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Pi[x] >=loglogx

  1. Sep 4, 2006 #1
    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. jcsd
  3. Sep 4, 2006 #2

    shmoe

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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.
     
  4. Sep 4, 2006 #3

    CRGreathouse

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    Bertrand's postulate?
     
  5. Sep 4, 2006 #4

    shmoe

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    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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