + 1 - \dfrac{1}{lcm\{B\}}
\le 2 - \dfrac{2}{lcm\{X\}}

And boom goes the dynamite.

Solving a third Erd√∂sproblem (?)

November 2, 2010

In their monograph on problems in combinatorial number theory, Erdös and Graham ask the following question:

Is it true that if 1 < x_1 < x_2 < .. < x_k and \displaystyle \sum_{i=1}^k \dfrac{1}{x_i} < 2 then there exist \epsilon_i = 0 or 1 so that \displaystyle \sum_{i=1}^k \dfrac{\epsilon_i}{x_i} < 1 and \displaystyle \sum_{i=1}^k \dfrac{1-\epsilon_i}{x_i} < 1?

And I can’t find a way to do this where the x_i‘s are \{2,3,4,5,6,8,9,11,12,13,16\}. And, if this indeed is a counterexample, I’m bold enough to conjecture that it’s the easiest. \{2,3,4,5,6,7,8,11,12,13,33\} could very well be a candidate too, but I haven’t checked that one thoroughly.

Edit: I now proved that the first one of these is indeed a counterexample, but of course, the proof is¬†a case-by-case-argument (how else could it be?), so I won’t put it here, because¬†of uglyness.

Weighed cointossing

November 2, 2010

Assume S = \{s_1, s_2, .. \}¬†is a set with 1 \le¬†s_1¬†\le s_2 \le .. ,¬†such that \displaystyle \sum \dfrac{1}{s_i} diverges.¬† Let a_i = \pm 1, each with probability \dfrac{1}{2}. What can we say about \displaystyle \sum_{i=1}^n \dfrac{a_i}{s_i} when n goes to infinity? Must there, for example,¬†be infinitely many sign changes with probability 1? It can’t be hard to show this for bounded s_i, but what about S = \aleph¬†for example? I honestly have no clue whatsoever.

EDIT 22 Februari 2011
I can show that there is a positive probability that \left|\displaystyle \sum_{i=1}^n \dfrac{a_i}{s_i}\right| > c for any c, for n large enough; Let k = k(c) be the smallest integer such that \displaystyle \sum_{i=1}^k \dfrac{1}{s_i} > c (note that this k exists since the sum of \dfrac{1}{s_i} is assumed to go to infinity).
Then either the probability that \displaystyle \sum_{i=k+1}^n \dfrac{a_i}{s_i} \ge 0 is at least \dfrac{1}{2} or the probability that \displaystyle \sum_{i=k+1}^n \dfrac{a_i}{s_i} \le 0 is at least \dfrac{1}{2} (*). Assume the first case (second case goes analogous), then we have: P\left(\displaystyle \sum_{i=1}^n \dfrac{a_i}{s_i} > c\right) \ge P\left(\displaystyle \sum_{i=k+1}^n \dfrac{a_i}{s_i} \ge 0\right) \displaystyle \prod_{j=1}^k P(a_j = 1) \ge \dfrac{1}{2}\dfrac{1}{2^k} = \dfrac{1}{2^{k+1}} > 0

(*) It is fun to note that both are the case, independent of S. This can be used to show that P\left(\left|\displaystyle \sum_{i=1}^n \dfrac{a_i}{s_i}\right| > c\right) \ge \dfrac{1}{2^k} for all n \ge k