Write a Python procedure to implement basic Newton\'s Method. The definition sta

Question

Write a Python procedure to implement basic Newton's Method. The definition statement of your program should be def Newt(f, fp, x0, TOL, N): where f and fp are the function and derivative, x0 is the initial guess, TOL is the stopping tolerance, and N is the maximum number of iterations allowed. Implement the stopping condition where the difference of the iterates is less than the tolerance. Test your program on the function f(x) = x - cos x on [0, 1]. Print out each iteration and compare with the numbers from class or Excel. Use 20 as the maximum number of iterations. Your output statement should be of the form: The solution to the equation is XXXX in NNNN iterations. or The solution wasn't found in N iterations. Where X0X0X0 is the initial value x_0, XXXX is the value the sequence converges to and NNNN is the number of iterations it takes to converge.

Explanation / Answer

CODE:

# Newton Raphson Method.

import math
"""
* Variable Description:
*
* f           :   Given function
* fp       :   Derivative of f
* x0       :   Initial value
* TOL       :   Tolerance
* N           :   Max number of iterations
"""

def Newt(f, fp, x0, TOL, N):
   """
   Takes a function f, its derivative fp, initial value x0, tolerance value TOL and
   max number of iterations N and returns the root of the equation
   using the newton-raphson method.
   """
   n=1
   while n<=N:
       # use the Newton Raphson formula to calculate the root
       x1 = x0 - (f(x0)/fp(x0))
       print x1
      
       # Compare with the Tolerance value.
       if x1 - x0 < TOL:
           return (x1,n)
       else:
           x0 = x1
          
       # Increment the iteration count.
       n += 1
   return (False,n)

if __name__ == "__main__":
              
   # f(x) = x - cos(x)
   def func(x):
       return x-math.cos(x)

   # fp(x) = 1 + sin(x)
   def func_(x):
       return 1+math.sin(x)
  
   #Invoking Newton Raphson Method
   res,iter = Newt(func,func_,0.1,0.0001,20)
  
   if res != False:
       print "The solution to the equation is " + str(res) + " in " + str(iter) + " iterations."
  
   else:
       print "The solution wasn't found in " + str(iter) + " iterations."

OUTPUT:

$python newton.py
0.913763386101
0.744664241982
The solution to the equation is 0.744664241982 in 2 iterations.

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