The Analytical Approach

Analytically, period one fixed points are determined by solving the equation $x_{n+1} = x_n$. Denoting the solution as $x^{*}$ and using Eq.(3.1), one obtains

$$x^{*} = f(x^{*}) .$$ (3.3)

For the mapping function $f(x)=x^2$ the solutions are $x^{*}=0, 1$ as expected. The stability of fixed points can also be analysed analytically. Consider a nearby point $x_n = x^{*} + \epsilon$ with $\epsilon$ a small parameter. Then using a Taylor expansion and to leading order in $\epsilon$, one obtains

$$x_{n+1} = f(x^{*} + \epsilon)$$ (3.4)

$$= f(x^{*}) + \epsilon {d f \over dx}\vert _{x^{*}}$$ (3.5)

$$= x^{*} + (x_n - x^{*}) {d f \over dx}\vert _{x^{*}} ,$$ (3.6)

and hence

$$\vert x_{n+1} -x^{*}\vert = \vert x_n - x^{*}\vert \vert f'(x^{*})\vert .$$ (3.7)

Therefore if $\vert f'(x^{*})\vert < 1$ an iteration brings the starting point $x_n$ closer to the fixed point, implying that the fixed point is stable, at least locally. In other words, a period-one fixed point is stable if the slope of the mapping function at that point has magnitude less than one. Similarly, if the slope of the mapping function at a period-one fixed point is greater than one, the point is unstable. Finally, if the slope is exactly equal to one the fixed point is locally neutral, meaning that small perturbations lead to another equilibrium position ( the mechanical analogy here is that of a ball lying on a flat surface). For the mapping function $f(x)=x^2$, we had determined a fixed point at $x^{*} =0$ and another at $x^{*}=1$. Since $f'(0) =0$, this implies that the origin is a stable fixed point, while since $f'(1)=2$ the other fixed point is unstable. These results of course agree with the earlier qualitative discussion.