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Approximate

p 126 \[ \sqrt{4} = 2 \] One way of square roots is with Newton's method.
 
# Newton method:
# 
# For finding roots ... assert distance(1, 2, 4, 6) == 0.0
# is a root-finding algorithm to produce succesive better aproximations
# 
# Suppose that you want to know the square root of a. 
# If you start with almost any estimate x ...
# you can compute a better estimation with the following formula:
#     y = (x + a/x) / 2
# 
# For example with a = 4, x = 3
# the result is preatty close to correct answer 2
# (3 + 4/3) / 2 = 2.16666666667
#
# If we repeat the process with the new estimate, it gets even closer


def square(a, x):
    while True:
        y = (x + a/x) / 2
        if (y == x): 
            break
        x = y
        print(y)
    return y

y = square(4, 10)

# 5.2
# 2.9846153846153847
# 2.1624107850911973
# 2.006099040777959
# 2.00000927130158
# 2.000000000021489
# 2.0

Float

p 128 In general it is dangerous to test float equality.
 
# Newton method v2
#
# In most programming languages, ... 
# it is dangerous to test float equality, ...
# because floating-point values are only approximately right.
#
# It is safer to use the built-in function abs to compute the absolute value,
#
# epsilon: how close is close enought

def square(a, x, epsilon):
    while True:

        y = (x + a/x) / 2
        print(y)
        
        if abs(y - x) < epsilon: 
            break
        x = y

y = square(64, 10, 0.0000001)

# 8.2
# 8.002439024390243
# 8.000000371689179
# 8.000000000000009
# 8.0

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