Let $I=\{1,2,...,n\}$ for any $n\in \Bbb N$. Assume that $\forall i\in I, \alpha_i\in\Bbb R$.
What is the series expansion for $$p_n(x)=\prod_{i\in I}(x+\alpha_i)$$ I've noticed that $p_n(x)=xp_{n-1}(x)+\alpha_np_{n-1}(x)$, because $p_n(x)=(x+\alpha_n)p_{n-1}(x)$. And I've expanded the first 4, and I can tell there's a pattern, but I can't pin it down. Thanks.
You can expand the full product actually: \begin{align} p_n(x)&=(x+\alpha_1)(x+\alpha_2)...(x+\alpha_n) \\ &= \sum_{J\subset I}x^{n-|J|}\prod_{j\in J} \alpha_j \\ &= \sum_{k=0}^n \beta_k x^k \end{align} Where the first sum is over all subsets of $I$ (including $\emptyset$ and $I$ itself). The coefficients $\beta_k$ can be written as \begin{align} \beta_k = \sum_{J\subset I, |J|=n-k}\prod_{j\in J} \alpha_j \end{align} for example: \begin{align} \beta_n &= 1 \\ \beta_{n-1} &= \sum_{i\in I} \alpha_i \\ \beta_{n-2} &= \sum_{i<j\in I} \alpha_i \alpha_j \\ \beta_{n-3} &= \sum_{i<j<k\in I} \alpha_i \alpha_j \alpha_k \\ &...\\ \beta_0 &= \alpha_1\alpha_2...\alpha_n \end{align}
Is this the kind of series expansion you were looking for?