text:
User:Begoon/vector.js

//////////////////////////////////////////////// //--My familiar Gadgets to make me comfy...---// //////////////////////////////////////////////// //UTC Clock/Purge link in toolbar importScript("User:Begoon/UTCLiveClock.js"); //Open external links in new tab/window importScript("User:Begoon/exlinks.js"); //embeds ACE Cloud9 editor for editing JS/CSS importScript("User:Begoon/CodeEditor.js"); // Tools/Navigation popups mw.loader.load('https://en.wikipedia.org/w/index.php?title=MediaWiki:Gadget-popups.js&action=raw&ctype=text/javascript'); mw.loader.load('https://en.wikipedia.org/w/index.php?title=MediaWiki:Gadget-navpop.css&action=raw&ctype=text/css', 'text/css'); //////////////////////////////////////////////// //---END "GADGETS"// ////////////////////////////////////////////////

//Read some configuration values for WikiEditor toolbar /////////////////////////////////////////////////////// importScript("User:Begoon/ConfWikiEditor.js");

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: User:Begoon/vector.js


text:
User:Begoon/vector.css

@import url("https://en.wikipedia.org/w/index.php?title=User:Begoon/override-typography-refresh.css&action=raw&ctype=text/css"); /* Preceding line includes code from User:Begoon/override-typography-refresh.css to override the typography refresh @import lines must be at very beginning of css page, preceding even comments... */

/* Ace Code Editor background not horrible blue, and font bigger */ div.ace_content {background:white; font-size: 120%;}

/* Syntax highlighting stuff... */ div.mw-geshi { padding: 1em; margin: 1em 0; border: 1px dashed #2f6fab; background-color: #f9f9f9; } /* ...more... Syntax highlighting stuff... */ pre.css, div.css, pre.source-css, div.source-css, pre.php, div.php, pre.source-php, div.source-php, pre.javascript, div.javascript, pre.source-javascript, div.source-javascript { font-size:1.1em; white-space: pre-wrap; }

title: User:Begoon/vector.css


text:
User talk:OpenScientist/DataONE Workshop on Data Governance 2011

I hope this is the talk page (?) I'm still fairly new to this. What style do you want references in, or do you even want formal references?

I'll try and contribute bits here and there where I think I can. Metacladistics 15:03, 29 November 2011 (UTC)


text:
User:Metacladistics

http://about.me/rossmounce

title: User:Metacladistics


text:
Arduino/Junk pile steeper motor survey

The purpose of this page is to record the wiring of junked stepper motors to arduino motor shields.

Stepper motors have 4 or more wires going into them. Not all motors with 4 or more wires are stepper motors. Typically these motors do not settle down and display a constant ohm value. Instead the resistance reading go to zero or infinity. Switching the polarity of the probes measuring resistance causes different resistance values.

Next Steps

 * Attach motor shield to larger power source .. 9-12volts .. perhaps up to 1.5amps. See how motor strength improves and how chips on motor shield heat up.
 * Get spark fun motor shields working .. suppose to handle more power
 * Count teeth on the gears to see if there are standards that can be leveraged
 * Research PC power supply specifications to see if they are going to be useful
 * Figure out how fast the motors can rotate .. some seem pretty slow .. this may be due to the stepper motor program or sketch.

title: Arduino/Junk pile steeper motor survey


text:
User:Сергей Киже

Немного о себе: Я - Наблюдатеть. Большего о себе пока не скажу:) Я на Facebook

title: User:Сергей Киже


text:
User talk:Сергей Киже

 Hello Сергей Киже, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. All users are expected to abide by our Privacy policy, Civility policy, and the Terms of Use while at Wikiversity.

To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.
 * Explore Wikiversity with the links to your left.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
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You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --HappyCamper 16:29, 29 November 2011 (UTC)

title: User talk:Сергей Киже


text: TESOL/Recommending clothing

title: TESOL/Recommending clothing


text:
Spanish History

Welcome / Bienvenido (under construction)
Welcome to the Spanish History topic page, where slowly but surely a whole suite of courses will be developed which will take you through the history of this Southern European country. The first course to debut will be "Spanish History - a rough guide", which will be followed by more developed courses on "20th Century Spain" and "The Spanish Second Republic".

Spanish History - A Rough Guide
"Spain is different". In just three words one can start to unpack the multifaceted history of a country that encompasses huge deserts, snowcapped mountains, bustling towns and a persevering linguistic diversity. You may know a lot about Spain, you may know nothing, but below you will find a short introduction to Spain and its history, in Spanish: España, from the pre-romanic period to the present day.

title: Spanish History


text:
User talk:Alexander.Llewelyn

 Hello Alexander.Llewelyn, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. All users are expected to abide by our Privacy policy, Civility policy, and the Terms of Use while at Wikiversity.

To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.
 * Explore Wikiversity with the links to your left.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.
 * Follow Wikiversity on twitter (http://twitter.com/Wikiversity) and identi.ca (http://identi.ca/group/wikiversity).

You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --HappyCamper 20:24, 29 November 2011 (UTC)

title: User talk:Alexander.Llewelyn


text:
Controversies in Science/RNA world vs metabolism first/A Critique of Ribozyme catalysis of metabolism in the RNA world

(Review Paper) Cited in RNA world vs metabolism first

The RNA world could have all the things it needed to survive, including the ability to manufacture nucleotides and perform other necessary metabolic reactions.


text:
Controversies in Science/RNA world vs metabolism first/A Critique of Hydrothermal vents and the origin of life

(Review Paper) Cited in RNA world vs metabolism first

The origin of life is believed to have started from CO2. There is evidence that life was able to start from the elements and proteins present in the deep sea vents. Carbon and hydrogen were also important elements in the creation of life.

Results
There is a vent system that have carbonate chimneys that rise 60 meters above the sea floor. They have a similar composition to that of very early volcanoes, and they get extremely hot. This is named the Lost City hydrothermal field (LCHF). There has been evidence of ribosomal RNA in the venting carbonated chimneys in the LCHF. There are some microbial communities also within the vents. The bacteria that is present in the Lost City is able to synthesize it's own ATP.


text:
Controversies in Science/RNA world vs metabolism first/A Critique of Origin of Self-Replicating Biopolymers: Autocatalytic Feedback Can Jump-Start the RNA World

(Journal article) Cited in RNA world vs metabolism first

Life as we know was formed from an RNA world with life based on self-replicating biopolymers. If life was created from biopolymers, the metabolism theory would be incorrect because biopolymers are not rocks. Biopolymers are molecules that could have been found in places like ponds or beneath a desert floor where there is moisture.

Points Made
Life as we know was formed from an RNA world with life based on self-replicating biopolymers. If life was created from biopolymers, the metabolism theory would be incorrect because biopolymers are not rocks. Biopolymers are molecules that could have been found in places like ponds or beneath a desert floor where there is moisture. When RNA begins reproducing it starts out as a steady state and then the polymers concentration greatly increases. The first ribozymes that are created do not need to be perfect. A fully autocatalytic metabolic system does not work as the information in those molecules are not stored in sequence.

Methods
A set of chemical reaction equations describing synthesis of RNA polymers was used. Wu and Higgs used a stochastic simulation in a finite volume to back the points made in the article.

Potential Biases
The math used in the equations to back the evidence could have been done incorrectly. There could have been an error in a program or process in stochastic simulation. The numbers used are hypothetical and not fact biasing the calculations. Also, the stochastic simulation is done in a finite volume where the volume is hypothetical. There are also the personal biases such as money, recognition, and an extremist view against the metabolism theory.


text:
User talk:Dfazz11

 Hello Dfazz11, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. All users are expected to abide by our Privacy policy, Civility policy, and the Terms of Use while at Wikiversity.

To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.
 * Explore Wikiversity with the links to your left.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.
 * Follow Wikiversity on twitter (http://twitter.com/Wikiversity) and identi.ca (http://identi.ca/group/wikiversity).

You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --HappyCamper 03:48, 30 November 2011 (UTC)

title: User talk:Dfazz11


text:
Research in education: Open and networked practices/Full paper

This is a work in progress, articulating a full article about open and networked method of research for education.

Example of citizen research: Surfacestations
The leaked email records of researchers connected to the Climate Research Unit at the University of East Anglia revealed, among other things, a lack of openness and transparency. Temperature data sets could not be made available to wide peer review - despite Freedom of Information requests, due to a range of legal and subjective decisions on the part of the defending scientists, ultimately resulting in wide questioning of the reliability of their findings and recommendations. In response to, and in direct contrast to those closed practices, was a 'citizen science' project called Surfacestations.org, where large numbers of volunteers networked online to coordinate independent checks of the instrument stations used to collect surface temperature data in the USA (Watts 2011). Better known citizen science projects include NASA's Stardust@home, Zooniverse, Galaxy Zoo, and many others, such as the Wikimedia projects.

Open data is critical to a open, networked research. Open management practices are arguably desirable as well, if volunteers are to be attracted to work on large data sets and to trust lead researchers enough to help organise, gather and interpret information. Such are the motivations behind the Wikileaks project, yet another project pitching some hard questions to established publishers of investigative journalism. What if these methods and processes where to extend to other fields? Could we see research institutes being challenged in much the same way as Wikipedia has challenged encyclopedia projects, or the way activist meteorologists have challenged climate science methods and findings?

Complimenting these participatory developments are web service businesses like Kickstarter which aims to assist people and organisations build investment and social capital around their proposals. What is the potential in the challenges and opportunities that these participatory research practices, and support web service businesses, present to traditional researchers and their institutions? There are examples where new and independent groups have attracted human resources quickly and far beyond the capacities of traditional institutions, associations or even consortiums with significantly more funding.

Arguably these participatory projects, and the capacity of some to source crowds of volunteers, is a culture emerging from the internet. Alternatively, it is a culture being projected through the internet (Blackall 2009). Or perhaps the massive online cooperation projects we are seeing are evolving via a combination of internet culture and the collaborative, open spirit of science and research. Key elements of the emergent open, networked research culture are openness, transparency, digital formats, asynchronicity, and a tolerance for anonymity and open participation. But under these more amiable qualities are the presence of bullying and trolling, fickleness, new and opache hierarchies and processes, prejudices, group think and cultish behaviours (Keen 2007).

Edgelessness - research as learning
From Keith - What a thought-provoking abstract.

On reading it I thought about 1. Edgeless institutions 2. MOOCs

... and how both are linked to function rather than form.

I read the abstract after looking at:

About HREA's Distance Learning Programme

Since 2002, over 4,000 human rights defenders, development workers, staff members of international organisations and graduate students have successfully participated in Human Rights Education Associates (HREA)'s e-learning courses.

Further information about HREA's Distance Learning Programme can be found at: http://www.hrea.org/DLP/

 * Leigh's suggestion:
 * Edgeless institutions could be added in relation to para 1 where Rob writes: "If we see research as a form of learning and the dissemination of its results, a specific instance of teaching (Wagner, 1993), research can be seen as an educational practice." This could be expanded by us, where Keith introduces MOOCs and the HREA example. So, if Keith could write a paragraph of two, we could expand that edgelessness between research, practice and learning that Rob introduces.

