text:
University of Canberra/RCC2011/What key facilitation capacities help create successful wikis?

Outcome - We can come up with a list of key facilitation capacities, tell some success stories and look at common themes for good facilitation. We could also define what is a successful wiki? Put outcome on RCC Canberra wiki and all can use it and develop it further

What Principles should we use for this conversation?

 * Create opportunity for everyone to contribute
 * Our diversity is captured in some ways
 * Be aware of when we stray from central topic/keep focused on the goal
 * Use the 3 Open Space principles - who ever comes is the right people, whenever it starts and finishes is the right time, whatever happens is the only thing that could of happened
 * step away if you become too emotionally charged (one strategy that we explored a bit more - see below)

How are we dealing when emotions come up?
 * Wikipedia experience - sometimes is best to walk away.
 * If someone gets emotional, let's put ourselves in their shoes.
 * Give feedback to the frustrated person.
 * Get very specific about the difference. Help people unpack their thinking.

Hot areas in wikipedia: When you visit the article, you can smell that the editor is emotional. Step away if you become too emotionally charged vs let's engage with it

Facilitator: Role of a guardian. Also as a supporter /gardener/nurturer of editors

Challenges: - dealing with how to keep a volunteer in as an editor - just because they are acting obnoxiously doesn't mean that they are not working in good faith

Bring out contentious ideas knowledge levels are different

What key facilitation capacities help create successful wikis?
 * awareness of emotional content
 * read between the lines of online text
 * don't trust your own judgement - check or test your assumptions
 * assume good faith and treat people consistently as if they are not evil
 * identify when people get labelled or excluded from the system
 * acknowledge people who want to belong to the system
 * respect everyone
 * turn articles for deletion into a good faith process
 * allow things to happen and let the community manage it

One key facilitation capacity from each participant (to get diversity)
 * harnessing/being comfortable with conflict
 * staying focused on the dynamics
 * try to read between lines (get the smell)
 * ok to be wrong. Wrong doesn't mean not respected.
 * understand the drivers of the people in dispute
 * be completely neutral in my own opinions (is this possible?)
 * Can I remove my own emotional input?
 * listening to all perspectives
 * read between the lines, carefully read between the lines (be instinctive but not react subjectively, react objectively)
 * calmness, distance, pathos - empathy
 * explain the reasoning behind my thinking and support others to do the same
 * laser focus on understanding the difference
 * democratic community ownership
 * avoid making decisions unilaterally
 * understand the cultural norms

Successful Examples: Wikihow has a positive culture
 * conscious effort to build a friendly community
 * less academic
 * fair bit of tolerance
 * good open leadership, encouraging others to be good leaders

Wikipedia:
 * is a successful wiki
 * vastly different opinions creating high quality output
 * Finely tuned dispute resolution process works really well, but a lot of people don't last long enough to go through the process
 * 1) bring in third opinion
 * 2) further backup plans
 * 3) informal mediation
 * 4) formal mediation
 * 5) than the committee

Other points: Incentive Conservapedia - working in the right place

Most valuable insights from this conversation Next steps/directions
 * good hearing diverse opinions
 * hero's story in mythology: wrestling with the troll is worth it
 * this conversation encourages the isolated editor
 * open leadership is key
 * good to be reminded of things I should be doing
 * importance of context
 * good to hear from all our experience
 * great to hear and celebrate how amazing people are
 * understand how to create successful wikis should be a default
 * what happens long term especially to small wikis? Do they get archived?
 * use more language of open leadership every day

What is a successful wiki?


text:
University of Canberra/RCC2011/The Bruce Declaration

Based on discussions the day before, a small group decided to have a go at drafting a declaration that could be used for submissions into public consultations regarding education in Australia. The group used the white board to brainstorm, will continue to use this MeetingWords to draft sentences and links, and is structuring the declaration on the Wikiversity project Workshop for Australian education policy.

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


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title: File:Fe1.s11.mtg14.djvu


title: File:Nm1.s11.mtg14.djvu


text:
User:Bilby

meta:User:Bilby, attended University of Canberra/RCC2011.

title: User:Bilby


title: File:Egm6611.s11.team6.hw2.9.1.png


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text:
University of Canberra/RCC2011/RecentChangesCamp Expanded/San Francisco

Organizing on a MeetingWords Etherpad

Calling it RCCx - which is nice :-)

Also, see: https://www.noisebridge.net/wiki/Wiki_meetup (possible SF venue)

RCCx San Francisco, February 2011

RCCx (RecentChangesCamp eXpanded) events are in addition to full-scale RecentChangesCamp events, and include smaller meetups or other wiki get-togethers.

rccx-sf 2011-02-01 raw notes venue: noisebridge sf

Attendees

 * Mark Dilley
 * Pete F
 * Brion
 * Stephen
 * Peter Kaminski - kaminski@istori.com
 * eekim
 * Kaliya
 * Joe
 * Arlen Abraham - localwiki board member
 * Paul -- Internet Archive, FSF connections. OpenLibrary.org

RecentChangesCamp 2011, Canberra Update
Mark's first Open Space facilitation. He found people very interested in discussing wiki culture.

