Finding error of a Gaussian-Chebyshev quadrature rule

Question

Suppose we want to integrate $$I(f) := \int_{-1}^1{f(x)\over\sqrt{1-x^2}}\,dx.$$ If I have some quadrature formula given by $$Q_2(f) = {\pi\over 2}\sum_{i=1}^2f(x_i),$$ I want to put an upper bound on the error given by $$\begin{align*}|E_2(f)| &= |I(f) - Q_2(f)|,\end{align*}$$ where $x_i = \pm{\sqrt2\over2}$. I have shown that if $f$ is a cubic polynomial, then the error is zero. However, when $f$ is any (differentiable) function in general, I'm not entirely sure what can be done. I know that we can do the following: $$\begin{align*}I(f) &= \int_{-1}^1{f(x)\over\sqrt{1-x^2}}\,dx\\ &= \int_0^\pi f(\cos\theta)\,d\theta \tag{Let $x=\cos\theta$}\\ &\approx \sum_{i=1}^n\alpha_if(x_i),\end{align*}$$

which $$\begin{align*}|E_2(f)| &= |I(f) - Q_2(f)|\\ &= \left|\int_0^\pi f(\cos\theta)\,d\theta - \sum_{i=1}^2{\pi\over2}f(x_i)\right|.\end{align*}$$ but I'm not sure where I can go from here to start talking about the error bound. I think we could use say the Lagrange interpolating polynomial for $f$, but the different indices seem like this isn't what we should be doing.

Answer 1

I don't think that it is possible to bound the error purely from the differentiability of $f$. You need a stronger condition.

I can choose a function such that $f\left(\pm \frac{1}{\sqrt{2}}\right) = 0$ so that $Q_2(f) = 0$ e.g. $$ f(x) = a\cos\left(\frac{\pi}{\sqrt{2}} x\right). $$

Now since the integral is non-zero and proportional to $a$ I can make the error as big as I like by increasing $a$.

What you could do is use a Taylor expansion for $f$ to compute the error and take the leading term as an estimate of the error and then use that to bound the result in terms of say $\max_x{|f''(x)}|$ or something like that.

Answer 2

Apparently here is the intended solution to the problem:

Let $\pi_n$ be the Lagrange interpolating polynomial to $f$, $\displaystyle \omega_{n+1}(x) = \prod_{i=0}^n(x-x_i)$, and $W(x) = {1\over\sqrt{1-x^2}}$ be a positive weight function. We know that $$\begin{align*}f(x) - \pi_n(x) &= {f^{(n+1)}(\eta(x))\over(n+1)!}\omega_{n+1}(x) \\ \implies f(x) - \sum_{i=0}^nf(x_i)L_i(x) &= {f^{(n+1)}(\eta(x))\over(n+1)!}\omega_{n+1}(x) \\ \implies W(x)f(x) - \sum_{i=0}^nf(x_i)L_i(x)W(x) &= {f^{(n+1)}(\eta(x))\over(n+1)!}\omega_{n+1}(x)W(x)\\ \implies \int_{-1}^1W(x)f(x)\,dx - \sum_{i=0}^nf(x_i)\int_{-1}^1L_i(x)W(x)\,dx &= {f^{(n+1)}(\xi)\over(n+1)!}\int_{-1}^1\omega_{n+1}(x)W(x)\,dx \\ \implies |I(f) - Q(f)| &= \left|{f^{(n+1)}(\eta(x))\over(n+1)!}\int_{-1}^1\omega_{n+1}(x)W(x)\,dx\right|\\ &\le {\|f^{(n+1)}\|_\infty\over(n+1)!}\int_{-1}^1\left|\omega_{n+1}(x)W(x)\right|\,dx\\ &\le {\|f^{(n+1)}\|_\infty\over(n+1)!}\|\omega_{n+1}\|\int_{-1}^1W(x)\,dx\\ & = {2\pi\over(n+1)!}\|f^{(n+1)}\|_\infty.\end{align*}$$

Hence, whatever $Q$ is, we have that the quadrature rule works.