Finding $g(N)$ for $T(N)= \frac{\exp(N^3)}{\lg N}$ such that $T(N) = \Theta(g(N))$

Question

Could a correct answer be $$g(n)=\frac{N\exp(N^3)}{N\lg N}$$ for $T(N)=\Theta(g(N))$ if $T(N)= \frac{\exp(N^3)}{\lg N}$?

Answer 1

Sure this will work, it's the same function. Another one is a different way to write it, note that $\log N = e^{\log \log N}$, so $$ \frac{\exp\left(N^3\right)}{\exp(\log \log N)} = \exp\left(N^3 - \log \log N\right) $$

You can scale by any constant and add any lower order terms, for example $$ 25 \pi\exp\left(N^3 - \log \log N\right) + N^{100} e^N $$ will do fine as well.