expressing as a polynomial

Question

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.

Answer 1

You can use Vieta's formulas since $P_n(x)$ has roots $r_i=-\frac{1}{a_i}$.

Answer 2

As an example:

$$ (1 + a_1 x) (1 + a_2 x) (1 + a_3 x) (1 + a_4 x) \equiv b_0 + b_1 x + b_2 x^2 + b_3 x^3 + b_4 x^4 $$

if and only if:

$$ \begin{aligned} & b_0 = 1 \\ & b_1 = a_1 + a_2 + a_3 + a_4 \\ & b_2 = a_1 a_2 + a_1 a_3 + a_1 a_4 + a_2 a_3 + a_2 a_4 + a_3 a_4 \\ & b_3 = a_1 a_2 a_3 + a_1 a_2 a_4 + a_1 a_3 a_4 + a_2 a_3 a_4 \\ & b_4 = a_1 a_2 a_3 a_4 \\ \end{aligned} $$

where we note combinations without repetition of $4$ objects of class $k=0,1,2,3,4$.

Answer 3

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 $b_k$.