
Re: on real part of [(1+isqrt(7))/2]^n
Posted:
Feb 11, 2014 1:59 PM


On Tuesday, February 11, 2014 7:16:10 PM UTC+1, Robin wrote: > On 11/02/2014 16:54, konyberg wrote: > > > > > Hm. I try this again. You might be correct :) > > > > I am usually. > > > > > a^1 = (1 + i*sqrt(7))/2 > > > r = sqrt(2) > > > a_1 = 1/2 (the error I made) > > > sqrt(2)*cos(1*u) = 1/2 > > > cos(u) = sqrt(2)/4. You are correct! Thank you! (cos^(1)(sqrt(2)/4) = tan^(1)(sqrt(7)) > > > > > > However. I dont think this changes the conclusion. > > > > Well cos(nu) hasn't a limit when n > infinity, but that doesn't > > mean that cos(nu)2^{n/2} doesn't tend to infinity as n > infinity. > > > > It's a bit like that old chestnut: prove (cos(n))/n > 0 > > as n > infinity. My undergrads can usually do that, but if > > you ask them to prove that (cos(n)) is a divergent sequence ... :(
Hi.
This problem is not the same as lim cos(n)/n as n goes to infinity, which is {1,+1}/"infinity" = 0. In this problem we have "infinity" *{1,+1}. A completly different problem. Undefined (< infinity) / Infinity = 0, but Infinity * Undefined (< infinity) = Undefined
KON

