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Let $f = x+\sum_{n=2}^{\infty} a_n x^n$ and $g = x+\sum_{k=2}^{\infty} b_n x^n$ be two formal series without constant term. Then $$ f \circ g(x) =x+ \sum_{k=2}^{\infty} b_k x^k+\sum_{n=2}^{\infty} a_n (x+\sum_{k=2}^{\infty} b_k x^k)^n, $$ Similarly, $$ g \circ f(x) =x+ \sum_{n=2}^{\infty} a_n x^n+\sum_{k=2}^{\infty} b_k (x+\sum_{n=2}^{\infty} a_n x^n)^k. $$ Now, we have $$ \left(\sum_{k=1}^{\infty} b_k x^k \right)^n = \sum_{i_1 + \dotsm + i_k = n} b_{i_1} \dotsm b_{i_k} x^n, \ i_k \geq 1. $$ I want to get the conditions when $f(g(x))=g(f(x))$ i.e., when the two power series commutes each other. Then, $$x+ \sum_{k=2}^{\infty} b_k x^k+\sum_{n=2}^{\infty} a_n (x+\sum_{k=2}^{\infty} b_k x^k)^n=x+ \sum_{n=2}^{\infty} a_n x^n+\sum_{k=2}^{\infty} b_k (x+\sum_{n=2}^{\infty} a_n x^n)^k,$$ i.e., $$\sum_{k=2}^{\infty} b_k x^k+\sum_{n=2}^{\infty} a_n (x+\sum_{k=2}^{\infty} b_k x^k)^n=\sum_{n=2}^{\infty} a_n x^n+\sum_{k=2}^{\infty} b_k (x+\sum_{n=2}^{\infty} a_n x^n)^k$$, i.e, $$\sum_{k=2}^{\infty} b_k x^k+\sum_{n=2}^{\infty} a_n x^n+\cdots+\color{blue}{\sum_{n=2}^{\infty} a_n (\sum_{k=2}^{\infty}b_kx^k)^n}=\sum_{n=2}^{\infty} a_n x^n+\sum_{k=2}^{\infty} b_k x^k+\cdots+\color{blue}{\sum_{k=2}^{\infty} b_k(\sum_{n=2}^{\infty}a_nx^n)^k}. $$ The first two terms of both sides cancels out and comparing the last term of both sides i.e, coefficients of $x^{kn}$, we get $$ a_n \left(\sum_{i_1 + \dotsm + i_k = n} b_{i_1} \dotsm b_{i_k}\right) = b_n \left(\sum_{i_1 + \dotsm + i_k = n} a_{i_1} \dotsm a_{i_k}\right),\ i_k \geq 2. $$ But how to deal with middle $\cdots$ terms ??

That is, what is criteria so that $f(g(x))=g(f(x))$. ?

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    $\begingroup$ Your lower summation index should be $2$ instead of $1$ since you already have $x$ by itself. $\endgroup$
    – Somos
    May 8 '20 at 19:56
  • $\begingroup$ @Somos, oh, yes,thanks $\endgroup$
    – Why
    May 8 '20 at 20:04
  • $\begingroup$ I don't understand your question "But how to deal with middle ... terms??" What are you trying to do here? Please explain what you want to do. $\endgroup$
    – Somos
    May 8 '20 at 22:57
  • $\begingroup$ @Somos, I want to find the criteria when $f(g(x))=g(f(x))$. I arranged the equality but having problem to compare the coefficients of $x$ from both sides. As you in the last equation,I compared the coefficient of $x^{kn}$. What about the other coefficients so that $f(g(x))=g(f(x))$ ? $\endgroup$
    – Why
    May 9 '20 at 6:40
  • $\begingroup$ When you equate the terms of $f(g(x))=g(f(x))$ you get, for example, the coefficient of $x^4$ giving $a_2^2b_2+a_2b_3=b_2^2a_2+b_2a_3$. You get similar equations for the higher coefficients. $\endgroup$
    – Somos
    May 9 '20 at 10:57
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No idea why you got stuck.

$$\sum_{n\ge 1} a_n (\sum_{m\ge 1} b_m x^m)^n=\sum_{n\ge 1} a_n \sum_{r \in \Bbb{Z}_{\ge 1}^n} \prod_{j=1}^n b_{r_j} x^{r_j} = \sum_{l\ge 1} x^l \sum_{n=1}^l a_n \sum_{r \in \Bbb{Z}_{\ge 1}^n, \sum_{j=1}^n r_j = l}\prod_{j=1}^n b_{r_j} $$ and $\sum_{n\ge 1} a_n (\sum_{m\ge 1} b_m x^m)^n=\sum_{n\ge 1} b_n (\sum_{m\ge 1} a_m x^m)^n$ iff $$\sum_{n=1}^l a_n \sum_{r \in \Bbb{Z}_{\ge 1}^n, \sum_{j=1}^n r_j = l}\prod_{j=1}^n b_{r_j}=\sum_{n=1}^l b_n \sum_{r \in \Bbb{Z}_{\ge 1}^n, \sum_{j=1}^n r_j = l}\prod_{j=1}^n a_{r_j}$$

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  • $\begingroup$ Thanks. But I think the indices will start from $n=2$ instead of $n=1$. $\endgroup$
    – Why
    May 9 '20 at 12:35
  • $\begingroup$ No, $a_1=1$, there is no reason to separate $x^1$ $\endgroup$
    – reuns
    May 9 '20 at 12:37
  • $\begingroup$ ok, It is clear now $\endgroup$
    – Why
    May 9 '20 at 12:37

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