Compare the growth of the following 2 functions:

$$n^{\log \log \log n}$$ $$(\log n)!$$

My solution:

Let $n=2^m$

$$n^{\log \log \log n}=(2^m)^{\log \log \log_2 m}$$ $$=(2^m)^{\log \log m)}$$ $$=(\log m)^{\log 2^m}$$ $$=(\log m)^m$$

$$(log n)! = (log_2 2^m)!$$ $$= m!$$

Now I don't understand as to how to compare between

$$(\log m)^m$$ AND $$ m!$$

It would be a great if someone could help me with this. Thank you.

  • $\begingroup$ What does $(\log n)!$ even mean? $\endgroup$ – Randall Feb 5 at 15:32
  • 1
    $\begingroup$ You may use Stirling's Approximation with $\log n$ as the input inside of the Factorial. $\endgroup$ – Paras Khosla Feb 5 at 15:36
  • $\begingroup$ Have you tried Stirling as suggested yesterday? It will solve your problem. Again, let $m=2^k$ so the log comes out even. $\endgroup$ – Ross Millikan Feb 7 at 15:05
  • $\begingroup$ Thanks @RossMillikan. Stirling's approximation did solve my problem. $\endgroup$ – Kivtas Feb 7 at 16:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.