I found the following task in a book and I would be interested if someone has an idea to solve it:

Find all the functions $f$ that satisfy $f(\frac{x+4}{1-x}) + f(x) = x$.

My ideas:

Assuming that $f$ is a power series or making a substitution.

I tried several approaches, but without success.

  • $\begingroup$ Are you sure there wasn't any other information? For example, I assume this is a real-valued function $\endgroup$ – JasonM Jul 27 '16 at 18:19
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    $\begingroup$ Oh, I' m sorry. The only other Information is that the functions are real-valued and defined for $x \neq \pm 1$. $\endgroup$ – Peter123 Jul 27 '16 at 18:27
  • $\begingroup$ The functions and why wouldn't it (they) be defined for $x=-1$? Are you sure there's not some details you've left out? $\endgroup$ – Henrik supports the community Jul 27 '16 at 19:21
  • $\begingroup$ @Henrik I think it's not defined for $x=-1$ so future manipulations can be made (see my partial answer) $\endgroup$ – JasonM Jul 27 '16 at 19:25
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    $\begingroup$ Can you give the source of this problem? $\endgroup$ – user261263 Jul 27 '16 at 19:35

The equation reads $$\tag1f(g(x))+f(x)=x$$ where $g(z)=\frac{z+4}{-z+1}$ is a Möbius transformation (a nice bijection of $\Bbb C\cup\{\infty\}$ to itself) with fix points $\pm2i$. The Möbius transform $h(z)=\frac{z-2i}{z+2i}$ maps $2i\mapsto 0$ and $-2i\mapsto \infty$ and has an inverse transform $k(z)=\frac{2iz+2i}{-z+1}$. Note that $$G(z):=h(g(k(z))) = \underbrace{\frac{-3+4i}{5}}_{=:\alpha}\cdot z$$ is just a rotation by an irrational (see below) angle. Now $$ f(k(G(z)))+f(h^{-1}(z))=k(z)$$ so that $F(z):=f(k(z))$ (which leads to $f(z)=F(h(z))$) obeys the functional equation $$\tag2F(\alpha z)+F(z)=k(z).$$ If we do not postulate continuity, we only find $F(0)=\frac12k(0)=i$ and $F(\infty)=\frac12k(\infty)=-i$, and beyond that we can pick an arbitrary value per equivalence class of $\Bbb C^\times$ modulo multiplication with $\alpha$.

From $F(1)+F(\alpha)=k(1)=\infty$, we see that at least one of $F(1)$, $F(\alpha)$ must be infinite. As the set $\{\alpha^n\}_{n\in\Bbb N}$ is dense in $S^1$, we conclude that $F(z)=\infty$ for a set that is dense in $S^1$. If $F$ is continuous (as a map of $\Bbb C\cup \{\infty\}$ to itself), this implies $F(z)=\infty$ for all $z$ with $|z|=1$. This means that $f(z)=\infty$ for all $z\in\Bbb R$, so we better consider $F$ only inside or outside the unit circle, $f$ only on the upper or lower half plane.

One may argue that the instances of $(1)$ that involve infinity, i.e., the cases $x=1$ and $x=\infty$, do not apply; this means we split this one orbit into two half-orbits. However, this changes the situation only when we allow non-continuous $f$ in the first place.

If we demand $F$ to be smooth, then we find $\alpha F'(\alpha z)+F'(z)=k'(z)=\frac{4i}{(z-1)^2}$, in particular $F'(0)=\frac{4i}{1+\alpha}$. Next, $\alpha^2 F''(\alpha z)+F''(z)=-8i(z-1)^{-3}$ and more generally $$ \alpha^nF^{(n)}(\alpha z)+F^{(n)}(z)=4in!(1-z)^{-n}$$ and in particular $$ F^ {(n)}(0)=\frac{4in!}{1+\alpha^n}$$ fo $n>0$. Therefore, we make an analytic "attempt" $$ F(z)=i+4i\sum_{n=1}^\infty\frac{z^n}{1+\alpha^n}.$$ The convergence seems to be non-trivial, though, as $\alpha^n+1$ becomes arbitrarily small. (As mentioned in the comments, this is related to the irrationality measure of $\alpha$, so probably ew are now way beyond the level where the original question occured - if we have not left that level a lot earlier).

Remark: That $\alpha$ is an irrational rotation ultimately follows from the number-theoretical fact that $1+2i$ and $1-2i$ are non-associate primes in the ring $\Bbb Z[i]$

Based on the niveau and context of the problem source, it may be more appropriate to show the following simpler

Claim. There is no continuous function $f\colon \Bbb R\to\Bbb R$ such that $(1)$ holds for all $x\in\Bbb R\setminus\{1\}$.