Some additional references to consider

 * 1) Australian Research Council supports open access for research pubs & data, at up to 2% of grant http://alexholcombe.wordpress.com/2011/01/07/finally-australian-research-council-supports-open-access/
 * 2) Should scientific data be shared more openly – and how would that help science? Have your say: http://blogs.nature.com/news/2011/05/opening_up_scientific_data.html?WT.mc_id=TWT_natureblogs
 * 3) Open science in The Guardian: http://www.guardian.co.uk/education/2011/may/22/open-science-shared-research-internet

P2P production
''This essay, written in a manifesto form, addresses some crucial issues related to the timely topic of the distributed or Peer-to-Peer (P2P) energy production. It uses the emerging mode of the P2P production in the immaterial field of production (information, culture, knowledge) as a point of departure to realize the dynamics of this new energy technology and shed light on its socio-economic aspects.'' Bauwens (2011)

Wikipedia and the Death of the Expert
This article looked like it was going to lend intellectual weight to the idea that crowd sourced knowledge disembodies expertise, but instead it gave another sales pitch for Wikipedia. Still, I list it here as a resource to stimulate thought and discussion. I would like us to more honestly represent the concerns of the nay sayers, understand them and present their truth as the real yet timid challenge it is. Some of the comments give pointers to a deeper perspective.


text:
User talk:Esak7

My User Talk!

Hello everyone

title: User talk:Esak7


text:
User talk:Nabvedr1

Preferred Perfumes - Precisely What Helps Make All Of Them So Well Liked? Parfum definitely seems to be significant business as of late. Just about every full week that passes by I notice in the news that yet another celebrity has brought out their very own innovative brand of scent. It tends to make me question exactly how much feedback they will really have about the fragrance or whether or not they are simply just presented with a range to pick from!

There are some timeless colognes in existence that have continued to sell for a long time with hardly any and sometimes virtually no marketing and advertising at all. Hanae Mori perfume is one such fragrance which has been a highly regarded seller for pretty much fifteen years. The first cologne which Hanae Mori launched was actually the Butterfly Hanae Mori perfume long ago in 1996 which continues to sell and sell without any marketing and advertising.

So what on earth makes somebody go out and buy a parfum from a public figure. Is it truly the indisputable fact that it smells wonderful? For some reason I do not think so. The truth is I would place a wager that many folks who purchase a parfum from their own favored celebrity do not know what it has the scent of! Many individuals will still only buy it simply because they like that particular star when in reality the actual cologne might truly smell like cheap candle fragrance!

Quite a few colognes which are launched at present have a limited shelf life. In fact, the shelf-life is really as long as that certain star is actually well-liked, after that you can just about wager it's going to become a collectors item. Now compare and contrast that to the actual Hanae Mori Butterfly parfum. That will most definitely continue to be a number one seller in ten years time!

title: User talk:Nabvedr1


text: File:Ponytrussfb.jpg

title: File:Ponytrussfb.jpg


text:
Collaborative play writing/The Countess of Challand

The countess of Challand

Dramatic characters (18)

Bianca, countess of Challand

Count Mansino, lover of Bianca

Count Baizzo, lover of Bianca

The duke of Bourbon

Pietro di Cardona, gentleman

Carlo di Cardona, gentleman

Riccardatto, steward of Bianca

Agostino, steward of Mansino

Clara, wife of Agostino

Noce-Moscata, daughter of Agostino

Torbido, master of a brothel

Voga, mistress of the brothel

Vago, client in the brothel

Decio, suitor to Noce-Moscata

Pompina, servant to Mansino

Alicia, servant to Mansino

Servilio, servant to Bianca

Michela, grocer's daughter

An officer-of-law

Time: Early 17th century

Place: Italy

/Act_1/

/Act_2/

/Act_3/

/Act_4/

/Act_5/


text: Template:Watchduck

title: Template:Watchduck


title: File:Function.Overview.1.A.20111201.pdf


text:
File:Escher triangle veiw 1.JPG

Licensing
for school project

title: File:Escher triangle veiw 1.JPG


text:
File:Escher triangle veiw 2.JPG

Licensing
for school project

title: File:Escher triangle veiw 2.JPG


text:
File:Escher uphill roll 1.JPG

Licensing
for school project

title: File:Escher uphill roll 1.JPG


text:
File:Escher uphill roll 2.JPG

Licensing
for school project

title: File:Escher uphill roll 2.JPG


text:
File:Escher uphill roll 3.JPG

Licensing
for school project

title: File:Escher uphill roll 3.JPG


text:
File:Escher uphill roll 4.JPG

Licensing
for school project

title: File:Escher uphill roll 4.JPG


text:
User:Egm6322.s12

$$\displaystyle \color{red}{\clubsuit}$$ EGM 6322 Principles of Engineering Analysis 2, Spring 2012 (syllabus and other important info) Prof. Loc Vu-Quoc

University of Florida

Interesting news: $$\displaystyle \color{red}{\clubsuit}$$ Advancing the Open Front (about MITx), By Steve Kolowich, Inside Higher Ed, December 20, 2011. $$\displaystyle \color{red}{\clubsuit}$$ M.I.T. Expands Its Free Online Courses (about MITx), By TAMAR LEWIN, New York Times, Published: December 19, 2011.

 Licensing info ,  Interesting news

= Important links =

PEA2 S12 is a continuation of  PEA1 F11.

PEA1 S12 Lecture plan. Other versions: Spring 2009.

$$\displaystyle \color{red}{\clubsuit}$$

PEA1 F11 Lecture plan.

PEA2 S12 Report table. Other versions: Spring 2009.

$$\displaystyle \color{red}{\clubsuit}$$

PEA1 F11 Report table.

Writing tools

Other courses

Time: New York time, full screen Stopwatch full screen

title: User:Egm6322.s12


text:
File:Escher uphill roll 6.JPG

Licensing
for school project

title: File:Escher uphill roll 6.JPG


text:
File:Escher uphill roll 7.JPG

Licensing
for school project

title: File:Escher uphill roll 7.JPG


text:
User:EGM6321.f10.team1.Zhichao Gong/Mtg2

Mtg 2: Thu, 27 Jan 11

 * {| style="width:100%" border="0"

 * style="width:95%" |
 * style="width:95%" |

page2-1

$${ \color{blue} {P.1-3,}} \ {\color{red}{Eq. (2)}} \ : f=u^{'} \ or \ u^{2} \ {\color{red} (1)}$$

$$Dynamics \ : \ F=m \underset{ \underset{accel \rightarrow { { \overset{}{u}}^{1} \ and \ { \overset{}{u}}^{2} \ and \ { \overset{''}{Y}}^{1} }}{ \uparrow}}{a}$$

HW*: Derive(3)-(4) p.1-3 *:Improvements over existing solution on course wiki: - Better explanations, figures etc. - Theoretical background related to problem -(4)p.1-3: similar to  Coriolis force Dynamics Reynold(Material Time derivate) Transport Thm
                                                          Continum Mech

(heat, solids, fluids, electronag)

Similarities and differences Montesquieu

Equation of motion (EOM) of wheel/magnet

$${\color{blue} \underset{c_{3}:{\color{red}mass} \ this \ term \ is  \ N2-ODE}{ \underbrace}}+c_{2}(Y^{'},t){\color{red} { \left( {\color{black}{ \overset{'}{Y}}^{1} }\right)}^{2}}+c_{1}(Y^{'},t){ \overset{'}{Y}}^{1}+{\color{blue} \underset{ \underbrace}}=0 \ {\color{red}(3)}$$

page2-2

$${\color{blue}\underline {Note:}} \ Terms \ with \ { \overset{'}{Y}}^{1} \ are \  \underset{not} \ due \ to damping \ {\color{,}} \ but \ to \ {\color{blue}\underline \ {\color{black}convection}}$$

$${\color{blue}Ref: \ VQ4O \ 1989 \ CMAME \ {\color{red}Eqs \ (2.1 \ bc)}}$$

$${\color{blue} \underset{"There \ exists "}{ \underbrace}} \ a \ mispring \ in \ {\color{red}(2.16)}$$

$$c_{0}(Y^{1},t)=-{\color{green} \overset{(A)}{ \overbrace}}-F^{2}u^{2}_{,s}- \frac{T}{R}+ \underset{M}{\color{red}[{\color{black}[1- \overset{-}{R}u^{2}_{,ss}][u^{'}_{,tt}- \overset{-}{R}u^{2}_{,stt}]+u^{2}_{,s}u^{2}_{,tt}}]} \ {\color{red}(1)}$$

From (3) p.2-1:-c0 = horizontal force acting on wheel and magnet

$${\color{green}(A)} \ := \F^{1}[1-\overset{-}{R}u^{2}_{,ss}]$$

$${\color{blue}Dimen Anal:} \ {\color{blue} \underset{ \overset{ \uparrow}{"dimens.  \ of"}}}=F \ {\color{blue}(force)}$$


text:
User:Julius2012

''' I AM THE GREATEST FAN OF THE BEATLES IN THE WORLD ! '''

title: User:Julius2012


text:
TAO/Overview

This page contains links and information to get an quick overview of the resources of the project within Wikiversity.

Please remember to use categories. And think about a good place for a new “resource” / page as you create it.

Index of all pages starting with TAO
You can get an overview of all pages by following this link.

Content with subsites (for pdf-Version)
 Introduction

Chapter
 * subsites 1
 * Subsites 2

Chapter
 * subsites 1
 * Subsites 2

Categories used

 * Category:TAO → all pages belonging to the project.
 * Category:TAO Handbook → all pages of the handbook.

title: TAO/Overview


text:
TAO/Templates

For the start simply keep the plain list updated:
 * Template:TAO/Infobox Activity
 * Template:TAO/LayoutMainpage
 * Template:TAO/LayoutEntrypoint
 * Template:TAO/LayoutCoP
 * General Drafts

title: TAO/Templates


text: Template:TAO/Infobox Activity

title: Template:TAO/Infobox Activity


text:
User:Beat Estermann/sandbox-1

test

title: User:Beat Estermann/sandbox-1


text:
User:Sammethod

Definitions

Where E is Erdos-Borwein constant

Problem 7.

-

Show that,

$$\frac{1}{9}=\frac{2}{(2^1-1)(2^2-1)(2^3-1)}+\frac{2^2}{(2^2-1)(2^3-1)(2^4-1)}+\frac{2^3}{(2^3-1)(2^4-1)(2^5-1)}+\cdots$$

I found this series through error and trial.

---

Problem 6.

