adivarshnei / approximating_pi_graphs

Approximating $\pi$ graphically using various formulae

The following formulae were compared:

• Double Factorial Formula

$$ \sum_{k=0}^{\infty}\frac{k!}{(2k+1)!!}=\sum_{k=0}^{\infty}\frac{2^k{k!}^2}{(2k+1)!}=\frac{\pi}{2} $$

• Bailey-Borwein-Plouffe Formula

$$ \sum_{k=0}^{\infty}\frac{1}{16^k}(\frac{4}{8k+1}-\frac{2}{8k+4}-\frac{1}{8k+5}-\frac{1}{8k+6})=\pi $$

• Basel Problem

$$ \zeta(2)=\sum_{k=1}^{\infty}\frac{1}{k^2}=\frac{\pi^2}{6} $$

• Leibniz Formula

$$ \sum_{k=0}^{\infty}\frac{(-1)^k}{2k+1}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\ldots=\arctan{1}=\frac{\pi}{4} $$

• Wallis Product

$$ \prod_{k=1}^{\infty}\frac{(2k)(2k)}{(2k-1)(2k+1)}=\frac{\pi}{2} $$

• Viète's Formula

$$ \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{2+\sqrt{2}}}{2}\cdot\frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2}\cdot\ldots=\frac{2}{\pi} $$

Formulae to add:

• Chudnovsky Algorithm

$$ \sum_{k=0}^{\infty}\frac{(-1)^k(6k)!(13591409+545140134k)}{(3k)!(k!)^3{640320}^{3k}}=\frac{4270934400}{\sqrt{10005}\pi} $$

• Ramanujan-Sato Series

$$ \sum_{k=0}^{\infty}\frac{(4k)!(1103+26390k)}{(k!)^4{396}^{4k}}=\frac{9801}{2\sqrt{2}\pi} $$

• Plouffe's series

$$ \sum_{k=1}^{\infty}k\frac{2^k{k!}^2}{(2k)!}=\pi+3 $$

• Riemann zeta function

$$ \zeta(4)=\sum_{k=1}^{\infty}\frac{1}{k^4}=\frac{\pi^4}{90} $$

• Newton, Second Letter to Oldenburg

$$ \sum_{k=0}^{\infty}\frac{(-1)^\frac{k^2-k}{2}}{2k+1}=1+\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{9}+\frac{1}{11}-\ldots=\frac{\pi}{2\sqrt{2}} $$

• Madhava series

$$ \sum_{k=0}^{\infty}\frac{(-1)^{k}}{3^k(2k+1)}=1-\frac{1}{3^1\cdot3}+\frac{1}{3^2\cdot5}-\frac{1}{3^3\cdot7}+\frac{1}{3^4\cdot9}-\ldots=\sqrt{3}\arctan\frac{1}{\sqrt{3}}=\frac{\pi}{2\sqrt{3}} $$

• Nilakantha series

$$ \sum_{k=0}^{\infty}\frac{(-1)^{k+1}}{k(k+1)(2k+1)}=\pi-3 $$

• $n$-th Fibonacci number

$$ \sum_{n=1}^{\infty}\frac{F_{2n}}{n^2\left({{2n}\atop{}n}\right)}=\frac{4\pi^2}{25\sqrt{5}} $$

• Euler's infinite product

$$ \left(\prod_{p\equiv1\text{ (mod }4)}\frac{p}{p-1}\right)\cdot\left(\prod_{p\equiv3\text{ (mod }4)}\frac{p}{p+1}\right)=\frac{3}{4}\cdot\frac{5}{4}\cdot\frac{7}{8}\cdot\frac{11}{12}\cdot\frac{13}{12}\cdot\ldots=\frac{\pi}{4} $$

• Euler's continued fraction

$$ \frac{4}{1+\cfrac{1^2}{2+\cfrac{3^2}{2+\cfrac{5^2}{2+\cfrac{7^2}{2+\ddots}}}}}=\pi $$

• Archimedes' algorithm

$$ a_0=2\sqrt3,b_0=3,a_{n+1}=\text{hm}(a_n,b_n),b_{n+1}=\text{gm}(a_{n+1},b_n),\pi=\lim_{n\rightarrow\infty}a_n=\lim_{n\rightarrow\infty}b_n $$

• Riemann sum to evaluate the area of a unit circle

$$ \lim_{n\rightarrow\infty}\frac{4}{n^2}\sum_{k=1}^{n}\sqrt{n^2-k^2}=\pi $$

• Remainder formula

$$ \lim_{n\rightarrow\infty}\frac{1}{n^2}\sum_{k=1}^{n}({n}\text{ mod }{k})=1-\frac{\pi^2}{12} $$

• Summing a circle's area

$$ \lim_{r\rightarrow\infty}\frac{1}{r^2}\sum_{x=-r}^{r}\sum_{y=-r}^{r}\left{\begin{matrix}1&\mbox{if}&\sqrt{x^2+y^2}\le{r}\0&\mbox{if}&\sqrt{x^2+y^2}\gt{r}\\end{matrix}\right.=\pi $$

• Combination of Stirling's approximation and Wallis product

$$ \lim_{n\rightarrow\infty}\frac{2^{4n}n!^4}{n(2n)!^2}=\lim_{n\rightarrow\infty}\frac{2^{4n}}{n\left({{2n}\atop{n}}\right)^2}=\lim_{n\rightarrow\infty}\frac{1}{n}\left({\frac{(2n)!!}{(2n-1)!!}}\right)^2=\pi $$

The comparison was against math.pi from Python's math package, and matplotlib was used to plot the graph.