0
$\begingroup$

For $\alpha \in (0,1)$ let the sequence $\{x_n\}$ be such that $x_0 = 0, x_1 = 1 $ and $x_{n+1} = \alpha x_n + (1-\alpha)x_{n-1},\quad n\geq1$. Find $\displaystyle \lim_{n\rightarrow \infty}x_n$.

My try:

Since $\alpha \in (0,1)\implies (\alpha-1)\in (-1,0)$

$\begin{align}|x_{n+1}-x_n|=&|\alpha -1||x_n-x_{n-1}|\\ &\vdots\\ &|\alpha -1|^n|x_1-x_0|\end{align} \implies \{x_n\} \rightarrow 0 $

Is there any alternative proof ?

$\endgroup$
  • $\begingroup$ ${x_n}\to0$ not true. $\endgroup$ – Samvel Safaryan Nov 12 '18 at 8:06
  • $\begingroup$ Can you please suggest what is wrong in my proof? $\endgroup$ – Yadati Kiran Nov 12 '18 at 8:07
  • 1
    $\begingroup$ Oh! I guess I have shown that the sequence is convergent but not the limit. $\endgroup$ – Yadati Kiran Nov 12 '18 at 8:09
  • $\begingroup$ $|x_{n+1}-x_n|=|\alpha-1|^n$, but it doesn't follow that $lim_{n\to+\infty}{x_n}=0$. $\endgroup$ – Samvel Safaryan Nov 12 '18 at 8:13
  • $\begingroup$ Below is my solution, you can prove by induction that $x_n$ satisfies all the conditions of the problem. $\endgroup$ – Samvel Safaryan Nov 12 '18 at 8:17
1
$\begingroup$

Let's look for $x_n$ in the form $\lambda^n$ $$ \\\lambda^2=\alpha\lambda+(1-\alpha) \\\lambda^2-1-\alpha(\lambda-1)=0 \\(\lambda-1)(\lambda+1-\alpha)=0 $$ $=>\lambda=1,\;\alpha-1=>x_n=A+B(\alpha-1)^n\\ 0=x_0=A+B\\ 1=x_1=A+B(\alpha-1) \\B(\alpha-2)=1=>B=\frac{1}{\alpha-2}=>x_n=\frac{1}{2-\alpha}\Big(1-(\alpha-1)^n\Big)=>\lim_{n\to+\infty}{x_n}=\frac{1}{2-\alpha} $

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.