2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
7
$\begingroup$

Solve the system of questions:

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.