$\displaystyle\int_a^b f(x)\,dx - \sum_{i=1}^n w_i\,f(x_i) =\frac{(b-a)^{2n+1} (n!)^4}{(2n+1)[(2n)!]^3} f^{(2n)} (\xi) , \qquad a < \xi < b.$
• Expand f(x) into a power series up to terms $x^{2n}$. Show that all lower order powers when approximated by Gaussian quadrature give the exact integral. – herb steinberg Apr 23 '18 at 20:57
• $\sum_{i=1}^n w_if(x_i)$ is the Gauss quadrature approximation. – herb steinberg Apr 25 '18 at 0:12
• Yes, I understand that as well. The part I am trying to derive is $\frac{(b-a)^{2n+1} (n!)^4}{(2n+1)[(2n)!]^3}$ – BaronFiner Apr 25 '18 at 1:35