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单词 CompositionOfMultiplicativeFunctions
释义

composition of multiplicative functions


Theorem.

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

Proof.

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

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:

(ττ)(23)=(ττ)(6)=τ(τ(6))=τ(4)=3
(ττ)(2)(ττ)(3)=τ(τ(2))τ(τ(3))=τ(2)τ(2)=22=4
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