Here, I am trying to find the close form for series $$X_1$$,

[1]

$$X_1=1-2\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}\right)+3\left(\frac{1}{3\cdot4}-\frac{1}{ 4\cdot5}+\frac{1}{5\cdot6}-\frac{1}{6\cdot7}\right)-4\left(\frac{1}{7\cdot8}-\cdots-\frac{1}{14\cdot15}\right)+\cdots$$

and through the process end up with these two series as shown below,

[2]

$$\frac{X_1+E+2\ln2-3}{2}+\frac{11}{18}=\frac{3\cdot2^2}{(2^2-1)(2^3-1)}+\frac{5\cdot2^4}{(2^4-1)(2^5-1)}+ \frac{7\cdot2^6}{(2^6-1)(2^7-1)}+\cdots$$

[3]

$$\frac{E-2\ln2-1-X_1}{2}+\frac{43}{18}=\frac{2\cdot2^1}{(2^1-1)(2^2-1)}+\frac{4\cdot2^3}{(2^3-1)(2^4-1)}+ \frac{6\cdot2^5}{(2^5-1)(2^6-1)}+\cdots$$

and we combined [2] and [3] we have which is already known (From problem 1).

$$\frac{X_1+E+2\ln2-3}{2}+\frac{E-2\ln2-1-X_1}{2}+\frac{11}{18}+\frac{43}{18}=\frac{2\cdot2}{(2^1-1)(2^2-1)}+\frac{3\cdot2^2}{(2^2-1)(2^3-1)}+ \frac{4\cdot2^3}{(2^3-1)(2^4-1)}+\cdots$$

$$\frac{2E-4}{2}+3=\frac{2\cdot2^1}{(2^1-1)(2^2-1)}+\frac{3\cdot2^2}{(2^2-1)(2^3-1)}+ \frac{4\cdot2^3}{(2^3-1)(2^4-1)}+\cdots$$

therefore we have (from problem 1.)

$$E+1=\frac{2\cdot2^1}{(2^1-1)(2^2-1)}+\frac{3\cdot2^2}{(2^2-1)(2^3-1)}+ \frac{4\cdot2^3}{(2^3-1)(2^4-1)}+\cdots$$

Here is how I worked it out, for [2] and [3]

---

Let $$X$$ and $$X_1$$ be the two series as shown below,

$$X=1-2\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}\right)-3\left(\frac{1}{3\cdot4}-\frac{1}{ 4\cdot5}+\frac{1}{5\cdot6}-\frac{1}{6\cdot7}\right)-4\left(\frac{1}{7\cdot8}-\cdots-\frac{1}{14\cdot15}\right)-\cdots$$

$$X_1=1-2\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}\right)+3\left(\frac{1}{3\cdot4}-\frac{1}{ 4\cdot5}+\frac{1}{5\cdot6}-\frac{1}{6\cdot7}\right)-4\left(\frac{1}{7\cdot8}-\cdots-\frac{1}{14\cdot15}\right)+\cdots$$

and

Let $$Y_0$$ and $$X_1$$ be the two series as shown below,

$$Y_0 = \frac{1}{1\cdot2}-2\left(\frac{1}{2\cdot3}-\frac{1}{3\cdot4}\right)-3\left(\frac{1}{4\cdot5}-\frac{1}{5\cdot6}+\frac{1}{6\cdot7}-\frac{1}{7\cdot8}\right)-4\left(\frac{1}{8\cdot9}-\cdots-\frac{1}{15\cdot16}\right)-\cdots$$

$$Y_1 = \frac{1}{1\cdot2}-2\left(\frac{1}{2\cdot3}-\frac{1}{3\cdot4}\right)+3\left(\frac{1}{4\cdot5}-\frac{1}{5\cdot6}+\frac{1}{6\cdot7}-\frac{1}{7\cdot8}\right)-4\left(\frac{1}{8\cdot9}-\cdots-\frac{1}{15\cdot16}\right)+\cdots$$

These two series [A] and [B] it is known

[A]

$$-\frac{25}{18}+\frac{1}{2}=-\frac{2}{2^2}-\frac{4}{2^4}-\frac{6}{2^6}-\cdots$$

[B]

$$-\frac{11}{18}=-\frac{3}{2^3}-\frac{5}{2^5}-\frac{7}{2^7}-\cdots$$

The reason I leaves these series in this form, because it makes the calculating process easy to do.

We also known the following (from solving other problems)

$$X=2\gamma-2\ln2+2-E\cdot$$

$$Y_0=2\ln2-2\gamma\cdot$$

$$Y_1=2\ln2-1\cdot$$

Solution 6 to [3]

-

$$X+X_1=2-4\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}\right)-8\left(\frac{1}{7\cdot8}-\cdots-\frac{1}{14\cdot15}\right)- 12\left(\frac{1}{31\cdot32}-\cdots-\frac{1}{62\cdot63}\right)-\cdots$$

[C]

$$\frac{X+X_1}{2}=1-2\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}\right)-4\left(\frac{1}{7\cdot8}-\cdots-\frac{1}{14\cdot15}\right)- 6\left(\frac{1}{31\cdot32}-\cdots-\frac{1}{62\cdot63}\right)-\cdots$$

________________

$$Y_0+Y_1=1-4\left(\frac{1}{2\cdot3}-\frac{1}{2\cdot3}\right)-8\left(\frac{1}{8\cdot9}-\cdots-\frac{1}{15\cdot16}\right)- 12\left(\frac{1}{32\cdot33}-\cdots-\frac{1}{63\cdot64}\right)-\cdots$$

[D]

$$\frac{Y_0+Y_1}{2}=\frac{1}{2}-2\left(\frac{1}{2\cdot3}-\frac{1}{2\cdot3}\right)-4\left(\frac{1}{8\cdot9}-\cdots-\frac{1}{15\cdot16}\right)- 6\left(\frac{1}{32\cdot33}-\cdots-\frac{1}{63\cdot64}\right)-\cdots$$

Add [C] + [D], which give us

[E]

$$\frac{X+X_1+Y_0+Y_1}{2}=\frac{3}{2}-2\left(\frac{1}{1\cdot2}-\frac{1}{3\cdot4}\right)-4\left(\frac{1}{7\cdot8}-\frac{1}{15\cdot16}\right)- 6\left(\frac{1}{31\cdot32}-\frac{1}{63\cdot64}\right)-\cdots$$

[A]

$$-\frac{25}{18}+\frac{1}{2}=-\frac{2}{2^2}-\frac{4}{2^4}-\frac{6}{2^6}-\cdots$$

Add [E] + [A] and we will have

$$\frac{X+X_1+Y_0+Y_1}{2}-\frac{43}{18}=-2\left(\frac{2}{3}\right)-4\left(\frac{8}{105}\right)-6\left(\frac{32}{1953}\right)-\cdots$$

Simplyfing and we have

$$-\frac{X+X_1+Y_0+Y_1}{2}+\frac{43}{18}=\frac{2\cdot2}{(2^1-1)(2^2-1)}+\frac{4\cdot2^3}{(2^3-1)(2^4-1)}+ \frac{6\cdot2^5}{(2^5-1)(2^6-1)}+\cdots$$

Substitute in Y_0, X, and Y_1, therefore we have

$$\frac{E-2\ln2-1-X_1}{2}+\frac{43}{18}=\frac{2\cdot2^1}{(2^1-1)(2^2-1)}+\frac{4\cdot2^3}{(2^3-1)(2^4-1)}+ \frac{6\cdot2^5}{(2^5-1)(2^6-1)}+\cdots$$

Solution 6 to [2]

--

$$X-X_1=-6\left(\frac{1}{3\cdot4}-\frac{1}{6\cdot7}\right)-10\left(\frac{1}{15\cdot16}-\cdots-\frac{1}{30\cdot31}\right)- 14\left(\frac{1}{63\cdot64}-\cdots-\frac{1}{126\cdot127}\right)-\cdots$$

[ C' ]

$$\frac{X-X_1}{2}=-6\left(\frac{1}{3\cdot4}-\frac{1}{6\cdot7}\right)-10\left(\frac{1}{15\cdot16}-\cdots-\frac{1}{30\cdot31}\right)- 14\left(\frac{1}{63\cdot64}-\cdots-\frac{1}{126\cdot127}\right)-\cdots$$

________________

$$Y_0-Y_1=-6\left(\frac{1}{4\cdot5}-\frac{1}{7\cdot8}\right)-10\left(\frac{1}{16\cdot17}-\cdots-\frac{1}{31\cdot32}\right)- 14\left(\frac{1}{64\cdot65}-\cdots-\frac{1}{127\cdot128}\right)-\cdots$$

[ D' ]

$$\frac{Y_0-Y_1}{2}=-6\left(\frac{1}{4\cdot5}-\frac{1}{7\cdot8}\right)-10\left(\frac{1}{16\cdot17}-\cdots-\frac{1}{31\cdot32}\right)- 14\left(\frac{1}{64\cdot65}-\cdots-\frac{1}{127\cdot128}\right)-\cdots$$

Add [C'] + [D'], which give us

[ E' ]

$$\frac{X-X_1+Y_0-Y_1}{2}=-3\left(\frac{1}{4\cdot5}-\frac{1}{7\cdot8}\right)-5\left(\frac{1}{16\cdot17}-\frac{1}{31\cdot32}\right)- 7\left(\frac{1}{63\cdot64}-\frac{1}{127\cdot128}\right)-\cdots$$

[B]

$$-\frac{11}{18}=-\frac{3}{2^3}-\frac{5}{2^5}-\frac{7}{2^7}-\cdots$$

Add [E'] + [B] and we will have

$$\frac{X-X_1+Y_0-Y_1}{2}-\frac{11}{18}=-3\left(\frac{4}{21}\right)-5\left(\frac{16}{465}\right)-7\left(\frac{64}{8001}\right)-\cdots$$

Simplyfing and we have

$$-\frac{X-X_1+Y_0-Y_1}{2}+\frac{11}{18}=\frac{3\cdot2^2}{(2^2-1)(2^3-1)}+\frac{5\cdot2^4}{(2^4-1)(2^5-1)}+ \frac{7\cdot2^6}{(2^6-1)(2^6-1)}+\cdots$$

Substitute in $$Y_0$$, $$X$$, and $$Y_1$$, therefore we have

$$\frac{X_1+E+2\ln2-3}{2}+\frac{11}{18}=\frac{3\cdot2^2}{(2^2-1)(2^3-1)}+\frac{5\cdot2^4}{(2^4-1)(2^5-1)}+ \frac{7\cdot2^6}{(2^6-1)(2^7-1)}+\cdots$$

--

Problem 1.

Show that,

$$E+1=\frac{2\cdot2}{(2^1-1)(2^2-1)}+\frac{3\cdot2^2}{(2^2-1)(2^3-1)}+ \frac{4\cdot2^3}{(2^3-1)(2^4-1)}+\cdots$$

Solution 1.

---

$$X=1-2\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}\right)-3\left(\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+\frac{1}{5\cdot6}-\frac{1}{6\cdot7}\right)-\cdots$$

$$2\gamma-2\ln2+2-X=1-\frac{1}{3}+2\left(\frac{1}{3}-\frac{1}{7}\right)+3\left(\frac{1}  {7}-\frac{1}{15}\right)+\cdots$$

$$2\gamma-2\ln2+2-X=E\cdot$$

$$X=2\gamma-2\ln2+2-E\cdot$$

$$Y_0 = \frac{1}{1\cdot2}-2\left(\frac{1}{2\cdot3}-\frac{1}{3\cdot4}\right)-3\left(\frac{1}{4\cdot5}-\frac{1}{5\cdot6}+\frac{1}{6\cdot7}-\frac{1}{7\cdot8}\right)-4\left(\frac{1}{8\cdot9}-\cdots-\frac{1}{15\cdot16}\right)-\cdots$$

$$Y_0=2\ln2-2\gamma\cdot$$

$$2=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+\cdots$$

---

$$X+Y_0\cdot$$

We have

$$2-E=\frac{3}{2}-2\left(\frac{1}{1\cdot2}-\frac{1}{3\cdot4}\right)- 3\left(\frac{1}{3\cdot4}-\frac{1}{7\cdot8}\right)-4\left(\frac{1}{7\cdot8}-\frac{1}{15\cdot16}\right)-\cdots$$

$$-2=-\frac{1}{2}-\frac{2}{2^2}-\frac{3}{2^3}-\frac{4}{2^4}-\cdots$$

Now add these two together and we have

$$-E=1-2\left(\frac{2}{3}\right)-3\left(\frac{4}{21}\right)-4\left(\frac{8}{105}\right) -\cdots$$

Neaten it up, we have

$$E+1=\frac{2\cdot2}{(2^1-1)(2^2-1)}+\frac{3\cdot2^2}{(2^2-1)(2^3-1)}+ \frac{4\cdot2^3}{(2^3-1)(2^4-1)}+\cdots$$

Problem 2.

