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Apr 13, 2017 at 12:58 history edited CommunityBot
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Nov 16, 2010 at 1:57 vote accept JSE
Nov 2, 2010 at 17:26 answer added user6976 timeline score: 15
Nov 2, 2010 at 16:18 comment added Franz Lemmermeyer Dirichlet density was not chosen because it had any nice properties, but because density results usually are proved with the help of zeta functions; the fact that the behaviour of such functions in the vicinity of poles is closely connected to densities of primes made Marcus, in his excellent textbook "Number Fields", introduce the name "polar density". Introducing weaker notions of density would make perfect sense if there were methods of proof which would produce results related to different notions of density.
Nov 2, 2010 at 13:59 comment added JSE I have done so.
Nov 2, 2010 at 13:59 history edited JSE CC BY-SA 2.5
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Nov 2, 2010 at 11:31 answer added Simon Lyons timeline score: 10
Nov 2, 2010 at 6:20 comment added KConrad In the new part, where you say (for Dir. density) that p_i is the prob. measure on Z "assigning C(n^{-1-1/i}) to i, where 1/C = zeta(1+1/i)" you meant that number to be the value p_i assigns to n, not to i. Why not write that p_i(n) = 1/(n^{1+1/i}zeta(1+1/i))?
Nov 2, 2010 at 4:57 history edited JSE CC BY-SA 2.5
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Nov 2, 2010 at 4:53 comment added JSE But that's the case only for the naive density -- for the Dirichlet density p_i is supported on all the natural numbers (but is still a probability measure.) I will change, let me know if it's clear.
Nov 2, 2010 at 4:31 comment added KConrad Oh, then please edit your question to say p_i is a probability measure on 1,2,...,i.
Nov 2, 2010 at 4:15 comment added JSE Sorry, what I meant was that in the case of natural density p_i would be "uniform distribution on 1..i" and in the case of Dirichlet density p_i would be "n gets probability proportional n^{-1-1/i}, normalized to make the total mass 1" or something like that.
Nov 2, 2010 at 4:07 comment added KConrad Your clarification of what "why stop" means doesn't link the different p_i's together. Did you mean to say the p_i's are finitely additive measures, p_1 is natural density, p_2 is Dirichlet density, and for each i if p_i(S) exists then so does p_j(S) for all smaller j but there's a choice of S for which p_i(S) exists while p_{i-1}(S) does not exist?
Nov 2, 2010 at 4:05 comment added KConrad Number theorists don't use probability measures on the (positive) integers; all those densities (natural, Dirichlet, logarithmic) are finitely additive, not countably additive. Take a look at Tenenbaum's book "Introduction to Analytic and Probabilistic Number Theory". There is a section in there where he compares several concepts of density on Z.
Nov 2, 2010 at 3:05 comment added user6976 Oops. I did not notice that you intersect with $[1,..n]$ and not with $[-n,...,n]$. Is there a difference?
Nov 2, 2010 at 3:01 comment added user6976 One way is to choose an ultrafilter $\omega$, and take the limit with respect to the ultrafilter. The result is a (finitely additive) probability measure on all subsets of $\mathbb Z$. As a free bonus you get invariance with respect to shifts by integers. If the limit exists, then it coincides with the $\omega$-limit. Is that what your philosopher wants? This is how one proves that $\mathbb Z$ is amenable.
Nov 2, 2010 at 3:00 comment added David Hansen What about the "logarithmic density" used in Rubinstein-Sarnak?
Nov 2, 2010 at 2:57 comment added JSE He didn't know about Dirichlet density -- he is a philosopher of probability who was asking me about what pure mathematicians mean when they talk about the "probability that a random integer" has this property or that. He had in mind the notion of naive density, and when I told him about Dirichlet density he asked this natural question.
Nov 2, 2010 at 2:56 comment added BCnrd Dear JSE: Where did you find a philosopher who knows about Dirichlet density?
Nov 2, 2010 at 2:52 comment added Victor Miller @JSE: I'm not sure of the specific answer, but a good place to look would be in Persi Diaconis' Ph.D. theses which was about probability measures on the integers.
Nov 2, 2010 at 2:46 history asked JSE CC BY-SA 2.5