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

# # Example of finite precision

# Let us evaluate
# $$
# y = x^3 - 3 x^2 + 3 x - 1
# $$
# in single precision. Note that it can also be written as 
# $$
# y = (x-1)^3
# $$

# In[18]:


get_ipython().run_line_magic('matplotlib', 'inline')
get_ipython().run_line_magic('config', "InlineBackend.figure_format = 'svg'")
import numpy as np
import matplotlib.pyplot as plt


# We evaluate in the interval $[0.99,1.01]$.

# In[19]:


x = np.linspace(0.99,1.01,50,dtype=np.float32)
y = x**3 - 3.0*x**2 + 3.0*x - 1.0
plt.plot(x,y,'-o',x,(x-1)**3)
plt.xlabel('x')
plt.ylabel('y')
plt.legend(('Single precision','Exact'));


# Now, let us do it in double precision

# In[20]:


x = np.linspace(0.99,1.01,50,dtype=np.float64)
y = x**3 - 3.0*x**2 + 3.0*x - 1.0
plt.plot(x,y,'-o',x,(x-1)**3)
plt.xlabel('x')
plt.ylabel('y')
plt.legend(('Double precision','Exact'));