Expressing products of sum as sum of products

Question

Given two sequences $a_k$ and $b_k$, can anybody help me find a closed form for the product $$P=\prod_{k=1}^N (a_k+b_k)$$ as a sum of products of $\{a_k\}$ and $\{b_k\}$? For example, when $N=3$ we have \begin{align*} P&=(a_1+b_1)(a_2+b_2)(a_3+b_3) \\ &=a_1a_2a_3+a_1a_2b_3+a_1b_2a_3+a_1b_2b_3+b_1a_2a_3+b_1b_2a_3+b_1a_2b_3+b_1b_2b_3 \end{align*} which are all the possible triples (with no repetitions) out of a set of six symbols. Is there a convenient way to express this?

Answer 1

You could write it this way : $$ P = \sum_{\underset{\forall k,\ \varepsilon(k) \in \{a_k,b_k\}}{\varepsilon \in \mathcal{F}\left(\mathbb{N}_N,\{a_1,b_1,\ldots,a_n,b_n\}\right)}} \prod_{k=1}^N \varepsilon(k)$$ but I'm not sure this answer is what you're looking for.

Answer 2

One possible way to simplify is to write $$c_k^{(i)} = \begin{cases} a_k & \text{if } i=0 \\ b_k &\text{if } i=1\end{cases}$$ Then you can write $$P = \sum_{A \in \mathcal P(\{1, \dots, N\})} \prod_{k=1}^N c_k^{(\chi_A(k))}$$ where $\chi$ is the characteristic function.

Again, whether an expression like this is useful or not depends on your situation, but the original expression as a product is going to be simpler than any other expression you invent

Answer 3

Let $[N]=\{1,2,\ldots,N\}$ denote the natural numbers starting with $1$ and being less or equal $N$. When multiplying out the $N$ factors in \begin{align*} \prod_{k=1}^N (a_k+b_k) \end{align*} we select for each $k\in[N]$ either $a_k$ or $b_k$. We can equivalently say for each subset $S\subseteq [N]$ we select $a_k$ with $k\in S$ and $b_k$ with $k\in [N]\setminus S$. This can be written in sum notation as \begin{align*} \color{blue}{\prod_{k=1}^N (a_k+b_k)}&=\sum_{S\subseteq [N]}\left(\prod_{k \in S} a_k\right)\left(\prod_{j \in [N]\setminus S} b_j\right)\\ &\,\,\color{blue}{=\sum_{q=0}^{N}\sum_{{S\subseteq [N]}\atop {|S|=q}}\left(\prod_{k \in S} a_k\right)\left(\prod_{j \in [N]\setminus S} b_j\right)} \end{align*} where the summands in the last line are sorted with increasing size of $S$.