I have recently encountered a series of perturbation problems in which the Big Oh notation is used frequently. Since I have not encountered this notation before, I am a little bit confused about it. I have read various websites about it, and I get the idea behind it (I also can quite easily look at a function and determine what order the function has), but a few statements in my text book still leave me confused.
For instance, in my book it says the following at one point:
Consider
$q(x,\epsilon) = y_{0} + y_{1} = e^{1-x} + e(1 - e^{-x/ \epsilon})$
If $x = O(1)$, then
$q(x, \epsilon) = e^{1-x} + e + O(\epsilon)$
I am a little bit confused about the whole "If $x = O(1)$, then. . ." part of the problem. Why is it here necessary to state this? Is this because, if $x = O(1)$, then we have $x < A$, where $A$ is a constant? Thus $x$ does not approach infinity, and then the last estimation above follows? Is this correct reasoning? This is what I assume based on how I've interprerted the definition of Big Oh, but I could be wrong here.
I would greatly appreciate it if someone could explain this to me.
Answer
I think there is confusion because usually when we talk about Big-Oh, it is assumed that $n\rightarrow\infty$. However, when we are talking about 'small quantities' like $\epsilon$, then we mean that $\epsilon\rightarrow0$ instead. You can think of it as $\epsilon=\frac{1}{n}\rightarrow 0$.
You can see the term $e^{1-x/\epsilon}$ is smaller than $\epsilon$. So we say it is $O(\epsilon)$. A graph on wolframalpha may convince you by ploting $e^{1-1/x}-x$ on x=0..1. See here
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