On Sat, 19 Jul 2008 23:58:32 -0700, William Elliot
<marsh@[EMAIL PROTECTED]
> wrote in
<news:Pine.BSI.4.58.0807192335480.2040@[EMAIL PROTECTED]
> in
alt.algebra.help:
[...]
> Never mind. I'll guess what you're trying to say.
> Let P be a set of primes.
> Let n(y,P) be the number of integers n, in [1,y]
> with for all p in P and p <= sqr y, p not divide n.
> Then n(y,P) converges on being negligibly different from
> the number of primes found in that interval. Hm. Mostly
> gibberish. Let's see.
> Let p(y) be the number of primes in [1,y]
> Then lim(y->oo) n(y,P)/p(y) = 1.
> Hm. n(y,P) = o(p(y)) or O(p(y)) ?
No: n(y, P) = o(p(y)) means that the limit in the previous
line is 0, and n(y, P) = O(p(y)) means that there is a
positive bound C such that for all sufficiently large y,
n(y, P)/p(y) < C.
There is a notation abbreviating the statement that
lim_{y --> oo} {n(y, P)/p(y)} = 1: n(y, P) ~ p(y).
The claim is obviously false. Take P = {2}. For all
y >= 4, n(y, P) is simply the number of odd integers in the
interval [1, y], which is ceil(y/2) ~ y/2. By the prime
number theorem, however, p(y) ~ y/ln(y) Thus,
n(y, {2})/p(y) = ~ (y/2)/(y/ln(y)) = ln(y)/2 --> oo as
y --> oo.
More generally, let P be any finite set of primes, and let
m = prod(P). From basic properties of Euler's phi function
we know that n(m, P) = phi(m) = prod{p - 1 : p in P}. It's
also clear that for any k in N, the number of integers in
the interval [km + 1, (k+1)m] that are divisible by none of
the primes in P is also phi(m). Thus, for any y in Z+ we
have n(km, P) = k * phi(m). But p(km) ~ km/ln(km), so
n(km, P)/p(km) ~ k * phi(m)/[km/ln(km)] =
[phi(m)/m] * ln(km), which increases without bound as k
increases, since phi(m)/m is constant. In particular, the
limit isn't 1.
This shows that Jack's claim is false, assuming that your
translation of it is correct.
[...]
Brian
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