1. ## functional equation

$solve f(x)+2x=f(f(x))$

2. Solve using some knowledge of "involuntary functions." For example, if for nonzero $x,$ $f(x)=1/x,$ then $f(f(x))=x$ again. Similarly, if $f(x)=-x,$ then $f(f(x))=x$ again. This last one is what we want. So

$f(x)=-x \implies f(x)+2x=x=f(f(x)),$

Q.E.D.

3. Unless you were forced to come up with an answer, I think the word should be "involutary".

By the way, $f(x)=2x$ solves the equation as well. Maybe there are other solutions?

4. Originally Posted by halbard
Unless you were forced to come up with an answer, I think the word should be "involutary".

By the way, $f(x)=2x$ solves the equation as well. Maybe there are other solutions?
Freudian slip, I guess...

I'd be quite curious to know if anyone can come up with other solutions. I would conjecture that they would be significantly more complicated.

5. Just some stuff I've found out about the equation...

Observation 1 : We write: $f^k(x)=f(f(...f(x)))$ ( where f appears k times )

Your equation then turns into: $f^1(x)+2f^{0}(x)=f^2(x)$ very suggestive, isn't it?

This clearly generalises to: $f^{n+1}(x)+2f^{n}(x)=f^{n+2}(x)$ now fix $x$ and set: $a_n=f^n(x)$ solving the recurrence relation: $f^k(x)=A(x)\cdot{(-1)^k}+B(x)\cdot{2^k}$

Setting: $k=0$ we have the relation: $x=A(x)+B(x)$ ( in particular your solutions correspond to the cases $A(x)=0$ and the other $B(x)=0$ )

Now, this implies: $f(x)=-A(x)+2B(x)=-x+3B(x)$ and going to the original equation we get the following restriction $2B(x)=B(3B(x)-x)=B(f(x))$

So it seems it all comes down to solving that functional equation.

Observation 2: $f$ is injective. To see this, suppose $f(x)=f(y)$ then $f(f(x))=f(f(y))$ thus: $f(x)+2x=f(y)+2y$ but then $2x=2y$ and so $x=y$

Observation 3: It follows from 2 that since $f(f(0))=f(0)$ we must have $f(0)=0$