composition of multiplicative functions


If f is a completely multiplicative functionMathworldPlanetmath and g is a multiplicative function, then fg is a multiplicative function.


First note that (fg)(1)=f(g(1))=f(1)=1 since both f and g are multiplicative.

Let a and b be relatively prime positive integers. Then

(fg)(ab) =f(g(ab))
=f(g(a)g(b)) since g is multiplicative
=f(g(a))f(g(b)) since f is completely multiplicative

Note that the assumptionPlanetmathPlanetmath that f is completely multiplicative (as opposed to merely multiplicative) is essential in proving that fg is multiplicative. For instance, ττ, where τ denotes the divisor functionDlmfDlmfMathworldPlanetmath, is not multiplicative:

Title composition of multiplicative functions
Canonical name CompositionOfMultiplicativeFunctions
Date of creation 2013-03-22 16:09:50
Last modified on 2013-03-22 16:09:50
Owner Wkbj79 (1863)
Last modified by Wkbj79 (1863)
Numerical id 6
Author Wkbj79 (1863)
Entry type Theorem
Classification msc 11A25