Which is asymptotically larger $n^2 \log(n)$ or $n (\log(n))^{10}$?

Question

Which is asymptotically larger $n^2 \log(n)$ or $n ( \log(n))^{10}$? I have tried by plugging in the values and $n^2 \log(n)$ turns out to be bigger. How can this be done analytically?

Answer 1

We wish to compare $n$ to $\log(n)^k$. Notice that $$\lim_{n\to\infty}\frac{n}{\log(n)^k}=\lim_{n\to\infty}\frac{1}{k\log(n)^{k-1}\frac{1}{n}}=\lim_{n\to\infty}\frac{n}{k\log(n)^{k-1}}$$

We see that by applying l'hopitals rule $k$ times, we find out that the limit diverges, for any constant $k$. Thus we find that, for all $k$, $n\log(n)^k=O(n^2\log(n))$

Answer 2

Comparing the two equations: $\frac{n^2\log(n)}{n(\log n)^{10}}$ is equivalent to $\frac{n}{(\log n)^9}$.

Taking the derivative gives: $\frac{1}{(\log n)^9}-\frac{9}{(\log n)^{10}}$, which simplifies to: $\frac{\log n - 9}{(\log n)^{10}}$ For large enough $n$ this will be positive so we can see that the fraction will always increase so $n^2\log n$ must be bigger.

Answer 3

$$ Let ~ f(n) = n^2 log(n) - n(log(n))^{10} $$ $$ = n~log(n)(n - (log~n)^9) $$ We are interested in the behavior of the function $f(n)$ for large values of $n$. Since $n~log(n) > 0 ~for~ n>0$ , it can be removed. We are left with: $$ f(n) = n - (log~n)^9 $$

Since $n$ grows exponentially faster than $log~n$ , meanwhile $(log~n)^9$ grows polynomially faster than $log~n$ , $n$ is therefore expected to grow faster than $(log~n)^9$. Therefore $n^2 log(n)$ is asymptotically larger than $n(log(n))^{10}$.

It will be good to know for which values of $n$ does $f(n) > 0$ . Below are some plots of $f(n)$. You can interact with the graph by entering f(x)=x-(ln(x))^9 under the graph section of Math Hotseat. It can be seen from the plots that $f(n)<0$ for $3.0 < n < 2.54385 \times 10^{13}$ (approximately).