Are the iterates of the cosine linearly independent?

Solution 1:

If you can use the fact that $\cos(x)$ and its iterates are entire functions of a complex variable, you can use the following idea (I use your notations): We proceed by inductioon, the case $n=1$ is obvious.

Let $n\geq 2$, and suppose that $a_1f_1(x)+a_2f_2(x)+\cdots+a_nf_n(x)=0$ for all $x\in \mathbb{R}$. Then this imply that $g(z)=a_1z+a_2f_1(z)+\cdots+a_nf_{n-1}(z)$ is zero for all $z\in [-1,1]$ (because $g(\cos(x))=0$, we have put $z=\cos(x)$). As $g$ is entire, this imply that $g(z)=0$ for all $z\in \mathbb{C}$, and that $a_1z$ is periodic with period $2\pi$. Hence $a_1=0$. Now , putting $b_1=a_2,...$ etc, we have $b_1f_1(x)+\cdots+b_{n-1}f_{n-1}(x)=0$ for all $x$. The induction hypothesis apply, and we are done. .