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Can the statement that $f$ is a $C^n$ function (say for $\mathbb{R}\to\mathbb{R}$ functions) be written in terms of big/small O?

Maybe it is equivalent to existence of numbers $f(a),\dots,f^{(n)}(a)$ such that one of the following holds?

  1. $f(x) = f(a) + \frac{f'(a)}{1!}(x-a) + \dots + \frac{f^{(n)}(a)}{n!}(x-a) ^n + O((x-a)^{n+1})$

  2. $f(x) = f(a) + \frac{f'(a)}{1!}(x-a) + \dots + \frac{f^{(n)}(a)}{n!}(x-a) ^n + o((x-a)^n)$

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No they are both weaker; both properties can only encode the existence of the first derivative when $n=1$, but cannot got any further. Take for instance the function $$ f(x) = \exp\left(-\frac1{|x|}\right) \sin\left(\frac1{|x|}\right). $$ It satisfies both properties for every $a$ and every $n$, but $f'$ is unbounded near $0$ (hence discontinuous, because it vanishes at the origin).

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  • $\begingroup$ It is. With all $f^{(i)}=0$. Because $|f(x)|\leq|e^{-1/x}|=o(x^n)$ for every $n$. $\endgroup$ – Federico Oct 31 '18 at 21:48
  • $\begingroup$ You are right, it is almost smooth $\endgroup$ – porton Oct 31 '18 at 21:53

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