"er" is the Danish equivalent of "is" in English. I need some help with the square root one. Additionally, it would be nice to know if the other ones are correct.
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You can use this fact that $\log(n) = O(n^k)$ for any constant $k > 0$ and $k \in \mathbb{R}$. $k$ in your case is $\frac{1}{2}$. Hence, $\sqrt{n}$ is not $O(2\log(n))$.
OmG
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But I think I remember that sqrt(x) isn't a polynomial. In other words $x^{0.5}$ isn't a polynomial. – Alexander Nielsen Sep 01 '19 at 14:47
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@AlexanderNielsen it doesn't matter. It is true for any $k > 0$ (not just for integers). – OmG Sep 01 '19 at 14:48
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Okay, thanks. By the way, do you happen to know whether my other answers are correct? – Alexander Nielsen Sep 01 '19 at 14:48
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+1 for posting that it is homework. -10 for posting homework. – gnasher729 Sep 01 '19 at 15:01
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I don't get it, what is wrong with asking for help with homework? @gnasher729 – Alexander Nielsen Sep 01 '19 at 15:02