# Extracurricular Elementary Mathematics Marathon (1 Viewer)

##### -insert title here-
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:

#### seanieg89

##### Well-Known Member
I'm doubt the validity of my solution, and I'm certain there are massive logical gaps, but here's what I have:

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?

Note also that you have used continuity nowhere. This is essential, as we have a vast array of solutions to the functional equation if continuity is not required:

Partition the reals into an uncountable union of sets, each with either 1 or 3 elements.
Define f to map elements of singleton sets to themselves and to cycle the three elements in each of the other sets.

Any such function f will satisfy the functional equation, but almost all of them will be highly discontinuous.

#### seanieg89

##### Well-Known Member
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.

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.

##### -insert title here-
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.

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 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

#### seanieg89

##### Well-Known Member
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
No, singularities are not allowed, because the function is continuous and has domain R. (We cannot have any domain "holes".)

##### -insert title here-

• kawaiipotato

##### -insert title here-

Last edited:
• KingOfActing