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Math Help - A proof

  1. #1
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    A proof

    Can you show that, please?

    lg(n!)=O(nlgn)
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  2. #2
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    Quote Originally Posted by scofield View Post
    Can you show that, please?

    lg(n!)=O(nlgn)

    I'm guessing you wanted to write \log(n!)=O(n\log n), and this follows from \log(n!)=\sum\limits_{k=1}^n\log n\le n\log n ...

    Tonio
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  3. #3
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    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?
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  4. #4
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    Quote Originally Posted by scofield View Post
    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 \log_2, is it?

    Tonio
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  5. #5
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    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.
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  6. #6
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    By the way,

    Thank you so much for your solution...
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