Buga1234

28 reputation

	28 reputation	bio	website		visits	member for	5 months
3 badges		location			seen	Dec 3 '12 at 15:04

10 Actions

Dec 1	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	awarded	Supporter
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 $a_{n+2} = \sqrt{a_{n+1}\cdot a_{n}}$?