Möbius Inversion with Additional Term Suppose I have two arithmetic functions $g(n)$ and $f(n)$, where $$g(n) = 1 + \sum_{d|n}f(d). $$ Is it possible to perform a Mobius Inversion and obtain a formula for $f(n)$? If so, how might one proceed? I do not see how to manipulate the above expression to allow a proper inversion to take place. 
 A: We can apply Möbius inversion if we rewrite the relation
$$g(n) = 1 + \sum_{d \mid n} f(d) \tag{1}$$
by modifying $f$ or $g$. We can look at $\tilde{g} \colon n \mapsto g(n) - 1$, so that $(1)$ becomes
$$\tilde{g}(n) = \sum_{d\mid n} f(d), \tag{$1'$}$$
and apply Möbius inversion. In the resulting sum, we can separate the part arising from $g$ and the part arising from the constant $1$ (which simplifies greatly). The same result is obtained - in a very slightly easier manner - if we modify $f$ rather than $g$ and write $(1)$ as
$$g(n) = \sum_{d \mid n} \tilde{f}(d), \tag{$1''$}$$
where $\tilde{f}(n) = f(n) + e(n)$ with
$$e(n) = \begin{cases} 1 &, n = 1 \\ 0 &, n \neq 1\end{cases}.$$
Möbius inversion gives
$$\tilde{f}(n) = \sum_{d\mid n} \mu(d) g\biggl(\frac{n}{d}\biggr),$$
which becomes
$$f(n) = \Biggl(\sum_{d\mid n} \mu(d) g\biggl(\frac{n}{d}\biggr)\Biggr) - e(n).$$
Naturally, since $(1)$ involves a term not coming from $f$, the formula for $f(n)$ in terms of $\mu \ast g$ involves a term not coming from $g$. In this case, the extra terms are very simple.
In the comments, the question of multiplicativity was raised. Since $(1)$ in particular yields $g(1) = 1 + f(1)$, and for a nonzero [whether the zero function is considered multiplicative depends on the author] multiplicative function $h$ one has $h(1) = 1$, not both of $f$ and $g$ can be nonzero multiplicative functions: if $f$ is a nonzero multiplicative function, then $g(1) = 2$, so $g$ cannot be multiplicative, and if $g$ is a nonzero multiplicative function, then $f(1) = 0$, so $f$ cannot be a nonzero multiplicative function.
