Rank the following functions by order of growth; that is, find an arrangement 91(n), 92(n), ..., 928 (n) of functions satisfying gi(n) = O(gi+1(n)) for every E {1, ..., 27}. Partition your list into equivalence classes such that f(n) and g(n) are in the same class if and only if f(n) = (g(n)). Make sure it is clear what your equivalence classes are. You do not have to prove your answers. n logn n2/3 n3/2 logn 1000 logio n n 21 2n+1 log logn n! n+n2/1020 221 2logn logn log(n?) Vn Vlogn n2n (n + 1)! log 2 i 31 n + logn 4n 10100 Remarks: In this class we use log n to denote the logarithm base 2. Use Stirling's formula to figure out how to rank n!., Stirling's formula is: n! = V2^n (~)*(1+0(5) Remember that log2 n and 2are inverse functions of each other. That is, 2log f(n) = f(n); also log 2f(n) = f(n). Use also this fact: for any constants b, b2 > O: log i n = O(nb) and nb2 = O(logi n) In short, logarithm of n raised to any constant power grows slower than any constant power of n