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Math Help - Big oh and theta calculation

  1. #1
    Junior Member
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    Big oh and theta calculation

    How could one show that:

    3n 2n + 1 O(n).

    3n 2n + 1 Θ(n).
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  2. #2
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    by referring to their definitions.

    3n – 2n + 1 O(n).
    in the definition, just let the constant be 4

    3n – 2n + 1 Θ(n).
    in the definition, let the constant be 2 in omega and let the constant be 4 in big-oh.


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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by vexiked View Post
    How could one show that:

    3n^3  2n^2 + 1 \in O(n^3)
    For n>1:

    n^3<3n^3-2n^3+1<3n^3-2n^2+1<3n^3+2n^3+n^3=6n^3

    CB
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