find-the-sequence-u-n-wich-verify-u-n-1-u-n-u-n-1-real-

Question Number 142987 by mathmax by abdo last updated on 08/Jun/21

find the sequence u_{n} wich verify u_{n + 1} = u_{n} - λ u_{n - 1}

λ real

Answered by Dwaipayan Shikari last updated on 08/Jun/21

u_{n} = {m r}^{n}

{m r}^{n + 1} = {m r}^{n} - λ {m r}^{n - 1}

\Rightarrow r^{2} - r + λ = 0 \Rightarrow r = \frac{1 \pm \sqrt{1 - 4 λ}}{2} λ ⩽ \frac{1}{4} r \in R

i f r \in C t h e n - \infty < λ < \infty

u_{n} = Λ {(\frac{1 + \sqrt{1 - 4 λ}}{2})}^{n} - Φ {(\frac{1 - \sqrt{1 - 4 λ}}{2})}^{n}

u_{0} = Λ - Φ

u_{1} = \frac{Λ - Φ}{2} + \frac{Λ + Φ}{2} \sqrt{1 - 4 λ} \Rightarrow \frac{1}{\sqrt{1 - 4 λ}} (2 u_{1} - u_{0}) = Λ + Φ

Λ = \frac{1}{2} . \frac{(2 u_{1} - u_{0})}{\sqrt{1 - 4 λ}} + \frac{u_{0}}{2}

Φ = \frac{1}{2} . \frac{(2 u_{1} - u_{0})}{\sqrt{1 - 4 λ}} - \frac{u_{0}}{2}

u_{n} = \frac{(2 u_{1} - u_{0})}{2 \sqrt{1 - 4 λ}} {{(\frac{1 + \sqrt{1 - 4 λ}}{2})}^{n} - {(\frac{1 - \sqrt{1 - 4 λ}}{2})}^{n}} + \frac{u_{0}}{2} {{(\frac{1 + \sqrt{1 - 4 λ}}{2})}^{n} - {(\frac{1 - \sqrt{1 - 4 λ}}{2})}^{n}}