Draft:Pi

π is a mathematical constant defined as the ratio between the circumference of a circle and its diameter. Its value is approximately 3.1415926535897932385 but let's derive an exact value:

Consider the unit circle $x^{2}+y^{2}=1$:

$y^{2}=1-x^{2}$

${\frac {d_{y}}{d_{x}}}\cdot 2y=2x$

${\frac {d_{y}}{d_{x}}}=-{\frac {2y}{2x}}$

${\frac {d_{y}}{d_{x}}}=-{\frac {x}{y}}$

Also note that for the top half of the circle $y={\sqrt {1-x^{2}}}$.

Therefore, for the top half of the circle,

Now let’s consider the vector perpendicular to the unit circle at specific point on the top part of the circle and has an x-component of 1:

It is given by ${\vec {v}}(x)=<1,{\frac {d_{y}}{d_{x}}}>$

Therefore, ${\vec {v}}(x)=<1,-{\frac {x}{\sqrt {1-x^{2}}}}>$

$|{\vec {v}}(x)|=|<1,-{\frac {x}{\sqrt {1-x^{2}}}}>|$

$|{\vec {v}}(x)|={\sqrt {1^{2}+\left(-{\frac {x}{\sqrt {1-x^{2}}}}\right)^{2}}}$

$|{\vec {v}}(x)|={\sqrt {1^{2}+{\frac {x^{2}}{1-x^{2}}}}}$

$|{\vec {v}}(x)|={\sqrt {{\frac {1-x^{2}}{1-x^{2}}}+{\frac {x^{2}}{1-x^{2}}}}}$

$|{\vec {v}}(x)|={\sqrt {\frac {1-x^{2}+x^{2}}{1-x^{2}}}}$

$|{\vec {v}}(x)|={\sqrt {\frac {1}{1-x^{2}}}}$

$|{\vec {v}}(x)|={\frac {1}{\sqrt {1-x^{2}}}}$

File:20251102161222!chunkedupload c9b80ee5afe0.png

This table is used to derive a Taylor's series for u(x).

Let $u(x)={\frac {1}{\sqrt {1-x^{2}}}}$.

From the table to the right, we can conclude that the Taylor's series for u(x) is $u(x)=1+\sum _{n=0}^{\infty }\left({\frac {x^{\left(n+1\right)}\prod _{m=0}^{n}\left(2m+1\right)}{\left(n+1\right)!\left(2^{\left(n+1\right)}\right)}}\right)$

$|{\vec {v}}(x)|=u(x^{2})$

Therefore, $|v(x)|=1+\sum _{n=0}^{\infty }\left({\frac {\left(x^{2}\right)^{\left(n+1\right)}\prod _{m=0}^{n}\left(2m+1\right)}{\left(n+1\right)!\left(2^{\left(n+1\right)}\right)}}\right)$

$|{\vec {v}}(x)|=1+\sum _{n=0}^{\infty }\left({\frac {x^{\left(2n+2\right)}\prod _{m=0}^{n}\left(2m+1\right)}{\left(n+1\right)!\left(2^{\left(n+1\right)}\right)}}\right)$.

Since ${\vec {v}}(x)$ is a vector perpendicular to a point on the unit circle whose x-component is 1, $\arcsin(x)=\int _{0}^{x}|{\vec {v}}(a)|d_{x}$:

$\arcsin {x}=\int _{0}^{x}1+\sum _{n=0}^{\infty }\left({\frac {a^{\left(2n+2\right)}\prod _{m=0}^{n}\left(2m+1\right)}{\left(n+1\right)!\left(2^{\left(n+1\right)}\right)}}\right)d_{x}$

$\arcsin {x}=x+\sum _{n=0}^{\infty }\left({\frac {x^{\left(2n+3\right)}\prod _{m=0}^{n}\left(2m+1\right)}{\left(n+1\right)!\left(2n+3\right)\left(2^{\left(n+1\right)}\right)}}\right)$

Because $\sin \left({\frac {\pi }{6}}\right)={\frac {1}{2}}$, we can deduce that $\arcsin \left({\frac {1}{2}}\right)={\frac {\pi }{6}}$ which means that $\pi =6\arcsin {\left({\frac {1}{2}}\right)}$. If we substitute $\arcsin {x}=x+\sum _{n=0}^{\infty }\left({\frac {x^{\left(2n+3\right)}\prod _{m=0}^{n}\left(2m+1\right)}{\left(n+1\right)!\left(2n+3\right)\left(2^{\left(n+1\right)}\right)}}\right)$ into $\pi =6\arcsin {\left({\frac {1}{2}}\right)}$ we get:

$\pi =6\left({\frac {1}{2}}+\sum _{n=0}^{\infty }\left({\frac {\left({\frac {1}{2}}\right)^{\left(2n+3\right)}\prod _{m=0}^{n}\left(2m+1\right)}{\left(n+1\right)!\left(2n+3\right)\left(2^{\left(n+1\right)}\right)}}\right)\right)$

$\pi =3+6\sum _{n=0}^{\infty }\left({\frac {\prod _{m=0}^{n}\left(2m+1\right)}{\left(n+1\right)!\left(2n+3\right)\left(2^{\left(3n+4\right)}\right)}}\right)$

Therefore, the exact value of π is $3+6\sum _{n=0}^{\infty }\left({\frac {\prod _{m=0}^{n}\left(2m+1\right)}{\left(n+1\right)!\left(2n+3\right)\left(2^{\left(3n+4\right)}\right)}}\right)$.