PI day

Pi Day is March 14. This is an opportunity to test all sorts of ideas for calculating this number.
I loved David Shiftman’s song on the youtube channel “The coding train”.
This is the only number you can eat ! 🥮

Coding Challenge 169: Pi in the Sky Game

The Discovery That Transformed Pi

The Veritasium video “The discovery that transformed Pi” is extraordinary.

This video has 11M views, and it’s no coincidence.

It shows Isaac Newton as a real genius, in the sense that he invented a revolutionary method of calculation starting from the trivial “binomial expansion”, related to Pascal’s triangle and known for centuries.

The binomial expansion is below.

\[(1+x)^n = 1 + nx + \frac{n\times(n-1)}{2!}x^2 + \frac{n\times(n-1)\times(n-2)}{3!}x^3 + \frac{n\times(n-1)\times(n-2)\times(n-3)}{4!}x^4 + ...\]

This formula was well known with “n” as an integer.
Newton discovered it works also with negative and fractional “n” values.
When n is an integer, the serie ends because a zero appears when k=n and all remaining terms are then multiplied by zero. When n is negative or fractional the sequence turns to be infinite but eventually converges to the expected value !

I wanted to compute PI with the formula described in the Video.

\[\pi = \frac{3\times\sqrt{3}}{4} - 24\times\sum_{n=0}^{\infty}\frac{(2n)!}{2^{4n+2}\times(n!)^2\times(2n-1)(2n+3)}\]

I wanted also compute the square root of 3 as Newton would do, using the following formula. $(x+1)^\frac{1}{2}$ with $x=-\frac{1}{4}$

\[sqrt{3} = 2\times(1-\frac{1}{4})^\frac{1}{2}\] \[(1-\frac{1}{4})^\frac{1}{2} = 1 - \frac{1}{8} - \frac{1}{128} - \frac{1}{2^{10}} + \frac{5}{2^{15}} ...\]

The code below is not optimized at all.

from math import *
# factorial function
def fact(n):
    f=1
    for k in range(n):
        f=f*(k+1)
    return f
def expansion(x,n,c):
    s=1
    num=n
    for k in range(1,c):
        s+=(num/fact(k))*(x**k)
        num*=n-k
    return s 
def sqr3(c):
    return 2*expansion(-1/4,1/2,c)
def newton_pi(c):
    npi=3*sqr3(c)/4
    for k in range(c):
        npi-=24*fact(2*k)/((2**(4*k+2))*(fact(k)**2)*(2*k-1)*(2*k+3))
    return npi
for i in range(1,21):
    print(str(i)+" iterations.")
    print(newton_pi(i))
    print(pi)
    print()

This gives after 20 iterations, a number very close to the pi of the math library, that is great for a first try !

20 iterations.
3.1415926535897993
3.141592653589793

References

The Discovery That Transformed Pi - Veritasium
Calculating π by hand the Isaac Newton way: Pi Day 2020
NEWTON’S APPROXIMATION FOR PI
TEACHER NOTES AND SOLUTIONS