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If $f: \mathbb{R} \to\mathbb{R} $, $x \ne 0,1$

Find all functions $f(x)$ such that $f(x)+f\left(\frac{1-x}{x}\right)=1-x$

My try:

Letting $$g(x)=x+f(x)$$ we get

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

Replacing $x \to 1-x$, we get:

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

any clue here?

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  • $\begingroup$ Also replace $x$ with $x/(1-x)$, $1/(1-x)$ and whatever else comes up. Soon you will get back to what you started with, and can make progress. $\endgroup$ – Jyrki Lahtonen Mar 28 at 17:21
  • $\begingroup$ In other words, my guess (didn't check) is that the tricks used here will work on your problem as well. May be close to a duplicate actually? $\endgroup$ – Jyrki Lahtonen Mar 28 at 17:29
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    $\begingroup$ If $x=\frac{-1\pm \sqrt{5}}{2}$ then $\frac{1-x}{x}=x.$ So in those cases, $$f\left(\frac{-1\pm \sqrt{5}}{2}\right)=\frac{3\mp\sqrt{5}}{4}.$$ $\endgroup$ – Thomas Andrews Mar 28 at 17:31
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    $\begingroup$ This Find $f(x)$ where $ f(x)+f\left(\frac{1-x}x\right)=x$ seems to be more closely related $\endgroup$ – Sil Mar 28 at 19:13
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    $\begingroup$ That explains why it has no nice solutions, original problem did not ask to find all $f$, it only asked about $f(x)=P(x)/Q(x)$ with polynomials of degree $3$ and $2$ respectively. This is important information that should have been included in original post... $\endgroup$ – Sil Mar 29 at 18:08
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Chasing that idea in the comments: $f(x)+f\left(\frac{1-x}{x}\right) = 1-x$, then $f\left(\frac{1-x}{x}\right) + f\left(\frac{2x-1}{1-x}\right) = \frac{2x-1}{x}$, then $f\left(\frac{2x-1}{1-x}\right) + f\left(\frac{2-3x}{2x-1}\right) = \frac{2-3x}{1-x}$ and so on. It doesn't loop back.

In fact, there are only two points for which the sequence we get from iterating $T(x) =\frac{1-x}{x}$ ever repeats at all; its two fixed points $\frac{-1\pm\sqrt{5}}{2}$. We have $T^n(x)=\frac{F_n-F_{n+1}x}{F_nx -F_{n-1}}$ where $F_n$ is the Fibonacci sequence; solving $T^n(x)=x$, we get the quadratic equation $F_nx^2 -F_{n-1}x = F_n-F_{n+1}x$, which simplifies to $F_n(x^2+x-1)=0$.

Because of this behavior, we can construct infinitely many very badly behaved functions that satisfy the functional equation. Choose some arbitrary $x$, choose $f(x)$ arbitrarily, define $f(Tx) = 1-x-f(x)$, define $f(T^{-1}x)=1-T^{-1}x-f(x)$, and keep iterating in both directions. Repeat this process on new values of $x$ until everything is filled in.

There's one place that demands special treatment - the sequence $$\dots,\frac58,\frac35,\frac23,\frac12,1,0,\infty,-1,-2,-\frac32,-\frac53,-\frac85,\dots$$ Here, we can't iterate in both directions, because of that $\infty$. Instead, choose an arbitrary value for $f(-1)$ and iterate forward only, and an arbitrary value for $f(1)$ and iterate backward only. We do need that value of $f(1)$ to make sense of the functional equation $f\left(\frac12\right)+f(1)=\frac12$ at $\frac12$; this problem doesn't lend itself to neatly cutting off the domain of $f$ around the bad values.

That key sequence can be used to cut $\mathbb{R}$ into regions that each only appear once in each sequence of iterates of $T$. Our choices can be condensed to a choice of an arbitrary function on $[1,\infty)$ and a choice of an arbitrary value at $-1$.

This problem simply doesn't have a nice answer. It's quite likely that the question is in error, and the transformation $T$ was meant to be something of finite order like $\frac{x-1}{x}$.

