Polynomial invariant — from product formula to monomial expansion

Question

Context

This question deals with the polynomial invariant denoted by $ H_{n} $ in Maksym Fedorchuk and Igor Pak's 2004 paper Rigidity and polynomial invariants of convex polytopes (sections 7.6 and 9). The paper also introduces the associated polynomial invariant, denoted by $ \tilde H_{n} $, which is the polynomial invariant relating the circumradius and sides of a cyclic polygon, which is also known in the literature as the (generalized) Heron r-polynomial.

Product formula

Let $ n $ be an integer such that $ n \geq3 $. Denote $ \left[ n \right] \equiv \{1,2, \dotsc ,n \} $. The polynomial $ H_{n} \left(x_1, \dotsc ,x_n \right) $ is (supposedly) given by the product $$ \prod {\left( x_{n} - \frac{1}{2} \left(1 - \sum_{\begin{aligned} I \subseteq \left[ n-1 \right] \\ \lvert I \rvert \, \text{even} \, \end{aligned}}{\left( \left( -1 \right)^{\frac{\lvert I \rvert}{2}} \prod_{i \in I}{\kappa_i} \right) \prod_{i \in I}{\sqrt{4x_{i} \left( 1-x_{i} \right)}} \prod_{j \in \left[ n-1\right] \setminus I}{\left( 1-2x_{j}\right)}} \right) \right)} $$ where $ \kappa_{n-1} = 1 $ and the product is taken over all possible ($2^{n-2}$) choices of $$ \kappa_{1}, \dotsc , \kappa_{n-2} \in \{-1,1\}. $$

With some work it's possible to show that this product is a polynomial in $ x_1, \dotsc ,x_n $ with integer coefficients.

This explicit formula is my own and does not appear in the literature as far as I know. Here I do not ask for a proof of this formula for $ H_n $. My question is as follows.

Question

Can one expand the product in such a way which will give "nice"/"concrete" functions for the coefficients of its monomial expansion?

"Some work"

In each factor of the product, the part $$ \frac{1}{2} \left(1 - \sum_{\begin{aligned} I \subseteq \left[ n-1 \right] \\ \lvert I \rvert \, \text{even} \, \end{aligned}}{\left( \left( -1 \right)^{\frac{\lvert I \rvert}{2}} \prod_{i \in I}{\kappa_i} \right) \prod_{i \in I}{\sqrt{4x_{i} \left( 1-x_{i} \right)}} \prod_{j \in \left[ n-1\right] \setminus I}{\left( 1-2x_{j}\right)}} \right) $$ can be written as $$ \frac{1}{2} \left(1 - \prod_{j \in \left[ n-1\right]}{\left( 1-2x_{j}\right)} \right) - \\ \frac{1}{2} \left( \sum_{\begin{aligned} \emptyset \neq I \subseteq \left[ n-1 \right] \\ \lvert I \rvert \, \text{even} \; \; \end{aligned}}{\left(\left( -1 \right)^{\frac{\lvert I \rvert}{2}} \prod_{i \in I}{\kappa_i} \right) \prod_{i \in I}{\sqrt{4x_{i} \left( 1-x_{i} \right)}} \prod_{j \in \left[ n-1\right] \setminus I}{\left( 1-2x_{j}\right)}} \right). $$ Also $$ \begin{aligned} \frac{1}{2} \left(1 - \prod_{j \in \left[ n-1\right]}{\left( 1-2x_{j}\right)} \right) &= \sum_{j \in \left[ n-1\right]}{\left(-2\right)^{j-1} \sum_{\begin{aligned} K \subseteq \left[ n-1 \right] \\ \lvert K \rvert = j \; \; \end{aligned}}{\prod_{k \in K}{x_k}}} \\ &= \sum_{j \in \left[ n-1\right]}{\left(-2\right)^{j-1} \operatorname{e}_{j} \left( x_1, \ldots ,x_{n-1} \right)} \end{aligned} $$ where $ \operatorname{e}_{j} $ is the elementary symmetric polynomial. This is a polynomial in $ x_1, \ldots, x_n $ with integer coefficients.

For any $ \ell \in \left[ n-2 \right] $ the factors of the product can be paired off where in each pair the factors only differ in the choice for $ \kappa_{\ell} $, so only the terms of the sums which have $ \sqrt{4x_{\ell} \left( 1-x_{\ell} \right) }$ as a factor change sign. Now each pair is a difference of two squares, and $ \sqrt{4x_{\ell} \left( 1-x_{\ell} \right) }$ can be factored out from inside one square, so to obtain that the appearance of $ x_{\ell} $ in the expansion depends only on powers of $ 4x_{\ell} \left( 1-x_{\ell} \right) $ and $ 1-2x_{\ell} $. The product is a symmetric function in $ x_1, \ldots, x_{n-1} $ hence the last argument also applies for $ \ell = n-1 $. Thus the product is a polynomial in $ x_1, \dotsc, x_n $. Furthermore, the factor of $ \frac{1}{2} $ is absorbed so the coefficients are integers.

Answer 1

Just to get the ball rolling. Let's introduce $t_j$ such that $1-2x_j=\cos(t_j)$ and $\sqrt{4x_j(1-x_j)}=\sin(t_j)$. Then \begin{split} H_n(x_1,\dots,x_n) &= \prod_{\kappa_1,\dots,\kappa_{n-2}\in\{-1,+1\}}\left( x_n - \frac12\left( 1 - \frac12\prod_{j=1}^{n-1} \big(\cos(t_j) + I\kappa_j\sin(t_j)\big)-\frac12\prod_{j=1}^{n-1} \big(\cos(t_j) - I\kappa_j\sin(t_j)\big)\right)\right) \\ &=\prod_{\kappa_1,\dots,\kappa_{n-2}\in\{-1,+1\}}\left( x_n - \frac12\left( 1 - \cos\big(\sum_{j=1}^{n-1} \kappa_jt_j\big)\right)\right) \\ &=\prod_{\kappa_1,\dots,\kappa_{n-2}\in\{-1,+1\}}\left( x_n - \sin\big(\frac12\sum_{j=1}^{n-1} \kappa_jt_j\big)^2\right) \end{split} By taking logarithm is may be possible to view the corresponding sum as $2^{n-2}$ times the even part of the function in $t_1,\dots,t_{n-2}$.