pi[x] >=loglogx


by AlbertEinstein
Tags: >loglogx
AlbertEinstein
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#1
Sep4-06, 10:45 AM
P: 113
I am going through Hardy's book on number theory.The following theorem I do not understand.

theorem 10: pi[x] >= loglog x
where pi[x] is the prime counting function
and >= stands for greater than or equal to

The arguments written in the book are very compact.please help .
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shmoe
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#2
Sep4-06, 11:28 AM
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Do you follow any of it?

Do you understand how they derived [tex]p_{n}<2^{2^n}[/tex] ?

this is an important step. The rest just follows from pi(x) being increasing, and also [tex]\pi(p_n)=n[/tex] which they use but don't explicitly mention.
CRGreathouse
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#3
Sep4-06, 01:34 PM
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Bertrand's postulate?

shmoe
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#4
Sep4-06, 01:46 PM
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pi[x] >=loglogx


Nope! the bound of p_n above is much weaker than Bertrand's will give you. It's correspondingly simpler to prove though, it follows from a slight adaptation of Euclid's proof there are infinitely many primes (in case anyone who hasn't seen it wants to give it a stab)


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