# Moebius function

1. Jul 14, 2004

### 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 fucntion and the prime number coutnign function.

2. Jul 14, 2004

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.

3. Jul 14, 2004

### 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.

4. Jul 15, 2004

### matt grime

5. Jul 17, 2004

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)

Last edited: Jul 17, 2004
6. Jul 22, 2004

### 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.

Last edited: Jul 22, 2004
7. Jul 22, 2004

### Janitor

Good deal.

8. Jul 23, 2004

### eljose79

Thanks a lot for your replies...

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

9. Jul 23, 2004

### 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.