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If $f(x)$ and $g(x)$ are formal power series, i.e. $$f(x)=\sum_{n\ge 0} a_n x^n, g(x)=\sum_{n\ge 0} b_n x^n,$$ and $$f(g(x))=x,$$ can it be proved that always have $$g(f(x))=x?$$ It seems intuitive, then $f(x)$ will be the inverse function of $g(x)$ and vice versa. But I could not prove it.

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    $\begingroup$ In general, it's not true that $fg=\mathrm{id}$ (the identity function) implies that $gf=\mathrm{id}$. For example, $f:\{0\}\to\{0,1\}$ and $g:\{0,1\}\to\{0\}$; it follows that $gf=\mathrm{id}$, but $fg\neq\mathrm{id}$. In this specific context, I don't know. $\endgroup$
    – amrsa
    May 15 at 15:52
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    $\begingroup$ In "formal power series": in general, if $b_0 \ne 0$, then substitution $f(g(x))$ need not even be defined. Example $f(x) = \sum n x^n$ and $g(x) = 2+x$. $\endgroup$
    – GEdgar
    May 15 at 15:59
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    $\begingroup$ If $a_0$ and $b_0$ are supposed to be $0$, then $n \ge 0$ should be replaced with either $n \ge 1$ or $n > 0$. $\endgroup$ May 15 at 16:46

1 Answer 1

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It is true if we assume that the ring of coefficients $R$ is an integral domain. In this case it holds: let $h = \sum b_i X^i,g = \sum a_i X^i \in R[[X]]$ where $g \neq 0$ has no constant term, i.e. $a_0 = 0$.

If $h\circ g = 0$ then $h = 0$.

This can be seen by the formula for the coefficients of the composition: $$c_n = \sum_{k \in \mathbb{N}_0, I \in \mathbb{N}^k} b_k a_I$$ where $a_I = a_{i_1}...a_{i_k}$ for $I \in \mathbb{N}^k$. It follows inductively: if $c_n = 0$ for all $n \in \mathbb{N}_0$ then $b_i = 0$ for all $i \in \mathbb{N}_0$ (because $R$ has no zero divisors).

Now, let $f\circ g = X$. Then $f$ has no constant term and $g \circ f$ is defined. Therefore $(g\circ f) \circ g = g$ thus $(g\circ f - X)\circ g = 0$ and by the above $g\circ f - X = 0$, i.e. $g \circ f = X$.

Probably there is some easy counterexample if $R$ has zero divisors, but I don't have one right now.

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    $\begingroup$ You mean $a_0 = 0$, not $a_0 \ne 0$. $\endgroup$ May 15 at 16:44
  • $\begingroup$ @RobertIsrael yes. $\endgroup$
    – psl2Z
    May 15 at 16:45
  • $\begingroup$ @psl2Z sorry I don’t understand the part about the $c_n$ and $a_I$, with the notations like $N_0$, $N^k$. Do you happen to know some links that I can read that up pls? $\endgroup$
    – athos
    May 17 at 21:53
  • $\begingroup$ @athos $\mathbb{N}$ natural numbers (without $0$) and $\mathbb{N}_0$ natural numbers with $0$. $\mathbb{N}^k$ are $k$-tuples of natural numbers. Basically, you get the formula if you formally plug $g$ in $h$ and multiply all out, like for polynomials. $\endgroup$
    – psl2Z
    May 21 at 14:24

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