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I need to show that

$$f(x)= \begin{cases} \sin(1/x),& x \neq 0 \\ 0,& x = 0\end{cases}$$

on $[0,1]$ has an infinite arc length.

I've tried to prove that $(f'(x))^2$ is unbounded on $[0,1]$.

I also tried to use the fact that it's not uniformly continuous (maybe it's not relevant).

Thank you.

Clarinetist
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Shirly Geffen
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  • It does not suffice that $(f'(x))^2$ is unbounded. For instance $g(x)=\sqrt{1-x^2}$ also has an unbounded $(g'(x))^2$ on [0,1], but it still has a finite arc length. Similarly uniform continuity is not relevant. – Klaas van Aarsen Mar 22 '14 at 18:50

2 Answers2

4

The function is not continuous at $0$, so it isn't even a curve on $[0,1]$.

It is a curve with infinite length on $(0,1]$. The simplest way to see this is probably to notice that the curve length between each maximum and the neighboring minimum must be at least $2$. Since there are infinitely many wiggles and each of them contributes a curve length of at least $2$ ...

hmakholm left over Monica
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We can find a lower bound on the length of the curve by approximating it with straight lines. For every natural number $n$ the point $$ p_n=\Big(\frac{1}{(n+1/2)\,\pi},(-1)^n\Big) $$ lies on the curve. And the distance between $p_n$ and $p_{n+1}$ is at least $2$. That's enough to do it!

user134824
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