# Expanding a product of linear combinations with coefficients $1$ and $-1$

For any odd natural number $$n$$, denote $$t \equiv \frac{n-1}{2}$$. Let $$K$$ be a field such that $$\operatorname{char} K \neq 2$$. Working over the polynomial ring $$K\left[x_1,x_2,...,x_{n} \right]$$, denote by $$\Pi_n$$ the product of all possible linear combinations of the indeterminants $$x_1,x_2,...,x_{n}$$, where each linear combination has $$i$$ coefficients which are $$-1$$ and its remaining $$n-i$$ coefficients are $$1$$, and $$0 \le i \le t$$.

## Problem

Find an expression for the coefficient of the monomial $$\prod_{j=1}^{n}x_{j}^{p_j}$$ in the expansion of the aforementioned product $$\Pi_n$$.

## Examples of $$\Pi_n$$

• For $$n=1$$ obtain that $$t=0$$ and that $$i \in \{0\}$$; then $$\Pi_1$$ is $$x_1$$
• For $$n=3$$ obtain that $$t=1$$ and that $$i \in \{0,1\}$$; then $$\Pi_3$$ is $$\left(x_1+x_2+x_3 \right)\left(-x_1+x_2+x_3 \right)\left(x_1-x_2+x_3 \right)\left(x_1+x_2-x_3 \right)$$
• For $$n=5$$ obtain that $$t=2$$ and that $$i \in \{0,1,2\}$$; then $$\Pi_5$$ is \begin{align} & \left(x_1+x_2+x_3+x_4+x_5 \right)\\ & \left(-x_1+x_2+x_3+x_4+x_5 \right)\left(x_1-x_2+x_3+x_4+x_5 \right)\left(x_1+x_2-x_3+x_4+x_5 \right)\\& \left(x_1+x_2+x_3-x_4+x_5 \right)\left(x_1+x_2+x_3+x_4-x_5 \right)\\ &\left(-x_1-x_2+x_3+x_4+x_5 \right)\left(-x_1+x_2-x_3+x_4+x_5 \right)\left(-x_1+x_2+x_3-x_4+x_5 \right)\\& \left(-x_1+x_2+x_3+x_4-x_5 \right)\left(x_1-x_2-x_3+x_4+x_5 \right)\left(x_1-x_2+x_3-x_4+x_5 \right)\\& \left(x_1-x_2+x_3+x_4-x_5 \right)\left(x_1+x_2-x_3-x_4+x_5 \right)\left(x_1+x_2-x_3+x_4-x_5 \right)\\& \left(x_1+x_2+x_3-x_4-x_5 \right)\\ \end{align}

## Quick observations

• In general, $$\Pi_n$$ has $$\binom{n}{i}=\binom{2t+1}{i}$$ factors which correspond to each $$0 \le i \le t$$; hence the total number of factors is $$\sum_{i=0}^{\frac{n-1}{2}} \binom{n}{i}=\sum_{i=0}^{t} \binom{2t+1}{i}=2^{2t+1-1}=2^{2t}=4^{t}$$.
• Any one of the indeterminants $$x_1,x_2,...,x_{n}$$ has coeffient $$-1$$ in the linear combinations appearing in exactly $$\sum_{j=0}^{\frac{n-1}{2}-1} \binom{n-1}{j}=\sum_{j=0}^{t-1} \binom{2t}{j}=2^{2t-1}-\frac{1}{2}\binom{2t}{t}$$ of the factors of $$\Pi_n$$.
• Any $$k$$ of the indeterminants $$x_1,x_2,...,x_{n}$$ all have coeffient $$-1$$ in the linear combinations appearing in exactly $$\sum_{j=0}^{\frac{n-1}{2}-k} \binom{n-k}{j}=\sum_{j=0}^{t-k} \binom{2t+1-k}{j}$$ of the factors of $$\Pi_n$$.

It seems that Inclusion-Exclusion is in order to solve this problem, as there is a matter of choice of "how many minus signs" each of the indeterminants $$x_1,x_2,...,x_{n}$$ "gets".

• Further observations: $\Pi_n^2 = \prod (\pm x_1 \pm x_2 \pm \cdots)$. The sum of the coefficients can be found by setting all the variables to $1$, giving $\prod_{i=0}^t (n-2i) \binom n i$ – Peter Taylor Oct 4 '19 at 6:45