Hey,
How do you prove that log (n!) is big theta of n log n? I tried using the definition method and the shortcut rules but I keep getting stuck.
shakunni
-3
Light Poster
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Jump to PostI already provided an answer in the thread you made in the Java forum. But logically n! < n^n and log(n^n) = n log n.
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BestJewSinceJC
700
Posting Maven
I already provided an answer in the thread you made in the Java forum. But logically n! < n^n and log(n^n) = n log n.
SebKom
0
Newbie Poster
I have this on my lecture notes but I don't get it:
If g ∈ O(f (n)) then for some N, c we have that for i ≥ N , g(i) ≤ c.f (n). So for any m we have m.g(i) ≤ m.c.f (i) and hence m.g(n) ∈ O(f (n)).
SebKom
0
Newbie Poster
(My bad, please delete it) :)
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