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Dec 1, 2010 at 7:32 comment added Alex B. @Elizabeth arccos is smooth on $[-1,1]$ and cos is smooth everywhere, so $f$ is smooth on $[-1,3]$. Moreover, the values of $f$ in that interval also land in the same interval, so all iterates of $f$ are smooth there. Am I missing something obvious?
Dec 1, 2010 at 6:57 comment added Elizabeth S. Q. Goodman No, it's more interesting than that. The suggested path traverses all the graphs on the interval [-1, 3], but not in the order you suggest, because the original paths aren't smooth. Rather, it appears that you can trace a curve that is $C^1$ at least by switching from one iterate of $f$ to another at singular points which conveniently meet. I have no idea how to find other examples, but for the purposes of creating a $C^1$ or smooth curve here, I notice the velocities don't match with the original parametrizations. To make them match a constant-speed path would be better.
Dec 1, 2010 at 6:04 history edited Alex B. CC BY-SA 2.5
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Dec 1, 2010 at 5:54 comment added Alex B. What do you mean by "where the curve intersects itself"? I took your question to mean "let's look at the original curve as parametrised by the interval $[0,1]$, say, and the iterate as parametrised by $[1,2]$ and so on, append them to each other to get something that is parametrised by $[0,\infty)$ and see whether all derivatives with respect to $t$ exist, where $t$ is the parameter". Is that not what you meant?
Dec 1, 2010 at 5:43 comment added Anixx And what about analiticity? Can we say that all integer iterates of this function are just different branches of the same analytic curve?
Dec 1, 2010 at 5:41 comment added Anixx Thank you, but what about smoothness in other points (not just $a$ and $b$)? For example, in points where the curve intersects itself?
Dec 1, 2010 at 5:29 history edited Alex B. CC BY-SA 2.5
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Dec 1, 2010 at 5:24 history answered Alex B. CC BY-SA 2.5