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I have come upon the curve with the following parametric equations:

$$x(t)=\log(2+2\cos(t))/2$$ $$y(t)=t/2$$

for $-\pi<t<\pi$. It gives the image in the complex plane under $\log(1+z)$ of the unit circle. Does anyone know whether it has a name? It seems like this must have been studied at some point before.

Greg

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@DJC: I added the (geometry) tag back since I think the "terminology" tag shouldn't be on its own, if possible. –  t.b. Jul 28 '11 at 1:33
    
@Theo: I didn't know that. Thank you. –  JavaMan Jul 28 '11 at 1:54
    
@DJC: That's just my opinion. But (geometry) makes it clear what this thread is about, while (terminology) could be just about anything. –  t.b. Jul 28 '11 at 1:55
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For a non-parametric form, observe $x = \log(2+2\cos t)/2=\log(4 \cos^2(t/2))/2=\log|2\cos y|$. The bounds on $t$ make $\cos y$ non-negative, so that we can drop the absolute value, and we can go on to write $e^x = 2\cos y$. –  Blue Jul 28 '11 at 2:20
    
I think this curve has to do with $f(\phi)=\sum_{k=1}^\infty \cos(k\varphi)/k=-\log(2\sin\frac{\varphi}{2})$. Note that $\sum_{k=1}^\infty \sin(k\varphi)/k=\frac{\pi-\varphi}{2}$ is the sawtooth function. –  Peter Sheldrick Jul 28 '11 at 11:23
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1 Answer

up vote 6 down vote accepted

This is more or less a reflected and shifted version of the so-called "catenary of equal resistance" (en français, sorry). Here is the paper where they were first studied.

Wikipedia gives a derivation for the equation of the catenary of equal resistance; in some references, this is also called the "catenary of uniform strength". See this for instance.

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I'm somewhat curious to how you found/knew this answer. –  Willie Wong Jul 28 '11 at 11:54
    
@Willie: I will admit to spending (wasting?) the better part of my teenage years studying plane curves (mucking around in BASIC/Logo, scrounging books, etc.)... :) ...so yes, I still have those names from memory. I only went to that French site to check if my memory was correct. –  J. M. Jul 28 '11 at 14:37
    
Impressive! =) –  Willie Wong Jul 28 '11 at 16:01
    
Yes, thank you! –  Greg Markowsky Jul 29 '11 at 11:31
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