Results 1 to 2 of 2

Thread: Big O notation.Show using the definition of O(f)?

  1. #1
    Junior Member
    Feb 2010

    Big O notation.Show using the definition of O(f)?

    Hi I'm having some trouble with big O notation. I know how to do the simple ones but this is above my head. Any help would be appreciated.

    Show using only the definition of O(f) that:
    1) 5^n is O(n!)
    2)n! is not O(5^n)

    I know that 5^n<=n! for n=12 but I don't know how to show this. The book says to use induction 'but I don't know how and for 2 as well. Any ideas?

    Thanks in advance.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Oct 2009
    1) 5^n is O(n!)
    Intuitive idea: in 5^n, all factors are equal to 5, whereas in n!, the factors start with 1 but soon overtake 5 and continue to increase. So, 5^5>5!, but this won't last long. Calculate 5^n and n! for n=6,7,...,12. As you said, 5^{12} is already less than 12!. This can be established by direct calculation (e.g., in Windows' calculator).

    Now what happens when we move from n to n+1 for n\ge 12? 5^n is multiplied by 5 to get 5^{n+1}; but n! is multiplied by n+1>5 to get (n+1)!. So, it is clear that 5^n<n! implies 5^{n+1}<(n+1)! for n\ge12.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: Mar 10th 2011, 02:23 AM
  2. Replies: 1
    Last Post: Jan 23rd 2011, 12:07 AM
  3. show violation of metric definition
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: Apr 1st 2010, 03:28 PM
  4. Show definition of derivative
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Jan 30th 2010, 11:22 AM
  5. Replies: 1
    Last Post: Aug 26th 2009, 12:40 PM

Search Tags

/mathhelpforum @mathhelpforum