Copyright (C) 2020 Andreas Kloeckner
$p$-norms can be computed in two different ways in numpy:
import numpy as np
import numpy.linalg as la
x = np.array([1.,2,3])
First, let's compute the 2-norm by hand:
np.sum(x**2)**(1/2)
3.7416573867739413
Next, let's use numpy
machinery to compute it:
la.norm(x, 2)
3.7416573867739413
Both of the values above represent the 2-norm: $\|x\|_2$.
np.sum(np.abs(x)**5)**(1/5)
3.0773848853940629
la.norm(x, 5)
3.0773848853940629
The $\infty$ norm represents a special case, because it's actually (in some sense) the limit of $p$-norms as $p\to\infty$.
Recall that: $\|x\|_\infty = \max(|x_1|, |x_2|, |x_3|)$.
Where does that come from? Let's try with $p=100$:
x**100
array([ 1.00000000e+00, 1.26765060e+30, 5.15377521e+47])
np.sum(x**100)
5.1537752073201132e+47
Compare to last value in vector: the addition has essentially taken the maximum:
np.sum(x**100)**(1/100)
3.0
Numpy can compute that, too:
la.norm(x, np.inf)
3.0
Once you know the set of vectors for which $\|x\|=1$, you know everything about the norm, because of semilinearity. The graphical version of this is called the 'unit ball'.
We'll make a bunch of vectors in 2D (for visualization) and then scale them so that $\|x\|=1$.
alpha = np.linspace(0, 2*np.pi, 2000, endpoint=True)
x = np.cos(alpha)
y = np.sin(alpha)
vecs = np.array([x,y])
p = 5
norms = np.sum(np.abs(vecs)**p, axis=0)**(1/p)
norm_vecs = vecs/norms
import matplotlib.pyplot as pt
pt.grid()
pt.gca().set_aspect("equal")
pt.plot(norm_vecs[0], norm_vecs[1])
pt.xlim([-1.5, 1.5])
pt.ylim([-1.5, 1.5])
(-1.5, 1.5)