Convergence of Newton's Method

Copyright (C) 2020 Andreas Kloeckner

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import numpy as np
import matplotlib.pyplot as pt

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

def df(x):
    return np.exp(x)

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

[<matplotlib.lines.Line2D at 0x7fec13f69048>]

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

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

0.69314718056
0.0

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

errors = []
x = 2

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

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

0.69314718056
0.0

for err in errors:
    print(err)

0.577523385913
0.13881012318
0.00920340345722
4.22216905669e-05
8.91323015395e-10
0.0

• Do you have a hypothesis about the order of convergence?

# Doubles number of digits each iteration: probably quadratic.

---

Let's check:

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

0.416180751055
0.47764604026
0.498469622214
0.499992953401
0.0