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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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 AP Posts: 137 Registered: 3/4/09
Re: on real part of [(1+isqrt(7))/2]^n
Posted: Feb 11, 2014 2:58 AM

On Mon, 10 Feb 2014 20:49:53 -0800, William Elliot <marsh@panix.com>
wrote:

>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

marvelous : (1+isqrt(7))/2=1+isqrt(7)
so real part of ( (1+isqrt(7))/2)^n is the real part of (1+isqrt(7)
)^n
!!!!!
>> first values of 2a_n , n>=0, :
>> 2/1/-3/-5/1/11/9/-13/-31/-5/57/

>
>Means nothing.

of course : my 2a_i is your ni/2^(i-1) , i>=1 :

1=1/1 ; -3=-6/2 ; -5=-20/4 ; 1=8/8

you understand sir?
>> 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+...

>

bye

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