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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: 338
Registered: 12/6/04
Re: on real part of [(1+isqrt(7))/2]^n
Posted: Feb 11, 2014 4:08 PM
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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:
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> > On 11/02/2014 16:54, konyberg wrote:
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> > > Hm. I try this again. You might be correct :)
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> > I am usually.
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> > > a^1 = (1 + i*sqrt(7))/2
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> > > r = sqrt(2)
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> > > a_1 = 1/2 (the error I made)
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> > > sqrt(2)*cos(1*u) = 1/2
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> > > cos(u) = sqrt(2)/4. You are correct! Thank you! (cos^(-1)(sqrt(2)/4) = tan^(-1)(sqrt(7))
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> > > 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
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> > 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
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> > you ask them to prove that (cos(n)) is a divergent sequence ... :-(
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> Hi.
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> 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}.
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> A completly different problem. Undefined (< infinity) / Infinity = 0, but Infinity * Undefined (< infinity) = Undefined
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> 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




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