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,...
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Needs to be explained also, like what method you used to compare
the growth rate. Thank you
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