Exercise 4.3.6 (If $F(n) = \sum_{d \mid n} f(d)$ then$f(n) = \sum_{d \mid n} \mu(n/d) f(d)$)

Question

If $F(n) = \sum_{d \mid n} f(d)$ for every positive integer $n$, prove that $f(n) = \sum_{d \mid n} \mu(n/d) f(d)$.

Richard Ganaye · Accepted Answer

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\begin{proof} By Theorem 4.8 (Möbius inversion formula), 
$$f(n) = \sum_{d \mid n} \mu(d) F(n/d).$$
For every positive integer $n$, let
$$A_n = \{d \in \N^*: d \mid n\}$$
denote the set of divisors of $n$. Consider
$$\varphi
\left\{
\begin{array}{ccl}
A_n &	o &A_n\
d & \mapsto & n/d.
\end{array}
ight.
$$
Then $\varphi$ is an involution, i.e. $\varphi \circ \varphi = 1_{A_n}$, thus $\varphi$ is a bijection. Then the change of indices $\delta = \varphi(d)$ gives
\begin{align*}
f(n) &= \sum_{d \mid n} \mu(d) F(n/d)\
&= \sum_{d \in A_n} \mu(d) F(n/d)\
&=\sum_{\delta  \in A_n} \mu(n/ \delta) F(\delta)&(\delta = n/d)\
&=  \sum_{d \mid n} \mu(n/d) f(d).
\end{align*}
 If $F(n) = \sum_{d \mid n} f(d)$ for every positive integer $n$, then $f(n) = \sum_{d \mid n} \mu(n/d) f(d)$.
\end{proof}

Exercise 4.3.6 (If $F(n) = \sum_{d \mid n} f(d)$ then$f(n) = \sum_{d \mid n} \mu(n/d) f(d)$)

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Exercise 4.3.6 (If $F(n) = \sum_{d \mid n} f(d)$ then$f(n) = \sum_{d \mid n} \mu(n/d) f(d)$)

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