Exercise 6.1.13

We can also apply Newton’s method to $f(x)=\mathit{sin}(x)$. Depends on the initial value, we get different solutions each time.

from scipy import optimize 
import numpy as np 
import math 
import matplotlib.pyplot as plt 
 
def f(x): 
   return x**2+1 
 
def df(x): 
   return 2*x 
 
#root = optimize.newton(f, 1.5, fprime = lambda x: math.cos(x)) 
#root 
 
def newton(f, df, x0, maxit=500): 
   x = x0 
   xs, ys = [x0], [f(x0)] 
 
   for it in np.arange(maxit): 
       x = x - f(x)/df(x) 
       if np.abs(f(x)) <= 1.0e-12: 
           print(’Find␣the␣root:’, x, f(x), ’␣at␣iteration:␣’, it) 
           break 
       xs.append(x) 
       ys.append(f(x)) 
   print("Exhausted␣all␣", maxit, "␣iterations.") 
   return x, f(x), xs, ys 
 
print("####␣Problem␣VI.1.12") 
v, fv, xs, ys = newton(f, df, 2) 
 
print("####␣Problem␣VI.1.13") 
v13, fv13, xs13, ys13 = newton(math.sin, math.cos, 7)

\#\#\#\# Problem VI.1.12
Exhausted all  500  iterations.
\#\#\#\# Problem VI.1.13
Find the root: 6.283185307179586 -2.4492935982947064e-16  at iteration:  3
Exhausted all  500  iterations.

plt.plot(xs, ys, ’.’)

[<matplotlib.lines.Line2D at 0x292ac0c5128>]