I would like to get a standard / precise definition of the phrase "up to the order $O(\epsilon^{n})$ " when used in perturbation theory. For example, suppose I have a curve $Q$ given implicitly $Q(x,y, \epsilon)$ also parametrically in terms of $ Q(x(t, \epsilon)$, $y(t,\epsilon))$ where $t$ is a parameter. If I expand both forms of $Q$ in $\epsilon$ and then say "the parametric form of $Q$ coincides with the implicit form of $Q$ up to $O(\epsilon^{n})$ terms", does this mean:
a) the terms at $O(\epsilon^{n})$ differ
or
b) the terms at $O(\epsilon^{n+1})$ differ?
Similarly, if I say "expand $Q$ up to $O(\epsilon^{n})$ ", does this mean I write:
a) $Q = Q_{0} + \epsilon Q_{1} +... \epsilon^{n-1} Q_{n-1} + O(\epsilon^{n})$
or
b) $Q = Q_{0} + \epsilon Q_{1} +...+\epsilon^{n}Q_{n} + O(\epsilon^{n+1})$ ?
The phrase "coincides up to $O(\epsilon^n)$" means that if you remove all the terms that are the same in both expressions then the terms you have left are $O(\epsilon^n)$. Note: it is an important but subtle point that, by the definition of $O(\cdots)$, the expressions a) and b) at the end of your question are indeed equal "up to order $O(\epsilon^n)$" since $O(\epsilon^{n+1}) \subset O(\epsilon^n)$ in the limit $\epsilon \to 0$.
I would interpret "expand $Q$ up to order $O(\epsilon^n)$" to correspond to expression a) since it includes the actual symbol $O(\epsilon^n)$ and the terms outside the symbol are strictly larger than $\epsilon^n$ asymptotically.