import numpy as np
import matplotlib.pyplot as pt
def f(x):
return np.exp(x) - 2
xgrid = np.linspace(-2, 3, 1000)
pt.grid()
pt.plot(xgrid, f(xgrid))
[<matplotlib.lines.Line2D at 0x7f44dd2925c0>]
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
xbefore = 3
At each iteration, print the current guess and the error.
slope = (f(x)-f(xbefore))/(x-xbefore)
xbefore = x
x = x - f(x)/slope
print(x)
errors.append(abs(x-xtrue))
print(errors[-1])
nan nan
-c:1: RuntimeWarning: invalid value encountered in double_scalars
for err in errors:
print(err)
0.882400077493 0.411823511031 0.147482044876 0.0276859623403 0.00198268429064 2.73106724006e-05 2.70651508982e-08 3.69593244898e-13 1.11022302463e-16 1.11022302463e-16 nan
# Does not quite double the number of digits each round--unclear.
Let's check:
for i in range(len(errors)-1):
print(errors[i+1]/errors[i]**1.618)
0.504224909965 0.619635842142 0.612688428557 0.657169643929 0.644727394358 0.655276572424 0.648759771781 14482.1405299 7243300082.99 nan