자연계열 2번

1.아래 그림에서, 임의의 $\displaystyle 0<t<\frac\pi4$에 대해, $\displaystyle \cos t=\frac{1+\cos2t}{\sqrt{2+2\cos2t}}=\sqrt{\frac{1+\cos2t}2}$이다.따라서 $\displaystyle f(x)=\sqrt{\frac{1+x}2}$로 할 수 있다.이때, $\displaystyle f(\cos0)=\sqrt{\frac{1+1}2}=1=\cos0,\ f\left(\cos\frac\pi2\right)=\sqrt{\frac{1+0}2}=\frac1{\sqrt2}=\cos\left(\frac\pi4\right)$이므로, $\displaystyle 0\le t\le\frac\pi4$일 때, $\displaystyle \cos t=f(\cos2t)$가 성립한다.

2.$\displaystyle a_1=\sqrt2=2\cos\frac\pi4=2\cos\frac\pi{2^2},$$\displaystyle a_2=\sqrt{2+\sqrt2}=\sqrt{2+a_1}=2\sqrt{\frac{1+\cos\frac\pi4}2}=2f\left(\cos\frac\pi4\right)=2\cos\frac\pi8=2\cos\frac\pi{2^3},$$\displaystyle a_3=\sqrt{2+\sqrt{2+\sqrt2}}=\sqrt{2+a_2}=2\sqrt{\frac{1+\cos\frac\pi8}2}=2f\left(\cos\frac\pi8\right)=2\cos\frac\pi{16}=2\cos\frac\pi{2^4},\quad\cdots$로부터 $\displaystyle a_n=2f\left(\cos\frac\pi{2^n}\right)=2\cos\frac\pi{2^{n+1}}$임을 알 수 있다.

따라서 극한값은 $\displaystyle \lim_{n\to\infty}a_n=\lim_{n\to\infty}2\cos\frac\pi{2^{n+1}}=2\cdot1=2$이다.

3. $\displaystyle \begin{aligned}b_m&=\lim_{n\to\infty}\sum_{k=1}^n\frac1n\sqrt{1+\cos\frac{\pi k}{2^mn}}\\&=\frac{\sqrt2}\pi\lim_{n\to\infty}\sum_{k=1}^n f\left(\cos\frac{\pi k}{2^mn}\right)\frac\pi n=\frac{\sqrt2}\pi\lim_{n\to\infty}\sum_{k=1}^n\cos\left(\frac{\pi k}{2^{m+1}n}\right)\frac\pi n\\&=\frac{\sqrt2}\pi\lim_{n\to\infty}\sum_{k=1}^n\cos\left(\frac{\frac\pi{2^{m+1}}k}n\right)\frac{\frac\pi{2^{m+1}}}n2^{m+1}\\&=\frac{\sqrt2}\pi2^{m+1}\int_0^{\frac\pi{2^{m+1}}}\cos t\,dt=\frac{\sqrt2}\pi2^{m+1}\sin\frac\pi{2^{m+1}}\end{aligned}$이다. 따라서 $\displaystyle b_1$은 $\displaystyle \frac{\sqrt2}\pi2^{1+1}\sin\frac\pi{2^{1+1}}=\frac4\pi$이고

$\displaystyle \lim_{m\to\infty}b_m=\lim_{m\to\infty}\frac{\sqrt2}\pi2^{m+1}\sin\frac\pi{2^{m+1}}=\lim_{m\to\infty}\sqrt2\frac{\sin\frac\pi{2^{m+1}}}{\frac\pi{2^{m+1}}}=\sqrt2\cdot1=\sqrt2$이다.