Problems

Solution: Claim: \(\displaystyle \sum_{r=1}^n \frac1{2^r} \tan \tfrac{\theta}{2^r} = \frac1{2^n}\cot \tfrac{\theta}{2^n} - \cot \theta\) Proof: (By Induction) Base case: \(n = 1\) \begin{align*} && LHS &= \sum_{r=1}^1 \frac1{2^r} \tan \frac{\theta}{2^r} \\ &&&= \frac1{2} \tan \frac{\theta}{2}\\ \\ && RHS &= \frac12 \cot \frac{\theta}{2} - \cot \theta \\ &&&= \frac12 \frac{1}{\tan \frac{\theta}{2}} - \frac{1-\tan^2 \frac{\theta}{2}}{2 \tan \frac{\theta}{2}} \\ &&&= \frac{1}{2} \tan \frac{\theta}{2} = LHS \end{align*} Therefore our base case is true. Assume our statement is true for some \(n=k\), then consider \(n = k+1\), ie \begin{align*} \sum_{r=1}^{k+1} \frac1{2^r} \tan \tfrac{\theta}{2^r} &= \sum_{r=1}^{k} \frac1{2^r} \tan \tfrac{\theta}{2^r} + \frac1{2^{k+1}} \tan \frac{\theta}{2^{k+1}} \\ &= \frac{1}{2^k} \cot \frac{\theta}{2^k} - \cot \theta + \frac{1}{2^{k+1}}\tan \frac{\theta}{2^{k+1}} \\ &= \frac{1}{2^{k+1}} \left (2 \cot \frac{\theta}{2^k} +\tan \frac{\theta}{2^{k+1}} \right) - \cot \theta \\ &= \frac{1}{2^{k+1}} \left (2\frac{1-\tan^2 \frac{\theta}{2^{k+1}}}{2 \tan \frac{\theta}{2^{k+1}}} + \tan \frac{\theta}{2^{k+1}} \right) - \cot \theta \\ &= \frac{1}{2^{k+1}} \cot \frac{\theta}{2^{k+1}} - \cot \theta \\ \end{align*} Therefore, since as \(x \to 0, x\cot x \to 1\) or \(x \cot \theta x \to \frac{1}{\theta}\) \begin{align*} \sum_{r=1}^{\infty}\frac{1}{2^{r}}\tan\frac{\theta}{2^{r}} &= \lim_{k\to \infty} \sum_{r=1}^{k}\frac{1}{2^{r}}\tan\frac{\theta}{2^{r}} \\ &= \lim_{k\to \infty} \left ( \frac{1}{2^{k+1}} \cot \frac{\theta}{2^{k+1}} - \cot \theta \right) \\ &= \lim_{k\to \infty}\frac{1}{2^{k+1}} \cot \frac{\theta}{2^{k+1}} - \cot \theta \\ &= \lim_{k\to \infty}\frac{1}{2^{k+1}} \cot \frac{\theta}{2^{k+1}} - \cot \theta \\ &= \lim_{k\to \infty}\frac{1}{2^{k+1}} \cot \frac{\theta}{2^{k+1}} - \cot \theta \\ &= \frac{1}{\theta} - \cot \theta \end{align*}