Hopefully this is the right section. I need to prove that $2^{(log(n)^{1/2})}$ is $O(n^a)$. From the basic principle of Big-O notation, I know I need to find two numbers $c$ and $N$ so that $f(n) \le c\cdot g(n)$ for all values of $n \ge N$. In this case, $f(n)$ is $2^{(log(n)^{1/2})}$, and $g(n)$ is $n^a$.The first problem in the set had me prove that $2^{n+a}$ is $O(2^n)$. For this problem, I had separated the $f(n)$ to $2^n \cdot 2^a$. I was guided by another user to set $c = 2^a$. This way, with $g(n) = 2^n$, $cg(n) = 2^a2^n$, which is obviously $\ge f(n)$, $2^{n+a}$, for all values of $n \ge 0$. This problem, however, I don't find as easy.Could someone point me in the right direction? I've been trying to rearrange the problem and solve it for a while now.
Prove that $2^{(log(n)^{1/2})}$ is $O(n^a)$
asymptotics;landau notation