How to apply Multinomial Theorem here?

Question

I know that if $I_1, \dots, I_n$ are finite index sets and for every $i \in \bigcup_{j=1}^n I_j$ there exists $v_i \in \mathbb{R}$, then holds

$$ \prod_{j=1}^n \sum_{i \in I_j} v_i = \sum_{(i_1, \dots, i_n) \in I_1 \times \dots, \times I_n} \prod_{j=1}^n v_{i_j}. $$

I am trying to use this statement to prove the following claim: Let $N \in \mathbb{N}$, $x \in \mathbb{R}^n$. Then holds

$$ \sum_{v \in \big\{0, \frac{1}{N}, \frac{2}{N}, \dots, 1\big\}^n} \prod_{j=1}^r (1 - N|x_j - v_j|) = \prod_{j=1}^n \sum_{i=0}^N (1 - N|x_j - i/N|), $$

but I'm unsure where to start or how to translate the notation. I don't think that I can take $I_1, \dots, I_n := \{0, \frac{1}{N}, \dots, 1\}$, since in this case, we can't define $v_i$ meaningfully.

Answer 1

Given finite index sets $I_1,I_2,\ldots,I_n$ we can apply the generalised distributive law \begin{align*} \prod_{j=1}^n \sum_{i \in I_j} v_{i_j} = \sum_{(i_1, \dots, i_n) \in I_1 \times \dots, \times I_n} \prod_{j=1}^n v_{i_j}\tag{1} \end{align*}

We start with the right-hand side and obtain \begin{align*} \color{blue}{\prod_{j=1}^n}&\color{blue}{\sum_{i=0}^N\left(1-N\left|x_j-\frac{i}{N}\right|\right)}\\ &=\prod_{j=1}^n\sum_{q\in\{0,1,\ldots,n\}}\left(1-N\left|x_j-\frac{q}{N}\right|\right)\tag{2}\\ &=\prod_{j=1}^n\sum_{v_{j}\in\left\{0,\frac{1}{N},\ldots,\frac{n}{N}\right\}}\left(1-N\left|x_j-v_{j}\right|\right)\tag{3}\\ &\,\,\color{blue}{=\sum_{\left(v_1,\ldots,v_n\right)\in\left\{0,\frac{1}{N},\ldots,\frac{n}{N}\right\}^n} \prod_{j=1}^n\left(1-N\left|x_j-v_{j}\right|\right)}\tag{4} \end{align*} according to the claim.

Comment:

• In (2) we just use a different representation of the index set of the inner sum and do not longer necessarily use the summation order $0\leq i\leq n$.

• In (3) we introduce $v_j=\frac{q}{N}$ as new index variable for the inner sum. Note we use the sub-index $j$ in $v_j$ to indicate the relationship with the scope of the index variable $j$.

• In (4) we apply the generalised distributive law (1) and write $(v_j)_{1\leq j\leq n}=\left(v_1,\ldots,v_n\right)$.