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How can I parametrize the trajectory

this guy

so that it is a smooth path $h:[-1,1]\rightarrow \mathbb{C}$?

I think that I should use $$h=\left\{\begin{array}{ccl}t+i |t|&:&-1\leq t \leq 0\\ ? &:&0\leq t \leq 1\end{array}\right.$$ but I'm not sure what to put for $?$ to make it smooth.

I apologize in advance if this is a duplicate. I couldn't find it in the search.

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2 Answers 2

up vote 3 down vote accepted

The parametrization $h : t \mapsto \left\{ \begin{matrix} \exp \left( 1- \frac{1}{t^2} \right) (1,-1) & \text{if} \ t<0 \\ \exp \left( 1- \frac{1}{t^2} \right) (1,1) & \text{if} \ t>0 \\ 0 & \text{if} \ t=0 \end{matrix} \right.$ is even $\mathcal{C}^{\infty}$. However, the derivative of any parametrization is zero at $(0,0)$ since the slope of the tangents is discontinuous at this point.

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That's the one, nice, +1. –  1015 Feb 20 '13 at 14:28

Your path is not smooth. Under no parametrization will it become smooth.

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@julien It depends on the exact definition of smooth. (In some books, it is required that the derivative of the paramatrization is non-zero.) –  mrf Feb 20 '13 at 14:05
    
@mrf Yes, and now that I think about it, they're right. In a manifold context, one would understand smooth parametrization by smooth diffeormorphism. –  1015 Feb 20 '13 at 14:50
    
@julien I also think that is the better definition. But there seems to be no consensus among textbook authors. –  mrf Feb 20 '13 at 14:52

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