I'm trying to describe a counterexample for a theorem which includes the figure eight or "infinity" symbol, but I'm having trouble finding a good piecewise function to draw it. I need it to be the symbol, except at the "crossing point" the function jumps (not continuous) so that we still have a manifold.
The function that draws a figure eight
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3Odds are you'll get a much better response if you accept a response from your previous five questions! – 2010-10-27
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0I'm sorry, I don't know how to "accept" a response on here :( – 2010-10-27
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2See [the FAQ](http://math.stackexchange.com/faq#howtoask): "When you have decided which answer is the most helpful to you, mark it as the accepted answer by clicking on the check box outline to the left of the answer. This lets other people know that you have received a good answer to your question. Doing this is helpful because it shows other people that you're getting value from the community." – 2010-10-27
3 Answers
You can use the function $$t\in(-\tfrac12\pi,\tfrac32\pi)\mapsto(\cos t,\sin t\cos t)\in\mathbb R^2.$$
The resulting curve is (I'm omitting a little bit on the left and a little bit on the right from the domain in this picture):
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1This is the lemniscate of Gerono (alias "eight curve"). – 2010-10-27
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1Well, it is the one you get if play 1 minute with a CAS which can do parametric plots... I guess Gerono did not have one, so +1 to him! – 2010-10-27
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0But wouldn't you hit (0,0) at both pi/2 and -pi/2? – 2010-10-27
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0@JimJones: indeed. It is trivial to fix that. – 2010-10-27
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1@Jim: To be more explicit, you can exploit the periodicity of the underlying trigonometric functions such that when traversing the curve, you only hit the origin once. – 2010-10-27
There are no functions that describe the "lemniscate", but there are parametric and polar equations for the lemniscate of Bernoulli and the lemniscate of Gerono, to name two of the more famous lemniscates.
Might as well share this, since the question already has an accepted answer. Here is my favorite way of generating the lemniscate of Bernoulli, as an envelope of circles centered on a rectangular hyperbola, and passing through the origin:
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0There's also the generalization of the lemniscate of Bernoulli: the [lemniscate of Booth](http://www.mathcurve.com/courbes2d/booth/booth.shtml). – 2010-10-27
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0...and the [hippopede of Proclus](http://mathworld.wolfram.com/Hippopede.html) as well. – 2010-10-27
Is the lemniscate what you want? I don't know what you need as a jump at the crossing point, but maybe you can get that with a trigonometric parameterization.