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Although it might be a bit of newbie question, my question is, How can I apply the Master theorem to the following relation:

T(n) = 99T(n/100) + log(n!)

I'm trying to learn about algorithms, but I'm not really comfortable with logarithms(haven't studied in school yet), so I'd really appreciate a bit of help. Thanks.

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    $\begingroup$ Try using $\log (n!) = \Theta(n \log n)$. $\endgroup$ – Inuyasha Yagami Jan 30 at 16:53
  • $\begingroup$ Thanks very much, but could please link to a proof of this? $\endgroup$ – kasra Jan 30 at 18:34
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    $\begingroup$ en.wikipedia.org/wiki/Stirling%27s_approximation. You asked something similar a few days before also. :) $\endgroup$ – Inuyasha Yagami Jan 30 at 18:35
  • $\begingroup$ yes, I'm really new to this stuff. Thanks very much :) $\endgroup$ – kasra Jan 30 at 18:39
  • $\begingroup$ You can answer your own question if you have solved it. Just for the sake of completeness. $\endgroup$ – Inuyasha Yagami Feb 3 at 8:07
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We know that $n! \le n^n$, therefore $\log n! \le n \log n$. Then $$ T(n) \le 99T( \frac{n}{100} ) + n \log n $$

Let $c=\log_{100} 99 < 1$ and notice that $n \log n$ is polynomially larger than $n^c$. Indeed: $n \log n \in \Omega(n) \subset \Omega(n^c).$ By the master theorem we have $T(n) = O(n \log n)$.

This is tight because $\Omega(n \log n)$ is also a lower bound since: $$ T(n) \ge \log n! \ge \log ( \lfloor n/2 \rfloor^{(n/2)} ) = (n/2) \log ( \lfloor n/2 \rfloor) = \Omega(n \log n). $$

To summarize, you have $T(n) = \Theta(n \log n)$.

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  • $\begingroup$ Thanks a lot, I tried to answer it myself but my answer was wrong so I deleted it. Thank you. $\endgroup$ – kasra Feb 5 at 12:22

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