[Math] Weights for Gauss-Radau Quadrature

Question

Gauss-Radau-(Legendre) quadrature rules are defined by having one of the abscissa prescribed at $\pm 1$. Thus the abscissa can be expressed as the roots of
$$
\Psi_m(x) = (x_0 – x)r_{m-1}(x).
$$
WLOG we take the prescribed abscissa to be at $x_0 = 1$. The polynomial $r_m$ is defined such that $\Psi_m$ is orthogonal to all polynomials of degree $m-2$ or less. Hence
$$
\int_{-1}^{1} \Psi_m r_{n} dx = \int_{-1}^{1} (1 – x)r_{n}r_m dx = \delta_{ij}.
$$
This requires $r_m(x) = P^{(1,0)}_m(x)$ which are the Jacobi polynomials that are orthogonal with respect to the weight function $(1-x)$. The quadrature weights are given by
$$
\omega_i = \int_{a}^{b} \ell_i(x) dx,
$$
where $\ell_i(x)$ is the $i$th Lagrange basis associated with the abscissa. For $i > 1$ we have
$$
\omega_i = \frac{1}{1 – x_i} \tilde{\omega}_{i-1},
$$
where the $\tilde{w}_i$ are the quadrature weights associated with $P^{(1,0)}_{m-1}(x)$. The weight associated with $x_0$ can be given either by straight integration of the associated basis Lagrange basis function or via
$$
w_0 = 2 – \sum_{i=1}^m \omega_i.
$$
In most references it is stated that
$$
w_0 = \frac{2}{m^2}.
$$
My question is how can this be shown?

Answer 1

This is shown in Hildebrand's book "Introduction to Numerical Analysis". See page 340. How did you get the formula for $\omega_i$ in terms of $\tilde{\omega_{i-1}}$?