big-o and big-omega related to big-theta recursion

Question

Let's suppose that a recursive formula is a big-o(n^2), and at the same time a big-omega(n^2). Does this imply that the recursion is a big-Theta(n^2)?

Answer 1

To make the long story short: the answer is Yes, it does. See proof below.

Though everybody have heard about big-o notation lets recall what exactly does these notations mean with a help of Introduction to Algorithms. For a general case it is said Ο(g(n)), Ω(g(n)), Θ(g(n)), but we will consider yours.

Ο(n²)

Ο(n²) notation defines a set of functions for each of which the following statement holds: There exists such positive constants c and n₀ that 0 ≤ f(n) ≤ cn² holds for all n ≥ n₀.

So f(n) is just a function from Ο(n²). Examples 13n, -5, 4n² + 5. All these pertain to Ο(n²).

Ω(n²)

Ω(n²) notation defines a set of functions for each of which the following statement holds: There exists such positive constants c and n₀ that 0 ≤ cn² ≤ f(n) holds for all n ≥ n₀.

So f(n) is just a function from Ω(n²). Examples n⁴ + n - 1, 3ⁿ, n² - 12. All these pertain to Ω(n²).

Θ(n²)

Θ(n²) notation defines a set of functions for each of which the following statement holds: There exists such positive constants c₁, c₂ and n₀ that 0 ≤ c₁n² ≤ f(n) ≤ c₂n² holds for all n ≥ n₀.

Again f(n) is just a function from Θ(n²). These are its representatives n²/2 + 3, 5n².

Proof

I bet saying that a recursive formula is a big-o(n^2), and at the same time a big-omega(n^2) you meant there is a function (lets call it) f(n) pertaining to Ω(n²) and Ο(n²).

From Ω(n²) we have existence of c₁ that c₁n² ≤ f(n) holds. From Ο(n²) we have existence of c₂ that f(n) ≤ c₂n² holds. Consequently we have existence of c₁ and c₂ that c₁n² ≤ f(n) ≤ c₂n², that is exactly what Θ(n²) is about.