Möbius Inversion Formula

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Theorem

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


Then:

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

if and only if:

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

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.