Kluyver's Formula for Ramanujan's Sum
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Theorem
Let $q \in \N_{>0}$.
Let $n \in \N$.
Let $\map {c_q} n$ be Ramanujan's sum.
Let $\mu$ denote the Möbius function.
Then:
- $\displaystyle \map {c_q} n = \sum_{d \mathop \divides \gcd \set {q, n} } d \map \mu {\frac q d}$
where $\divides$ denotes divisibility.
Proof
Let $\alpha \in \R$.
Let $e: \R \to \C$ be the mapping defined as:
- $\map e \alpha := \map \exp {2 \pi i \alpha}$
Let $\zeta_q$ be a primitive $q$th root of unity.
Let:
- $\displaystyle \map {\eta_q} n := \sum_{1 \mathop \le a \mathop \le q} \map e {\frac {a n} q}$
By Complex Roots of Unity in Exponential Form this is the sum of all $q$th roots of unity.
Therefore:
- $\displaystyle \map {\eta_q} n = \sum_{d \mathop \divides q} \map {c_d} n$
By the Möbius Inversion Formula, this gives:
- $\displaystyle \map {c_q} n = \sum_{d \mathop \divides q} \map {\eta_d} n \map \mu {\frac q d}$
Now by Sum of Roots of Unity, we have:
- $\displaystyle \map {c_q} n = \sum_{d \mathop \divides q} d \map \mu {\frac q d}$
as required.
$\blacksquare$
Source of Name
This entry was named for Jan Cornelis Kluyver.