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log(n!) = Theta(nlogn)
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How to explain this equation? It seems to be clear that log(n!) = O(n log n), since log (1.2.3.4.5....n) <= log (n.n.n.n...n) (n times).
However, how do we prove that log (n!) = Omega (n log n)
asymptoticnotation
asked
May 9, 2017
in
Asymptotic Analysis
by
shakexin
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180
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edited
Jun 3, 2017
by
Amrinder Arora
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1 Answer
+1
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Best answer
log(n!) = log(1) + log(2) + ... + log(n1) + log(n)
You can get the upper bound by
log(1) + log(2) + ... + log(n) <= log(n) + log(n) + ... + log(n)
= n*log(n)
And you can get the lower bound by doing a similar thing after throwing away the first half of the sum:
log(1) + ... + log(n/2) + ... + log(n) >= log(n/2) + ... + log(n)
>= log(n/2) + ... + log(n/2)
= n/2 * log(n/2)
answered
Jun 15, 2017
by
Chris
AlgoStar
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408
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selected
Jul 18, 2017
by
Amrinder Arora
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