-

Show that,

$$E=\frac{1\cdot2^1}{(2^1-1)(2^2-1)}+\frac{2\cdot2^2}{(2^2-1)(2^3-1)}+ \frac{3\cdot2^3}{(2^3-1)(2^4-1)}+\cdots$$

Solution 2.

---

[1]

$$\gamma=\frac{1}{2}-\frac{1}{3}+2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{6}-\frac{1}{7}\right)+3\left(\frac{1}{8}-\cdots-\frac{1}{15}\right)+\cdots$$

[2]

$$E-\gamma=1-\frac{1}{2}+2\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}\right)+3\left(\frac{1}{7}-\cdots-\frac{1}{14}\right)+\cdots$$

[3]

$$2\gamma-1=\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+2\left(\frac{1}{4\cdot5}-\frac{1}{5\cdot6}+\frac{1}{6\cdot7}-\frac{1}{7\cdot8}\right)+ 3\left(\frac{1}{8\cdot9}-\frac{1}{9\cdot10}-\cdots-\frac{1}{15\cdot16}\right)+\cdots$$

[4]

$$1=\frac{1}{2^2}+\frac{2}{2^3}+\frac{3}{2^4}+\frac{4}{2^5}+\cdots$$

[2] - [1]

We have

$$E-2\gamma=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+2\left(\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+\frac{1}{5\cdot6}-\frac{1}{6\cdot7}\right)+ 3\left(\frac{1}{7\cdot8}-\frac{1}{8\cdot9}-\cdots-\frac{1}{14\cdot15}\right)+\cdots$$

Now we add this to [3]

We have

$$E-1=\frac{1}{1\cdot2}-\frac{1}{3\cdot4}+2\left(\frac{1}{3\cdot4}-\frac{1}{7\cdot8}\right)+3\left(\frac{1}{7\cdot8}-\frac{1}{15\cdot16}\right)+\cdots$$

Now add this to [4]

We have

$$E=\frac{2}{3}+\frac{4}{21}+\frac{8}{105}+\cdots$$

Neaten it up we have

$$E=\frac{1\cdot2^1}{(2^1-1)(2^2-1)}+\frac{2\cdot2^2}{(2^2-1)(2^3-1)}+ \frac{3\cdot2^3}{(2^3-1)(2^4-1)}+\cdots$$

Problem 3.

--

Show that,

$$1=\frac{2^1}{(2^1-1)(2^2-1)}+\frac{2^2}{(2^2-1)(2^3-1)}+ \frac{2^3}{(2^3-1)(2^4-1)}+\cdots$$

Solution 3,

-- [1]

$$E=\frac{1\cdot2^1}{(2^1-1)(2^2-1)}+\frac{2\cdot2^2}{(2^2-1)(2^3-1)}+ \frac{3\cdot2^3}{(2^3-1)(2^4-1)}+\cdots$$

[2]

$$E+1=\frac{2\cdot2}{(2^1-1)(2^2-1)}+\frac{3\cdot2^2}{(2^2-1)(2^3-1)}+ \frac{4\cdot2^3}{(2^3-1)(2^4-1)}+\cdots$$

[2] - [1]

We have

$$1=\frac{2^1}{(2^1-1)(2^2-1)}+\frac{2^2}{(2^2-1)(2^3-1)}+ \frac{2^3}{(2^3-1)(2^4-1)}+\cdots$$

Problem 4.

-

Show that,

$$E-1=\frac{1}{2(2^1-1)}+\frac{1}{2^2(2^2-1)}+\frac{1}{2^3(2^3-1)}+\frac{1}{2^4(2^4-1)}\cdots$$

--

$$E-1=\frac{1}{1\cdot2}-\frac{1}{3\cdot4}+2\left(\frac{1}{3\cdot4}-\frac{1}{7\cdot8}\right)+3\left(\frac{1}{7\cdot8}-\frac{1}{15\cdot16}\right)+\cdots$$

Here we need to expand and collect the terms togther

$$E-1=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{7\cdot8}+\frac{1}{15\cdot16}+\frac{1}{31\cdot32}+\cdots$$

Neaten it up and we have

$$E-1=\frac{1}{2(2^1-1)}+\frac{1}{2^2(2^2-1)}+\frac{1}{2^3(2^3-1)}+\frac{1}{2^4(2^4-1)}\cdots$$

Problem 5.

-

Show that,

$$E-\gamma=1-\frac{1}{2}+2\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}\right)+3\left(\frac{1}{7}-\cdots-\frac{1}{14}\right)+\cdots$$

Solution 5.

- [1]

$$\gamma-2\ln2+2-X=1-\frac{1}{1\cdot2}+2\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}\right)+\cdots$$

[2]

$$2\gamma-2\ln2+2-X=1-\frac{1}{3}+2\left(\frac{1}{3}-\frac{1}{7}\right)+3\left(\frac{1}{7}-\frac{1}{15}\right)+\cdots$$

From [2], we have

$$2\gamma-2\ln2+2-X=E\cdot$$

Substitute X into [1], we have

$$E-\gamma=1-\frac{1}{2}+2\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}\right)+3\left(\frac{1}{7}-\cdots-\frac{1}{14}\right)+\cdots$$

-

$$2E-1=\sum_{n=1}^{\infty}\frac{1}{2^n}\frac{2^n+1}{2^n-1}$$

Proposed problem, find the closed-form of the series Z,

$$Z=1-\frac{1}{3}+2\left(\frac{1}{3}-\frac{1}{7}\right)+3\left(\frac{1}{7}-\frac{1}{15}\right)+4\left(\frac{1}{15}-\frac{1}{31}\right)+\cdots$$

and also it is equal to Erdos-Borwein constant

Where Z = Z'

$$Z'=\frac{1}{2^1-1}+\frac{1}{2^2-1}+\frac{1}{2^3-1}+\frac{1}{2^4-1}+\cdots=\sum_{n=1}^{\infty}\frac{1}{2^n-1}$$

I encountered it, when trying to solve the above problem.

How did I get from the above series to Erdos-Borwein constant is as show below

(1)

$$1-\gamma=\frac{1}{3}-\frac{1}{4}+2\left(\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}\right)+3\left(\frac{1}{9}-\cdots-\frac{1}{16}\right)+\cdots$$

(2)

$$\gamma-2\ln2+2-X=1-\frac{1}{2}+2\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}\right)+3\left(\frac{1}{7}-\cdots-\frac{1}{14}\right)+\cdots$$

(2) - (1)

and we have

$$Z=\frac{1}{2^1-1}+\frac{1}{2^2-1}+\frac{1}{2^3-1}+\frac{1}{2^4-1}+\cdots=\sum_{n=1}^{\infty}\frac{1}{2^n-1}$$

____________________________________________________________________________________________

Definition

(1)

$$X=1-2\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}\right)-3\left(\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+\frac{1}{5\cdot6}-\frac{1}{6\cdot7}\right)-\cdots$$

(2)

$$-2\ln2+2=1-\frac{1}{1\cdot2}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\cdots$$

(3) Vacca's formula

$$\gamma=\frac{1}{2}-\frac{1}{3}+2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{6}-\frac{1}{7}\right)+3\left(\frac{1}{8}-\cdots-\frac{1}{15}\right)+\cdots$$

________________

Work out

(3) + (2) - (1)

We have

$$\gamma-2\ln2+2-X=1-\frac{1}{1\cdot2}+2\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}\right)+\cdots$$

2(3) + (2) - (1)

We have

$$2\gamma-2\ln2+2-X=1-\frac{1}{3}+2\left(\frac{1}{3}-\frac{1}{7}\right)+3\left(\frac{1}{7}-\frac{1}{15}\right)+\cdots$$

_____________

Curiosity

1st, closed-form is not known

$$\gamma-2\ln2+2-X=1-\frac{1}{2}+2\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}\right)+3\left(\frac{1}{7}-\cdots-\frac{1}{14}\right)+\cdots$$

2nd, Vacca's formula

$$\gamma=\frac{1}{2}-\frac{1}{3}+2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{6}-\frac{1}{7}\right)+3\left(\frac{1}{8}-\cdots-\frac{1}{15}\right)+\cdots$$

3rd, known

$$1-\gamma=\frac{1}{3}-\frac{1}{4}+2\left(\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}\right)+3\left(\frac{1}{9}-\cdots-\frac{1}{16}\right)+\cdots$$

I want to find the closed-form for Z, so I can work out the close-form for the 1st.

______________________________________________________________________________

$$\sum_{n=1}^{\infty}\frac{N_1(n)N_0(n)}{n(n+1)}=\frac{1}{1\cdot2}+2\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}\right)+3\left(\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+ \frac{1}{6\cdot7}+\frac{1}{7\cdot8}\right)+\cdots$$

$$\gamma+\frac{3}{2}-\ln2=\frac{1}{1\cdot2}+2\left(\frac{1}{2\cdot3}\right)+3\left(\frac{1}{4\cdot5}+\frac{1}{6\cdot7}\right)+\cdots$$

$$\frac{1}{2}-\gamma+\ln2=2\left(\frac{1}{3\cdot4}\right)+3\left(\frac{1}{5\cdot6}+\frac{1}{7\cdot8}\right)+\cdots$$

---

Prove that,

$$\gamma=\frac{1}{1\cdot2}+3\left(\frac{1}{4\cdot5}-\frac{1}{5\cdot6}+\cdots+\frac{1}{7\cdot8}\right)+5\left(\frac{1}{16\cdot17}-\frac{1}{17\cdot18}+\cdots+\frac{1}{31\cdot32}\right)+\cdots$$

Let

$$2\ln2-1=\frac{1}{1\cdot2}-2\left(\frac{1}{2\cdot3}-\frac{1}{3\cdot4}\right)+ 3\left(\frac{1}{4\cdot5}-\cdots-\frac{1}{7\cdot8}\right)+\cdots$$

and

$$2\gamma+1-2\ln2=\frac{1}{1\cdot2}+2\left(\frac{1}{2\cdot3}-\frac{1}{3\cdot4}\right)+ 3\left(\frac{1}{4\cdot5}-\frac{1}{5\cdot6}+\frac{1}{6\cdot7}-\frac{1}{7\cdot8}\right)+\cdots$$

Add both of them and we have

$$\gamma=\frac{1}{1\cdot2}+3\left(\frac{1}{4\cdot5}-+\cdots+\frac{1}{7\cdot8}\right)+5\left(\frac{1}{16\cdot17}-\cdots+\frac{1}{31\cdot32}\right)+ 7\left(\frac{1}{64\cdot65}-\cdots+\frac{1}{127\cdot128}\right)+\cdots$$

Prove that,

$$2\gamma+1-2\ln2=\frac{1}{1\cdot2}+2\left(\frac{1}{2\cdot3}-\frac{1}{3\cdot4}\right)+ 3\left(\frac{1}{4\cdot5}-\frac{1}{5\cdot6}+\frac{1}{6\cdot7}-\frac{1}{7\cdot8}\right)+\cdots$$

