- Prove that f(n) = O(g(n)) implies that g(n) = Omega(f(n)).
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*Note:*All we can assume about f and g is that they are asymptotically positive functions.] Now, suppose that f(n) is O(g(n)). Then, by definition of O(g(n)), there exists c > 0 and n0 > 0 such that for all n > n0, 0 <= f(n) <= c g(n). Let c' = 1/c. [*Note:*we actually used the fact c != 0 here.] Then, for all n > n0, we know that 0 <= c' f(n) <= g(n). Therefore, g(n) = Omega(f(n)). - Disprove that f(n) = Theta(f(n/2)).
f(n) = 2^n is a counter-example. In this case, f(n) = 2^(n/2) f(n/2). So, for all c2 > 0 and for all n0 > 0, there is an n > n0, such that f(n) > c2 f(n/2). In particular, this holds for all n > 2 log(c2). Thus, f(n) cannot be O(f(n/2)). Hence, f(n) is not Theta(f(n/2)).

Last Modified: Thu Feb 1 17:06:42 EST 1996 Richard Chang, chang@gl.umbc.edu