Let the functions $f$, $g$ and $D$ be defined on $\mathbb{R} \backslash \mathbb{Z}$ by: $$f(x) = \pi \operatorname{cotan}(\pi x) = \pi \frac{\cos(\pi x)}{\sin(\pi x)}, \quad g(x) = \frac{1}{x} + \sum_{n=1}^{+\infty} \left(\frac{1}{x+n} + \frac{1}{x-n}\right).$$ We set $D = f - g$. Justify the existence of $\alpha \in [0,1]$ such that $\widetilde{D}(\alpha) = M$, where $M = \sup_{t \in [0,1]} \widetilde{D}(t)$, then show that: $$\forall n \in \mathbb{N}, \quad \widetilde{D}\left(\frac{\alpha}{2^n}\right) = M.$$
Let the functions $f$, $g$ and $D$ be defined on $\mathbb{R} \backslash \mathbb{Z}$ by:
$$f(x) = \pi \operatorname{cotan}(\pi x) = \pi \frac{\cos(\pi x)}{\sin(\pi x)}, \quad g(x) = \frac{1}{x} + \sum_{n=1}^{+\infty} \left(\frac{1}{x+n} + \frac{1}{x-n}\right).$$
We set $D = f - g$.
Justify the existence of $\alpha \in [0,1]$ such that $\widetilde{D}(\alpha) = M$, where $M = \sup_{t \in [0,1]} \widetilde{D}(t)$, then show that:
$$\forall n \in \mathbb{N}, \quad \widetilde{D}\left(\frac{\alpha}{2^n}\right) = M.$$