Completely multiplicative function

Examples

The easiest example of a completely multiplicative function is a monomial with leading coefficient 1: For any particular positive integer n, define f(a) = aⁿ. Then f(bc) = (bc)ⁿ = bⁿcⁿ = f(b)f(c), and f(1) = 1ⁿ = 1.

The Liouville function is a non-trivial example of a completely multiplicative function as are Dirichlet characters, the Jacobi symbol and the Legendre symbol.

Properties

A completely multiplicative function is completely determined by its values at the prime numbers, a consequence of the fundamental theorem of arithmetic. Thus, if n is a product of powers of distinct primes, say n = p^a q^b ..., then f(n) = f(p)^a f(q)^b ...

While the Dirichlet convolution of two multiplicative functions is multiplicative, the Dirichlet convolution of two completely multiplicative functions need not be completely multiplicative. Arithmetic functions which can be written as the Dirichlet convolution of two completely multiplicative functions are said to be quadratics or specially multiplicative multiplicative functions. They are rational arithmetic functions of order (2, 0) and obey the Busche-Ramanujan identity.

There are a variety of statements about a function which are equivalent to it being completely multiplicative. For example, if a function f is multiplicative then it is completely multiplicative if and only if its Dirichlet inverse is ${\displaystyle \mu \cdot f}$ where ${\displaystyle \mu }$ is the Möbius function.

Completely multiplicative functions also satisfy a distributive law. If f is completely multiplicative then

${\displaystyle f\cdot (g\ast h)=(f\cdot g)\ast (f\cdot h)}$

where * represents the Dirichlet product and ${\displaystyle \cdot }$ represents pointwise multiplication. One consequence of this is that for any completely multiplicative function f one has

${\displaystyle f\ast f=\tau \cdot f}$

which can be deduced from the above by putting both ${\displaystyle g=h=1}$, where ${\displaystyle 1(n)=1}$ is the constant function. Here ${\displaystyle \tau }$ is the divisor function.

Proof of distributive property

${\displaystyle \begin{array}{cc}f\cdot \left(g\ast h\right)(n)& =f(n)\cdot \sum _{d|n}g(d)h\left(\frac{n}{d}\right)=\\ & =\sum _{d|n}f(n)\cdot (g(d)h\left(\frac{n}{d}\right))=\\ & =\sum _{d|n}(f(d)f\left(\frac{n}{d}\right))\cdot (g(d)h\left(\frac{n}{d}\right))\text{ (since }f\text{ is completely multiplicative) }=\\ & =\sum _{d|n}(f(d)g(d))\cdot (f\left(\frac{n}{d}\right)h\left(\frac{n}{d}\right))\\ & =(f\cdot g)\ast (f\cdot h).\end{array}}$

Dirichlet series

The L-function of completely (or totally) multiplicative Dirichlet series ${\displaystyle a(n)}$ satisfies

${\displaystyle L(s,a)=\sum _{n=1}^{\mathrm{\infty }}\frac{a(n)}{n^s}=\prod _p(1-\frac{a(p)}{p^s}{)}^{-1},}$

which means that the sum all over the natural numbers is equal to the product all over the prime numbers.