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.
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