I'm in the process of learning Big-Oh and came across an equality presented as a fact without justification:
$\sqrt{n + \mathcal O(n^2)} = \sqrt{n} + \mathcal O(n^\frac{3}{2})$
Is this so? What is the best way to approach a problem like this?
I'm in the process of learning Big-Oh and came across an equality presented as a fact without justification:
$\sqrt{n + \mathcal O(n^2)} = \sqrt{n} + \mathcal O(n^\frac{3}{2})$
Is this so? What is the best way to approach a problem like this?
This is true in the limit $n \to 0^+$. One very useful tool for problems like this is Taylor series.
Hint: Write $n+O(n^2)$ as $n(1+O(n))$, bring a $\sqrt{n}$ outside of the square root, then try to use the binomial series.