Show that the quadrature formula
$$\int_{-1}^{1}\frac{f(x)}{\sqrt{1-x^2}}dx=\frac{\pi}{n}\sum_{k=0}^{n-1}f\bigg(\cos\bigg(\frac{2k+1}{2n}\pi\bigg)\bigg)$$
is exact for all polynomials of degree up to and including $2n-1$.$$$$
I know that this is the Chebyshev polynomial and $\frac{1}{\sqrt{1-x^2}}$ is its weight function, $\frac{2k+1}{2n}\pi$ are the zeros of the corresponding Chebyshev polynomial, and this knowledges must consist of at least $90\%$ of the proof. What I'm asking is, why with only n-th Chebyshev polynomial we can obtain the precision degree of $2n-1$? How is the bridge to higher degrees? I don't fully understand and I'm stuck here.
Prove that $\int_{-1}^{1}\frac{f(x)}{\sqrt{1-x^2}}dx=\frac{\pi}{n}\sum_{k=0}^{n-1}f(\cos(\frac{2k+1}{2n}\pi))$ for polynomials with degree $\le 2n-1$.
chebyshev polynomialsnumerical methods
Best Answer
Apply a change of independent variable, $x=-\cos(s)$, $s\in[0,\pi]$. Name the composition $g(s)=f(-\cos(s))$. If $f$ is a polynomial of degree $2n-1$ or smaller, then $g(s)$ is a periodic function with even symmetry at $s=0$ and $s=\pi$ and moreover a trigonometric polynomial with highest frequency $2n-1$, can thus be expanded as $$ g(s)=\sum_{f=0}^{2n-1} a_f\cos(fs) $$
Then the equation transforms to $$ \int_0^{\pi}g(s)\,ds = \frac{\pi}{n}\sum_{k=0}^{n-1} g\left(\pi-\frac{2k+1}{2n}\pi\right) = \frac{\pi}{n}\sum_{k=0}^{n-1} g\left(\frac{2k+1}{2n}\pi\right) $$ the point locations $s_k=(2k+1)\frac{f\pi}{n}$ being symmetric in the interval, $s_k+s_{n-1-k}=\pi$.
It remains to compute the sum for the basis functions for $f>0$ \begin{align} \sum_{k=0}^{n-1}\cos(fs_k) &=Re\left(\sum_{k=0}^{n-1}q^{2k+1}\right) =Re\left(q\,\frac{q^{2n}-1}{q^2-1}\right) \\ &=Re\left(q^{n}\frac{q^{n}-q^{-n}}{q-q^{-1}}\right) =Re(e^{if\frac\pi2})\frac{\sin(f\frac\pi2)}{\sin(f\frac{\pi}{2n})} =\frac{\sin(f\pi)}{2\sin(f\frac{\pi}{2n})}=0 \end{align} with $q=e^{i\frac{f\pi}{n}}$. The integral value for these terms is likewise zero.
For $f=0$ the summation is over the constant term $1$, so results in $n$, with the factor giving also the same value as the integral.