Prove by showing:

Question

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

Answer 1

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

Answer 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

Answer 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)

Answer 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!

Answer 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.

Answer 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)

Answer 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?

Answer 8

Thanks anyway! :)

Answer 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)