Math
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>
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \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:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \sum_{k=1}^{\infty} \dfrac{1}{k^2} = \dfrac{\pi^2}{6} }
This proof extends to other even powers as well:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \sum_{k=1}^{\infty} \dfrac{1}{k^4} = \dfrac{\pi^4}{90} }
and
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \sum_{k=1}^{\infty} \dfrac{1}{k^6} = \dfrac{\pi^6}{945} }
Then, in 1744, obtained:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \sum_{k=1}^{\infty} \dfrac{1}{k^26} = \dfrac{2^{24} 76977927 \pi^{26} }{27!} }
by the same method.
This principle solves
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \sum_{k=1}^{\infty} = \dfrac{1}{k^{2n}} }
for natural numbers Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle n} .
The corresponding set of problems for odd powers,
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \sum_{k=1}^{\infty} \dfrac{1}{k^3} }
is still an open problem. The best Euler could do was:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \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