Did I do this big-Omega proof correctly?

Question

Prove or disprove: 6n^3 – 4n^2 + 3n +2 is in Ω (5n^3 – n^2 + n +1). So I'm not sure if I did this right or not, any pointers or the correct steps would be helpful

Ǝc ∈ ℝ+, ƎB ∈ ℕ, ∀n ∈ ℕ, n ≥ B ⇒ 6n^3 – 4n^2 + 3n +2 ≥ c (5n^3 – n^2 + n +1).

Let c = 1, Then c

Let B = 1, Then B

Assume n ∈ ℕ and n ≥ B  # generic real numbers and domain assumption

    Then 6n^3 – 4n^2 + 3n +2    ≥ 6n^3 – 4n^2 + 2
                ≥ 6n^3 + n^3 
                ≥ 7n^3  
                ≥ 7n^3 – 2n^3
                ≥ 5n^3 – n^2    
                ≥ 5n^3 – n^2 + n + 1
                ≥ c(5n^3 – n^2 + n + 1) # c = 1, B = 1
Then ∀n ∈ ℕ, n ≥ B ⇒  6n^3 – 4n^2 + 3n +2 ≥ c (5n^3 – n^2 + n +1)   # introduce ∀ and ⇒

Then Ǝc ∈ ℝ+, ƎB ∈ ℕ, ∀n ∈ ℕ, n ≥ B ⇒ 6n^3 – 4n^2 + 3n +2 ≥ c(5n^3 – n^2 + n +1) #introduce Ǝ Then 6n^3 – 4n^2 + 3n +2 is in Ω (5n^3 – n^2 + n +1)

Answer 1

I have a problem with the step where you say $6n^{3} -4n^{2} + 3n + 2 \geq 6n^3 + n^3$. This just isn't true in the long run.

You can cut to the chase a little bit faster:

$6n^{3} -4n^{2} + 3n + 2 \geq 5n^{3} - n^{2} + n + 1$ for all $n \geq 0$ if and only if $n^{3} - 3n^{2} + 2n + 1 \geq 0$ for all $n \geq 0$

Observe that this is true for $n = 0, 1$. You need to show by induction this holds true for $n > 1$.

Alternatively, you can use the limit test. We have:

$$L = \lim_{n \to \infty} \frac{6n^{3} -4n^{2} + 3n + 2}{5n^{3} - n^{2} + n + 1} = \frac{6}{5}$$

And so $6n^{3} -4n^{2} + 3n + 2 \in \Theta(5n^{3} - n^{2} + n + 1) \implies$

$6n^{3} -4n^{2} + 3n + 2 \in \Omega(5n^{3} - n^{2} + n + 1)$.