Given the following four functions, which function will have the largest value for sufficiently large values of n (1

Question

Given the following four functions, which function will have the largest value for sufficiently large values of n (1 < n -> infinity)?

A. f_1(n) = n^(sqrt(n))
B. f_2(n) = 2^n
C. f_3(n) = n^(log_2 n)
D. f_4(n) = n^10

Accepted Answer

Function A (n^(sqrt(n)) or equivalent super-polynomial form) has the largest value for sufficiently large n. We compare four functions: a super-polynomial n^(sqrt(n)), an exponential 2^n, another super-polynomial n^(log_2 n), and a high-degree polynomial n^10. The key insight is that n^(sqrt(n)) = 2^(sqrt(n) * log_2 n). Since sqrt(n) * log_2 n grows without bound and is eventually larger than n itself, n^(sqrt(n)) eventually surpasses 2^n. So n^(sqrt(n)) is the largest for sufficiently large n.

Given the following four functions, which function will have the largest value for sufficiently large values of n (1 < n -> infinity)? A. f_1(n) = n^(sqrt(n)) B. f_2(n) = 2^n C. f_3(n) = n^(log_2 n) D. f_4(n) = n^10