Fixed points of an iterated function

Question

I recently came across an interesting problem: Find all the real roots of $$ x^3+1=2\sqrt[3]{2x-1}. $$

I was able to hack through and find $3$ real roots, namely: $$ x=1,\frac{-1\pm\sqrt{5}}{2} $$ but my method didn't help me decide if there were more or not, but I was reasonably confident that there were no more.

So I thought about it some more and realized that if I let $$ f(x)=\frac{x^3+1}{2} $$ then the original problem reduced to finding all $x$ such that $$ f(f(x))=x $$ i.e. find all the real fixed points of the iterated function.

Is there a "slick" way to take advantage of this observation? I have been trying to read up on iterated functions, but I can't find much useful information about their fixed points.

Answer 1

First note that your three solutions are the three solutions of $f(x)=x$. Also note that $f'$ is positive so $f$ is increasing. Now if $f(a)=b$ and $f(b)=a$ where $a \neq b$ then one must be less than the other, so WLOG assume $a < b$. However, if we apply $f$ to both sides it should preserve the inequality, but we get $$a < b \implies f(a) < f(b) \implies b < a$$

So we get a contradiction and therefore the only fixed points of $f \circ f$ are the fixed points of $f$

Answer 2

Consider the set $\text{Fix}(f^2)$ of solutions to $f^2(x)=x$. Notice that it must be finite; in particular, since $f$ is a polynomial of degree $3$, $f^2$ is a polynomial of degree $9$ and hence $|\text{Fix}(f^2)|\leq 9$.

Write then $\text{Fix}(f^2)=\{x_1,\dots,x_m\}$ for some $m\leq 9$, and label the $x_i$ so that they are increasing. For each $i$, let $y_i=f(x_i)$.

Notice that $f^2(y_i)=f^2\Big(f(x_i)\Big)=f\Big(f^2(x_i)\Big)=f(x_i)=y_i$. In other words, each $y_i$ is also a solution to $f^2(x)=x$, and hence belongs to $\text{Fix}(f^2)$.

Claim: $y_i=x_i$ for all $i$.

Indeed, $f$ is monotone increasing. Since $y_m\in \text{Fix}(f^2)$ and $y_m=f(x_m)\geq x_m$, we must have that $y_m=x_m$. The claim follows from induction.

The claim implies that all fixed points of $f^2$ are fixed points of $f$, that is, $\text{Fix}(f^2)\subset\text{Fix}(f)$.

On the other hand, clearly we have $|\text{Fix}(f)|\leq 3$. After all, $f$ is a polynomial with degree $3$. Indeed, you can check that the solutions you found are the roots of $f(x)-x$.