YT Math: Fixed points of a recursive function

Don’t square? Why not? Okay, fine, we’ll do something else.

The big expression in the question has an obvious recursive structure. If we define $f(t) = \sqrt{10 – \frac{3}{t}}$ , the question’s equation becomes $x = f(f(x))$ .

Generally speaking, the solutions to an equation of the form $x = f(x)$ are known as the fixed points of $f$ . Clearly, the fixed points of any $f$ will also be fixed points of $f \circ f , f \circ f \circ f, …$ This follows immediately from the definition of a fixed point. We can exploit this observation to simplify the problem of finding fixed points of $f \circ f$ to finding the fixed points of $f$ . So let’s find the fixed points of $f$ by simply writing the equation and solving it:

x = \sqrt{10 – \frac{3}{x}}

With a little massaging, this becomes the cubic

x^3 – 10x + 3 = 0

By inspection, $x – 3$ is a factor. Dividing by this monomial, we’re left with $x^2 + 3x – 1$ . After using the quadratic formula, we end up with 3 roots:

\left\{ 3, -\frac{3}{2} \pm \frac{\sqrt{13}}{2} \right\}

These are solutions to the original question, which I confirmed with Octave¹. $\blacksquare$

Three solutions are probably already two more than this question deserves, but I feel compelled to admit that I glossed over something when I mentioned $f \circ f$ inheriting the fixed points of $f$ . It is possible, in general, that $f \circ f$ has additional fixed points that $f$ does not. This is pretty easy to see. Imagine some $f$ sends $1$ to $2$ and $2$ to $1$ . In that case, $1$ and $2$ are not fixed points of $f$ , but they will be fixed points of $f \circ f$ . That is, $f(a) = b$ and $f(b) = a$ implies $f(f(a)) = a$ and $f(f(b)) = b$ . The same kind of situation applies to higher-order iterations too.

There is also a graph-theoretic way to view this. Associate with $f$ a graph $G(f)$ whose vertex set is the union of the domain and range of $f$ , and which has an edge $u \to v$ wherever $f(u) = v$ . Then 2-cycles in $G(f)$ will correspond with loops in $G(f \circ f)$ . Specifically, every vertex involved in a 2-cycle in $G(f)$ will have a loop on it in $G(f \circ f)$ . In terms of $G(f)$ ‘s adjacency matrix $\mathbf{A}$ , the statement “ $f \circ f$ may have fixed points that $f$ does not” is the same as “ $\mathbf{A}^2$ may have nonzero diagonal entries where $\mathbf{A}$ does not”.

Do any such additional fixed points exist for this particular problem’s $f \circ f$ ? This addendum has already dragged on, so I leave it as an exercise.

https://octave.org/ ↩︎