The first 5000 decimals of the π
The value of π
The Ratio method
A simple method based on the experience. The ratio of a circle's attributes is nearly constant.
Steps to calculate:
You will need: something in a circle shape, ruler, string
- Measure the circumference of the circle.
- Measure the diameter of the circle.
- Divide the circumference with the diameter.
- Repeat the process on diferent size of circles.
Examples
- IKEA glass: 28.3cm / 9cm = 3.1444
- Glass at ESS: 28.8cm / 9.5cm = 3.0315
- Bycicle wheel: 223cm / 69.8cm = 3.1948
The Gregory-Leibniz Series
The derivation begins with the formula for the geometric series, which is as follows: $$ \frac{1}{1 - x} = 1 + x + x^2 + x^3 + \dots = \sum_{n=0}^{\infty} x^n \quad \text{where } |x| < 1 $$
Substitute the \(-x^2\) value for the \(x\) value in the formula:
$$ \frac{1}{1 - (-x^2)} = 1 + (-x^2) + (-x^2)^2 + (-x^2)^3 + \dots $$ $$ \frac{1}{1 + x^2} = 1 - x^2 + x^4 - x^6 + \dots = \sum_{n=0}^{\infty} (-1)^n x^{2n} $$Now we integrate both sides of the equation by \(dx\):
$$ \int \frac{1}{1 + x^2} \,dx = \int (1 - x^2 + x^4 - x^6 + \dots) \,dx $$On the left, \(\frac{1}{1 + x^2}\) is integrated by the \(\arctan(x)\) (arctangent) function:
$$ \int \frac{1}{1 + x^2} \,dx = \arctan(x) + C $$On the right-hand side, we integrate the series term by term:
$$ \int (1 - x^2 + x^4 - x^6 + \dots) \,dx = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \dots + C $$Combining the two sides, we get the power series of the arctangent:
$$ \arctan(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \dots $$Finally, we use the mathematical fact that \(\arctan(1) = \frac{\pi}{4}\). Substitute \(x=1\) into the power series:
$$ \arctan(1) = 1 - \frac{1^3}{3} + \frac{1^5}{5} - \frac{1^7}{7} + \dots $$ $$ \frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \dots $$This has led to the Gregory-Leibniz formula, which gives the value of Pi.
\[ \frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \dots = \sum_{k=0}^{\infty} \frac{(-1)^k}{2k+1} \]
Let's give it a try: