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

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?

• 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. – Jyrki Lahtonen Mar 28 at 17:21
• 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? – Jyrki Lahtonen Mar 28 at 17:29
• If $x=\frac{-1\pm \sqrt{5}}{2}$ then $\frac{1-x}{x}=x.$ So in those cases, $$f\left(\frac{-1\pm \sqrt{5}}{2}\right)=\frac{3\mp\sqrt{5}}{4}.$$ – Thomas Andrews Mar 28 at 17:31
• This Find $f(x)$ where $f(x)+f\left(\frac{1-x}x\right)=x$ seems to be more closely related – Sil Mar 28 at 19:13
• 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... – Sil Mar 29 at 18:08

## 2 Answers

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 $$\frac{x-1}{x}$$.

• 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 – Sil Mar 29 at 18:11

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

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

$$\because$$ 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