K time differentiable function Is there any k time differentiable function such that $$f(f'(f''(f'''(......f^{(k)}(x))))=x$$ for all $x$ belongs to $\mathbb R$?
EDIT:- What will the case be when the order of the functions taken are reversed?
 A: There is no such $f$.
Let $J$ be the range of $f' \circ f'' \circ \ldots \circ f^{(k)}$, which is an interval.  Then $f$ must be injective on $J$ and maps it to $\mathbb R$.  But we can't have 
$f(x) \to \pm \infty$ as $x$ approaches any point of $J$ or its closure, because then $f^{(k)}$ wouldn't be defined there.  So $J$ must be the whole real line.
But if $f$ is differentiable and injective on $\mathbb R$, it is either increasing or decreasing, so either $f' \ge 0$ everywhere or $f' \le 0$ everywhere.  But then
$J$ is contained in either $[0,\infty)$ or $(-\infty, 0]$, contradiction!
EDIT: Similarly, consider the order-reversed problem
$$ f^{(k)}(\ldots(f'(f(x)))\ldots) = x$$
Let $J$ be the range of $f^{(k-1)} \circ \ldots \circ f$, again an interval $J$,
which $f^{(k)}$ must map one-to-one onto $\mathbb R$.  Again, this should imply that $J$ is all of $\mathbb R$.  The same agument then implies that the range of
$f^{(k-2)} \circ \ldots \circ f$ is all of $\mathbb R$, and $f^(k-1)$ is one-to-one.  But that would require $f^{(k)}$ to be always $\ge 0$ or always $\le 0$.  
