Given $n \in\mathbb{N}$, find all continuous functions $f : \mathbb{R}\to \mathbb{R}$ such that for all $x\in\mathbb{R},$
$$\sum_{k=0}^{n}\binom{n}{k}f(x^{2^{k}})=0. $$
Setting in the binomial theorem x=1 and y = -1 yields zero on the right side and f= (-1)^k. This staircase function 1,-1,1,-1,1,-1 is a solution but ist not continous.
Setting in the binomial theorem x=1 and y = -1 yields zero on the right side and f= (-1)^k. This staircase function 1,-1,1,-1,1,-1 is a solution but ist not continous.