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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 $[0,1]$


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?


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$$ f\left(\frac1k\right) = \frac1k\cos\frac{\pi k}{2}, $$ 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.

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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

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