Firstly $f \circ f = g \circ g$ has many counterexamples such as involutions. Also in case of $f \circ f \circ f = g \circ g \circ g$ with $f, g \in \mathbb{C}[x]$ there is an evident counterexample: $f(x) = x, g(x) = \omega x$ where $\omega$ is any complex cube root of unity. I made the following observations:

  1. $\deg f = \deg g$
  2. The leading term of $f$ and $g$ must be equal.
  3. The term before the leading term has same coefficient. Let $\deg f = \deg g = n$. The $(n^3 - 1)$-degree terms of $f \circ f \circ f$ and $g \circ g \circ g$ can only be produced by multiplying $(n - 1)$ $n^2$-degree terms and one $n^2 - 1$ terms of $f \circ f$ and $g \circ g$, respectively, where the $(n^2 - 1)$-degree terms of $f \circ f$ and $g \circ g$ can be only produced by multiplying $n - 1$ $n$-degree terms and one $(n - 1)$-degree term of $f$ and $g$, respectively. Since the $n^2$-degree terms and $n$-degree terms are equal by 2, $(n - 1)$-degree terms of $f$ and $g$ must be also equal. Namely, let $f(x) = a_nx^n + a_{n - 1}x^{n - 1} + \ldots$ then $f(f(x)) = a_n^{n + 1}x^{n^2} + na_n^na_{n - 1}x^{n^2 - 1} + \ldots$, finally $f(f(f(x))) = a_n^{n^2 + n + 1}x^{n^3} + n^2a_n^{n^2 + n}a_{n - 1}x^{n^3 - 1} + \ldots$ Induction like this may lead to the solution for the problem.
  4. If $a$ is a fixed point, possibly complex, of $f$, then $a$ is a fixed point of $g \circ g \circ g$. Thus the orbit of $a$ with respect to $g$ has a period of $1$ or $3$. Proving that the period is always $1$ will ensure that $f(x) - x$ and $g(x) - x$ have the same roots up to multiplicity of roots.
  5. Since $f'(x)f'(f(x))f'(f(f(x))) = g'(x)g'(g(x))g'(g(g(x)))$ there are some relations between the critical points of $f$ and $g$.
  6. For arbitrary affine polynomials $\lambda$ the conjugates $\lambda \circ f \circ \lambda^{-1}, \lambda \circ g \circ \lambda^{-1}$ also satisfies the condition.
  7. The case when $f, g$ is constant or affine is trivial. For $f, g$ quadratic, it is enough to consider when $f$ and $g$ is monic by 2 and 6. Let $$p = \arg \min f(x), q = \arg \min g(x), r = \min f(x), s = \min g(x)$$ If $p \ne q$ and $r \ne s$, WLOG $r < s$, with conjugation by $\lambda = x \mapsto x - r$ we can assume that $r = 0$. Assume that $p < 0$. Since $g'(q) = 0$ $f'(q)f'(f(q))f'(f(f(q)) = 0$. Then $p = q$ or $f(q) = p$ or $f(f(q)) = p$ but $p \ne q$ so $p \in \operatorname{ran} f$, which is a contradiction. Thus $p \ge 0$. Then $[p, \infty) \subset \operatorname{ran} f$, so $\operatorname{ran} f \supset f(\operatorname{ran} f) \supset f([p, \infty)) = \operatorname{ran} f$. Thus $\operatorname{ran} f \circ f \circ f = f(f(\operatorname{ran} f)) = f(\operatorname{ran} f) = \operatorname{ran} f \supsetneq \operatorname{ran} g \supset \operatorname{ran} g \circ g \circ g$ which is a contradiction. Thus $p = q$ or $r = s$.
    1. If $p = q$, with conjugation by $\lambda = x \mapsto x - p$ we only have to consider when $f(x) = x^2 + a$ and $g(x) = x^2 + b$. But in $((x^2 + a)^2 + a)^2 + a$ the sextic term is $4ax^6$ thus $f = g$.
    2. If $r = s$, $f(x) = (x - p)^2$ and $g(x) = (x - q)^2$. But in $(((x - p)^2 - p)^2 - p)^2$ the septic term is $-8px^7$ thus $f = g$.

Are there any ideas to solve this problem? Or if there is a counterexample, how strong should be the restrictions to $f, g$ so that the statement holds?


