Asymptotic relation between $n^{lg (c)}$ and $c^{lg (n)}$?

55 Views Asked by At

Assume that $c > 0$ is a constant and that $n$ is a positive variable. Then why is $n^{lg (c)} = O(c^{lg (n)})$? Furthermore, can you explain why $n^{lg (c)}$ is not $o(c^{lg (n)})$, that is $n^{lg (c)} \neq o(c^{lg (n)})$?

1

There are 1 best solutions below

1
On BEST ANSWER

Taking the lg of both gives $\operatorname{lg}(c)\operatorname{lg}(n)$ so they are actually equal. Hence the provided asymptotics follow from $f\in\mathcal O(f)$ but $f\notin o(f)$.