I cannot understand about this section
||AX||2=||A||2||X||2
please explain why
I cannot understand about this section ||AX||2=||A||2||X||2 please explain why Bookmark Show all steps: Chapter 7.1, Pr...
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,...