From charlesreid1

Note: https://www.encyclopediaofmath.org/index.php/Main_Page

MediaWiki Math

To make a multiline equation in MediaWiki, use the following syntax:

<math>
\begin{align}
\pi^{-1} &=& \dfrac{\sqrt{8}}{99^2} \sum_{k \geq 0} \dfrac{ (4k)! }{ (4^k k!)^4 } \dfrac{1103 + 26390 k}{ 99^4k } \\
\pi^{-1} &=& \dfrac{1}{\pi}
\end{align}
</math>


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

Floating Point Numbers

What every computer scientist should know about floating point numbers: http://docs.oracle.com/cd/E19957-01/806-3568/ncg_goldberg.html

Euler

Infinite Series of Surprises: https://plus.maths.org/content/infinite-series-surprises

Basel Problem:


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

This proof extends to other even powers as well:


\sum_{k=1}^{\infty} \dfrac{1}{k^4} = \dfrac{\pi^4}{90}

and


\sum_{k=1}^{\infty} \dfrac{1}{k^6} = \dfrac{\pi^6}{945}

Then, in 1744, obtained:


\sum_{k=1}^{\infty} \dfrac{1}{k^26} = \dfrac{2^{24} 76977927 \pi^{26} }{27!}

by the same method.

This principle solves


\sum_{k=1}^{\infty} = \dfrac{1}{k^{2n}}

for natural numbers n.

The corresponding set of problems for odd powers,


\sum_{k=1}^{\infty} \dfrac{1}{k^3}

is still an open problem. The best Euler could do was:


\sum_{k=0}^{\infty} \dfrac{ (-1)^k }{ (2k+1)^3 } = 1 - \frac{1}{27} + \frac{1}{125} - \dots = \dfrac{ \pi^3 }{32}

Translation of Euler's paper: Remarques sur un beau rapport entre les series des puissances tant directes que reciproques

http://eulerarchive.maa.org//docs/translations/E352.pdf