3 ms·
This is a bit misleading. If the integrand is analytic, then the Gauss quadrature converges geometrically. For example with f(x) = sqrt(1 + sin(x)) and you app
by stabbles 8y ago
This is a bit misleading. If the integrand is analytic, then the Gauss quadrature converges geometrically.
For example with f(x) = sqrt(1 + sin(x)) and you approximate integral f(x) dx from -1 to 1 with n Gauss points, then the the convergence is geometric:
n | approx
1 | 2.0-------------
2 | 1.9172----------
3 | 1.917703--------
4 | 1.917702153-----
5 | 1.917702154417--
6 | 1.91770215441681
But with a function that is not differentiable such as g(x) = |x|, there is definitely not geometric convergence.
n | approx
1 | 0.0-----------
10 | 1.007---------
100 | 1.00008-------
1000 | 1.0000008-----