Accuracy of Newton-Cotes¶

Copyright (C) 2020 Andreas Kloeckner

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In [1]:
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.)

In [2]:
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:

In [3]:
# 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 $[0,1]$:

In [4]:
(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)

[0.16666667 0.66666667 0.16666667]

Here is a function and its definite integral from $0$ to $x$:

$$\text{int_f}(x)=\int_0^x f(\xi)d\xi$$

In [5]:
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:

In [6]:
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")
Out[6]:
<matplotlib.legend.Legend at 0x7fb962f0bd00>
No description has been provided for this image

This here plots the function, the interpolant, and the area under the interpolant:

In [7]:
# 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")
Out[7]:
<matplotlib.legend.Legend at 0x7fb95faaef70>
No description has been provided for this image

Compute the following:

  • The true integral as true_val (from int_f)

  • The quadrature result as 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?

In [8]:
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)

1 0.6456349206349207 0.6106770833333334 0.0349578373015873
0.5 0.4054346478174603 0.4055010477701823 6.639995272200716e-05
0.25 0.22314353367638967 0.22314811622103056 4.582544640890918e-06
0.125 0.11778303564688754 0.11778322438355342 1.887366658792633e-07
0.0625 0.06062462181642996 0.06062462863932557 6.822895609603918e-09

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.

In [9]:
for i in range(len(errors)-1):
    print(np.log(errors[i+1]/errors[i])/np.log(1/2))

9.040218004272907
3.856961374376886
4.601702307158235
4.789846769147522
In [ ]: