
Re: on real part of [(1+isqrt(7))/2]^n
Posted:
Feb 11, 2014 1:16 PM


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 ... :(

