[math-fun] Re: math-fun Digest, Vol 22, Issue 7
This suprises me about S. The primes become arbitrarily sparse, on average, so their natural density should exist and = 0. No?
The # of primes <= x is asymptotic to x/ln(x), so the density of primes <= x is asymptotic to 1/ln(x), which -> 0 as x -> oo,
If the primes' density is 0, then it seems that must also be true of any subset of the primes, such as the one menitoned above.
Apologies - I meant to say the natural density in the set of primes! (and not the density in the positive integers) To repeat the question, let S be the set of primes with first digit 1. Define the natural density of S in the primes to be the limit of number of elements in S that are <= n -------------------------------------- as n --> infinity. number of primes that are <= n This does not exist (stated in Serre "A course in arithmetic"). Why so? Gary McGuire
Gary McGuire <gmg@maths.nuim.ie> wrote:
To repeat the question, let S be the set of primes with first digit 1. Define the natural density of S in the primes to be the limit of
number of elements in S that are <= n -------------------------------------- as n --> infinity. number of primes that are <= n
This does not exist (stated in Serre "A course in arithmetic"). Why so?
S is as far ahead as possible when x=2*10^n, after it's just gotten a whole bunch of primes starting with 1. At that point the ratio gets up to nearly 1/2, since there are about as many primes in [x,2x] as in [0,x]. (Of course, only "about" in the limit -- the ratio is close to 2( ln x / (ln x + ln 2) ) - 1, and for big enough x, that's 1ish.) But S is as far behind as possible when x=10^n, when there have been no primes starting with 1 for a long time. By similar logic, by then the ratio is down to about 1/9. Since the ratio keeps getting near those two far-apart marks, it can't possibly tend to a limit. --Michael Kleber -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.
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Gary McGuire -
Michael Kleber