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    $\begingroup$ OP posted source of the problem in comments, original problem actual asked about very specific form to $f(x)$ - that should explain why it does not have nice solution $\endgroup$ – Sil Mar 29 at 18:11
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Similar to Find $f(x)$ where $ f(x)+f\left(\frac{1-x}x\right)=x$:

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

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

$\because$ The general solution of $T(x+1)=\dfrac{1}{T(x)}-1$ is $T(x)=\dfrac{(\sqrt5-1)^{x+1}+\Theta(x)(-\sqrt5-1)^{x+1}}{2(\sqrt5-1)^x+2\Theta(x)(-\sqrt5-1)^x}$ , where $\Theta(x)$ is an arbitrary periodic function with unit period

$\therefore f\left(\dfrac{(\sqrt5-1)^{x+1}+(-\sqrt5-1)^{x+1}}{2(\sqrt5-1)^x+2(-\sqrt5-1)^x}\right)+f\left(\dfrac{2(\sqrt5-1)^x+2(-\sqrt5-1)^x}{(\sqrt5-1)^{x+1}+(-\sqrt5-1)^{x+1}}-1\right)=1-\dfrac{(\sqrt5-1)^{x+1}+(-\sqrt5-1)^{x+1}}{2(\sqrt5-1)^x+2(-\sqrt5-1)^x}$

$f\left(\dfrac{(\sqrt5-1)^{x+1}+(-\sqrt5-1)^{x+1}}{2(\sqrt5-1)^x+2(-\sqrt5-1)^x}\right)+f\left(\dfrac{(\sqrt5-1)^x(3-\sqrt5)+(-\sqrt5-1)^x(3+\sqrt5)}{(\sqrt5-1)^{x+1}+(-\sqrt5-1)^{x+1}}\right)=\dfrac{2(\sqrt5-1)^x+2(-\sqrt5-1)^x-(\sqrt5-1)^x(\sqrt5-1)-(-\sqrt5-1)^x(-\sqrt5-1)}{2(\sqrt5-1)^x+2(-\sqrt5-1)^x}$

$f\left(\dfrac{(\sqrt5-1)^{x+1}+(-\sqrt5-1)^{x+1}}{2(\sqrt5-1)^x+2(-\sqrt5-1)^x}\right)+f\left(\dfrac{\dfrac{(\sqrt5-1)^x(\sqrt5-1)^2}{2}+\dfrac{(-\sqrt5-1)^x(\sqrt5+1)^2}{2}}{(\sqrt5-1)^{x+1}+(-\sqrt5-1)^{x+1}}\right)=\dfrac{(\sqrt5-1)^x(3-\sqrt5)+(-\sqrt5-1)^x(3+\sqrt5)}{2(\sqrt5-1)^x+2(-\sqrt5-1)^x}$

$f\left(\dfrac{(\sqrt5-1)^{x+1}+(-\sqrt5-1)^{x+1}}{2(\sqrt5-1)^x+2(-\sqrt5-1)^x}\right)+f\left(\dfrac{(\sqrt5-1)^{x+2}+(-\sqrt5-1)^{x+2}}{2(\sqrt5-1)^{x+1}+2(-\sqrt5-1)^{x+1}}\right)=\dfrac{(\sqrt5-1)^x(3-\sqrt5)+(-\sqrt5-1)^x(3+\sqrt5)}{2(\sqrt5-1)^x+2(-\sqrt5-1)^x}$

$f\left(\dfrac{(\sqrt5-1)^{x+1}+(-\sqrt5-1)^{x+1}}{2(\sqrt5-1)^x+2(-\sqrt5-1)^x}\right)=\Theta(x)(-1)^x+\sum\limits_x\dfrac{(\sqrt5-1)^x(3-\sqrt5)+(-\sqrt5-1)^x(3+\sqrt5)}{2(\sqrt5-1)^x+2(-\sqrt5-1)^x}$

, where $\Theta(x)$ is an arbitrary periodic function with unit period

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