From b34e01f05431160736287227c40533119c2e9904 Mon Sep 17 00:00:00 2001 From: Andrew Cumming Date: Sun, 17 Sep 2023 11:04:07 -0400 Subject: [PATCH] Updates integration.md --- integration.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/integration.md b/integration.md index af16ffd..4f881c7 100644 --- a/integration.md +++ b/integration.md @@ -95,7 +95,7 @@ For a suitable choice of the weights $w_i$ and the locations $x_i$ it can be sho This means that with 2 evaluations of $f(x)$ we can exactly evaluate the integral of a cubic polynomial! How does this compare with Simpson's method? ``` -The proof that this works and the calculation of the values of $w_i$ and $x_i$ is quite involved. [One thing to mention is that for this particular form of the integral, the values of $x_i$ are the roots of the Legendre Polynomial $P_N(x)$.] However, we can look up $w_i$ and $x_i$ using [`numpy.polynomial.legendre.leggauss`](https://numpy.org/doc/stable/reference/generated/numpy.polynomial.legendre.leggauss.html) +The proof that this works and the calculation of the values of $w_i$ and $x_i$ is quite involved. [It uses the theory of orthogonal polynomials, so for example for this particular form of the integral, the values of $x_i$ are the roots of the Legendre Polynomial $P_N(x)$.] However, we can look up $w_i$ and $x_i$ using [`numpy.polynomial.legendre.leggauss`](https://numpy.org/doc/stable/reference/generated/numpy.polynomial.legendre.leggauss.html) For example,