direct computation of inverse of Dirichlet convolution

Question

For an arithmetical function $f(n)$ with $f(1)\neq 0$ we have the recursive formula for the inverse (under Dirichlet convolution) of $f$ : $$f^{-1}(n)=(-1/f(1))\sum_{d|n, d<n}f(d)f^{-1}(n/d)$$ such that $(f*f^{-1})(n)=\begin{cases}1,\ n=1\\ 0,\ n>1\end{cases}$ . Though it is clear from a formal calculation that for two arithmetical function $f,g$ with $f(1), g(1)$ both nonzero we have $(f*g)^{-1}=f^{-1}*g^{-1}$ from the commutativity and associativity of Dirichlet convolution $*$, I'm confused when I verify it by a direct computation from the definition as below: (for notational convenience, let $h=f*g$) $h(1)=f(1)g(1)\neq 0\Rightarrow$\begin{align} h^{-1}(n)&=(-1/h(1))\sum_{d|n, d<n}h(d)h^{-1}(n/d)\\&=\frac{-1}{f(1)g(1)}\sum_{d|n, d<n}\left(\sum_{c|d}f(c)g\left(\frac{d}{c}\right)\right)\left(\sum_{e|(n/d)}f^{-1}(e)g^{-1}\left(\frac{n}{de}\right)\right)\end{align} where the second paratheses in the last row follows by induction. On the other hand, \begin{align}\left(f^{-1}*g^{-1}\right)(n)&=\sum_{d|n}f^{-1}(n/d)g^{-1}(d)\\ &=\sum_{d|n}\left(\frac{-1}{f(1)}\sum_{e|(n/d),e<(n/d)}f(e)f^{-1}\left(\frac{n}{de}\right)\right)\left(\frac{-1}{g(1)}\sum_{c|d,c<d}g(c)g^{-1}\left(\frac{d}{c}\right)\right)\end{align} I cannot figure out why $h^{-1}(n)=(f^{-1}*g^{-1})(n)$, in particular about the apparent difference in the minus sign

Answer 1

I'm going to ignore the minus sign, the values at 1, and the $<$ inequality signs. But otherwise, your first sum is \[ \sum _{de|n}\sum _{c|d}f(c)f^{-1}(e)g(d/c)g^{-1}(n/de)=\sum _{cde|n}f(c)f^{-1}(e)g(d)g^{-1}(n/cde)=\sum _{cde|n}f(c)f^{-1}(e)g(n/ced)g^{-1}(d)=\sum _{cde|n}f(e)f^{-1}(c)g(n/ecd)g^{-1}(d)\] whilst the second is \[ \sum _{de|n}\sum _{c|d}f(e)f^{-1}(n/de)g(c)g^{-1}(d/c)=\sum _{cde|n}f(e)f^{-1}(n/cde)g(c)g^{-1}(d)=\sum _{cde|n}f(e)f^{-1}(c)g(n/dec)g^{-1}(d)\] so both sums are equal.

The manipulations for the first equation array, if needed (second is similar): First equality - write the $d$'s as multiples of $c$. Second equality - replace $d$ by $(n/ce)/d$. Third equality - just relabelling $e$ and $c$.

Answer 2

To verify this, I recommend using Dirichlet series as it is a more powerful device to study Dirichlet convolution. Suppose we define the symbol

$$ D(s;f)=\sum_{n\ge1}{f(n)\over n^s} $$

It is not difficult to show that $D(s;f)D(s;g)=D(s;f*g)$. Using the properties of multiplication, we easily see that this indicates that Dirichlet convolutions are commutative and associative. If $f(1)\ne0$, then we see that

$$ D(s;f^{-1})=[D(s;f)]^{-1} $$

This indicates that if $f(1)g(1)\ne0$ then

$$ D(s;f^{-1}*g^{-1})=[D(s;f)D(s;g)]^{-1}=D(s;(f*g)^{-1}) $$

Hope this can address your concern!