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I am considering the expression $$ n\cdot e^{\mathcal{O}(n\cdot z_n^2)}$$

where the big-O-Notation is meant for $n\to\infty$; $(z_n)$ is a Zero sequence, i.e.$z_n\to 0$ as $n\to\infty$.

I am wondering what happens with this Expression as $n$ is large, i.e. what is $$ \lim_{n\to\infty} n\cdot e^{\mathcal{O}(n\cdot z_n^2)}. $$

Let $f(n)\in\mathcal{O}(n\cdot z_n^2)$ for $n\to\infty$

by definition, this means that there exists some positive real number $M$ and some $n_0$ such that $$ \lvert f(n)\rvert\leq M\lvert nz_n^2\rvert~\forall n\geq n_0. $$

So I think we have to use this in order to compute $$ \lim_{n\to\infty}n e^{f(n)} $$

Rhjg
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1 Answers1

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The limit could be anything $\geq 0$. Consider $n e^{-\sqrt{n}}$, $n e^{c-\log n}$, and $n e^{\sqrt{n}}$.

Antonio Vargas
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  • What I am really aiming it: I have some function $f$ which has a helpful asymptotic expansion for large arguments. Hence, in order to expand $f(n\cdot e^{\mathcal{O}(nz_n^2)})$ with this large argument expansion it would be interesting for me, whether the argument $n\cdot e^{\mathcal{O}(nz_n^2)}$ is large as $n\to\infty$. Is this the case? – Rhjg Jan 23 '18 at 12:54
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    @Rhjg, My answer does address "whether the argument is large as $n \to \infty$". Please consider the three examples of functions of the form $ne^{O(nz_n^2)}$ that I gave. – Antonio Vargas Jan 23 '18 at 20:08
  • I can’t understand your examples since e.g. $\sqrt{n}$ is not in $O(nz_n^2)$ in General. – Rhjg Jan 23 '18 at 22:30
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    @Rhjg, You are mistaken. Take $z_n = n^{-1/4}$ to get $nz_n^2 = \sqrt n$. – Antonio Vargas Jan 24 '18 at 00:03
  • Ok. Then I come to the conclusion that the proof given here; https://math.stackexchange.com/questions/2149381/asymptotic-to-a-sequence-of-algebraic-numbers?noredirect=1&lq=1 by Mark Viola does not work since the argument of the W-function in (3) needs not to large as you explained to me. Or do I overlook something there? – Rhjg Jan 24 '18 at 10:44
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    @Rhjg in the case of Mark's answer, the value represented by $O(n z_n^2)$ is eventually $\geq 0$ for large enough $n$, so $ne^{O(nz_n^2)} \geq n \to \infty$. The extra context makes the claim true in his answer, whereas my examples show that it's false in general. – Antonio Vargas Jan 24 '18 at 19:42
  • Could you please explain to me where the context comes into play in order to have that the value represented by $O(nz_n^2)$ is eventually non-negative? I do not see this, although I tried very hard. – Rhjg Jan 25 '18 at 20:53
  • I guess you mean the following: Suppose $f(z_n)=O(z_n^2)$ as $n\to\infty$ which in particular means that $f(z_n)\to 0$ for $n\to\infty$ since $(z_n)$ is a zero sequence. Then $n\cdot f(z_n)\geq\lvert f_n(z_n)\rvert\to 0$, i.e. eventually, $nf(z_n)\geq 0$ and hence $n\cdot\exp(nf(z_n))\geq n\to\infty$ as $n\to\infty$. - - At least I guess from the context that $n\cdot f(z_n)$ is the expression meant with $O(nz_n^2)$. – Rhjg Sep 12 '18 at 14:14
  • @Rhjg It's not true that every sequence $a_n$ which is $O(nz_n^2)$ has $a_n \geq 0$ for $n$ large enough. But certainly some sequences which are $O(nz_n^2)$ are eventually positive. The specific sequence in Mark's post is eventually positive, but you can't deduce that from the fact that it's $O(n z_n^2)$. – Antonio Vargas Sep 12 '18 at 21:02
  • Maybe you can say which explicit sequence that is which is meant in the answer, I cannot read it off it. – Rhjg Sep 12 '18 at 21:05
  • @Rhjg, It's been 9 months, I'm moving on. – Antonio Vargas Sep 12 '18 at 23:14
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    I thought good mathematics are timeless. (; – Rhjg Sep 13 '18 at 07:29