Is there any sense in which Dirichlet density is "optimal?" - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T15:41:51Z http://mathoverflow.net/feeds/question/44512 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/44512/is-there-any-sense-in-which-dirichlet-density-is-optimal Is there any sense in which Dirichlet density is "optimal?" JSE 2010-11-02T02:46:38Z 2010-11-05T13:10:49Z <p>A philosopher asked me an interesting math question today! We know that there are sets S of integers which don't have a "natural" or "naive" density -- that is, the quantity (1/n)|S intersect [1..n]| doesn't approach a limit. The "analytic" or "Dirichlet" density exists whenever the naive density does, and is equal to it -- but sometimes is well-defined even when the naive density is not, for instance when S is the "Benford set" of integers whose first digit is 1. (See <a href="http://mathoverflow.net/questions/9143/analytic-density-of-the-set-of-primes-starting-with-1" rel="nofollow">this related MO question</a>.)</p> <p>Anyway, the philosopher asked: why stop at Dirichlet density? Is there a sequence of probability measures p_1, p_2, p_3, .... on the natural numbers such that</p> <ul> <li><p>Whenever S has a Dirichlet density, the limit of p_i(S) exists and is equal to the Dirichlet density, but</p></li> <li><p>there are subsets of the natural numbers such that p_i(S) approaches a limit but S has no Dirichlet density?</p></li> </ul> <p>(Possibly clarifying addition: to obtain naive density, one can take p_i to be the measure assigning probability 1/i to each integer in [1..i] and 0 to integers greater than i. To get Dirichlet density, take p_i to be the measure with p_i(n) = 1/(n^{1+1/i}zeta(1+1/i)). In either case, the corresponding density of S is the limit in i of p_i(S), whenever this limit exists. So what I have in mind is densities which can be thought of as limits of probability measures, though perhaps there are reasons to have yet more general entities in mind?)</p> <p>If so, are there any examples which are interesting or which are used for anything in practice?</p> http://mathoverflow.net/questions/44512/is-there-any-sense-in-which-dirichlet-density-is-optimal/44548#44548 Answer by Simon Lyons for Is there any sense in which Dirichlet density is "optimal?" Simon Lyons 2010-11-02T11:31:46Z 2010-11-02T11:31:46Z <p>Tenenbaum's book on analytic and probabilistic number theory has something to say on this subject. He mentions that it is "easy" to construct a set without Dirichlet density, but unfortunately he doesn't provide such a construction.</p> <p>A good place to start is with the logarithmic density: Tenenbaum shows that the Dirichlet density exists if and only if</p> <p>$\lim_{n \rightarrow \infty} \frac{1}{\log n} \sum_{n \in S} \frac{1}{n}$</p> <p>exists.</p> <p>The reason naive density fails to exist for the Benford set (call it $B$) is that oscillations between elements in $B$ and elements not in $B$ grow exponentially with $n$. I would conjecture that a set with superexponential gaps does not have a Dirichlet density - for example the set of integers $n$ where the number of digits of $n$ begins with a 1. One might have to sum over a series like $e^{-\log(\log(n))}$ to get convergence.</p> http://mathoverflow.net/questions/44512/is-there-any-sense-in-which-dirichlet-density-is-optimal/44590#44590 Answer by Mark Sapir for Is there any sense in which Dirichlet density is "optimal?" Mark Sapir 2010-11-02T17:26:01Z 2010-11-05T13:10:49Z <p>The best way to impress a philosopher is to tell him/her about ultrafilters. A (non-principle) filter on $\mathbb N$ is a set of infinite subsets of $\mathbb N$ closed under intersections and taking super-sets. A maximal filter (under inclusion) is called an ultrafilter. There are plenty of those but nobody saw them since their existence depends on the axiom of choice. For every filter $\omega$ one can define the concept of convergence of sequences of real numbers: a sequence $b_n$ converges to $b$ if for every $\epsilon$ the set of $i$'s such that $|b-b_i|\le \epsilon$ is in $\omega$. If $\omega$ is an ultrafilter, then every bounded sequence of real numbers has unique $\omega$-limit. It is not true if $\omega$ is not an ultrafilter. The smallest interesting filter (called Fréchet filter) consists of all sets with finite complements. The limit corresponding to that filter is the ordinary limit studied in Calculus. You can start from the Fréchet filter and add sets to it to produce bigger and bigger filters. Each filter gives you a Dirichlet-like density. If $\omega$ is an ultrafilter, then all sets will have density (between 0 and 1). Otherwise there will always be sets without density assigned. </p>