Does it have a valid mathematical meaning?

Question

Does the following expression:

$$2^{O(\log m)}$$

have any mathematical meaning? Is it even correct to write this? I have certain doubts.

$O(\log m)$ is not a well-defined number, so it's difficult to say what is meant by raising $2$ to this power.

Answer 1

It means by definition that it's equal to

$2^{a_m\log(m)}$, with $a_m$ a bounded sequence.

In other world, there exists $C$ such that

$0 \leq 2^{O(\log(m))} \leq 2^{C\log(m)}$

Also, notice that $2^{C\log(m)} = \exp( C \ln(2) \log(m) ) = m^{C'}$, so it's at most polynomial