Prove by showing:

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

I am not quite sure what you mean by O(n)

f(n) is of order n^2, not n, as n-> infinity

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

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)

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!

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.

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)

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?

Thanks anyway! :)

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)

To prove that f(n) = n^2 + 4n is O(n), we need to show that there exist positive constants c and k such that for all values of n greater than or equal to k, f(n) ≤ c * n.

First, let's simplify the expression of f(n) = n^2 + 4n:

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

Now, let's choose c = 5 and k = 1. We need to show that for all values of n greater than or equal to 1, f(n) ≤ 5n.

For n ≥ 1:
f(n) = n(n + 4) ≤ n(2n) = 2n^2
≤ 5n^2

So, we have shown that for all values of n greater than or equal to 1, f(n) ≤ 5n^2.

This implies that f(n) = n^2 + 4n is O(n) with c = 5 and k = 1, because for all values of n greater than or equal to 1, f(n) ≤ 5n.