I would like to find all functions $f:\mathbb{R}\backslash\{0,1\}\rightarrow\mathbb{R}$ such that

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

I do not know how to solve the problem. Can someone explain how to solve it?

In one of my attempts I did the following, which is confusing to me: By the substitution $y=1-\frac{1}{x}$ one gets

$f(y)+f\left( \frac{1}{1-y}\right)=\frac{1}{1-y}$. So with $x=y$ it follows that $0=x-\frac{1}{1-x}$. So it would follow that there is no solution. Is that possible or is there a mistake?

Best regards

  • $\begingroup$ You need to substitute for all x not only one x $\endgroup$ – Archis Welankar May 4 '17 at 17:28
  • $\begingroup$ Didn't I do that? $\endgroup$ – Sammyy Delbrin May 4 '17 at 17:29
  • $\begingroup$ Your second bracket seems incorrect $\endgroup$ – Archis Welankar May 4 '17 at 17:34
  • $\begingroup$ By substituiting $y=1-\frac 1x$ you don't get $$f(y)+f\left( \frac{1}{1-y}\right)=\frac{1}{1-y}$$ $\endgroup$ – Jaideep Khare May 4 '17 at 17:37
  • 2
    $\begingroup$ Possible duplicate of Find $f$ if $ f(x)+f\left(\frac{1}{1-x}\right)=x $ $\endgroup$ – Hanno May 28 '17 at 7:32

make $x:= \frac{1}{1-x}$ then

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

do it again in the last equation:

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

now make $(1)-(2)$ and get:

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

Subtract the equation is the statement and this last one.

$$2f(x)=x-\frac{1}{1-x}+1-\frac{1}{x}\to f(x)=\frac12\left(x-\frac{1}{1-x}+1-\frac{1}{x}\right)$$

| cite | improve this answer | |
  • 1
    $\begingroup$ [+1] I am amazed at the way you have found your way with a machete toward the hidden unique solution. My answer, which in fact parallels yours, is guided by a group attached to the functional equation. $\endgroup$ – Jean Marie May 4 '17 at 21:23
  • $\begingroup$ Thanks for the comments, @JeanMarie . $\endgroup$ – Arnaldo May 4 '17 at 21:29
  • 1
    $\begingroup$ Wow... what a beautiful and unique solution. This will definitely help me in the future when I am working with functional equations. I think I'm going to give you a +50 bounty for this one. $\endgroup$ – Franklin Pezzuti Dyer May 20 '17 at 21:40
  • $\begingroup$ @Frpzzd: Thank you for the comment. I am glad to help. $\endgroup$ – Arnaldo May 20 '17 at 21:44

I would like to shed some light on this issue by taking a more abstract point of view.

In my answer to this recent question : (How to solve an equation of the form $f(x)=f(a)$ for a fixed real a.), I used the following group of functions (with the algebraic meaning of the word "group")

$$\begin{cases}\phi_1(x)=x, & \ \ \ \ \phi_2(x)=1-x, & \ \ \ \ \ \phi_3(x)=\tfrac{1}{x},\\ \phi_4(x)=1-\tfrac{1}{x}, & \ \ \ \ \phi_5(x)=\tfrac{1}{1-x}, & \ \ \ \ \ \phi_6(x)=\tfrac{x}{x-1}.\end{cases}$$

Here also, the presence of this group is natural because it provides all the potentially fruitful changes of variables leading ultimately to the solution.

Let us take the following notation:


Thus, the given functional equation can be written:

$$\tag{1} f(x)+f(\phi_5(x))=x \ \ \ \iff \ \ \ \color{red}{f(x)+\psi_5(x)=x},$$

Substitution $x \to \phi_4(x)$ in (1) gives:

$$\tag{2}f(\phi_4(x))+f(\underbrace{\phi_5(\phi_4(x))}_{\phi_1(x)=x})=\phi_4(x) \ \iff \ \color{red}{\psi_4(x)+f(x)=1-\tfrac{1}{x}},$$

Substitution $x \to \phi_5(x)$ in (1) gives:

$$\tag{3}f(\phi_5(x))+f(\underbrace{\phi_5(\phi_5(x))}_{\phi_4(x)})=\phi_5(x) \ \iff \ \color{red}{\psi_5(x)+\psi_4(x)=\tfrac{1}{1-x}}.$$

It suffices now to make the following combination of equations (1)+(2)-(3) (the parts in red) to obtain:


Remark: the group of functions $\phi_k$ has been recognized by Kummer in the mid-nineteenth century in connection with hypergeometric differential equations. See p. 306 of (http://www.springer.com/la/book/9781461457244), a fascinating book about the rise of complex function theory.

This group has also an interest in projective geometry; for this reason, it is sometimes called the "cross-ratio group". For a modern presentation of the projective invariant called the cross-ratio, take a look for example at (http://www.maths.gla.ac.uk/wws/cabripages/klein/pinvariant.html).

| cite | improve this answer | |
  • $\begingroup$ (+1), very instructive solution. I didn't know about "cross-ratio group". $\endgroup$ – Arnaldo May 4 '17 at 21:51

By replacing $x$ with $\frac{1}{1-x}$ and $\frac{x-1}{x}$ sequentially, you obtain a system of 3 equations. Then, you can get the solution.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.