pi[x] >=loglogx

AlbertEinstein

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 .

shmoe

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

Bertrand's postulate?

shmoe

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)

AlbertEinstein	pi[x] >=loglogx Tweet 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 .

shmoe Recognitions: Homework Help Science Advisor	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 Recognitions: Homework Help Science Advisor	Bertrand's postulate?

shmoe Recognitions: Homework Help Science Advisor	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)