이 글의 내용은 성균관대학교 권순학 교수님의 2019년 5월 9일 수업 내용을 재구성한 것입니다.

Möbius function

\\mu(x) = \\begin{cases} 1 &\\text{if } n = 1 \\\\ 0 &\\text{if } p^2 \\mid n \;\; \\text{for some prime p} \\\\ -1^t &\\text{if } n = \\prod_i^t p_i \\end{cases}

Möbius function is multiplicative

증명

$$(m, n) = 1$$

case 1. \(m = 1\) or \(n = 1\)

자명하므로 증명은 생략한다.

case 2. \(p^2 \mid m\) or \(p^2 \mid n\)

$$\\mu(mn) = 0 \;\; (\\because p^2 \\mid mn)$$ $$\\mu(m)\\mu(n) = 0 \;\; (\\because p^2 \\mid m or p^2 \\mid n)$$

case 3

$$m = \\prod_i^s p_i$$ $$n = \\prod_i^t q_i$$ $$\\mu(mn) = (-1)^{s + t}$$ $$\\mu(m)\\mu(n) = (-1)^s (-1)^t = (-1)^{s + t}$$

Theorem 7.8: Summation of \(f\) is multiplicative if \(f\) is multiplicative

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

증명

$$(m, n) = 1$$ $$F(mn) = \\sum_{d \\mid mn} {f(d)}$$ $$d = gh \\\\ g \\mid m, h \\mid n$$ $$F(mn) = \\sum\_{d \\mid mn} {f(d)} = \\sum\_{g \\mid m} {\\sum\_{h \\mid n} f(d)} \\\\ = \\sum\_{g \\mid m} {\\sum\_{h \\mid n} f(g)f(h)} \\\\ = \\sum\_{g \\mid m} {f(g)} \\sum\_{h \\mid n} {f(h)} \\\\ = F(m) F(n)$$

Theorem 7.15: Summation of Möbius function

\\sum_{d \\mid n} \\mu(n) = \\begin{cases} 1 &\\text{if } n = 1 \\\\ 0 &\\text{if } n > 1 \\end{cases}

증명

$$F(p^a) = \\sum_{d \\mid p^a} {\\mu(d)} \\\\ = \\mu(1) + \\mu(p) + \\cdots + \\mu(p^a) \\\\ = 1 + -1 \\\\ = 0$$ $$F: \\text{multiplicative} \\\\ \\therefore F(n) = 0 \;\; \\text{ if } n > 1$$