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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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 William Elliot Posts: 2,637 Registered: 1/8/12
Re: on real part of [(1+isqrt(7))/2]^n
Posted: Feb 10, 2014 11:49 PM

On Mon, 10 Feb 2014, AP wrote:

> Be a=(1+isqrt(7))/2 and a_n=the real part of a^n
>
> question : show lim |a_n | is +inf ? (it-is not a home-work..)
>

Let r = sqr 7, b = 1 + ir.

(1 + ir)(nj + i.mj) = nj + i(r.nj + mj) - r.mj
n_(j+1) = nj - r.mj; m_(j+1) = r.nj + mj

n1 = 1; m1 = r
n2 = 1 - 7 = -6; m2 = 2r
n3 = -6 - 14 = -20; m3 = -6r + 2r = -4r
n4 = -20 + 28 = 8; m4 = -20r - 4r = -24r

> first values of 2a_n , n>=0, :
> 2/1/-3/-5/1/11/9/-13/-31/-5/57/

Means nothing.

> we have
> (1-2z)/(2-z+z^2)=sum_{n>=0 } (a_{n+1}/2^n)z^n for |z|<sqrt(2)
> =1/2-(3/4)z-(5/8)z^2+(1/16)z^3+...

> (if b=(1-isqrt(7))/2, 1/(a-z)+1/(b-z)=...)
>
> the radius of convergence is R>=sqrt(2)
> but if R>sqrt(2) we obtain a contradiction because (1-2z)/(2-z+z^2) is
> not define for a (|a|=sqrt(2))
> so R=sqrt(2)
>
> hence, if z=2 , the series diverges and |a_n| is not bounded.
> But , after ...
>
>
>
>
> ---
> Ce courrier Ã©lectronique ne contient aucun virus ou logiciel malveillant parce que la protection avast! Antivirus est active.
> http://www.avast.com
>
>

Date Subject Author
2/10/14 AP
2/10/14 Brian Q. Hutchings
2/10/14 Pubkeybreaker
2/10/14 Karl-Olav Nyberg
2/10/14 William Elliot
2/11/14 AP
2/11/14 Karl-Olav Nyberg
2/11/14 Robin Chapman
2/11/14 Karl-Olav Nyberg
2/11/14 Robin Chapman
2/11/14 Karl-Olav Nyberg
2/11/14 Karl-Olav Nyberg
2/15/14 Karl-Olav Nyberg
2/15/14 quasi
2/15/14 Karl-Olav Nyberg
2/15/14 quasi
2/11/14 William Elliot
2/11/14 Robin Chapman
2/11/14 William Elliot
2/13/14 quasi
2/14/14 Brian Q. Hutchings