Calculating Pi to the Nth digit

Question

I'm trying to enter in a number and calculate pi to that digit input. I managed to be able to calculate Pi, however no matter what number I type it will still generate the same amount of Pi numbers.

I'm a bit confused at what point it's causing to do that

from math import factorial
from decimal import Decimal, getcontext
# Chudnovsky algorithm for figuring out pi
getcontext().prec=100

pi_input = input('How many digits of pi would you like?')
n = int(pi_input)

def calc(n):
    t= Decimal(0)
    pi = Decimal(0)
    deno= Decimal(0)

    for k in range(n):
        t = ((-1)**k)*(factorial(6*k))*(13591409+545140134*k)
        deno = factorial(3*k)*(factorial(k)**3)*(640320**(3*k))
        pi += Decimal(t)/Decimal(deno)
    pi = pi * Decimal(12) / Decimal(640320 ** Decimal(1.5))
    pi = 1/pi
    return pi

print calc(n)

Here is my output

How many digits of pi would you like? 5 

3.141592653589793238462643383279502884197169399375105820974944592307816346
94690247717268165239156011

Solution 1

Using the Chudnovsky algorithm, the calculation produces about 14.18 decimal digits per iteration: log10((640320^3)/(24*6*2*6)) ~= 14.18. This can be more clearly seen in the formula for a_k / a_k-1 as shown on this web page:

https://www.craig-wood.com/nick/articles/pi-chudnovsky

For n = 5, the result has about 70 digits of precision.

Solution 2

I just added round function in the return statement in your code and hope and it works for you as it for me.

from math import factorial
from decimal import Decimal, getcontext
# Chudnovsky algorithm for figuring out pi
getcontext().prec=1000

pi_input = input('How many digits of pi would you like?')
n = int(pi_input)

def cal(n):
    t= Decimal(0)
    pi = Decimal(0)
    deno= Decimal(0)

    for k in range(n):
        t = ((-1)**k)*(factorial(6*k))*(13591409+545140134*k)
        deno = factorial(3*k)*(factorial(k)**3)*(640320**(3*k))
        pi += Decimal(t)/Decimal(deno)
    pi = pi * Decimal(12) / Decimal(640320 ** Decimal(1.5))
    pi = 1/pi
    return round(pi,n)

print(cal(n))

Comments

[klutt]

What makes you think it will produce n decimals? As far as I can see it will produce better results with a higher n but certainly not exactly n digits. And no, the result is different if you change n.

[sascha]

Without analyzing the code: what do you mean with your output? Show an example! It is possible, that the output is just a floating-point print and therefore the number of digits seen is not dependent on how many digits you computed? Be more verbose

[klutt]

To be clear, you will output the same amount of digits regardless of what n you pick, but the result will be more accurate with higher n.

[Blooze]

Updated my output, I guess I am confused how I can make my function perform the calculation based on the input

[il_raffa]

Along with the code, you should also add a description of the solution you're proposing.

[Luo Hongshuai]

not work for an N greater than 50 3.1415926535897931159979634685441851615905761718750000000000‌​0000000

[PirateNinjas]

rather than iterating you could use a slice, given that you're not calculating pi, per se.

[Ruli]

Such sulution was already proposed by multiple answers, why to add one more?

[OctopuSS7]

But this is calculating the whole thing first, so is very resourcce inneficient.

[Sreekant Shenoy]

This works only up to a limit of 15 decimal places since pi value imported from numpy or math consists of 3.141592653589793.

[Mutex]

nope after 48 digits it just shows 0000000000........

[Jules G.M.]

it's in the comment