# Functions satisfying $f(x)+f(\frac{1}{1-x})=x$ with $x\in\mathbb{R}\setminus\{0,1\}.$ [duplicate]

I have used this identity: if $g(x)=1/(1-x),$ then $$g^{-1}(x)=1-\frac{1}{x},$$ to get all functions satisfying: $f(x)+f(\frac{1}{1-x})=x$ with $x\in\mathbb{R}\setminus\{0,1\},$ but I didn't get a general form of its solution. My question here is:

Is there any simple method to solve the titled functional equation?

## marked as duplicate by user236182, José Carlos Santos, Community♦Sep 14 at 17:43

• What is your equation? – Dr. Sonnhard Graubner Sep 14 at 13:22
• What do you mean by a function satisfying an expression? Functions, or any type of objects can satisfy formulas, not expressions. E.g., it makes sense to ask what functions satisfy $f(x)+f(1/x)=x^2$. This is a formula. – A. Pongrácz Sep 14 at 13:23
• Thanks , i have fixed it – zeraoulia rafik Sep 14 at 13:23
• Could you please tell me how $\left( \frac{1}{1-x} \right)^{-1}=1-\frac{1}{x}$? – Math Lover Sep 14 at 13:45
• The exponent $-1$ stands for the reverse. If you substitute $1/(1-x)$ and $1-1/x$ in the functional equation and solve three simultaneous equations may we can get the general solution form – zeraoulia rafik Sep 14 at 13:48

Let $S:=\mathbb{R}\setminus\{0,1\}$. Let $g:S\to S$ be defined by $$g(x):=\frac{1}{1-x}\text{ for all }x\in S\,.$$ Prove that $g\circ g\circ g$ is the identity function $\text{id}_S$ on $S$.

Thus, we have $$f(x)+f\big(g(x)\big)=x\,,$$ $$f\big(g(x)\big)+f\big((g\circ g)(x)\big)=g(x)\,,$$ and $$f\big((g\circ g)(x)\big)+f(x)=(g\circ g)(x)\,,$$ for all $x\in S$.

This shows that $$f(x)=\frac{1}{2}\,\Big(x+(g\circ g)(x)-g(x)\Big)\text{ for every }x\in S\,.$$ In other words, $$f(x)=\frac{x^3-x+1}{2x(x-1)}\text{ for all }x\in S\,.$$ In fact, if $h:S\to S$ is arbitrary, then the solution $f:S\to S$ to the functional equation $$f(x)+f\big(g(x)\big)=h(x)\text{ for all }x\in S$$ is $$f(x)=\frac{1}{2}\,\Big(h(x)+(h\circ g\circ g)(x)-(h\circ g)(x)\Big)\text{ for each }x\in S\,.$$ (See, for example, Determine all functions $f$ satisfying the functional relation $f(x)+f\left(\frac{1}{1-x}\right)=\frac{2(1-2x)}{x(1-x)}$.)

• Notice that similar reasoning applies whenever $g^{2k+1} = \text{id}$ for any $k\geqslant 0$. – Fimpellizieri Sep 14 at 14:09

We will prove that $$f(x)=x-\frac{1}{2(1-x)}$$ when $$f(x)+f\left(\frac{1}{1-x}\right)=x$$ At first we get $$f\left(\frac{1}{1-x}\right)+f\left(\frac{1}{1-\frac{1}{1-x}}\right)=\frac{1}{1-x}$$ and we get

$$f\left(\frac{1}{1-x}\right)+f\left(\frac{x-1}{x}\right)=\frac{1}{1-x}$$ Now we get

$$f\left(\frac{x-1}{x}\right)+f\left(\frac{1}{1-\frac{x-1}{x}}\right)=\frac{x-1}{x}$$ so $$f\left(\frac{x-1}{x}\right)+f(x)=\frac{x-1}{x}$$ and we obtain

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

and $$f(x)=\frac{-x^3+x-1}{x}$$

• The equation after "Now weget" is wrong. The right-hand side should be $1-\dfrac{1}{x}$. – Batominovski Sep 14 at 14:09
• Ok i will check it! – Dr. Sonnhard Graubner Sep 14 at 14:11
• You should fix the lines after the correction, though. – Batominovski Sep 14 at 14:23
• Can you show it to me, i have lost it. – Dr. Sonnhard Graubner Sep 14 at 14:27
• The line starting with $x-f(x)+x-f(x)$ is clearly a carried-over mistake. – Batominovski Sep 14 at 14:28