从零开始的莫比乌斯反演(函数)[详细推导]

也许更好的阅读体验


\(前置技能\)

  • 学会莫比乌斯函数必须要先知道狄利克雷函数
  • 以及什么是逆元(一本正经胡说八道)

\(狄利克雷卷积\)

两个函数 $f,g$
<font size="3">$\begin{aligned}f*g(n)=\sum_{d|n}^nf(d)g(\frac{n}{d})\end{aligned}$
* 性质
__交换律__$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ f*g=g*f$
__结合律__$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ f*g*h=f*(g*h)$
__加法分配率__$\ \ \ \ \ \ \ \ \ \ f*(g+h)=f*g+f*h$

\(几个定理\)

* ① 定义单位函数$I(n)=[n=1]$,即只有当n等于1时,$I(1)=1$,其余情况都为0
* ② $I * f=f$ 
* ③ 已知函数$f$,定义函数$f$的逆$f^{-1}$,满足$f * f^{-1}=I$,且积性函数的逆也是积性函数
* ④ 由③可得$f^{-1}(1)=\frac{1}{f(1)}$
* ⑤ 令$\xi(n)=1$

\(莫比乌斯函数\)

  • 莫比乌斯函数 \(\mu=\xi^{-1}\)
    \(n=p_1^{a_1}·p_2^{a_2}·\cdots p_k^{a_k}\)
    \(\mu(n)=\xi^{-1}(n)= \left\{\begin{matrix}1 &n=1 \\(-1)^k &a1=a2=\cdots =ak=1&\\ 0&otherwise \end{matrix}\right.\)

\(如何推导\)

  • \(如何求逆\)

    ③等价于

    \(\begin{aligned}\forall n: \sum_{d|n} f(d)·f^{-1}(\frac nd)=[n=1]\end{aligned}\)

    \(\begin{aligned}\sum_{d|n}f(d)·f^{-1}(\frac nd)=I(n) (n=\not 1)=0\end{aligned}\)

    \(f^{-1}(n)·f(1)=-\sum_{d|n,d=\not1}f(d)·f^{-1}(\frac nd)\)

    \(\begin{aligned}f^{-1}(n)=\frac{-\sum_{d|n,d=\not1}f(d)·f^{-1}(\frac nd)}{f(1)}\end{aligned}\)

  • \(莫比乌斯函数\)

    \(f=\xi\),\(p\)为质数

    \(f^{-1}(1)=1\)

    \(f^{-1}(p)=\frac{-f(p)·f^{-1}(1)}{f(1)}=-1\)

    \(f^{-1}(p^2)=-(f(p^2)·f^{-1}(1)+f(p)·f^{-1}(p))=0\)

    \(f^{-1}(p^3)=-(f(p^3)·f^{-1}(1)+f(p^2)·f^{-1}(p)+f(p)·f^{-1}(p^2))=0\)

    \(\begin{aligned}f^{-1}(p^k)=-\sum_{i=1}^k f(p^i)·f^{-1}(p^{k-i})\end{aligned}\)

    其中只有 \(f(p^k)·f^{-1}(1)=1,f(p^{k-1})·f^{-1}(p)=-1\),其余项都是0

    \(f^{-1}(p^k)=\left\{\begin{matrix}-1 & k=1\\0 & k>1 \end{matrix}\right.\)

    \(n=p_1^{a_1}·p_2^{a_2}·\cdots p_k^{a_k}\)

    所以莫比乌斯函数可以推出来了

    \(\mu(n)=f^{-1}(n)= \left\{\begin{matrix}1 &n=1 \\(-1)^k &a1=a2=\cdots =ak=1&\Leftarrow 积性函数推出 \\ 0&otherwise \end{matrix}\right.\)


\(莫比乌斯函数的性质\)

  • \(\mu*\xi=I\)
    因为\(\mu=\xi^{-1}\)
    所以有\(\begin{aligned}\sum_{d|n}^n\mu(d)=\sum_{d|n}^n\mu·1=\mu*\xi(n)=I(n)=[n=1]\end{aligned}\)
    记住这句,下面的证明要用到
  • \(I*f=f\)
  • \(f*\mu*\xi=f\)
  • \(f*\xi=g \Rightarrow f=g*\mu\)

\(线性筛\)

