Paradoxica
-insert title here-
I'm doubt the validity of my solution, and I'm certain there are massive logical gaps, but here's what I have:Find with proof, all continuous functions f:R->R such that:
f(f(f(x)))=x for all real x.
I'm doubt the validity of my solution, and I'm certain there are massive logical gaps, but here's what I have:Find with proof, all continuous functions f:R->R such that:
f(f(f(x)))=x for all real x.
Yeah, you have done something fishy in deducing that f=f^{-1}, care to explain your reasoning if you still believe this fact after thinking more?I'm doubt the validity of my solution, and I'm certain there are massive logical gaps, but here's what I have:
Composition cannot create parity? Why not? Eg f(x)=1-x is not odd or even, but composed with itself is the identity which is odd.
No discontinuities allowed, this includes singularities of rational functions? for example, 1/(1-x) is clearly a solution to the functional equation, but has a simple pole at x=1Composition cannot create parity? Why not? Eg f(x)=1-x is not odd or even, but composed with itself is the identity which is odd.
It is true that the only polynomial function that works is the identity and yes degree considerations give you this quickly as you note.
This isn't too fruitful a way of thinking about what general continuous functions can solve the equation though.
No, singularities are not allowed, because the function is continuous and has domain R. (We cannot have any domain "holes".)No discontinuities allowed, this includes singularities of rational functions? for example, 1/(1-x) is clearly a solution to the functional equation, but has a simple pole at x=1