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I am looking for functions $ f:\Bbb R \to \Bbb R $ satisfying $$f\Big(\frac{x-3}{x+1}\Big)+f\Big(\frac{x+3}{1-x}\Big)=x$$

I used the substitution $ x=\cos(2t) $ for $ x\in (0,2\pi) $, to use the identities $$1+x=2\cos^2(t) \text{ and } 1-x=2\sin^2(t)$$

The new equation will be $$f\Big(1-\frac{2}{\cos^2(t)}\Big)+f\Big(\frac{2}{\sin^2(t)}-1\Big)=$$ $$\cos^2(t)-\sin^2(t)$$

I think that this approach won't allow me the get the answer. Any idea will be appreciated.

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  • $\begingroup$ Hint: Let $g(x) = (x-3)/(x+1)$. Find $g^{-1} (x)$. $\endgroup$
    – Calvin Lin
    Commented Jan 6, 2022 at 20:06
  • $\begingroup$ @CalvinLin $ g^{-1}(x)=\frac{x+3}{1-x} $ $\endgroup$ Commented Jan 6, 2022 at 20:09
  • $\begingroup$ Your functional equation should hold only for $x \neq \pm1$. $\endgroup$
    – WhatsUp
    Commented Jan 6, 2022 at 20:11
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    $\begingroup$ Further hints in white font (highlight to view): 1) $\color{white} {\text{Find }g^2(x)} $ OR 2) $\color{white} {\text{Find } f(2)} $ $\endgroup$
    – Calvin Lin
    Commented Jan 6, 2022 at 20:14
  • $\begingroup$ @CalvinLin the thing that makes it go is your $g(g(g(x))) = x$ $\endgroup$
    – Will Jagy
    Commented Jan 6, 2022 at 23:10

3 Answers 3

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Substitute in $x\mapsto\frac{x-3}{x+1}$. Then we have $$f\left(\frac{x+3}{1-x}\right)+f\left(x\right)=\frac{x-3}{x+1}$$ Now substitute again $x\mapsto \frac{x-3}{x+1}$,

$$f(x)+f\left(\frac{x-3}{x+1}\right)=\frac{x+3}{1-x}$$

Now add these two to get

$$2f(x)+f\left(\frac{x+3}{1-x}\right)+f\left(\frac{x-3}{x+1}\right)=\frac{x-3}{x+1}+\frac{x+3}{1-x}$$ $$2f(x)=\frac{x-3}{x+1}+\frac{x+3}{1-x}-x$$ $$f(x)=\frac{1}{2}\left(\frac{x-3}{x+1}+\frac{x+3}{1-x}-x\right)$$

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  • $\begingroup$ Why the downvote? $\endgroup$ Commented Jan 6, 2022 at 20:15
  • $\begingroup$ I didn't downvote your answer, but I am having trouble following here. In particular, how did you get the last 2 lines? $\endgroup$
    – Mike
    Commented Jan 6, 2022 at 20:41
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    $\begingroup$ @Mike I used the original F.E. $\endgroup$ Commented Jan 6, 2022 at 20:44
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    $\begingroup$ the Mobius transformation $m(x) = \frac{x-3}{x+1}$ cubes to the identity transformation, which is why the initial lines work out nicely. $f(m(x)) + f(m^2(x) ) = m^3 (x) $ where $m^2(x) $ means $m(m(x))$ and $m^3(x) $ means $m(m(m(x))),$ while $m(m(m(x))) =x.$ $\endgroup$
    – Will Jagy
    Commented Jan 6, 2022 at 23:07
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    $\begingroup$ As @WillJagy mentioned above, the important thing is $g^3(x) = x$. Once we demonstrate the function is periodic (for a given $x$), we're just solving the system of linear equations (which has a unique solution for an odd period, infinitely many solutions for an even period). $\quad$ If the function is non-periodic (for a given starting value), then we have an infinite number of solutions by setting $f(x) = a, f(g(x)) = b$ and iterating accordingly (EG $f( g^{-1} (x) ) = x - b$). $\quad$ In particular, for "nice" olympiad questions, you're almost guaranteed to have a periodic $g(x)$. $\endgroup$
    – Calvin Lin
    Commented Jan 6, 2022 at 23:23
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Let $h(x)=f(\frac{x-3}{x+1})$. Then the functional equation becomes $$\tag1 h(x)+h(-x)=x\qquad\text{for }x\ne\pm1.$$ There are obviously many such $h$. In fact, if $k\colon(0,\infty)\setminus\{1\}\to\Bbb R$ is arbitrary, we can set $$h(x)=\begin{cases}k(x)&x>0\\x-k(x)&x<0\\0&x=0\end{cases} $$ and obtain a solution for $(1)$. As the inverse of $x\mapsto \frac{x-3}{x+1}$ is (also) $x\mapsto \frac{x+3}{1-x}$, we obtain a solution $$ f(x)=h(\tfrac{x+3}{1-x})$$ of the original functional equation.

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    $\begingroup$ Your $h(-x)$ might not be the thing you want it to be. BTW you can read the answers/comments posted by others. $\endgroup$
    – WhatsUp
    Commented Jan 6, 2022 at 20:36
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Set $u= \frac{x-3}{x+1}$ the equation becomes $f(u)+f(\frac{u+3}{1-u})=\frac{u+3}{1-u}$

Set $v=\frac{x+3}{1-x}$ the equation becomes $f(\frac{v-3}{v+1})+f(v)=\frac{v-3}{v+1}$

Now replace all dummy variables with $y$ and solve for $f(x)$,

  1. $$A+B=y$$

2.$$B+C=\frac{y+3}{1-y}$$

  1. $$A+C=\frac{y-3}{y+1}$$ Can you finish it?
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    $\begingroup$ Could you check your equations. It seems to me that the very first sentence is wrong. $\endgroup$
    – WhatsUp
    Commented Jan 6, 2022 at 20:14
  • $\begingroup$ @WhatsUp Should be right now $\endgroup$ Commented Jan 6, 2022 at 20:16
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    $\begingroup$ Your answer could be improved with additional supporting information. Please edit to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Commented Jan 6, 2022 at 20:33

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