1. An ordinary linear system
For p(x) ∝ exp(−Σⱼ λⱼxʲ), the score is s(x) = −Σⱼ jλⱼxʲ⁻¹. The normalizing constant disappears after differentiation. With j,k ∈ {1,2,3,4}, the score-matching system is Aⱼₖ = jkmⱼ₊ₖ₋₂ and dⱼ = j(j−1)mⱼ₋₂. We write the right-hand side as d to distinguish it from the prior’s shape parameter b. A stable, scaled pivoted solve returns λ. Paper Eqs. (8)–(9).