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Math Help - Asymptotic analysis

  1. #1
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    Asymptotic analysis

    Which of the following conjecture is true? Justify

    10n = O(n) 10n^2  = O(n)  10n^{55}  = O(n^2 )


    1) 10n \leq cn => n=1 c=10

    f(n) \leq 10g(n) for all n \geq 1. f(n) \in O(g(n))

    How can I solve the other two?

    I found this problem on the Internet:
    It is true that n^2 + 200n + 300  =   O(n^2 ) ? And n^2 -200n -300  =   O(n) ?
    Last edited by Apprentice123; March 17th 2009 at 08:26 AM.
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  2. #2
    MHF Contributor matheagle's Avatar
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    I don't know what you mean by n2?
    Is that 2 times n or a sequence n_2?

    O(n) just means that the term is bounded about when divided by n,
    limsup is finite.
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  3. #3
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    Now the order of the exercise is correct
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  4. #4
    MHF Contributor matheagle's Avatar
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    DIVIDE by n^2 and n and take the limits.

    Quote Originally Posted by Apprentice123 View Post
    Which of the following conjecture is true? Justify

    10n = O(n) 10n^2 = O(n) 10n^{55} = O(n^2 )


    1) 10n \leq cn => n=1 c=10

    f(n) \leq 10g(n) for all n \geq 1. f(n) \in O(g(n))

    How can I solve the other two?

    I found this problem on the Internet:

    It is true that n^2 + 200n + 300 = O(n^2 )
    YES

    And n^2 -200n -300 = O(n)
    NO
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  5. #5
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    Sorry, I do not understand
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  6. #6
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    Please look

    1.1) 10n \leq cn => n \geq 1 and c \geq 10


    1.2) 10n^2 \leq cn^2 n \geq 1 and c \geq 10

    1.3) 10n^{55} \leq c2^n

    \frac{10n^{55}}{2^2} \leq c How can I continue?

    2.1) n^2+200n+300 \leq cn^2

    1+\frac{200}{n}+\frac{300}{n^2} \leq c

    n \geq 1 and c \geq 501

    2.2) n^2-200n-300 \leq cn

    n-200-\frac{300}{n} \leq c How can I continue?


    What I done is correct?
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