# Differentiable functions satisfying $f'(f(x))=f(f'(x))$

I am wondering whether or not there is a reasonable characterization of differentiable functions $f: \mathbb{R}\to \mathbb{R}$ such that $f'(f(x))=f(f'(x))$ for each $x\in\mathbb{R}$. (Or, if you like the composition sign, $f'\circ f=f\circ f'$).

I could only come up with trivial examples of such functions: $f(x)=0$ and $f(x)=e^{x}$.

This reminds me of a recent Putnam problem (2010), which asked whether or not there exists a strictly increasing function $f:\mathbb{R}\to\mathbb{R}$ satisfying $f'(x)=f(f(x))$. (The answer is: No).

Note: I see that a question of similar type has been asked here.

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$f(x) = x$ is also a solution since $f'(x) = 1$. –  Cameron Williams Jun 2 '13 at 2:58
@CameronWilliams: Right. :) –  Prism Jun 2 '13 at 3:06
Actually there is a larger class of functions than just $f(x) = x$ amongst monomial solutions. Let $n\in\mathbb{N}$, then $f(x) = n^{-n+1}x^n$ is a solution to the functional equation. –  Cameron Williams Jun 2 '13 at 3:10
As for functions of the form $ax^2 + b$, the only nontrivial solution is $a = \frac{1}{2}, b = 1$. As for functions of the form $ax^2 + bx + c$ (with $c\neq 0$), there are no solutions. There are also no solutions for $ax^n + b$ when $n>2$. I think some binomial theorem or (abstract) algebraic arguments can be used to state what kinds of restrictions are needed for polynomial functions. –  Cameron Williams Jun 2 '13 at 3:47
$f(x) = \frac{1}{x}$ is another solution. In fact, we can adapt the solution from before to get that $f(x) = (-1)^{n+1}n^{n+1}x^{-n}$ is also a solution. –  Cameron Williams Jun 2 '13 at 4:07