
Re: on real part of [(1+isqrt(7))/2]^n
Posted:
Feb 11, 2014 4:08 PM


On Tuesday, February 11, 2014 7:59:37 PM UTC+1, konyberg wrote: > 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
Hi
a_n = (sqrt(2))^n * cos(n*acos(sqrt(2)/4)), where acos = cos^(1), the inverse of cos.
What do we know?
1. As n goes to infinity: (sqrt(2))^n goes to infinity. 2. As n goes to infinity: cos(n*acos(sqrt(2)/4)) is not defined. 3. 0<= cos(v) <= 1. 4. infinity * cos() = infinity * 0 when n goes to infinity can give an answer. 5. The probality that a_n (n goes to infinity) = infinity, is really large. 6. But do we know? No!
KON

