$p$-norms can be computed in two different ways in numpy:

In [1]:

```
import numpy as np
import numpy.linalg as la
```

In [2]:

```
x = np.array([1.,2,3])
```

First, let's compute the 2-norm by hand:

In [3]:

```
np.sum(x**2)**(1/2)
```

Out[3]:

Next, let's use `numpy`

machinery to compute it:

In [4]:

```
la.norm(x, 2)
```

Out[4]:

Both of the values above represent the 2-norm: $\|x\|_2$.

In [5]:

```
np.sum(np.abs(x)**5)**(1/5)
```

Out[5]:

In [6]:

```
la.norm(x, 5)
```

Out[6]:

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 $p=100$:

In [7]:

```
x**100
```

Out[7]:

In [8]:

```
np.sum(x**100)
```

Out[8]:

Compare to last value in vector: the addition has essentially taken the maximum:

In [9]:

```
np.sum(x**100)**(1/100)
```

Out[9]:

Numpy can compute that, too:

In [10]:

```
la.norm(x, np.inf)
```

Out[10]:

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 $\|x\|=1$.

In [12]:

```
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
%matplotlib inline
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])
```

Out[12]: