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Let $n$ be an integer greater than $1$. The notation $f^n$ is notoriously ambiguous: it means either the $n$-th iterate of $f$ or its $n$-th power.

I was wondering when the two interpretations are in fact the same. In other words, if we write $f^n(x)$ for $f(f(\dotsb f(x) \dotsb))$ and $f(x)^n$ for $f(x) \cdot f(x) \dotsb f(x)$:

For which functions $f \colon I \to \mathbb R$ is $f^n(x) = f(x)^n$ for all $x \in I$?

So far I have been able to show that:

  • The constant functions $f(x) = 0$ and $f(x) = 1$ satisfy the condition for all $n$ and $f(x) = -1$ satisfies the condition for all odd $n$. Also, if $f$ satisfies the condition for $n$, then the only fixed points of $f$ can be either $0$, $1$, or if $n$ is odd also $-1$.
  • The squaring function $f(x) = x^2$ satisfies the condition for $n = 2$ and is essentially the only non-constant differentiable function to do so. Indeed, $$f(f(x)) = f(x)^2 \implies f'(f(x)) f'(x) = 2 f(x) f'(x)$$ so if we assume $f'(x) \neq 0$ and let $y = f(x)$, we have that $$f'(y) = 2 y \implies f(y) = y^2 + c$$ and furthermore $$(y^2 + c)^2 + c = (y^2 + c)^2 \implies c = 0.$$ (Of course we could then consider also, e.g., $f(x) = x^2$ for $x \ge 0$ and $f(x) = 0$ for $x < 0$.)
  • More generally, $f(x) = x^\sqrt[n-1]{n}$ satisfies the condition for any $n > 2$. These functions are only defined on $\mathbb R^+$: this is why I chose an interval $I$ instead of all $\mathbb R$ as the domain, but I actually don't care if a solution is defined on any other non-trivial subset.

Are there any other solutions? If not, how can we prove so?


Remark: The question can be generalized to $n \in \mathbb Z$ if we assume $f$ to be invertible and denote by $f^{-n}$ either the $n$-th iterate of its compositional inverse or the $n$-th power of its multiplicative inverse (the cases $n = 0, 1$ are trivial). But this seems to be an even harder problem. For the case $n = -1$ see this question.

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    $\begingroup$ Isn't $f^n(x)$ sometimes used for the nth derivative ? $\endgroup$ – herb steinberg Aug 3 '18 at 23:44
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    $\begingroup$ That would usually be with parentheses: $f^{(n)}(x)$. $\endgroup$ – Zachary Aug 4 '18 at 0:59
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    $\begingroup$ I'll mention an identity. If we assume $f$ is invertible, then $f^{n-1}(y)=y^n$. Letting $y=z^p$ and exponentiating both sides by $q$ gives $$[f^{n-1}(z^p)]^q=z^{npq}=z^{nqp}=[f^{n-1}(z^q)]^p$$. $\endgroup$ – user254433 Aug 4 '18 at 1:52
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Something non-constant and discontinuous everywhere:

$$f(x) = \left\{\begin{array}{ll} x^2 & \text{if } x \text{ is an integer} \\ \lceil x \rceil & \text{if } x \text{ is rational but not an integer}\\ \lfloor x \rfloor & \text{if } x \text{ is irrational} \end{array} \right.$$

seems to satisfy $f(f(x))=f(x)\cdot f(x)$

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  • $\begingroup$ should it be $f(x)$ not $f(m)$? $\endgroup$ – user254433 Aug 4 '18 at 0:54
  • $\begingroup$ @user254433 Yes it should - and now does - thank you $\endgroup$ – Henry Aug 4 '18 at 7:26
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    $\begingroup$ I like this example because together with user254433's comment it suggests a way to make a new function $g$ out of a known invertible one $f$. First we need to find a subset $A$ such that $f(A) \subseteq A$. Then we need to find a subset $B$ disjoint from $A$ and a function $g \colon B \to A$. Finally we extend $g$ to $A \cup B$ by letting $g(x) = f(x)$ for $x \in A$. It follows that $g^n(x) = g^{n-1}(g(x)) = f^{n-1}(g(x)) = g(x)^n$ for all $x \in A \cup B$. $\endgroup$ – Luca Bressan Aug 4 '18 at 8:08
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If for all $x$, $f(\cdots f(f(x)))=(f(x))^n$ then $f(\cdots f(y))=(y)^n$, and the function $f$ is a functional $n-1^{th}$ root of the $n^{th}$ power function.

In particular, for $n=2$, $f$ is the square function. For $n=3$, a solution is $f(x)=x^{\sqrt 3}$ and more generally

$$f(x)=x^{\sqrt[n-1]n}.$$

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  • $\begingroup$ I don't see how this provides an answer to the question, since I had already pointed out the solution $f(x) = x^{\sqrt[n-1]{n}}$ in my attempt. $\endgroup$ – Luca Bressan Aug 14 '18 at 11:08
  • $\begingroup$ @LucaBressan: I am turning your question to a more classical one, that of the functional root. $\endgroup$ – Yves Daoust Aug 14 '18 at 11:58
  • $\begingroup$ This only shows that $f^{n-1}(y)=y^n$ for $y\in f(X)$, $X$ the domain of $f$. The relation may not hold on all of $X$, for example the zero function satsfies OP's condition on all $\mathbb R$ but only satisfies your condition on $\{0\}$. $\endgroup$ – Jack M Aug 15 '18 at 20:51

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