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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: $\displaystyle 4n^3 + n^2\sqrt{n}-4n \in \Theta (n^3)$

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

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


    Thank you,
    Last edited by Rapha; Apr 25th 2009 at 12:32 PM.
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  2. #2
    Nov 2006
    Let $\displaystyle f(n)=4n^3+n^2\sqrt{n}-4n$. Clearly for $\displaystyle n$ large enough you have that $\displaystyle 10n^3>f(n)$ and $\displaystyle n^3<f(n)$. 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 $\displaystyle an^3$ for both inequalities, but it follows the same idea.
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