A simple convergence comparison between Leibniz's and mine pi formula

Author: Kazi Abu Rousan

Here there is $\pi$, there is circle.

Today's blog will be a bit different. This will not discuss any formula or any proof, rather it will just contain a python program to compare a $\pi$ formula given by Leibniz and a one discovered by me (blush).

How does Leibniz formula looks like?

The Leibniz formula for $\pi$ is named after Gottfried Leibniz( published in 1676). Sometimes it is also called Leibniz-Madhava series as this is actually a special case of a more general formula discovered by Madhava of Sangamagrama in 14th century.

The formula is:

$\frac{\pi}{4}$ = $1 - \frac{1}{3}$ + $\frac{1}{5}$ - $\frac{1}{7}$ + $\frac{1}{9}$ -$\cdots$ + $\frac{1}{4n+1}$ - $\frac{1}{4n+3}$ + $\cdots$

If you want a visual proof, read my article : Article- click

or this video: Geometric proof

So, how can you find $\pi$ using this?, Just calculate each fractions and add.

As an example, $4, 4(1-\frac{1}{3}) = 2.6667$, $4(1-\frac{1}{3}$ + $\frac{1}{5}) = 3.4664$, $4(1-\frac{1}{3}$ + $\frac{1}{5}$ - $\frac{1}{7}) =2.895238 $, just like this.

But it is nowhere close to the value of $\pi = 3.141592653\cdots$

The conclusion is that, it converges very slowly. Let's ask a computer to find the value of $\pi$ using the series.

pi_val = 3.1415926535897932#right value upto 17 decimals
calculated_pi_from_series = 0
n = 1000
for i in range(0,n):
    calculated_pi_from_series = calculated_pi_from_series + 1/(4*i+1) - 1/(4*i+3)
calculated_pi_from_series = 4*calculated_pi_from_series
print(calculated_pi_from_series)
print("error = ",pi_val - calculated_pi_from_series)
#output: 3.1410926536210413
#output: error =  0.0004999999687518297

For a output of $3.141592153589724$ and error $= 5.000000689037165e-07$, we need $n$ = $10,00,000$. As you can see slow this series is. But even then it has great mathematical significance. It even has a beautiful geometric relationship with Fermat's two square theorem and prime numbers.

How does my formula looks like?

As it is my own discovery, it doesn't have any name but you may call it Rousan's Formula (🤣🤣). This one is almost identical to Leibniz's formula. It is so as it also uses 2d complex numbers as it's basis just like Leibniz's formula.

The main difference is that it mine uses Eisenstein Integers, i.e., triangular grid while Leibniz's uses Gaussian Integers, i.e., regular square grid.

This formula can be written as :

$\frac{\pi}{3\sqrt{3}}$ = $1 - \frac{1}{2}$ + $\frac{1}{4}$ - $\frac{1}{5}$ + $\frac{1}{7}$ - $\cdots$ + $\frac{1}{3n+1}$ - $\frac{1}{3n+2}$ + $\cdots$

Almost same right?, now let's use python to see how fast it converges.

import math as mp
pi_val = 3.1415926535897932#right value upto 17 decimals
calculated_pi_from_series = 0
n = 1000
for i in range(0,n):
    calculated_pi_from_series = calculated_pi_from_series + 1/(3*i+1) - 1/(3*i+2)
calculated_pi_from_series = 3*calculated_pi_from_series*mp.sqrt(3)
print(calculated_pi_from_series)
print("error = ",pi_val - calculated_pi_from_series)
#output: 3.141015303363362
#output: error =  0.0005773502264312391

For $n = 10,00,000$, we get $3.1415920762392955$ with an error of $5.773504976325228e-07$. So, This shows mine converges a bit more slowly.

A direct comparison and reason

So, why mine converge more slowly?, well it is hidden in it's shape. Mine uses triangular grid and leibniz's uses squares. This is the main reason. But fear not, I have actually found a more generalized version of fermat's two square theorem, by using it i can get another similar series which converges faster than leibniz. Maybe I will discuss that in some later blog.

For now, let's try to plot both series's error. To we will plot the number of terms taken in x axis and error in y axis.

import matplotlib.pyplot as plt
import math as mp
pi_val = 3.1415926535897932
sqrt_3_3 = 3*mp.sqrt(3)

def leib_err(n):
    result = 0
    for i in range(n):
        result = result + 1/(4*i+1) - 1/(4*i+3)
    result = 4*result
    error = pi_val - result
    return result

def rou_err(n):
    result = 0
    for i in range(n):
        result = result + 1/(3*i+1) - 1/(3*i+2)
    result = result*sqrt_3_3
    error = pi_val - result
    return result
n_vals = [i for i in range(10_000+1)]
leibniz = [leib_err(i) for i in n_vals]
# print(leibniz)
rousan  =  [rou_err(i) for i in n_vals]
#
plt.plot(n_vals,leibniz,c='red',label='Leibniz Error')
plt.plot(n_vals,rousan,c='black',label='My formula Error')
plt.legend(loc='best',prop={'size':7})
plt.ylim(3.1375,3.1425)
plt.title("Error vs No. of terms for Leibniz and my $\\pi$ formula.")
plt.show()

The output is:

This is all for today, see you guys next time with something new.