March 6th, 2014, 04:34 AM  #1 
Newbie Joined: Jun 2012 Posts: 9 Thanks: 0  proving big O
so, I have a question that is stated: prove 2^n=O(n!) here is my attempt to prove that the statement is true or false: 2^1=2 1!=1 false 2 is not <= 1 2^2=4 2!=2 false 4 is not <=2 2^3=8 3!=6 false 8 is not <= 6 2^4=16 4!=24 true 16<=24 2^5=32 5!=120 it looks like the statement is true only when n is greater then or equal to 4. am I following this correctly? 
March 6th, 2014, 06:44 AM  #2 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: proving big O
"It looks like" isn't going to help here, since you're asked to prove the inclusion. Here's a thought. Suppose 2^n < n! for some n > 1. Then 2^(n+1) = 2^n * 2 < n! * 2 < n! * (n+1) = (n+1)!. 

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