find the sequence (v_n ) wich verify v_(n+2) =(√(v_n .v


Question and Answers Forum

All Questions Topic List
Relation and Functions Questions
Previous in All Question Next in All Question
Previous in Relation and Functions Next in Relation and Functions
Question Number 32996 by abdo imad last updated on 09/Apr/18

$f i n d t h e s e q u e n c e (v_{n}) w i c h v e r i f y v_{n + 2} = \sqrt{v_{n} . v_{n + 1}} .$
Commented by prof Abdo imad last updated on 11/Apr/18

$v_{n} m u s t b e p o s i t i f a n d l n (v_{n + 2}) = \frac{1}{2} l n (v_{n}) + \frac{1}{2} l n (v_{n + 1})$ $l e y p u t u_{n} = l n (v_{n}) \Rightarrow u_{n + 2} = \frac{1}{2} u_{n} + \frac{1}{2} u_{n + 1} \Rightarrow$ $2 u_{n + 2} - u_{n + 1} - u_{n} = 0 c a r s c t e r i s t i c e q u a t i o n$ $2 x^{2} - x - 1 = 0 Δ = 1 - 4 (2) (- 1) = 9$ $x_{1} = \frac{1 + 3}{4} = 1 a n d x_{2} = \frac{1 - 3}{4} = - \frac{1}{2} \Rightarrow u_{n} = α + β {(- \frac{1}{2})}^{n}$ $u_{0} = α + β a n d u_{1} = α - \frac{β}{2} \Rightarrow u_{0} - u_{1} = \frac{3}{2} β \Rightarrow$ $β = \frac{2}{3} (u_{0} - u_{1})$ $α = u_{0} - β = u_{0} - \frac{2 u_{0} - 2 u_{1}}{3} = \frac{u_{0} + 2 u_{1}}{3} \Rightarrow$ $u_{n} = \frac{u_{0} + 2 u_{1}}{3} + \frac{2 u_{0} - 2 u_{1}}{3} {(- \frac{1}{2})}^{n}$ $v_{n} = e^{u_{n}} .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum