3
$\begingroup$

I'm looking for a method to solve:

$$f(f(x))=x$$

Where $f$ is defined for $x \in R$

So far by inverting both sides I have:

$f(x)=f^{-1}(x)$

Which means that my function should be symmetrical over $y=x$. I may "guess" the functions:

$y=x$

$y=c-x$

However I'm wondering is there a way to solve this without "guessing".

$\endgroup$
4
  • $\begingroup$ $f(x)=c-x$ is also a solution. $\endgroup$
    – Bernard
    Sep 27, 2015 at 15:04
  • 1
    $\begingroup$ @AhmedS.Attaalla It's not - point $(0,1)$ belongs to this line, but $(1,0)$ doesn't. $\endgroup$
    – Wojowu
    Sep 27, 2015 at 15:07
  • 1
    $\begingroup$ Related: math.stackexchange.com/questions/46635/… $\endgroup$
    – MonadBoy
    Sep 27, 2015 at 15:08
  • $\begingroup$ Wow how did I miss that. I was thinking parrellel $\endgroup$ Sep 27, 2015 at 15:08

1 Answer 1

6
$\begingroup$

Functions satisfying this property are known as involutions. See the article for many examples in various fields. You are correct about the symmetry condition.

Edit: I should add that there are unaccountably infinitely many of such functions (even restricting to continuous functions).

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .