9

I would like to prove that $O(n^3)$ is bigger than $O((\log n)^4)$.

I thought that I can divide both powers with 4 so it is $$O\left(n^{\frac{3}{4}}\right)$$ vs $$O(\log n)$$ but then I don't know how I can prove that $$O(n^k)$$ is bigger than $O(\log n)$ for $k > 0$.

Robert Soupe
  • 14,663
Complex1238877
  • 153
  • Can you show that

    $$\lim_{n \to \infty} \frac{n^3}{(\ln n)^4} = \infty$$

    Or alternatively the limit of the reciprocal is zero?

     – Simon S Apr 26 '15 at 13:22
  • Is it possible to prove it without using lim, I'm told to prefer to prove without lim whenever it is possible, sorry for not adding it in the description above. – Complex1238877 Apr 26 '15 at 13:23

1 Answers1

11

Put $n=e^s$. Then $n^k=e^{ks}$ and $\log(n)=s$.

We know that $$n^k=e^{ks}>2^{ks}=(1+1)^{ks}\geq 1+ks>ks=k\log(n),$$ where the $\geq$ is Bernoulli's inequality. Therefore $\frac{1}{k}n^k>\log(n)$ for $n$ large.

Hence $O(n^k)\supset O(\log(n))$.

Complex1238877
  • 153
Alamos
  • 1,175