why are all the weights of the Gaussian quadrature formula non zero

Question

Let us assume that we want integrate $f(x)$ in the interval $[-1,1]$ and obtain the approximation (Gauss quadrature)

\begin{eqnarray} \int_{-1}^1 f(x) dx \approx \sum_{i=0}^n w_i f(x_i) \end{eqnarray} where $x_i$ are the solutions the solutions of the equation $P_n(x)=0$ where $P_n$ is the $n^{th}$ Legendre polinomial and the weights $w_i$ can be obtained from Lagrange interpolation from the formula \begin{eqnarray} w_i = \int_{-1}^1 \prod_{i=0, i \ne j}^n \frac{x-x_j}{x_i-x_j} dx. \end{eqnarray}

Is there a simple way to show that $w_i \ne 0$, $i=0,1, \cdots, n$?

Thanks.

Answer 1

There are other formulas for finding the weights.

For example $$w_i = \frac {-2}{(n+1)P'_n(x_i)P_{n+1}(x_i)} $$ Which clearly shows $$ w_i\ne 0 $$

For derivation of this formula see Atkinson,$1989, p.276$;Ralston and Rabinowitz,$1978, p. 105.$

Answer 2

I borrowed this proof from Wikipedia.

Consider \begin{eqnarray} f(x) = \prod_{j=0, j \ne i}^n \frac{(x - x_j)^2}{(x_i-x_j)^2}. \end{eqnarray} Then $f(x_j)=\delta_{ij}$ and $f$ is a polynomial.

To compute this formula we use Gauss quadrature. Since the order of $f$ is $2n$, it is smaller than $2n+2$ so the Gauss quadrature is exact (0 error). Then the evaluation of $\int_{-1}^1 f(x)$ is

\begin{eqnarray} \int_{-1}^1 f(x) dx = \sum_{j=0}^n w_j f(x_j) = \sum_{j=0}^n \delta_{ij} w_j = w_i > 0. \end{eqnarray}