sqrt.py

# Find sqrt(2)
# Get a graph of y^2 - 2, find the 0. Use newtons's method.
import math

def iterate_root(start, n, p):
    dx = .01
    dy = ((start + dx) ** p) - (start ** p)
    return start - ((start**p) - n)/(dy/dx)

def get_sqrt(n):
    return reduce(lambda cur, y: iterate_root(cur, n, 2), [10] * 10)

def get_cube(n):
    return reduce(lambda cur, y: iterate_root(cur, n, 3), [10] * 10)

def test_sqrt():
    assert int(get_sqrt(2) * 1e5) == int(math.sqrt(2) * 1e5)
    assert int(get_sqrt(3) * 1e5) == int(math.sqrt(3) * 1e5)
    assert int(get_cube(2) * 1e5) == int(2 ** (1/3.0) * 1e5)