How do I go about proving that $$n\sqrt{n}∈O(n^2)$$? I think I get the idea of $O$-Notation, whereby it shows if a function grows faster than or at the same rate compared to another function. The idea seems pretty simple and this question is listed as an example of an $O$ function without an explanation, along with many others. Also what's another simpler example of an $O$ function.
Could you say that $a(n)=n^2+3n+4$, $a(n^2)∈O(n^2)$?
On
$\dfrac{n\sqrt{n}}{n^{2}}=\dfrac{1}{\sqrt{n}}\leq 1$, so $n\sqrt{n}\in O(n^{2})$. But it is not the case that $a(n^{2})\in O(n^{2})$.
On
Note that by definition $$ f\in O(g)\text{ as }x\rightarrow a $$ whenever $$ \limsup_{x\rightarrow a}\left|\frac{f(x)}{g(x)}\right|<\infty $$ in this case since
$$\dfrac{n\sqrt{n}}{n^2}=\dfrac{1}{\sqrt{n}}\to0$$
we have that $n\sqrt{n}∈O(n^2)$.
As $n$ goes to infinity, $$ \dfrac{n\sqrt{n}}{n^2}=\dfrac{1}{\sqrt{n}}$$ is absolutely bounded by the constant $1$ (or any constant at all for that matter, since it tends to $0$, but the point is that there is such a constant).
In your last example, $a(n)$ is $O(n^2)$, but not $a(n^2)$ if that is really what you meant.