# Möbius Inversion Formula

From ProofWiki

## Theorem

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

Then:

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

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

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.