# Find function $f$ such that $f(\frac{x-3}{x+1})+f(\frac{x+3}{1-x})=x$

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.

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

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

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

• Why the downvote? Commented Jan 6, 2022 at 20:15
• I didn't downvote your answer, but I am having trouble following here. In particular, how did you get the last 2 lines?
– Mike
Commented Jan 6, 2022 at 20:41
• @Mike I used the original F.E. Commented Jan 6, 2022 at 20:44
• 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 $m(m(m(x))) =x.$ Commented Jan 6, 2022 at 23:07
• 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 $g(x)$. Commented Jan 6, 2022 at 23:23

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.

• Your $h(-x)$ might not be the thing you want it to be. BTW you can read the answers/comments posted by others. Commented Jan 6, 2022 at 20:36

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?
• Could you check your equations. It seems to me that the very first sentence is wrong. Commented Jan 6, 2022 at 20:14
• @WhatsUp Should be right now Commented Jan 6, 2022 at 20:16