Can you show that, please? lg(n!)=O(nlgn)
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Originally Posted by scofield Can you show that, please? lg(n!)=O(nlgn) I'm guessing you wanted to write , and this follows from ... Tonio
No, actually lg n is correct. lg n is log2 (n).I mean log base 2 of n. However this proof is the same for lg n too.Isn't it?
Originally Posted by scofield No, actually lg n is correct. lg n is log2 (n).I mean log base 2 of n. However this proof is the same for lg n too.Isn't it? Of course: the above works for any basis...and lg is not standard notation for , is it? Tonio
I know that; In mathematics, (log2 n) is called the binary logarithm (the logarithm for base 2).It can be written like lg n.The binary logarithm is often used in computer science.Hence, maybe u didn't see this notation.
By the way, Thank you so much for your solution...
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