What is the big 0 of (2^n+n^2)(n^3+3^n).

I get that I foil this out. If I did this correctly, I get that this is O(6^n). Is this right or did I mess up somewhere?

Printable View

- Oct 14th 2013, 08:35 AMAlucard2487Big O problem
What is the big 0 of (2^n+n^2)(n^3+3^n).

I get that I foil this out. If I did this correctly, I get that this is O(6^n). Is this right or did I mess up somewhere? - Oct 14th 2013, 09:17 AMemakarovRe: Big O problem
First, the question is probably what this function is big O of. Second, there are infinitely many answers, and you are probably looking for the slowest-growing one. Then 6^n is correct.

- Oct 14th 2013, 01:11 PMAlucard2487Re: Big O problem
My apologies. Here is the question: Give a big-O estimate for each of these functions. For the function g in your estimate f(x) is O(g(x)),

use a simple function g of smallest order. - Oct 14th 2013, 01:25 PMemakarovRe: Big O problem
Your estimate is still correct. :)

- Oct 16th 2013, 08:25 AMemakarovRe: Big O problem
The function log(log(n)) grows very, v e r y slowly. If the logs are based 10, it returns 2 for n=10¹⁰⁰, which is greater than the number of particles in the universe. Nevertheless, this function tends to infinity as n → ∞.

I am not sure what exactly you difficulty with log(log(n)) is. Feel free to post concrete questions about these functions. - Oct 16th 2013, 07:49 PMAlucard2487Re: Big O problem
I tried the problem out. 2n!, 2^2^n,2^n^2, 2^100n, n^4/3 (logn)^2, n^3/2, n(logn)^3/2, nlogn loglogn. How far off am I?