On Tue, 22 Jul 2008 01:37:36 +0100, Jack <jj@[EMAIL PROTECTED]
>
wrote in <news:hZ9hk.17700$A42.13504@[EMAIL PROTECTED]
> in
alt.algebra.help:
[...]
>> So for y in N we let P(y) be the set of primes not exceeding
>> sqrt(y) and m(y) = prod(P(y)); p(y) is the number of primes
>> not exceeding y, i.e., the number usually denoted by pi(y).
>> You say that you want to be sure that
>> lim_{y --> oo}{[p(y)/y]/[y * phi(m(y))/m(y)]} = 1,
>> where phi is the Euler totient function.
>> This appears not to be true. Heuristically speaking, the
>> denominator y * phi(m(y))/m(y) ought to be roughly equal to
>> the number of integers in [1, y] that are not divisible by
>> any member of P(y). The integers in [1, y] that are not
>> divisible by any member of P(y) are precisely the primes in
>> the interval (sqrt(y), y], of which there are
>> p(y) - p(sqrt(y)). Thus, the ratio
>> [p(y)/y]/[y * phi(m(y))/m(y)]
>> ought to be about
>> [p(y)/y]/[p(y) - p(sqrt(y))]
>> for large y. Consider the reciprocal, which on your view
>> should also approach 1:
>> [p(y) - p(sqrt(y))]/[p(y)/y] =
>> y * [1 - p(sqrt(y))/p(y)].
>> I noted before that p(sqrt(y))/p(y) is about 2/sqrt(y) for
>> large y, so y * [1 - p(sqrt(y))/p(y)] is about
>> y * [1 - 2/sqrt(y)] = y * [sqrt(y) - 2]/sqrt(y) =
>> sqrt(y) * [sqrt(y) - 2],
>> which is on the rough order of y, not 1.
> As you know I find the equations difficult to follow. But
> I do not see in them any mention of ln(n).
So what? In any case, it's there, in the statement 'I noted
before that p(sqrt(y))/p(y) is about 2/sqrt(y) for large y'.
> I should say also that the number theorist I had worked
> with was so adamant that the claim holds that I wonder
> whether we are speaking at cross purposes.
I'm simply responding to what you tell me. Unfortunately,
that changes from post to post. As a result, at this point
I have no idea what the number theorist actually told you.
> As y increases, each new prime, p, of value <sqrt(y) is,
> on average, further apart in absolute terms than the
> previous (i.e. the prime that is adjacent but of lower
> value than p), but closer in pro****tionate terms to the
> previous; and the value of 1/p, p in P gets smaller and
> smaller; furthermore, for any p,q in P, where q < p and
> p and q are adjacent primes, the pro****tion of n in
> [1,y] whose factors are exclusively q and 1 gets ever
> closer to the pro****tion of n whose factors are exclusive
> p and 1.
Bollocks. First, the only positive integer whose only
factors are 1 and q is q; I assume that you mean numbers
whose only prime factor is q. Those are simply the powers
q^n for n > 0. Now q^n <= y if and only if n ln(q) <=
ln(y), so for large y there are about ln(y)/ln(q) such
powers of q; the pro****tion of such numbers is therefore
about ln(y)/[y ln(q)]. Similarly, for large y the
pro****tion of integers in [1, y] having only p as prime
factor is about ln(y)/[y ln(p)]. The ratio of these
pro****tions is ln(p)/ln(q), which is not 1.
> This is found in both the case for the phi
> function and also for the approximate number of primes
> found as by the prime number theorem. So surely the two
> values, the one found by phi(m) y/(prod{p : p inP})) and
> the other found by n/log(n) (if I am correct in thinking
> that is the formula usually quoted for obtaining pi(n)
> by the PNT) should converge on one another.
Apparently you paid no attention to my comments on notation
and terminology two posts back; oh, well. I will simply
point out two things. First, this is not the claim that you
made last time. You have replaced the p(y)/y that you had
last time with an n/log(n) that is almost certainly supposed
to be y/ln(y) and probably is really supposed to be p(y).
Secondly, even if your attempted justification didn't
contain an obvious error, it wouldn't justify your
conclusion: two things can vary in qualitatively similar
fa****on without being asymptotic to each other.
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