We've been given a function and asked to find whether it is a rectifiable path or not. Let $\gamma : [0,1]\to \mathbb{C} $ defined as
$\gamma (t)= t + \iota t \sin(\frac{1}{t})$
and $\gamma (0)=0$
Ofcourse it is a path (as by definition, it is continuous in the domain, so a path)
Now we need to show it is not rectifiable (this is given in the answer key)
I have a definition that a function $f$ is rectifiable if it is a function of bounded variation.
So we need to find a partition P of $[0,1]$ such that
$V(\gamma , P)=\sum_{k=1}^n |\gamma (t_k)-\gamma (t_{k-1})|$
is not bounded
I am not able to find such a partition.
Please help me if there is some other approach. Please use basic techniques as I am new to this course of complex analysis. Thanks!
Take the partition $t_n=\frac{2}{n\pi}$ Then for $n$ odd we have the point, $(\frac{2}{n\pi},\frac{2}{n\pi})$ and for $n$ even we have $(\frac{2}{n\pi},0)$ So
$$|\gamma(t_n)-\gamma(t_{n-1})|=\frac{2}{n\pi}\sqrt{(\frac{1}{n-1})^2+1}$$ and it is easy to see that $$\sum_n\frac{2}{n\pi}\sqrt{(\frac{1}{n-1})^2+1}$$ diverges by a limit comparison with $\frac{1}{n}$.