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Asymptotic bound of n lgn

  1. Oct 12, 2015 #1
    When considering the asymptotic bounds for n lgn, for what value of a in na does n lgn satisfy O(na) and for what value does it satisfy Ω(na)?

    For example n lgn = O(n1.261) but n lgn = Ω(n0.797). Can someone please tell me where for what a does Ω change to O? Also how does the answer change when you consider general logb instead of lg. Thanks!!!
     
  2. jcsd
  3. Oct 13, 2015 #2
    A good bound that has been found for the logarithm (in a form that might be useful to you) is the following,

    ##\ln x \leq x^{\frac{1}{e}}##

    That of course implies,

    ##|x \ln x| \leq |x^{\frac{e+1}{e}}|##

    Now review the mathematical definition of ##f(x)=O(g(x))##. You'll get the answer for your first question.

    If you switch to ##\log_{b} n## you can still express it in terms of the natural logarithm and that will help you determine an asymptotic bound for logarithms with different bases.

    Hope this helped.
     
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