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Exercise:

Find the limit of the perimeter of a broken line $M_0M_1...M_n$ inscribed in a logarithmic spiral $t = e^{-\phi}$ (as $n \to \infty$), if the vertices of this broken line have, respectively, the polar angles $\phi_0 = 0$, $\phi_1 = \frac{\pi}{2}$, $\cdots$, $\phi_n = \frac{n\pi}{2}$.

Attempt:

I have no idea how to go about this. All I've been able to do is verify that there is a defined limit:

$\lim\limits_{n \to \infty}{e^{\frac{-\pi}{2}n}} = 0$, so the sum of smaller and smaller lengths will result in a defined number.

Request:

Can I get a kickstart? Hints are welcome. (If I'm still lost I'll ask for the solution.)

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    Can you find the length of the line segment joining $M_n$ and $M_{N+1}$, for any $n$? These vertices all lie either on the $x$ or $y$ axis, which makes the computation not too messy. Also, the fact that the segments are getting smaller does not tell you that the limit exists! (The infinite series you get in this case will converge, but that is not a given.) – 2017-02-28

1 Answers 1

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I've worked this out and tested it. So let me get you started. We can express this logarithmic curve in the complex plane as

$$z=e^{(i-1)\theta}$$

So first we identify the discrete points on the line whose length you would like to determine, i.e,

$$z_n=e^{n(i-1)\pi/2}$$

Then the length we wish to determine is given simply by

$$\sum_{n=1}^\infty|z_n-z_{n-1}|$$

And you can take it from there. Just bear in mind that $|e^{i\theta}|=1$. I imagine this can generalized for any flair coefficient and $\Delta\theta$.