# Completely multiplicative function

In number theory, functions of positive integers which respect products are important and are called completely multiplicative functions or totally multiplicative functions. A weaker condition is also important, respecting only products of coprime numbers, and such functions are called multiplicative functions. Outside of number theory, the term "multiplicative function" is often taken to be synonymous with "completely multiplicative function" as defined in this article.

## Definition

A completely multiplicative function (or totally multiplicative function) is an arithmetic function (that is, a function whose domain is the natural numbers), such that f(1) = 1 and f(ab) = f(a) f(b) holds for all positive integers a and b.[1]

Without the requirement that f(1) = 1, one could still have f(1) = 0, but then f(a) = 0 for all positive integers a, so this is not a very strong restriction.

## 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.

## 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 = pa qb ..., 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.

There are a variety of statements about a function which are equivalent to it being completely multiplicative. For example, if a function f multiplicative then is completely multiplicative if and only if the Dirichlet inverse is $\mu f$ where $\mu$ is the Möbius function.[2]

Completely multiplicative functions also satisfy a pseudo-associative law. If f is completely multiplicative then

$f \cdot (g*h)=(f \cdot g)*(f \cdot h)$

where * represents the Dirichlet product and $\cdot$ represents pointwise multiplication.[3] One consequence of this is that for any completely multiplicative function f one has

$f*f = \tau \cdot f$

which deduced from the latter/above for [both] $g = h = 1$, where $1(n) = 1$ is well-known constant function. Here $\tau$ is the divisor function.

### Proof of pseudo-associative property

\begin{align} f \cdot \left(g*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)*(f \cdot h). \end{align}

### Dirichlet series

Moreover, The L-function of completely (or totally) multiplicative Dirichlet series a(n) satisfies

$L(s,a)=\sum^\infty_{n=1}\frac{a(n)}{n^s}=\prod_p\biggl(1-\frac{a(p)}{p^s}\biggr)^{-1},$

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