Newton's square root This is the code I wrote to compute the newton's square root of a number.

I aimed for extreme clarity over efficiency and I have put an abundant docstring at the start of the newton_sqrt function where I describe the main formula and explain all the parameters.

This function is not meant to be fast (standard library is 4x times faster and more accurate), I wrote just to learn some good math coding techniques in Python.


This was written in Python 3 but is fully compatible with Python 2.
I created constants for the error messages. Is that a good practice? Or should I just write the errors down verbatim when need occurs?
I chose n / 2 as my initial guess. Do you know a better initial guess?
Are all the option parameters that I am using necessary or should I delete some of them?




from __future__ import division
import time


def newton_sqrt(n,x0=None,tolerance=10 ** (-3),max_iter=15,epsilon=10 ** (-5)):
    """
   This function computes the sqare root of a number
   using Newton's method.
   After chosing a starting approximate value for x runs:
   x = x - f(x) / derivative_of_f(x)
   After a given number of iterations or if the result is good enough
   the programme will stop and return x.

   Parameters:
       Compulsory:
           @ n: the number of which the square root must be taken.
       Optional:
           @ x0: the initial approximation of n.
           @ tolerance: the maximum error permitted.
           @ max_iter: the maximum number of iterations.
           @ epsilon: a number so small that dividing by it would break
           @        the programme, giving vastly inapproximate results.
   """
    DIVISION_BY_TINY_DERIVATIVE_ERROR = """The derivative of f is {} and therefore under the {} limit chosen.
   Dividing by it would give a result with a vast error."""
    NOT_ENOUGH_ACCURACY_IN_ITER_ERROR = """Solution of {} accuracy for {} was not found in {} iterations"""

    if x0 is None:
        x0 = n / 2

    f = lambda x: x ** 2 - n
    fprime = lambda x: 2 * x  # The derivative of f(x)

    for i in range(max_iter):
        y = f(x0)
        yprime = fprime(x0)
        if(abs(yprime) &lt; epsilon):
            raise Exception(DIVISION_BY_TINY_DERIVATIVE_ERROR.format(
            yprime, epsilon))
        x1 = x0 - y / yprime
        if(abs(x1 - x0) / abs(x1) &lt; tolerance):
            return x1
        x0 = x1

    raise Exception(NOT_ENOUGH_ACCURACY_IN_ITER_ERROR.format(
    tolerance,n,max_iter))


def test_and_timing(max_):
    start = time.time()
    n_sqrt = [newton_sqrt(i) for i in range(1, max_)]
    print("Newton's sqrt took {} seconds to execute for i 1:{}".format(
        time.time() - start, max_))

    start = time.time()
    b_sqrt = [i ** 0.5 for i in range(1, max_)]
    print("Built-in sqrt took {} seconds to execute for i 1:{}".format(
        time.time() - start, max_))

    print("""Sum of all newton sqrts is: {}
   Sum of all built-in sqrts is: {}""".format(sum(n_sqrt), sum(b_sqrt)))

if __name__ == "__main__":
    test_and_timing(100000)
Looks good overall. Some suggestions:


The newton_sqrt is written in a quite generic manner. Unfortunately it may only calculate square roots - because lambdas are defined inside the function. I would rewrite it along the lines of

def newton_raphson(x, x0, func, derivative, governing_options):
    your code verbatim, except the lambda definitions 

def newton_sqrt(x, governing_options):
    return newton_raphson(x, x/2, lambda y: y*y - x, lambda y: 2*y, governing_options)

In any case, you most likely want to test for n &gt; 0 and raise InvalidArgiment accordingly.
I am not sure about the epsilon. If an abs(prime) &lt; epsilon condition triggers, the iteration would not converge anyway. I don't see an extra value here, but maybe it is just me.