import numpy as np
import matplotlib.pyplot as plt
%matplotlib inline

params = {'legend.fontsize': 20,
          'figure.figsize': (12, 6),
         'axes.labelsize': 20,
         'axes.titlesize': 20,
         'xtick.labelsize': 20,
         'ytick.labelsize': 20,
         'lines.linewidth': 3}
plt.rcParams.update(params)

Make some data

Let's make 3 pieces of data:

• error1 might represent error that looks like $e\sim n^{-1}$
• error15 might represent error that looks like $e\sim n^{-1.5}$
• error2 might represent error that looks like $e\sim n^{-2}$

What should this look like in linear, log-linear, and log-log?

This is called algebraic convergence, with an algebraic index of convergence of $\alpha = 1.0, 1.5, 2.0$, where $$e \sim n^{-\alpha}$$

n = np.logspace(1, 6, 100) # evenly distribute numbers over logspace

error1 = 1 / n**1
error15 = 1 / n**1.5
error2 = 1 / n**2

p = plt.plot
p(n, error1, label=r'$n^{-1}$')
p(n, error15, label=r'$n^{-1.5}$')
p(n, error2, label=r'$n^{-2}$')

plt.xlabel('$n$')
plt.ylabel('error')
plt.grid()
plt.legend(frameon=False)

<matplotlib.legend.Legend at 0x11e502400>

Think about faster convergence than algebraic

Let's make 3 pieces of data:

• error21 might represent error that looks like $e\sim 2^{-n}$
• error23 might represent error that looks like $e\sim 2^{-3n}$
• error2e might represent error that looks like $e\sim e^{-2n}$

What should this look like?

Here the algebraic index is unbounded (the error decays fastter than $n^{-\alpha}$ for any $\alpha$). So we call this exponential or spectral or a form of geometric convergence.

That is $$e \sim e^{-\mu n}$$ for some rate $\mu$ of exponential convergence.

n = np.logspace(0, 1, 100) # evenly distribute numbers over logspace

error21 = 2**-n
error23 = 2**-(3*n)
error2e  = np.exp(-2*n)

p = plt.plot
p(n, error21, label=r'$2^{-n}$')
p(n, error23, label=r'$2^{-3n}$')
p(n, error2e, label=r'$e^{-2n}$')

plt.xlabel('$n$')
plt.ylabel('error')
plt.grid()
plt.legend(frameon=False)

<matplotlib.legend.Legend at 0x11e357400>

n = np.logspace(0, 2, 100) # evenly distribute numbers over logspace

error2e  = np.exp(-2*n)
error2 = 1 / n**2

p = plt.plot

p(n, error2e, label=r'$e^{-2n}$')
p(n, error2, label=r'$n^{-2}$')

plt.xlabel('$n$')
plt.ylabel('error')
plt.grid()
plt.legend(frameon=False)

<matplotlib.legend.Legend at 0x11f5766a0>