ISI2021-MMA: 14

Question

Given a real number $\alpha \in (0, 1),$ define a sequence $\{ x_{n}\}_{n \geq 0}$ by the following recurrence relation:

$$x_{n+1} = \alpha x_{n} + (1 – \alpha) x_{n – 1}, n \geq 1.$$

If $\lim_{n \rightarrow \infty} x_{n} = \ell$ then the value of $\ell$ is

• A. $\frac{\alpha x_{0} + x_{1}} {1 – \alpha}$
• B. $\frac{(1 – \alpha) x_{0} + x_{1}} {2 – \alpha}$
• C. $\frac{\alpha x_{0} + x_{1}} {2 – \alpha}$
• D. $\frac{(1 – \alpha) x_{1} + x_{0}} {2 – \alpha}$

Answer 1

Initial part read from here: https://math.stackexchange.com/questions/4412959/if-lim-x-to-infty-x-n-l-then-find-the-value-of-l-for-a-reccuring-sequen

Next part:

$l =(1-\tilde{\alpha})x_0+\tilde{\alpha}x_1$
$l=(1-\frac{\alpha}{1-(1-\alpha)^2})x_0+\frac{\alpha}{1-(1-\alpha)^2}x_1$
Solve this you will get $\frac{(1-\alpha)x_0+x_1}{2-\alpha}$

$Ans: B$