Solving an asymptotic equivalence for x

Question

This might be a very dumb question, but I have googled around and looked in asymptotic analysis reference texts and cannot seem to find what I am looking for. So here we go:

Suppose I know that there exists some $x$ such that $xn \sim n^2$ as $n \to \infty$. Can I conclude that $x \sim n$?

Welcome to Maths SX! I suppose $x$ is an expression depending on $n$? — Bernard, Jul 04 '18 at 16:58
There are various equivalence relations that you might have in mind. In any case a proof will proceed (perhaps indirectly) from the definition of that equivalence, so stating it in the Question is certainly expeditious. — hardmath, Jul 04 '18 at 16:59

score 1 · Answer 1 · answered Jul 04 '18 at 17:01

1

Yes, because equivalence is compatible with multiplication, so $$xn\sim n^2\quad\text{and}\quad \frac1n\sim\frac1n\quad\text{imply}\quad xn\cdot \frac1n=x\sim n^2\frac1n=n.$$

answered Jul 04 '18 at 17:01

Bernard

175,478

Ah, thank you. Is there a similar thing I can do for an equation like $x^2 - nx \sim n^2$ — again with $n$ arbitrary large and $x = x(n)$? – H. Löw Jul 04 '18 at 17:11
I'm afraid not (I don't know the context), as equivalence is not compatible with addition nor subtraction. – Bernard Jul 04 '18 at 17:27
For that you would use the method of dominant balance @H.Löw – Antonio Vargas Jul 05 '18 at 00:19

Solving an asymptotic equivalence for x

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