# The criteria of commuting two formal power series,

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 $$f(g(x))=g(f(x))$$. ?

• Your lower summation index should be $2$ instead of $1$ since you already have $x$ by itself. May 8 '20 at 19:56
• @Somos, oh, yes,thanks
– Why
May 8 '20 at 20:04
• 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. May 8 '20 at 22:57
• @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 $f(g(x))=g(f(x))$ ?
– Why
May 9 '20 at 6:40
• 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. May 9 '20 at 10:57

$$\sum_{n\ge 1} a_n (\sum_{m\ge 1} b_m x^m)^n=\sum_{n\ge 1} a_n \sum_{r \in \Bbb{Z}_{\ge 1}^n} \prod_{j=1}^n b_{r_j} x^{r_j} = \sum_{l\ge 1} x^l \sum_{n=1}^l a_n \sum_{r \in \Bbb{Z}_{\ge 1}^n, \sum_{j=1}^n r_j = l}\prod_{j=1}^n b_{r_j}$$ 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 $$\sum_{n=1}^l a_n \sum_{r \in \Bbb{Z}_{\ge 1}^n, \sum_{j=1}^n r_j = l}\prod_{j=1}^n b_{r_j}=\sum_{n=1}^l b_n \sum_{r \in \Bbb{Z}_{\ge 1}^n, \sum_{j=1}^n r_j = l}\prod_{j=1}^n a_{r_j}$$
• Thanks. But I think the indices will start from $n=2$ instead of $n=1$.
• No, $a_1=1$, there is no reason to separate $x^1$ May 9 '20 at 12:37