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 Moebius 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 Sum of Moebius Function over Divisors: Lemma:
 * $\mu * u = \iota$


 * By Properties of Dirichlet Convolution, Dirichlet convolution is commutative, associative and $h * \iota = h$ for all $h$.

We have:

Conversely:

Hence the result.