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Thread: Big Theta

  1. #1
    Senior Member
    Nov 2008

    Big Theta

    Hello everyone!

    No need to answer this question, I solved it.

    prove or disprove: 4n^3 + n^2\sqrt{n}-4n \in \Theta (n^3)

    I don't have any clue. Does anyone know how to solve this?

    I think 4n^3 + n^2\sqrt{n}-4n \in \Theta (n^3) is true, because


    Thank you,
    Last edited by Rapha; Apr 25th 2009 at 01:32 PM.
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  2. #2
    Nov 2006
    Let f(n)=4n^3+n^2\sqrt{n}-4n. Clearly for n large enough you have that 10n^3>f(n) and n^3<f(n). I'll leave it to you to show that these are true.

    *note: I am sure you can come up with better bounds on the constants for an^3 for both inequalities, but it follows the same idea.
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