Advanced Math

Prove by showing:
f(n) = n^2 + 4n is O(n)

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asked by VP
  1. I am not quite sure what you mean by O(n)
    f(n) is of order n^2, not n, as n-> infinity

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    posted by drwls
  2. Yes it is suppose to be
    f(n) = n^2 + 4n is O(n^2)

    This is what I have:
    n^2 = n^2 + 4n^2 = 5n, but that all I have and don't know what else to do, can you help. Thanks

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    posted by VP
  3. I am as puzzled by your notation as drwls was.

    how did you get n^2 = n^2 + 4n^2 = 5n

    from f(n) = n^2 + 4n is O(n^2) ???


    BTW, in the original you had O(n)

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    posted by Reiny
  4. Oh okay, I am sorry for typing error. It is definely suppose to be
    f(n) = n^2 + 4n is O(n^2)

    I am typing to fast bc my answer was suppose to read
    n^2 = n^2 + 4n <= n^2 + 4n^2 = 5n^2


    Does this make any sense. Please Help!

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    posted by VP
  5. nope, sorry, makes no sense at all

    how did f(n) which is a function notation, become n^2 ??

    how did the term 4n suddenly turn into 4n^2 ??

    I have certainly never seen a notation such as O(n^2) used in that kind of setting.

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    posted by Reiny
  6. okay, well it is suppose to read Big O of n^2

    how about this question
    what is the theta class for
    f(n) = n^2 + 4n x lg(n)

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    posted by VP
  7. I am sorry not to be able to help you.
    I have never seen that kind of question before.

    What level of math is this supposed to be, and whose curriculum are you studying?

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    posted by Reiny
  8. Thanks anyway! :)

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    posted by VP
  9. Yes it is suppose to be
    f(n) = n^2 + 4n is O(n^2) :

    Hi,

    I assume n>1.
    let g(n)=n^2

    This is straight forward according to big O notation.

    f(n) = n^2 + 4n

    <= n^2 + 4n^2 = 5n^2

    That means |f(n)|<=C|g(n)| where C is a constant.

    According to big O definition

    f(n) is O(n^2)

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    posted by Qun

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