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.

Consider the plane curve whose vector equation is $r(t) = ti +f(t)j$, where $$f(t)=t\cos\bigg(\frac{\pi}{2t}\bigg)$$ if $t$ is not equal $0$, and $0$ otherwise.

Consider the following partition of the interval $[0,1]$

$$P=\bigg\{0,\frac{1}{2n},\frac{1}{2n-1},\ldots,\frac{1}{2},1\bigg\}$$

Show that corresponding inscribed polygon $$|\pi(P)|=\sum_{k=1}^n||r(t_k)-r(t_{k-1})||$$ has length $$|\pi(P)|>1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2n}$$

And deduce that the curve is nonrectifiable.

Now this would follow if every member of the sum $|\pi(P)|$ is bigger than every member of the harmonic series. Tried to see directly - does not work.

Another idea is to find a sequence so that the sum of it would be less than of the inscribed polygon but bigger than the harmonic series, but could not find one.

How should I approach this problem? any hints?

Thanks!

share|improve this question

1 Answer 1

up vote 0 down vote accepted

$$ f\left(\frac1k\right) = \frac1k\cos\frac{\pi k}{2}, $$ so we get $$ \begin{array}{c|cccccccccccccccccccc} k & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & \cdots \\[12pt] f\left(\frac1k\right) & 0 & \frac{-1}{2} & 0 & \frac14 & 0 & \frac{-1}6 & 0 & \frac18 & 0 & \frac{-1}{10} & 0 & \frac{1}{12} & \cdots \end{array} $$ The vertical components of the arc length form a harmonic series, so that diverges to infinity.

share|improve this answer
    
oh crap! made a stupid algebraic mistake and was not able to spot the sequence the function was making. thanks! –  Sarunas Feb 6 '13 at 20:18

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.