Möbius Inversion Formula

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


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

if and only if:

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


$d \divides n$ 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.


$\ds \map f n = \prod_{d \mathop \divides n} \map g d$

if and only if:

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


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 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:

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


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

Hence the result.


Source of Name

This entry was named for August Ferdinand Möbius.