Multiplicative arithmetic function

2020 Mathematics Subject Classification: Primary: 11A25 [MSN][ZBL]

$\def\Epsilon{\mathrm{E}}$ An arithmetic function of one argument, $f(m)$, satisfying the condition $$\begin{equation} f(mn) = f(m) f(n) \label{mult} \end{equation}$$ for any pair of coprime integers $m,n$. It is usually assumed that $f$ is not identically zero (which is equivalent to the condition $f(1)=1$). A multiplicative arithmetic function is called strongly multiplicative if $f(p^a) = f(p)$ for all prime numbers $p$ and all natural numbers $a$. If $\eqref{mult}$ holds for any two numbers $m,n$, and not just for coprime numbers, then $f$ is called totally multiplicative; in this case $f(p^a) = f(p)^a$.

Examples of multiplicative arithmetic functions. The function $\tau(m)$, the number of divisors of a natural number $m$; the function $\sigma(m)$, the sum of divisors of a natural number $m$; the Euler function $\phi(m)$; and the Möbius function $\mu(m)$. The function $\phi(m)/m$ is a strongly multiplicative arithmetic function, a power function $m^k$ is a totally multiplicative arithmetic function.

Comments

The Dirichlet convolution product

$$(f*g)(n) = \sum_{d\vert n} f(d) g(n/d)\ $$

yields a commutative group structure on the multiplicative functions. The unit element is given by the function $e$, where $e(1)=1$ and $e(m) = 0$ for all $m > 1$. Another standard multiplicative function is the constant function $\Epsilon(n)$ with $\Epsilon(m) = 1$ for all $m$ and its inverse $\mu$, the Möbius function. Note that $\phi = \mu * N_1$, where $N_1(n) = n$ for all $n$, and that $\tau = \Epsilon * \Epsilon$, $\sigma = \Epsilon * N_1$. In this context, the Möbius inversion formula states that if $g = \Epsilon * f$ then $f = \mu * g$.

Formally, the Dirichlet series of a multiplicative function $f$ has an Euler product:

$$\sum_{n=1}^\infty f(n) n^{-s} = \prod_p \left({1 + f(p) p^{-s} + f(p^2) p^{-2s} + \cdots }\right) \ ,$$

whose form simplifies considerably if $f$ is strongly or totally multiplicative: if $f$ is strongly multiplicative then $$\sum_{n=1}^\infty f(n) n^{-s} = \prod_p \left({1 + f(p) p^{-s} (1 - p^{-s})^{-1}} \right) \ ,$$ and if $f$ is totally multiplicative then $$\sum_{n=1}^\infty f(n) n^{-s} = \prod_p \left({1 - f(p) p^{-s}}\right)^{-1} \ ,$$

Dirichlet convolution of functions corresponds to multiplication of the associated Dirichlet series.

References

[HaWr]

G.H. Hardy, E.M. Wright, "An introduction to the theory of numbers", Clarendon Press (1960) pp. Chapts. XVI-XVII MR2445243 MR1561815 Zbl 0086.25803