the following is an exercise from P. Do Carmo's "Differential Geometry of Curves and Surfaces" (Ch.1):
- b) (A Nonrectifiable Curve) The following example shows that, with any reasonavle definition, the arc length of a $C^0$ curve in a closed interval may be unbounded. Let $\alpha:[0,1] \to R^2$ be given as $\alpha(t)=\big(t,t\sin\left(\frac{\pi}{t}\right)\big)$ if $t\neq0$, and $\alpha(0)=(0,0)$. Show, geometrically, that the arc length of the portion of the curve corresponding to $\frac{1}{n+1}$ $\leqslant t \leqslant$ $\frac{1}{n}$ is at least $\frac{1}{2n+1}$.
Use this to show that the length of the curve in the interval $\frac{1}{N}$$\leqslant t\leqslant$$1$ is greater than $$2\sum_{n=1}^N \frac{1}{n+1}$$
I'm having trouble answering the first part of this question. My approach has been to determine the lengths between the most convenient points contained within the interval -- i.e. the endpoints of the interval and the critical point, $\frac{1}{n+\frac{1}{2}}$.
$$ \begin{split} \left|\alpha\Big(\frac{1}{n}\Big)-\alpha\Big(\frac{2}{2n+1}\Big)\right|&+\left|\alpha(2n)-\alpha\Big(\frac{2}{2n+1}\Big)\right|\\ &= \frac{1}{n(n+1)(2n+1)}(n+1)\sqrt{1+4n^2}+n\sqrt{1+4(n+1)^2} \end{split} $$
Which is long and intractable. My sense is that there is a more elegant approach. Where have I gone wrong?
Between $t_{n+1}={1\over n+1}$ and $t_n={1\over n}$, where $|\sin|=0$, there is the value $\tau={1\over n+{1\over2}}$, where $|\sin|=1$. It follows that the arc length of $\alpha$ within $[t_{n+1},t_n]$ is $$\geq|\alpha(t_n)-\alpha(\tau)|+|\alpha(\tau)-\alpha(t_{n+1})|\geq 2|\alpha_2(\tau)|= 2\tau={4\over 2n+1}\ .$$ The sum of these terms is $\infty$.
The $\alpha(2n)$ in your last displayed formula should be replaced by $\alpha\bigl({1\over n+1}\bigr)$, but even then the carrying of the first component $\alpha_1(t)$ along would create computational difficulties.