Solving the Functional Equation $f(f(x))=x$ gone wrong

Question

For a function to be its inverse (i.e. an involution), it needs to satisfy the functional equation $f(x)=f^{-1}(x)$ or $f(f(x))=x$. I expressed $f^{\circ n}(x)$ (the composition of $f(x)$ to itself $n$ times, with $n=0$ corresponding to $x$) as a recurrence relation $a_n$. So $a_n=a_{n-2}$. The characteristic equation of this recurrence is $r^2-1=0$ and $r=\pm1$. Therefore, $a_n=a(x)(1)^n+b(x)(-1)^n=a(x)+b(x)(-1)^n$, where $a(x)$ and $b(x)$ are functions to be determined. When $n=0$, we get that $a(x)+b(x)=x$, and when $n=1$ we get that $a(x)-b(x)=f(x)$. So I thought that, if $a(x)+b(x)=x$, then $a(x)-b(x)$ is an involution. But this is easily contradicted by $a(x)=2x$ and $b(x)=-x$. Where did I go wrong? If we let $n=2$, we get $a(x)+b(x)=f(f(x))$ so it seems alright.

Answer 1

Your error was to assume that only involutions $f$ define a sequence $(a_n)$ such that $a_0=x$, $a_1=f(x)$, and $a_n=a_{n-2}$ for $n\geq 2$. As your construction shows, if you define the functions $a(x)=\frac{1}{2}(x+g(x))$ and $b(x)=\frac{1}{2}(x-g(x))$, where $g$ is an arbitrary real function, then the sequence $(b_n)$ with elements $b_n=a(x)+(-1)^nb(x)$ $(n\in\mathbb{N})$ also satisfies $b_0=x$, $b_1=g(x)$, and $b_n=b_{n-2}$ for $n\geq 2$.

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Addendum

Based on the title of your question, I suppose that you want to solve the functional equation $f(f(x))=x$. Here is the most general solution: write the domain $D$ of $f$ as a disjoint union of subsets of $D$ containing only one or two elements (the Axiom of Choice allows you to do that). For subsets $\{x\}$ with a single element, define $f(x)=x$; for subsets $\{x_1,x_2\}$ with two elements, define $f(x_1)=x_2$ and $f(x_2)=x_1$.