$O( n^3)$ vs $O(n^2 \ log n)$

Question

I was wondering how $n^3$ compares to $2n^2 \log n$ as I thought that $n^3$ is $\Omega(n^2 \log n)$

but there is the fact that $n$ is $O(n \log n)$ so I wasn't sure whether it is bigO or $\Omega$

Answer 1

Yes, $n^3$ grows asymptotically faster than $2n^2\log n$, so $n^3$ is $\Omega(n^2\log n)$.

This is the same as saying that $n^2\log n$ is $O(n^3)$, which should be well known -- since $n>\log n$ for all $n>0$, we have $n^3 \ge n^2\log n$ even before taking asymptotics.

Answer 2

I don't know what is $\Omega$, but a classical result says that $n^2\log n=o(n^\alpha)$ for all $\alpha>2$ (hence $n^2\log n=O(n^\alpha)$.

As for your assertion, $n=o(n\log n)$, not $O$.

Answer 3

$n^3$ is $\Omega(n^2\log n)$ and $\Omega(n^{2.9995}\log^7n$ and $O(n^3)$ and $O(n^{17})$ and $O(n^3\log n).$ Remember that $O$ essentially means "not faster than" and $\Omega$ essentially means "not slower than".