Prove Summation of Mobius Inversion w/ Sigma Function

[math_grl]

Would like to show $$\sum_{d \mid n} \mu (d) \sigma_0 (d) = (-1)^{\omega (d)}$$.

This proof is just left out of text I'm looking at and I can't seem to piece how $$F(n/d) = \sigma_0 (d)$$, where $$F(x) = \sum_{s \mid x} f(x)$$.

 

[dodo]

Hi,
can you make the problem a little clear? Who are omega and sigma_0, and how do they relate to little f or to big F. Maybe sigma_0(d) = F(n/d) is how sigma_0 is defined to begin with. Or not.

 

[math_grl]

$$\sigma _0 (n) = \sum_{d \mid n} 1$$ and $$\omega (n) = \sum_{p \mid n} 1$$ where $$p$$ is a prime and $$d$$ is a divisor.

I thought the notation was standard for arithmetic functions.

 

[dodo]

Ok. I knew your sigma_0 as "tau", but never heard of omega.

I have a proof of the statement (I believe), but it is somewhat wordy. Here is a sketch:

First, realize that the value of that sum for n=(p1^whatever)(p2^whatever)(p3^whatever)... is the same as the value of the sum for another n, constructed as n=p1.p2.p3... (which is squarefree). This occurs because the mu() function will set to 0 the terms of the sum where the divisor d is not squarefree. So actually you only need to prove the statement for a squarefree n.

That said, the divisors of a squarefree n are just combinations of distinct primes; their corresponding mu values are either 1 or -1. And there will be two cases: one where all the mu(d) are equal to mu(n/d), and other when all the mu(d) are equal to minus mu(n/d). And these two cases occur precisely when omega(n) is even or odd, respectively.

 

[math_grl]

Now you've given me something to think about...

very interesting way of proving it as I thought being in the same section as that of Mobius inversion, was supposed to follow direct from that or something...

 

[dodo]

It sort of follows... it is based on
$$\sum_{d \mid n} \mu (\frac n d) \sigma_0 (d) = 1$$
which I forgot to mention in my previous post.