-

Let

$$Y_0 = \frac{1}{1\cdot2}-2\left(\frac{1}{2\cdot3}-\frac{1}{3\cdot4}\right)-3\left(\frac{1}{4\cdot5}-\frac{1}{5\cdot6}+\frac{1}{6\cdot7}-\frac{1}{7\cdot8}\right)-4\left(\frac{1}{8\cdot9}-\cdots-\frac{1}{15\cdot16}\right)-\cdots$$

Vacca

$$Y_1=\gamma=\frac{1}{2}-\frac{1}{3}+2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{6}-\frac{1}{7}\right)+3\left(\frac{1}{8}-\cdots-\frac{1}{15}\right)+\cdots$$

$$Y_2=2\ln2-1=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{3\cdot4}-\cdots$$

$$Y_2-Y_0\cdots$$

$$Y_2-Y_0=\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+ 2\left(\frac{1}{4\cdot5}-\frac{1}{5\cdot6}+\frac{1}{6\cdot7}-\frac{1}{7\cdot8}\right)+ 3\left(\frac{1}{8\cdot9}-\cdots-\frac{1}{15\cdot16}\right)+\cdots$$

$$Y_1-Y_2+Y_0=\left(\frac{1}{3}-\frac{1}{4}\right)+ 2\left(\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}\right)+ 3\left(\frac{1}{9}+\cdots-\frac{1}{16}\right)+\cdots$$

$$Y_1-Y_2+Y_0+Y_1=\left(\frac{1}{2}-\frac{1}{4}\right)+ 2\left(\frac{1}{4}-\frac{1}{8}\right)+ 3\left(\frac{1}{8}-\frac{1}{16}\right)+\cdots$$

$$2Y_1-Y_2+Y_0=\frac{1}{2^2}+\frac{2}{2^3}+\frac{3}{2^4}+\cdots$$

$$2\gamma-2\ln2+1+Y_0=1\cdot$$

$$Y_0=2\ln2-2\gamma\cdot$$

Therefore

$$ 2\ln2-2\gamma= \frac{1}{1\cdot2}-2\left(\frac{1}{2\cdot3}-\frac{1}{3\cdot4}\right)-3\left(\frac{1}{4\cdot5}-\frac{1}{5\cdot6}+\frac{1}{6\cdot7}-\frac{1}{7\cdot8}\right)-4\left(\frac{1}{8\cdot9}-\cdots-\frac{1}{15\cdot16}\right)-\cdots$$

Neaten it up and we have

$$2\gamma+1-2\ln2=\frac{1}{1\cdot2}+2\left(\frac{1}{2\cdot3}-\frac{1}{3\cdot4}\right)+ 3\left(\frac{1}{4\cdot5}-\frac{1}{5\cdot6}+\frac{1}{6\cdot7}-\frac{1}{7\cdot8}\right)+\cdots$$

--

Result from above formula

$$2\gamma+1-2\ln2=\frac{1}{1\cdot2}+2\left(\frac{1}{2\cdot3}-\frac{1}{3\cdot4}\right)+ 3\left(\frac{1}{4\cdot5}-\frac{1}{5\cdot6}+\frac{1}{6\cdot7}-\frac{1}{7\cdot8}\right)+\cdots$$

and

$$2\ln2-1=\frac{1}{2}-\frac{1}{2\cdot3}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+\cdots$$

add both and we have

$$2\gamma-1=\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+2\left(\frac{1}{4\cdot5}-\frac{1}{5\cdot6}+\frac{1}{6\cdot7}-\frac{1}{7\cdot8}\right)+ 3\left(\frac{1}{8\cdot9}-\frac{1}{9\cdot10}-\cdots-\frac{1}{15\cdot16}\right)+\cdots$$

Vacca

$$\gamma=\frac{1}{2}-\frac{1}{3}+2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{6}-\frac{1}{7}\right)+3\left(\frac{1}{8}-\cdots-\frac{1}{15}\right)+\cdots$$

Using Vacca with the above, we obtain

$$1-\gamma=\frac{1}{3}-\frac{1}{4}+2\left(\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}\right)+3\left(\frac{1}{9}-\cdots-\frac{1}{16}\right)+\cdots$$

Prove that,

$$1-\gamma=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n(n+1)}\sum_{k=1}^{\infty}\frac{1}{(k+1)k^n}$$

$$\ln\left(\frac{2}{1}\right)=\frac{1}{2}+\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n(n+1)}\left(\frac{1}{1^n\cdot2}\right)$$

$$\ln\left(\frac{3}{2}\right)=\frac{1}{3}+\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n(n+1)}\left(\frac{1}{1 2^n\cdot3}\right)$$

Now I combined them to get

$$\ln\left(\frac{3}{1}\right)=\frac{1}{2}+\frac{1}{3}+\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n(n+1)}\left(\frac{1}{1^n\cdot2}+\frac{1}{2^n\cdot3}\right)$$

$$\ln\left(\frac{5}{1}\right)=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+ \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n(n+1)}\left(\frac{1}{1^n\cdot2}+\frac{1}{2^n\cdot3}+\frac{1}{3^n\cdot4}+\frac{1}{4^n\cdot5}\right)$$

$$1+\ln\left(\frac{k}{1}\right)=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots{\frac{1}{k}}+ \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n(n+1)}\left(\frac{1}{1^n\cdot2}+\frac{1}{2^n\cdot3}+\frac{1}{3^n\cdot4}+\cdots+{\frac{1}{(k+1)^n\cdot{k}}}\right)$$

$$1=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots{\frac{1}{k}}-\ln\left(\frac{k}{1}\right)+ \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n(n+1)}\left(\frac{1}{1^n\cdot2}+\frac{1}{2^n\cdot3}+\frac{1}{3^n\cdot4}+\cdots+{\frac{1}{(k+1)^n\cdot{k}}}\right)$$

$$1=\gamma+ \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n(n+1)}\left(\frac{1}{1^n\cdot2}+\frac{1}{2^n\cdot3}+\frac{1}{3^n\cdot4}+\cdots+{\frac{1}{(k+1)^n\cdot{k}}}\right)$$

$$1-\gamma= \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n(n+1)}\left(\frac{1}{1^n\cdot2}+\frac{1}{2^n\cdot3}+\frac{1}{3^n\cdot4}+\cdots+{\frac{1}{(k+1)^n\cdot{k}}}\right)$$

$$1-\gamma=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n(n+1)}\sum_{k=1}^{\infty}\frac{1}{(k+1)k^n}$$

Now Prove that,

$$\ln\left(\frac{b^{b+1}a!}{a^{a+1}b!}\right)=b-a+\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n(n+1)}\left[\frac{1}{(b-1)^n}+\frac{1}{(b-2)^n}+\cdots+\frac{1}{a^n}\right]$$

Let b = 2 and a = 1, we have

$$\ln\left(\frac{2}{1}\right)=\frac{1}{2}+\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n(n+1)}\left(\frac{1}{1^n\cdot2}\right)$$

--

Prove that,

$$\sum_{n=1}^{\infty}\frac{N_1(n)+N_0(n)}{2n(n+1)}=1$$

and

$$\sum_{n=1}^{\infty}\frac{N_1(n)+N_0(n)}{2(n+1)(2n+1)}=1-\gamma$$

$$\sum_{n=1}^{\infty}\frac{N_1(n)+N_0(n)}{n(n+1)}=X_2$$

$$X_2 = \frac{1}{1\cdot2}+2\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}\right)+ 3\left(\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\right)+... $$

$$X_2 = \frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{n}{2^n}$$

Using Arithmetic and geometric formulas to work X_2

$$\lim_{n\to\infty}{X_2}=2$$

$$\sum_{n=1}^{\infty}\frac{N_1(n)+N_0(n)}{n(n+1)}=2$$

$$\sum_{n=1}^{\infty}\frac{N_1(n)+N_0(n)}{2n(n+1)}=1$$

--

Sondow's formula

$$\sum_{n=1}^{\infty}\frac{N_1(n)+N_0(n)}{2n(2n+1)}=\gamma$$

$$\sum_{n=1}^{\infty}\frac{N_1(n)+N_0(n)}{2n(n+1)}-\sum_{n=1}^{\infty}\frac{N_1(n)+N_0(n)}{2n(2n+1)}=1-\gamma$$

and we have

$$\sum_{n=1}^{\infty}\frac{N_1(n)+N_0(n)}{2(n+1)(2n+1)}=1-\gamma$$

Prove that,

$$\sum_{n=1}^{\infty}\frac{N_1(n)+N_0(n)}{2n(2n+1)(2n+2)}=\gamma-\frac{1}{2}$$

-

Using Sondow's Formula

$$\sum_{n=1}^{\infty}\frac{N_1(n)+N_0(n)}{2n(2n+1)}-\frac{1}{2}\sum_{n=1}^{\infty}\frac{N_1(n)+N_0(n)}{2n(n+1)}=\gamma-\frac{1}{2}$$

$$\sum_{n=1}^{\infty}\frac{N_1(n)+N_0(n)}{2n(2n+1)(2n+2)}=\gamma-\frac{1}{2}$$

Prove that

$$\frac{1}{1-\frac{1}{4-\frac{3}{8-\frac{7}{48-\frac{47}{2208-\cdots}}}}}=\prod_{n=1}^{\infty}\left(1+\frac{1}{L_{2^n+1}}\right)=3-\phi$$

Millin series

$$\sum_{n=0}^{\infty}\frac{1}{F_{2^n}}=\frac{7-\sqrt{5}}{2}$$

Euler's continued fraction formula

$$1+\frac{1}{a_0}+\frac{1}{a_0a_1}+\frac{1}{a_0a_1a_2}+...=\frac{1}{1-\frac{1}{1+a_0-\frac{a_0}{1+a_1-\frac{a_1}{1+a_2-\cdots}}}}$$

---

$$\sum_{n=0}^{\infty}\frac{1}{F_{2^n}}=\frac{1}{F_1}+\frac{1}{F_2}+\frac{1}{F_4}+\frac{1}{F_8}+\frac{1}{F_{16}}+\cdots=\frac{7-\sqrt{5}}{2}$$

$$\frac{1}{F_1}+\frac{1}{F_2}+\frac{1}{F_4}+\frac{1}{F_8}+\frac{1}{F_{16}}+\cdots=\frac{7-\sqrt{5}}{2}$$

$$\frac{1}{F_2}+\frac{1}{F_4}+\frac{1}{F_8}+\frac{1}{F_{16}}+\cdots=3-\phi$$

Now I used

$$F_{2^n}=\prod_{n=0}^{k-1}L_{2^n}$$

to change the following to Ln

$$\frac{1}{L_1}+\frac{1}{L_1L_2}+\frac{1}{L_1L_2L_4}+\frac{1}{L_1L_2L_4L_8}+\cdots=3-\phi$$