 * Wikimedia is the big gorilla of wiki culture.
 * Wikimedia Australia chapter was able to get lots of people there. UCNISS was also able to get a fair amount of people there because we utilized our local networks including educators.
 * Facilitation this time may have been better than the first RCC Australia, led to richer outcome
 * Reminisces of Wikipedia 10, which was well-facilitated by eekim (and also, kickoff by wardc was a great focusing)
 * session: "why wikis don't work"
 * Brion says MediaWiki installer is getting better! (yay)
 * Interesting discussion of WikiTravel: a site that was bought by an absentee company. They don't interfere, but they don't upgrade either!
 * Lonely Planet publisher is doing a cool OpenOffice/MediaWiki mashup
 * A lot of wikispaces users attended
 * white walls are fantastic for this type of event

Future Stuff

 * How does the Australian RCC community plug in and work with the RCC community in the USA? Australian perspective is a bit of "ignore the Americans" as the Americans are ignoring us. The American in Australian is just RAR! WE ARE AWESOME! we listen to you! Why no return favor?  (Also, desire to get Mark out for New Zealand three day some time in 2012.)
 * discussion: maybe RCC USA is not as organized as we seem :-) maybe we're not ignoring, we're just stumbling around
 * maybe wikimedia vs. rcc issue?
 * Possibly? Could be non-profit/educator vs WMF & commercial wiki communities?
 * larger question: what about the wiki ecosystem as a whole
 * kaliya shares story about IIW -- see section below
 * I think having a, you know, wiki will help (you know, one you can edit ;)
 * Seriously: Issue for me. RCC in Australia and New Zealand have the advantage of being a bit more topical.  And they are a lot more WMF and educator centric than the ones in the USA.  We also like the idea of doing one day RCCs.  I've yet to be able to feel out how others feel about that.
 * This 'topical' RCC was also a factor for Wikimedia Australia; I doubt the WMAu board would have agreed to get behind the RCC if it didnt have some credibility in that lots of people engaged in the (real) education sector had attended the one-day RCC last year.
 * And yes, the wiki being functional would be good. I think its failure to work is going to push the RCC community in Australia to be more focused on Wikiversity.  (Probably as part of a grand experiment to see if Wikiversity can work.)

Forking

 * discussion about wikitravel fork discussion
 * There was some discussion of forking wikiTravel. Or maybe WikiVoyage.org wants to take add the other languages they didnt fork originally.
 * has there ever been a successful fork of a wiki community?
 * wired article about Spanish Wikipedia Fork
 * http://www.wired.co.uk/news/archive/2011-01/20/wikipedia-spanish-fork
 * http://en.wikipedia.org/wiki/Wikipedia:Wikipedia_Signpost/2011-01-24/In_the_news
 * Essay on the Spanish fork: http://meta.wikimedia.org/wiki/Transwiki:Wikimania05/Paper-AS1
 * Polish Wikipedia did too, smaller
 * both re-merged later
 * John Vandenberg (rocks) and Mark are going to discuss the idea of forking communities over on Meatball, in a general way
 * citizendium
 * wikia? conservapedia?
 * somebody should make a wiki family tree! Would this be possible on wikindex?
 * Where the inspiration came from for each? Would be fun.
 * the inspiration came from beer (there was some cider involved too).

 * Other recent Open Source forks: LibreOffice (from OpenOffice), MariaDB (from MySQL)

Jyte -- JanRain -- AboutUs

 * note the difference between forking an entire dataset+community and diverting particular sub-areas out of a larger community to a smaller one -- can be related though, and there may be a fuzzy boundary
 * citizendium -- can this be considered a fork? same _content_ goals as wikipedia, but "forking" the creation process itself
 * define successful? A few of the clones/forks/spawns were moderately successful, but one key element (for me) is whether the fork can replace the original.  Google Juice is one of the problems that a fork has to contend with : how does a new wiki compete with the old one which has millions of incoming links and a network of 'follow' links that Google loves.
 * many wikitravel, wikia communities (as well as the non-wikipedia WMF projects) started at least in part by diverting people interested in  doing certain kinds of things from wikipedia to their own smaller  projects
 * define fork? I consider Fan History a Wikipedia fork. We have no content that we really borrowed from there but we left Wikipedia because it did not work with our goals.  It was an intentional response to Wikipedia.
 * also consider cases where a community leaves their content behind and has to recreate it elsewhere due to licensing or other lockdown issues

family tree
nice idea from Kaliya - a family tree/visualization of how wikis have evolved, on three dimensions:
 * wiki family/people
 * content
 * software

Internet Identity Wolkshop
Kaliya brings this up as a different community dealing with international issues.
 * iiw.idcommons.net

Blogs
How can we get better blogging about how to use wikis?

