Find the fixed points of the difference equation $ \ a_n=\frac{2}{7} a_{n-1}-1 \ $ . Classify the fixed points whether stable, unstable or Neutral.
Answer:
Let $ \ x \ $ be the fixed point.
Then,
$ x=\frac{2}{7} x-1 \\ \Rightarrow 7x=2x-7 \\ \Rightarrow 5x=-7 \\ \Rightarrow x =-\frac{7}{5} $
But how to decide whether $ \ x =-\frac{7}{5} $ is stable or unstable.
Help me out.
Hint:
Let $a_n=b_n-\frac75$. The recurrence is
$$b_n-\frac75=\frac27\left(b_{n-1}-\frac75\right)-1=\frac27b_{n-1}-\frac75,$$ or
$$b_n=\frac27b_{n-1}.$$
You should be able to conclude.