Exercise 4.4.9 ($\lim_{n o \infty} u_n$, where $u_n = (u_{n-1} + u_{n-2})/2$)

Question

Let $u_0$ and $u_1$ be given, and for $n\geq 2$ put $u_n = (u_{n-1} + u_{n-2})/2$. Show that $\lim_{n	o \infty} u_n$ exists, and that it is a certain weighted average of $u_0$ and $u_1$.

Richard Ganaye · Accepted Answer

ewcommand{\N}{\mathbb{N}}

\begin{proof} 
The characteristic polynomial associated to the sequence is
\begin{align*}
p(x) &= x^2 - \frac{1}{2} x - \frac{1}{2}\
&= \left(x- \frac{1}{4}ight)^2 - \left( \frac{3}{4} ight)^2\
&= (x-1) \left( x + \frac{1}{2}ight).
\end{align*}
The roots of $p(x)$ are
$$\lambda= -\frac{1}{2},\qquad \mu = 1.$$
By Theorem 4.10, there are real numbers $\alpha, \beta$ such that for all $n \in \N$,
$$u_n = \alpha + \beta \lambda^n = \alpha + \beta \left ( -\frac{1}{2}ight)^n.$$
Since $|\lambda|<1$, $\lim_{n	o \infty} \lambda^n = 0$, so
$$\lim_{n 	o \infty}u_n = \alpha.$$
Moreover, $\alpha$ and $\beta$ are the roots of the system
$$
\left\{
\begin{array}{ll}
u_0 &= \alpha + \beta,\
u_1 &= \alpha - \frac{1}{2} \beta,
\end{array}
ight.
$$
thus 
$$\lim_{n 	o \infty}u_n = \alpha = \frac{1}{3} u_0 + \frac{2}{3} u_1$$
 is a weighted average of $u_0$ and $u_1$, with weights $\frac{1}{3}, \frac{2}{3} $ such that $\frac{1}{3}  + \frac{2}{3} = 1$.
\end{proof}

Exercise 4.4.9 ($\lim_{n\to \infty} u_n$, where $u_n = (u_{n-1} + u_{n-2})/2$)

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Exercise 4.4.9 ($\lim_{n\to \infty} u_n$, where $u_n = (u_{n-1} + u_{n-2})/2$)

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