#!/usr/bin/env python
# coding: utf-8

# # Vector Norms

# ## Computing norms by hand
# 
# $p$-norms can be computed in two different ways in numpy:

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import numpy as np
import numpy.linalg as la

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

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

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

# Next, let's use `numpy` machinery to compute it:

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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:

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

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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 $p=100$:

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x**100

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np.sum(x**100)

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

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

# Numpy can compute that, too:

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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 $\|x\|=1$.

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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])

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