### Newton-Raphson method

Newton-Raphson is a very popular method for the numerical calculation of an equation’s root. Its’ basic concepts for formulation originate from the Taylor theorem and of course the fact that function value becomes zero at the root point. You can read more about the method in the Wikipedia entry.

### SymPy

SymPy is a Python library which enables the use of symbolic mathematical operations. It can solve equations, differentiate or integrate, simplify complex expressions or evaluate mathematical functions.

In various distributions you can install the package directly with the help of the corresponding package manager. For example in Ubuntu:

apt-get install python-sympy

For the latest version, you can use `pip`

:

pip install sympy

### Code

Our “solver” will receive the precision as an optional parameter, with default value of 5⋅10^{-6}, the function representation in a SymPy compatible format and letting **x** be the symbolic variable (check the docs) and the starting guess value.

#!/usr/bin/env python import argparse import math import sys import time import sympy def get_parameters(): parser = argparse.ArgumentParser() parser.add_argument( "-f", "--function", help="Define a function" ) parser.add_argument("-s", "--starting", help="Starting point value", type = float, default = 0.0) parser.add_argument("-p", "--precision", help="Convergence precision", type = float, default = 5*10**(-6)) return parser.parse_args() if __name__ == "__main__": parameters = get_parameters() sym_x = sympy.Symbol('x') # convert the given function to a symbolic expression try: fx = sympy.S(args.function) except sympy.core.sympify.SympifyError: sys.exit('Unable to convert function to symbolic expression.') # calculate the differential of the function dfdx = diff(fx, Symbol('x')) # e is the relative error between 2 consecutive # estimations of the root e = 1 x0 = parameters.starting iterations = 0 start_time = time.time() while e > args.precision: # new root estimation try: r = x0 - fx.subs({sym_x: x0}) / dfdx.subs({sym_x: x0}) except ZeroDivisionError: print "Function derivative is zero. Division by zero, program will terminate." raise # relative error e = abs((r - x0) / r) iterations += 1 x0 = r total_time = time.time() - start_time print 'Function:' sympy.pprint(fx) print 'Derivative:' sympy.pprint(dfdx) print 'Root %10.6f calculated after %d iterations' % (r, iterations) print 'Function value at root %10.6f' % fx.subs({sym_x : r}) print 'Finished in %10.6f seconds'% total_time

### Example

We want to solve the equation

^{2}= 400

near the initial value 910 (the positive root).

This is the output of a test run.

$ ./nr.py -f '6*x**2 - 400' -s 910 Function: 6⋅x - 400 Derivative: 12⋅x Root 8.164966 calculated after 11 iterations Function value at root 0.000000 Finished in 0.035108 seconds