# Floating point vs Finite Differences¶

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In [5]:
import numpy as np
import numpy.linalg as la
import matplotlib.pyplot as pt

Define a function and its derivative:

In [17]:
c = 20*2*np.pi

def f(x):
return np.sin(c*x)

def df(x):
return c*np.cos(c*x)

n = 2000
x = np.linspace(0, 1, n, endpoint=False).astype(np.float32)

pt.plot(x, f(x))
Out[17]:
[<matplotlib.lines.Line2D at 0x7f0d292a0cc0>]

Now compute the relative $l^\infty$ norm of the error in the finite differences, for a bunch of mesh sizes:

In [16]:
h_values = []
err_values = []

for n_exp in range(5, 24):
n = 2**n_exp
h = (1/n)

x = np.linspace(0, 1, n, endpoint=False).astype(np.float32)

fx = f(x)
dfx = df(x)

dfx_num = (np.roll(fx, -1) - np.roll(fx, 1)) / (2*h)

err = np.max(np.abs((dfx - dfx_num))) / np.max(np.abs(fx))

print(h, err)

h_values.append(h)
err_values.append(err)

pt.rc("font", size=16)
pt.title(r"Single precision FD error on $\sin(20\cdot 2\pi)$")
pt.xlabel(r"$h$")
pt.ylabel(r"Rel. Error")
pt.loglog(h_values, err_values)
0.03125 4.8089
0.015625 1.24653
0.0078125 0.314495
0.00390625 0.0789223
0.001953125 0.0200939
0.0009765625 0.00580978
0.00048828125 0.003088
0.000244140625 0.00217628
0.0001220703125 0.00588608
6.103515625e-05 0.0104866
3.0517578125e-05 0.0216255
1.52587890625e-05 0.0410156
7.62939453125e-06 0.0843182
3.814697265625e-06 0.166426
1.9073486328125e-06 0.416862
9.5367431640625e-07 0.586523
4.76837158203125e-07 1.42031
2.384185791015625e-07 3.42323
1.1920928955078125e-07 7.42658
Out[16]:
[<matplotlib.lines.Line2D at 0x7f0d295ce908>]
In [12]: