Big-O Notation?

Question

The problem is to to evaluate the following contour integral along a path $C$ defined/parameterized as $z(t)=εe^{it}$:

$ \int_C \frac{e^{iz}}{z} dz$

The solution for the problem proceeds to say that $e^{iz}=1+O(ε)$.

I am somewhat familiar with asymptotics and big-O notation, but I have no clue what this means in this context. This substitution appears to be much simpler than direct substitution for the integral, but I just don't know what they're up to.

Any clarification would be much appreciated.

Answer 1

Pedantically, $f=O(\epsilon)$ usually means $\underset{t \to \infty}{\limsup} \frac{f(t)}{\epsilon(t)} < \infty$, or in a hand-wavey manner, f is asymptotically bounded above by epsilon, or is "on the order of epsilon".

My recommendation would be to evaluate the integral explicitly until you get a closed form expression in terms of $e^{iz}$ and then at that point try to use the fact that it is "asymptotically bounded above by 1" as well as any limiting procedure which presents itself.

EDIT: this answer assumes OP is familiar with how to compute contour integrals, and is primarily concerned about how to consider $O(\epsilon)$ in an analytic setting.

EDIT EDIT: Two more subtle points I forgot to mention originally --

1. O notation can be defined with respect to any point. The most common application is for considering the limit of the ratio as some parameter goes to positive (or negative) infinity, but it could also be possible that one wants to know whether or not, for example, "f=O(g) at x=0", i.e. whether $\underset{x \to 0}{\limsup} \frac{f}{g} < \infty$. In other words the point where the asymptotics are being evaluated can vary and depends on the context. For this specific problem this might be a very pertinent point, since obviously it is possible to take many more different kinds of limits on the complex plane than on the real line.

   2. As a further note, little o notation is defined as $f=o(g)$ at $x=c$ (or $\pm \infty$) if and only if $\underset{x \to c}{\lim} \frac{f}{g}=0$. This clearly implies big O, but is also much more stronger and hence provides more information. One could describe it as "f is asymptotically dominated by g" or "f is asymptotically negligible compared to g". For instance, arguments using little o allow one to ignore higher order terms in the taylor expansion when proving the central limit theorem (well not exactly but almost). Such strong conclusions are in general not possible with big O alone.

Thus whether or not $e^{iz}-1$ is little o or big O of $\epsilon$ depends most likely (I think, I haven't quite thought about it) on where the limit point is, i.e. the setting of the asymptotic comparison, and the difference between the two could substantially affect how useful the information is in proving the desired result.

UPDATE: I read the document OP linked to -- the idea here is that we are taking the limit as $\epsilon$ goes to 0. Then $e^{iz}-1$ corresponds to the remainder term which we would like to ignore, and thus the fact that it is asymptotically bounded above by $\epsilon$ as $\epsilon \to 0$ means that it must also go to 0 as $\epsilon$ goes to 0, and hence that we are in a rigorous sense justified in ignoring that term as we go to the limit.