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Math Help - Big-Oh Proofs

  1. #1
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    Big-Oh Proofs

    Last time I had some questions on Big-Oh proofs, this set of problems is similar, I was wondering if the different notation still means the same thing, except now f(n) is always above 0? So "there exists some c so that if n is above the threshold B, then no matter what n is cg(n) will always be equal or greater than f(n)"?



    I'm not really familiar with the new notation.
    Thanks.
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  2. #2
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    Which exactly notation is new?
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  3. #3
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    Oh the |-> (arrow) and the R>0.
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  4. #4
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    Quote Originally Posted by Selena View Post
    Oh the |-> (arrow) and the R>0.
    the bar is read as "in which" and the arrow is "if" and the R>0 is real numbers that are greater than 0.
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  5. #5
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    Cool

    Ok, so it translates to "if f(n) results in a real number greater to or equal to 0, then ... ..."?
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  6. #6
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    Ok, so it translates to "if f(n) results in a real number greater to or equal to 0, then ... ..."?
    It's not clear what n you are referring to.

    The notations here have nothing to do with big-Oh; they are standard mathematical notations.

    \{x:A\mid P(x)\} -- the set of elements x of A for which P(x) holds. This is set-builder notation.

    f:\mathbb{N}\mapsto\mathbb{R}^{\ge0} -- f is a function with domain \mathbb{N} and codomain \mathbb{R}^{\ge0} (the set of nonnegative real numbers). Actually, I think \to should be used instead of \mapsto. One writes x\mapsto y to mean a function maps x (an element of the domain) into y.

    A\Rightarrow B -- if A, then B. See the list of logic symbols.
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  7. #7
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    Hmm so it has a domain of natural numbers, and codomain of non-negative real numbers... isn't that just the same as saying it's in a set of non-negative real numbers? Don't they overlap?
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  8. #8
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    isn't that just the same as saying it's in a set of non-negative real numbers?
    What is in the set of non-negative real numbers?
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