积性函数
Def.
积性函数：
\(1)f(1)=1\)
\(2)x,y\in N^{+},gcd(x,y)=1,都有f(xy)=f(x)f(y)\)
完全积性函数:
\(1)f(1)=1\)
\(2)x,y\in N^{+},都有f(xy)=f(x)f(y)\)
e.g.
完全积性函数：
\(1）单位函数：\varepsilon(n)=[n==1]\)
\(2）恒等函数：id_k(n)=n^k,id_1简记id\)
\(3)常数函数：1(n)=1\)
积性函数：
\(1)欧拉函数：\phi(n)=\sum_{i=1}^{n}[gcd(i,n)==1]\)
\(2)除数函数:\sigma_k(n)=\sum_{d|n}d^k\)
\(\quad k=0,约数个数函数\sigma_0\)
\(\quad k=1,约数和函数\sigma_1，简记为\sigma\)
\(3)Möbius ：\mu(n)=\begin{cases} 1&n =1\\ 0&\text{n含有平方因子}\\ (-1)^k& \text{k为n的本质不同的质因子个数} \end{cases}\)
性质：
\(若f(x)和g(x)均为积性函数，则以下函数也都是积性函数:\)
\(1)h(x)=f(x^p)\)
\(2)h(x)=f^p(x)\)
\(3)h(x)=f(x)g(x)\)
\(4)h(x)=\sum_{d|x}f(d)g(\frac{x}{d})\)
卷积
Def.\((f\ast g)(n)=\sum_{d\mid n}f(d)g(\frac{n}{d})\)
\(性质：\)
\(\cdot 交换律:f\ast g=g\ast f\)
\(\cdot 结合律:(f\ast g)\ast h=f\ast (g\ast h)\)
\(\cdot分配律:f\ast (g+h)=f\ast g +f\ast h\)
\(\cdot f\ast \varepsilon=f,其中\varepsilon 为Dirichlet 卷积的单位元 (任何函数卷\varepsilon 为其本身)\)
Theorem.若f,g为两个积性函数，则f*g为积性函数
prove.n⊥m(gcd(n,m)=1)
\(要证f*g为积性函数，即证：(f*g)(n)\cdot(f*g)(m)=f*g(nm)\)
\(根据定义，\sum_{d\mid n}f(d)g(\frac{n}{d})\)
\(左=(\sum_{d_1\mid n}f(d_1)g(\frac{n}{d_1}))(\sum_{d_2\mid m}f(d_2)g(\frac{m}{d_2}))\)
\(\quad=\sum_{d_1\mid n}(f(d_1)g(\frac{n}{d_1})(\sum_{d_2\mid m}f(d_2)g(\frac{m}{d_2}))\)
\(\quad=\sum_{d_1\mid n}\sum_{d_2\mid m}f(d_1)g(\frac{n}{d_1})f(d_2)g(\frac{m}{d_2})[]\)
\(\quad=\sum_{d_1\mid n}\sum_{d_2\mid m}f(d_1d_2)g(\frac{nm}{d_1d_2})\)
\(\quad=\sum_{d\mid nm}f(nm)g(\frac{nm}{d})\)
\(\because n\perp m\)
\(\therefore d_1\perp d_2,\forall d|nm,则\exists d=d_1\cdot d_2唯一分解，s.t. d_1|n且d_2|m\)
\(\therefore d_1不同或d_2不同,d一定不同[分解唯一性]，d不同，d_1或d_2也不同[d_1*d_2=常数]\)
\([像极了线性代数空间]\)
定义
\(n=\prod_{i=1}^kp_i^{c_i}，其中p_i为质因子，c_i\ge1\)
\(1.n=1,\mu(n)=1;\)
\(2.对于n\neq 1时\)
\(\quad a.\exists i\in [1,k] ,s.t.c_i>1 ,则\mu(n)=0\)
\(\quad b.\forall i\in [1,k],c_i=0,则\mu(n)=(-1)^k\)
性质
\(1.积性函数\)
\(2.f(n)=\sum_{d\mid n}\mu(d)=\begin{cases}1&n=1\\0&n\neq 1\\\end{cases}.\)
\(3.f=\varepsilon\)
\(4.即\varepsilon=\mu*1\)
\(prove.\)
\(2.设n=\prod_{i=1}^kp_i^{c_i},m=\prod_{i=1}^kp_i\)
\(\therefore \sum_{d\mid n}\mu(d)=\sum_{d\mid m}\mu(d)=\sum_{i=0}^kC_k^i\cdot(-1)^k=(1+(-1))^k\)
\(\therefore \sum_{d\mid n}\mu(d)=\begin{cases}1&n=1\\0&n\neq 1\\\end{cases}\)
\(\therefore \sum_{d\mid n}\mu(d)=[n==1]\)
