# coding: utf-8

# # Newton's method in $n$ dimensions

# In[1]:

import numpy as np
import numpy.linalg as la


# In[3]:

def f(xvec):
    x, y = xvec
    return np.array([
        x + 2*y -2,
        x**2 + 4*y**2 - 4
        ])

def Jf(xvec):
    x, y = xvec
    return np.array([
        [1, 2],
        [2*x, 8*y]
        ])


# Pick an initial guess.

# In[20]:

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


# Now implement Newton's method.

# In[27]:

x = x - la.solve(Jf(x), f(x))
print(x)


# Check if that's really a solution:

# In[30]:

f(x)


# * What's the cost of one iteration?
# * Is there still something like quadratic convergence?

# --------------------
# Let's keep an error history and check.

# In[32]:

xtrue = np.array([0, 1])
errors = []
x = np.array([1, 2])


# In[41]:

x = x - la.solve(Jf(x), f(x))
errors.append(la.norm(x-xtrue))
print(x)


# In[42]:

for e in errors:
    print(e)


# In[43]:

for i in range(len(errors)-1):
    print(errors[i+1]/errors[i]**2)