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Thread: n*log(n) asymptotic behaviour

  1. June 29th 2010, 06:58 AM #1
    courteous
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    Question n*log(n) asymptotic behaviour

    Quote from a textbook:
    ... n^{1.26} is asymptotically still greater than n\log n.
    Well, if n\log n is presented as n^x, then what is x?
    Or, if there is no such general representation, what would x be for n = 100, 1000, ... ( \log is base 2)?

    I should've plotted first. "Solved" I guess.
    Last edited by courteous; June 29th 2010 at 09:08 AM.
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