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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

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  • $\begingroup$ You need to substitute for all x not only one x $\endgroup$ Commented May 4, 2017 at 17:28
  • $\begingroup$ Didn't I do that? $\endgroup$ Commented May 4, 2017 at 17:29
  • $\begingroup$ Your second bracket seems incorrect $\endgroup$ Commented May 4, 2017 at 17:34
  • $\begingroup$ By substituiting $y=1-\frac 1x$ you don't get $$f(y)+f\left( \frac{1}{1-y}\right)=\frac{1}{1-y}$$ $\endgroup$ Commented May 4, 2017 at 17:37
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    $\begingroup$ Possible duplicate of Find $f$ if $ f(x)+f\left(\frac{1}{1-x}\right)=x $ $\endgroup$
    – Hanno
    Commented May 28, 2017 at 7:32

3 Answers 3

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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:

$$\psi_k(x):=f(\phi_k(x))$$

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:

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

enter image description here

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).

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  • $\begingroup$ (+1), very instructive solution. I didn't know about "cross-ratio group". $\endgroup$
    – Arnaldo
    Commented May 4, 2017 at 21:51
  • $\begingroup$ @Jean Marie How to check obtained solution is only solution satisfying given condition? $\endgroup$
    – mathophile
    Commented Feb 2, 2022 at 16:56
  • $\begingroup$ @mathophile I don't understand your question: we have seen that necessarily we are "driven" to this function: it is the only one. See the same process in the solution by Arnaldo. $\endgroup$
    – Jean Marie
    Commented Feb 2, 2022 at 17:56
  • $\begingroup$ @JeanMarie Like in this question math.stackexchange.com/questions/4371595/… . It seemed to be there is only one function at start. But later Reinhard pointed out that there are many such function. $\endgroup$
    – mathophile
    Commented Feb 3, 2022 at 6:35
  • $\begingroup$ @mathophile I understand your concern. In the reference you give, there is an issue whether the function is assumed continuous or not. Here , my solution (and that of Arnaldo) do not use continuity arguments. $\endgroup$
    – Jean Marie
    Commented Feb 3, 2022 at 6:40
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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.

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

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    $\begingroup$ [+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
    Commented May 4, 2017 at 21:23
  • $\begingroup$ Thanks for the comments, @JeanMarie . $\endgroup$
    – Arnaldo
    Commented May 4, 2017 at 21:29
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    $\begingroup$ 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. $\endgroup$ Commented May 20, 2017 at 21:40
  • $\begingroup$ @Frpzzd: Thank you for the comment. I am glad to help. $\endgroup$
    – Arnaldo
    Commented May 20, 2017 at 21:44
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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.

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