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Suppose that I have a sequence of formal power series $\{f_i(x)\}_{i=1}^n$. I don't know what $n$ is and I want to do this in general for any positive integer $n$. Let $$ f_i(x) = c_0^{(i)} + c_1^{(i)}x + \dots. $$ I need a formula for the coefficients of $\prod_{i=1}^nf_i(x)$. For example, let $(d_0, d_1,\dots)$ be the coefficients of the product. Then, \begin{align*} d_0 &= \prod_{i=1}^nc_0^{(i)},\\ d_1 &= d_0\sum_{i=1}^n\frac{c_1^{(i)}}{c_0^{(i)}} & \text{(assuming $c_0^{(i)} \neq 0$) }. \end{align*} Is there any similar formula for $d_2, d_3,\dots$ that involves only $\sum_{i=1}^n$ and $\prod_{i=1}^n$?

I was able to obtain the formula I needed for the specific problem I was working on using Maple, but I don't know how it did it.

Thank you.

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This is a generalisation of the Cauchy-product of power-series, which can be written as \begin{align*} \prod_{j=1}^2f_j(x)&=\left(\sum_{i_1=0}^\infty a_{i_1}^{(1)}x^{i_1}\right)\left(\sum_{i_2=0}^\infty a_{i_2}^{(2)}x^{i_2}\right)\\ &=\sum_{k=0}^\infty\left(\color{blue}{\sum_{{i_1+i_2=k}\atop{i_k\geq 0,\,k=1,2}}a_{i_1}^{(1)}a_{i_2}^{(2)}}\right)x^k\tag{1}\\ &=\sum_{k=0}^\infty\left(\sum_{i_1=0}^ka_{i_1}^{(1)}a_{k-i_1}^{(2)}\right)x^k \end{align*}

We obtain for general $n$ \begin{align*} \prod_{j=1}^nf_j(x)&=\prod_{j=1}^n\left(\sum_{i_j=0}^\infty a_{i_j}^{(j)}x^{i_j}\right)\\ &=\sum_{k=0}^\infty\left(\sum_{{i_1+i_2+\cdots+i_n=k}\atop{i_k\geq 0,\,k=1,2,\dots,n}}a_{i_1}^{(1)}a_{i_2}^{(2)}\cdots a_{i_n}^{(n)}\right)x^k\tag{2}\\ &=\sum_{k=0}^\infty\left(\color{blue}{\sum_{{i_1+i_2+\cdots+i_n=k}\atop{i_k\geq 0,\,1\leq k\leq n}}\prod_{j=1}^n a_{i_j}^{(j)}}\right)x^k \end{align*}

We see (2) is a generalisation of (1).

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