# Möbius Inversion Formula

Jump to navigation Jump to search

## Theorem

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

Then:

$(1): \quad \ds \map f n = \sum_{d \mathop \divides n} \map g d$
$(2): \quad \ds \map g n = \sum_{d \mathop \divides n} \map f d \, \map \mu {\frac n d}$

where:

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

### Abelian Group

Let $G$ be an abelian group.

Let $f, g: \N \to G$ be mappings.

Then

$\ds \map f n = \prod_{d \mathop \divides n} \map g d$
$\ds \map g n = \prod_{d \mathop \divides n} \map f d^{\mu \paren {\frac n d} }$

## 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:

$\mu * u = \iota$

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

We have:

 $\ds f = g * u$ $\leadsto$ $\ds f * \mu = \paren {g * u} * \mu$ $\ds$ $\leadsto$ $\ds f * \mu = g * \paren {u * \mu}$ $\ds$ $\leadsto$ $\ds f * \mu = g$

Conversely:

 $\ds g = f * \mu$ $\leadsto$ $\ds g * u = \paren {f * \mu} * u$ $\ds$ $\leadsto$ $\ds g * u = f * \paren {\mu * u}$ $\ds$ $\leadsto$ $\ds g * u = f$

Hence the result.

$\blacksquare$

## Source of Name

This entry was named for August Ferdinand Möbius.