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Thread: Size of functions

  1. #1
    Jan 2010

    Size of functions

    Hello Everyone!

    I have a couple of questions:

    (1) is (\log n!)^3 = O(3n^3+3) true?

    (2) is n^{\log _2 n} = O(2^{\sqrt{n}}) true?

    I'm going for yes for (2), but have no clue regarding (1)
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  2. #2
    Senior Member
    Dec 2010
    For the first one,
    if (\log n!)^3 \in O(3n^3+3) \subseteq O(n^3)
    then (\log n!)^3 \leq C^3n^3 for some C and n sufficiently large.
    So \log n! \leq Cn or n! \leq a^{n} which is not true for n much larger than a. Contradiction.

    For the second one,
    n^{\log _2 n} = 2^{(\log_2 n)^2}
    Now show that (\log_2 n)^2 \in O(\sqrt{n}).
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