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My studies lead me to the following parametrization of perhaps a new class of plane curves ( which are similar in shape to the classical sinusoidal spirals but not identical ). If the curves are not yet known I humbly dare to call them $T$-curves.

\begin{align} &x(t)= \cos t + \cos \left(\frac{n+1}{n-1} t \right), \\ &y(t)=-\sin t + \sin \left(\frac{n+1}{n-1} t \right). \end{align}

As you see there is a connection to Chebyshev polynomials of the first and second kind.

If you plot for odd values of $n>1$ the graph shows spirals with $n$ sheets as expected. But if you plot for even values $n \geq 2$ the graph shows strangely $2n$ leafs. ( the parameter domain is $0$ to $2 (n-1) \pi$ )

For e.g. $n=2$ I would have liked to see the Bernoulli lemniscate.

Q1 : is there a parametrization which ideally would resulting correctly in $n$ leafs for all values of $n$ (perhaps only $n \geq 3$)?
Q2 : how to tackle the problem whether these $T$-curves are algebraic (as the classical sinusoidal spirals are)?

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Q2 can at least be answered in the positive for n=3.
( this is then the analog of the so called Kiepert curve which is
the classical sinusoidal spiral for n=3 )

I found by trial and error under heavy use of a CAS system
the implicit equation

(x^2+y^2)^2=2x(x^2-3y^2)

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Edit to correct a previous faulty version. These curves are related to special cases of epicycloids for the special case $R=1$

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  • $\begingroup$ In Wikipedia I read that epicycloids have "cusp" which these curves dont have. I dont see what to set for r in the equations there x(t)=r(n+1)cos(t)-rcos((n+1)t) y(t)=r(n+1)sin(t)-rsin((n+1)t). $\endgroup$ – Wolfgang Tintemann Apr 3 '14 at 16:26
  • $\begingroup$ The case of cusps which one usually sees are with $R\neq 1$. $\endgroup$ – jena Apr 3 '14 at 16:36
  • $\begingroup$ But the curves go through the origin. How can an epicycloid go through the origin if the small circle ( r=? ) is rolling outside the large circle with radius R(=1) ? $\endgroup$ – Wolfgang Tintemann Apr 3 '14 at 17:07
  • $\begingroup$ Sorry. Made a mistake in my computations. Have edited to acknowledge this. (my response is of course now void of content but I can't delete it). $\endgroup$ – jena Apr 3 '14 at 17:25

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