Question

What's the general form of f(x)?

Answer 1

$$\underbrace{f(f(...f}_{\text{n times}}(x)))= x$$

Answer 2

I think that we have the subsequent equation:
\[\Large {f^n} \equiv I\]
where I is the identity function

Answer 3

I have found a function that isn't the identity function that satisfies this condition, but wasn't sure if it was something perhaps known and I was curious to see how people try to tackle this! :D

Answer 4

please keep in mind that my formula above is a functional writing, keep in mind if we substitute f(x)=sin(x), for example

Answer 5

Ahhh yes ok I see, I misread it and thought you were saying that $$f= I$$ but I see now that that's not what you're saying!

Answer 6

if f(x)= sin(x), i get:
\[\Large \left( {{{\sin }^n}} \right)\left( x \right) = x\]
where:
\[\Large \left( {{{\sin }^n}} \right)\left( x \right) \ne {\left( {\sin x} \right)^n}\]

Answer 7

Here's a hint and little story, while playing around with these things called Rational Tangles (part of Knot Theory), there's a certain function that represents rotating a tangle of two strings by 90 degrees and the fraction it represents gets altered by this function:

$$g(x)=\frac{-1}{x}$$

Which is its own inverse. So this is my hint and also how I guessed at a general solution to the question I asked, but I feel like there is an even more general solution than the one I've found so far.