# coding: utf-8

# # Convergence of Newton's Method

# In[2]:

import numpy as np
import matplotlib.pyplot as pt


# In[3]:

def f(x):
    return np.exp(x) - 2


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xgrid = np.linspace(-2, 3, 1000)
pt.grid()
pt.plot(xgrid, f(xgrid))


# What's the true solution of $f(x)=0$?

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xtrue = np.log(2)
print(xtrue)
print(f(xtrue))


# Now let's run Newton's method and keep track of the errors:

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errors = []
x = 2
xbefore = 3


# At each iteration, print the current guess and the error.

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slope = (f(x)-f(xbefore))/(x-xbefore)

xbefore = x
x = x - f(x)/slope
print(x)
errors.append(abs(x-xtrue))
print(errors[-1])


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for err in errors:
    print(err)


# * Do you have a hypothesis about the order of convergence?

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# Does not quite double the number of digits each round--unclear.


# ------------
# Let's check:

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for i in range(len(errors)-1):
    print(errors[i+1]/errors[i]**1.618)