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Topic: Primes...
Replies: 14   Last Post: Feb 11, 2003 7:41 AM

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Robert Israel

Posts: 11,902
Registered: 12/6/04
Re: Primes...
Posted: Jan 30, 2003 4:19 PM
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"r.e.s." <rs.1@mindspring.com> wrote in message news:<b1arp4$nmd$1@slb6.atl.mindspring.net>...

> Letting p(n) be the nth prime, I understand that the
> following is equivalent to the Prime Number Theorem:
> p(n)/(n log(n)) -> 1 as n -> infinity.


> That makes it easy to show, if I'm not mistaken, that
> for every integer k>=0, as n -> infinity,


> p(n+k)/p(n) -> 1
> p(n*k)/p(n) -> k
> p(n^k)/p(n) -> k*n^(k-1)


> The pattern in those cases is
> p(f(n))/p(n) -> f'(n),
> where f' is the derivative of f.


> I wonder if there's a nice way to generalize this,
> e.g., to arbitrary polynomial f. (?)


Yes, if f is increasing on some (N, infinity). We then have
f(n) ~ c n^d for some positive c and positive integer d as n -> infinity
(~ meaning asymptotic to). Then p(f(n)) ~ f(n) log f(n), and
log f(n) ~ d log n, so p(f(n)) ~ cd n^d log n
p(f(n))/p(n) ~ c d n^d log n /(n log n) = c d n^(d-1) ~ f'(n).

Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2




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