Exercise 4.3.23* (Möbius multiplicative inversion formula)

Question

Suppose that $f(n)$ is an arithmetic function whose values are all nonzero, and put $F(n) = \prod_{d \mid n} f(d)$. Show that
$$f(n) = \prod_{d \mid n} F(n/d)^{\mu(d)}$$
for all positive integers $n$.

Richard Ganaye · Accepted Answer

ewcommand{\N}{\mathbb{N}}

ewcommand{\C}{\mathbb{C}}

In fact the Möbius inversion formula is valid when $f$ takes values in any abelian group. The sentence is the same as Theorem 4.8 in multiplicative notations, where the group $(\C, +)$ is replaced by the group $(\C^*, 	imes)$, so the sentence does not require a new proof. Nevertheless I rewrite this proof in multiplicative notations.

\begin{proof} 
Suppose that $F(n) = \prod\limits_{d \mid n} f(d)$.

We know by Theorem 4.7 that for all positive integers $j$,
$$\sum_{l \mid j} \mu(l) = \delta(j)  =
\begin{cases}
 1 & 	ext{if } j= 1\
 0 & 	ext{otherwise.}
 \end{cases}
$$
Then
\begin{align*}
\prod_{d \mid n} F\left(\frac{n}{d} ight)^{\mu(d)} &= \prod_{d \mid n} \left [\prod_{k \mid \frac{n}{d}} f(k) ight]^{\mu(d)}\
&= \prod_{dk \mid n} f(k)^{\mu(d)}\
&= \prod_{k \mid n} \prod_{d \mid \frac{n}{k}}  f(k)^{\mu(d)}\
&= \prod_{k \mid n} f(k)^{\sum_{d \mid \frac{n}{k}} \mu(d)}\
&=\prod_{k \mid n} f(k)^{\delta(\frac{n}{k})}\
&= f(n).
\end{align*}
If $F(n) = \prod\limits_{d \mid n} f(d)$ for all positive integers $n$, then for all $n \in \N^*$,
$$f(n) = \prod_{d \mid n} F\left(\frac{n}{d} ight)^{\mu(d)}.$$
\end{proof}

Exercise 4.3.23* (Möbius multiplicative inversion formula)

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Exercise 4.3.23* (Möbius multiplicative inversion formula)

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