Question

Prove (n²) is in ( θ ) of ( 1/2n² - 3n ).

a) (n²) is in (O(12n² - 3n))

b) (n²) is in ( Ω (1/2n² - 3n))

c) (n²) is in ( θ (1/2n² - 3n))

d) (n²) is not in ( θ (1/2n² - 3n)

Answer 1

By demonstrating that n² is both in O(1/2n² - 3n) and Ω(1/2n² - 3n), we can conclude that n² is in θ(1/2n² - 3n), which fulfills the Big Theta condition of being bounded both above and below by 1/2n² - 3n.

Step-by-step explanation:

To prove that n² is in θ of (1/2n² - 3n), we must show that it is bounded above and below by this function, up to constant factors, for sufficiently large n. First, we need to establish that:

1. n² is in O(1/2n² - 3n), meaning there exists a constant c > 0 such that n² ≤ c(1/2n² - 3n) for all n greater than some n0.
2. n² is in Ω(1/2n² - 3n), implying there is a constant k > 0 such that n² ≥ k(1/2n² - 3n) for all n larger than some n1.

By showing both (a) and (b), we can conclude that n² is, by definition, in θ(1/2n² - 3n).