포항공대 수학경시대회 7회 4번 (1)

$\underset{k\to \mathrm{\infty }}{lim}{\displaystyle \frac{{\sin }_k(2)}{{\sin }_k(1)}}$을 구하여라.  (${\sin }_1=\sin ,{\sin }_{n+1}=\sin \circ {\sin }_n$)

---

임의의 실수 $x\in (0,\pi )$에 대하여, $x_n={\sin }_n(x)$로 두자.

 

$x_{n+1}=\sin x_n$이고 $0<x_{n+1}<x_n$이 성립하므로 Banach fixed-point Thm에 의해 $x_n\to 0$

 

$x\to 0$이면, ${\displaystyle \frac{1}{{\sin }^{2}x}}-{\displaystyle \frac{1}{x^2}}={\displaystyle \frac{x^2-{\sin }^{2}x}{x^2{\sin }^{2}x}}={\displaystyle \frac{x-\sin x}{x^3}}{\displaystyle \frac{x+\sin x}{x}}{\displaystyle \frac{x^2}{{\sin }^{2}x}}\to {\displaystyle \frac{1}{6}}\cdot 2\cdot 1={\displaystyle \frac{1}{3}}$이므로

 

$n\to \mathrm{\infty }$이면, $x_n\to 0$이므로 ${\displaystyle \frac{1}{x_{n+1}^2}}-{\displaystyle \frac{1}{x_n^2}}={\displaystyle \frac{1}{{\sin }^2{x}_n}}-{\displaystyle \frac{1}{x_n^2}}\to {\displaystyle \frac{1}{3}}$이다.

 

$a_n\to a$일 때, ${\displaystyle \frac{1}{n}}\sum _{i=1}^n{a}_i\to a$이므로 ($\because$ Cesàro mean)

 

${\displaystyle \frac{1}{n}}({\displaystyle \frac{1}{x_{n+1}^2}}-{\displaystyle \frac{1}{x_1^2}})={\displaystyle \frac{1}{n}}\sum _{i=1}^n({\displaystyle \frac{1}{x_{i+1}^2}}-{\displaystyle \frac{1}{x_i^2}})\to {\displaystyle \frac{1}{3}}$이다.

 

따라서 ${\displaystyle \frac{1}{x_n^2}}\sim {\displaystyle \frac{n}{3}}$이고 $x_n\sim \sqrt{{\displaystyle \frac{3}{n}}}$

 

즉, $\sqrt{n}{x}_n\to \sqrt{3}$이다.

 

$\underset{k\to \mathrm{\infty }}{lim}{\displaystyle \frac{{\sin }_k(2)}{{\sin }_k(1)}}=\underset{k\to \mathrm{\infty }}{lim}{\displaystyle \frac{\sqrt{k}{\sin }_k(2)}{\sqrt{k}{\sin }_k(1)}}={\displaystyle \frac{\underset{k\to \mathrm{\infty }}{lim}\sqrt{k}{\sin }_k(2)}{\underset{k\to \mathrm{\infty }}{lim}\sqrt{k}{\sin }_k(1)}}={\displaystyle \frac{\sqrt{3}}{\sqrt{3}}}=1$