Solutions to functional equation $f(f(x))=x$

Question

Is there any more solutions to this functional equation $f(f(x))=x$?

I have found: $f(x)=C-x$ and $f(x)=\frac{C}{x}$.

Answer 1

If you don't make any niceness assumptions about $f$, there are lots. Partition $\Bbb{R}$ (or whatever you want $f$'s domain to be) into $1$- and $2$-element subsets, in any way you like. Then define $f(x)=y$, if $\{x, y\}$ is in your partition, or $f(x)=x$, if $\{x\}$ is in your partition.

Moreover, any such $f$ yields such a partition, into the sets $\{x, f(x)\}$. So this is a complete description of all solutions.

Of course, $f$ will be wildly discontinuous for most choices of partition.

Answer 2

This answer will only deal with continuous maps.

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A map $f : X \to X$ such that $f \circ f = \operatorname{id}_X$ is called an involution of $X$. Two involutions of $X$, $f$ and $g$, are said to be equivalent if there is a self-homeomorphism $h$ such that $f\circ h = h\circ g$ (or written differently, $f = h\circ g \circ h^{-1}$). With this terminology at our disposal, we can state the following result.

Every non-identity involution of $\mathbb{R}$ is equivalent to the involution $g(x) = -x$.

Therefore, $f$ is an involution of $\mathbb{R}$ if and only if $f = h\circ g\circ h^{-1}$ for some homeomorphism $h : \mathbb{R} \to \mathbb{R}$, or $f = \operatorname{id}_X$.

For example, if we let $f(x) = C - x$, then $f = h\circ g\circ h^{-1}$ where $h : \mathbb{R} \to \mathbb{R}$ is the homeomorphism given by $h(x) = x + \frac{C}{2}$.

Answer 3

$f(x) = c-x$

$f(x) = c/x$

$f(x) = \frac{c_{1}-x}{c_{2}x+1}$

$f(x) = \frac{1}{2}\left ( \sqrt[]{{c_{1}^2}+c_{2}-4x^2} \right )+ c_{1}x$

$f(x) = \sqrt[3]{c-x^3}$

http://www.wolframalpha.com/input/?i=f%28f%28x%29%29%3Dx

Answer 4

Any function whose graphic is symmetric against the line y=x.

Answer 5

In fact this belongs to a functional equation of the form http://eqworld.ipmnet.ru/en/solutions/fe/fe2315.pdf.

Let $\begin{cases}x=u(t)\\f=u(t+1)\end{cases}$ ,

Then $u(t+2)=u(t)$

$u(t)=\theta(t)$ , where $\theta(t)$ is an arbitrary periodic functions with period $2$

$\therefore\begin{cases}x=\theta(t)\\f=\theta(t+1)\end{cases}$ , where $\theta(t)$ is an arbitrary periodic functions with period $2$