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单词 CriterionForAMultiplicativeFunctionToBeCompletelyMultiplicative
释义

criterion for a multiplicative function to be completely multiplicative


Theorem.

Let f be a multiplicative functionMathworldPlanetmath with convolution inverse g. Then f is completely multiplicative if and only if g(pk)=0 for all primes p and for all kN with k>1.

Proof.

Note first that, since f(1)=1 and f*g=ε, where ε denotes the convolution identity function, then g(1)=1. Let p be any prime. Then

0=ε(p)=(f*g)(p)=f(1)g(p)+f(p)g(1)=g(p)+f(p).

Thus, g(p)=-f(p).

Assume that f is completely multiplicative. The statement about g will be proven by inductionMathworldPlanetmath on k. Note that:

0=ε(p2)=(f*g)(p2)=f(1)g(p2)+f(p)g(p)+f(p2)g(1)=g(p2)+f(p)(-f(p))+(f(p))2=g(p2)

Let m with m>2 such that, for all k with 1<k<m, g(pk)=0. Then:

0=ε(pm)=(f*g)(pm)=f(1)g(pm)+f(pm-1)g(p)+f(pm)g(1)=g(pm)+(f(p))m-1(-f(p))+(f(p))m=g(pm)

Conversely, assume that g(pk)=0 for all k with k>1. The statement f(pk)=(f(p))k will be proven by induction on k. The statement is obvious for k=1. Let m such that f(pm-1)=(f(p))m-1. Then:

0=ε(pm)=(f*g)(pm)=f(pm-1)g(p)+f(pm)g(1)=(f(p))m-1(-f(p))+f(pm)=-(f(p))m+f(pm)

Thus, f(pm)=(f(p))m. It follows that f is completely multiplicative.∎

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