Fixed points that are NOT convergent points

Question

Are there any fixed points that are NOT converget (aka attractig fixed points) in the sequence $x_n = 5\ln x_{n-1}$? How do you determine this?

Answer 1

The first step is to find the fixed points of $x_n = 5\ln x_{n-1}$. These points are the solutions to $x = 5\ln x$. As it turns out, there are two such fixed points: $x^* \approx 1.295$ and $x^* \approx 12.713$. You can express the exact values using the Lambert-W function.

The next step is to determine if these points are attracting fixed points or not.

A useful theorem is the following:

If $x^*$ is a fixed point of the iteration $x_n = f(x_{n-1})$, then:

1. If $|f'(x^*)| < 1$, then $x^*$ is an attracting fixed point.

2. If $|f'(x^*)| > 1$, then $x^*$ is not an attracting fixed point.

There is no conclusion of this theorem if $f'(x^*) = \pm 1$, but that's not an issue here.

For the recursion $x_n = 5\ln x_{n-1}$, the function $f$ is $f(x) = 5\ln x$. So all you have to do is compute $f'(x^*)$ for $x^* \approx 1.295$ and $x^* \approx 12.713$ and see if $|f'(x^*)| < 1$ or $|f'(x^*)| > 1$ for each fixed point.

Answer 2

The equation $x=5\ln x$ has two solutions: one between $1$ and $e$ and one between $e$ and $e^3$. (There can't be more than two solutions because the derivative of $5\ln x$ is monotonic).

At the smaller solution, $\frac d{dx}5\ln x$ is larger than $1$ (just draw a sketch to convince yourself of this!) and therefore the fixpoint is repelling rather than attracting.