Proposition 1. Let $n,m\ge1$. Let $$\begin{align}p(x)&=x^n+a_1x^{n-1}+\ldots+a_n,\\\tilde p(x)&=x^n+\tilde a_1x^{n-1}+\ldots+\tilde a_n,\\ q(x)&=x^m+b_1x^{m-1}+\ldots +b_m,\\\tilde q(x)&=x^m+\tilde b_1x^{m-1}+\ldots +\tilde b_m\end{align}$$ be polonomials with $p\circ q=\tilde p\circ \tilde q$. Then $q-\tilde q$ is constant.

Proof. The behaviour near $\infty$ tells us about the higher coefficients of $q$. More precisely, we have $$\tag1p(q(z))=q(z)^n+ O(z^{nm-m})$$ so that we can read off the highest coefficients of $$q(z)^n=z^{nm}+c_1z^{nm-1}+c_2z^{nm-2}+\ldots$$ where $$c_k=nb_k+(\text{polynomial in }b_1,\ldots, b_{k-1}).$$ Now from $c_1,\ldots, c_{m-1}$, we can determine one by one the coefficients $b_1,\ldots, b_{m-1}$, and of course obtain the same result when we use the same method to obtain $\tilde b_1,\ldots,\tilde b_{m-1}$. $\square$

Note that we cannot expect more since in we can always replace $q(x)$ with $q(x)+\delta$ and $p(x)$ with $p(x-\delta)$.

Proposition 2. In the situation of proposition 1, assume additionally $a_1=\tilde a_1$. Then $q=\tilde q$.

Proof. We can strengthen $(1)$ to $$\tag2 p(q(z))=q(z)^n+a_1q(z)^{n-1}+O(z^{nm-2m}). $$ The coefficient of $z^{nm-m}$ is $$ a_1+nb_m+(\text{polynomial in }b_1,\ldots, b_{m-1})$$ and allows us to also determine $b_m$. $\square$.

Now assume that $f,g$ are univariate polynomials with $(f^{\circ r}=g^{\circ r}$ for some $r\ge 2$. Then from $\deg(f^{\circ r})=(\deg f)^r$ and same for $g$, we conclude $\deg f=\deg g=:m$. If $m=1$, we have $(x+c)^{\circ r}=x+rc$ and so $f=g$ is immediate. Assume therefore that $m\ge2$. Apply proposition 1 to $n:=m^{r-1}$, $p=f^{\circ (r-1)}$, $q=f$, $\tilde p=g^{\circ (r-1)}$, $\tilde q=g$, to find that $f-g=\text{const}$. In particular, we know enough of $f$ to use proposition 2 and conclude $f=g$.

  • $\begingroup$ Assumption that $f, g$ is monic is just for nice equations. Right? Also based on your proof, I think this can be generalized to any iteration with an additional constraint: the sign of the leading coefficients are equal. $\endgroup$ – Ris Feb 9 at 11:26

I don't have a solution (yet), but I can show that the derivatives $f'$ and $g'$ have the same sign.

Suppose $f$ is strictly monotone on an interval $[a, b]$. Then $f \circ f \circ f = g \circ g \circ g$ is also strictly monotone on $[a, b]$. This implies $g \circ g \circ g$ is injective, which in turn implies $g$ must be injective (if $h$ is a left-inverse of $g \circ g \circ g$, then $h \circ h \circ h \circ g \circ g$ is a left-inverse of $g$). Since $g$ is injective, $g$ too must be strictly monotone.

Moreover, it's not difficult to verify that $g$ is monotone increasing if and only if $g \circ g \circ g$ is monotone increasing, i.e. $g$ and $g \circ g \circ g$ have the same monotonicity. The same can therefore be said of $g$ and $f$.

This means that $$f'(x) > 0 \iff g'(x) > 0 \text{ and } f'(x) < 0 \iff g'(x) < 0.$$

EDIT: Actually, this is not quite true. I'm implicitly assuming that the derivative has no repeated roots, and so each root of the derivative is accompanied by a sign change.

  • $\begingroup$ Why is $f\circ f\circ f$ strictly monotone? Are you assuming that $f$ maps $[a,b]$ into $[a,b]$? $\endgroup$ – user1551 Feb 9 at 10:19
  • $\begingroup$ @user1551 Ugh, you're right. $\endgroup$ – user744868 Feb 9 at 10:20

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.