According to Chapter 5 of Numerical Methods and Software by Kahaner, et al. (1989), it can be shown that the error associated with Gaussian quadrature is

$\displaystyle\int_a^b f(x)\,dx - \sum_{i=1}^n w_i\,f(x_i) =\frac{(b-a)^{2n+1} (n!)^4}{(2n+1)[(2n)!]^3} f^{(2n)} (\xi) , \qquad a < \xi < b.$

How is this error derived?

  • $\begingroup$ 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. $\endgroup$ – herb steinberg Apr 23 '18 at 20:57
  • $\begingroup$ I understand the presence of the derivative order but not the expression that precedes it. $\endgroup$ – BaronFiner Apr 23 '18 at 21:13
  • $\begingroup$ $\sum_{i=1}^n w_if(x_i)$ is the Gauss quadrature approximation. $\endgroup$ – herb steinberg Apr 25 '18 at 0:12
  • $\begingroup$ Yes, I understand that as well. The part I am trying to derive is $\frac{(b-a)^{2n+1} (n!)^4}{(2n+1)[(2n)!]^3}$ $\endgroup$ – BaronFiner Apr 25 '18 at 1:35
  • $\begingroup$ I have not worked it out, but I suspect it is the approximation to the next term in a power series. $\endgroup$ – herb steinberg Apr 26 '18 at 16:05

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