Suppose I give you a function $f: \mathcal{P}\to \mathbb{C}.$ There are at least three ways to construct a multiplicative function extending $f.$ The first is "completely multiplicative" (to compute $f(n),$ factor $n$ and multiply $f$ of prime factors, with multiplicity), in the second, you compute $f$ of the square free part of $n,$ and the third (a la Moebius), is to set $f$ of a non-square free $n$ to $0,$ (and otherwise do the obvious thing). The question is, what is the formal relationship between these three extensions? And are there other natural extensions which have escaped me?
$\begingroup$
$\endgroup$
6
asked Jul 27, 2017 at 2:35
-
$\begingroup$ In the context of L-functions, you'll want to write $$F(s) = \prod_p (1+f(p) p^{-s}) \prod_p (1+\sum_{k \ge 2} g(p^k)p^{-sk})$$ Such that (the Dirichlet series for) $\sum_p \log (1+\sum_{k \ge 2} g(p^k)p^{-sk})$ converges (absolutely) for $\Re(s) > \sigma$ the abscissa of convergence of $\sum_{p} \log(1+f(p) p^{-s})$ or $\sum_p f(p) p^{-s}$. $\endgroup$– reunsCommented Jul 27, 2017 at 4:14
-
$\begingroup$ The second is a la von Mangoldt. $\endgroup$– Lior Bary-SorokerCommented Jul 27, 2017 at 6:38
-
1$\begingroup$ The Euler phi-function is pretty natural, and falls into none of your three categories. Ditto for the number of divisors, the sum of divisors,.... $\endgroup$– Gerry MyersonCommented Jul 27, 2017 at 12:46
-
$\begingroup$ @GerryMyerson Yes, granted. $\endgroup$– Igor RivinCommented Jul 27, 2017 at 13:17
-
$\begingroup$ @GerryMyerson ... but if you notice, I am phrasing this as an extension problem, and presumably some extensions are a priori more natural than others (the Euler phi is extending $p-1,$ the divisor function $p+1,$ the number of divisors, $2,$ and if you did not know about their deeper meanings none of the extensions look particularly natural (notice that once you define the function on prime powers, there is no choice, so maybe the question is about the extension to prime powers...) $\endgroup$– Igor RivinCommented Jul 27, 2017 at 14:33
|
Show 1 more comment
1 Answer
$\begingroup$
$\endgroup$
Let $\ f_n:\mathcal P\rightarrow\mathbb C\ $ be arbitrary for $\ n=1\ 2\ ...\ $ Let $\ \sigma(x)\ $ be the square-free part of $x,\ $ and $\ \rho(x):= \frac x{\sigma(x)}.\ $ Then define:
$$ F(x)\,\ :=\,\ \prod_n\ f_n(x_n) $$
where \begin{eqnarray} y_0\ &:=&\ x \nonumber\\ \forall_{n=1\ 2\ \ldots}\quad x_n\ &:=& \sigma(y_{n-1}) \nonumber\\ \forall_{n=1\ 2\ \ldots}\quad y_n\ &:=&\ \rho(y_{n-1}) \end{eqnarray}
This is a pretty general thing (possibly just general?).
