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Math Help - Asymptotics.

  1. #1
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    Asymptotics.

    I'm trying to order some numbers....For n very large, I need to order:

    2^n, 4^n, 2^(n^2), n!, and n^n

    I know 2^n<4^n<n^n, but I'm not sure where the 2^(n^2) and the n! come in...I want to say it goes 2^n<4^n<2^(n^2)<n!<n^n, but I'm not too sure.

    Is this correct?
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  2. #2
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    Logarithms will buy you the first three.

    n\cdot log(2)

    2n\cdot log(2)

    n^{2}\cdot log(2)

    That is now rather easy to order.

    2^{n}\;<\;4^{n}\;<\;2^{n^{2}}

    The last to are easy enough by themselves. One is n factors of n and the other is n factors, but only the largest is as great as n.

    n!\;<\;n^{n}

    Two logs will settle everything but the factorial. You do that.

    All we've left is the factorial vs. the 2^{n^{2}}

    No ideas? Search the brain.

    Note: The log is useful because it is monotonic increasing. This is a useful transformation that does not change the order of anything.
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  3. #3
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    Actually, I think I have it. 2^(n^2) is the biggest of all of them since 2^(n^2)=2^(n*n)=(2^n)^n, which is larger than n^n.
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  4. #4
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    n^{2} \cdot log(2)

    2\cdot log(n) + log(log(2))

    vs.

    n \cdot log(n)

    log(n) + log(log(n))

    Determination:

    log() >> log(log()), so essentially, we have 2 \cdot log(n) > log(n) and I think you have it. (But I am still a little worried about just how big 'n' is.)
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