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$S$ must be positive-definite, not just positive-semidefinite, else $AS^{-1}$ does not exist. Suppose $A$ is positive-definite, and let $A^{1/2}$ be its positive-definite square root. Then the suprem …

The minimum must occur at vectors $x,y$ where $x_i$ and $y_i$ take only two values each. This should make it easy to check Neil Strickland's experimental result (where $x$ and $y$ are indeed of this f …

To answer Question Q2: Yes, as the special case $u_j = \alpha p_j$ of the following result. We use $j,k$ for the indices rather than $i,j$ because we need $i = \sqrt{-1}$. Proposition. For pairwise di …

Q1 The determinant is $\prod_{n=1}^{N-1} (1 - e^{-2\alpha(p_{n+1}-p_n)})$. Q2 Yes, using the answer to Q1. Q3 Yes, using the answer to Q1. The formula for Q1 is proved by induction on $N$. The ba …

Since you know already that the optimal $a_i$ have $2a_i + 1 = x_i = \pm 1$ and the roots of $P'_{n-1}$, the calculation of $V_n$ comes down to the discriminant of $P'_{n-1}$, its leading coefficient, …

Robert Israel proved a lower bound proportional to $n^2 (c/\beta)^{2n}$. I claim an upper bound of much the same shape, $O(n(Bc)^{2n})$, for some constant $B > \beta^{-1}$, namely $\sqrt{\pi e/2} = 2. …