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Problem: Find all continuous real-valued functions $f$ such that $$f(x)=\frac{1}{1+f(\frac{1}{1+f(x)})}.\tag{1}$$

Here $f$ is allowed to be defined only on a subset of $\mathbb{R}$.

The only solutions I found are the constant functions $$f(x)=\frac{1}{\varphi},\quad\text{and}\quad f(x)=-\varphi,\tag{2}$$ where $\varphi$ is the golden ratio. I also proved that the only linear functions $f(x)=a+bx$ satisfying $(1)$ are the two above ones. I am tempting to conjecture that these are the only solutions but I failed to either prove or disprove it.

Patrick
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  • Does $f$ have to be defined for $1/\big(1+f(x)\big)$ whenever it is defined for $x$? Otherwise I can come up with lots of solutions... –  May 13 '15 at 19:37
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    @Patrick Setting $g(x)=\dfrac1{1+f(x)}$ might be a good start: $g \circ g = \dfrac1g-1$. – Olivier Oloa May 13 '15 at 19:40
  • Starting with $y:=\dfrac 1{1+f(x)};$ and some work should give you $f(y)=\dfrac {2y-1}{1-y}$. (Oups Olivier was first here... :-)) – Raymond Manzoni May 13 '15 at 19:44
  • Where did you find this problem? (I'm asking just out of curiosity) – A.P. May 13 '15 at 21:15
  • $f(x)=\frac{1}{1+f(\frac{1}{1+f(x)})}=\frac{1}{1+f(\frac{1}{1+f(\frac{1}{1+f(\frac{1}{1+\ldots})})})}=\frac{1}{1+f(f(x))}$ say $f(x)=a$ then $a=\frac{1}{1+f(a)}$ then $ f(x)=\frac{1}{x}-1$ where is the mistake? – user198613 Jan 16 '16 at 02:03

1 Answers1

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let $ 1+ f(x)= \frac{1}{y}$------[1]

by adding 1 on both side of given equation $$ 1+f(x)=1 + \frac{1}{1+f(\frac{1}{1+f(x)})}$$

by [1] $$\frac{1}{y}=1+ \frac{1}{1+f(y)}$$

solving ahead will get us $f(y)=\frac{2y-1}{1-y}$

vikas meena
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