What is $\mu * f?$
How can it be found?
- $f(n) = 1/n$.
What is $\mu * f?$
How can it be found?
Hint: Consider $\mu*f$ at the prime powers. $\mu*f(p)=\frac{1}{p}-1=\frac{1-p}{p}$ $\mu*f(p^\alpha)=\frac{1}{p^\alpha}-\frac{1}{p^{\alpha-1}}=\frac{1-p}{p^\alpha}.$Then if $n=p_1^{r_1}\cdots p_m^{r_m}$ we have that $\mu*f(n)=\frac{1}{n}\prod_{p|n} \left(1-p\right).$ Now, be expanding the product, you can write write $\frac{1}{n}\prod_{p|n} \left(1-p\right)=\frac{1}{n}\sum_{d|n} \mu(d) d.$ I don't think it can be simplified further.
Hope that helps,
$(\mu * f)(n)=\sum_{d|n} \mu(\frac{n}{d})f(d)=$ $\frac{1}{n} \sum_{d|n}\mu(\frac{n}{d})\frac{n}{d}=$ $\frac{1}{n} \sum_{d|n} \mu(d) d$
When $n=p^k$, then $(\mu*f)(n) = \frac{1-p}{p^k}$
So if $n=p_1^{k_1}p_2^{k_2}...p_t^{k_t}$, then:
$(\mu*f)(n) = \frac{(1-p_1)(1-p_2)...(1-p_t)}{n}$
Typically, when $f \colon \mathbb{N} \to \mathbb{C}$ is an arithmetic function, Möbius inversion is defined as the Dirichlet convolution \begin{align} (\mu * f)(n) = \sum_{d \mid n} \mu(d) f(\tfrac{n}{d}). \end{align} The arithmetic function $g = \mu * f$ is said to be the Möbius inverse of $f$. For your example, \begin{align} (\mu * f)(n) = \frac{1}{n} \sum_{d \mid n} \mu(d) d. \end{align}