12 Prove and disprove (a) Prove or disprove the following claim: 16n = (b) Prove or disprove the following claim: = O(16n)

(a) This claim is incomplete as there is no value or expression given after the equals sign. Therefore, it cannot be proven or disproven without additional information.

(b) This claim is true. To prove it, we need to show that there exists a constant c and an n0 such that 16n ≤ c16n for all n ≥ n0. Let c = 1 and n0 = 1. Then, 16n ≤ 16n for all n ≥ 1, which satisfies the definition of O(16n). Therefore, the claim is proven.

相关问题

prove or disprove:f(n) = O(g(n)) implies 2^f(n) = O (2^g(n))

Prove:

Assume that f(n) = O(g(n)), which means there exist constants C and n0 such that for all n≥n0, f(n) ≤ Cg(n). Then, we have:

2^f(n) = 2^(Cg(n))
= (2^C)^g(n)

Let's assume that there exist constants C' and n1 such that for all n≥n1, 2^f(n) ≤ C'2^g(n).

Then we have:
(2^C)^g(n) ≤ C'2^g(n)

Take the logarithm of both sides, we get:
g(n) * log(2^C) ≤ log(C') + g(n) * log(2)

Simplifying the above equation, we get:
g(n) ≤ (log(C') + C * log(2)) / (log(2^C))

As the right-hand side is a constant, there exists some n2 such that for all n≥n2 , g(n) ≤ (log(C') + C * log(2)) / (log(2^C)), which contradicts the initial assumption that f(n) = O(g(n)).

Therefore, 2^f(n) is not O(2^g(n)).

Hence, we have proved that f(n) = O(g(n)) does not imply 2^f(n) = O(2^g(n)).

prove of disprove f(n) = O(g(n)) implies g(n) = O(f(n))

This statement is false.

To disprove it, we can provide a counterexample. Let f(n) = n and g(n) = n^2.

Then, f(n) = O(g(n)) because n <= n^2 for all n >= 1. However, g(n) = n^2 is not O(f(n)) because there does not exist a positive constant c such that n^2 <= c*n for all n >= 1.

Therefore, f(n) = O(g(n)) does not imply g(n) = O(f(n)).