Question

2^n + 6n^2 + 3n

I guess it’s just O(2^n), using the highest order term, but what is the formal approach to go about proving this?

Answer 1

You can prove that 2^n + n^2 + n = O(2^n) by using limits at infinity. Specifically, f(n) is O(g(n)) if lim (n->inf.) f(n)/g(n) is finite.

lim (n->inf.) ((2^n + n^2 + n) / 2^n)

Since you have inf/inf, an indeterminate form, you can use L’Hopital’s rule and differentiate the numerator and the denominator until you get something you can work with:

lim (n->inf.) ((ln(2)*2^n + 2n + 1) / (ln(2)*2^n))
lim (n->inf.) ((ln(2)*ln(2)*2^n + 2) / (ln(2)*ln(2)*2^n))
lim (n->inf.) ((ln(2)*ln(2)*ln(2)*2^n) / (ln(2)*ln(2)*ln(2)*2^n))

The limit is 1, so 2^n + n^2 + n is indeed O(2^n).