Can/should wiki communities use blogging and social networking in more integrated ways? What does it change about the community when people communicate outside the wiki? (cf existing mailing lists, irc chat) Does this need to be more two-way, or is that already the case?

Keeping track of published research about what sort of interventions make wiki communities/online communities work best

 * Wikihow did research. Where is that?

WikiOhana
How to generate more WikiOhana?

 * group blog
 * signpost
 * wikiplanet
 * #wikipintouch
 * "energy of the we"
 * part of why we like wiki is because we can participate when we have energy, and not when we don't
 * Conference sponsorship from wiki companies (good faith participation in wider community, encouraging their members to engage)

WikiFactor
[Soft ways of directing traffic & activity to interesting things]

Automated activity-detection for promotion of cool stuff? "This page is now at wiki factor 8, come check it out!"

Universal Edit Button

 * Help file built in, that explains syntax for whatever software you're in?

She's Geeky

 * Kaliya reports…they just had their 4th bay area one
 * a lot of women doing amazing stuff -- but mostly they want to do their own stuff
 * How do large institutions (e.g., Anita Borg Institute, Wikimedia, ...) relate to grassroots organizations in the same space in a healthy way, without "pretending" they're on the same level?
 * e.g., keep a community database of grassroots orgs, so people (and they) can find each other

We want MetaOrganizing

 * Larger organizations should play a role in helping smaller orgs be networked
 * Wiki Planet
 * Wiki weblog?
 * What is the role of Wikimedia Foundation to do meta-organizing?
 * Is this a role for Wikimedia member organizations?
 * Pete K feels it's important for the Foundation to be more "conscious" of interfacing with grassroots, non-Wikimedia wiki communities.
 * Some interesting thoughts about how this needs to work; problems vs benefits of having explicit interfaces.
 * If someone comes and asks for something to be done, sometimes the right answer is "no we're not going to do it -- but YOU can!" If nobody actually says it, then the interested party might never do it. There's as much value in saying "go ahead" as there is in actually being able to do it for someone.

Membership organizations
Would some of the issues we're discussing be better served in the U.S. by a membership organization (like Wikimedia chapters)? Is there a possible U.S. Wikimedia chapter?

 * There have been a number of discussions about a U.S. chapter.
 * There's a Wikimedia New York City
 * There have been rumblings about Washington DC, California, and Oregon

 * Eugene: Is what we need "another organization" or is it simply "more organizing" irrespective of what the specific orgs are?
 * Piggyback on Canada?
 * Not as useful as a geographically overlapping chapter with seperate leadership for the reasons in http://talknicer.com/wm10ca.pdf
 * Advantage of chapters is money (it's easier to do fundraising with the Wikimedia brand than without) Money can help with organizing. Other advantage is credibility. RCC Canberra had more credibility because of ties to UCNISS and Wikimedia Australia.  Wikiversity might have more credibility down here if it looked like we had WMF support.

 * Would it be worthwhile to have a "wiki" foundation apart from WMF?
 * WMF exists to "protect" the content/servers, at the expense of volunteers to some extent
 * i.e. they avoid legal battles that involve volunteers, or it appears that way. A member-based organisation, which doesnt 'own' the servers, can have more focus on the community's needs.  A US member based organisation would have clout (money, resources, etc) in court, and could also provide (real) dispute resolution services for issues which involve people within its member base and member vs non-member.  services such as teleconferencing and engaging professional mediators.
 * US organization doing that might not be viewed favorably internationally as it could be seen as another attempt by Americans to impose their will on the world. This is problem enough on Wikipedia.
 * What's the one thing we all could do, or have other people do, that would help?
 * Agile Manifesto is an example (link?)
 * Some reflection on 10 years of the manifesto: http://www.pragprog.com/magazines/2011-02/agile--

discussions tend to focus on things that don't work; sometimes necessary to keep perspective
 * you've got to accentu-ate the positive!

