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Is it known that for every epsilon there is N_0 such that all intervals of the form [N, (1+\epsilon)*N], where N > N_0, contain prime numbers?

flag	asked Oct 26 2009 at 23:32 Ilya Nikokoshev 9,019●5●30●73

Michael Lugo · Accepted Answer · 2009-10-26 23:41:49Z UTC

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This follows from the Prime Number Theorem. Let π(n) be the number of primes less than n. Then π(n) ~ n/log(n); it follows π((1+ε)n)-π(n) -> ∞ as n -> ∞.

answered Oct 26 2009 at 23:41

Michael Lugo
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	How string is the statement that π(n) ~ n/log(n)? One certainly imagines it to be strong enough to say something specific about pi(n)-pi(n-1), but this is certainly not the typical Prime Number Theorem people study. – Ilya Nikokoshev Oct 26 2009 at 23:55
	~ has a precise definition: f(n) ~ g(n) means that lim_{n -> infty} f(n)/g(n)=1. That's strong enough to derive the desired conclusion. – David Speyer Oct 26 2009 at 23:59
	Mm, yes it is so. – Ilya Nikokoshev Oct 27 2009 at 0:04

David Eppstein · Answer 2 · 2009-10-26 23:52:55Z UTC

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If one wants an explicit bound on N₀, apparently this can be gleaned from a Ph.D. thesis by Pierre Dusart (in French) which contains the result that for all x > 3275 there is a prime between x and x(1 + 1/(2 ln² x)). So we can take N₀ to be max(3275,exp((2 epsilon)^−1/2)).

answered Oct 26 2009 at 23:52

David Eppstein
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(Google Translate translates the thesis but the equations get messed up.) The statement can be found on Wikipedia:en.wikipedia.org/wiki/PNT and the same is derived near the end of the paper:ams.org/journals/mcom/1999-68-225/…. – To be cont'd Feb 14 2011 at 0:00

Strong Bertrand postulate

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