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

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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

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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

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

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posted by VP
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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