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Math Help - Order according to big O

  1. #1
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    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)
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Rola66 View Post
    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
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