"Brian M. Scott" <b.scott@[EMAIL PROTECTED]
> wrote in message
news:83moctyfxgbp.wjxllm2oemie.dlg@[EMAIL PROTECTED]
> 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.
But if you look at the graph here:
http://mathworld.wolfram.com/PrimeNumberTheorem.html
are you denying that the lines defining the bounds Li(n) and L/Ln(n) do
not
serve also as bounds within which the pro****tion of integers n in any
interval [1,y] for which p in P does not divide n diverges from the value
found by your formula n(m, P) = phi(m) = prod{p - 1 : p inP}?
Cheers.
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