22
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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 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
  • 1
    $\begingroup$ Can you give the source of this problem? $\endgroup$ – user261263 Jul 27 '16 at 19:35
14
+50
$\begingroup$

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 $|x|>2(M+6)$. $\square$

$\endgroup$
  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ 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
3
$\begingroup$

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 $$\left[f\left(1-\frac{5}{x+1}\right)+f\left(\frac{5}{1+x}-1\right)\right]+\left[f(x)+f(-x)\right]=1-\frac{5}{x+1}-x\implies \\$$

$$\boxed{\left[f(y)+f(-y)\right]+\left[f(x)+f(-x)\right]=y-x}$$

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.

$\endgroup$
0
$\begingroup$

Hint:

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

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

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

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

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

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

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

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

$V(x+2)+2V(x+1)+5V(x)=0$

$V(x)=\theta_1(x)(-1+2i)^x+\theta_2(x)(-1-2i)^x$ , 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 $f(2-2\tan((\tan^{-1}2)(x+1)))+f(2-2\tan((\tan^{-1}2)x))=2-2\tan((\tan^{-1}2)x)$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.