Find $f(x)$ where $ f(x)+f\left(\frac{1-x}x\right)=x$ What function satisfies $ f(x)+f\left(\frac{1-x}x\right)=x$ ?
 A: 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.
A: $f(x)+f\left(\dfrac{1-x}{x}\right)=x$
$f(x)+f\left(\dfrac{1}{x}-1\right)=x$
$\because$ The general solution of $T(x+1)=\dfrac{1}{T(x)}-1$ is $T(x)=\dfrac{(\sqrt5-1)^{x+1}+\Theta(x)(-\sqrt5-1)^{x+1}}{2(\sqrt5-1)^x+2\Theta(x)(-\sqrt5-1)^x}$ , where $\Theta(x)$ is an arbitrary periodic function with unit period
$\therefore f\left(\dfrac{(\sqrt5-1)^{x+1}+(-\sqrt5-1)^{x+1}}{2(\sqrt5-1)^x+2(-\sqrt5-1)^x}\right)+f\left(\dfrac{2(\sqrt5-1)^x+2(-\sqrt5-1)^x}{(\sqrt5-1)^{x+1}+(-\sqrt5-1)^{x+1}}-1\right)=\dfrac{(\sqrt5-1)^{x+1}+(-\sqrt5-1)^{x+1}}{2(\sqrt5-1)^x+2(-\sqrt5-1)^x}$
$f\left(\dfrac{(\sqrt5-1)^{x+1}+(-\sqrt5-1)^{x+1}}{2(\sqrt5-1)^x+2(-\sqrt5-1)^x}\right)+f\left(\dfrac{(\sqrt5-1)^x(3-\sqrt5)+(-\sqrt5-1)^x(3+\sqrt5)}{(\sqrt5-1)^{x+1}+(-\sqrt5-1)^{x+1}}\right)=\dfrac{(\sqrt5-1)^{x+1}+(-\sqrt5-1)^{x+1}}{2(\sqrt5-1)^x+2(-\sqrt5-1)^x}$
$f\left(\dfrac{(\sqrt5-1)^{x+1}+(-\sqrt5-1)^{x+1}}{2(\sqrt5-1)^x+2(-\sqrt5-1)^x}\right)+f\left(\dfrac{\dfrac{(\sqrt5-1)^x(\sqrt5-1)^2}{2}+\dfrac{(-\sqrt5-1)^x(\sqrt5+1)^2}{2}}{(\sqrt5-1)^{x+1}+(-\sqrt5-1)^{x+1}}\right)=\dfrac{(\sqrt5-1)^{x+1}+(-\sqrt5-1)^{x+1}}{2(\sqrt5-1)^x+2(-\sqrt5-1)^x}$
$f\left(\dfrac{(\sqrt5-1)^{x+1}+(-\sqrt5-1)^{x+1}}{2(\sqrt5-1)^x+2(-\sqrt5-1)^x}\right)+f\left(\dfrac{(\sqrt5-1)^{x+2}+(-\sqrt5-1)^{x+2}}{2(\sqrt5-1)^{x+1}+2(-\sqrt5-1)^{x+1}}\right)=\dfrac{(\sqrt5-1)^{x+1}+(-\sqrt5-1)^{x+1}}{2(\sqrt5-1)^x+2(-\sqrt5-1)^x}$
$f\left(\dfrac{(\sqrt5-1)^{x+1}+(-\sqrt5-1)^{x+1}}{2(\sqrt5-1)^x+2(-\sqrt5-1)^x}\right)=\Theta(x)(-1)^x-\sum\limits_x\dfrac{(\sqrt5-1)^{x+1}+(-\sqrt5-1)^{x+1}}{2(\sqrt5-1)^x+2(-\sqrt5-1)^x}$ , where $\Theta(x)$ is an arbitrary periodic function with unit period
