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:

With a little massaging, this becomes the cubic

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:

These are solutions to the original question, except the negative root $-\frac{3}{2} – \frac{\sqrt{13}}{2}$ which turns out not to work when substituted, so we discard it. $\blacksquare$

One solution to this problem is probably already enough, 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 $f(u) = v$. In this setting, fixed points are loops, that is, edges from a vertex to the same vertex. If we construct $G(f \circ f)$ in the same fashion, 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 adjacency matrices, notice that if $G(f)$‘s is $\mathbf{A}$, $G(f \circ f)$‘s is $\mathbf{A}^2$. Then 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.

---