Finding the fixed points of a recurrence relation (and systems of) analytically?

Question

How would I go about finding the fixed points of the following recurrence?

$$X_n = 2X_{n-1}(2- 3X_{n-1}) + X_{n-1}$$

And therein, determining their stability analytically? Also, how does one find the fixed points among a system of recurrences, i.e.:

$$\Delta a_n = a_n - a_{n-1} = 3a_{n-1} - 6$$

$$\Delta b_n = b_n - b_{n-1} = -3a_{n-1} + 2b_{n-1}$$

Is there a simple way to do so for both that I am unaware of?

Answer 1

We can write the recurrence as $X_n = F(X_{n-1})$, where $F(X_{n-1}) = 2X_{n-1}(2- 3X_{n-1}) + X_{n-1}$. You can think of $F$ as an iterative map, i.e. if you start with an initial iterate $X_{0}$, $F$ will generates $X_{1}, X_{2}, \dots$ so that $X_{n}$ converges to the fixed point $X^{*}$ of $F$ (i.e. that $F(X^{*}) = X^{*}$).

In view of above, one simple way to find the fixed points are to simply solve for $X^{*}$ in $X^{*} = F(X^{*})$ (assuming that $X_{n}$ converges to the fixed points $X^{*})$. In this case, we get $X = 0, 2/3$. To determine the stability of the fixed points, look at the Jacobian of the map $F$.

To find the solution of the system of recurrences, note that you can recast the problem as:

$$\left[ \begin{array}{c} a_{n} \\ b_{n} \end{array} \right] = \begin{bmatrix} 4 & 0 \\ -3 & 3 \end{bmatrix} \left[ \begin{array}{c} a_{n-1} \\ b_{n-1} \end{array} \right] + \left[ \begin{array}{c} -6 \\ 0 \end{array} \right].$$

Note that the above iteration won't converge to a unique solution since the spectral radius of the matrix $$\begin{bmatrix} 4 & 0 \\ -3 & 3 \end{bmatrix} $$ is greater than $1$.

Answer 2

Also note that you can use the derivative of $$f(x^*)$$ to analyze the stability of points. If $$|f(x^*)|< 1$$ We have a stable point, and if $$|f(x^*)|> 1$$ we have a unstable point, where $$x^*$$ can be found by solving $$x^*=f(x^*)$$ In your case, both points are unstable. This is a slightly more strait forward approach to your problem, however, the first post is correct as well.