Is there any more solutions to this functional equation $f(f(x))=x$?
I have found: $f(x)=C-x$ and $f(x)=\frac{C}{x}$.
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2It likely matters what assumptions of continuity you make. But beyond that I think those are be the only ones. – Semiclassical Jul 26 '14 at 05:50
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If these are the only continuous ones, interesting to see what discontinuous solutions are like. – Somnium Jul 26 '14 at 05:52
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2$f(x) = x$ is also a "nice" solution. – JimmyK4542 Jul 26 '14 at 05:55
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@JimmyK4542 Oops, missed the simplest one) – Somnium Jul 26 '14 at 05:56
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1And there is a good case against $\frac{C}{x}$. – André Nicolas Jul 26 '14 at 05:58
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@AndréNicolas you mean division by zero? That's only one point) – Somnium Jul 26 '14 at 06:00
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One thing I notice is that all of the nice cases (accepting $1/x$ for the moment) are all Mobius transforms which happen to be their own inverse. – Semiclassical Jul 26 '14 at 06:04
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functional square root: http://en.wikipedia.org/wiki/Functional_square_root – user136920 Jul 26 '14 at 06:25
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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.
Micah
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This answer will only deal with continuous maps.
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 $h(x) = x + \frac{C}{2}$.
Michael Albanese
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$f(x) = c-x$
$f(x) = c/x$
$f(x) = \frac{c_{1}-x}{c_{2}x+1}$
$f(x) = \frac{1}{2}\left ( \sqrt[]{{c_{1}^2}+c_{2}-4x^2} \right )+ c_{1}x$
$f(x) = \sqrt[3]{c-x^3}$
Shabbeh
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1The last one can be generalized (will include first one): $f(x)=\sqrt[a]{b-x^a}$ – Somnium Jul 26 '14 at 06:47
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Any function whose graphic is symmetric against the line y=x.
Anixx
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1And therefore the implicit form of the function should be invariant when swapping coordinates, $F(x,y)=F(y,x)=0$. The examples in the question are the simplest possible choices $F_1(x,y) = x+y-C=0$ and $F_2(x,y) = x\cdot y-C=0$ – DenDenDo Aug 06 '14 at 22:59
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-1: flippant, incomplete. Your answer does not show that the class with this symmetry has any members other than those in the question – Charles Stewart Jan 09 '15 at 19:28
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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 $2$
doraemonpaul
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