# A polynomial implicitization

This is a cross-post from MSE where it did not receive a response.

For integer $$n\geq 2$$, consider a parameterization of the coordinates $$(x_1, x_2, ..., x_{n})$$ in terms of the parameters $$(s_{1},s_{2}, ..., s_{n-1})$$ given by $$x_{j} = \displaystyle\sum_{i=1}^{n-1}(-1)^{i+1}s_{i}^{n-j+1}, \qquad j=1, ..., n.$$ We want to find the implicitization $$p(x_1, x_2, ..., x_{n}) = 0$$ by eliminating $$s_{1},s_{2}, ..., s_{n-1}$$.

For example, when $$n=2$$, the parameterization becomes $$x_{1} = s_{1}^{2},\\ x_{2} = s_{1},$$ and the implicit equation is $$p(x_1,x_2) \equiv x_{2}^{2}-x_{1} = 0$$.

When $$n=3$$, the parameterization becomes $$x_{1} = s_{1}^{3} - s_{2}^{3},\\ x_{2} = s_{1}^{2} - s_{2}^{2},\\ x_{3} = s_{1} - s_{2},$$ and elementary algebra gives the implicit equation $$p(x_1,x_2,x_3) \equiv x_{3}^{4}-4x_{3}x_{1}+3x_{2}^{2} = 0$$.

When $$n=4$$, the parameterization becomes $$x_{1} = s_{1}^{4} - s_{2}^{4} + s_{3}^{4},\\ x_{2} = s_{1}^{3} - s_{2}^{3} + s_{3}^{3},\\ x_{3} = s_{1}^{2} - s_{2}^{2} + s_{3}^{2},\\ x_{4} = s_{1} - s_{2} + s_{3}.$$

Question: How to systematically derive a general formula for $$p(x_1,x_2,...,x_n)$$?

(i) I have tried looking into the Macaulay resultant but it is not clear to me if this helps to get a general formula in terms of $$n$$.
(ii) If we had all plus signs in the $$s$$ monomials, then the parameterization would reduce to power sums, and we could use Newton's identities to get the implicitization. But I am not sure if that can be generalized here.
Let me slightly change your notation so that we have $$p_k(s_1,s_2,s_3,\dots, s_{n-1})=s_1^k-s_2^k+s_3^k+\cdots.$$ and we are interested in the relation between $$p_1,p_2,\dots,p_{n}$$. Let us introduce the sequence $$A_k(s_1,s_2,\dots)$$ via the generating function $$F(t)=\sum_{k\geq 0} A_k t^k=\frac{(1-s_1t)(1-s_3t)\cdots}{(1-s_2t)(1-s_4t)\cdots}.$$ Since the generating functions is a rational function with denominator of degree $$m=\lfloor \frac{n-1}{2}\rfloor$$ then the following Hankel determinant of vanishes $$\det[A_{n-2m+i+j}]_{i,j=0}^{m}=0 \tag{*}$$ (This is a famous relation for the coefficients of a rational function. See this answer for an explanation.) Moreover, it is easy to see that the logarithmic derivative of $$F(t)$$ is the appropriate generating function of the $$p_k$$'s, so that $$F(t)=e^{-\left(\sum_{k\geq 1}\frac{p_k}{k}t^k\right)}$$ which implies that every $$A_k$$ can be written as a polynomial of degree $$k$$ in the $$p$$'s (considering $$p_i$$ of degree i). Substituting each such polynomial expression into $$(*)$$ gives the desired relation. This will be a relation of degree $$(m+1)(n-m)$$, which matches your examples. For $$n=4$$ this produces a relation of degree $$6$$ and for $$n=5$$ a relation of degree $$9$$.