実数 $u, \theta$ は $0 < u < \frac{\pi}{4}, 0 < \theta < \frac{\pi}{2}$ を満たす定数とする。 (1) $t = \tan u$ とおくとき、$\frac{\tan u}{\tan 2u}$ を $t$ を用いて表す。 (2) 無限級数 $\sum_{n=1}^{\infty} \log(1 - \tan^2 \frac{\theta}{2^n})$ の和を求める。

1. 問題の内容

実数 $u, \theta$ は $0 < u < \frac{\pi}{4}, 0 < \theta < \frac{\pi}{2}$ を満たす定数とする。

(1) $t = \tan u$ とおくとき、$\frac{\tan u}{\tan 2u}$ を $t$ を用いて表す。

(2) 無限級数 $\sum_{n=1}^{\infty} \log(1 - \tan^2 \frac{\theta}{2^n})$ の和を求める。

2. 解き方の手順

(1) $\tan 2u = \frac{2\tan u}{1 - \tan^2 u}$ なので、

$\frac{\tan u}{\tan 2u} = \frac{\tan u}{\frac{2\tan u}{1 - \tan^2 u}} = \frac{1 - \tan^2 u}{2} = \frac{1 - t^2}{2}$

(2)

$S_N = \sum_{n=1}^{N} \log(1 - \tan^2 \frac{\theta}{2^n})$ とする。

$\log(1 - \tan^2 \frac{\theta}{2^n}) = \log(\frac{\cos^2 \frac{\theta}{2^n} - \sin^2 \frac{\theta}{2^n}}{\cos^2 \frac{\theta}{2^n}}) = \log(\frac{\cos \frac{\theta}{2^{n-1}}}{\cos^2 \frac{\theta}{2^n}})$

したがって、

$S_N = \sum_{n=1}^{N} \log(\frac{\cos \frac{\theta}{2^{n-1}}}{\cos^2 \frac{\theta}{2^n}}) = \sum_{n=1}^{N} (\log(\cos \frac{\theta}{2^{n-1}}) - 2\log(\cos \frac{\theta}{2^n}))$

$S_N = (\log(\cos \theta) - 2\log(\cos \frac{\theta}{2})) + (\log(\cos \frac{\theta}{2}) - 2\log(\cos \frac{\theta}{4})) + \dots + (\log(\cos \frac{\theta}{2^{N-1}}) - 2\log(\cos \frac{\theta}{2^N}))$

$S_N = \log(\cos \theta) - \log(\cos \frac{\theta}{2}) - \log(\cos \frac{\theta}{4}) - \dots - \log(\cos \frac{\theta}{2^N}) - \log(\cos \frac{\theta}{2^N})$

$S_N = \log(\cos \theta) - \sum_{n=1}^{N} \log(\cos \frac{\theta}{2^n}) - \log(\cos \frac{\theta}{2^N})$

$S_N = \log \cos \theta - \log (\prod_{n=1}^N \cos \frac{\theta}{2^n}) - \log (\cos \frac{\theta}{2^N})$

$\prod_{n=1}^N \cos \frac{\theta}{2^n} = \cos \frac{\theta}{2} \cos \frac{\theta}{4} \dots \cos \frac{\theta}{2^N} = \frac{\sin \theta}{2^N \sin \frac{\theta}{2^N}}$ より

$S_N = \log \cos \theta - \log(\frac{\sin \theta}{2^N \sin \frac{\theta}{2^N}}) - \log (\cos \frac{\theta}{2^N}) = \log (\cos \theta \cdot \frac{2^N \sin \frac{\theta}{2^N}}{\sin \theta} \cdot \frac{1}{\cos \frac{\theta}{2^N}}) = \log(\frac{\cos \theta}{\sin \theta} \cdot \frac{2^N \sin \frac{\theta}{2^N}}{\cos \frac{\theta}{2^N}})$

$\lim_{N \to \infty} 2^N \sin \frac{\theta}{2^N} = \theta$ かつ $\lim_{N \to \infty} \cos \frac{\theta}{2^N} = 1$ なので、

$\lim_{N \to \infty} S_N = \log (\frac{\theta \cos \theta}{\sin \theta}) = \log (\theta \cot \theta)$

3. 最終的な答え

(1) $\frac{\tan u}{\tan 2u} = \frac{1 - t^2}{2}$

(2) $\sum_{n=1}^{\infty} \log(1 - \tan^2 \frac{\theta}{2^n}) = \log (\theta \cot \theta)$