L1 = 1

$$1+\frac{1}{L_2}+\frac{1}{L_2L_4}+\frac{1}{L_2L_4L_8}+\cdots=3-\phi$$

$$\frac{1}{1-\frac{1}{1+L_2-\frac{L_2}{1+L_4-\frac{L_4}{1+L_8-\cdots}}}}=3-\phi$$

$$\frac{1}{1-\frac{1}{4-\frac{3}{8-\frac{7}{48-\cdots}}}}=3-\phi$$

--

Definition

$$\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}$$

$$\eta(s)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^s}$$

--

$$\gamma+\ln\left(\frac{\pi}{2^2}\right)=2\sum_{n=2}^{\infty}\frac{(-1)^n\zeta(n)}{2^nn}$$

$$\gamma+\ln\left[\frac{\Gamma^4\left(\frac{1}{4}\right)}{4^4}\right]= 4\sum_{n=2}^{\infty}\frac{(-1)^n\zeta(n)}{4^{n}n}$$

$$\ln2-\gamma=\sum_{n=1}^{\infty}\frac{\zeta(2n+1)}{(2n+1)2^{2n}}$$

$$2\ln2-\gamma-2\beta^'(0)=\sum_{n=1}^{\infty}\frac{\zeta(2n+1)}{(2n+1)2^{4n}}$$

Where

$$\beta^'(0)=\sum_{n=0}^{\infty}\frac{\eta(2n+1)}{(2n+1)2^{2n+1}}$$

$$\ln{\pi}-\gamma=\sum_{n=2}^{\infty}\frac{\zeta(n)-\eta(n)}{n}$$

$$\ln{\frac{\pi}{4}}+\gamma=\sum_{n=2}^{\infty}(-1)^n\left[\frac{\zeta(n)-\eta(n)}{n}\right]$$

$$\gamma+\ln2=\sum_{n=1}^{\infty}\left(\frac{\zeta(2n)}{n}-\ln\frac{n+1}{n}\right)$$

-

$$1-\gamma=\sum_{n=1}^{\infty}\frac{2\zeta(2n)+\zeta(2n+1)}{2^{2n}(2n+1)}$$

Date:31\1\12

$$1-\gamma=\sum_{n=1}^{\infty}\frac{N_1(n)+N_0(n)}{2(n+1)(2n+1)}$$

--

Date:24/1/12

Tachiya's algebraic infinite product and its equilvalent continued fraction

n :  0  1  2  3  4  5  6 ...

Fn: 0 1 1 2 3 5 8 13 ...

Ln: 2 1  3  4  7  11 18 ...

$$\frac{1}{1-\frac{1}{4-\frac{3}{8-\frac{7}{48-\frac{47}{2208-\cdots}}}}}=\prod_{n=1}^{\infty}\left(1+\frac{1}{L_{2^n+1}}\right)=3-\phi$$

Date:28/1/12

$$3-\frac{9}{8-\frac{7}{48-\frac{47}{2208-\cdots}}}=\prod_{n=1}^{\infty}\left(1+\frac{1}{F_{2^n+1}}\right)=\frac{3}{\phi}$$

Date:22/1/12

$$\prod_{n=0}^{k-1}L_{2^n}=F_{2^k}$$

Date:25/1/12

$$\frac{1}{1-\frac{1}{8-\frac{7}{48-\frac{47}{2208-\frac{2207}{4870848-\cdots}}}}}=\prod_{n=1}^{\infty}\left(1+\frac{1}{F_{2^n+1}L_{2^n+1}}\right)=\frac{3}{\phi^2}$$

Date:26/1/12

$$F_{k+1}=\frac{\phi^k}{\sqrt5}\prod_{n=1}^{\infty}\left(1+\frac{L_k}{L_{2^n+k}}\right)$$

$$L_{k+1}=\phi^k\prod_{n=1}^{\infty}\left(1+\frac{F_k}{F_{2^n+k}}\right)$$

Date:27/1/12

$$F_{2k+2}=\phi^{2k}\prod_{n=1}^{\infty}\left(1+\frac{L_kF_k}{L_{2^n+k}F_{2^n+k}}\right)$$

--

$$\frac{1-\phi^{-4a-4b}}{\left(1-\phi^{-2a}\right)^2} \times\left(\frac{\sqrt{5}F_{a}}{L_{a}}\right)^{(-1)^a}= \prod_{n=1}^{\infty}\left(1+\frac{L_bF_b}{L_{2^na+b}F_{2^na+b}}\right)$$

Date:26/1/12

n greater than 0

$$\frac{L_{2^n}L_{2^n+1}-2}{2L_{2^n+1}-L_{2^n}}=F_{2^n+1}$$

$$L_{2^n}L_{2^n+1}L_{2^n+2}+\left(L_{2^n}-L_{2^n+2}\right)\left(1+L_{2^n+1}^2\right)-4L_{2^n+1}=0$$

$$L_{2^n}F_{2^n+1}L_{2^n+2}-\left(L_{2^n}+L_{2^n+2}\right)\left(1+F_{2^n+1}^2\right)+4F_{2^n+1}=0$$

$$F_{2^n+1}\left(L_{2^n}-L_{2^n+2}\right)\left(L_{2^n+1}^2-1\right)+L_{2^n+1}\left(L_{2^n}+L_{2^n+2}\right)\left(F_{2^n+1}^2-1\right)=0$$

---

$$F_{n+1}=\frac{\phi^n+(-1)^n\phi^{-n-2}}{3-\phi}$$

---

Date:14/1/12

$$\tan{A}\prod_{j=0}^{i}\tan\left(60-3^jA\right)\tan\left(60+3^jA\right)= \tan\left(3^{i+1}A\right)$$

Let

{x = tan(A), y = tan(60-A) and z = tan(60+A)}

$$\frac{1}{x}\left(z-y\right)-\frac{1}{y}\left(x+z\right)+\frac{1}{z}\left(x-y\right)=6$$

$$\frac{(yz)^2+1}{yz}-\frac{(xz)^2+1}{xz}+\frac{(xy)^2+1}{xy}=6$$

Date:27/3/11

where n ≥ 0

$$\ln2=\sum_{k=0}^{\infty}(-1)^k\sum_{j=1}^{2n+1}\frac{(-1)^{j+1}}{(2n+1)k+j}$$

$$\frac{\pi}{4}=\arctan\left(\frac{1}{7}\right)+4\left[\arctan\left(\frac{1}{13}\right) +\arctan\left(\frac{1}{21}\right)\right]+2\left[\arctan\left(\frac{1}{31}\right)+ \arctan\left(\frac{1}{43}\right)+\arctan\left(\frac{1}{57}\right)\right]$$

$$\frac{\pi}{4}=\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\arctan \left(\frac{1}{n^2F_{2k-1}+nF_{2k+2}+F_{2k+2}}\right)$$

Date:25/12/11

$$\frac{\pi}{4}+(k+1)\arctan\left(\frac{1}{2}\right)= \sum_{n=1}^{k}(k+1-n)\arctan\left(\frac{1}{n^2+3n+3}\right)+ \sum_{n=1}^{k+2}\arctan\left(\frac{1}{n}\right)$$

$$\frac{\pi}{4}-\arctan\left(\frac{1}{2}\right)= \sum_{n=1}^{2n}(-1)^{n+1}\arctan\left(\frac{1}{n}\right)- \sum_{n=1}^{n-1}\arctan\left(\frac{1}{4n^2+6n+3}\right)$$

$$\frac{\pi}{4}=\arctan\left(\frac{1}{3}\right)+ \sum_{n=1}^{\infty}\arctan\left(\frac{1}{n^2+3n+3}\right)$$

$$\frac{\pi}{4}=\arctan\left(\frac{1}{3}\right)+ \sum_{n=1}^{\infty}\arctan\left(\frac{1}{5n^2-n-1}\right)$$

$$\frac{\pi-\sqrt{5}\left[J_3^2(\phi^{-1})-J_3^2(\phi^{-2})\right]}{4}= \sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\sum_{k=0}^{\infty}\frac{1}{F_{2k+1}^{2n+1}}$$

Date:26/12/11

$$\frac{\pi}{4}=\arctan\left(\frac{1}{\phi}\right)+ \arctan\left(\frac{1}{\phi^3}\right)$$

$$\frac{\pi}{2}=3\arctan\left(\frac{1}{\phi}\right)- \arctan\left(\frac{1}{\phi^5}\right)$$

$$\arctan\left(\frac{1}{5}\right)=\arctan\left(\frac{1}{\phi}\right)- \arctan\left(\frac{1}{\phi^3}\right)-\arctan\left(\frac{1}{\phi^5}\right)- \arctan\left(\frac{1}{\phi^7}\right)$$

$$\arctan\left(\frac{1}{239}\right)=3\arctan\left(\frac{1}{\phi}\right)- 5\arctan\left(\frac{1}{\phi^3}\right)-4\arctan\left(\frac{1}{\phi^5}\right)- 4\arctan\left(\frac{1}{\phi^7}\right)$$

$$\arctan\left(\frac{1}{239}\right)=2\arctan\left(\frac{1}{\phi}\right)- 3\arctan\left(\frac{1}{\phi^3}\right)-3\arctan\left(\frac{1}{\phi^5}\right)- 4\arctan\left(\frac{1}{\phi^7}\right)$$

Date: 4/1/12

$$\frac{\pi}{2}=\arctan\left(\frac{1}{\phi^{\phi}}\right)+ \arctan\left(\frac{1}{\phi^{-\phi+1}}\right)+ \arctan\left(\frac{1}{\phi^{-\phi+2}}\right)- \arctan\left[\frac{1}{(\phi^{-2\phi}+2\phi^{-2})\phi^{-\phi+3}}\right]$$

Definition
$$\phi=\frac{1+\sqrt{5}}{2}$$

Where

$$F_1=1$$

$$F_2=1$$

$$F_{n+2}=F_n+F_{n+1}$$

Where

$$K_0=1$$

$$K_1=1$$

$$K_{n+2}=K_n+K_{n+1}+F_{n+1}$$

Date:1/12/11

Where

x ≥ 1

$$\left(-1\right)^{x-1}\sum_{n=1}^{x}{(-\phi)}^{x+1-n}F_{x+1-n} =-K_{x-1}\phi^{x+1}+\phi^{x-1}\sum_{n=0}^{x-1}K_{n}$$

Date:10/12/11

$$4\phi^4+3\phi^3+2\phi^2+1\phi=\phi^8$$

$$2\phi^2+1\phi=\phi^4$$

Date: 16/12/11

$$F_m\phi^n-F_n\phi^m=\frac{F_mF_{n+x}-F_nF_{m+x}}{F_x}$$

Date:18/12/11

$$\frac{\phi^n-2F_{n+1}}{\phi^k-2F_{k+1}}= \frac{\phi{F_{n-1}}+F_{n+1}}{\phi{F_{k-1}}+F_{k+1}}$$

Date:23/12/11

Where

n : 0, 1, 2, 3, 4,  5,  6,...

Tn : 0, 1, 3, 6, 10, 15, 21,...

$$\frac{\pi^2}{6}=1-\frac{1}{4T_0+1-\frac{1^4}{4T_1+1-\frac{2^4}{4T_2+1-\frac{3^4}{4T_3+1-...}}}}$$

title: User:Sammethod


title: File:DigitalDesign.timing.20111201.pdf


text: File:DigitalDesign.register.20111201.pdf


text: File:DigitalDesign (HW).20111201.pdf

title: File:DigitalDesign (HW).20111201.pdf


text:
Talk:Motivation and emotion/Assessment/Multimedia

Comments
Hi, there are tools now available for free that are even better then the ones mentioned in the recommendation section ("If you're not sure and want to keep it simple, the recommended approach is a screencast using http://screenr.com" etc).

For example the better quality of screen recording can be achieved if you use the following free tool that is mentioned in the video I created it with, see http://www.youtube.com/user/kinestheticlearning?feature=mhee#p/c/039ABF3351606F24/0/Nh_caz2ziJY

Prezi, that is mentioned in the video is another tool that can be recommended for making slideshows that look much better then the ones made by Powerpoint or Slideshare.

