Why is the computing of the value of pi using the Machin Formula giving a wrong value?

Question

For my school project I was trying to compute the value of using different methods. One of the formula I found was the Machin Formula that can be calculated using the Taylor expansion of arctan(x).

I wrote the following code in python:

import decimal count = pi = a = b = c = d = val1 = val2 = decimal.Decimal(0) #Initializing the variables decimal.getcontext().prec = 25 #Setting percision while (decimal.Decimal(count) <= decimal.Decimal(100)): a = pow(decimal.Decimal(-1), decimal.Decimal(count)) b = ((decimal.Decimal(2) * decimal.Decimal(count)) + decimal.Decimal(1)) c = pow(decimal.Decimal(1/5), decimal.Decimal(b)) d = (decimal.Decimal(a) / decimal.Decimal(b)) * decimal.Decimal(c) val1 = decimal.Decimal(val1) + decimal.Decimal(d) count = decimal.Decimal(count) + decimal.Decimal(1) #The series has been divided into multiple small parts to reduce confusion count = a = b = c = d = decimal.Decimal(0) #Resetting the variables while (decimal.Decimal(count) <= decimal.Decimal(10)): a = pow(decimal.Decimal(-1), decimal.Decimal(count)) b = ((decimal.Decimal(2) * decimal.Decimal(count)) + decimal.Decimal(1)) c = pow(decimal.Decimal(1/239), decimal.Decimal(b)) d = (decimal.Decimal(a) / decimal.Decimal(b)) * decimal.Decimal(c) val2 = decimal.Decimal(val2) + decimal.Decimal(d) count = decimal.Decimal(count) + decimal.Decimal(1) #The series has been divided into multiple small parts to reduce confusion pi = (decimal.Decimal(16) * decimal.Decimal(val1)) - (decimal.Decimal(4) * decimal.Decimal(val2)) print(pi)

The problem is that I am getting the right value of pi only till 15 decimal places, no matter the number of times the loop repeats itself.

For example:

at 11 repetitions of the first loop

pi = 3.141592653589793408632493

at 100 repetitions of the first loop

pi = 3.141592653589793410703296

I am not increasing the repetitions of the second loop as arctan(1/239) is very small and reaches an extremely small value with a few repetitions and therefore should not affect the value of pi at only 15 decimal places.

EXTRA INFORMATION:

The Machin Formula states that:

 π = (16 * Summation of (((-1)^n) / 2n+1) * ((1/5)^(2n+1))) - (4 * Summation of (((-1)^n) / 2n+1) * ((1/239)^(2n+1)))

I didn't check your code, but isn't there a problem with writing Decimal(1/5) ? It doesn't provides 0.2 but 0.2000....1110223024... because 1/5 is first converted to a float and a float can't store 0.2 exactly. Writing Decimal(1)/Decimal(5) provides 0.2 exactly — Jal
– Jal, Commented Sep 27, 2016 at 15:00

PM 2Ring · Accepted Answer · 2016-09-27 15:47:59Z

That many terms is enough to get you over 50 decimal places. The problem is that you are mixing Python floats with Decimals, so your calculations are polluted with the errors in those floats, which are only precise to 53 bits (around 15 decimal digits).

You can fix that by changing

c = pow(decimal.Decimal(1/5), decimal.Decimal(b))

to

c = pow(1 / decimal.Decimal(5), decimal.Decimal(b))

or

c = pow(decimal.Decimal(5), decimal.Decimal(-b))

Obviously, a similar change needs to be made to

c = pow(decimal.Decimal(1/239), decimal.Decimal(b))

You could make your code a lot more readable. For starters, you should put the stuff that calculates the arctan series into a function, rather than duplicating it for arctan(1/5) and arctan(1/239).

Also, you don't need to use Decimal for everything. You can just use simple Python integers for things like count and a. Eg, your calculation for a can be written as

a = (-1) ** count

or you could just set a to 1 outside the loop and negate it each time through the loop.

Here's a more compact version of your code.

import decimal decimal.getcontext().prec = 60 #Setting precision def arccot(n, terms): base = 1 / decimal.Decimal(n) result = 0 sign = 1 for b in range(1, 2*terms, 2): result += sign * (base ** b) / b sign = -sign return result pi = 16 * arccot(5, 50) - 4 * arccot(239, 11) print(pi)

output

3.14159265358979323846264338327950288419716939937510582094048

The last 4 digits are rubbish, but the rest are fine.

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