Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $f$ be a function defined on $[0,1]$ by

$$f(x) = { 0, \text{ if } x = 0} $$ $$f(x) = { x \sin \frac 1 x , \text{ if } 0 < x \leq 1} $$

Prove that the curve $\{(x, f(x)) : x \in [0,1]\}$ is not rectifiable.

I'm not sure how to approach this. The general idea seems logical, we're proving that the length of the curve is infinite, but the method seems difficult to find.

share|improve this question
    
You know how to form the arclength integral, no? –  J. M. Dec 12 '11 at 6:01
    
it's the integral of the absolute value of the derivative of any parametrization. i'm not sure where to go from there. –  uwuw Dec 12 '11 at 6:11
    
Form the arclength integral and see if you can evaluate it. –  J. M. Dec 12 '11 at 6:20
add comment

2 Answers

The length of a curve is by definition the $\sup$ of the lengths of inscribed chord-polygons. For your curve define

$$x_0:=1,\qquad x_k:={2\over k\pi}\quad(1\leq k< N),\qquad x_N:=0$$

and consider the chord-polygon $\gamma_N$ through the points $\bigl(x_k, f(x_k)\bigr)$ $\ (0\leq k\leq N)$. As $\bigl|\sin{1\over x_k}\bigr|$ is alternatively $0$ and $1$ one has $|f(x_k)-f(x_{k-1})|\geq{1\over k}$; therefore the individual chords (apart from the first and the last one) have a length $\geq{1\over k}$. It follows that $\gamma_N$ has a total length $\geq\sum_{k=2}^{N-1}{1\over k}$. Since this sum is unbounded for $N\to\infty$ the considered curve $\gamma$ is not rectifiable.

share|improve this answer
add comment

The arclength $L$ of a curve $(x,f(x))$ from $x=a$ to $x=b$ is defined as $$L=\int_a^b\sqrt{1+(f'(x))^2}dx.$$ Therefore, in this case, $f(x)=\displaystyle x\sin(\frac{1}{x})$, which implies that $$(f'(x))^2=\Big[\sin(\frac{1}{x})-\frac{1}{x}\cos(\frac{1}{x})\Big]^2\geq-\frac{2}{x}\sin(\frac{1}{x})\cos(\frac{1}{x})+\frac{1}{x^2}\cos^2(\frac{1}{x}).$$ Hence, if $x\in\displaystyle[\frac{1}{2\pi n+\pi/3},\frac{1}{2\pi n}]$ where $n\in\mathbb{N}$, then $$(f'(x))^2\geq -2(2\pi n+\frac{\pi}{3})+4\pi^2 n^2\cos^2(2\pi n+\pi/3)=\pi^2n^2-4\pi n-\frac{2\pi}{3}.$$ Therefore, the arclength $L$ of $(x,f(x))$ from $x=0$ to $x=1$ can be estimated as follows: $$L=\int_0^1\sqrt{1+(f'(x))^2}dx\geq \sum_{n=1}^\infty\int^{\frac{1}{2\pi n}}_{\frac{1}{2\pi n+\pi/3}}\sqrt{1+(f'(x))^2}dx$$ $$\geq \sum_{n=1}^\infty\int^{\frac{1}{2\pi n}}_{\frac{1}{2\pi n+\pi/3}}\sqrt{1+\pi^2n^2-4\pi n-\frac{2\pi}{3}}dx$$ $$=\sum_{n=1}^\infty\sqrt{1+\pi^2n^2-4\pi n-\frac{2\pi}{3}}\cdot\frac{\frac{\pi}{3}}{(2\pi n)(2\pi n+\pi/3)}.$$ It's easy to see that the last series in $n$ diverges to infinity by using limite comparison test with the harmonic series $\displaystyle\sum_{n=1}^\infty\frac{1}{n}$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.