# Accuracy of Newton-Cotes¶

In [21]:
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 [34]:
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 [129]:
# 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 [130]:
(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 [131]:
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 [132]:
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[132]:
<matplotlib.legend.Legend at 0x7fe907aaa978>

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

In [133]:
# 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[133]:
<matplotlib.legend.Legend at 0x7fe907a37d30>

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 [134]:
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)

errors.append(error)

1 0.6456349206349207 0.610677083333 0.0349578373016
0.5 0.4054346478174603 0.40550104777 6.6399952722e-05
0.25 0.22314353367638967 0.223148116221 4.58254464089e-06
0.125 0.11778303564688754 0.117783224384 1.88736665879e-07
0.0625 0.06062462181642996 0.0606246286393 6.8228956096e-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 [135]:
for i in range(len(errors)-1):
print(np.log(errors[i+1]/errors[i])/np.log(1/2))

9.04021800427
3.85696137438
4.60170230716
4.78984676915

In [ ]: