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

 Messages: [ Previous | Next ]
 Karl-Olav Nyberg Posts: 1,575 Registered: 12/6/04
Re: on real part of [(1+isqrt(7))/2]^n
Posted: Feb 11, 2014 11:12 AM

On Monday, February 10, 2014 6:00:19 PM UTC+1, AP wrote:
> Be a=(1+isqrt(7))/2 and a_n=the real part of a^n
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> question : show lim |a_n | is +inf ? (it-is not a home-work..)
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> first values of 2a_n , n>=0, : 2/1/-3/-5/1/11/9/-13/-31/-5/57/
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> we have
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> (1-2z)/(2-z+z^2)=sum_{n>=0 } (a_{n+1}/2^n)z^n for |z|<sqrt(2)
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> =1/2-(3/4)z-(5/8)z^2+(1/16)z^3+...
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> (if b=(1-isqrt(7))/2, 1/(a-z)+1/(b-z)=...)
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> the radius of convergence is R>=sqrt(2)
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> but if R>sqrt(2) we obtain a contradiction because (1-2z)/(2-z+z^2) is
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> not define for a (|a|=sqrt(2))
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> so R=sqrt(2)
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> hence, if z=2 , the series diverges and |a_n| is not bounded.
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> But , after ...
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> ---
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> Ce courrier Ã©lectronique ne contient aucun virus ou logiciel malveillant parce que la protection avast! Antivirus est active.
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> http://www.avast.com

Hi.

a_n = Re(a^n)
a^n = (x + iy)^n = r^n (cos(nu) + isin(nu))
a_n = r^n cos(nu)
a = (1 + isqrt(7))/2 gives r = sqrt(2), and u = pi/4 (45 deg)
a_n = (sqrt(2))^n * cos(n*pi/4)

From this I think that lim (n goes to inf) of (a_n) is undefined, because cos(n*pi/4) is undefined when n goes to infinity. It can be "any" value from 0 to 1 (defined by the expression).

KON

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