Finite Differences
In [1]:
import numpy as np
import numpy.linalg as la
import matplotlib.pyplot as pt
In [29]:
degree = 2
h = 0.25
# Assume even degree so that there's a well-defined middle node.
assert degree % 2 == 0
nodes = np.linspace(-h/2, h/2, degree+1)
nodes
Out[29]:
Now construct V (the generalized Vandermonde) and Vprime (the generalized Vandermonde for the derivatives):
In [30]:
V = np.array([
nodes**i
for i in range(degree+1)
]).T
In [31]:
def monomial_deriv(i, x):
if i == 0:
return 0*x
else:
return i*nodes**(i-1)
Vprime = np.array([
monomial_deriv(i, nodes)
for i in range(degree+1)
]).T
Combine them to form the derivative matrix:
In [32]:
diff_mat = Vprime.dot(la.inv(V))
Let's say we only care about the derivative at the middle node:
In [33]:
finite_difference_weights = diff_mat[degree//2]
finite_difference_weights
Out[33]:
- What have we learned?
- What formula does this amount to?
- How do these weights scale with \(h\)?
- What formula does this amount to, really?
- What happens if we change the degree?
- What happens if we shift all nodes?
- What happens if we shift only one of the nodes?
- How do we apply this formula?
In [36]:
# * We could have left the middle point out. :)
# * -4*f(x-0.25) + 4*f(x+0.25)
# * They scale with 1/h, as you might expect.
# * (f(x-h/2) + f(x+h/2))/h
# * We get a more complicated (but more accurate) formula.
# * Nothing.
# * We get a different formula (that's valid for those nodes).
# * See below.
In [55]:
def f(x):
return np.sin(4*x)
def df(x):
return 4*np.cos(4*x)
In [56]:
x = np.arange(10) * 0.125
pt.plot(x, f(x), "o-")
Out[56]:
Now use the weights to compute the finite difference derivative as deriv:
In [57]:
fdw = finite_difference_weights
fx = f(x)
deriv = np.zeros(len(x)-2)
for i in range(1, 1+len(deriv)):
deriv[i-1] = fx[i-1]*fdw[0] + fx[i]*fdw[1] + fx[i+1]*fdw[2]
Now plot the finite difference derivative:
In [59]:
pt.plot(x[1:-1], df(x[1:-1]), label="true")
pt.plot(x[1:-1], deriv, label="FD")
pt.legend()
Out[59]:
