How should I approach this one a(n) = $\frac{n^3}{log^{3}(n)}$. As We can tell that $n^3$ grow much faster than $log^{3}(3)$. All of sudden, not sure what to do, found this [post][1], which is also in the fraction form, Do we calculate out $\frac{n^3}{log^{3}(n)}$ or just pick $n^3$. Eventually, the accurate number is not needed, but I somehow understand what it can simplify to and compare with other Big O, such as \begin{equation} 2^{n^{0.008}} \end{equation} Any hint is appreciated

Sorry for my first typo while using LaTex it is \begin{equation} 2^{n^{0.008}} \end{equation} instead of $(2^{n})^{0.0008}$

  • $\begingroup$ One simple way to check if $O(f(n))$ is the same as $O(g(n))$ is to take the limit $f(n)/g(n)$ as n approaches infinity. If the limit is zero or infinity, you know that the two are not equivalent. For instance $a(n)/n^3 \rightarrow 0$, so you know that $O(\frac{n^3}{log^3(n)})$ is strictly smaller than $O(n^3)$. $\endgroup$
    – Alex Smart
    Dec 12, 2018 at 3:49

1 Answer 1


It is indeed confusing if you are expecting some kind of simpler form.

However, ${n^3}/{\log^{3}n}$ is already in the simplest form in term of big-$\Theta$ notation. Consequently, it is also sort of the best form in term of big-$\Omega$ notation or big-$O$ notation. It is certainly true that $n^3/\log^3n = O(n^3)$. However, that kind of big-$O$ bound are usually frowned upon since it is obviously not tight, $n^3/\log^3n = o(n^3)$. In fact, in a similar vein, many time-complexities is described by $\Theta(n\log n)$, $O(n\log n)$ or $\Omega(n\log n)$, where the factor $\log n$ cannot be or is not removed.

The point is that although you can ignore the asymptotically insignificant terms in a sum to arrive at a simpler big-$O$ (or big-$\Theta$, big-$\Omega$) bound, you cannot do away with any unbounded factors without losing significant asymptotical information. For example, it is correct that $n^3+\log^{3}n=O(n^3)$ or $n^3+\log^{3}n=\Omega(n^3)$ or $n^3+\log^{3}n=\Theta(n^3)$, but neither $n^3/\log^{3}n=\Omega(n^3)$ nor $n^3\log^{3}n=O(n^3)$ is correct.

If you want to compare $n^3/\log^3n$ with $(2^{n})^{0.008}$, you can proceed as the following. $$\text{Since } n^3/\log^3n = O(n^3) \text { and } n^3=o((2^{0.008})^n), \text{ so } n^3/\log^3n =o((2^{0.008})^n)$$ where the second equality comes from the fact $2^{0.008}>1$ and the fact that any polynomial function grows slower than any increasing exponential function, which is proven in another answer of mine. Please note that I am using the little $o$-notation, which means basically "asymptotically growing slower than".

If you want to compare $n^3/\log^3n$ with $2^{n^{0.008}}$, you can proceed as the following. $$\text{Since } n^3/\log^3n = O(n^3) \text { and } n^3=2^{3\log_2n} = o\left(2^{(n^{0.008})}\right), \text{ so } n^3/\log^3n =o\left(2^{n^{0.008}}\right)$$ where the second equality comes from the fact $\log_2n=o(n^{0.008})$, which is also proven in that answer of mine since $\log_2n=(\log_2e)\log n$.

You can check the reference question and answers on sorting functions for good advices and more detailed and systematic treatment.

  • $\begingroup$ I am bit lost when you claim that $n^3$ =o($(2^{0.008})^n)$. can you please elaborate a little? Yes I have also update(correct) the \begin{equation} 2^{n^{0.008}} \end{equation}. to match exactly same as the book I am reading. $\endgroup$
    – Maxfield
    Sep 30, 2018 at 18:25
  • $\begingroup$ I just update my answer as well. $\endgroup$
    – John L.
    Sep 30, 2018 at 23:56
  • $\begingroup$ interesting. We actually need to compare which one is bigger \begin{equation} 2^{n^{0.008}} \end{equation} or $n^3/\log^3(n)$. but any both case, you all claim they are equal. $\endgroup$
    – Maxfield
    Oct 1, 2018 at 2:49
  • $\begingroup$ Apparently you did not see my usage of the small $o$-notation. That is NOT the big-$O$ notation as you understood. For example, $n^3/\log^3n =o((2^{0.008})^n)$ means $n^3/\log^3n$ grows slower than $(2^{0.008})^n$. $\endgroup$
    – John L.
    Oct 2, 2018 at 2:07

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