$$f(x)=\sum_{n=1}^\infty a_n x^n$$

be a formal power series with no constant term such that $f(f(x))=x$. We find that


so $a_1^2=1$. If $a_1=1$, you need all the other terms to be zero, however if $a_1=-1$ we get a family of nontrivial solutions. Let $a_2=a$, and requiring the higher coefficients of $f(f(x))$ to be zero we can find $a_3=2a^2$, $a_4=-\frac{11}2a^3$, $a_5=\frac{11}2a^4$, $a_6=\frac{105}4a^5$, $a_7=-\frac{279}2 a^6$...

Is there a closed form for these numbers?

  • $\begingroup$ @J.G. Why did you consider that form? $\endgroup$
    – FShrike
    Sep 28, 2022 at 18:57
  • $\begingroup$ @FShrike Because $a_n\propto a^{n-1}$. $\endgroup$
    – J.G.
    Sep 28, 2022 at 18:58
  • $\begingroup$ Apologies, correction: write $f(x)=-x+xg(ax)$ with $(g(x)-1)g(x(g(x)-1))=g(x)$. $\endgroup$
    – J.G.
    Sep 28, 2022 at 19:14
  • $\begingroup$ @Derivative why "$a_3=2a^2$"? and even why "$+a_3a_3)x^3$"? I find $+a_3a_1^3)x^3$ and $a_3=a^2$. $\endgroup$ Sep 28, 2022 at 19:34
  • $\begingroup$ Oups sorry $a_3=-a^2$. $\endgroup$ Sep 28, 2022 at 19:49

4 Answers 4


The claim that the coefficients are unique given the quadratic term is incorrect. Before going into that, here's some background on Anne's answer. We consider the simpler question of how to determine solutions to $f(f(x)) = x$ where $f$ is a Mobius transformation $f(x) = \frac{ax + b}{cx + d}$. Over a field $K$ the group of Mobius transformations is isomorphic to the projective general linear group $PGL_2(K)$, with the isomorphism given by sending a $2 \times 2$ matrix $\left[ \begin{array}{cc} a & b \\ c & d \end{array} \right]$ to $\frac{ax + b}{cx + d}$. So the problem reduces to finding matrices $M$ squaring to a scalar multiple of the identity.

So write $M^2 = c$ for some scalar $c$, so that the eigenvalues of $M$ are $\pm \sqrt{c}$ (over an algebraically closed field, say $\mathbb{C}$). Since we only need to work up to scale we might as well divide $M$ by $c$, or equivalently assume WLOG that $c = 1$, so $M^2 = 1$ and $M$ has eigenvalues $\pm 1$. If $M$ is conjugate to a nontrivial Jordan block then it squares to another nontrivial Jordan block so can't square to $1$; hence $M$ is diagonalizable. If $M$ has eigenvalues $\{ 1, 1 \}$ or $\{ -1, -1 \}$ then it's a scalar multiple of the identity; otherwise its eigenvalues are $\{ 1, -1 \}$. This implies that $\text{tr}(M) = 0$ and $\det(M) = -1$, so $M$ has the form $\left[ \begin{array}{cc} a & b \\ c & -a \end{array} \right]$ where $\det(M) = -a^2 - bc = -1$. This gives $bc = 1 - a^2$.

Now we add the constraint that as a formal power series $f(x)$ should have no constant term. This means $f(0) = 0$ which gives $b = 0$. Then $a = \pm 1$ and we can take $a = 1$ WLOG, which gives $M = \left[ \begin{array}{cc} 1 & 0 \\ c & -1 \end{array} \right]$, hence

$$f(x) = \frac{x}{cx - 1} = -x - cx^2 - c^2 x^3 - \dots $$

as in Anne's answer (with $a = -c$).

On the other hand Klaus's answer implies that the resulting power series cannot be unique, since we can conjugate by an invertible (with respect to composition) formal power series. With a little more effort (showing that we can even arrange for the first $n$ coefficients of this series past $x$ to vanish) we can show that the first $n$ coefficients never uniquely determine the others. So you did something funny with your calculations but I'm not sure what.

Generally, it's known that every formal power series of finite order (with respect to composition) is conjugate to $f(x) = \zeta x$ for $\zeta$ some root of unity (above we have $\zeta = -1$, and in general we probably need to work over an algebraically closed field). This implies:

Classification: If $f(x)$ is a formal power series satisfying $f(0) = 0$ and $f(f(x)) = x$, then either $f(x) = x$ or $\boxed{ f(x) = g(-g^{-1}(x)) }$ where $g(x) = x + \dots$.

Given $g(x)$, the coefficients of $g^{-1}(x)$ can be computed using Lagrange inversion. This gives solutions depending on an infinite number of parameters, namely the higher coefficients $g_i$ of $g(x)$, and the first $n$ coefficients of $f$ depend only on the first $n$ coefficients of $g$. For example, if $g(x) = x + g_2 x^2 + \dots$ then $g^{-1}(x) = x - g_2 x^2 + \dots$ which gives $f(x) = -x + 2 g_2 x^2 + \dots$.

To give a relatively explicit example, take $g(x) = x - ax^2$. Then $g^{-1}(x) = \frac{1 - \sqrt{1 - 4ax}}{2a}$ by the quadratic formula (we have to take the minus sign so that $g^{-1}(0) = 0$); this is a version of the generating function of the Catalan numbers. If $y - ay^2 = x$ then $g(-y) = -y - ay^2 = x - 2y$, which gives

$$\boxed{ \begin{align*} f(x) &= x - \frac{1 - \sqrt{1 - 4ax}}{a} \\ &= -x - 2ax^2 - 4a^2 x^3 - 10a^3 x^4 - \dots \\ &= -x - \sum_{n=2}^{\infty} \frac{2}{n} {2n-2 \choose n-1} a^{n-1} x^n. \end{align*} }$$


Warning: this answer is wrong. See Qiaochu Yuan's comment and answer.

The coefficients are uniquely determined recursively, hence they are those of the involutive series $$-\frac x{1+ax}=-x+ax^2-a^2x^3+a^3x^4-\dots.$$

  • 3
    $\begingroup$ This is a valid solution but the coefficients are not uniquely determined recursively and there are many others; see my answer. $\endgroup$ Sep 28, 2022 at 21:22

A result of Marshall Cohen posted in the arxiv answers the question. Their result is:

If $n$ is a positive integer and $\omega$ is a primitive $n$-th root of unity in the field $\mathbb F$, then for every infinite sequence $\{a_n\}_{1<k\neq nj+1}$ of elements of $\mathbb F$ there exists a unique sequence $\{a_{nj+1}\}_{n\in\mathbb N}$ such that the formal power series

$$f(z)=\omega z+\sum_{h=2}^\infty a_h z^h$$

has order $n$ as a power series over $\mathbb F$.

The question asked for the case $n=2$ and $\mathbb F=\mathbb C$, and made a specific choice for the $\{a_n\}_{1<k\neq nj+1}$.

  • $\begingroup$ The OP only made a specific choice for $a_2$; this result, specialized to this case, says that one needs to make a specific choice for $a_{2k}$ for all $k$. $\endgroup$ Sep 29, 2022 at 17:00

I don't think your approach will work. If $f$ is subject to $f \circ f = \mathrm{id}$, then every bijective map $g$ induces another solution $h := g^{-1} \circ f \circ g$: $$h \circ h = g^{-1} \circ f \circ g \circ g^{-1} \circ f \circ g = g^{-1} \circ f \circ f \circ g = g^{-1} \circ g = \mathrm{id}.$$


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