# Möbius inversion of $1/n$

What is $$\mu * f?$$

How can it be found?

• $f(n) = 1/n$.
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What is a better way to write this question? –  quanta Apr 25 '11 at 15:36
Is your main question "what is a better way to write 'what is the moebius inverse of 1/n?' ?" ? Or are you more concerned with what is the moebius inverse of 1/n? –  Mitch Apr 25 '11 at 15:41
Speaking as a non-expert, doesn't $f$ need to be of the form $\sum_{d|n} g(d)$ for this question to make sense? –  Grumpy Parsnip Apr 25 '11 at 15:44
@Jim: $f$ actually is in that form! We can write $$\frac{1}{n}=\sum_{d|n} \frac{1}{d}\sum_{r|d}\mu(r)r.$$ That was partly a joke, but Dirichlet Convolution does makes sense for any two functions on the natural numbers. Given $f,g$ we just define $$f*g(n)=\sum_{d|n}f(d)g(n/d).$$ –  Eric Naslund Apr 25 '11 at 15:56
@Eric: I see. Thanks! –  Grumpy Parsnip Apr 25 '11 at 18:21

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,

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$$(\mu * f)(n)=\sum_{d|n} \mu(\frac{n}{d})f(d)=$$ $$\frac{1}{n} \sum_{d|n}\mu(\frac{n}{d})\frac{n}{d}=$$ $$\frac{1}{n} \sum_{d|n} \mu(d) d$$

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:

$$(\mu*f)(n) = \frac{(1-p_1)(1-p_2)...(1-p_t)}{n}$$

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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}

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You don't really need $f$ to be multiplicative to define the Moebius inversion, nor do you need $f(0)$ defined. The Dirichlet convultion makes a ring out of the set of functions from $\mathbb{N}$ to a ring $R$, and $\mu$ is just the multiplicative inverse of the constant function $f(n)=1$ in that ring of functions –  Thomas Andrews Apr 25 '11 at 15:58
@Thomas. Nice Addendum. Thanks. –  user02138 Apr 25 '11 at 16:45