Copyright (C) 2020 Andreas Kloeckner
import numpy as np
import numpy.linalg as la
import scipy.special as sps
import matplotlib.pyplot as pt
Here's a utility routine to do biseciton, to use below:
def bisection(f, a, b, tol=1e-14):
assert np.sign(f(a)) != np.sign(f(b))
while b-a > tol:
m = a + (b-a)/2
fm = f(m)
if np.sign(f(a)) != np.sign(fm):
b = m
else:
a = m
return m
Set the number of nodes:
n = 5
Gauss nodes are the roots of the $n$th Legendre polynomial $P_n$:
nodes = sps.legendre(n).weights[:, 0]
mesh = np.linspace(-1, 1, 300)
pt.plot(mesh, sps.eval_legendre(n, mesh))
pt.plot(nodes, 0*nodes, "o")
/usr/local/lib/python3.5/dist-packages/IPython/core/formatters.py:92: DeprecationWarning: DisplayFormatter._ipython_display_formatter_default is deprecated: use @default decorator instead. def _ipython_display_formatter_default(self): /usr/local/lib/python3.5/dist-packages/IPython/core/formatters.py:669: DeprecationWarning: PlainTextFormatter._singleton_printers_default is deprecated: use @default decorator instead. def _singleton_printers_default(self):
[<matplotlib.lines.Line2D at 0x7f4a1c40d438>]
def eval_legendre_deriv(n, x):
return (
(x*sps.eval_legendre(n, x) - sps.eval_legendre(n-1, x))
/
((x**2-1)/n))
brackets = sps.legendre(n-1).weights[:, 0]
nodes = np.zeros(n)
nodes[0] = -1
nodes[-1] = 1
from functools import partial
# Use the fact that the roots of P_{n-1} bracket the roots of P_{n-1}':
for i in range(n-2):
nodes[i+1] = bisection(
partial(eval_legendre_deriv, n-1),
brackets[i], brackets[i+1])
mesh = np.linspace(-1, 1, 300)
pt.plot(mesh, eval_legendre_deriv(n-1, mesh))
pt.plot(nodes, 0*nodes, "o")
/usr/local/lib/python3.5/dist-packages/ipykernel/__main__.py:5: RuntimeWarning: invalid value encountered in true_divide
[<matplotlib.lines.Line2D at 0x7f4a13d9ad30>]
For Gauss-Radau (with the left endpoint included), the nodes are the roots of the following function:
def radau_func(n, x):
return (
(sps.eval_legendre(n-1, x) + sps.eval_legendre(n, x))
/
(1+x))
nodes = None
# Root finding left as an exercise for the reader. :)
mesh = np.linspace(-1, 1, 300)
pt.plot(mesh, radau_func(n, mesh))
/usr/local/lib/python3.5/dist-packages/ipykernel/__main__.py:5: RuntimeWarning: invalid value encountered in true_divide
[<matplotlib.lines.Line2D at 0x7f4a13b73588>]
Use method of undetermined coefficients to find the interpolatory quadrature rule for nodes
:
max_degree = len(nodes) - 1
powers = np.arange(max_degree+1)
Vt = nodes ** powers.reshape(-1, 1)
a, b = -1, 1
rhs = 1/(powers+1) * (b**(powers+1) - a**(powers+1))
weights = la.solve(Vt, rhs)
Now compare the approximate integrals of monomials from our rule with the true answers:
for i in range(2*n + 1):
approx = weights @ nodes**i
true = 1/(i+1)*(1. - (-1)**(i+1))
print("Error at degree %d: %g" % (i, approx-true))
Error at degree 0: 0 Error at degree 1: -1.66533e-16 Error at degree 2: 1.11022e-16 Error at degree 3: -2.498e-16 Error at degree 4: 2.22045e-16 Error at degree 5: -3.21965e-15 Error at degree 6: -8.88178e-16 Error at degree 7: -4.4964e-15 Error at degree 8: 0.0145125 Error at degree 9: -5.02376e-15 Error at degree 10: 0.0339253