자연계열 3번

3-1) $\displaystyle |\vec b_i|=2\sin\frac{\pi i}{n+1}$이고 두 벡터 $\displaystyle \vec b_i,\ \vec b_j$의 사이각은 $\displaystyle \left|\frac{\pi(i-j)}{n+1}\right|$이다. 따라서$\displaystyle \begin{aligned}\vec b_i\cdot\vec b_j&=\left(2\sin\frac{\pi}{n+1}i\right)\left(2\sin\frac{\pi}{n+1}j\right)\cos\frac{\pi}{n+1}(i-j)\\&=\left(2\sin\frac{\pi}{n+1}i\right)\left(2\sin\frac{\pi}{n+1}j\right)\left(\cos\frac{\pi}{n+1}i\cos\frac{\pi}{n+1}j+\sin\frac{\pi}{n+1}i\sin\frac{\pi}{n+1}j\right)\end{aligned}$이다. 따라서, $\displaystyle n=7$일 때,$\displaystyle \begin{aligned}\vec b_2\cdot\vec b_5&=\left(2\sin\frac\pi4\right)\left(2\sin\frac{5\pi}8\right)\cos\frac{3\pi}8\\&=\left(2\sin\frac\pi4\right)\left(2\cos\frac\pi8\right)\sin\frac\pi8\\&=\left(2\sin\frac\pi4\right)\sin\frac\pi4\\&=1\end{aligned}$을 얻는다.

3-2) 3-1)에서 얻어진 공식$\displaystyle \vec b_1\cdot\vec b_i=\left(2\sin\frac{\pi}{n+1}\right)\left(2\sin\frac{\pi i}{n+1}\right)\left(\cos\frac{\pi}{n+1}\cos\frac{\pi i}{n+1}+\sin\frac{\pi}{n+1}\sin\frac{\pi i}{n+1}\right)$로 부터$\displaystyle \vec b_1\cdot\left(\sum_{i=1}^n\vec b_i\right)=\left(\sum_{i=1}^n\left(\sin\frac{2\pi}{n+1}\right)\sin\frac{2\pi i}{n+1}\right)+4\left(\sum_{i=1}^n\left(\sin\frac{\pi}{n+1}\right)^2\left(\sin\frac{\pi i}{n+1}\right)^2\right)$을 얻는다. 따라서$\displaystyle \lim_{n\to\infty}\vec b_1\cdot\left(\sum_{i=1}^n\vec b_i\right)=\lim_{n\to\infty}\left(\sum_{i=1}^n\left(\sin\frac{2\pi}{n+1}\right)\sin\frac{2\pi i}{n+1}\right)+4\lim_{n\to\infty}\left(\sum_{i=1}^n\left(\sin\frac{\pi}{n+1}\right)^2\left(\sin\frac{\pi i}{n+1}\right)^2\right)$이다. 여기서,$\displaystyle \begin{aligned}\lim_{n\to\infty}\left(\sum_{i=1}^n\left(\sin\frac{2\pi}{n+1}\right)\sin\frac{2\pi i}{n+1}\right)&=\left(\lim_{n\to\infty}\left(\frac{\sin\frac{2\pi}{n+1}}{\frac1{n+1}}\right)\right)\left(\lim_{n\to\infty}\sum_{i=1}^n\left(\sin\frac{2\pi i}{n+1}\right)\frac1{n+1}\right)\\&=2\pi\int_0^1\sin2\pi x\,dx=0\end{aligned}$이다. 또한 $\displaystyle \left(\sum_{i=1}^n\left(\sin\frac{\pi}{n+1}\right)^2\left(\sin\frac{\pi i}{n+1}\right)^2\right)\le n\left(\sin\frac{\pi}{n+1}\right)^2$로 부터$\displaystyle \lim_{n\to\infty}\left(\sum_{i=1}^n\left(\sin\frac{\pi}{n+1}\right)^2\left(\sin\frac{\pi i}{n+1}\right)^2\right)=0$을 얻는다. 따라서, $\displaystyle \lim_{n\to\infty}\vec b_1\cdot\left(\sum_{i=1}^n\vec b_i\right)=0$이다.