Lemma. Barycentric cancellation below the top degree [cn-0031]

    Status:

    Statement. For the barycentric weights [cn-0030] on \(n\) distinct nodes, every scalar polynomial \(A\) of degree at most \(n-2\) satisfies

    \[\sum _{p=1}^n w_p A(x_p)=0.\]

    Proof. Lagrange interpolation gives

    \[A(t)=\sum _p A(x_p)w_p\frac {F(t)}{t-x_p}.\]

    Both sides have degree at most \(n-1\) and agree at all \(n\) nodes, so the identity follows from the root bound for polynomials. Each quotient \(F(t)/(t-x_p)\) is monic of degree \(n-1\), whereas the coefficient of that degree in \(A\) is zero. Comparing those coefficients proves the assertion.

    Proof dependencies
    Barycentric weights [cn-0030]
    Provenance
    OpenAI, 23 September 2026 [cn-0042], equation (4.5); reconstructed as a reading account on 9 October 2026. Lean source: NodeMoments.