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December 2nd, 2018, 04:33 PM  #1 
Member Joined: Oct 2017 From: Rumba Posts: 34 Thanks: 0  BigO question  Algorithms 
December 2nd, 2018, 08:06 PM  #2 
Senior Member Joined: Sep 2016 From: USA Posts: 521 Thanks: 293 Math Focus: Dynamical systems, analytic function theory, numerics 
There is already a hint included in the exercise and it basically gives you the answer. What more do you want?

December 3rd, 2018, 05:59 AM  #3 
Member Joined: Oct 2017 From: Rumba Posts: 34 Thanks: 0  
December 3rd, 2018, 04:11 PM  #4 
Senior Member Joined: May 2016 From: USA Posts: 1,210 Thanks: 497 
I am going to use a slightly more general definition. $g(n) = O(f(n)) \text { as } n \rightarrow \infty \iff \exists \ c > 0 < n_0 \text { such that}$ $f(n) \le c * g(n) \text { for all } n \ge n_0.$ As I said at the other site, you first need to identify f(n). In this problem, $f(n) = 2n  1 \text { for } n \ge 1.$ Can you think of a simple function that always exceeds f(n) if n is at least 1? Sure you can. $n \ge 1 \implies 2n > 2n  1.$ So what do you think are $c$ and $n_0$? 
December 3rd, 2018, 04:22 PM  #5  
Member Joined: Oct 2017 From: Rumba Posts: 34 Thanks: 0  Quote:
http://prntscr.com/lqc7a7  
December 3rd, 2018, 04:30 PM  #6  
Senior Member Joined: May 2016 From: USA Posts: 1,210 Thanks: 497  Quote:
If so, c = 2, g(n) = n, and n_0 = 1 satisfies the definition. You seem to be making something fairly simple really difficult. STUDY the definition until you understand it. Last edited by JeffM1; December 3rd, 2018 at 05:12 PM.  
December 3rd, 2018, 05:19 PM  #7 
Member Joined: Oct 2017 From: Rumba Posts: 34 Thanks: 0  Thanks i understand it now, please respond to your private message.


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