jconorgrogan/Slice-of-Pi

Unique way to Find Pi via repeated application of angle bisection using only a compass and straightedge

Slice of Pi

TLDR; Geomtric way of representing a derivation of Viete's formula for pi that involves nested square roots of two

Definitions and Assumptions

Let ( R = 2^n ) for some positive integer ( n ). Consider a circle of radius ( R ) centered at the origin. Divide the circle into ( 2R ) equal slices. The ( y )-coordinate of the first slice's intersection point above the ( x )-axis approaches ( \pi ) as ( n \rightarrow \infty ).

1. Define ( R = 2^n ) where ( n ) is a positive integer.
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2. Consider a circle of radius ( R ) centered at the origin.
3. Divide the circle into ( 2R ) equal slices.
4. The angle for each slice is ( \theta = \frac{2\pi}{2R} = \frac{\pi}{R} ).
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Y-Coordinate Expression

The ( y )-coordinate for the first slice's intersection point above the ( x )-axis in Cartesian coordinates is ( y = R \sin(\theta) ). Substituting ( \theta = \frac{\pi}{R} ) and ( R = 2^n ), we get ( y = 2^n \sin\left(\frac{\pi}{2^n}\right) ). -

Limit Calculation

We want to find ( \lim_{n \to \infty} 2^n \sin\left(\frac{\pi}{2^n}\right) ). -

Substitute ( u = \frac{\pi}{2^n} ) and ( 2^n = \frac{\pi}{u} ), so ( u \to 0 ) as ( n \to \infty ). The limit becomes ( \lim_{u \to 0} \frac{\pi}{u} \sin(u) ). Utilizing the limit identity ( \lim_{u \to 0} \frac{\sin(u)}{u} = 1 ), the limit is ( \pi ). -

Conclusion:

• Therefore, the ( y )-coordinate of the first slice's intersection point above the ( x )-axis approaches ( \pi ) as ( n \rightarrow \infty ).

Generalized Slice of Pi Theorem for k-th Intersection Point

Theorem

Let ( R = 2^n ) for some positive integer ( n ). Consider a circle of radius ( R ) centered at the origin and divide it into ( 2R ) equal slices. The ( y )-coordinate of the ( k )-th slice's intersection point above the ( x )-axis approaches ( k\pi ) as ( n \rightarrow \infty ).

Definitions and Assumptions

• Define ( R = 2^n ) where ( n ) is a positive integer.
• Consider a circle of radius ( R ) centered at the origin.
• Divide the circle into ( 2R ) equal slices.
• The angle for the ( k )-th slice is ( \theta_k = k \left( \frac{2\pi}{2R} \right) = k \left( \frac{\pi}{R} \right) ).

Y-Coordinate Expression

The ( y )-coordinate for the ( k )-th slice's intersection point above the ( x )-axis in Cartesian coordinates is ( y_k = R \sin(\theta_k) ).

Substituting ( \theta_k = k \frac{\pi}{R} ) and ( R = 2^n ), we get: [ y_k = 2^n \sin\left(k \frac{\pi}{2^n}\right) ]

Limit Calculation

We want to find ( \lim_{n \to \infty} 2^n \sin\left(k \frac{\pi}{2^n}\right) ).

Substitute ( u = k \frac{\pi}{2^n} ) and ( 2^n = \frac{k \pi}{u} ), so ( u \to 0 ) as ( n \to \infty ).

The limit becomes ( \lim_{u \to 0} \frac{k \pi}{u} \sin(u) ).

Utilizing the limit identity ( \lim_{u \to 0} \frac{\sin(u)}{u} = 1 ), the limit is ( k \pi ).

Therefore, as ( n \rightarrow \infty ), the ( y )-coordinate of the ( k )-th intersection point will approach ( k \pi ).

5. Visualizations:

Each of these horizontal lines approaches a multiple of pi as R goes to infinity

As R increases, each 2nd node y coordinate moves towards 2pi, 3rd node 3pi, etc. How each point converges to these multiples of pi creates some extremely interesting fractals which are worthy of future research

The infinite prime ladder, and the Grogan Rollercoaster