Nifty Trick to Prove f(n)∉cn^2

Let's use this as an example to show this little nifty trick (Not really a trick but I thought it was interesting)

                     6n⁵-4n³+2n ∉ O(5n⁴-3n²+1)

Now, normally, people would try to prove this is a normal way by converting the 6n⁵-4n³+2n ∉ O(5n⁴-3n²+1) into it's definition form then proving the statement. That is totally fine and it is not a wrong method to prove the statement.

The nifty trick that I learned from my tutorial is to switch the sides of the polynomials and prove that 5n⁴-3n²+1 ∈ O(6n⁵-4n³+2n), because if 6n⁵-4n³+2n is not upper bounded by 5n⁴-3n²+1 then that means 6n⁵-4n³+2n grows faster than 5n⁴-3n²+1. Therefore, that means 5n⁴-3n²+1 is upper bounded by 6n⁵-4n³+2n.


∃c∈R², ∃B∈N, such that ∀n∈N, n≥B → f(n)≤cn²
           Pick c=1/3, then c∈R²
Pick B=1, then B∈N
                     Assume n∈N
                                         Assume n≥1
                                                             Then, 6n⁵-4n³+2n ≥ 6n⁵-4n³
                                                                                              ≥ 6n⁵-4n⁵
                                                                                              ≥ 2n⁵
                                                                        2n⁵ ≥ (1/3)(6n⁴)    # n≥1                                                                                 ≥ (1/3)(5n⁴+n⁴)
                                                                                ≥ (1/3)(5n⁴+1)
                                                                                ≥ (1/3)(5n⁴-3n²+1)
                                                              Then, 6n⁵-4n³+2n ≥ c(5n⁴-3n²+1)
                      Then, ∀n∈N, n≥B → 6n⁵-4n³+2n ≥ c(5n⁴-3n²+1)
 Then, ∃c∈R², ∃B∈N, such that ∀n∈N, n≥B → f(n)≤cn²
Then, 5n⁴-3n²+1 ∈ O(6n⁵-4n³+2n)
Then, 6n⁵-4n³+2n ∉ O(5n⁴-3n²+1)