import numpy as np
import numpy.linalg as la
import matplotlib.pyplot as pt
A function to make Vandermonde matrices:
(Note that the ordering of this matrix matches the convention in our class but disagrees with np.vander
.)
def vander(nodes, ncolumns=None):
if ncolumns is None:
ncolumns = len(nodes)
result = np.empty((len(nodes), ncolumns))
for i in range(ncolumns):
result[:, i] = nodes**i
return result
Fix a set of nodes:
# nodes = [0.5] # Midpoint
# nodes = [0]
#nodes = [0, 1] # Trapezoidal
nodes = [0, 0.5, 1] # Simpson's
#nodes = [0, 1/3, 1]
Find the weights for the Newton-Cotes rule for the given nodes on $[0,1]$:
(a, b) = (0, 1)
nodes = np.array(nodes)
n = len(nodes)
degs = np.arange(n)
rhs = 1/(degs+1)*(b**(degs+1 - a**(degs+1)))
weights = la.solve(vander(nodes).T, rhs)
print(weights)
Here is a function and its definite integral from $0$ to $x$:
$$\text{int_f}(x)=\int_0^x f(\xi)d\xi$$fdeg = 9
def f(x):
return sum((-x)**i for i in range(fdeg + 1))
def int_f(x):
return sum(
(-1)**i*1/(i+1)*(
(x)**(i+1)-0**(i+1)
)
for i in range(fdeg + 1))
Plotted:
plot_x = np.linspace(0, 1, 200)
pt.plot(plot_x, f(plot_x), label="f")
pt.fill_between(plot_x, 0*plot_x, f(plot_x),alpha=0.3)
pt.plot(plot_x, int_f(plot_x), label="$\int f$")
pt.grid()
pt.legend(loc="best")
This here plots the function, the interpolant, and the area under the interpolant:
# fix nodes
h = 1
x = nodes * h
# find interpolant
coeffs = la.solve(vander(x), f(x))
# evaluate interpolant
plot_x = np.linspace(0, h, 200)
interpolant = vander(plot_x, len(coeffs)) @ coeffs
# plot
pt.plot(plot_x, f(plot_x), label="f")
pt.plot(plot_x, interpolant, label="Interpolant")
pt.fill_between(plot_x, 0*plot_x, interpolant, alpha=0.3, color="green")
pt.plot(x, f(x), "og")
pt.grid()
pt.legend(loc="best")
Compute the following:
true_val
(from int_f
)quad
(using x
and weights
and h
)The error as err
(the difference of the two)
(Do not be tempted to compute a relative error--that has one order lower.)
Compare the error for $h=1,0.5,0.25$. What order of accuracy do you observe?
errors = []
for h in [1, 0.5, 0.25, 0.125, 0.125*0.5]:
true_val = int_f(h)
quad = h * weights @ f(h * nodes)
error = abs(quad - true_val)
print(h, true_val, quad, error)
errors.append(error)
Estimate the order of accuracy:
We assume that the error depends on the mesh spacings $h$ as $E(h)\approx C h^p$ for some unknown power $p$. Taking the $\log$ of this approximate equality reveals a linear function in $p$: $$ E(h) \approx C h^p \quad \iff \quad \log E(h) \approx \log(C) + p\log(h). $$ You can now either do a least-squares fit for $\log C$ and $p$ from a few data points $(h,E(h))$ (more accurate, more robust), or you can use just two grid sizes $h_1$ and $h_2$, and estimate the slope: (less accurate, less robust) $$ p \approx \frac{ \log(\frac{E(h_2)}{E(h_1)}) } {\log(\frac{h_2}{h_1})}. $$ This is called the empirical order of convergence or EOC.
for i in range(len(errors)-1):
print(np.log(errors[i+1]/errors[i])/np.log(1/2))