I've proven that $n\log_{2}\log_{2}n = \Omega (n\log_{3}\log_{3}n)$ but is $n\log_{2}\log_{2}n = O(n\log_{3}\log_{3}n)$ also true? Looks like it's not and actually $n\log_{2}\log_{2}n = \omega(n\log_{3}\log_{3}n)$. A problem of CLRS led me ask this.

  • $\begingroup$ We have "$\log_{a}b$" or "$\log_{a}(b)$" for $\log$ function with base "$a$" and argument "$b$", but what is $$\log_{a}^{b}$$ Is this power of $\log$? if yes, then what is argument? fix, please. $\endgroup$
    – zkutch
    Aug 12 '21 at 12:07
  • $\begingroup$ @zkutch It is just a logarithm with base 2 and argument $\log(n)$ and the other one is a logarithm as well. I'll fix it. $\endgroup$
    – Emad
    Aug 12 '21 at 12:19
  • $\begingroup$ math.stackexchange.com/a/37379/890149 does this answer your question? $\endgroup$
    – nir shahar
    Aug 12 '21 at 12:47
  • $\begingroup$ @nirshahar Yes. My question got solved. The idea of changing the base of the logarithm and computing the limit had stricken my mind but I was lazy to compute the limit! I computed it and it got solved. $\endgroup$
    – Emad
    Aug 12 '21 at 15:46

Yes, also we can get a tighter result that show us $ n\log_2\log_2n=\Theta(n\log_3\log_3n)$ .Suppose $c',c''>0$ is a constant, hence

$$\lim_{n\to\infty}\frac{n\log\log n}{n\log_{3}\log_{3}n} = \frac{n\log\log n}{cn\frac{\log\left(\frac{\log n}{\log 3}\right)}{\log 3}}=c'$$ Also $$\lim_{n\to\infty}\frac{n\log_{3}\log_{3}n}{n\log\log n} = \frac{cn\frac{\log\left(\frac{\log n}{\log 3}\right)}{\log 3}}{n\log\log n}=c''.$$

Note that constant factor have no effect on the growth rate of asymptotic functions, therefore


  • $\begingroup$ I also solved it using the limit but I did some more calculations to show that indeed the limit exists: Things like the L'Hopital's rule. $\endgroup$
    – Emad
    Aug 12 '21 at 15:49

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