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Homework Help: Big-Oh proof

  1. Mar 9, 2016 #1
    • Moved from a non-homework section, so missing the template
    To prove that n log n is big oh of log(n!), I did:
    n log n <= C log(n!)
    n log n/ log(n!) <= C

    Let k = 1
    n > k, so for n = 2
    2 log 2 / log 2 <= C
    2 <= C
    C is an element of [2, infinity)
    Taking C = 2 and k = 1
    can we say, n log n <= 2 log(n!)
    and hence n log n is big oh of log(n!) ?
  2. jcsd
  3. Mar 9, 2016 #2


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  4. Mar 10, 2016 #3
    You can use Stirling's approximation as follows:
    log(n)≅nlog(n)-n ∈O(nlog(n)) .
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