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The total cost of our operations are: Σ(i=1 to n) log(i).

Prove that this sum is Ω(n log(n)).

I'm a little bit stuck on how to go about proving this. I realize the summation comes out to be log(n!), since log(1) + log(2) + log(3) = log(3!) (and so forth, etc)

But then I'm stuck on where to go for a formal proof. Any help would be appreciated!

user677786
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2 Answers2

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Are you sure , that it is big omega and not big O , coz I thought every log(i) , 0 <= i <= n , could be represented as O(log(n)) , so summation would give you O(nlog(n))

Prashant
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your easiest attack is to argue that \sum{i=1}^{n}\log(i) < \sum{i=1}^{n}\log(n), then show your summand is independent of the index. alternately, you can show that n! < n^n, then apply properties of logs to get your answer.

NQZ
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