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I need your help in the next question:

Prove directly from the definition the Möbius inversion formula.

( Möbius function defined as follows: μ(n) = 1 if n is a square-free positive integer with an even number of prime factors. μ(n) = −1 if n is a square-free positive integer with an odd number of prime factors. μ(n) = 0 if n is not square-free.)

I'm not sure what I need to do and how

Thanks !

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If we don't have an idea of how to do this, we'll need your reference to know how you're expected to work it out. If we do have an idea of how to do this, we'll need to know your reference to know how to explain it to you. If you were the author of this reference, why would you expect someone to be able to work it out by this point? –  Loki Clock Jun 16 '13 at 13:40
    
Do you already know that the set of arithmetic functions which don't vanish at $\;1\;$ is an abelian group wrt the Dirichlet product $\,*\,$ ? –  DonAntonio Jun 16 '13 at 13:53
    
Yes , if we let G be the group consisting of all arithmetic functions under the "convolution"operation ∗, defined as (f∗g)(n):=∑ab=nf(a)g(b),n∈N. so we can say G is a torsion-free abelian group and G is not finitely generated and in particular is infinite. –  bar ben yair Jun 16 '13 at 14:03

1 Answer 1

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Well, since you already know about the abelian group structure, things are way easier:

Moebius Inversion Formula: For arithmetic functions $\,f\,,\,g\,$ we have

$$f(n)=\sum_{d\mid n}g(d)\iff g(n)=\sum_{d\mid n}f(d)\mu\left(\frac nd\right)$$

Proof: With the functions

$$u(n)=1\;\;\forall n\in\Bbb N\;,\;\;I(n)=\begin{cases}1&\,\;\;n=1\\0&,\,\,n>1\end{cases}$$ we get that

$$f=g*u\stackrel{\text{mult. by $\,\mu\,$}\;}\implies\,f*\mu=(g*u)*\mu=g*(u*\mu)=g*I=g\;\;\;\;\;\square$$

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thank you very much ! –  bar ben yair Jun 16 '13 at 14:36

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