Newton's Formula for Pi

Theorem
$\pi$ (pi) can be approximated using the formula:
 * $\pi = \dfrac {3 \sqrt 3} 4 + 24 \paren {\dfrac 2 {3 \times 2^3} - \dfrac 1 {5 \times 2^5} - \dfrac 1 {28 \times 2^7} - \dfrac 1 {72 \times 2^9} - \dfrac 5 {704 \times 2^{11} } - \dfrac 7 {1664 \times 2^{13} } - \cdots}$

Proof
Let $\AA$ denote the area of the shaded region in the following diagram:


 * Newtons-Approximation-to-Pi.png

Consider the semicircle embedded in the cartesian coordinate plane:
 * whose radius is $\dfrac 1 2$

and
 * whose center is the point $\tuple {\dfrac 1 2, 0}$.

We have:

We calculate $\AA$ in $2$ different ways.

First we calculate it as the definite integral between $x = 0$ and $x = \dfrac 1 4$:

Then we calculate $\AA$ using the techniques of plane geometry.

From the construction, we have that:
 * $AC = CD$
 * $AB = BC$
 * $BD$ is common

so by Triangle Side-Side-Side Equality:
 * $\triangle ABD = \triangle CBD$

Thus:
 * $AD = AC = CD$

and so $\triangle ACD$ is equilateral.

Thus we have that $\triangle BCD$ has angles $30 \degrees$, $60 \degrees$ and $90 \degrees$.

Hence by Pythagoras's Theorem:

Then we observe that $\AA$ is:
 * the area $\AA_S$ of the sector of the semicircle whose central angle is $\dfrac \pi 3$

minus:
 * the area $\AA_T$ of the right triangle $\triangle BCD$.

$\AA_S$ is $\dfrac 1 6$ of the area of the circle whose radius is $\dfrac 1 2$.

Thus from Area of Circle:
 * $\AA_S = \dfrac 1 6 \paren {\pi \paren {\dfrac 1 2}^2}$

$\AA_T$ is given from Area of Triangle in Terms of Side and Altitude as:
 * $\AA_T = \dfrac 1 2 \paren {\dfrac 1 4} \paren {\dfrac {\sqrt 3} 4}$

Hence:

Finally, setting $(1)$ equal to $(2)$: