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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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Karl-Olav Nyberg

Posts: 461
Registered: 12/6/04
Re: on real part of [(1+isqrt(7))/2]^n
Posted: Feb 11, 2014 1:59 PM
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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.
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>
>
> 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.
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>
>
> It's a bit like that old chestnut: prove (cos(n))/n -> 0
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> 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



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