利用积性函数的性质以及\(\mu\)的表达式

int cnt;
int prime[maxn],mu[maxn];
bool vis[maxn];
void mu_sieve (int n)
{
	mu[1]=1;
	for (int i=2;i<=n;++i){
		if (!vis[i]){	prime[++cnt]=i,mu[i]=-1;}
		for (int j=1;j<=cnt&&i*prime[j]<=n;++j){
			vis[i*prime[j]]=true;
			if (i%prime[j]==0){
				mu[i*prime[j]]=0;
				break;
			}
			mu[i*prime[j]]=-mu[i];
		}
	}
}


\(莫比乌斯反演\)

  • 若有 \(\begin{aligned}g(n)=\sum_{d|n}f(d)\end{aligned}\)
    则有 \(\begin{aligned}f(n)=\sum_{d|n}\mu(d)g(\frac{n}{d})=\sum_{d|n}\mu(\frac{n}{d})g(d)\end{aligned}\)

    • \(证明\)

      法一 \(\begin{aligned} \sum_{d|n}{\mu(d)g(\frac{n}{d})}=\sum_{d|n}{\mu(d)\sum_{x|\frac{n}{d}}{f(x)}}=\sum_{x|n}{f(x)}\sum_{d|\frac{n}{x}}{\mu(d)}=\sum_{x|n}{f(x)[\frac{n}{x}=1]}=f(n) \end{aligned}\)
      法二 \(g=f*\xi,f=g*\mu\)
  • 若有\(\begin{aligned}g(n)=\sum_{n|d}f(d)\end{aligned}\)
    则有\(\begin{aligned}f(n)=\sum_{n|d}\mu(\frac{d}{n})g(d)\end{aligned}\)

    • \(证明\)

      \(\begin{aligned}\sum_{n|d}\mu(\frac{d}{n})g(d)=\sum_{k=1}^{+\infty}\mu(k)g(kn)=\sum_{k=1}^{+\infty}\mu(k)\sum_{kn|d}f(d)=\sum_{n|d}f(d)\sum_{k|\frac{d}{n}}\mu(k)=\sum_{n|d}f(d)[\frac{d}{n}=1]=f(n) \end{aligned}\)

    这一种一般用的比较多


\(莫比乌斯反演的应用\)

莫比乌斯函数一般用于\(gcd\)
\(\begin{aligned} \sum_{i=1}^n\sum_{j=1}^m\gcd(i,j)&=\sum_d^{min(n,m)} d\sum_{i=1}^n\sum_{j=1}^m[gcd(i,j)=d] \\ &=\sum_{d}^{min(n,m)}d\sum_{i=1}^{\frac nd}\sum_{j=1}^{\frac md}[gcd(i,j)=1] \\ &=\sum_{d}^{min(n,m)}d\sum_{i=1}^{\frac nd}\sum_{j=1}^{\frac md}\sum_{k\mid gcd(i,j)}\mu(k) \\ &=\sum_d^{min(n,m)} d\sum_k^{min(\frac{n}{d},\frac{m}{d})}\mu(k)\sum_{k\mid i}^{\frac nd}\sum_{k\mid j}^{\frac md} \\ &=\sum_{d}^{min(n,m)}d\sum_k^{min(\frac{n}{d},\frac{m}{d})}\mu(k)\lfloor \frac n{kd}\rfloor \lfloor\frac m{kd}\rfloor \end{aligned}\)
\(T=kd\)
\(\begin{aligned} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\sum_{T}^{min(n,m)}\lfloor\frac nT\rfloor\lfloor\frac mT\rfloor\sum_{k\mid T}\frac Tk\mu(k)\end{aligned}\)
以上是莫比乌斯反演推出来的式子

\(\begin{aligned}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\sum_{T}^{min(n,m)}\lfloor\frac nT\rfloor\lfloor\frac mT\rfloor\varphi(T)\end{aligned}\)
这一步是欧拉反演,证明请移步欧拉函数|(扩展)欧拉定理|欧拉反演
当然,这里这个例子不如直接用欧拉反演
\(\begin{aligned}\sum_{i=1}^n\sum_{j=1}^m\gcd(i,j)&=\sum_{i=1}^n\sum_{j=1}^m\sum_{d|gcd(i,j)}\varphi(d)=\sum_{d=1}^{min(n,m)}\lfloor\frac{n}{d}\rfloor\lfloor\frac{m}{d}\rfloor\varphi(d)\end{aligned}\)


如有哪里讲得不是很明白或是有错误,欢迎指正
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posted @ 2019-07-11 20:37  Morning_Glory  阅读(1524)  评论(1编辑  收藏  举报
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