= Solution
Work with the usual commutative unital <ring> convention. Write $f(T)=a_1T+a_2T^2+\cdots$, where $a_1=f'(0)$ is a <unit>, and seek $g(T)=\sum_{n\ge1}b_nT^n$. Since $g$ has zero constant term, every coefficient of the composition of these <formal power series> involves only finitely many terms.
The linear coefficient forces $b_1=a_1^{-1}$. Suppose $b_1,\ldots,b_{n-1}$ have been chosen. The coefficient of $T^n$ in $f(g(T))$ is
$$
a_1b_n+C_n(a_2,\ldots,a_n;b_1,\ldots,b_{n-1}).
$$
Indeed the term involving $b_n$ can only come from $a_1g$: in $g^j$ with $j\ge2$, using a degree-$n$ term forces total degree at least $n+j-1>n$. Choose $b_n=-a_1^{-1}C_n$. This <recursive construction of a compositional inverse> uniquely gives $f\circ g=T$.
To check the other composition rather than assume it, construct $h(T)=\sum_{n\ge1}c_nT^n$ satisfying $h\circ f=T$. At degree $n$, its new coefficient is $c_na_1^n$ plus terms already determined; $a_1^n$ is a <unit>, so the same recursion works. Composition is associative, as can be checked coefficientwise, or after truncation modulo $T^{N+1}$ for every $N$. Therefore
$$
h=h\circ(f\circ g)=(h\circ f)\circ g=g.
$$
Consequently the <compositional inverse of a formal power series> is unique and satisfies
$$
\boxed{f(g(T))=g(f(T))=T.}
$$
Back to article page
