# Möbius Inversion Formula

## Theorem

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

Then:

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

iff:

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

where:

$d \mathop \backslash n$ 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:

$\mu * u = \iota$

We have:

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

Conversely:

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

Hence the result.

$\blacksquare$

## Source of Name

This entry was named for August Ferdinand Möbius.