Möbius Inversion Formula

From ProofWiki
Jump to: navigation, search

Theorem

Let $f$ and $g$ be arithmetic functions.


Then:

$(1): \quad \displaystyle f \left({n}\right) = \sum_{d \mathop \backslash n} g \left({d}\right)$

if and only if:

$(2): \quad \displaystyle g \left({n}\right) = \sum_{d \mathop \backslash n} f \left({d}\right) \mu \left({\frac n d}\right)$

where:

$d \mathrel \backslash n$ denotes that $d$ is a divisor of $n$
$\mu$ is the Möbius function.


Proof

Let $u$ be the unit arithmetic function and $\iota$ the identity arithmetic function.

Let $*$ denote Dirichlet convolution.

Then equation $(1)$ states that $f = g * u$ and $(2)$ states that $g = f * \mu$.

The proof rests on the following facts:

By the lemma to Sum of Möbius Function over Divisors:

$\mu * u = \iota$

By Properties of Dirichlet Convolution, Dirichlet convolution is commutative, associative and $h * \iota = h$ for all $h$.


We have:

\(\displaystyle f = g * u\) \(\implies\) \(\displaystyle f * \mu = \left({g * u}\right) * \mu\)                    
\(\displaystyle \) \(\implies\) \(\displaystyle f * \mu = g * \left({u * \mu}\right)\)                    
\(\displaystyle \) \(\implies\) \(\displaystyle f * \mu = g\)                    


Conversely:

\(\displaystyle g = f * \mu\) \(\implies\) \(\displaystyle g * u = \left({f * \mu}\right) * u\)                    
\(\displaystyle \) \(\implies\) \(\displaystyle g * u = f * \left({\mu * u}\right)\)                    
\(\displaystyle \) \(\implies\) \(\displaystyle g * u = f\)                    

Hence the result.

$\blacksquare$


Source of Name

This entry was named for August Ferdinand Möbius.