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# 定理内容

$$F(n)=\sum_{d|n}f(d)$$

$$f(n)=\sum_{d|n}F(d)\mu(\frac{n}{d})$$

# 证明

## 狄利克雷卷积

### 运算法则

$$(f*g)(n)=\sum_{d|n}f(d)g(\frac{n}{d})$$

### 单位元

$$[n=1]= \begin{cases} 1&n=1\ 0&n\not=1 \end{cases}$$

$$(\epsilon*f)(n)=\sum_{d|n}\epsilon(d)f(\frac{n}{d})=\ \sum_{d|n}f(\frac{n}{d})[d=1]$$

### 其它的奇妙的函数

#### 1函数

$1(n)=1$

$$(1*f)(n)=\sum_{d|n}f(d)$$

#### $\mu$函数

$\mu$是莫比乌斯函数，它的定义是这样的：

$$\mu= \begin{cases} 1&n=1\ (-1)^k&n=p_1p_2\cdots p_k\ 0&otherwise \end{cases}$$

For any positive integer n, define μ(n) as the sum of the primitive nth roots of unity. It has values in {−1, 0, 1} depending on the factorization of n into prime factors:

• $\mu (n)=1$ if n is a square-free positive integer with an even number of prime factors.
• $\mu (n)=-1$ if n is a square-free positive integer with an odd number of prime factors.
• $\mu (n)=0$ if n has a squared prime factor.

From Wikipedia

$$\sum_{d|n}\mu(d)=[n=1]$$

## 证明

$$F(n)=\sum_{d|n}f(d)$$

$$f(n)=\sum_{d|n}F(d)\mu(\frac{n}{d})$$

$$F=1*f$$

$$f=\mu*F$$

$$F=1*\mu*F$$

# 参考资料

qwq

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