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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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