# Taking Derivatives with Vandermonde Matrices¶

MIT License Permission is hereby granted, free of charge, to any person obtaining a copy of this software and associated documentation files (the "Software"), to deal in the Software without restriction, including without limitation the rights to use, copy, modify, merge, publish, distribute, sublicense, and/or sell copies of the Software, and to permit persons to whom the Software is furnished to do so, subject to the following conditions: The above copyright notice and this permission notice shall be included in all copies or substantial portions of the Software. THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY, FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM, OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE SOFTWARE.
In [1]:
import numpy as np
import numpy.linalg as la
import matplotlib.pyplot as pt


Here are a few functions:

In [2]:
if 1:
def f(x):
return np.sin(5*x)
def df(x):
return 5*np.cos(5*x)
elif 0:
gamma = 0.15
def f(x):
return np.sin(1/(gamma+x))
def df(x):
return -np.cos(1/(gamma+x))/(gamma+x)**2
else:
def f(x):
return np.abs(x-0.5)
def df(x):
# Well...
return -1 + 2*(x<=0.5).astype(np.float)

In [3]:
x_01 = np.linspace(0, 1, 1000)
pt.plot(x_01, f(x_01))

Out[3]:
[<matplotlib.lines.Line2D at 0x7f5f98d5dd30>]
In [4]:
degree = 4
h = 1

nodes = 0.5 + np.linspace(-h/2, h/2, degree+1)
nodes

Out[4]:
array([0.  , 0.25, 0.5 , 0.75, 1.  ])

Build the gen. Vandermonde matrix and find the coefficients:

In [5]:
V = np.array([
nodes**i
for i in range(degree+1)
]).T

In [6]:
coeffs = la.solve(V, f(nodes))


Evaluate the interpolant:

In [7]:
x_0h = 0.5+np.linspace(-h/2, h/2, 1000)

In [8]:
interp_0h = 0*x_0h
for i in range(degree+1):
interp_0h += coeffs[i] * x_0h**i

In [9]:
pt.plot(x_01, f(x_01), "--", color="gray", label="$f$")
pt.plot(x_0h, interp_0h, color="red", label="Interpolant")
pt.plot(nodes, f(nodes), "or")
pt.legend(loc="best")

Out[9]:
<matplotlib.legend.Legend at 0x7f5f98d180f0>

Now build the gen. Vandermonde matrix $V'=$Vprime of the derivatives:

In [10]:
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


Compute the value of the derivative at the nodes as fderiv:

In [16]:
fderiv = Vprime @ la.inv(V) @ f(nodes)


And plot vs df, the exact derivative:

In [15]:
pt.plot(x_01, df(x_01), "--", color="gray", label="$f$")
pt.plot(nodes, fderiv, "or")
pt.legend(loc="best")

Out[15]:
<matplotlib.legend.Legend at 0x7f5f98bc9d68>
• Why don't we hit the values of the derivative exactly?
• Do an accuracy study.
In [13]:
print(np.max(np.abs(df(nodes) - fderiv)))

1.8560489763862222

• Can we assign a meaning to the entries of the matrix $D=V'V^{-1}$?
• What happens to the entries of $D$ if we...
• change $h$?
• shift the nodes?
• Using this, how would you construct methods for finding $f''$?
In [18]:
(
Vprime @ la.inv(V)
).round(3)

Out[18]:
array([[ -8.333,  16.   , -12.   ,   5.333,  -1.   ],
[ -1.   ,  -3.333,   6.   ,  -2.   ,   0.333],
[  0.333,  -2.667,   0.   ,   2.667,  -0.333],
[ -0.333,   2.   ,  -6.   ,   3.333,   1.   ],
[  1.   ,  -5.333,  12.   , -16.   ,   8.333]])
In [19]:
nodes

Out[19]:
array([0.  , 0.25, 0.5 , 0.75, 1.  ])
In [ ]: