Let $f: \mathbb{Q} \cap [0,1] \to K$ and set $F(n) = \sum_{k = 1}^n f(\frac k n)$, $F^*(n) = \sum_{k = 1, (k,n) = 1}^n f(\frac k n)$.

Show that $F^* = \mu * F$ where $*$ is the Dirichlet product.

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


This is a previous step to obtain $\mu(n) = \sum_{k = 1, (k,n) = 1} exp(2 \pi i\frac k n)$ which can be concluded from the fact that $F(n) = e(n) = \sum_{d | n} \mu(d)$ using Möbius inversion formula. This result is also done in: The Möbius function is the sum of the primitive $n$th roots of unity..


Tom Apostol, Introduction to Analytic Number Theory, page 48.

  • 1
    $\begingroup$ $$\sum_{k=1,gcd(k,n)=1}^n f(k/n) = \sum_{k=1}^n f(k/n)\sum_{d | gcd(k,n)} \mu(d)= \sum_{d | n} \mu(d) \sum_{k=1,d | k}^n f(k/n)$$ $\endgroup$ – reuns Apr 18 at 22:38

$$F^*(n) = \sum_{k = 1, (k,n) = 1}^n f\Big(\frac k n\Big) = \sum_{k = 1}^n f\Big(\frac k n\Big) \Big(\sum_{d | (k,n)} \mu(d)\Big) = \sum_{k = 1}^n \sum_{d|k,d|n} \mu(d)f\Big(\frac k n\Big) = \ldots$$

For each $k$ such that $d|k$ we write $k = qd$ and $1 \le k \le n$ is equivalent to $1 \le q \le \frac n d$ and we write:

$$\ldots = \sum_{d|n} \mu(d) \sum_{k = 1,d|k} f\Big(\frac k n\Big) = \sum_{d|n} \mu(d) \sum_{q = 1}^{n/d} f\Big(\frac {q}{n/d}\Big) = \mu * F(n)$$

Let's see how we prove the consequence:

Take $f(x) = exp(\frac{2 \pi i x}{n})$, then observe that $F(n) = e(n)$ where $e(n) = 1$ if $n = 1$ and $0$ otherwise. This is the neutral element for Dirichlet convolution. Therefore, $\mu * F = \mu$ and by the above $\mu * F = F^*$, so we conclude $\mu = F^*$. This proves the proposition.


Thanks to @reuns for his comment.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.