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. Especially in number theory, 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 multiplicative function is a monomial with leading coefficient 1: For any particular positive integer n, define f(a) = aⁿ.

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 = 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.

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 μf where μ is the Mobius 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

$f*f = \tau \cdot f.$

Here τ is the divisor function.

Proof of pseudo-associative property

$f \cdot \left(g*h \right)(n) = f(n) \sum_{d|n} g(d) h \left( \frac{n}{d} \right)$

$f \cdot \left(g*h \right)(n) = \sum_{d|n} f(n) g(d) h \left( \frac{n}{d} \right)$

$f \cdot \left(g*h \right)(n) = \sum_{d|n} f(d) f \left( \frac{n}{d} \right) g(d) h \left( \frac{n}{d} \right)$ (since f is completely multiplicative)

$f \cdot \left(g*h \right)(n) = \sum_{d|n} f(d) g(d) f \left( \frac{n}{d} \right) h \left( \frac{n}{d} \right)$

$f \cdot \left(g*h \right)(n) = (f \cdot g)*(f \cdot h).$

See also

References

1. ^ Apostol, Tom (1976). Introduction to Analytic Number Theory. Springer. pp. 30. ISBN 0-387-90163-9. 
2. ^ Apostol, p. 36
3. ^ Apostol pg. 49