# $C^n$ function with big/small O notation

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

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).
• It is. With all $f^{(i)}=0$. Because $|f(x)|\leq|e^{-1/x}|=o(x^n)$ for every $n$. – Federico Oct 31 '18 at 21:48