Moebius Function: Prime Values & Relationship

[eljose79]

I,am looking for several information about the moebius function...specially its values for x equal to prime and if there is a relationship between this function and the prime number coutnign function.

 

[Ad Infinitum NAU]

Moebius mu function

For n in Z+: mu(1)=1, mu(n)=0 if n is not square-free, and mu(p1p2...pj)=(-1)^j, where the pj are distinct positive primes.

 

[Janitor]

I don't know what a prime number counting function is. But the definition of the Moebius function given by Ad Infinitum Lumberjack seems to say that mu(prime)=-1.

I am about half sure that John Baez discusses the Moebius function somewhere in his extensive website.

 

[matt grime]

a suitably refined google search for mobius function will provide the answers to all your queries.

 

[Ad Infinitum NAU]

But the definition of the Moebius function given by Ad Infinitum Lumberjack seems to say that mu(prime)=-1.

Yea... Sorry I wasn't clear enough there.. Hopefully this will clear it up a bit more:

mu(n) =

{ 1 ... if n=1
{ 0 ... if p^2|n (p^2 divides n) for some prime p
{ (-1)^j ... if n=p1*p2*...*pj where the pj are distinct primes (n is prime factored)

 

[shmoe]

Hi, there is most definitely a relation to the mobius function and the prime counting function. It can be shown that the statement $$\sum_{n\leq x}\mu (n)=O(x^{1/2+\epsilon})$$ is equivalent to the Riemann hypothesis, which dictates the error term in the prime number theorem. You should be able to find more infor on RH and the PNT easily enough.

ps. for Janitor, the prime counting function is $$\pi(x)=\sum_{p\ prime,\ p\leq x}1$$, in other words, $$\pi(x)$$ is the number of primes less than or equal to x.

 

[Janitor]

Good deal.

 

[eljose79]

Thanks a lot for your replies...

so knowing moebius function would be equivalent to solve Riemann hypothesis?..how interesting.

 

[shmoe]

Yes, actually $$\frac{1}{\zeta(s)}=\sum_{n\geq1}\frac{\mu(n)}{n^s}$$ for real part s greater than 1. If the bound I gave for the mobius function were true, you could use this to show this Dirichlet series is absolutely convergent on $$Re(s)>1/2+\epsilon$$, (any $$\epsilon>0$$), which means $$1/{\zeta(s)}$$ has no poles in this region and therefore zeta has no zeros here.