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I have a problem : Let $a$ be a real number in $(0,2)$ and the sequence $x_n$ is defined by : $x_{n+1}=ax_{n}+(1-a)x_{n-1}$. Find the limit of $x_n$ as $n\rightarrow \infty$.

Please give me a hint to solve it or a solution is very appreciated.

Thanks

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    Hint: Let $b = a-1 \in (-1, 1)$. Then $x_{n+1} - x_{n} = b(x_n - x_{n-1}).$2012-11-14

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The characteristic polynomial for this recurrence is $\lambda^2 - a\lambda + a-1 =0$ which has roots $\lambda = 1, a-1$ so there exist constants $\alpha$ and $\beta$ (which can be determined given initial values of the sequence) such that $x_n = \alpha (a-1)^n + \beta.$ Now since $a\in (0,2)$, we see that $x_n \to \beta.$ Put $n=0$ so that $x_0 = \alpha + \beta$ and then $n=1$ so that $x_1 = \alpha (a-1) + \beta$ and solve to express $\beta$ in terms of $x_0$ and $x_1.$

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    @sos440 Thanks :).2012-11-14