Math

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>

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

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:

$\sum _{k=1}^{\infty }{\dfrac {1}{k^{2}6}}={\dfrac {2^{24}76977927\pi ^{26}}{27!}}$

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