Find the simplest Big O estimate for this log function?

Question

$$f(n) = \log^4(n)+\sqrt{n}$$

What I've did:

$\sqrt(n)\in O(n^{1/2})$

$\log^4(n)\in O(n^{4})$

$f(n)\in O(n^{1/2} * n^4) = O(n^{9/2})$

Is this correct? Question also says to find C and K to justify my answer, how can I do this?

Answer 1

$f(n)$ certainly belongs to $O(n^{9/2})$, but I'm afraid by 'simplest' the question really means the smallest power of $n$ possible. Note that $(\log n)^m\in O(n^k),m\in\Bbb R,k\in\Bbb R^+$, which can be easily observed through the divergence of $$\lim_{n\to\infty}\frac{n^k}{(\log n)^m}$$This means $(\log n)^4\in O(\sqrt n)$. Clearly, $\sqrt n\in O(\sqrt n)$ and thus $f(n)\in O(\sqrt n)$. Indeed, $1/2$ is the smallest power of $n$ that is an asymptotic upper bound of $f(n)$. Moreover, $f(n)\in\Theta(\sqrt n)$ because$$\lim_{n\to\infty}\frac{f(n)}{\sqrt n}=1$$Note here that instead of multiplying the upper bounds of $(\log n)^4$ and $\sqrt n$, I added them up for a tighter bound. This is because of the result$$f(n)\in O(F(n)),g(n)\in O(G(n))\implies kf(n)+lg(n)\in O(F(n)+G(n))$$