Proof. Assume that $f$ is continuous. Then in particular, $f$ is bounded by some $M$ on the interval $(-3,6)$. Then $f$ is bounded by $M+6$ on $\{\,g(x)\mid -2<x<4,x\ne1\,\}$, i.e., on both $(\tfrac 74,\infty)$ and $(-\infty,-2)$. Hence $f$ is bounded by $M+6$, which gives a contradiction with $(1)$ for $x$ with $|x|>2(M+6)$. $\square$

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    $\begingroup$ Just a small correction; we have $F(0) = k(0)/2 = i$ not $2i$ so your formula for $F(z)$ has the wrong constant term. $\endgroup$ – Kibble Jul 28 '16 at 5:49
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    $\begingroup$ Thank you very much for this solution. This is a great answer, but I think there also exists a simpler solution without Möbius transformation as the book in which I found this Task is about a beginners lecture in calculus. $\endgroup$ – Peter123 Jul 28 '16 at 12:40
  • $\begingroup$ Convergence of the $F$-series seems to be related to the irrationality measure of $\zeta = \frac{1}{2} + \frac{\arctan(3/4)}{\pi}$ ($\alpha = e^{i\pi \zeta}$). If $\zeta$ has a finite irrationality measure (which it’s almost surely has) then the series should converge for $|z| < 1$. $\endgroup$ – Winther Jul 30 '16 at 15:29
  • $\begingroup$ @Winther Thanks $\endgroup$ – Hagen von Eitzen Aug 2 '16 at 6:37
  • $\begingroup$ @Peter123 Well, at a beginner's level it might be sufficient to show: "There does not exist a continuous function $f\colon D\to\Bbb R$ where domain $D\subseteq \Bbb R$ with $\Bbb R\setminus D$ nowhere dense and $f(\tfrac {x+4}{1-x})+f(x)0x$ holds for all $x\ne 1$ with $x,\frac{x+4}{1-x}\in D$" (just to turn the task into an explicit statement). This might be doable without using too much knowledge about Möbius transforms $\endgroup$ – Hagen von Eitzen Aug 2 '16 at 6:43

This was the closest I got. Feel free to point out any mistakes.

We have $$f\left(\frac{x+4}{1-x}\right)+f(x)=x$$

The inside of the first one can be simplified to $\frac{5}{1-x}-1$, so the expression becomes $$f\left(\frac{5}{1-x}-1\right)+f(x)=x\tag{1}$$

Making the transformation $\frac{5}{1-x}-1\rightarrow x$ yields $$f(x)+f\left(1-\frac{5}{x+1}\right)=1-\frac{5}{x+1}\tag{2}$$

Now making the substitution $x\rightarrow -x$ in $(1)$ gives us $$f\left(\frac{5}{1+x}-1\right)+f(-x)=-x$$

Adding this to $(2)$ gives us $$\left[f\left(1-\frac{5}{x+1}\right)+f\left(\frac{5}{1+x}-1\right)\right]+\left[f(x)+f(-x)\right]=1-\frac{5}{x+1}-x\implies \\$$


subject to $y=1-\frac{5}{x+1}$

This looked really good because the lefthand side is the sum of two even functions, while the righthand side is a difference of two odd functions. However, this isn't necessarily a contradiction, since there are two variables. Either way, I think this fully describes the symmetry of the function. I hope this helps someone come to the solution.



Consider $T(x+1)=\dfrac{T(x)+4}{1-T(x)}$ ,

Let $T(x)=U(x)+1$ ,

Then $U(x+1)+1=\dfrac{U(x)+5}{-U(x)}$



Let $U(x)=\dfrac{V(x+1)}{V(x)}$ ,

Then $\dfrac{V(x+2)}{V(x+1)}=-2-\dfrac{5V(x)}{V(x+1)}$



$V(x)=\theta_1(x)(-1+2i)^x+\theta_2(x)(-1-2i)^x$ , where $\theta_1(x)$ and $\theta_2(x)$ are arbitrary periodic functions with unit period

$V(x)=\theta_1(x)e^{x\ln(-1+2i)}+\theta_2(x)e^{x\ln(-1-2i)}$ , where $\theta_1(x)$ and $\theta_2(x)$ are arbitrary periodic functions with unit period

$V(x)=\theta_1(x)e^{\frac{x\ln5}{2}+(\pi-\tan^{-1}2)ix}+\theta_2(x)e^{\frac{x\ln5}{2}-(\pi-\tan^{-1}2)ix}$ , where $\theta_1(x)$ and $\theta_2(x)$ are arbitrary periodic functions with unit period

$V(x)=\Theta_1(x)5^\frac{x}{2}\sin((\tan^{-1}2)x)+\Theta_2(x)5^\frac{x}{2}\cos((\tan^{-1}2)x)$ , where $\Theta_1(x)$ and $\Theta_2(x)$ are arbitrary periodic functions with unit period

$\therefore T(x)=\dfrac{\Theta_1(x+1)5^\frac{x+1}{2}\sin((\tan^{-1}2)(x+1))+\Theta_2(x+1)5^\frac{x+1}{2}\cos((\tan^{-1}2)(x+1))}{\Theta_1(x)5^\frac{x}{2}\sin((\tan^{-1}2)x)+\Theta_2(x)5^\frac{x}{2}\cos((\tan^{-1}2)x)}+1$ , where $\Theta_1(x)$ and $\Theta_2(x)$ are arbitrary periodic functions with unit period

$T(x)=\dfrac{\Theta(x)\sqrt5\sin((\tan^{-1}2)(x+1))+\sqrt5\cos((\tan^{-1}2)(x+1))}{\Theta(x)\sin((\tan^{-1}2)x)+\cos((\tan^{-1}2)x)}+1$ , where $\Theta(x)$ is an arbitrary periodic function with unit period

Hence $f(2-2\tan((\tan^{-1}2)(x+1)))+f(2-2\tan((\tan^{-1}2)x))=2-2\tan((\tan^{-1}2)x)$


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