Is $2^{log(n^2)}$ = $\Omega (\sqrt{n^3}) $?

Question

$$\text{Is }\;2^{log(n^2)} = \Omega (\sqrt{n^3}) \;?$$

If I take $n = 1$, I would get $1 = 1$, and if I'd take $n = 2$, I would get $1.52 = 2.82$.

Is that enough to prove that the statement is wrong?

Answer 1

Absolutely not. Big-O and similar notations are not meant to be seen as equalities. They are statements about the behaviour of functions when an argument approaches a value, usually (as in your case) $\infty$.

The question you initially asks means to find out if there exists a positive constant $c$ and a natural number $n_0$ such that

$$2^{\log(n^2)} \ge c \sqrt{n^3}, \forall n > n_0$$

As the formulation about "$\forall n > n_0$" should make clear, looking at one or two values for $n$ can give you absolutely no confirmation either way.

As a hint, try looking at the limit (which may or may not exist)

$$\lim_{n\to\infty}\frac{2^{\log(n^2)}}{\sqrt{n^3}}$$