I've looked around for this and I can't find anything close to what I'm looking for.
Usually when I saw Big O notation in my classes it was something like:
$n^2 + 2n + 1 = O(n^2)$
However, the question I'm being issued now is:
If $f(n)$ is $O(g(n))$, prove or disprove this statement:
$f(n^2)$ is $O(g(n^2))$
Now, I don't know where to go from here beyond:
$f(n^2) = c * g(n^2)$
It's driving me absolutely nuts.
$f(n)=O(g(n))$ means the ratio $\dfrac{\lvert f(n)\rvert}{\lvert g(n}\rvert$ is bounded from above for all $n$ large enough such that $g(n)\ne 0$.
The same is clearly true for any function $u(n)$ such that $\lim\limits_{n\to\infty}=+\infty$, since, if the above bound is valid for any $n\ge N_0$, there exists an interger $N_1$ such that $u(n)\ge N_0$ for all $n\ge N_1$.