Möbius Inversion Formula

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


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

if and only if:

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


$d \mathrel \backslash n$ denotes that $d$ is a divisor of $n$
$\mu$ is the Möbius function.


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:

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


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

Hence the result.


Source of Name

This entry was named for August Ferdinand Möbius.