Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I was reading on the big O/little O notation etc. and I understand the definitions, but how exactly would I use it to find the order of an expression/function?

I am asked to determine the order of $\sqrt{\epsilon(1-\epsilon)}$ and $4\pi^2\epsilon$ as $\epsilon \rightarrow 0$. But how would I do that?

share|cite|improve this question
You should divide by the $\epsilon^p$ and see for which $p$ the expression goes to zero as $\epsilon$ goes to zero. Then choose an $\epsilon$ near the supremo of such p's. – checkmath Feb 20 '12 at 20:36
Thanks @chessmath. May I ask -- where does the $\epsilon ^p$ come from? – Heijden Feb 20 '12 at 20:59
The main idea is that $\epsilon^p$ is an infinitesimal of order $p$ and an infinitesimal of "larger" order goes to zero "fast". – checkmath Feb 20 '12 at 21:31

For $\sqrt{\epsilon(1-\epsilon)}$, note that $(1-\epsilon)$ isn't small at all as $\epsilon \to 0$, so all that matters is the $\epsilon$.

For $4\pi^2\epsilon$, the $4 \pi^2$ is just a constant, so it does not affect the order of the zeroness.

share|cite|improve this answer

Big-Oh means that, simply speaking, asymptotically, $ \forall \epsilon < \epsilon' \ \exists \ C>0$ s.t. the ratio of the functions $f,g$ are upper-bounded by $C$. In such case we write $f=O(g)$. In your case:

$$ \lim_{\epsilon \to 0}\frac{\sqrt{\epsilon(1-\epsilon)}}{4 \pi^2 \epsilon} = \frac{1}{4 \pi^2} \lim_{\epsilon \to 0}\sqrt{\frac{\epsilon(1-\epsilon)}{\epsilon^2}} = \infty $$
this means that there are no such $ \epsilon', \ C$ that $ \forall \epsilon <\epsilon' \ f(\epsilon) \leq C g(\epsilon)$. Since the limit of the ratio is $\infty$, we write $f=\omega(g)$ or equivalently $g=o(f)$.

share|cite|improve this answer
Thanks. Sorry, perhaps I should make the question clearer --- I am trying to find the order of $\sqrt{\epsilon(1-\epsilon)}$ and the order of $4\pi^2\epsilon$. Its 2 separate questions. – Heijden Feb 20 '12 at 21:01
the order of the first function is $o(\sqrt{\epsilon})$, the second one $o(\epsilon)$ as $\epsilon \to 0$ – sigma.z.1980 Feb 20 '12 at 21:09
Thanks. May I ask, how did you get/what is your method to get $o(\sqrt{\epsilon})$ for $\sqrt{\epsilon(1-\epsilon)}$? – Heijden Feb 20 '12 at 21:18
for $\epsilon \to 0 \ \sqrt{\epsilon - \epsilon^2}$ is dominated by the first term (since squaring a value between 0 and 1 gives a lower value), therefore the largest term in this expression is $\sqrt{\epsilon}$ and the order of convergence is $o(\sqrt{\epsilon})$ – sigma.z.1980 Feb 20 '12 at 21:54

You may do as in the following way:


Which goes to zero when $\epsilon$ goes to zero iff $p<0.5$ in this case is $o(\epsilon)$. When $p=0.5$ we have $O(\epsilon)$. The other case is analogous.

share|cite|improve this answer
I'm not sure what you mean. $\sqrt{\epsilon (1-\epsilon)}$ is $\Theta(\epsilon^{0.5})$, but $w(\epsilon) = \epsilon^{0.5-p} (1-\epsilon)^{0.5}$ isn't $o(\epsilon)$ for $p<0.5$, because it would mean that $w(\epsilon)/ \epsilon$ is bounded for small $\epsilon$, which is not true. I would rather say that it is $o(1)$. – savick01 Feb 20 '12 at 21:25
Hi savicko, could you explain to me how you got $o(1)$ as your answer? – Heijden Feb 20 '12 at 21:31
No, my answer for $\sqrt{\epsilon (1-\epsilon)}$ is $\Theta(\epsilon^{0.5})$ as I stated above. But $w(\epsilon)$ is $o(1)$ if $p<0.5$. – savick01 Feb 20 '12 at 21:34
Ok sorry, my bad. – Heijden Feb 20 '12 at 21:44

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.