Final answer:
To prove f(n) = Ο(g(n)) is equivalent to g(n) = Ω(f(n)), one must understand that Big O provides an upper bound and Big Omega provides a lower bound. These bounds indicate that the functions grow in a way that if one is bounded above by another function, the other is bounded below by the first, and both bounds must be satisfied for any mutually bounding relationship.
Step-by-step explanation:
To prove that f(n) = Ο(g(n)) if and only if g(n) = Ω(f(n)), we need to understand the definitions of Big O and Big Omega notations. Big O notation, Ο(g(n)), describes an upper bound on the growth rate of a function, indicating that function f(n) does not grow faster than g(n) times a constant factor past a certain point. Conversely, Big Omega notation, Ω(f(n)), provides a lower bound, suggesting that g(n) grows at least as fast as f(n) times a constant factor.
Formally, f(n) = Ο(g(n)) means there exist positive constants c and n₀ such that 0 ≤ f(n) ≤ c*g(n) for all n ≥ n₀. Meanwhile, g(n) = Ω(f(n)) means there are positive constants c' and n₀' such that 0 ≤ c'*f(n) ≤ g(n) for all n ≥ n₀'. The implication of these relationships is that as f(n) grows smaller, g(n) must grow larger to satisfy the inequality and vice versa, thereby demonstrating the inverse nature of the two conditions.
Hence, proving that f(n) is bounded above by g(n) (Ο(g(n))) inherently proves that g(n) is bounded below by f(n) (Ω(f(n))) since the two statements are dependent on the product of the functions being bounded by a constant, just approached from different inequality directions. Therefore, f(n) = Ο(g(n)) if and only if g(n) = Ω(f(n)).