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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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Pubkeybreaker

Posts: 1,416
Registered: 2/12/07
Re: on real part of [(1+isqrt(7))/2]^n
Posted: Feb 10, 2014 2:08 PM
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On Monday, February 10, 2014 12:00:19 PM UTC-5, 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..) first values of 2a_n , n>=0, : 2/1/-3/-5/1/11/9/-13/-31/-5/57/ 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 électronique ne contient aucun virus ou logiciel malveillant parce que la protection avast! Antivirus est active. http://www.avast.com

Norms are multiplicative. Norm(z) = Norm(conj(z)). What is the norm of your
number......?



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