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How can I find the limit of the following recursive formula?

$a, b \in \mathbb{R}$

$a_0=a; a_1=b; a_n= \frac{1}{2}(a_{n-1}+a_{n-2})$

$\lim\limits_{n \rightarrow \infty}{a_n} = ?$

Many thanks in advance.

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closed as off-topic by Did, Shailesh, heropup, iadvd, Claude Leibovici Sep 16 '16 at 10:10

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Think what happens if $a=0$, $b=1$. You can draw a line and see what happens, by drawing the points of the sequence $$0,1,1/2,3/4,5/8,11/16,\ldots$$ In other words, we start at zero, go $1$ unit to the rigth, $1/2$ unit to the left, $1/4$ of unit to the right, $1/8$ unit to the left, $\ldots$, $1/2^n$ unit to the left, is $n$ is odd and to the right if $n$ is odd.

This gives us the formula (in the case $a=0$, $b=1$) $$a_0=0,\qquad a_n=\sum_{i=0}^{n-1}(-2^{-1})^i$$ therefore the limit point is $\lim a_n=\sum_{i=0}^{\infty}(-2^{-1})^i=\frac{1}{1-(-2^{-1})}=\frac{2}{3}$.

In the case of general $a$ and $b$, we are not working on the unit interval, but rather on an interval of length $(b-a)$, and $a$ units to the right (left? Anyway...), so scaling gives us, for general $a$ and $b$, $$\lim a_n=a+\frac{2}{3}(b-a)=\frac{a+2b}{3}$$

If you want to be formal in the general case, show by induction that $a_n=a+(b-a)\sum_{i=0}^{n-1}(-2^{-1})^i$.

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I learned to solve these using linear algebra.

$\begin {bmatrix} a_{n+1} \\ a_n\end{bmatrix} = \begin{bmatrix} \frac 12 &\frac 12\\1&0\end{bmatrix}\begin{bmatrix}a_n\\a_n-1\end{bmatrix}$

$\begin {bmatrix} a_{n+1} \\ a_n\end{bmatrix} = \begin{bmatrix} \frac 12 &\frac 12\\1&0\end{bmatrix}^n\begin{bmatrix}a_1\\a_0\end{bmatrix}$

diagonalize the matrix

$\begin {bmatrix} a_{n+1} \\ a_n\end{bmatrix} = \frac 13\begin{bmatrix}1 &-1\\1&2\end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0& (-\frac 12)^n\end{bmatrix}\begin{bmatrix}2 &1\\-1&1\end{bmatrix}\begin{bmatrix}b\\a\end{bmatrix}$

We an apply the limit here, and and take $n$ to infinty. It will reduce the algebra that follows, but it is worth seeing how to come up with a general fromula for $a_n$

$\begin {bmatrix} a_{n+1} \\ a_n\end{bmatrix} = \frac 13\begin{bmatrix}1 &-1\\1&2\end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0& (-\frac 12)^n\end{bmatrix}\begin{bmatrix}2b+a\\-b+a\end{bmatrix}$

$\begin {bmatrix} a_{n+1} \\ a_n\end{bmatrix} = \frac 13\begin{bmatrix}1 &-1\\1&2\end{bmatrix}\begin{bmatrix}2b+a\\(-\frac 12)^n(-b+a)\end{bmatrix}$

$a_n = \frac 13 (2b+a + 2(-\frac12)^n(-b+a))$

As $n$ goes to infnity $\frac 13 (2b+a)$

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  • $\begingroup$ In the end it should be $\frac{1}{3}(2b+a)$ (it's just a typo). $\endgroup$ – Luiz Cordeiro Sep 15 '16 at 19:01

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