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Topic: on real part of [(1+isqrt(7))/2]^n
Replies: 20   Last Post: Feb 15, 2014 5:52 AM

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Robin Chapman

Posts: 412
Registered: 5/29/08
Re: on real part of [(1+isqrt(7))/2]^n
Posted: Feb 11, 2014 1:16 PM
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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 ... :-(

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