Wallis' formula for pi

Question

def wallis(n):
    pi = 0.0

    for i in range(n):
        left = (2 * i)/(2 * i - 1)
        right = (2 * i)/(2 * i + 1)
        total = left * right
        pi = pi + total

    return pi

print wallis(1000)
print wallis(10000)
print wallis(100000)

I copied the formula exactly but I keep getting 0 as the output. Can someone please tell me what I am doing wrong. Python 2.7.

The link to the formula is here

Answer 1

Python is doing integer division, and truncates the decimals. This is because both values to division are integers. Convert one of the numbers to a float to get a floating point value in return.

left = float(2 * i)/(2 * i - 1)
right = float(2 * i)/(2 * i + 1)

OR, as @kindall points out, you can change the constants to floats directly and avoid the call to the float function:

left = (2.0 * i)/(2 * i - 1) # just 2. works, too
right = (2.0 * i)/(2 * i + 1)

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If/when you switch to python 3.x, you won't need to do this. In fact, you need to explicitly request integer division with //.

As per a comment by @Serdalis, you could also add from __future__ import division at the top of your file to get the same behavior as python 3.x (i.e. you won't need to add the float in your equation.)

Answer 2

Apart from problem highlighted by @SethMMorton your formula is wrong. First it is a product not sum, second it gives pi/2 not pi. At last there is no reason to loop from 0.

def wallis(n):
    pi = 2.
    for i in xrange(1, n):
        left = (2. * i)/(2. * i - 1.)
        right = (2. * i)/(2. * i + 1.)
        pi = pi * left * right
    return pi