# The function that draws a figure eight

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.

-
Odds are you'll get a much better response if you accept a response from your previous five questions! – Hooked Oct 27 '10 at 3:29
I'm sorry, I don't know how to "accept" a response on here :( – JimJones Oct 27 '10 at 3:32
See the FAQ: "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." – Rahul Oct 27 '10 at 3:40

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):

-
This is the lemniscate of Gerono (alias "eight curve"). – J. M. Oct 27 '10 at 3:39
Well, 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! – Mariano Suárez-Alvarez Oct 27 '10 at 3:41
But wouldn't you hit (0,0) at both pi/2 and -pi/2? – JimJones Oct 27 '10 at 3:46
@JimJones: indeed. It is trivial to fix that. – Mariano Suárez-Alvarez Oct 27 '10 at 3:47
@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. – J. M. Oct 27 '10 at 4:54

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:

-
There's also the generalization of the lemniscate of Bernoulli: the lemniscate of Booth. – J. M. Oct 27 '10 at 3:38
...and the hippopede of Proclus as well. – J. M. Oct 27 '10 at 3:43

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.

-