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

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.

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,$1978, p. 105.$

• Thanks for your prompt response. I was looking for something simpler.... – Herman Jaramillo Apr 18 '18 at 23:48
• For the first few weights, we can use consistency requirement to find the actual values, but in general we need a formula to work with. – Mohammad Riazi-Kermani Apr 19 '18 at 0:02
• I am in the process of doing this using a Vandermonde system, for which there is an explicit and positive set of weights. – Herman Jaramillo Apr 19 '18 at 0:16
• The wikipedia page en.wikipedia.org/wiki/Gaussian_quadrature has some interesting formulas. $w_i=2/(1-x_i^2)[ P_n'(x_i)]^2$ among others..... – Herman Jaramillo Apr 19 '18 at 1:10

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 $f(x_j)=\delta_{ij}$.

To compute this formula we use Gauss quadrature. Since the order of the polynomial is $2n-2$, 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}