Questions about Big Oh notation

[AxiomOfChoice]

I'm sure I could think my way through these, but I'm sick and on a tight schedule, so I was hoping someone here could help me out. I would appreciate a verification, with or without proof, of the following assertions:

<br /> (\mathcal O(\epsilon))^2 = \mathcal O(\epsilon^2)<br />

and

<br /> \sqrt{1 + \mathcal O(\epsilon^2)} = 1 + \mathcal O(\epsilon^2)<br />

Thanks so much.

 

[AxiomOfChoice]

I think I've managed to show the first one. Suppose f(\epsilon) = \mathcal O(\epsilon) (as \epsilon \searrow 0). Then there exists C &gt;0, \delta &gt; 0 such that 0 &lt; \epsilon &lt; \delta implies

<br /> \left| \frac{f(\epsilon)}{\epsilon} \right| \leq C.<br />

To show that (\mathcal O(\epsilon))^2 = \mathcal O(\epsilon^2), one simply observes that

<br /> \left| \frac{f^2(\epsilon)}{\epsilon^2} \right| \leq C^2.<br />

 

[mathman]

The second approximation can be gotten by using the binomial expansion of the left side.