My two cents only :)

Best regards Youtuniverity 15:00, 1 December 2011 (UTC)

 * Great suggestions, thank-you. I've added these links and will try the tools. -- Jtneill - Talk - c 20:36, 1 December 2011 (UTC)

 * Hi again, I just found out you and Leigh wrote the general articles (Wikiversity "resources") about Screencasting and Session recording, but I haven't seen them being linked by you from the "Emotion and Motivation" Assessment Multimedia. Youtuniverity 12:25, 8 December 2011 (UTC)

Grammar
I think there might be a minor grammar or spelling error in the multimedia marking criteria under HD "An excellent multimedia presentation which effectively how key concepts from motivation or emotion theory and research can be applied to a specific problem" -omitting the word 'shows' ? --Can Rogerthat (discuss • contribs) 02:49, 28 October 2014 (UTC)
 * Thanks, fixed. -- Jtneill - Talk - c 03:10, 28 October 2014 (UTC)


text: File:Hailong.2011 12 01 08 16 35.pdf

title: File:Hailong.2011 12 01 08 16 35.pdf


text:
Food Safety/Bacteriology

 * The Genus Enterococcus: a challenge for the food industry.

Brief Report

Enterococci are members of the lactic acid bacteria (LAB) group. The species of this genus are Gram positive cocci, facultative anaerobe organisms which and tolerant to a wide range of conditions: temperature (10 - 45°C), pH (4.5 - 10.0) and high sodium chloride concentrations [1, 2]. The use of LAB and/or their metabolites for food preservation is generally accepted by consumers as something “natural” and “health promoting”.

Most species of this genus are commensal bacteria of the gastrointestinal tracts of humans and other mammals, so these bacteria are generally considered as low pathogenic, however the isolation of species of Enterococcus that produce important nosocomial infections and acquired antibiotic resistance patterns have been increasing in recent years. Scientific data regarding the emergence of vancomycin-resistant enterococci revealed that antibiotic resistant species show an increased ability to colonize and/or cause disease in hospitalized patients worldwide [3, 4].

The hypothesis is that the species of enterococci with antimicrobial resistant and virulence factors [5, 6, 7] that colonize animal foods might enter into the human food chain and be a potential public health risk in inmunocompromised patients [8].

Another use of these organisms is as potential probiotics. As the bibliography expresses (FAO-WHO) [9], “…probiotics are additives containing live microbes that beneficially affect a host by improving the properties of the host intestinal microflora”. Enterococcus strains are used as probiotics but do not have virulence factors and antimicrobial resistance genes. Most bacteria are used as probiotics, however Enterococcus spp. have the ability to exchange resistance determinants intra-species and inter-species and become more virulent.

For these reasons and to differentiate secure species from pathogenic species, food microbiology has initiated investigations about the presence and characterization of these bacteria in animal foods and dairy products such as sausages, meat, cheese and milk.

In conclusion, the dualistic aspects of enterococci still represent a great challenge to the food industry and constitute a vast field for scientific research.

References

1. Foulquié Moreno MR, Sarantinopoulos P, Tsakalidou E, De Vuyst L. The role and application of enterococci in food and health. International Journal Food Microbiology. 2006;106:1-24.

2. Hardie JM, Whiley RA. Classification and overview of the genera Streptococcus and Enterococcus. Journal Applied Microbiology.1997;83:1S-11S.

3. Torres-Viera C, Dembry LM. Approaches to vancomycin-resistant enterococci. Current Opinion in Infectious Diseases. 2004;17:541-547.

4. de Bruin MA, Riley LW. Does vancomycin prescribing intervention affect vancomycin-resistant enterococcus infection and colonization in hospitals? A systematic review. BMC Infectious Diseases. 2007;7:24.

5. Eaton TJ, Gasson MJ. Molecular screening of Enterococcus virulence determinants and potential for genetic exchange between food and medical isolates. Applied Environmental Microbiology. 2001;67:1628-1635.

6. Robredo B, Singh KV, Baquero F, Murray BE, Torres C. Vancomycin resistant enterococci isolated from animals and food. International Journal Food Microbiology. 2000;54:197-204.

7. Gelsomino R, Vancanneyt M, Cogan TM, Condon S, Swings J. Source of enterococci in a farmhouse raw-milk cheese. Applied Environmental Microbiology. 2002;68:3560-3565.

8. Huys G, D'Haene K, Collard JM, Swings J. Prevalence and Molecular Characterization of Tetracycline Resistance in Enterococcus isolates from Food. Appl Environ Microbiol. 2004; 70(3): 1555-1562.

9. FAO-WHO 2001. Food and agriculture organization of the United Nations, World Health Organization. Report of a joint FAO-WHO expert consultation on evaluation of health and nutritional properties of probiotics in food including powder milk with live lactic acid bacteria. Córdoba, 2001. Available in: ftp://ftp.fao.org/docrep/fao/009/a0512e/a0512e00.pdf

title: Food Safety/Bacteriology


text:
Talk:TAO

Vorgehen zur Verbesserung des Wikiversity TAO-Projekts
Zielsetzung ist es im Dialog mit den Beteiligten aus dem TAO-Projekt bei Wikiversity zu überlegen, wie der Einstieg in den Leitfaden und das Projekt erleichtert (Abbau von Hemmschwellen, Überbrückung technischer Hürden, etc.) werden und wie man weitere Stakeholder in die Community of Practice (CoP) einbeziehen kann. Hierzu werden wir eine Umfrage unter dem TAO-Team starten, die erhebt:
 * Wie sollte die Startseite aussehen?
 * Welche Erfahrungen wurden bei der Erstellung der ertsen Lernaktivitäten/Module in Wikiversity gemacht?
 * Wie können weitere Personen in die CoP einbezogen werden?

Die Umfrage startet am 06.12.2011 per E-Mail; Rückmeldungen werden bis zum 12.12.2011 erwartet; diese werden ausgewertet und aufbereitet, danach trifft sich das engere TAO-Wikiversity-Team erneut zu einem Online-Treffen (14, 17 oder 18.12.2011) und bereitet weitere Schritte vor. --Marquard Markus 16:34, 1 December 2011 (UTC)

Umfrage zur Verbesserung des Wikiversity TAO-Projekts
Verbesserung der Startseite Was ist Eurer Meinung nach auf der Startseite des Projektes wichtig? Welche informationen sind zur Mitarbeit erforderlich? Wo kann man die Seite vereinfachen? Was könnte neue Mitstreiter/-innen zum Mitmachen motivieren? Welche Ansprüche/Erwartungen könnten die verschiednene Zielgruppen haben?

Erfahrungen aus der ersten Lernaktivität Habt Ihr Euche Lernaktivität bei wikiversity eingestellt? Wenn nein, warum nicht? Wenn ja, welche Schwierigkeiten hattet Ihr? Wo seht Ihr Verbesserungsbedarf? Konntet Ihre eine Zielgruppe und entsprechende Ansprüche/Bedrafe formulieren? Welche Bausteine waren einfach umzusetzen, wo gab es Probleme?

Einbindung in die Community of Practice Wie können wir neue Mitstreiter/-innen einbinden? Welche Voraussetzungen müssen die Mitmachenden mitbringen? Was könnte potenbtielle mitmachende motivieren/anregen mit zu machen? Wie kann man neue Mitmachende besser unterstützen? Welche Do's und Dont#s können formuliert werden? --Marquard Markus 19:53, 1 December 2011 (UTC)

Questions
1. What is your impression of the start page and the general presentation of the project? Wie findest Du die Startseite und die allgemeine Präsentation des Projekts bei Wikiversity?

 * 1.a What do you find positive? Was gefällt Dir?
 * 1.b What should be improved? Was sollte verbessert werden?

2. How is your experience in creating your first learning/research activity so far? Wie ist Deine Erfahrung bei der entwicklung der ersten Lern- und Forschungsaktivität auf Wikiversity?

 * 2.a What works out well? Was hat gut funktioniert?

 * 2.b What are the main drawbacks/obstacles from your point of view? Was sind aus Deiner Sicht die wichtigsten Nachteile und Hemmschwellen?

3. According to you, what minimal conditions need to be met before you will consider inviting more users to join the Community of Practice (CoP), for example to participate in a learning activity? Wenn Du an Dich denkst, was sind die minimalen Voraussetzungen und Bedingungen, bevor Du weitere Benutzer in die CoP einladen würdest, z.B. um an einer Lernaktivität bzw. einem Handbuchmodul mitzuarbeiten?

 * 3.a What could be done to make it more motivating for people to join the TAO Community of Practice? Was könnte gemacht werden, um mehr Menschen zur Mitarbeit in die TAO CoP zu motivieren?

 * 3.b What content could be provided to make the TAO Wikiversity project more interesting for people who are not directly involved in one of the learning activities? Welche Inhalte würden dazu beitragen, dass Menschen, die nicht direkt in das TAO Projekt einbezogen sind, bei dem TAO Wikiversity Projekt mitmachen würden?

4. General Remarks / Allgemeine Anmerkungen

--Beat Estermann 08:51, 2 December 2011 (UTC) --Marquard Markus 18:33, 7 December 2011 (UTC)

To-do list
''List for tasks that need no further discussion. Also routinous tasks, bits and pieces, beautifications, ...
 * Translate German text on this talk-page.
 * Complete and beautify TAO/Communication
 * ... with other ressources?
 * Check for corresponding links the respective other languages. Currently DE and EN.
 * Transfer completion status in template from percentages to traffic light colours.
 * proposal for states: RED first iedeas, sketch, brainstorm | YELLOW ready for (experimental) use, please ask and feed back experiences | GREEN really ready for use, as ready as a wiki-style ressource get's (i.e. it shall always be improved)   --Simon Lüke 16:05, 7 February 2012 (UTC)

title: Talk:TAO


text:
User talk:Celia~enwikiversity

 Hello Celia, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. All users are expected to abide by our Privacy policy, Civility policy, and the Terms of Use while at Wikiversity.

To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.
 * Explore Wikiversity with the links to your left.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.
 * Follow Wikiversity on twitter (http://twitter.com/Wikiversity) and identi.ca (http://identi.ca/group/wikiversity).

You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --HappyCamper 18:56, 1 December 2011 (UTC)

Your account will be renamed
Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

Unfortunately, your account clashes with another account also called Celia. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Celia~enwikiversity that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:29, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:04, 19 April 2015 (UTC)

title: User talk:Celia~enwikiversity


text:
User talk:Sistemx~enwikiversity

 Hello Sistemx, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. All users are expected to abide by our Privacy policy, Civility policy, and the Terms of Use while at Wikiversity.

To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
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You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --HappyCamper 19:00, 1 December 2011 (UTC)

 Hello! dear fellow Cubologist!

Search for "Speedcubing". Look on the "discussion" page. You should find a link to my small book. I plan to do major edits soon; good critical evaluation should help improve the work.

Happy Cubing!

