Editing Gaussian quadrature (section)

==== General formula for the weights ====
The weights can be expressed as

{{NumBlk|:|<math>w_{i} = \frac{a_{n}}{a_{n-1}} \frac{\int_{a}^{b} \omega(x) p_{n-1}(x)^2 dx}{p'_{n}(x_{i}) p_{n-1}(x_{i})}</math>|{{EquationRef|1}}}}

where <math>a_{k}</math> is the coefficient of <math>x^{k}</math> in <math>p_{k}(x)</math>. To prove this, note that using [[Lagrange interpolation]] one can express {{math|''r''(''x'')}} in terms of <math>r(x_{i})</math> as
<math display="block">r(x) = \sum_{i=1}^{n} r(x_{i}) \prod_{\begin{smallmatrix} 1 \leq j \leq n \\ j \neq i \end{smallmatrix}}\frac{x-x_{j}}{x_{i}-x_{j}}</math>
because {{math|''r''(''x'')}} has degree less than {{mvar|n}} and is thus fixed by the values it attains at {{mvar|n}} different points. Multiplying both sides by {{math|''ω''(''x'')}} and integrating from {{mvar|a}} to {{mvar|b}} yields
<math display="block">\int_{a}^{b}\omega(x)r(x)dx = \sum_{i=1}^{n} r(x_{i}) \int_{a}^{b}\omega(x)\prod_{\begin{smallmatrix} 1 \leq j \leq n \\ j \neq i \end{smallmatrix}} \frac{x-x_{j}}{x_{i}-x_{j}}dx</math>

The weights {{mvar|w<sub>i</sub>}} are thus given by
<math display="block">w_{i} = \int_{a}^{b}\omega(x)\prod_{\begin{smallmatrix}1\leq j\leq n\\j\neq i\end{smallmatrix}}\frac{x-x_{j}}{x_{i}-x_{j}}dx</math>

This integral expression for <math>w_{i}</math> can be expressed in terms of the orthogonal polynomials <math>p_{n}(x)</math> and <math>p_{n-1}(x)</math> as follows.

We can write
<math display="block"> \prod_{\begin{smallmatrix} 1 \leq j \leq n \\ j \neq i \end{smallmatrix}} \left(x-x_{j}\right) = \frac{\prod_{1\leq j\leq n} \left(x - x_{j}\right)}{x-x_{i}} = \frac{p_{n}(x)}{a_{n}\left(x-x_{i}\right)}</math>

where <math>a_{n}</math> is the coefficient of <math>x^n</math> in <math>p_{n}(x)</math>. Taking the limit of {{mvar|x}} to <math>x_{i}</math> yields using L'Hôpital's rule
<math display="block"> \prod_{\begin{smallmatrix} 1 \leq j \leq n \\ j \neq i \end{smallmatrix}} \left(x_{i}-x_{j}\right) = \frac{p'_{n}(x_{i})}{a_{n}}</math>

We can thus write the integral expression for the weights as

{{NumBlk|:|<math>w_{i} = \frac{1}{p'_{n}(x_{i})}\int_{a}^{b}\omega(x)\frac{p_{n}(x)}{x-x_{i}}dx</math>|{{EquationRef|2}}}}

In the integrand, writing
<math display="block">\frac{1}{x-x_i} = \frac{1 - \left(\frac{x}{x_i}\right)^{k}}{x - x_i} + \left(\frac{x}{x_i}\right)^{k} \frac{1}{x - x_i}</math>

yields
<math display="block">\int_a^b\omega(x)\frac{x^kp_n(x)}{x-x_i}dx = x_i^k \int_{a}^{b}\omega(x)\frac{p_n(x)}{x-x_i}dx</math>

provided <math>k \leq n</math>, because
<math display="block">\frac{1-\left(\frac{x}{x_{i}}\right)^{k}}{x-x_{i}}</math>
is a polynomial of degree {{math|''k'' − 1}} which is then orthogonal to <math>p_{n}(x)</math>. So, if {{math|''q''(''x'')}} is a polynomial of at most nth degree we have
<math display="block">\int_{a}^{b}\omega(x)\frac{p_{n}(x)}{x-x_{i}} dx = \frac{1}{q(x_{i})} \int_{a}^{b} \omega(x)\frac{q(x) p_n(x)}{x-x_{i}}dx </math>

We can evaluate the integral on the right hand side for <math>q(x) = p_{n-1}(x)</math> as follows. Because <math>\frac{p_{n}(x)}{x-x_{i}}</math> is a polynomial of degree {{math|''n'' − 1}}, we have
<math display="block">\frac{p_{n}(x)}{x-x_{i}} = a_{n}x^{n-1} + s(x)</math>
where {{math|''s''(''x'')}} is a polynomial of degree <math>n - 2</math>. Since {{math|''s''(''x'')}} is orthogonal to <math>p_{n-1}(x)</math> we have
<math display="block">\int_{a}^{b}\omega(x)\frac{p_{n}(x)}{x-x_{i}}dx=\frac{a_{n}}{p_{n-1}(x_{i})} \int_{a}^{b}\omega(x)p_{n-1}(x)x^{n-1}dx </math>

We can then write
<math display="block">x^{n-1} = \left(x^{n-1} - \frac{p_{n-1}(x)}{a_{n-1}}\right) + \frac{p_{n-1}(x)}{a_{n-1}}</math>

The term in the brackets is a polynomial of degree <math>n - 2</math>, which is therefore orthogonal to <math>p_{n-1}(x)</math>. The integral can thus be written as
<math display="block">\int_{a}^{b}\omega(x)\frac{p_{n}(x)}{x-x_{i}}dx = \frac{a_{n}}{a_{n-1} p_{n-1}(x_{i})} \int_{a}^{b}\omega(x) p_{n-1}(x)^{2} dx </math>

According to equation ({{EquationNote|2}}), the weights are obtained by dividing this by <math>p'_{n}(x_{i})</math> and that yields the expression in equation ({{EquationNote|1}}).

<math>w_{i}</math> can also be expressed in terms of the orthogonal polynomials <math>p_{n}(x)</math> and now <math>p_{n+1}(x)</math>. In the 3-term recurrence relation <math>p_{n+1}(x_{i}) = (a) p_{n}(x_{i}) + (b) p_{n-1}(x_{i})</math> the term with <math>p_{n}(x_{i})</math> vanishes, so <math>p_{n-1}(x_{i})</math> in Eq. (1) can be replaced by <math display="inline">\frac{1}{b} p_{n+1} \left(x_i\right)</math>.