Suppose I have an asymptotic estimate $K = O(\epsilon) - O(\delta)$, where $\epsilon \gg \delta$.
Is it correct then to write $K = O(\epsilon)$ ?
Additionally, what if I just have $\epsilon > \delta$? Would $K = O(\epsilon)$ still me correct?
Is it correct to replace $O(\epsilon) - O(\delta)$ by $ O(\epsilon)$ if $\epsilon \gg \delta$?
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asymptotics
approximation-theory
perturbation-theory
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1As soon as $\delta=O(\epsilon)$, $O(\epsilon)-O(\delta)\subseteq O(\epsilon)$. – 2017-01-30
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0@Alex Important to keep in mind, though: $O(\cdot)$ is just an *upper bound*-type statement. (While this is obvious from the definition, I've seen many people, including in papers, interpret it as a $\Theta(\cdot)$; that's not what you meant, is it?) – 2017-01-30
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0I am not quite certain what you mean by "Theta" @ClementC. – 2017-01-30
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0$f=\Theta(g)$ (at $\infty$, say) iff $f=O(g)$ and $f=\Omega(g)$ (using the [computer science definition of $\Omega$](https://en.wikipedia.org/wiki/Big_O_notation#Family_of_Bachmann.E2.80.93Landau_notations)); i.e., roughly speaking, $f$ and $g$ are "asymptotically equivalent up to constants." – 2017-01-30
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0No, I do not mean "asymptotic equivalence up to constants" - but how would the answer change then? @ClementC. – 2017-01-30
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0If you just have $\epsilon>\delta$, for instance $\epsilon = 2\delta$. Then $0=K=\epsilon - 2\delta = \Theta(\epsilon)-\Theta(\delta)$ does not satisfy $K=\Theta(\epsilon)$. – 2017-01-30