Möbius Inversion Formula
Theorem
Let $f$ and $g$ be arithmetic functions.
Then:
- $(1): \quad \displaystyle \map f n = \sum_{d \mathop \divides n} \map g d$
- $(2): \quad \displaystyle \map g n = \sum_{d \mathop \divides n} \map f d \, \map \mu {\frac n d}$
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:
| \(\displaystyle f = g * u\) | \(\leadsto\) | \(\displaystyle f * \mu = \paren {g * u} * \mu\) | |||||||||||
| \(\displaystyle \) | \(\leadsto\) | \(\displaystyle f * \mu = g * \paren {u * \mu}\) | |||||||||||
| \(\displaystyle \) | \(\leadsto\) | \(\displaystyle f * \mu = g\) |
Conversely:
| \(\displaystyle g = f * \mu\) | \(\leadsto\) | \(\displaystyle g * u = \paren {f * \mu} * u\) | |||||||||||
| \(\displaystyle \) | \(\leadsto\) | \(\displaystyle g * u = f * \paren {\mu * u}\) | |||||||||||
| \(\displaystyle \) | \(\leadsto\) | \(\displaystyle g * u = f\) |
Hence the result.
$\blacksquare$
Source of Name
This entry was named for August Ferdinand Möbius.
