Math - Revision history

Admin at 08:38, 14 August 2017

2017-08-14T08:38:42Z

← Older revision		Revision as of 08:38, 14 August 2017
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	\end{align}		\end{align}
	</math>		</math>

			==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==		==Euler==

Admin: /* MediaWiki Math */

2017-08-10T23:05:45Z

MediaWiki Math

← Older revision		Revision as of 23:05, 10 August 2017
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	<math>		<math>
	\begin{align}		\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{\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}		\pi^{-1} &=& \dfrac{1}{\pi}
	\end{align}		\end{align}

Admin at 23:05, 10 August 2017

2017-08-10T23:05:00Z

← Older revision		Revision as of 23:05, 10 August 2017
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			Note: https://www.encyclopediaofmath.org/index.php/Main_Page

			==MediaWiki Math==

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

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	<math>		<math>
	\begin{align}		\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{\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}		\end{align}
	</math>		</math>
	</pre>		</pre>

	~~Note: https:~~/~~/www.encyclopediaofmath.org/index.php/Main_Page~~		<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>

	==Euler==		==Euler==

Admin at 23:04, 10 August 2017

2017-08-10T23:04:12Z

← Older revision		Revision as of 23:04, 10 August 2017
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			To make a multiline equation in MediaWiki, use the following syntax:

			<pre>
			<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 }
			\end{align}
			</math>
			</pre>

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

Admin at 08:08, 15 July 2017

2017-07-15T08:08:09Z

← Older revision		Revision as of 08:08, 15 July 2017
Line 1:		Line 1:
			Note: https://www.encyclopediaofmath.org/index.php/Main_Page

	==Euler==		==Euler==

Admin at 07:41, 17 May 2017

2017-05-17T07:41:58Z

← Older revision		Revision as of 07:41, 17 May 2017
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	</math>		</math>

			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

	[[Category:Math]]		[[Category:Math]]

Admin: Created page with "==Euler== Infinite Series of Surprises: https://plus.maths.org/content/infinite-series-surprises Basel Problem: $\sum_{k=1}^{\infty} \dfrac{1}{k^2} = \dfrac{\pi^2}{6}..."$

2017-05-17T07:11:35Z

Created page with "==Euler== Infinite Series of Surprises: https://plus.maths.org/content/infinite-series-surprises Basel Problem: <math> \sum_{k=1}^{\infty} \dfrac{1}{k^2} = \dfrac{\pi^2}{6}..."

New page

==Euler==

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

Basel Problem:

<math>
\sum_{k=1}^{\infty} \dfrac{1}{k^2} = \dfrac{\pi^2}{6}
</math>

This proof extends to other even powers as well:

<math>
\sum_{k=1}^{\infty} \dfrac{1}{k^4} = \dfrac{\pi^4}{90}
</math>

and

<math>
\sum_{k=1}^{\infty} \dfrac{1}{k^6} = \dfrac{\pi^6}{945}
</math>

Then, in 1744, obtained:

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

by the same method.

This principle solves

<math>
\sum_{k=1}^{\infty} = \dfrac{1}{k^{2n}}
</math>

for natural numbers <math>n</math>.

The corresponding set of problems for odd powers,

<math>
\sum_{k=1}^{\infty} \dfrac{1}{k^3}
</math>

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

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

[[Category:Math]]