In article <lfsigpk3bklo$.1fqckprvpd1o9.dlg@[EMAIL PROTECTED]
>, Brian M. Scott
<b.scott@[EMAIL PROTECTED]
> wrote:
> On Mon, 21 Jul 2008 20:06:27 +0100, Jack <jj@[EMAIL PROTECTED]
>
> wrote in <news:R65hk.26526$gU4.11217@[EMAIL PROTECTED]
> in
> alt.algebra.help:
>
> >>> I want to be certain that p(y)/y tends towards phi(m)
> >>> \times y/(prod{p : p inP})).
>
> >> Your terminology is incorrect: one does not normally speak
> >> of a function f(y) tending towards another function g(y).
> >> Do you mean that you want to be certain that the ratio of
> >> the two function of y approaches 1 as y increases?
>
> > Yes.
>
> >> Next, you haven't defined your notation. What is m?
>
> > Same as the m you were using in your term phi (m).
>
> >> Is your P here the P(y) that I defined above?
>
> > The set of primes less than or equal to the square root of y.
>
> 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.
I don't know if this is of any interest - I haven't been paying as much
attention as I should, I guess and I'm surely no number theorist.
Let P be a (square free) product of primes and let A be a subset of |N.
Define S(A, P) = |{ n in A : n and P are relatively prime }|.
Now, as a special case, let A be the set of all n in |N less than or
equal to some real number X and let P be the product of all
primes <= z < X.
Finally, let pi(x) be the number of primes <= x.
It is a consequence of Legendre's identity that
S(A, P) >= pi(X) - pi(z) + 1.
Given y in |N, take z = sqrt(y) and X = y. S(A, P) counts the n <= y
which are divisible by none of the primes <= sqrt(y).
--
Paul Sperry
Columbia, SC (USA)
|
