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Suppose that $$P[x]=\sum_{n=0}^{\infty}a_nx^n$$ is a formal power series. Therefore, the issues of convergence are supposed to be ignored for now.

I'm looking for a direct formula that allows me to establish an identity of the form$$\sum_{n=0}^{\infty}b_nx^n=\left(\sum_{n=0}^{\infty}a_nx^n\right)^2$$

More precisely, I want $b_n$ to be the coefficient of $x^n$ in $P[x]^2$. No recurrence relations for $b_n$ are allowed. Express $b_n$ as a function $a_i$'s.

stressed out
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    That's exactly what the Cauchy product tells you. It gives you a closed form expression for $b_{n}$'s in terms of $a_{n}$'s. No recurrence needed. – eranreches Dec 07 '17 at 23:07
  • @eranreches: Yeah, I was confusing it with the case when someday I wanted to find the inverse of an infinite series and it gave me a recurrence relation. Here, it's pretty straightforward. – stressed out Dec 07 '17 at 23:18

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By definition, the product of two formal power series $\sum_{n=0}^\infty c_nx^n$ and $\sum_{n=0}^\infty d_n x^n$ is the formal power series $\sum_{n=0}^\infty b_nx^n$, where $b_n=\sum_{i=0}^nc_id_{n-i}$. To square your $P[x]$, you just want to multiply two copies of $P[x]$, so you want to take $c_n=d_n=a_n$ for all $n$. This gives $$b_n=\sum_{i=0}^na_ia_{n-i}.$$

Eric Wofsey
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  • Thanks. And I can generalize this to higher powers by double sums, triple sums and so on. Right? Out of curiosity, how does this relate to the identity I wrote in my question? edit: I can see how it relates to that. Thanks. – stressed out Dec 07 '17 at 23:19
  • Yeah, there are similar formulas for higher powers. This formula is related to the one you wrote by just grouping together all the terms with $x^k$ for a given value of $k$. For each $i<j$ such that $i+j=k$, you get terms $a_ia_j$ and $a_ja_i$ in the formula for $b_n$, giving $2a_ia_j$. If $2i=k$, you also get a term $a_ia_i=a_i^2$. – Eric Wofsey Dec 07 '17 at 23:21