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Consider a smooth plane curve $c:I\to \mathbb{R}^2$ with $I$ an open interval in the real line and $c' \neq 0$ everywhere. Assume $t_0 \in I$. My question is whether following statements are equivalent:

(1) $c''(t_0) \neq 0$

(2) There is a sequence $(h_n)_{n \in \mathbb{N}}$ with $h_n \to 0$ and all $h_n>0$ such that for each $n$, the three points $c(t_0 \pm h_n)$, $c(t_0)$ are not located on a straight line.

If this is true, what is a proof? If it is false, what is a counterexample? And if it is false, is there any modifiaction of statement (1) to make the equivalence true? Or a modification of statement (2) or both?

Edit: Thanks to the answers and comments, I learned that (1) $\Rightarrow$ (2) is false without further assumptions. But what about the direction (2) $\Rightarrow$ (1)?

russoo
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    Why is $c(t) = ((t+1)^2, (t+1)^2)$ not a counterexample? – Eric Towers Apr 30 '21 at 22:23
  • @EricTowers, what interval do you have in mind and what would you choose for $t_0$? – russoo Apr 30 '21 at 22:27
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    $I = [0,1] \subset \Bbb{R}$. There is no choice of $t_0$: the image of $c$ is a line segment, so all triples of points are collinear. – Eric Towers Apr 30 '21 at 22:28
  • I see. Thanks. I am completely confused right now. What I am really looking for is a definition of the osculating circle as a limit of circles through three non-collinear points of a curve. I am therefore looking for a condition on a curve $c$ such that (2) is satisfied so that we can build those circles and take the limit... Literature seems completely dead on this topic... – russoo Apr 30 '21 at 22:44
  • Sounds like you've posed an XY question. You might have more success asking for the criterion you want, rather than picking a criterion and asking if it works. – Eric Towers Apr 30 '21 at 22:48
  • Thanks for the suggestion. I already asked a question in this direction here: https://math.stackexchange.com/questions/4115922/limit-definition-of-the-osculating-circle but didn't get any useful answer. Maybe I try to reformulate the question... – russoo Apr 30 '21 at 22:54

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As Eric noted in the comments we can find a counterexample. It can be even as simple as: $$c(t) = \left(t^2,0\right).$$ Obviously, the curve $c$ is flat, so you can choose an arbitrary $t_0 \in \mathbb{R}$ and any real sequence $(h_n)_{n \in \mathbb{N}}$ convergent to $0$. On the other hand, we have $c''(t) = (2,0)$.

The key notion is how fast we move along the line. To make your statements equivalent you can add an assumption that the curve $c$ has to be parameterized by arc length i.e. $\lVert c'(t) \rVert = 1$.

mwt
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  • Thank you very much! Do you have any hints about how to prove it under the unit-speed assumption? – russoo Apr 30 '21 at 23:02
  • @russoo : $c(t) = (t+1, t+1)$ is unit speed. All triples of points on this line are collinear. I think you conflate the derivative(s) of $c$ with the slopes and curvature of the image in $\Bbb{R}^2$. – Eric Towers Apr 30 '21 at 23:12
  • @EricTowers, In your example, $c'=(1,1)$ and so $\Vert c' \Vert =\sqrt{2}$. Moreover $c''=0$... Or not? – russoo Apr 30 '21 at 23:17
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    @russoo : Right, right. $c(t) = \frac{1}{\sqrt{2}}(t+1, t+1)$. While $c'' = 0$ for this example, running at unit speed along $(t,|t|)$, $c''$ isn't even defined at $t = 0$. – Eric Towers Apr 30 '21 at 23:19