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I have to determine the order of loads of expressions as $\varepsilon \to 0$.

Can you help me by giving me an example of how to find the order of $\sqrt{\varepsilon(1-\varepsilon)}$.

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$\varepsilon(1-\varepsilon)=\varepsilon-\varepsilon^2\approx \varepsilon$ because $\varepsilon\to 0$. So $\sqrt{\varepsilon(1-\varepsilon)}\approx \sqrt{\varepsilon}$ –  userNaN Feb 23 '12 at 23:04
    
thank you Norbert. –  steven Feb 23 '12 at 23:05

2 Answers 2

The binomial theorem is applicable to arbitrary (even complex) exponents.

$$(1-x)^{r} = 1 - rx + \frac{r(r-1)}{2} x^2 - \frac{r(r-1)(r-2)}{6}x^3 + \dots $$

This series converges absolutely for any $r$, when $|x| \lt 1$.

You can use that, setting $r = \frac{1}{2}$.

See: Binomial Series.

In general, you can try getting the Taylor series at $0$ and see if that works. The above is nothing but the Taylor series of $(1-x)^r$.

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Thankyou very much for your help. –  steven Feb 24 '12 at 0:38
    
@steven: You are welcome. –  Aryabhata Feb 24 '12 at 0:50

Another approach is to use the big O notation and its "absorbing" properties, because then you can keep track of the orders: $$\sqrt{\epsilon\left(1-\epsilon\right)}=\sqrt{\epsilon-O(\epsilon^2)}=\sqrt{\epsilon+O(\epsilon^2)}=\sqrt{\epsilon}\left(\sqrt{1+O(\epsilon}\right)=\sqrt{\epsilon}\left(1+\frac{1}{2}O(\epsilon)\right)=\sqrt{\epsilon}+O(\epsilon^{3/2})$$

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