can you prove that n2fn?

From the premise for c = 2, there is an N0 > 0 such that f(n) >= 2 log(n) for all n > N0. By monotonicity of y = 2^x this is equivalent to 2^f(n) >= n^2 for all n > N0.

For any d > 0 there is an N1 > 0 such that n^2 >= n/d for all n > N1 since the quadratic n^2 grows faster than any linear n/d.

Combining the two inequalities, n/d <= n^2 <= 2^f(n) for all n > max(N0, N1).

CLICK HERE to find out more related problems solutions.

Leave a Comment

Your email address will not be published.

Scroll to Top