"Brian M. Scott" <b.scott@[EMAIL PROTECTED]
> wrote in message
news:ugkkzecxhz3c$.q5osvyzeigwb$.dlg@[EMAIL PROTECTED]
> On Sun, 20 Jul 2008 19:26:12 +0100, Jack <jj@[EMAIL PROTECTED]
>
> wrote in <news:6rLgk.13933$9S6.934@[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.
>
> [...]
>
>> I don't know why you are saying <<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>> and
>> subsequently <<More generally, let P be any finite set of
>> primes, and let m = prod(P)>> when I made it clear that
>> I want P to be the set of primes of value less than the
>> square root of y, [...]
>
> You did not make it clear to William. Look at his attempt
> to make sense of your claim, which is still present at the
> top of this post (and which you later said was correct): it
> does NOT say any such thing. It STARTS with a set P of
> primes, and lets y be arbitrary. And as I pointed out
> several times, I was refuting THAT claim (e.g., 'This shows
> that Jack's claim is false, assuming that your translation
> of it is correct').
>
> Apparently what you were really trying to say was something
> like this, which is another matter altogether:
>
> For any y in N let P(y) be the set of primes not
> exceeding sqrt(y), and let n(y) be the number of
> integers in [1, y] not divisible by any member of
> P(y). Then for sufficiently large values of y,
> n(y) is approximately p(y), the number of primes
> in [1, y].
>
> For any k in [1, y], either k is divisible by some prime in
> P(y), or k is a prime greater than sqrt(y). Thus, n(y) is
> simply the number of primes in [1, y] that are strictly
> greater than sqrt(y), and hence n(y) = p(y) - p(sqrt(y)).
>
> Now n(y)/p(y) = 1 - p(sqrt(y))/p(y). For large y we have
> p(y) approximately equal to y/ln(y) and p(sqrt(y))
> approximately equal to sqrt(y)/ln(sqrt(y)) =
> 2 * sqrt(y)/ln(y), so p(sqrt(y))/p(y) is approximately
> 2 * sqrt(y)/y = 2/sqrt(y), which does indeed tend to 0 as y
> increases.
But here there is no mention of the phi function. I want to be certain
that
p(y)/y tends towards phi(m) \times y/(prod{p : p inP})).
Cheers.
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