Section 8.1 Classification of fixed points
A fixed point of a function \(g\) is a number \(x^*\) such that \(g(x^*)=x^*\text{.}\) This is the same as a root of \(g(x)-x=0\) but we will see that writing an equation with \(x\) on the right provides a new approach to solving it.
Example 8.1.1. Find all fixed points of a function.
Find all fixed points of the function \(g(x) = \sqrt{1+x}\)
Squaring both sides of \(\sqrt{1+x} = x\) yields \(x^2 - x - 1\) which is a quadratic equation with roots \((1\pm \sqrt{5})/2 \text{.}\) However, the negative root does not satisfy the original equation \(\sqrt{1+x} = x\text{.}\)
Answer: one fixed point \(x^* = (1+\sqrt{5})/2\text{.}\)
The process of iterating some function \(g\) consists of the following: start from some initial value \(x_0\text{,}\) let \(x_1=g(x_0)\text{,}\) \(x_2 = g(x_1)\text{,}\) and generally \(x_{n+1} = g(x_n)\text{.}\) If the sequence \(x_n\) has a limit \(x^* \text{,}\) and \(g\) is a continuous function, it follows that \(g(x^*)=x^*\text{.}\)
However, can we expect the iteration to converge to a fixed point \(x^* \text{?}\) The answer depends on the value of \(g'(x^*)\text{.}\) Indeed, when \(x\approx x^*\text{,}\) the difference \(g(x)-x^*\) is approximately \(g'(x^*)(x-x^*)\) because the graph of a differential function is close to its tangent line. When \(|g'(x^*)|\lt 1\text{,}\) the distance to fixed point is decreasing but if \(|g'(x^*)| > 1\) it is increasing. This leads us to introduce the following concept.
Classification of fixed points. A fixed point \(x^*\) is called
- repelling if \(|g'(x^*)| > 1\)
- attracting if \(|g'(x^*)| \lt 1\)
- super-attracting if \(g'(x^*) = 0 \) (a special case of attracting fixed points)
- neutral if \(|g'(x^*)| = 1 \)
According to the above, iteration of \(g\) may converge to an attracting point but will not converge to a repelling one. The neutral points are a borderline case: some of them are limits of iterative process, some are not.
Example 8.1.2. Find and classify the fixed points.
Find and classify all fixed points of the function \(g(x) = x^3 + x^2 - x\text{.}\)
The equation \(x^3 + x^2 - x = x\) simplifies to \(x(x^2 + x - 2) =0\) which has roots \(0, 1, -2\text{.}\) These are the three fixed points. Plug each into the derivative \(g'(x) = 3x^2 + 2x - 1\) to find that
- \(|g'(0)| = |-1| = 1\text{,}\) a neutral fixed point
- \(|g'(1)| = |4| = 4\text{,}\) a repelling fixed point
- \(|g'(-2)| = |7| = 7\text{,}\) a repelling fixed point