Ray Calvin Baker 23:15, 1 December 2011 (UTC)

Followup
The way I think of the difference between Wikiversity and Wikibooks is that here we are trying to build online communities around online learning resources. This can come in many different forms, and often there is some overlap between what is done here and other wikis. Wikiversity is kind of the place where we want to try and push the boundaries of what a wiki can do for enabling online learning. I think a key part of this is creating high quality learning resources and communities to start. Does this help answer some of your questions? --HappyCamper 23:21, 1 December 2011 (UTC)
 * Actually I haven't made that scrapbook page yet, so it doesn't exist! --HappyCamper 07:34, 5 December 2011 (UTC)

Your account will be renamed
Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

Unfortunately, your account clashes with another account also called Sistemx. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Sistemx~enwikiversity that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:43, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:22, 19 April 2015 (UTC)

title: User talk:Sistemx~enwikiversity


text:
User:Egm6322.s12/Lecture plan

 Under construction throughout the semester. Egm6322.s12 21:28, 1 December 2011 (UTC)

= Lecture notes, report table =

Additional presentations (video, wiki, static html) made in class may not be recorded on these transparencies.

A mtg number n followed by a version letter in parentheses ([a-z]) indicates that this pdf file had been updated with a 2nd version "b", 3rd version "c", etc. The second letter "x" after a version letter designates an extra lecture; this letter became "m" after the extra lecture had become a make-up lecture. If you had downloaded this pdf file before, you want to clear the cache of your browser so to get the new version.

djvu: ( install the viewer DjView, or Sumatra PDF, or evince )

$$\displaystyle \bullet$$ Mtg 1, Tue, 2012.01.10: [[media:pea2.s12.sec0.djvu|Sec 0]]; $$\displaystyle \bullet$$ Mtg 2, Thu, 2012.01.12: [[media:pea1.f11.mtg42.djvu| Sec.42]]; $$\displaystyle \bullet$$ Mtg 3, Sat, 2012.01.14: [[media:pea1.f11.mtg42.djvu| Sec.42]]; $$\displaystyle \bullet$$ Mtg 4 (x), Mon, 2012.01.16: [[media:pea1.f11.mtg42.djvu| Sec.42]], kochmann.drugan.2009.pdf $$\displaystyle \bullet$$ Mtg 5, Tue, 2012.01.17: [[media:pea1.f11.mtg38.djvu| Sec.38]], [[media:pea1.f11.mtg39.djvu| Sec.39]], [[media:pea1.f11.mtg40.djvu| Sec.40]], [[media:pea1.f11.mtg41.djvu| Sec.41]], [[media:pea1.f11.mtg42.djvu| Sec.42]]; $$\displaystyle \bullet$$ Mtg 6, Thu, 2012.01.19: [[media:pea1.f11.mtg42.djvu| Sec.42 (c)]], [[media:pea2.s12.sec43.a.djvu| Sec.43.a]]; $$\displaystyle \bullet$$ Mtg 7, Tue, 2012.01.24: [[media:pea2.s12.sec43.a.djvu| Sec.43.a (b)]]; $$\displaystyle \bullet$$ Mtg 8, Tue, 2012.01.24: [[media:pea2.s12.sec43.b.djvu| Sec.43.b]]; $$\displaystyle \bullet$$ Mtg 9, Thu, 2012.01.26: [[media:pea2.s12.sec43.b.djvu| Sec.43.b (b)]]; $$\displaystyle \bullet$$ Mtg 10, Thu, 2012.01.26: [[media:pea2.s12.sec43.b.djvu| Sec.43.b (c)]], [[media:pea2.s12.sec43.c.djvu| Sec.43.c]]; $$\displaystyle \bullet$$ Mtg 11, Tue, 2012.01.31: [[media:pea2.s12.sec43.c.djvu| Sec.43.c (b)]], [[media:pea2.s12.sec43.d.djvu| Sec.43.d]]; $$\displaystyle \bullet$$ Mtg 12, Thu, 2012.02.03: [[media:pea2.s12.sec43.d.djvu| Sec.43.d (b)]]; $$\displaystyle \bullet$$ Mtg 13, Sat, 2012.02.05: [[media:pea2.s12.sec44.a.djvu| Sec.44.a]]; $$\displaystyle \bullet$$ Mtg 14, Tue, 2012.02.07: [[media:pea2.s12.sec44.b.djvu| Sec.44.b]]; $$\displaystyle \bullet$$ Mtg 15, Thu, 2012.02.09: [[media:pea2.s12.sec44.b.djvu| Sec.44.b (b)]]; $$\displaystyle \bullet$$ Mtg 16, Tue, 2012.02.14: [[media:pea2.s12.sec44.b.djvu| Sec.44.b (c)]], [[media:pea2.s12.sec45.a.djvu| Sec.45.a]]; $$\displaystyle \bullet$$ Mtg 17, Tue, 2012.02.14: [[media:pea2.s12.sec45.b.djvu| Sec.45.b]]; $$\displaystyle \bullet$$ Mtg 18, Thu, 2012.02.16: [[media:pea2.s12.sec45.c.djvu| Sec.45.c]]; $$\displaystyle \bullet$$ Mtg 19, Thu, 2012.02.16: [[media:pea2.s12.sec45.d.djvu| Sec.45.d]]; $$\displaystyle \bullet$$ Mtg 20, Tue, 2012.02.21: [[media:pea2.s12.sec45.c.djvu| Sec.45.c (b)]], [[media:pea2.s12.sec45.d.djvu| Sec.45.d (b)]]; $$\displaystyle \bullet$$ Mtg 21, Thu, 2012.02.23: [[media:pea2.s12.sec44.b.djvu| Sec.44.b (d)]], [[media:pea2.s12.sec45.d.djvu| Sec.45.d (c)]], [[media:pea2.s12.sec45.e.djvu| Sec.45.e]]; $$\displaystyle \bullet$$ Mtg 22, Sat, 2012.02.25: [[media:pea2.s12.sec45.e.djvu| Sec.45.e (b)]]; $$\displaystyle \bullet$$ Mtg 23, Mon, 2012.02.27: [[media:pea2.s12.sec45.d.djvu| Sec.45.d (d)]], [[media:pea2.s12.sec45.f.djvu| Sec.45.f]], [[media:pea2.s12.sec45.g.djvu| Sec.45.g]]; $$\displaystyle \bullet$$ Mtg 24, Tue, 2012.02.28: [[media:pea2.s12.sec45.f.djvu| Sec.45.f (b)]], [[media:pea2.s12.sec45.g.djvu| Sec.45.g (b)]]; $$\displaystyle \bullet$$ Mtg 25, Tue, 2012.03.06: [[media:pea2.s12.sec44.b.djvu| Sec.44.b (e)]], [[media:pea2.s12.sec45.f.djvu| Sec.45.f (c)]], [[media:pea2.s12.sec45.g.djvu| Sec.45.g (b)]], [[media:pea2.s12.sec45.h.djvu| Sec.45.h]]; $$\displaystyle \bullet$$ Mtg 26, Thu, 2012.03.08: [[media:pea2.s12.sec45.h.djvu| Sec.45.h (b)]]; $$\displaystyle \bullet$$ Mtg 27, Tue, 2012.03.13: [[media:pea2.s12.sec48.a.djvu| Sec.48.a]], [[media:pea2.s12.sec48.b.djvu| Sec.48.b]]; $$\displaystyle \bullet$$ Mtg 28, Thu, 2012.03.15: [[media:pea2.s12.sec48.a.djvu| Sec.48.a.(b)]], [[media:pea2.s12.sec48.b.djvu| Sec.48.b.(b)]], [[media:pea2.s12.sec48.c.djvu| Sec.48.c]]; $$\displaystyle \bullet$$ Mtg 29, Tue, 2012.03.20: [[media:pea2.s12.sec48.c.djvu| Sec.48.c]], [[media:pea2.s12.sec48.d.djvu| Sec.48.d]], [[media:pea2.s12.sec48.e.djvu| Sec.48.e]]; $$\displaystyle \bullet$$ Mtg 30, Thu, 2012.03.22: [[media:pea2.s12.sec48.e.djvu| Sec.48.e (b)]]; $$\displaystyle \bullet$$ Mtg 31, Sat, 2012.03.24: [[media:pea2.s12.sec48.e.djvu| Sec.48.e (c)]], [[media:pea2.s12.sec48.f.djvu| Sec.48.f]]; $$\displaystyle \bullet$$ Mtg 32, Tue, 2012.03.27: [[media:pea2.s12.sec48.g.djvu| Sec.48.g]]; $$\displaystyle \bullet$$ Mtg 33, Thu, 2012.03.29: [[media:pea2.s12.sec48.h.djvu| Sec.48.h]]; $$\displaystyle \bullet$$ Mtg 34, Sat, 2012.03.31: [[media:pea2.s12.sec48.h.djvu| Sec.48.h (b)]], [[media:pea2.s12.sec47.djvu| Sec.47]]; $$\displaystyle \bullet$$ Mtg 35, Tue, 2012.04.03: [[media:pea2.s12.sec48.i.djvu| Sec.48.i]]; $$\displaystyle \bullet$$ Mtg 36, Thu, 2012.04.05: [[media:pea2.s12.sec50.a.djvu| Sec.50.a]]; [[media:pea2.s12.sec48.j.djvu| Sec.48.j]]; $$\displaystyle \bullet$$ Mtg 37, Sat, 2012.04.07: [[media:pea2.s12.sec48.j.djvu| Sec.48.j (b)]], [[media:pea2.s12.sec48.k.djvu| Sec.48.k]]; $$\displaystyle \bullet$$ Mtg 38, Tue, 2012.04.10: [[media:pea2.s12.sec48.k.djvu| Sec.48.k (b)]]; $$\displaystyle \bullet$$ Mtg 39, Thu, 2012.04.12: [[media:pea2.s12.sec45.d.djvu| Sec.45.d (e)]], [[media:pea2.s12.sec48.l.djvu| Sec.48.l]]; $$\displaystyle \bullet$$ Mtg 40, Sat, 2012.04.14: [[media:pea2.s12.sec60.a.djvu| Sec.60.a]];

Report table

Inspiring video on teamwork and learning: A hole in the wall: How children learn without a teacher, (better video resolution from TED)

Mourning the Death of Handwriting, By Claire Suddath. Time Magazine, Monday, Aug. 03, 2009.

Op-Art: The Write Stuff, by Inga Dubay and Barbara Getty, NY Times, 8 Sep 2009.

= References =

Web references
$$\displaystyle \color{red} \clubsuit$$ Wolfram|Alpha is ... "learning resource available to your students at no cost that works as a computational knowledge engine. Wolfram|Alpha is not a search engine like Google or Yahoo!, because unlike a traditional search engine, Wolfram|Alpha has the capability to instantly compute the answer to previously unasked questions instead of scouring the web and returning links to pages that already exist. The results are displayed in an easy-to-read, understandable format that can be used as a primary source for educational and academic purposes." $$\displaystyle \color{red}{\spadesuit}$$

NEW:

Introduction to WolframAlpha: Part 1, Part 2. WolframAlpha syntax.

$$\displaystyle \spadesuit$$ Stephen Wolfram: Computing the theory of everything, video on TED.com

$$\displaystyle \spadesuit$$ Wolfram online integrator

$$\displaystyle \spadesuit$$ Wolfram Demonstration Project

title: User:Egm6322.s12/Lecture plan


text:
Talk:Microsoft Windows/Windows XP

I have noticed that I have cleaned up the articles of Windows XP. --Katarighe 00:23, 2 December 2011 (UTC)

title: Talk:Microsoft Windows/Windows XP


text:
Category:Fitness

See also Images in Category:Fitness of Commons

title: Category:Fitness