Compare the order of growth of the functions log(n) and sqrt (n). Which grows faster?
log(n) grows way more slower than sqrt(n) for example take n a very large number such as 2^100 (2 to the power of 100) log(n) is 100 where as sqrt(n) is 2^50 as we can see log(n) way way more less than sqrt(n). Hence log(n) grows slower and sqrt(n) grows faster.
Compare the order of growth of the functions log(n) and sqrt (n). Which grows faster?
I understand how it was simplified to n^(∈/(sqrt(logn))), but I'm trying to understand how to prove that logn grows faster for 0<∈<1. The derivative seems too complicated to prove this via Lhopital's Rule, so I tried using WolframAlpha to compare the two with logn as the numerator: http://www.wolframalpha.com/input/?i=limit+as+n+approaches+infinity+(logn)%2F(n%5E(0.5%2F(sqrt(logn))))&rawformassumption=%7B%22FunClash%22,+%22log%22%7D+-%3E+%7B%22Log10%22%7D However, this gives me a result of 0 for any value above 0, which would mean that n^(∈/(sqrt(logn))) grows at a faster rate, even when 0<∈<1. When I try to graph it,...
Order the following functions by asymptotic growth rate. 2n log n + 2n, 210, 2 log n, 3n + 100 log n, 4n, 2n, n2 + 10n, n3, n log n2
Order the following functions by growth rate: N, squrerootN, N1.5, N2, NlogN, N log logN, Nlog2N, Nlog(N2), 2/N,2N, 2N/2, 37, N2 logN, N3. Indicate which functions grow at the same rate.
Order the following functions by asymptotic growth rate: 4n, 2^log(n), 4nlog(n)+2n, 2^10, 3n+100log(n), 2^n, n^2+10n, n^3, nlog(n) You should state the asymptotic growth rate for each function in terms of Big-Oh and also explicitly order those functions that have the same asymptotic growth rate among themselves.
Needs to be explained also, like what method you used to compare the growth rate. Thank you 4) Order the following functions by growth rate. Indicate which functions grow at the same rate (15 points) N, N2, log N, N log N, log(N2), log2 N, N log2N, 2, 2N, 37, N2 log N, 5logN, N3, 10N log N2
Arrange the following functions in ascending order of asymptotic growth rate; that is if function g(n) immediately follows function f(n) in your list, then it should be the case that f(n) is O(g(n)): 2 Squareroot log n, 2^n, n^4/3, n(log n)^3, n log n, 2 2^n, 2^n^2. Justify your answer.
If a firm Explain grows faster than its sustainable growth rate, is that growth value decreasing? NOTE: This question has two parts, first: is the rate value decreasing, and second explain your answer.
76. Arrange the following functions in ascending or- der of growth rate: 4000 log n, 2n2 + 13n - 8, 1,036, 3n log n, 2" - n2, 2n! - n, n2 – 4n.
) Arrange the running time of functions mentioned below by the order (increasing) of their growth. n^3 , n, nlog(n) , log(n), 2^n, n^2, n! my guess without doing the math is: n nlog(n) log(n) n^2 n^3 2^n n!
Compare the asymptotic orders of growth of the following pairs of functions. log2 n and . n (n+1)/2 and n2. 2n and 3n