Linear map $f(x^n)=nf(x)$.

Question

I want to find a linear map $f$ with the property $f(x^n)=nf(x)$.

Suppose we have the recurrence $x_n+\frac{1}{x_n}=x_{n+1}$ with $x_1=a$. If such a map exists then $0=f(x_{n+1})$. So $f$ attains roots at $x_2,x_3,....$.

Does $\{x_2,x_3,....\}$ account for all real numbers if we are allowed to vary $a \in \mathbb{R}$?

Answer 1

Let $x_0\neq 0$ be a number so that $f(x_0)=0$. Then for any $c\in \Bbb R$, we have $f(cx_0)=cf(x_0)=0$.

Answer 2

we have $f(1^n)=nf(1)$ for all positive $n$, so $f(1)=0$. We also have $0=f((-1)^2)=2f(-1)$ and so $f(-1)=0$.

Since the function is linear and has two roots we have $f(x)=0$ for all $x$.