Difference between little o and big O notation in taylor expansion I know I can say that:
$$
y(t+\Delta t)=y(t)+y'(t)\Delta t+o(\Delta t)
$$
But can I say that:
$$
y(t+\Delta t)=y(t)+y'(t)\Delta t+O((\Delta t)^2)
$$
In both cases, do I have to add that $\Delta t \rightarrow 0$ or i do not have to?
 A: Yes, you should add $\Delta t \to 0$ to be explicit, but in general, $o(\Delta t)$ is understood to hold for $\Delta t \to 0$, so I think it is fine not to add it, but there is no harm in writing it. 
To both notations you give: They talk about different things. The first one is the definition of $y$ being differentiable at $t$, iff 
$$ y(t + \Delta t) = y(t) + y'(t)\Delta t + o(\Delta t), \qquad \Delta t \to 0$$
we have that 
$$ \frac{y(t+\Delta t) -y(t) -y'(t)\Delta t}{\Delta t} \to 0,\qquad \Delta t \to 0 $$
that is $y(t+\Delta t) - y(t) - y'(t)\Delta t$ goes to zero faster then linear. The second one tells us more about this term. Not only is the convergence faster than linear, but it is quadratic. That is the second one holds iff
$$ \limsup_{\Delta t \to 0} \frac{y(t+\Delta t) -y(t) -y'(t)\Delta t}{\Delta t^2} < \infty $$
Consider $y(t) = t^{3/2}$ at $t = 0$. Then $y'(0) = y(0) = 0$. Hence 
$$ y(0 + \Delta t) - y(0) - y'(0)\Delta t = (\Delta t)^{3/2} $$
which goes to zero faster than linear, but not quadratic. 
