1

I was browsing the Wikipedia article on Gaussian quadrature, and started to wonder whether the Gauss-Legendre rules could be seen as specific types of Riemann sums. (With non-uniform intervals.)

The Gauss-Legendre rules approximates an integral $\int_{-1}^1 f(x) \mathrm{d}x$ by a sum $\sum_{i=1}^n f(x_i) w_i$ for some suitably chosen $w_i$ and $x_i$, where $-1 < x_1 < x_2 \ldots <x_n < 1$.

For example, if you only use one point then the Gauss-Legendre rule is that $$ \int_{-1}^{1} f(x) \, \mathrm{d}x \approx 2f(0). $$

Clearly the right-hand side is a Riemann sum. But for two points we get $$ \int_{-1}^{1} f(x) \, \mathrm{d}x \approx f(-3^{1/2}) + f(3^{1/2})$$ and here the right-hand side can be written as the Riemann sum $$ f(-3^{1/2}) (0-(-1))+ f(3^{1/2})(1-0).$$

So I wonder whether this pattern continues. It would not hold if for example $-1+w_1<x_1$ because the point $x_1$ has to belong to the interval $[-1,-1+w_1].$

Another User
  • 5,048
Johan
  • 2,239
  • 1
    These are quadrature methods based on interpolation. They are not Riemann sums. – Mittens Jan 06 '23 at 15:17
  • 1
    I am not asking for the movation. I am asking whether they can be seen as Riemann sums. Clearly the first ones can, but I wonder whether the pattern continues. I think it would be an interesting motivation for Riemann sums if it turnede out to be true. – Johan Jan 09 '23 at 09:56
  • 1
    The answer is yes, nice question! – orangeskid Feb 12 '23 at 23:59

0 Answers0