0
$\begingroup$

Let $\alpha\in(0,2)$, and the sequence $$x_{n+1}=\alpha x_n +(1+\alpha)x_{n-1} \quad \forall n\geq 1$$ Find the limit in terms of $\alpha$, $x_0$ and $x_1$.

Check my work.

If $\alpha=1$, else $$\begin{align*} x_{n+1}&=x_n+2x_{n-1} \\ x_{2}&=x_1+2x_{0} \\ x_{3}&=x_2+2x_{1} \\ &\;\;\vdots \end{align*}$$ implies $\lim=x_1+2x_0$.

If $\alpha=2$, else $$\begin{align*} x_{n+1}&=2x_n+3x_{n-1} \\ x_{2}&=2x_1+3x_{0} \\ x_{3}&=2x_2+3x_{1}\\ &\;\;\vdots \end{align*}$$ implies $\lim=3x_0+x_1+x_{n-1}$.

$\endgroup$
  • $\begingroup$ Please, please don't use MathJax like that! Take a look at how I edited your question - try to follow that manner of formatting in the future! $\endgroup$ – Zev Chonoles Jul 8 '15 at 3:51
  • $\begingroup$ I thought $\alpha \in (0,2)$. So why can you take $\alpha = 2$? Do you mean $\alpha \in (0,2]$? $\endgroup$ – 0XLR Jul 8 '15 at 4:01
  • $\begingroup$ You're right! α∈(0,2) the intervals is open but I can't find the limit... $\endgroup$ – Sudo su Jul 8 '15 at 4:03
2
$\begingroup$

The recurrence relation being $$x_{n+1}=\alpha x_n +(1+\alpha)x_{n-1}$$ its characteristic equation is $$r^2=\alpha r+(1+\alpha)$$ the roots of which being $r_1=-1$ and $r_2=1+\alpha$; so the general solution is $$x_{n+1}=c_1 (\alpha +1)^n+c_2 (-1)^n$$ Applying the conditions for $n=0$ and $n=1$, this becomes $$x_{n+1}=\frac{(x_0+x_1) (\alpha +1)^n+(-1)^n (\alpha x_0+x_0-x_1)}{\alpha +2}$$

I am sure that you can take from here.

$\endgroup$
1
$\begingroup$

We can exploit the recurrence relation as follows (same thing as using the Characteristic polynomial, but a little neater) :

$$\left\{\begin{aligned} x_{n+1}+x_n&=(\alpha+1)(x_{n}+x_{n-1})=\cdots = (\alpha+1)^n(x_1+x_0)\\ x_{n+1}-(\alpha+1)x_n&=-x_n+(\alpha+1)x_{n-1}=\cdots = (-1)^{n}[x_1-(\alpha+1)x_0].\end{aligned}\right.$$

Multiply the first relation with $(\alpha+1)$ and add to get $$(\alpha+2)x_{n+1}=(\alpha+1)^{n+1}(x_1+x_0)+(-1)^n[x_1-(\alpha+1)x_0].$$ It is easy to see that the limit exists iff $x_1+x_0=0$ and $x_1-(\alpha+1)x_0=0$ in which case the sequence is constant.

$\endgroup$

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.