Consider the following functions where n is a positive integer: n1/3, log⁡(n),…

Step-by-Step Asymptotic Analysis

To determine the increasing order of asymptotic growth rates, we analyze the behavior of each function as n approaches infinity.

1. Analyze Individual Functions

• log(n): Grows logarithmically. This is the slowest growth among the given functions.

• n^(1/3): Grows as a fractional power of n. Polynomial growth (even with a small exponent) is faster than logarithmic growth.

• 2 log(n): This is a constant multiple of log(n). In asymptotic analysis, constant factors are ignored, so 2 log(n) = Θ(log(n)). It grows at the same rate as log(n).

• log(n!): Using Stirling's approximation, log(n!) ≈ n log(n). This grows faster than any polynomial n^k (where k < 1) and significantly faster than log(n).

2. Compare Growth Rates

Comparing the functions:

• log(n) vs n^(1/3): log(n) grows slower than n^(1/3).

• n^(1/3) vs 2 log(n): n^(1/3) grows faster than 2 log(n) because polynomial growth dominates logarithmic growth.

• 2 log(n) vs log(n!): Since 2 log(n) is Θ(log(n)) and log(n!) is Θ(n log(n)), log(n!) grows much faster.

3. Final Ordering

The correct increasing order of asymptotic growth rates is:

log(n) < n^(1/3) < 2 log(n) < log(n!)

Note: While log(n) and 2 log(n) are in the same complexity class, in terms of specific values for large n, 2 log(n) is strictly greater than log(n). However, the primary distinction is that n^(1/3) sits between the logarithmic terms and the factorial logarithm.

Thus, the list corresponding to this order is Option C.