7

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?

Umesh shankar
  • 11,335
  • 5
  • 33
  • 73
  • 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 '19 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 '19 at 17:29
  • 1
    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 '19 at 17:31
  • Maybe the given it's $f(x)+f\left(\frac{x-1}{x}\right)=1-x$? – Michael Rozenberg Mar 28 '19 at 17:37
  • @MichaelRozenberg its given $1-x$ only – Umesh shankar Mar 28 '19 at 17:42
  • This also means that if $x=\frac{1}{y+1}$ then $\frac{1-x}{x}=y$ So $$f\left(\frac1{y+1}\right)+f(y)=\frac{y}{y+1}.$$ – Thomas Andrews Mar 28 '19 at 17:46
  • 2
    This Find $f(x)$ where $ f(x)+f\left(\frac{1-x}x\right)=x$ seems to be more closely related – Sil Mar 28 '19 at 19:13
  • By the way what is the source of this problem? Perhaps you wrote it incorrectly from the source? There are variants of this problem that have nice solution, those are of form $f(x)+f(h(x))=g(x)$ where $h^n(x)=h^m(x)$ (i.e. $h(x)$ when $n$ times iterated loops eventually),which is not the case for $h(x)=(1-x)/x$, but is for $1/(1-x)$, $(x-1)/x$, and so on... – Sil Mar 28 '19 at 19:18
  • The source is https://brilliant.org/problems/so-many-compositions/ – Umesh shankar Mar 29 '19 at 06:48
  • 2
    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 '19 at 18:08

2 Answers2

4

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

jmerry
  • 19,943
  • 2
    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 '19 at 18:11
1

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

doraemonpaul
  • 16,488