Graphing Equations Is Useful, II

We already looked into a problem offered at the 1957 (XX) Moscow Mathematical Olympiad to the 8th graders. Here is a more difficult one that was given to high school seniors:

For $n\gt 0$, find all real solutions of the system

$\begin{array}{3,5} 1-x_{1}^{2}&=x_{2}, \\ 1-x_{2}^{2}&=x_{3}, \\ \cdots\cdots\cdots & \\ 1-x_{n-1}^{2}&=x_{n}, \\ 1-x_{n}^{2}&=x_{1}, \end{array}$

Solution

For $n\gt 0$, find all real solutions of the system

$\begin{array}{3,5} 1-x_{1}^{2}&=x_{2}, \\ 1-x_{2}^{2}&=x_{3}, \\ \cdots\cdots\cdots & \\ 1-x_{n-1}^{2}&=x_{n}, \\ 1-x_{n}^{2}&=x_{1}, \end{array}$

It is rather obvious that the system represent iterations of function $\displaystyle f(x)=1-x^{2}$ and the question is to find a starting point $x_{1}$ to which the iterations return after $n$ applications of function $f$. Let's graph $y=f(x)$. (The straight line is the graph of $y=x$.)

Obviously, there are two solutions defined by $f(x)=x$. Solving the quadratic equation we obtain two stationary points of function $f$:

$\displaystyle \frac{-1\pm\sqrt{5}}{2}$

On the graph these correspond to points A$\space=\left(\displaystyle\frac{-1+\sqrt{5}}{2},\displaystyle\frac{-1+\sqrt{5}}{2}\right)$ and B$\space=\left(\displaystyle\frac{-1-\sqrt{5}}{2},\displaystyle\frac{-1-\sqrt{5}}{2}\right)$.

Obviously this gives two sets of solutions for any $n\gt 0$, but this is not the whole story. If $n$ is even there are two more solutions: starting with $x_{1}=0$ and the other with $x_{1}=1$. In both cases the $0$s and $1$s then alternate. Are there any other solutions? The answer is No, but I am not sure how the high schoolers were supposed to prove that. The graph provides some clues, though.

By the construction, $(x_{1},x_{2})$ is a point on the graph of $y=f(x)$, as is any pair of successive iterations. The diagram above shows how two points $(x_{1},x_{2})$ and $(x_{2},x_{3})$ are connected graphically. There is an "intermediate" point $(x_{2},x_{2})$ that lies on the diagonal $y=x$. This observation helps to realize that, for a solution to exist, $x_{1}$ could not be outside the interval $\left(\displaystyle\frac{-1-\sqrt{5}}{2},\displaystyle\frac{1+\sqrt{5}}{2}\right)$. If it is, the iteration will certainly diverge to $\displaystyle -\infty$.

The applet below illustrates the behavior of the iterations in general. Click in the applet area to reset the initial value of $x_{1}$.

This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

 What if applet does not run?

Wherever you start in the interval $\left(\displaystyle\frac{-1-\sqrt{5}}{2},\displaystyle\frac{1+\sqrt{5}}{2}\right)$, the iteration settle into the 2-loop: $1,0,1,0,\ldots$. It helps to consider the graph of the iterated function $f(f(x))=1-(1-x^{2})^{2}=x^{2}(2-x^{2})$:

The iterated function has four stationary points: $0$ and $1$ are attractive, while $\displaystyle\frac{-1\pm\sqrt{5}}{2}$ are repelling. It may also help to graph the next two iterates:

No additional stationary points seem to appear (some would if there were other loops, i.e., solutions to the system for some $n$). By contrast, for the iterates of another quadratic function $y=4x(1-x)$, new stationary points pop up all the time:

Why not for $y=1-x^2$? To answer this question we have to prove that if $x\ne 1-x^2$ then also $x\ne 1-(1-x^{2})^{2}$. Let's do that. One will have to consider several cases. I shall point out to just one. Let $\alpha=(-1+\sqrt{5})/2$. One can see it from the graphs that, say, for $0\lt x\lt\alpha$, $x\lt 1-x^{2}$, but $x\gt 1-(1-x^{2})^{2}\gt 0$. As a consequence, $0\lt 1-(1-x^{2})^{2} \lt \alpha$ so that to this iterate applies whatever applied to $x$. Thus no new stationary points may appear in the interval $(0,\alpha)$.

Hopefully, the remaining cases could be dealt with similarly.