1. ## Dirichlet Inverse Problem

The Dirichlet Convolution is this:

$(f * g)(n) = \sum\limits_{d \mid n} f(d)g(\frac{n}{d})$

The $\iota(n)$ function:

$\iota(n) = \begin{cases} 1 & \text{if } n = 1 \\ 0 & \text{if } n > 1\\ \end{cases}$

The arithmetic function $g$ is said to be the inverse of the arithmetic function $f$ if $f * g = g * f = \iota$. Show
that the arithmetic function $f$ has an inverse if and only if $f(1) \neq 0$. Show that if $f$ has an inverse it is
unique. (Hint: When $f(1) \neq 0$, find the inverse $f^{-1}$ of $f$ by calculating $f^{-1}(n)$ recursively, using the
fact that $\iota(n) = \sum\limits_{d \mid n} f(d) f^{-1}(\frac{n}{d})$.)

I'm really stumped on how to approach this one. Obviously the inverse doesn't exist when $f(1) = 0$, but beyond that, how do I tackle this one?

2. ## Re: Dirichlet Inverse Problem

By definition, given an arithmetical function $f(n)$ with $f(1) \ne 0$, its inverse $f^{-1} (n)$ obeys to the relation...

$f^{-1} * f = \sum_{d|n} f^{-1} (d)\ f(\frac{n}{d}) = \iota(n)$ (1)

For n=1 the (1) gives us...

$f^{-1}(1)\ f(1) = 1 \implies f^{-1}(1) = \frac{1}{f(1)}}$ (2)

For n>1 we have...

$\sum_{d|n} f^{-1} (d)\ f(\frac{n}{d}) = f(1)\ f^{-1} (n) + \sum_{d|n, d (3)

... so that from (2) and (3) we derive...

$f^{-1} (n) = - \frac{1}{f(1)}\ \sum_{d|n, d (4)

... and this formula permits us to compute the $f^{-1}(n)$ recursively...

Kind regards

$\chi$ $\sigma$