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 a specific point on the top part of the circle which has an x-component of 1:

It is given by $\overrightarrow{v}(x)=<1,\frac{d_y}{d_x}>$

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

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

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

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

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

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

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

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

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

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

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

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

Since $\overrightarrow{v}(x)$ is a vector perpendicular to a point on the unit circle whose x-component is 1, $\arcsin (x)=\underset{0}{\overset{x}{\int }}|\overrightarrow{v}(a)|d_x$:

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

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

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 $\arcsin x=x+\sum _{n=0}^{\mathrm{\infty }}\left(\frac{x^{(2n+3)}\prod _{m=0}^n(2m+1)}{(n+1)!(2n+3)\left(2^{(n+1)}\right)}\right)$ into $\pi =6\arcsin \left(\frac{1}{2}\right)$, we get:

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

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

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

References

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