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 $\lim=3x_0+x_1+x_{n-1}$.

  • $\begingroup$ 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

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.


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.


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