积性加性函数运算的性质证明
目录
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 \(\forall n,m\in Z_+,gcd(n,m)=1\)
\(h(nm)=(f+g)(nm)=f(nm)+g(nm)=f(n)+f(m)+g(n)+g(m)=(\ f(n)+g(n)\ )+(\ f(m)+g(m)\ )=h(n)+h(m)\)
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
积性函数的相乘还是积性函数
设 \(\boldsymbol f,\boldsymbol g\) 为积性函数， \(h=\boldsymbol f\cdot \boldsymbol g\)
则对于 \(\forall n,m\in Z_+,gcd(n,m)=1\)
\(h(nm)=(\boldsymbol f\cdot \boldsymbol g)(nm)=\boldsymbol f(nm)\cdot \boldsymbol g(nm)=\boldsymbol f(n)\cdot \boldsymbol f(m)\cdot \boldsymbol g(n)\cdot \boldsymbol g(m)=(\ \boldsymbol f(n)\cdot \boldsymbol g(n)\ )\cdot (\ \boldsymbol f(m)\cdot \boldsymbol g(m)\ )=h(n)\cdot h(m)\)
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
积性函数的迪利克雷卷积还是积性函数
设 \(\boldsymbol f,\boldsymbol g\) 为积性函数，且 \(\boldsymbol h=\boldsymbol f*\boldsymbol g\)
则对于 \(\forall n,m\in Z_+,gcd(n,m)=1\)
\(\displaystyle \boldsymbol h(nm)=(\boldsymbol f*\boldsymbol g)(nm)=\sum_{d\mid nm}\boldsymbol f(d)\boldsymbol g({nm\over d})\)
考虑到 \(n,m\) 互质，因此没有质因数重合
故 \(\displaystyle \boldsymbol h(nm)=\sum_{d\mid nm}\boldsymbol f(d)\boldsymbol g({nm\over d})=\sum_{d_1\mid n\wedge d_2\mid m}\boldsymbol f(d_1d_2)\boldsymbol g({nm\over d_1d_2})=\sum_{d_1\mid n\wedge d_2\mid m}\boldsymbol f(d_1)\boldsymbol f(d_2)\boldsymbol g({n\over d_1})\boldsymbol g({m\over d_2})=(\sum_{d_1\mid n}\boldsymbol f(d_1)\boldsymbol g({n\over d_1})\ )\cdot (\sum_{d_2\mid n}\boldsymbol f(d_2)\boldsymbol g({m\over d_2})\ )=\boldsymbol h(n)\cdot \boldsymbol h(m)\)
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
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 \(\boldsymbol f\) 的逆元 \(\boldsymbol g\)
由上节课的知识我们可以得到： \(\displaystyle \boldsymbol g(n)=[n=1]-\sum_{d\mid n\wedge d\neq 1}\boldsymbol f(d)\boldsymbol g({n\over d})\)
首先 \(\boldsymbol g(1)=[1=1]-\sum_{d\mid 1\wedge d\neq 1}\boldsymbol f(d)\boldsymbol g({n\over d})=1-0=1\)
其次，假设 \(n,m\in Z_+,gcd(n,m)=1\) 且 \(nm>1,nm\) 以内 \(\boldsymbol g\) 都满足积性函数的性质
故：
\(\displaystyle \boldsymbol g(nm)=[nm=1]-\sum_{d\mid nm\wedge d\neq 1}\boldsymbol f(d)\boldsymbol g({nm\over d})\)
\(\displaystyle =0-\sum_{d_1\mid n\wedge d_2\mid m\wedge d_1d_2\neq 1}\boldsymbol f(d_1d_2)\boldsymbol g({nm\over d_1d_2})\)
由于 \(d_1d_2\neq 1\) 故 \({nm\over d_1d_2}<nm\) 使得 \(\boldsymbol g\) 满足积性函数的性质
\(\displaystyle =-\sum_{d_1\mid n\wedge d_2\mid m\wedge d_1d_2\neq 1}\boldsymbol f(d_1)\boldsymbol f(d_2)\boldsymbol g({n\over d_1})\boldsymbol g({m\over d_2})\)
\(\displaystyle =\boldsymbol f(1)\boldsymbol f(1)\boldsymbol g(n)\boldsymbol (m)-\boldsymbol f(1)\boldsymbol f(1)\boldsymbol g(n)\boldsymbol (m)-\sum_{d_1\mid n\wedge d_2\mid m\wedge d_1d_2\neq 1}\boldsymbol f(d_1)\boldsymbol f(d_2)\boldsymbol g({n\over d_1})\boldsymbol g({m\over d_2})\)
\(\displaystyle =1\cdot 1\cdot \boldsymbol g(n)\boldsymbol (m)-\sum_{d_1\mid n\wedge d_2\mid m}\boldsymbol f(d_1)\boldsymbol f(d_2)\boldsymbol g({n\over d_1})\boldsymbol g({m\over d_2})\)
\(\displaystyle =\boldsymbol g(n)\boldsymbol g(m)-(\sum_{d_1\mid n}\boldsymbol f(d_1)\boldsymbol g({n\over d_1})\ )\cdot (\sum_{d_2\mid m}\boldsymbol f(d_2)\boldsymbol g({m\over d_2})\ )\)
\(\displaystyle =\boldsymbol g(n)\boldsymbol g(m)-(\boldsymbol f*\boldsymbol g)(n)\cdot (\boldsymbol f*\boldsymbol g)(m)\)
\(\displaystyle =\boldsymbol g(n)\boldsymbol g(m)-\boldsymbol \varepsilon(n)\cdot \boldsymbol \varepsilon(m)\)
\(\displaystyle =\boldsymbol g(n)\boldsymbol g(m)-\boldsymbol \varepsilon(nm)\)
\(\displaystyle =\boldsymbol g(n)\boldsymbol g(m)-[nm=1]\)
\(\displaystyle =\boldsymbol g(n)\boldsymbol g(m)-0\)
\(\displaystyle =\boldsymbol g(n)\boldsymbol g(m)\)
由此，积性得证
