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Other then the logarithm, does anybody know of any other important completely additive functions, by completely additive I mean $f(ab)=f(a)+f(b)$, for any integers $a,b$ ?

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What do you mean by 'completely additive'? Functions with which domain and codomain? additive with respect to which operation? – Ittay Weiss Nov 22 '12 at 5:04
only for integers??? (try f(x)=0) – Ittay Weiss Nov 22 '12 at 5:09
Just look in Wikipedia... several are listed. The total number of prime factors of $n$, counting multiplicity, is one. – mjqxxxx Nov 22 '12 at 5:56
You want to exclude $0$ from the domain of $f$, or else one is forced to have $f(b)=f(0b)-f(0)=0$ for all $b\in\mathbf Z$. – Marc van Leeuwen Nov 22 '12 at 9:08
up vote 5 down vote accepted

There is a general description of such functions. For each prime p, choose $f(p)$ arbitrarily. Then for a general integer $n=\prod_ip_i^{n_i}$, we have $f(n)=\sum_i n_i f(p_i)$. Possible choices for $f(p)$ include log $p$ (which gives the logarthm) and 1 for every p (the weighted prime divisor counting function) but there are uncountably many more choices.

(PS: Gerry's example of the "$p$-adic order function" comes from setting $f(p) = 1$ for one specific prime $p$ and $f(q) = 0$ for all other primes $q$.)

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For any prime $p$, there's the $p$-adic order (or $p$-adic valuation) of $n$, written $\nu_p(n)$, defined by $\nu_p(n)=k$ if $p^k$ divides $n$ and $p^{k+1}$ doesn't. $\nu_p(mn)=\nu_p(m)+\nu_p(n)$

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Consider the fact that any positive rational number $q$ can be rewritten as $2^{e_1}3^{e_2}5^{e_3}7^{e_4}… p_n^{e_n}$. Then, taking the exponents $e_i$ of each prime, we can construct a vector as follows:


(for arbitrary constants $C_i$)

Our general function is then $f(q)=\sum_iC_ie_i$ (the sum of the vector components). You can think of this as a varation of the L1-norm where sign matters. It becomes equivalent to the logarithm function when $C_i =\ln p_i$.

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