A function with $k$ continuous derivatives is called a $\mathcal{C}^k$ function.
In order to specify a $\mathcal{C}^k $ function on a domain $X $, the notation $\mathcal{C}^k(X) $ is used.
Assume that a function $f: \R \to \R $ is $m $ times continuously differentiable (i.e. $f \in \mathcal{C}^m$) on an interval $[a, b] $. Denote $h = b - a $. Then,
$$f(b) = f(a) + \frac{ h }{ 1! } f^{(1)}(a) + \frac{ h^2 }{ 2! }f^{(2)}(a) + … + \frac{ h^m }{ m! } f^{(m)}(a) + o(h^m)$$
In particular, if $ f \in \mathcal{ C }^{m+1} $,
$$f(b) = f(a) + \frac{ h }{ 1! } f^{(1)}(a) + \frac{ h^2 }{ 2! }f^{(2)}(a) + … + \frac{ h^m }{ m! } f^{(m)}(a) + O(h^{m+1})$$
where $f^{\pare{i}} $ is the $i$th derivative of $f$.
Let $g $ be a real-valued function defined in some neighborhood of $\b{0} \in \R^n$, with $g(\b{x}) \neq 0 $ if $x \neq \b{0}$.
Let $\b{f} : \Omega \to \R^m$ be defined in a domain $\Omega \subseteq \R^n $ that includes $\b{0}$.
$f(\b{x}) = O(g(x))$ means that $\norm{ f(\b{x})} / \abs{ g(\b{x})} $ is bounded near $\b{0}$. i.e.
$$\underset{x \to 0, x \in \Omega}{\lim} \frac{ \norm{ f(\b{x})}}{ \abs{ g(\b{x})}} \leq K$$
$f(\b{x}) = o(g(x))$ means that
$$\underset{x \to 0, x \in \Omega}{\lim} \frac{ \norm{ f(\b{x})}}{ \abs{ g(\b{x})}} = 0$$
Note that:
$$f(\b{x}) = o(g(x)) \implies f(\b{x}) = O(g(x)) $$
However the converse is not necessarily true.
Please be careful that the big-O notation here is NOT the big-O notation that are commonly used in computer science, although they are similar. The big-O notation in computer science is about the limit when $x $ approaches infinity rather than zero.