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When looking for the approximate roots of $\epsilon^2x^6-\epsilon x^4-x^3+8=0$, since this is a single perturbation problem, we need to track down the three missing roots, so we consider all possible dominant balances between pairs of terms as $\epsilon\to 0$.

Now suppose $\epsilon^2x^6\sim\epsilon x^4$ ($\epsilon\to 0$) is the dominant balance. Then $x=O(\epsilon^{-1/2}) (\epsilon\to 0)$. It follows that the terms $\epsilon^2x^6$ and $\epsilon x^4$ are both $O(\epsilon^{-1})$. But $\epsilon x^4\ll x^3=O(\epsilon^{-3/2})$ as $\epsilon\to 0$.

Instead if writing the entire comparison, I will write the comparison using shorthand $\ll$. Now I am trying to understand why $O(\epsilon^{-1})\ll O(\epsilon^{-3/2})$ as $\epsilon\to 0$.

I am finding it hard to get the logic behind it or at least understand it in a more general setting.

I tried to reason as follows:
We know $O (\epsilon^{-1/3})\ll O(\epsilon^{-1})$ is obvious, then I guess it is also the case that $O (\epsilon^{-1/2})\ll O(\epsilon^{-1})$ because taking the root (square and cube) of $\epsilon$ makes it bigger for small $\epsilon$, since $\epsilon$ is in the denominator, the whole fraction gets smaller.

But $O (\epsilon^{-3})\gg O(\epsilon^{-1})$, because for small $\epsilon$ taking power of it makes it smaller, since $\epsilon$ is in the denominator, hence the whole fraction gets larger.

Then if we "combine" the operation of taking square root and raising to the power of 3. how can we conclude $O(\epsilon^{-1})\ll O(\epsilon^{-3/2})$? Why not $O(\epsilon^{-1})\gg O(\epsilon^{-3/2})$?

Is there any general rule to compare $O(\epsilon^{-1})$ with any $O(\epsilon^{-n/m})$? and to make it even more general to compare $O(\epsilon^{-n/m})$ with $O(\epsilon^{-r/s})$?

Thanks!

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    $\begingroup$ What's your definition of "O(something) $\,<<\;$ O(another something)" ? $\endgroup$
    – DonAntonio
    Commented Jun 2, 2013 at 23:27
  • $\begingroup$ @DonAntonio. I have added the definition, it is just the Big-O notation. hopefully it is clear. $\endgroup$
    – user71346
    Commented Jun 2, 2013 at 23:36
  • $\begingroup$ @user71346 but you did not add the definition of what it means when you write $O(f(\epsilon)) \ll O(g(\epsilon))$, which is what is unclear. I guess it might mean something like $h(\epsilon) \in O(f(\epsilon)) \Rightarrow h(\epsilon) \in o(g(\epsilon))$, but I'm not sure. I wrote my answer below with this interpretation in mind--does it help? $\endgroup$ Commented Jun 2, 2013 at 23:37
  • $\begingroup$ Oh, I know the big O definition, @user71346...hehe. I meant exactly what Antonio Vargas says: what does that "<<" mean here? $\endgroup$
    – DonAntonio
    Commented Jun 2, 2013 at 23:43
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    $\begingroup$ Yes you are right @DonAntonio. Thanks for your suggestions. I will think of the problem one more time and make sure I understand the answer by Antonio Vargas, I will edit the post afterwards. $\endgroup$
    – user71346
    Commented Jun 3, 2013 at 0:03

1 Answer 1

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Let $p$ and $q$ be positive real numbers with $p < q$. Then

$$ \frac{\epsilon^{-p}}{\epsilon^{-q}} = \epsilon^{q-p} \to 0 $$

as $\epsilon \to 0$ with $\epsilon > 0$.

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  • $\begingroup$ Thanks for the answer. Yes, your claim is correct. But I still cannot use your claim to compare $O(\epsilon^{-1})$ with any $O(\epsilon^{-n/m})$. If you understand what I mean, hopefully. $\endgroup$
    – user71346
    Commented Jun 3, 2013 at 0:30
  • $\begingroup$ Ignore my previous comment. I understand it now. If $0<\epsilon<1$, when it is raised to an exponent p, where 0<p<1, then it $\epsilon$ gets smaller, if raised to p for p>1, then it gets larger. I realised that I have asked a dumb question and complicated the problem too much. Thanks anyway for your help. $\endgroup$
    – user71346
    Commented Jun 3, 2013 at 0:44
  • $\begingroup$ @user71346, I think you mean the other way around. If $0 < \epsilon < 1$ then $\epsilon^p > \epsilon$ if $0 < p < 1$ and $\epsilon^p < \epsilon$ if $p > 1$. $\endgroup$ Commented Jun 3, 2013 at 2:58

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