The weights $A_j$ of the Gaussian quadrature rule are positive and furthermore it is $A_j=\int_a^b w(x)\prod_{i\neq j} \left(\frac{x-x_i}{x_j-x_i}\right)^2$
I am searching for a reference on the Gaussian quadrature rule, which might (or might not) contain the formula above, for the weights.
Do you know something useful, or might give an introduction yourself, especially on how to proof the formula above.
From my own lecture notes I can not even tell, how the Gaussian quadrature rule is defined.
Thanks in advance.
Your formula for the weights is wrong. If you have chosen the nodes $x_0,x_1,\cdots, x_n,$ then take $p(x)=\sum\limits_{i=0}^n f(x_i)\ell_i(x)$ to be the Lagrange interpolating polynomial, with $\ell_i(x)=\prod\limits_{j=0,\ j\neq i}^n\left(\frac{x-x_j}{x_i-x_j}\right).$ We want to find $A_i$ so that $$\int\limits_a^b f(x)w(x)dx\approx \sum\limits_{i=0}^n A_i f(x_i).$$ Well, $$\int\limits_a^b f(x)w(x)dx\approx \int\limits_a^b \sum\limits_{i=0}^n f(x_i)\ell_i(x)w(x)dx=\sum\limits_{i=0}^n \left(\int\limits_a^b w(x)\ell_i(x)dx \right)f(x_i).$$ So, that's where the formula for the weights comes from, with no square.
As for the positivity, consider the function $f(x)=[\ell_j(x)]^2$, for a fixed $j$. This is a polynomial of degree $2n$. A Gaussian quadrature rule with $n+1$ nodes is exact for polynomials up to degree $2n+1,$ so it is exact for $f$. In particular,
$$\int\limits_a^b f(x)w(x)dx=\sum\limits_{i=0}^n A_i f(x_i)=\sum\limits_{i=0}^n A_i\delta_{ij}=A_j.$$ But, $$\int\limits_a^b f(x)w(x)dx>0,$$ since $w$ is positive, and $f$ is non-negative, as well as positive on a set of positive measure.