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

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    $\begingroup$ Hint: Substitute (conveniently) different expressions of $x$ instead of $x$. You will get a linear system of equations in $f(x)$, $f(\frac{1-x}{x})$, $f(\frac{1}{x-1})$, $f(\frac{x}{x+1})$ ...etc, with the right-hand side being a rational function of $x$. Solve for $f(x)$. $\endgroup$ – OR. Dec 23 '13 at 13:20
  • $\begingroup$ If we assume $f$ is differentiable, perhaps the equation for the derivative is easier: $$ f'(x) - x^{-2} f'\left(\frac{1-x}{x}\right) = 1 $$ $\endgroup$ – gt6989b Dec 23 '13 at 13:25
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    $\begingroup$ @ABC : ok. but how? $\endgroup$ – shahrjar Dec 23 '13 at 13:41
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    $\begingroup$ Are you sure that it is $\frac{1-x}{x}$ and not $\frac{x-1}{x}$? Are you typing the problem statement right? $\endgroup$ – hhsaffar Dec 23 '13 at 20:44
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    $\begingroup$ What are the hypotheses on the function ? Where did you find this problem ? What have you tried ? $\endgroup$ – user37238 Jan 24 '14 at 9:05
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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.

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$f(x)+f\left(\dfrac{1-x}{x}\right)=x$

$f(x)+f\left(\dfrac{1}{x}-1\right)=x$

$\because$ 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

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