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There is a very simple proof for Möbius inversion formula through convolution:

If $A$ is a UFD and $B$ is a ring, $f,g:A\rightarrow B$ two functions, then

$$(f\ast g)(n) = \sum_{k\cdot l = n}f(k)\cdot g(l)$$ makes the set of functions into a ring with identity $\delta_1$.

In this case the Möbius function, assigning 0 to every $n\in A$ which is not square free, 1 to square free elements which are the products of an even number of primes and -1 to square free products of odd number of primes, is an inverse of the constant function 1. Then the Möbius inversion formula

$$f(n) = \sum_{d|n} g(d)\Longrightarrow g(n) = \sum_{d|n}f(d)\mu\left(\frac{n}{d}\right)$$

is simply another way of writing

$$f = g\ast 1\Longrightarrow g = f\ast \mu.$$

Is there a similarly general and elegant approach to the multiplicative formula

$$f(n) = \prod_{d|n}g(d)\Longrightarrow g(n) = \prod_{d|n} f(d)^{\mu(n/d)}?$$

Thank you.

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    I think they are the exact same theorem (no new proof needed), written in multiplicative notation rather than additive.2012-10-27
  • 0
    That does not seem to be the case, since in order to define the convolution you seem to need two different operations in $B$, and while fingin the inverse of the constant function $f\equiv 1$ both neutral elements 0 and 1 play a role. I didn't manage to define the convolution of functions into a group (rather than a ring).2012-10-28

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