RecentChangesCamp Canberra worked. :D Lots of positives coming out of it. We might have two to four wiki conferences come out of it. And we learned more about our own needs, how Wikimedia Australia can help us grow our national community, its mission, made contacts that we can use. We have less Wikiversity angst and think it could be more useful. (Lack of venue and funding issues probably helped made the experience a lot more positive than we have had in the USA and Canada.)

Community-Forming For Wiki

 * necessary components
 * networking
 * group identity
 * activation energy
 * what's one thing we can do now?
 * Wiki brand, wiki foundation
 * Agile Manifesto - ?? Wiki Manifesto
 * ward as spiritual leader
 * ask ward to come up with 5 or so core principles of WikiOhana
 * Ward needs more of an explanation. I don't know how well known Ward is. Leigh had no idea and wondered why you needed him to help establish credibility.
 * challenge statement: getting people who are clearly part of the wiki community to say "us" instead of "you guys"
 * a role for invitation? make people feel welcome, and make sure they know they'll be welcome with 'us' as well as off by themselves "let's go to the party together" not "hey come to MY party"
 * invitation -> WelcomeVisitors (see Welcoming section below)

to look up
There are pretty excellent connections betw. Wikisource, Openlibrary.org, and archive.org -- DJVU files from archive -> wikisource and OL.org… wikisource heavily links to archive.org - I think we would also heavily link to openlibrary.org but there hasnt been much call for this, esp. as openlibrary has been a bit fluid while they work out their automatic importing system (which is stablising) John, remind me to tell you about an idea IA has about linking to wikisource sounds good - if we dont get time now, we can take it offline via email or my talk page on en.ws Sure!
 * http://OpenLibrary.org - a wiki page for every book
 * http://wikisource.org - a wiki page for every PD page :P

Welcoming
Invitation (study on this done by WikiHow) -is there a power relationship with invitation

Not "come to my party" but "let's go to this fun party together"

Make event wikis more party like. Laura loves RCC Canberra wiki because it is so fun with animated gifs.

rcc-announcement list needed

 * now set up: http://groups.google.com/group/rcc-announcements
 * Announcement list for one day, expanded and events around the world? (yes)

---

PREVIOUS CONTENT FROM EVENT SETUP

Confirmed Event
Tuesday, February 1, 6:00pm at Noisebridge: https://www.noisebridge.net/wiki/Wiki_meetup

Please consider donating a few $ to this community-run hacker space when you attend!

Proposed Dates

 * Tuesday, February 1
 * Wednesday, February 2
 * Thursday, February 3 (morning)

Other Potential Venues
Hub SF HealthRally (SF) Palo Alto (SocialText? Methean?) Mountain View (Hacker Dojo?) Anywhere near Fremont?

Attendees
(please list any venue or date/time preferences)

Peter Kaminski - Tuesday evening at Noisebridge MarkDilley - where/whenever - http://MeetingWords.com/MarkDilley Geoff Dale - maybe (prefer Palo Alto, since I am rarely getting too far from  home base these days. At least during extended biz hours.) Pete F. http://peteforsyth.com Definitely in SF; would like to attend in Palo Alto, but would need a ride Eugene - Tuesday

Resources

 * wiki-sf-bay-area mailing list
 * Wikimedia Bay Area mailing list

Stuff to talk about
…
 * RecentChangesCamp Canberra
 * WikiCreole (common themes in wiki syntax from one wiki engine to another)
 * Gender and Wikipedia
 * wiki spam

Update RecentChangesCamp Wiki
http://en.wikiversity.org/wiki/University_of_Canberra/RCC2011/RecentChangesCamp_Expanded
 * When ready - http://RecentChangesCamp.org/wiki/RecentChangesCamp_Expanded (problems with spam prevent editing rrc.org)
 * Link this from main page/template at RCC wiki?


text:
Category:University of Canberra/RCC

Purpose statement here

title: Category:University of Canberra/RCC


text:
User:Egm6321.f10.team2.oztekin/New

=Problem 2.3:=

Given
The equivalency validation was asked in meeting 8.3. as a second case where there are orthonormal vectors. First case as given:
 * {| style="width:100%" border="0"

$$  \displaystyle \begin{align} &\underline{w}=\sum\limits_{i} &\forall \left\{ {{\alpha }_{1}},......,{{\alpha }_{n}} \right\}\in {{R}^{n}} \end{align} $$     (3.1)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$ \displaystyle \begin{align} &\underline{w}=\sum\limits_{i} &\forall \left\{ {{b}_{1}},.....,{{b}_{n}} \right\}\in {{R}^{n}} \end{align} $$     (3.2)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Where $$ \displaystyle a_{i} $$'s are orthonormal basis functions.We can identify orthonormal basis functions as,

