I'm looking for a continuous solution to the functional equation
$$f(2x) = N - \frac{2x}{f(x)^2}$$
where $N$ is a constant natural number and $x \in \mathbb{R}$ is nonnegative. I don't have much experience with functional equations so I haven't tried anything yet. If it helps I'm mostly interested near $x=0$. Any ideas?
This is a simple study of $f(x)$ as $x\to0$.
Let $N>0$.
First case, if $f(0)=0$, then $$\lim_{x\to0}\frac{2x}{f(x)^2}=N\implies f(x)=\sqrt\frac{2x}{N}+o(\sqrt x)$$
Second case, if $f(0)=N$,
Assuming $f(x)=N+ax+o(x)$, then \begin{align} f(x)^2 = \frac{2x}{N-f(2x)}&\implies N^2+o(1)=-\frac{2x}{2ax+o(x)}\\ &\implies a=-N^{-2} \end{align}
Assuming $f(x)=N-N^{-2}x+bx^2+o(x^2)$, then \begin{align} f(x)^2 = \frac{2x}{N-f(2x)}&\implies N^2-2N^{-1}x+o(x)=\frac{1}{N^{-2}-2bx+o(x)}\\ &\implies b=-N^{-5} \end{align}
And so on...