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$\begingroup$

Note: logs below are base 2. (Not sure how to do subscripts here)

Wondering if the below equation is true when thinking asymptotically (Computer Science)

$log_2((n!)^n) = O(n \sin(n \frac{\pi}{2}) + \log_2{n})$

But I'm not sure how to compute this.

I'm guessing we need to take the log of both sides of the following equation:

$log_2((n!)^n) < n sin(n (pi/2)) + log_2(n)$

getting us:

$log_2(log_2((n!)^n)) < log_2 (n sin(n (pi/2)) + log_2(n) )$

Not sure where to go from there.

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    Thanks, I tried to make it look nicer with your suggestions...hope it helps the readability.2010-09-08

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If you are interested in computer science, a very useful thing to remember is that

$ \log n! = \theta(n \log n) $

Now given this, can you tell what $f(n)$ is, if $ \log (n!^{n}) = \theta(f(n))$?

How would that compare to the right hand side?

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    @Sev: Yeah, unfortunately there could be different definitions of BigOh out there!2010-09-09