Why is the sum of chebyshev gaussian quadraure weights not $2$?

Question

When I sum the weights of the chebyshev gaussian quadrature over the interval $[-1,1]$. I get about $1.57$. I don't understand why it is not equal to the domain size of the integral (i.e. $2$). My question is why?

Answer 1

Chebyshev-Gauss quadrature, also called Chebyshev quadrature, is a Gaussian quadrature over the interval $[-1,1]$ with weighting function $W(x)=\frac{1}{\sqrt{1-x^{2}}}$. The abscissas for quadrature order $n$ are given by the roots of the Chebyshev polynomial of the first kind $T_n(x)$, which occur symmetrically about $0$. The weights are

$$w_i = -\frac{A_{n+1}\gamma_n}{A_nT'_n(x_i)T_{n+1}(x_i)}...(1) $$
$$ = \frac{A_n}{A_{n-1}}\frac{\gamma_(n-1)}{T_{n-1}(x_{i})T'_n(x_i)}...(2)$$, where $A_n$ is the coefficient of $x^n$ in $T_n(x)$,

$$ \gamma_n = A_n\pi(x)...(3)$$ and $\pi(x)$ the order-$n$ Lagrange interpolating polynomial for $T_n(x)$.

For Chebyshev polynomials of the first kind,

$$ A_n=2^{n-1}...(4)$$ so $$\frac{A_{n+1}}{A_n}=2...(5)$$ Additionally, $$\gamma_n=\frac{1}{2\pi}...(6)$$ so $$ w_i=-\frac{\pi}{T_{n+1}(x_i)T'_n(x_i)}...(7)$$ Since $$ T_n(x)= \cos(ncos^{-1}x)...(8)$$ the abscissas are given explicitly by $$ x_i=\cos[\frac{(2i-1)\pi}{2n}]...(9)$$ Since $$ T'_n(x_i) = \frac{(-1)^{i+1}n}{\sin\alpha_i}...(10)$$ and $$ T_{n+1}(x_i) = (-1)^{i}\sin\alpha_i...(11)$$ where $$ \alpha_i=\frac{(2i-1)\pi}{2n}...(12)$$ all the weights are $$w_i= \frac{\pi}{n}.$$ Hope it helps.