TITLE: Functions satisfying $\sum_{n=0}^k(-1)^n\binom{k}{k-n}f^{k-n}(x)=0$.
QUESTION [11 upvotes]: This question was motivated by this one. Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a continuous function. Define $f^n(x)=f\circ f\circ\cdot\cdot\cdot\circ f$, $n$ times, $f^0(x)=x$ and $k\geq 2$ a integer. Suppose that $f$ satisfies the equation: $$\tag{1}\sum_{n=0}^k(-1)^n\binom{k}{k-n}f^{k-n}(x)=0$$ 
Note that for each fixed $a\in\mathbb{R}$, $f_a(x)=x+a$ satisfies $(1)$. Are the only solutions of $(1)$ of the form $f_a$ for some $a$? Also note that $f_a$ is an solution of the equation $$\tag{2}(f-I)^k(x)=a$$
The same question applies here:  Is the only solution of $(2)$ of the form $f_a$? If we let $a$ vary in $(2)$, we get a family of equations. Is that family of equations equivalently to $(1)$?

REPLY [2 votes]: Let $P\in\Bbb R[X]$ be a polynomial of degree${}<k$, such that its polynomial function $p:\Bbb R\to\Bbb R$ is nowhere decreasing (in other words $p'(x)\geq0$ for all $x\in\Bbb R$; this obviously requires $\deg P$ to be odd). Then $p^{-1}$ is well defined and continuous. Define $f:x\mapsto p(p^{-1}(x)+1)$ which is also continuous, and satisfies $f^n(p(x))=p(x+n)$ for all $n$. Now
$$
  \sum_{i=0}^k(-1)^i\binom k{k-i}f^{k-i}(p(x))
 =\sum_{i=0}^k(-1)^i\binom kip(x+k-i)
 =\Delta^k(p)(x+k)
 =0
$$
for all $x\in\Bbb R$, where $\Delta$ is a finite difference operator defined by $\Delta g: x\mapsto g(x)-g(x-1)$, which decreases the degree of polynomial functions, so that $\Delta^k(p)=0$ identically.
Since $p$ is surjective this shows $f$ satisfies your requirement and that the answer to your first question is negative for $k\geq4$. The simplest counterexample is $f:x\mapsto (\sqrt[3]x+1)^3$.