Let $F (x)$ be the real-valued function defined for all real $x$ except for $x = 0$ and $x = 1$ and satisfying the functional equation $F (x) + F ((x − 1)/x) = 1 + x$. Find $F (x)$.

This functional equation looks like I could do an inverse substitution. Meaning, let $x = \dfrac{x-1}{x}$ then we have $F \left(\dfrac{x-1}{x} \right)+F\left( \dfrac{1}{x-1}\right) = \dfrac{2x-1}{x}$. Thus, $1+x-F(x) = \dfrac{2x-1}{x}-F\left( \dfrac{1}{x-1}\right)$. I am not sure how to proc

  • $\begingroup$ Such a function satisfies $F(2)=3/4$, $F(-1)=-3/4$, and $F(1/2)=9/4$. $\endgroup$ – StevenClontz Feb 5 '16 at 20:46

Solve the system of questions:

$$ \begin{align} F(x) + F\left(\frac{x-1}{x}\right) &= 1+x \\ F\left(\frac{x-1}{x}\right) + F\left(\frac{1}{1-x}\right) &= \frac{2x-1}{x} \\ F\left(\frac{1}{1-x}\right) + F(x) &= \frac{2-x}{1-x} \end{align} $$

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  • $\begingroup$ What do you get for $F(x)$? $\endgroup$ – Puzzled417 Feb 5 '16 at 21:23
  • $\begingroup$ To be honest, I did not solve the system myself. After my comment above I realized that the function $g(x)=(x-1)/x$ has a cycle length of three: $x\mapsto (x-1)/x \mapsto 1/(1-x) \mapsto x$. Using that trick, I got the above system, which is a routine linear algebra problem, letting $A=F(x)$, $B=F((x-1)/x)$, and $C=F(1/(1-x))$. $\endgroup$ – StevenClontz Feb 7 '16 at 0:33

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