\(3.f(n)=\sum_{d\mid n}\mu(d)\)
\(\mu\ast f=\sum_{d\mid n}\mu(d)f(\frac{n}{d})=\sum_{d\mid n}(\mu(d)\sum_{d_2\mid \frac{n}{d}}\mu(d_2))=\sum_{d\mid n}(\sum_{d_2\mid \frac{n}{d}}\mu(d)\mu(d_2))\)
\(考虑到只有gcd(d,d_2)=1，\mu(d)\neq 0，\mu(d_2)\neq 0\)
\(m=\prod_{i=1}^kp_i\)
\(\therefore 原式= \sum_{i=0}^kC_k^i\cdot(-1)^k=\mu(n)\)
\(\therefore \mu*f=\mu\)
\(\therefore f=\varepsilon\)
\(反演结论:\)
\(\varepsilon(gcd(i,j))=[gcd(i,j)==1]\Leftrightarrow \sum_{d\mid gcd(i,j)}\mu(d)\)
拓展
\(1）\phi*1=id\)
\(prove.\)
\(n=\prod_{i=1}^{k}p_i^{c_i},令m_i=p_i^{c_i}\)
\(\because 积性函数\)
\(\therefore \phi(n)=\prod_{i=1}^k\phi(p_i^{c_i})\)
\(若(\phi *1)(m_i)=id(m_i),则两边同乘(\phi*1)(m_j)=id(m_j)\)
\((\phi *1)(m_i)\cdot (\phi*1)(m_j)=id(m_i)\cdot id(m_j)\)
\(\Rightarrow (\phi *1)(m_i\cdot m_j)=id(m_i\cdot m_j)[\phi*1积性函数]\)
\(\Rightarrow 要证\phi(n)=id(n) ,只要证\phi(m_i)=id(m_i)\)
\(prove:\phi(m_i)=id(m_i)\)
\(因为p是质数，所以d=p^0,p^1,p^2,...,p^c\)
\(\phi(p^i)=ip^i-p^{i-1},因为一共有p^{i-1}个数是p的倍数，与p^i不互质\)
\(\phi*1=\sum_{d|m}\phi(\frac{m}{d})\)
\(\quad =\sum_{i=0}^{c}\phi(p^c)\)
\(\quad =p^0+(p^1-p^0)+(p^2-p^1)+...+(p^c-p^{c-1})\)
\(\quad =p^c\)
\(\quad =id\)
\(2)\phi=id*\mu\)
\(prove.\)
\(\phi*1=id,两边同时卷\mu\)
\(得\phi*(1*\mu)=id*\mu\)
\(得\phi*(\varepsilon)=id*\mu\)
\(得\phi=id*\mu\)
Möbius 反演
公式：
\(1）如果有f(n)=\sum_{d|n}g(d),那么有g(n)=\sum_{d|n}\mu(d)f(\frac{n}{d}),f=g*1\Rightarrow g=\mu*f\)
\(2）如果有f(n)=\sum_{n|d}g(d),那么有g(n)=\sum_{n|d}\mu(\frac{d}{n})f(d)\)
\(2)根据zx书写习惯：如果有f(d)=\sum_{d|n}g(n),那么有g(d)=\sum_{d|n}\mu(\frac{n}{d})f(n)\)
prove.
\(1)\)
\(方法一，利用\mu*1=\varepsilon\)
\(方法二，数论变换\)
\(\quad \sum_{d|n}\mu(d)f(\frac{n}{d})\)
\(\quad=\sum_{d|n}\mu(d)\sum_{k|\frac{n}{d}}g(k)\)
\(\quad=\sum_{d|n}\sum_{k|\frac{n}{d}}\mu(d)g(k)\)
\(\quad=\sum_{k|n}\sum_{d|\frac{n}{k}}\mu(d)g(k)\quad[d|n且k|\frac{n}{d} \Leftrightarrow k|n且d|\frac{n}{k},求和符号其实是矩阵元求和，元素(d,k)依然在矩阵中]\)
\(\quad=\sum_{k|n}g(k)\sum_{d|\frac{n}{k}}\mu(d)\)
\(\quad=g(n)\quad [\sum_{d|n}\mu(d)=[n==1]=\varepsilon ]\)
\(2)\)
\(利用上面方法二，考虑逆推这个式子[复习完黎曼重排继续推]\)
\(\quad\sum_{d|n}\mu(\frac{n}{d})f(n)\)
\(=\sum_{d|n}\mu(\frac{n}{d})\sum_{n|m}g(m)\quad 已知d枚举n，已知n枚举m\)
\(=(\sum_{d|n}\mu(\frac{n}{d}))(\sum_{n|m}g(m))\)
\(令\frac{n}{d}=i,\frac{m}{n}=j\)
\(=\sum_{i=1}^{\infty}\mu(i)\sum_{j=1}^{\infty}g(ijd)\)