Is there a neat way to express the polynomial $P_n(x)=(1+a_1x)(1+a_2x)\cdots(1+a_nx)$ as $P_n(x)=b_0+b_1x+b_2x+\cdots+b_nx^n$ ? I want to have the coefficients $b_i$ as functions of $a_i$. I do not want to use the McLaurin's expansion. Thanks.
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3The elementary symmetric polynomials of the $-1/a_k$ will give you the $b_j$ (up to sign and up to a multiplicative factor). – Anne Bauval Mar 23 '23 at 18:20
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2Does this answer your question? Relation betwen coefficients and roots of a polynomial – Anne Bauval Mar 23 '23 at 23:24
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We can multiply out the factors and obtain using $[n]:=\{1,2,\ldots,n\}$ \begin{align*} \prod_{j=1}^{n}\left(1+a_jx\right) &=\sum_{S\subseteq [n]}\left(\prod_{j\in S}a_j\right)x^{|S|}\\ &=\sum_{k=0}^{n}\left(\color{blue}{\sum_{{S\subseteq [n]}\atop{|S|=k}}\prod_{j\in S}a_j}\right)x^k =\sum_{k=0}^n\color{blue}{b_k}x^k \end{align*}
From each of the $n$ factors we select either $a_j$ or $1$. So, for each subset $S\subseteq [n]$ we have the product of $|S|$ factors with terms $a_j, j\in S$. In the last step we rearrange the summands according to increasing $|S|=k, 0\leq k \leq n$ which yields the wanted $b_k$.
Markus Scheuer
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