Limit by diagonalisation

Question

I want to compute the limit of the sequence $x_0=a$, $x_1=b$ and $x_n=\frac{x_{n-1}+x_{n-2}}{2}$.

I did it by setting:

$\begin{pmatrix} x_{n+1}\\ x_{n} \end{pmatrix}=$ $\begin{pmatrix} \frac{1}{2} &\frac{1}{2}\\ 1 &0\\ \end{pmatrix}$ $\begin{pmatrix} x_{n}\\ x_{n-1} \end{pmatrix}$

and calculating, that $$\lim\limits_{n\to\infty}\begin{pmatrix} \frac{1}{2} &\frac{1}{2}\\ 1 &0\\ \end{pmatrix}^n=\begin{pmatrix} \frac{2}{3} &\frac{1}{3}\\ \frac{2}{3} &\frac{1}{3}\\ \end{pmatrix}$$

And thus $\lim\limits_{n\to\infty}x_n=\frac{2}{3}b+\frac{1}{3}a$.

But this can't be true, because the result should be symmetric in $a$ and $b$. Where is my mistake?

Why do you think it should it be symmetric in $a$ and $b$? – angryavian Jun 15 '14 at 21:11 — angryavian, Jun 15 '14 at 21:11
Oh, I see my mistake, of course it isn't. Thank you. – Diz Jun 15 '14 at 21:12 — Diz, Jun 15 '14 at 21:12

score 1 · Answer 1 · answered Jun 15 '14 at 21:13

1

$x_2 = \dfrac{a + b}{2}$

$x_3 = \dfrac{a+3b}{4}$

It is already not symmetric in $a$ and $b$. There is no error.

answered Jun 15 '14 at 21:13

Brad

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Limit by diagonalisation

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