Why $w_i=\pi/n$ Chebyshev–Gauss quadrature?

Question

In Chebyshev–Gauss quadrature we have $$\int _{-1}^{+1}{\frac {f(x)}{\sqrt {1-x^{2}}}}\,dx\approx \sum _{i=1}^{n}w_{i}f(x_{i})$$ where $$x_{i}=\cos \left({\frac {2i-1}{2n}}\pi \right)$$ and the weight $w_i=\pi/n$. Why $w_i=\pi/n$?

Answer 1

If we change variables by $x=\cos\theta$, then the integral can be evaluated by dividing the interval into $n$ parts, using the midpoint rule on a circle, $$\int_{-1}^1\frac{f(x)}{\sqrt{1-x^2}}dx=\int_0^\pi f(\cos\theta)\,d\theta\approx\sum_{i=1}^nf(x_i)\frac{\pi}{n}$$ where the $x_i=\cos\theta_i$ are as written in the question.

---

For a fuller explanation of the weights:

The Cheybshev polynomials $T_n$ are orthogonal, that is $\int_{-1}^1\frac{T_i(x)T_j(x)}{\sqrt{1-x^2}}dx=\int_0^\pi \cos(i\theta)\cos(j\theta)\,d\theta=0$ for $i\ne j$.

To approximate the integral by $\sum_iw_if(x_i)$, we may require that the integral is exact on the $T_j$, $j=0,\ldots,n-1$, that is $$\sum_{i=1}^nw_iT_j(x_i)=\int_{-1}^1\frac{T_j(x)}{\sqrt{1-x^2}}dx$$ This is essentially the matrix equation $$\begin{pmatrix}1&\cdots&1\\ \cos\theta_1&\cdots&\cos\theta_n\\\cos2\theta_1&\cdots&\cos2\theta_n\\\vdots\end{pmatrix}\begin{pmatrix}w_1\\\vdots \\w_n\end{pmatrix}=\begin{pmatrix}\pi\\0\\\vdots\\0\end{pmatrix}$$

One can invert this matrix, or else notice that $$\begin{pmatrix}1&\cdots&1\\ \cos\theta_1&\cdots&\cos\theta_n\\\cos2\theta_1&\cdots&\cos2\theta_n\\\vdots\end{pmatrix}\begin{pmatrix}1\\\vdots\\1\end{pmatrix}=\begin{pmatrix}n\\0\\\vdots\\0\end{pmatrix}$$ to conclude that $w_i=\pi/n$.