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Pi[x] >=loglogxby AlbertEinstein
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#1
Sep406, 10:45 AM

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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 . 


#2
Sep406, 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. 


#3
Sep406, 01:34 PM

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Bertrand's postulate?



#4
Sep406, 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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