If for functions $f, g$ we have $f = \mathcal{O}(g)$, then does $g = \mathcal{O}(f)?$ Explain why this is true or give an example showing it is false.
Workings:
I believe this is false. if I take $g = n^3 +20 n +1$ then $f = \mathcal{O}(g) = \mathcal{n^3}$
But $g = \mathcal{n^3} = n^3 \neq n^3 +20n +1$
Is what I said correct? Any help will be appreciated.