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# ## Computing norms by hand
#
# $p$-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)
# Next, let's use `numpy` machinery to compute it:
# In[4]:
la.norm(x, 2)
# Both of the values above represent the 2-norm: $\|x\|_2$.
# --------------
#
# ## About the $\infty$-norm
#
# Different values of $p$ work similarly:
# In[5]:
np.sum(np.abs(x)**5)**(1/5)
# In[6]:
la.norm(x, 5)
# ---------------------
#
# 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
# In[8]:
np.sum(x**100)
# Compare to last value in vector: the addition has essentially taken the maximum:
# In[9]:
np.sum(x**100)**(1/100)
# Numpy can compute that, too:
# In[10]:
la.norm(x, np.inf)
# -------------
#
# ## Unit Balls
#
# 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[26]:
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])
# In[ ]: