1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

  4. Mar 10, 2016 #3
    Hi,
    You can use Stirling's approximation as follows:
    log(n)≅nlog(n)-n ∈O(nlog(n)) .
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Big-Oh proof
  1. Big Oh notation (Replies: 2)

  2. Big-Oh Notation (Replies: 1)

Loading...