积性加性函数运算的性质证明

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加性函数相加还是加性函数

设 (f,g) 为加性函数， (h=f+g)

则对于 (forall n,min Z_+,gcd(n,m)=1)

(h(nm)=(f+g)(nm)=f(nm)+g(nm)=f(n)+f(m)+g(n)+g(m)=( f(n)+g(n) )+( f(m)+g(m) )=h(n)+h(m))

且易证得，当且仅当两者都为完全加性时，相加也一定为完全加性

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积性函数的相乘还是积性函数

设 (oldsymbol f,oldsymbol g) 为积性函数， (h=oldsymbol fcdot oldsymbol g)

则对于 (forall n,min Z_+,gcd(n,m)=1)

(h(nm)=(oldsymbol fcdot oldsymbol g)(nm)=oldsymbol f(nm)cdot oldsymbol g(nm)=oldsymbol f(n)cdot oldsymbol f(m)cdot oldsymbol g(n)cdot oldsymbol g(m)=( oldsymbol f(n)cdot oldsymbol g(n) )cdot ( oldsymbol f(m)cdot oldsymbol g(m) )=h(n)cdot h(m))

且易证得，当且仅当两者都为完全积性时，相乘一定也为完全积性

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积性函数的迪利克雷卷积还是积性函数

设 (oldsymbol f,oldsymbol g) 为积性函数，且 (oldsymbol h=oldsymbol f*oldsymbol g)

则对于 (forall n,min Z_+,gcd(n,m)=1)

(displaystyle oldsymbol h(nm)=(oldsymbol f*oldsymbol g)(nm)=sum_{dmid nm}oldsymbol f(d)oldsymbol g({nmover d}))

考虑到 (n,m) 互质，因此没有质因数重合

故 (displaystyle oldsymbol h(nm)=sum_{dmid nm}oldsymbol f(d)oldsymbol g({nmover d})=sum_{d_1mid nwedge d_2mid m}oldsymbol f(d_1d_2)oldsymbol g({nmover d_1d_2})=sum_{d_1mid nwedge d_2mid m}oldsymbol f(d_1)oldsymbol f(d_2)oldsymbol g({nover d_1})oldsymbol g({mover d_2})=(sum_{d_1mid n}oldsymbol f(d_1)oldsymbol g({nover d_1}) )cdot (sum_{d_2mid n}oldsymbol f(d_2)oldsymbol g({mover d_2}) )=oldsymbol h(n)cdot oldsymbol h(m))

由于 (d_1,d_2) 分别是 (n,m) 的因数，故当两者均不为 (1) 的时候，对迪利克雷卷积的贡献是相互独立的，利用乘法原理可以分开来；当两者中存在 (1) 的时候，函数值为 (1) ，对迪利克雷卷积的贡献还是独立的

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积性函数的逆元还是积性函数

我们用归纳法证明。思路来源：铃悬的数学小讲堂——狄利克雷卷积与莫比乌斯反演

对于积性函数 (oldsymbol f) 的逆元 (oldsymbol g)

由上节课的知识我们可以得到： (displaystyle oldsymbol g(n)=[n=1]-sum_{dmid nwedge d eq 1}oldsymbol f(d)oldsymbol g({nover d}))

首先 (oldsymbol g(1)=[1=1]-sum_{dmid 1wedge d eq 1}oldsymbol f(d)oldsymbol g({nover d})=1-0=1)

其次，假设 (n,min Z_+,gcd(n,m)=1) 且 (nm>1,nm) 以内 (oldsymbol g) 都满足积性函数的性质

故：

(displaystyle oldsymbol g(nm)=[nm=1]-sum_{dmid nmwedge d eq 1}oldsymbol f(d)oldsymbol g({nmover d}))

(displaystyle =0-sum_{d_1mid nwedge d_2mid mwedge d_1d_2 eq 1}oldsymbol f(d_1d_2)oldsymbol g({nmover d_1d_2}))

由于 (d_1d_2 eq 1) 故 ({nmover d_1d_2}<nm) 使得 (oldsymbol g) 满足积性函数的性质

(displaystyle =-sum_{d_1mid nwedge d_2mid mwedge d_1d_2 eq 1}oldsymbol f(d_1)oldsymbol f(d_2)oldsymbol g({nover d_1})oldsymbol g({mover d_2}))

(displaystyle =oldsymbol f(1)oldsymbol f(1)oldsymbol g(n)oldsymbol (m)-oldsymbol f(1)oldsymbol f(1)oldsymbol g(n)oldsymbol (m)-sum_{d_1mid nwedge d_2mid mwedge d_1d_2 eq 1}oldsymbol f(d_1)oldsymbol f(d_2)oldsymbol g({nover d_1})oldsymbol g({mover d_2}))

(displaystyle =1cdot 1cdot oldsymbol g(n)oldsymbol (m)-sum_{d_1mid nwedge d_2mid m}oldsymbol f(d_1)oldsymbol f(d_2)oldsymbol g({nover d_1})oldsymbol g({mover d_2}))

(displaystyle =oldsymbol g(n)oldsymbol g(m)-(sum_{d_1mid n}oldsymbol f(d_1)oldsymbol g({nover d_1}) )cdot (sum_{d_2mid m}oldsymbol f(d_2)oldsymbol g({mover d_2}) ))

(displaystyle =oldsymbol g(n)oldsymbol g(m)-(oldsymbol f*oldsymbol g)(n)cdot (oldsymbol f*oldsymbol g)(m))

(displaystyle =oldsymbol g(n)oldsymbol g(m)-oldsymbol varepsilon(n)cdot oldsymbol varepsilon(m))

(displaystyle =oldsymbol g(n)oldsymbol g(m)-oldsymbol varepsilon(nm))

(displaystyle =oldsymbol g(n)oldsymbol g(m)-[nm=1])

(displaystyle =oldsymbol g(n)oldsymbol g(m)-0)

(displaystyle =oldsymbol g(n)oldsymbol g(m))

由此，积性得证