Question

home study engineering /computer science/computer architecture solutions manual data structures and algorithm analysis in c++ 4th edition Data Structures and Algorithm Analysis in C++ (4th Edition) E Chapter 2, Problem 3E Show all steps ON Bookmark Chapter 1 Problem Chapter 2 A Which function grows faster: N logN or N /Viog N e 0? 1E Step-by-step solution Step 1 of 1 A Growth rate of functions Compare the two functions N log Nand Ni e ViogN to understand which one grows faster Observe that ifN is extracted from both the functions, the real comparison depends between the growth of log N and Ne viopN. It is already known that linear and powered functions grow faster than logarithms. Here the value of e plays an important role. If e> then NevloEA 10E grows faster than log However, the only range where this does not hold true is when Therefore, It can be concluded Ni vlogN grows faster than N log N. 12E 13E Comment 14E 15E Was this solution helpful? 4 o 16E 17E

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, it seems like logn does grow faster when ∈ is a very small number between 0 and 1, but not for all values between 0 and 1. Here, I used ∈ = 0.5, and it still looks like n^(∈/(sqrt(logn))) grows faster.

Answer 1