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On my spare time, I'm trying to solve equations of the form $$f^{\circ n}(x)=f(x)^k,\quad n,k\in \mathbb{Z}$$ where $f^{\circ n}(x)=f\circ f\circ\dots\circ f$, $n$ times. I know $f(x)=x^{\sqrt[n-1]{k}}$ is a solution, but I cannot prove if there's a more general solution. How did I get the solution? I assumed that the equation had a solution of the form $x^t$, then I solved for $t$ in : $x^{tk}=f(x)^k = f^{\circ n}(x) = x^{t^n}$

For $n=2$ and any $k$, that is $f\circ f(x) = f(x)^k$, I am able to prove that the only solution is $f(x)=x^k$, without using my general solution.

My problem lies when $n>2$ or $n<0$ (inverse functions).

For instance, for $f\circ f\circ f(x)=f(x)$, I know the solution $f(x)=x^{\pm1}$ works (from my general solution), but I can't prove how to get it without using my general solution, and I can't prove its uniqueness.

Same goes for $n=k=-1$, in other words $f^{-1}(x) = \frac{1}{f(x)}$. I know $f(x)=x^{\pm i}$ is a solution (from my general solution), I can't get there without my general solution and I can't prove the uniqueness.

I tried taking the derivative and solving this way (it worked for $n=2$), but got nowhere.

NoetherianCheese
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  • Would it be simpler to let $y=f(x)$, and solve $f^{\circ (n-1)}(y)=y^k$ – Empy2 Apr 28 '20 at 05:15
  • Yes, that's another way I solved for $n=2$. – NoetherianCheese Apr 28 '20 at 06:55
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    Letting $y=f(x)$, I get $f^{\circ(n-1)}(y)=y^k$. If I apply $f$ on both sides : $f(y^k)=f^{\circ n}(x)=f(y)^k$. Or maybe I'm not headed in the right direction. – NoetherianCheese Apr 28 '20 at 16:26
  • From https://en.wikipedia.org/wiki/Functional_equation#Solving_functional_equations I found that $f(x)=\frac{b-x}{1+cx}$ is a more general solution for the case $n=3$ and $k=1$. Also, $f(x)=\frac ax$ is also a solution not included in the former. This proves that my general solution is not unique or not general enough. – NoetherianCheese Apr 28 '20 at 23:26
  • I also found this answer for the case $n=k=-1$, after substitution $y=f(x)$, therefor $x=f^{\circ -1}(y)$, you get eventually $f(1/y)=1/f(y)$. Then this question is relevant : https://math.stackexchange.com/questions/3371310/how-can-i-find-all-involutions-whose-reciprocals-are-also-involutions – NoetherianCheese Apr 29 '20 at 00:16
  • Also found this : http://eqworld.ipmnet.ru/en/solutions/fe/fe1208.pdf – NoetherianCheese Apr 29 '20 at 00:20

1 Answers1

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Let's try $f^{\circ 3} = f$. If $y = f(x)$, this says $f^{\circ 2}(y) = y$, i.e. $f$ is an involution on its range. There are a lot of solutions, which are all of the following form. I'll assume $f$ is supposed to be defined on domain $R$ (which could be any set).

Partition $R$ in four disjoint sets $A, B, C, D$, where $B$ and $C$ have the same cardinality ($A$ and $D$ could be anything, including the empty set). On $A$, let $f$ be any function into $B \cup C \cup D$. On $B$, let $f$ be any bijection from $B$ onto $C$. On $C$, let $f$ be the inverse of the restriction of $f$ to $B$. On $D$, let $f$ be the identity map.

Note that your $f(x) = x$ is the case $D = \mathbb R$, $A=B=C = \emptyset$. Your $f(x) = x^{-1}$ is not defined at $0$, but you could get a solution where $f(0)$ is any real value except $0$: you could then take $A = \{0\}$, $B = (-\infty, -1) \cup (1,\infty)$ and $C = (-1,0) \cup (0,1)$, and $D = \{-1,1\}$.

Robert Israel
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