A general formula for $\mathcal{F}_{n} = \prod_{i=1}^n (a_ix + b_iy)$

Question

So I am trying to simplify following product, $$\mathcal{F}_{n} = \prod_{i=1}^n \left(a_ix + b_iy\right)$$ in terms of products and summation. This is what I have come up with so far.

We see that for n=2, we get $$\mathcal{F}_{2} = (a_1x + b_1y) \times (a_2x + b_2y) = a_1a_2x^2 + xy(a_1b_2+a_2b_1) + b_1b_2y^2$$

Likewise for n=3, we get $$\mathcal{F}_{3} = a_1a_2a_3x^3 + x^2y(a_1a_2b_3+a_1a_3b_2+a_2a_3b_1) + xy^2(a_1b_2b_3+a_2b_1b_3+a_3b_1b_2) + b_1b_2b_3y^3$$

So the first and last term of $\mathcal{F}_{n}$ can be generalized to $\prod_{i=1}^n a_ix^n$ and $\prod_{i=1}^n b_iy^n$. Thus $\mathcal{F}_{n}$, can be given as, $$\mathcal{F}_{n} = \prod_{i=1}^n a_ix^n + \sum_{i=1}^{n-1}\left(x^{n-i}y^i\right)\sum_{k=1}^{\binom{n}{i}}\mathcal{f}(a_{1,2\dots,n}, b_{1,2\dots,n})+ \prod_{i=1}^n b_iy^n$$ So I am stuck here I have come, I need to find this function $\mathcal{f}(a_{1,2\dots,n}, b_{1,2\dots,n})$ which will surely have some product operator inside. I have come up with one solution. Let n=4, and for i=1, term we get is $$x^3y \sum_{k=1}^{\binom{4}{1} = 4} \prod_{m=1\backslash\{k\}}^{n=4}a_mb_k$$

Likewise for n=3, we get

$$xy^3 \sum_{k=1}^{\binom{4}{1} = 4} \prod_{m=1\backslash\{k\}}^{n=4}b_ma_k$$

I have not been able to come something similar for n=2, and then generalize it for all i and n.

Can some help, or provide some useful intuitions

Answer 1

$\def\ed{\stackrel{\text{def}}{=}}$ Will this do? \begin{align} \mathcal{F}_n&=\sum_{v\in\{0,1\}^n}\prod_{i=1}^n\big(a_ix\big)^{v_i}\big(b_iy\big)^{1-v_i}\\ &=\sum_{v\in\{0,1\}^n}x^\overline{v}y^{n-\overline{v}}\prod_{i=1}^na_i^{v_i}b_i^{1-v_i}\\ &=\sum_{k=0}^nx^ky^{n-k}\sum_{v\in W_k}\prod_{i=1}^na_i^{v_i}b_i^{1-v_i} \end{align} where $\ \overline{v}\ed\sum_\limits{i=1}^nv_i\ $ and $\ W_k\ed\big\{\,v\in\{0,1\}^n\,\big|\,\overline{v}=k\,\big\}\ $ is the set of $\ n$-tuples of $\ \{0,1\}^n\ $ of total weight $\ k\ $.

Answer 2

Another common way slightly different from the already given answer is using sets instead of $n$-tupels. We denote with $[n]:=\{1,2,\ldots,n\}$. We obtain \begin{align*} \prod_{i=1}^n\left(a_ix+b_iy\right) &=\sum_{S\subseteq [n]}x^{|S|}\prod_{i\in S}a_iy^{n-|S|}\prod_{i\in[n]\setminus S}b_i\tag{1}\\ &=\sum_{k=0}^nx^ky^{n-k}\sum_{{S\subseteq [n]}\atop{|S|=k}}\prod_{i\in S}a_i\prod_{i\in[n]\setminus S}b_i\tag{2} \end{align*}

Comment:

• In (1) we sum over subsets $S\subseteq [n]$ which contain the indices with selected terms $a_ix$.

• In (2) we rearrange the sums according to the size of $S$.