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Math Help - Determine whether x^3 is O(g(x)) for each of these functions g(x).

  1. #1
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    Determine whether x^3 is O(g(x)) for each of these functions g(x).

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    Determine whether x^3 is O(g(x)) for each of these functions g(x).
    
    d) g(x) = x2 + x4
    e) g(x) = 3x
    f) g(x) = x3/2
    d) Yes
    e) Yes
    f) No

    Are my answers correct? I took the problem to mean that g(x) < c * x3.
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    Re: Determine whether x^3 is O(g(x)) for each of these functions g(x).

    Hey lamentofking.

    I think you have it the other way around. 3^x is an exponential function with infinite numbers of positive powered terms and x^4 has a term higher than x^3. f should be correct (i.e. yes).
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    Re: Determine whether x^3 is O(g(x)) for each of these functions g(x).

    Quote Originally Posted by lamentofking View Post
    Determine whether x^3 is O(g(x)) for each of these functions g(x).

    I took the problem to mean that g(x) < c * x3.
    No, it means that eventually x3 < c * g(x).
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    Re: Determine whether x^3 is O(g(x)) for each of these functions g(x).

    Quote Originally Posted by emakarov View Post
    No, it means that eventually x3 < c * g(x).
    So then in terms of the question d), e), and f) are all yes? I thought f) would be no because if x3 < c * g(x) and (x^3)/2 means (1/2)x^3 with c=1/2 then x^3 would be greater than (>) c* g(x).
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    Re: Determine whether x^3 is O(g(x)) for each of these functions g(x).

    Quote Originally Posted by lamentofking View Post
    So then in terms of the question d), e), and f) are all yes?
    Yes.

    Quote Originally Posted by lamentofking View Post
    I thought f) would be no because if x3 < c * g(x) and (x^3)/2 means (1/2)x^3 with c=1/2 then x^3 would be greater than (>) c* g(x).
    Could you please re-formulate this clearly? "If x^3 < c * g(x)" eventually, then x^3 = O(g(x)), nothing more to prove. Are you sure you are assuming x^3 < c * g(x)? Next, by c * g(x) do you mean (1/4)x^3 since c = 1/2 and g(x) = (1/2)x^3? That's possible, but why not consider c = 1? Your quote above is not very clear.
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