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As much as I try, I can't seem to find in any book or paper how we obtain the error of the Gauss-Chebyshev quadrature formula of the first kind. I found only that the error is given by $$ R_n(f)=\frac{\pi}{2^{2n-1}(2n)!}f^{(2n)}(\xi), \ \xi\in \langle -1,1 \rangle. $$

The general error of quadrature formulae using orthogonal polynomials is

$$ E_n(f) = \frac{f^{(2n)}(\xi)}{(2n)!} \int\limits_a^b \! p_n^2(x)w(x) \mathrm{d}x, $$

where $p_n:[a,b]\to\mathbb{R}$ is the $n$-th orthogonal polynomial of the given sequence of orthogonal polynomials and $w:[a,b]\to\mathbb{R}$ is the weight function.

Here we have $p_n(x)= T_n(x)=\cos{(n \arccos{x})}$ (the $n$-th Chebyshev polynomial of the first kind), and $w(x)=\frac{1}{\sqrt{1-x^2}}$.

So now, plugging this into $E_n(f)$, I'm trying to calculate the integral (I left out the fraction which we can consider a constant) $$ I=\int\limits_{-1}^{1} \! \frac{T_n^2(x)}{\sqrt{1-x^2}}\,\mathrm{d}x. $$ I've tried putting $t=\cos{(n \arccos{x})}$, then $$ \mathrm{d}t = - \sin{(n\arccos{x})}\cdot n \cdot \frac{1}{\sqrt{1-x^2}}\mathrm{d}x, $$ which gives $$ I=-\int\limits_{\pi}^0 \! \frac{t^2}{n\sqrt{1-t^2}} \mathrm{d}t = \int\limits_0^{\pi} \! \frac{t^2}{n\sqrt{1-t^2}} \mathrm{d}t $$ but computing this integral doesn't get me anywhere near the desired result. I have no way of getting the term $1/2^{2n-1}$ if I continue in this manner. Does someone see an error or have a hint on how to try computing it differently or knows how I could get the error given by $R_n$?

EDIT: I just remembered that the Chebyshev polynomials satisfy

$$ \int\limits_{-1}^{1} \! \frac{T_m(x)T_n(x)}{\sqrt{1-x^2}} \mathrm{d} x = \begin{cases} 0, &m\neq n \\ \pi/2, &m=n\neq 0 \\ \pi, &m=n=0, \end{cases} $$ so I easily get the value of $I$, but this still isn't the same as $R_n$.

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The formula you've provided $$ E_n(f) = \frac{f^{(2n)}(\xi)}{(2n)!} \int_a^b p_n^2(x)w(x)dx $$ is true under the assumptions that the leading coefficient of $p_n(x)$ is unity, i.e. $p_n(x) = x^n + \dots$. Really, the error bound should not change if you simply scale the orthogonal polynomial family.

But the $T_n(x) = \cos n \arccos x$ Chebyshev polynomials do not satisfy this property, but $$ \tilde T_n(x) = 2^{1-n} \cos n \arccos x $$ do, so $$ E_n(f) = \frac{f^{(2n)}(\xi)}{(2n)!} \int_{-1}^{1} \frac{\tilde T_n^2(x)}{\sqrt{1-x^2}} dx = \frac{2\pi f^{(2n)}(\xi)}{4^n(2n)!} = \frac{\pi f^{(2n)}(\xi)}{2^{2n-1}(2n)!}. $$

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By substituting $x=\cos\theta$, $$\int_{-1}^{1}\frac{T_n^2(x)}{\sqrt{1-x^2}}\,dx = \int_{0}^{\pi}\cos^2(n\theta)\,d\theta =\frac{1}{2}\int_{0}^{\pi}(1+\cos(2n\theta))\,d\theta=\color{red}{\frac{\pi}{2}}.$$

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  • $\begingroup$ Thanks, do you perhaps also have an idea on how to get the much better bound $R_n$? Or should I post a new question for that? $\endgroup$ – implicati0n Sep 8 '15 at 19:50

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