# Weights for Gauss-Radau Quadrature

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?

## 1 Answer

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 $\tilde{\omega_{i-1}}$?