How does big-O notation relate to the actual error involved in a numerical differentiation? Suppose I have some position data ${x_1, x_2, ... x_n}$ that was sampled at an interval $h$. If I wanted the velocity data, I could apply a finite difference scheme:
$ v_1 = \frac{x_2 - x_1}{h} + O(h)$
$O(h)$ denotes that the error term is proportional to the step size. ..What exactly does this mean physically? For instance, say my $x$ terms were taken at 5 samples/second (so $h=0.2$). Does this mean the error in my velocity data is $\pm$0.2? How do I interpret the error in a physical sense, and can I write this like an uncertainty to a measurement? E.g., $v_1 = 10 \pm 0.2$ m/s?
 A: Look at the simplest functions $x(t)$ and compute the exact expressions
$v(h,t) = \frac{x(t+h)-x(t)}{h}$.
For $x(t) = at$ you have $v(h,t) = \frac{x(t+h)-x(t)}{h} = a$ and therefore
the error term $O(h)$ is zero.
For $x(t) = at^2$  you have $v(h,t) = 2 at+ah$ and  $O(h)=ah.\;$ Thus the
error is constant in time, it only depends on $a,h.$
For $x(t) = at^3$ you have $v(h,t) = 3 a t^2 + 3 a t h + a h^2$ and $O(h)=3 a
t h + a h^2.\;$ Once a again you have a constant term, here $a h^2,\;$ but now
the second error term $3 a t h$ is time-dependent ant its absolute value
increases.
In the two first cases you have an obvious error of the form $\pm x.y $ m/s but in the last case it is not so easy. You can give a maximum error for the
considered t range; or if you have to bound the error you have maximum
time interval for a valid approximation.
A: More generally, if $x$ is twice continuously differentiable, there exists a $\xi\in [0,1]$ for which $\frac{x(t+h)-x(t)}{h}=\frac{h}{2!}x''(t+\xi h)$, so that you need to estimate the second derivative: $v_1 = 10 \pm \frac{0.2}{2} v''_{max}$
