Brian,
>>>>> [...]
>
>>>>>> 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.
>
> [...]
>
>>>>> 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}?
>
>>> I'm not going to waste my time trying to disentangle all of
>>> those negatives. If you can recast that in a more
>>> intelligible form, I'll look at it. But whatever you're
>>> trying to say, the fact is that if William's translation of
>>> your claim is, as you say, accurate, then the claim is
>>> false. What's more, it's false *because of* the prime
>>> number theorem.
>
>> I made the claim not because of any hare-brained idea I
>> had, but because I was informed of it by a competent
>> number theorist.
>
> It's pretty clear that either you misunderstood him, he
> misunderstood you, or he made a mistake.
>
>> If you observe the limits defining, as by the PNT, the
>> number of primes that will be found in any interval
>> [1,y], by illustrated the graph in the link I referenced,
>> then I would be interested to know how the number of n,
>> found by phi(m) \times (y-x/(prod{p : p inP}))* can
>> possibly fail to fall within the same bounds, or at least
>> fall within limits that converge on on or other of the
>> values found by either of the two functions defining
>> such bounds.
>
> I don't know what x is.
>
> I showed above that n(km, P) = k * phi(m). From the
> Mathworld page that you cite, p(n)/(n/ln(n)) < 1.105 for
> large n, and hence p(n) < 1.105n/ln(n) for large n. Thus,
> for large k we have p(km) < 1.105km/ln(km).
>
> But then since n(km, P) = k * phi(m),
>
> n(km, P)/p(km) > k * phi(m)/[1.105km/ln(km)],
>
> n(km, P)/p(km) > phi(m) * ln(km)/(1.105m),
>
> and hence, setting c(m) = phi(m)/(1.105m),
>
> n(km, P)/p(km) > c(m) * ln(km)
>
> for large k. Since ln(km) grows without bound as k
> increases, and c(m) > 0, the ratio n(km, P)/p(km) also grows
> without bound as k increases.
>
>> * If I understand correctly, phi(m) is the number of n for
>> which no p in P divides n for n in an interval of length
>> equal to the product of all p in P.
>
> By definition it's the number of integers in the interval
> [1, m] that are relatively prime to m. In this case that's
> the number of integers in [1, m] that are not divisible by
> any member of P. If k is any integer, it can easily be
> shown that it's also the number of integers in the interval
> [k, k + m - 1] that are not divisible by any member of P.
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, and y in this case
is
what you call 'n' in the term ln(n). So how do you refute my claim with
relation to my own definition of P?
Cheers.
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