- #1

tan(pi/4) = 1

arctan(x) = [sum]

_{n=0}

^{[oo]}(-1)

^{n}* [ x

^{2n+1}/ (2n + 1) ]

pi = 4 * arctan(1) = 4 * [sum]

_{n=0}

^{[oo]}(-1)

^{n}* [ 1 / (2n + 1) ]

= 4 * ( 1 - 1/3 + 1/5 - 1/7 + 1/9 - 1/11 + ... )

= 4 * ( 2/3 + 2/35 + 2/99 + ... )

= 4 * [(1 * 2/3) + (4 * 1/70) + (8 * 1/396) + ...]

Remember the equation for the area of a circle.

A = pi * R

^{2}

= (2*R)

^{2}* [(1 * 2/3) + (4 * 1/70) + (8 * 1/396) + ...]

= D

^{2}* [(1 * 2/3) + (4 * 1/70) + (8 * 1/396) + ...]

Now, draw a circle with diameter D. In that circle, place a square with area (2/3)*D

^{2}. Notice that there is still some empty space left on the sides square (where the square and the circle do not both exist). Place four squares with area 1/70*D

^{2}in each of these four empty spaces. Continue to fill up the unoccupied space of the circle with progressively smaller squares ad infinitum.

Pi is simply an infinite series sum of scaling factors needed for filling a circle with squares in order to perform a

*numerical*integration method.

eNtRopY