# Thread: Order according to big O

1. ## Order according to big O

Order the functions by their Big-O time complexity.
order them by growth rate. such that the most slowly growing function comes first. If f(n) = theta(g(n)), put them on the same line)
2^(n), (2/5)n, 7n^(5) - n^(3) , log(n) , log(n^2) , log(log n) , 15^(451) ,
n/log n ,n + 50 log n, ln(n), n! , 2^(log n)

i'm not sure of my ordering

1)15^(451)

2)log(log n)

3)log(n) and log(n^2)

4)ln(n)

5)n/log n

6)(2/5)n and n + 50 log n

7)7n^(5) - n^(3)

8)n!

9)2^(log n)

10)2^(n)

2. Originally Posted by Rola66
Order the functions by their Big-O time complexity.
order them by growth rate. such that the most slowly growing function comes first. If f(n) = theta(g(n)), put them on the same line)
2^(n), (2/5)n, 7n^(5) - n^(3) , log(n) , log(n^2) , log(log n) , 15^(451) ,
n/log n ,n + 50 log n, ln(n), n! , 2^(log n)

i'm not sure of my ordering

1)15^(451)

2)log(log n)

3)log(n) and log(n^2)

4)ln(n)

5)n/log n

6)(2/5)n and n + 50 log n

7)7n^(5) - n^(3)

8)n!

9)2^(log n)

10)2^(n)
$\log(n^2)=2 \log(n)$ , so $\log(n) = \Theta(\log(n^2))$

Also from Strilings formula we know that $n!=\Theta(n^n e^{-n})$. So for any $k>0$; eventualy $n>e^{k+1}$, so $n!$ grows faster than $e^{kn}$. Which makes n! the fastest growing of your functions.

RonL