APOSTOL 해석적 정수론 CHAP 2. 12 ~ 19

1편 : http://drugstoreoftamref.tistory.com/15

Qkffl Qkffl dmdkdkdkdkdkd

---

D12. (Completely) Multiplicative Function

MF : $f \text{ s.t. } f(mn)=f(m)f(n) \ \forall m,n \text{ s.t. }gcd(m,n)=1$

CMF : $f \text{ s.t. } f(mn)=f(m)f(n) \ \forall m,n$

* 앞으로 MF들의 집합을 $\mathbb{M}$으로, CMF들의 집합을 $\mathbb{CM}$으로 denote.

C12-1. $$f \in \mathbb{M} \Rightarrow f(1)=1$$

T13. $$f \in \mathbb{M} \Leftrightarrow f(\prod_{k=1}^{\omega} p_k^{e_k}) = \prod_{k=1}^{\omega} f(p_k^{e_k}) \cdots (1)$$

$$\text{When } f \in \mathbb{M}, \\ f \in \mathbb{CM} \Leftrightarrow f(p^x)={f(p)}^x \cdots (2)$$

pf) $\text{Trivial.} \blacksquare$

T14.

$$f, g \in \mathbb{M} \Rightarrow f \ast g \in \mathbb{M}$$

pf)

$\text{let) } m,n \in \mathbb{N} \text{ s.t. } gcd(m,n)=1$

$(f \ast g)(m) \cdot (f \ast g)(n)$

$= \left( \sum_{x|m}f(x)g(\frac{m}{x}) \right) \cdot \left( \sum_{y|n}f(y)g(\frac{n}{y}) \right)$

$= \sum_{xy|mn} f(x)g(\frac{m}{x})f(y)g(\frac{n}{y}) \quad (\because gcd(x,y)=1)$

$= \sum_{xy|mn} f(x)f(y)g(\frac{m}{x})g(\frac{n}{y})$

$= \sum_{z|mn} f(z)g(\frac{mn}{z})$

$= (f \ast g)(mn) \ \blacksquare$

cf) $f, g \in \mathbb{CM} \not{\Rightarrow} f \ast g \in \mathbb{CM}$

$\text{ex) } \sigma = u \ast N \notin \mathbb{CM}$

T15. $$f, f \ast g \in \mathbb{M} \Rightarrow g \in \mathbb{M}$$

proof by contradiction)

$g(1) = 1 \ (\because f(1)=(f \ast g)(1) = 1)$

$\text{let) x be the minimal  value s.t. } \exists a|x \text{ s.t. } gcd(a,x/a)=1 \wedge g(x)\neq g(a)g(x/a)$ 

(즉, MF의 성질이 성립하지 않는 최솟값 가정)

$(f \ast g)(x) = (f \ast g)(a) \cdot (f \ast g) (x/a). \text{contradiction.} \blacksquare$

C15-1. $$f \in \mathbb{M} \Rightarrow f^{-1} \in \mathbb{M}$$

pf) $f, I(=f \ast f^{-1}) \in \mathbb{M}. \blacksquare$

T16. $$\text{When } f \in \mathbb{M}, \\ f \in \mathbb{CM} \Leftrightarrow f^{-1} = \mu \cdot f$$

$\Rightarrow )$

$f\ast (\mu f) (n)$

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

$= \sum_{d|n} f(n)\mu(d) (\because f \in \mathbb{CM} )$

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

$= f(n)I(n) = I(n) \ \blacksquare$

$\Leftarrow )$

$\text{claim) } f(p^k) = f(p)f(p^{k-1})$ (이후는 T13)

$(f \ast f^{-1})(p^k) = f(p^k) + f(p)\mu (p)f(p^{k-1}) = 0 \text{ for } \forall k>0$ 

$\therefore f(p^k)-f(p)f(p^{k-1})=0 \ \blacksquare$

E16-1. (Inversion of Euler Totient Function)

${\phi}^{-1} = (\mu \ast N)^{-1}$

$= \mu^{-1} \ast N^{-1}$

$= u \ast (\mu N) \ (\because N \in \mathbb{CM})$

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

$= \prod_{p|n} (1-p) \ \blacksquare$

T17. (E16-1에서 의식의 흐름)

$$\text{When } f \in \mathbb{M}, \\ \sum_{d|n}\mu(d)f(d) = \prod_{p|n} (1-f(p))$$

pf) 자명.

D18. Liouville's Function

$$n = \prod_{i=1}^{\omega} p_i^{e_i} \Rightarrow \lambda(n) := (-1)^{e_1 + \cdots e_{\omega}}$$

P18-1. $$\lambda \in \mathbb{CM}$$

P18-2. $$\sum_{d|n} \lambda(d) = \begin{cases} 1 & (\text{n : square}) \\ 0 & \text{otherwise} \end{cases}$$ 

pf) $\lambda(d) = -\lambda(n/d) \text{(n : square)}$

$\lambda(p^k\cdot x) = -\lambda(p^{e-k}\cdot x) \text{(e : odd)} \ \blacksquare$

P18-3. $$\lambda^{-1}(n) = |\mu(n)|$$

pf) $\lambda^{-1}(n) = \lambda(n)\mu(n) = |\mu(n)| \ \blacksquare$

D19. Divisors Function

$$\sigma_k  = u \ast N^k$$

P19-1. $$\sigma_k \in \mathbb{M}$$

P19-2. $$\sigma_k^{-1} = \ u^{-1} \ast {N^k}^{-1} = \mu \ast (\mu \cdot N^k) = \sum_{d|n} \mu(d)\mu(\frac{n}{d}) d^k$$

---

다음 포스팅에서 CH2 마무리.