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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.

Antonio Maria Di Mauro
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hamam_Abdallah
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3 Answers3

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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

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

TheBestMagician
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  • Why the downvote? – TheBestMagician Jan 06 '22 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 Jan 06 '22 at 20:41
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    @Mike I used the original F.E. – TheBestMagician Jan 06 '22 at 20:44
  • Thanks for clarifying. I see it now--I think! I do need to look at this more closely in a bit. But, an extra line or two of explanation would probably be helpful to the reader. – Mike Jan 06 '22 at 21:01
  • @TheBestMagician As you stay longer, you'll get used to random downvotes. Just ignore them if the downvoter doesn't leave a comment pointing out an issue in your post. – WhatsUp Jan 06 '22 at 21:03
  • Brilliant! This hinges not only on the fact that the second argument of $f$ is the inverse of the first argument, but that composing the first argument with itself equals the second argument. – John Wayland Bales Jan 06 '22 at 22:01
  • @JohnWaylandBales Yes, its a common technique in these problems. But I just found it by messing around, nothing sophisticated. – TheBestMagician Jan 06 '22 at 22:06
  • I suspect it will apply whenever the graphs of the arguments are not only symmetric wrt $y=x$ but also wrt $y=-x$. – John Wayland Bales Jan 06 '22 at 22:16
  • I didn't say that exactly right, but look at the graphs. – John Wayland Bales Jan 06 '22 at 22:18
  • Well it will certainly apply when they are inverses – TheBestMagician Jan 06 '22 at 22:20
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    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 $m(m(m(x))) =x.$ – Will Jagy Jan 06 '22 at 23:07
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    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 $g(x)$. – Calvin Lin Jan 06 '22 at 23:23
  • Thanks, that's very interesting! – TheBestMagician Jan 06 '22 at 23:44
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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.

Hagen von Eitzen
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    Your $h(-x)$ might not be the thing you want it to be. BTW you can read the answers/comments posted by others. – WhatsUp Jan 06 '22 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?
user1011183
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