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There exist smooth - but not analytic - closed curves without self-intersections. I just would like to see a simple example of such a curve.

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Press the "faq" and "how to ask" buttons on top please. –  BS. Apr 12 '11 at 13:11
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Let $\psi:\mathbb{R}\to\mathbb{R}$ be a $C^\infty$ function with compact support in $[0,\pi]$ and such that $\psi(t)>0$ if $0 < t < \pi$. Then $$ \gamma(t)=\psi(t)(\cos t,\sin t),\quad 0 < t <\pi,\quad \gamma(0)=\gamma(\pi)=(0,0) $$ is such a curve. You can take for instance $$ \psi(t)=\exp\Bigl(-\frac{1}{t^2(\pi^2-t^2)}\Bigr). $$

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Consider the curve

$$\gamma:\quad \phi\ \mapsto\ \Bigl(1+\exp{-1\over \pi^2 -\phi^2}\Bigr)\ (\cos\phi,\sin\phi)\qquad(-\pi< \phi< \pi)$$

with filled-in point $(-1,0)$.

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Isn't it the case that as $\phi$ approaches $\pi$ and $-\pi$, the expression approaches $(0,0)$, and is not smooth there...? –  Joseph O'Rourke Apr 12 '11 at 12:33
    
Here, the origin is a cusp point of this curve. I would like a curve without cusp singularities (but perhaps fold singularities). –  Louis Apr 12 '11 at 12:54
    
@Joseph and Louis: The 1+ went lost in the write-up. Sorry for the slip. –  Christian Blatter Apr 12 '11 at 13:39
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