Theorem: Newton's Formula for Pi

$\pi$ (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:
:500px
Consider the semicircle embedded in the cartesian plane:
:whose radius is $\dfrac 1 2$
and
:whose center is the point $\tuple {\dfrac 1 2, 0}$. 
We have:
{{begin-eqn}}
{{eqn | l = \paren {x - \frac 1 2}^2 + \paren {y - 0}^2
      | r = \frac 1 2 ^2
      | c = Equation of Circle
}}
{{eqn | ll= \leadsto
      | l = y
      | r = \sqrt {\paren {x - x^2} }
}}
{{eqn | r = \sqrt x \sqrt {\paren {1 - x} }
}}
{{eqn | r = \sqrt x \paren {1 - \frac 1 2 x - \frac 1 8 x^2 - \frac 1 {16} x^3 - \frac 5 {128} x^4 - \frac 7 {256} x^5 - \cdots}
      | c = General Binomial Theorem
}}
{{eqn | r = \paren {x^{\frac 1 2} - \dfrac 1 2 x^{\frac 3 2} - \dfrac 1 8 x^{\frac 5 2} - \dfrac 1 {16} x^{\frac 7 2} - \dfrac 5 {128} x^{\frac 9 2} - \dfrac 7 {256} x^{\frac {11} 2} - \cdots}
}}
{{end-eqn}}
We calculate $\AA$ in $2$ different ways.
First we calculate it as the definite integral between $x = 0$ and $x = \dfrac 1 4$:
{{begin-eqn}}
{{eqn | l = \AA
      | r = \int_0^\frac 1 4 \paren {x^{\frac 1 2} - \dfrac 1 2 x^{\frac 3 2} - \dfrac 1 8 x^{\frac 5 2} - \dfrac 1 {16} x^{\frac 7 2} - \dfrac 5 {128} x^{\frac 9 2} - \dfrac 7 {256} x^{\frac {11} 2} - \cdots} \rd x
      | c = 
}}
{{eqn | r = \intlimits {\paren {\frac 2 3 x^{\frac 3 2} - \frac 1 5 x^{\frac 5 2} - \frac 1 {28} x^{\frac 7 2} - \frac 1 {72} x^{\frac 9 2} - \frac 5 {704} x^{\frac {11} 2} - \frac 7 {1664} x^{\frac {13} 2} - \cdots } } {x \mathop = 0} {x \mathop = \frac 1 4}
      | c = 
}}
{{eqn | n = 1
      | r = \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}
      | c = 
}}
{{end-eqn}}
{{qed|lemma}}
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 Congruence:
:$\triangle ABD = \triangle CBD$
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:
{{begin-eqn}}
{{eqn | l = BD^2 + BC^2
      | r = CD^2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = BD
      | r = \sqrt {CD^2 - BC^2}
      | c = 
}}
{{eqn | r = \sqrt {\paren {\dfrac 1 2}^2 - \paren {\dfrac 1 4}^2}
      | c = 
}}
{{eqn | r = \dfrac 1 4 \sqrt {4 - 1}
      | c = 
}}
{{eqn | r = \dfrac {\sqrt 3} 4
      | c = 
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = \AA
      | r = \AA_S - AA_T
      | c = 
}}
{{eqn | r = \dfrac 1 6 \paren {\pi \paren {\dfrac 1 2}^2} - \dfrac 1 2 \paren {\dfrac 1 4} \paren {\dfrac {\sqrt 3} 4}
      | c = 
}}
{{eqn | n = 2
      | r = \dfrac {\pi} {24} - \dfrac {\sqrt 3} {32}
      | c = 
}}
{{end-eqn}}
{{qed|lemma}}
Finally, setting $(1)$ equal to $(2)$:
{{begin-eqn}}
{{eqn | l = \dfrac {\pi} {24} - \dfrac {\sqrt 3} {32}
      | r = \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}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \pi
      | r = \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}
      | c = 
}}
{{end-eqn}}
{{qed}}
{{Namedfor|Isaac Newton|cat = Newton}}