Find $f(x)$ where $ f(x)+f\left(\frac{1-x}x\right)=x$

Question

What function satisfies $ f(x)+f\left(\frac{1-x}x\right)=x$ ?

Answer 1

Let $$p:={1\over2}(\sqrt{5}-1),\quad q:=-{1\over2}(\sqrt{5}+1)$$ be the two fixed points of the Moebius transformation $$M:\quad x\mapsto x'=Mx:={1-x\over x}\ .$$ Introducing a new complex coordinate $y$ by means of $$y:={x-p\over x-q},\quad{\rm resp.}\quad x={p-qy\over 1-y}=:Ty$$ moves these two points to $y=0$ and $y=\infty$. It follows that in the $y$-domain the transformation $M$ appears as a simple scaling $y\mapsto y'=\lambda y\>$; see below.

We are given the functional equation $$f(x)+f(Mx)=x\ .$$ Writing $x=Ty$ here and introducing a dummy $TT^{-1}$ in front of the $M$ we obtain $$f(Ty)+f(TT^{-1}MTy)=Ty\ .\tag{1}$$ As announced above, after some computation it turns out that $$T^{-1}MT y=\lambda y,\quad \lambda:=-{3+\sqrt{5}\over2}\ .$$ Let $g:=f\circ T$ be the expression of $f$ in the new coordinate $y$. Then $(1)$ goes over into $$g(y)+g(\lambda y)=Ty=p+\sqrt{5}(y+y^2+y^3+y^4+\ldots)\quad.\tag{2}$$ Plugging the "Ansatz" $g(y):=\sum_{k=0}^\infty a_k y^k$ into $(2)$ gives $$a_0={\sqrt{5}-1\over 4}, \qquad a_k={\sqrt{5}\over 1+\lambda^k}\quad(k\geq1)\ .$$ It follows that $g$ is analytic at least in a disk of radius $|\lambda|\doteq2.618$ with center $0$ in the $y$-plane. Therefore $$f(x):=g\bigl(T^{-1}x\bigr)$$ is analytic at least in a certain disk with center $p$ in the $x$-plane and satisfies the given functional equation there.