From charlesreid1

Revision as of 21:22, 25 May 2026 by Admin (talk | contribs) (Add Wallis Product for Pi and Stirling's Approximation (via update-page on MediaWiki MCP Server))

The most beautiful formulas:

Ramanujan's inverse pi formula:

$ \pi^{-1} = \displaystyle{ \dfrac{\sqrt{8}}{99^2} } \displaystyle{ \sum_{k \geq 0} \dfrac{ (4k)! }{ (4^k k!)^4 } } \displaystyle{ \dfrac{1103 + 26390 k}{99^{4k}} } $

Wallis Product for Pi [1]:

$ \dfrac{\pi}{2} = \prod_{n=1}^{\infty} \left( \dfrac{2n}{2n-1} \cdot \dfrac{2n}{2n+1} \right) = \dfrac{2}{1} \cdot \dfrac{2}{3} \cdot \dfrac{4}{3} \cdot \dfrac{4}{5} \cdot \dfrac{6}{5} \cdot \dfrac{6}{7} \cdots $

Gaussian integral:

$ \int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{2 \pi} $

Ramanujan sum:

$ \sum_{k=1}^{\infty} k = - \frac{1}{12} $

Euler formula [2]:

$ e^{i \pi} + 1 = 0 $

Euler's sum of inverse squares:

$ \sum_{n=1}^{\infty} \dfrac{1}{n^2} = \dfrac{\pi^2}{6} $

Stirling's Approximation [3]:

$ n! \sim \sqrt{2 \pi n} \left( \dfrac{n}{e} \right)^n $

Archimedes' Recurrence Formula:

$ a_{2n} = \frac{2 a_n b_n}{a_n + b_n} $

$ b_{2n} = \sqrt{a_{2n} b_n} $