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seen Dec 3 '12 at 15:04

Dec
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comment How to solve this recurrence relation $a_{n+2} = \sqrt{a_{n+1}\cdot a_{n}}$?
is it the linear recurrence? Will be right if I'll do something like this $ 2\lambda^2 - \lambda - 1 = 0 $ or I must to do this $ \lambda^2 - 1/2 \lambda - 1/2 = 0 $
Nov
28
comment How to solve this recurrence relation $a_{n+2} = \sqrt{a_{n+1}\cdot a_{n}}$?
yeap... it must be $ 2b_n = b_{n-1} + b_{n-2} $
Nov
28
awarded  Scholar
Nov
28
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Nov
28
accepted How to solve this recurrence relation $a_{n+2} = \sqrt{a_{n+1}\cdot a_{n}}$?
Nov
28
comment How to solve this recurrence relation $a_{n+2} = \sqrt{a_{n+1}\cdot a_{n}}$?
oh, thank you very much! Now I got it, I think... $ a_n = 10^{b_n} $, $ 10^{b_n} = 10^{b_{n-1}} * 10^{b_{n-2}} $, $ 10^{b_n} = 10^{b_{n-1} + b_{n-2}} $, $ b_n = b_{n-1} + b_{n-2} $ is this a rigth way?
Nov
28
comment How to solve this recurrence relation $a_{n+2} = \sqrt{a_{n+1}\cdot a_{n}}$?
:) Now I've the same question. Where did you get that $ e $ ?
Nov
28
comment How to solve this recurrence relation $a_{n+2} = \sqrt{a_{n+1}\cdot a_{n}}$?
$$ b_n=\log a_n , a_n^{b_n} = k ? $$ What do you mean? How it works?
Nov
28
awarded  Student
Nov
28
asked How to solve this recurrence relation $a_{n+2} = \sqrt{a_{n+1}\cdot a_{n}}$?