# Möbius Inversion Formula

## Theorem

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

Then:

- $(1): \quad \displaystyle \map f n = \sum_{d \mathop \divides n} \map g d$

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

where:

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

## 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:

\(\displaystyle f = g * u\) | \(\leadsto\) | \(\displaystyle f * \mu = \paren {g * u} * \mu\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(\leadsto\) | \(\displaystyle f * \mu = g * \paren {u * \mu}\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(\leadsto\) | \(\displaystyle f * \mu = g\) | $\quad$ | $\quad$ |

Conversely:

\(\displaystyle g = f * \mu\) | \(\leadsto\) | \(\displaystyle g * u = \paren {f * \mu} * u\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(\leadsto\) | \(\displaystyle g * u = f * \paren {\mu * u}\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(\leadsto\) | \(\displaystyle g * u = f\) | $\quad$ | $\quad$ |

Hence the result.

$\blacksquare$

## Source of Name

This entry was named for August Ferdinand Möbius.