I am working on a review for a test and I'm trying to figure out how to explain the following problem:
Determine if the following statement is True or False. Briefly explain why:
If $\,f(n) = (2^{n} + 2n^{2})^{1/5}$ and $\,g(n) = 4n^{5} + 8n + 2\log(n)\,$ then $\,f = O(g)$
I just graphed the two equations and saw that the statement is true, however on a test with no calculator I have no idea how to explain this efficiently.
I know that $g(n)$ can be looked at as $g'(n) = n^{5}$ for large values of $n$, but I don't really know how to rewrite $f(n)$ so that I can show a comparison to $g'(n)$.
Any help would be greatly appreciated!
$\textbf{UPDATE:}$ Our teacher posted a solution and actually said the statement was False, which doesn't make sense because $f(10) \approx 4.15$ and $g(10) \approx 400,000$ so $f$ is growing no faster than $g$ or $f = O(g)$.
Let's look at $g$. As $n\to \infty$, the dominant thing in $g$ is $4n^5$. It's true that that's pretty big when $n=10$, and even bigger for larger $n$.
Now let's see $f$. It's perhaps a surprise, but for $f$ the dominant thing is $2^{n/5}$. This is smallish for $n=10$. It's only $4$ then. But it grows quite quickly. We don't care about $n$ as small as $10$. We want $n \to \infty$.
When $n=100$, we're looking at $4n^5 = 4*10^{10}$. And $2^{n/5}$ is about $10^6$. Let's try $n=1000$. $4n^5$ becomes $4*10^{15}$ while $2^{n/5}$ is about $10^{60}$
As a general rule if $a$ is a number greater than 1, then $a^n$ grows MUCH MUCH MUCH faster than $n^a$.