Mathematical Immaturity

14.13 Exercises

• Consider the plane curve whose vector equation is $\mathbf{r}(t) = t\mathbf{i} + f(t)\mathbf{j},$ where $$f(t) = t\cos\left(\frac{\pi}{2t}\right) \text{ if } t \neq 0, \quad f(0) = 0$$ Consider the following partition of the interval $[0,1]:$ $$P = \left\{0, \frac{1}{2n}, \frac{1}{2n-1}, \ldots, \frac{1}{3}, \frac{1}{2}, 1\right\}$$ Show that the corresponding inscribed polygon $\pi(P)$ has length $$|\pi(P)| > 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{2n}$$ and deduce that this curve is nonrectifiable.
• The length of the inscribed polygon, $|\pi(P)|$ can be expressed as the sum: $$ \begin{align*} \\ \left|\pi(P)\right| &= \sum_{k = 1}^{2n + 1}\left\|\mathbf{r}(t_k) - \mathbf{r}(t_{k-1})\right\| \\ &= \left\|\frac{1}{2n}\mathbf{i} + \frac{1}{2n}\cos(n\pi)\mathbf{j}\right\| \\ &+\sum_{k = 2}^{2n + 1}\left\|\left(t_k - t_{k-1}\right)\mathbf{i} + \left(t_k\cos\frac{\pi}{2t_k} - t_{k-1}\cos\frac{\pi}{2t_{k-1}}\right)\mathbf{j}\right\| \end{align*} $$ Looking at the right-hand side of the equation, we note that for $k \geq 2,$ if $k$ is odd, then the denominator of $t_k$ is even $(\text{eg}.\ t_{3} = \frac{1}{2n - 2},\ t_{2} = \frac{1}{2n - 1}),$ making $\left|\cos\frac{\pi}{2t_{k}}\right| = 1$ and $\cos\frac{\pi}{2t_{k-1}} = 0,$ which means that $$ \begin{align*} \\ \left\|\left(t_k - t_{k-1}\right)\mathbf{i} + \left(t_k\cos\frac{\pi}{2t_k} - t_{k-1}\cos\frac{\pi}{2t_{k-1}}\right)\mathbf{j}\right\| &= \sqrt{(t_k - t_{k-1})^2 + t_k^2} \\ &> |t_{k}| \end{align*} $$ (Note: in the partition as ordered above, $|t_k| > |t_{k - 1}|$ for all $k.$) If $k$ is even, then the denominator of $t_k$ is odd, making $\cos\frac{\pi}{2t_{k}} = 0$ and $\left|\cos\frac{\pi}{2t_{k-1}}\right| = 1,$ which means that $$ \begin{align*} \\ \left\|\left(t_k - t_{k-1}\right)\mathbf{i} + \left(t_k\cos\frac{\pi}{2t_k} - t_{k-1}\cos\frac{\pi}{2t_{k-1}}\right)\mathbf{j}\right\| &= \sqrt{(t_k - t_{k-1})^2 + t_{k-1}^2} \\ & > |t_{k-1}| \end{align*} $$ But if $\left\|\mathbf{r}(t_k) - \mathbf{r}(t_{k-1})\right\| \geq |t_{k - 1}|$ for $1 < k \leq 2n + 1,$ then the length of the inscribed polygon $|\pi(P)|$ is such that $$ |\pi(P)| > t_0 + t_1 + \cdots + t_{2n - 1} + t_{2n} $$ Which, after a reordering, is equivalent to $$ |\pi(P)| > 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{2n} $$ But as $n \rightarrow \infty,$ the sum $\sum_{k=1}^n 1/k$ diverges, which means that the length of the curve is nonrectifiable. $\quad \blacksquare$