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

$$  \displaystyle {{\underline{a}}_{i}}.{{\underline{a}}_{j}}={{\delta }_{ij}} $$     (3.3)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle \delta_{ij} =\left\{ \begin{matrix} \begin{matrix} \begin{matrix} 1 & for \\ \end{matrix} & i=j \\ \end{matrix} \\ \begin{matrix} \begin{matrix} 0 & for \\ \end{matrix} & i\ne j \\ \end{matrix} \\ \end{matrix} \right.
 * style="width:95%" |
 * style="width:95%" |

$$     (3.4)
 *
 * }

The operator of $$\displaystyle \underline{\underline{P}}(\underline{v})$$ was defined as Eq(4) in meeting 7-2. This operator would lead us to conclude with stiffness matrix,unknown matrix and force matrix.
 * {| style="width:100%" border="0"

$$  \displaystyle \underline{\underline{P}}(\underline{v}):=\sum\limits_{j=1}^{n}{{{\underline{b}}_{j}}{{v}_{j}}-\underline{v}=0} $$     (3.5)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle {{\underline{b}}_{i}}.\underline{\underline{P}}(\underline{v})=0 $$     (3.6)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle {{\underline{b}}_{i}}.\sum\limits_{j}{{{\underline{b}}_{j}}{{v}_{j}}={{\underline{b}}_{i}}.\underline{v}} $$     (3.7)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle {{\left[ {{K}_{ij}} \right]}_{n*n}}{{\left\{ {{v}_{j}} \right\}}_{n*1}}={{\left\{ {{F}_{i}} \right\}}_{n*1}} $$     (3.8)
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 * style="width:95%" |
 *
 * }

Find
Show $$ \displaystyle \underline{w}.\underline{\underline{P}}(\underline{v})=0$$ is equivalent to $$\displaystyle {{\underline{a}}_{i}}\underline{\underline{P}}(\underline{v})=0 $$

Solution
For i=1,..........n. We have n equations and n unknowns.Since the components of vector w are arbitrary we can decide what they are as our convenience in order to proof.

Choice 1 :$$ \displaystyle \beta _{1}=1,\beta _{2}=......=\beta _{n}=0$$ then we can observe ;

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

$$  \displaystyle \underline{w}={{\underline{a}}_{1}} $$     (3.9)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \left \{ b_{1},....,b_{n} \right \}=\left \{ \alpha _{1},......,\alpha _{n} \right \}\Rightarrow {{\underline{a}}_{1}} ={{\underline{b}}_{1}}
 * style="width:95%" |
 * style="width:95%" |

$$     (3.10)
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \underline{a}{}_{1}.\underline{\underline{P}}(\underline{v})=0 $$     (3.11) Choice 2 :$$ \displaystyle \left \{ \beta _{1} ,.......,\beta _{n}\right \}=\left \{ 0,1,0,......,0 \right \}$$ then ;
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle \underline{w}={{\underline{a}}_{2}} $$     (3.12)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \left \{ b_{1},....,b_{n} \right \}=\left \{ \alpha _{1},......,\alpha _{n} \right \}\Rightarrow {{\underline{a}}_{2}} ={{\underline{b}}_{2}}
 * style="width:95%" |
 * style="width:95%" |

$$     (3.13)
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle {{\underline{a}}_{2}}.\underline{\underline{P}}(\underline{v})=0 $$     (3.14)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle \begin{matrix} . \\   .  \\   .  \\ \end{matrix}
 * style="width:95%" |
 * style="width:95%" |

$$
 * }

Choice 3 : $$ \displaystyle \left \{ \beta _{1},....,\beta _{n} \right \}=\left \{ 0,.....0,1 \right \} $$

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

$$  \displaystyle \underline{w}={{\underline{a}}_{n}} $$     (3.15)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \left \{ b_{1},....,b_{n} \right \}=\left \{ \alpha _{1},......,\alpha _{n} \right \}\Rightarrow {{\underline{a}}_{n}} ={{\underline{b}}_{n}}
 * style="width:95%" |
 * style="width:95%" |

$$     (3.16)
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle {{\underline{a}}_{n}}.\underline{\underline{P}}(\underline{v})=0 $$     (3.17) So we can say that by selecting proper components for vector w we can conclude same equation which was given Eq(3.1).Actually Eq(3.1) is more  general casse than Eq(3.2). Also we can broaden combinations of components of vector w.For example;
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \left \{ \beta _{1},....,\beta _{n} \right \}=\left \{ 1,1,1,0,...,0 \right \} $$     (3.18)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle {{\underline{w}}_{1}}={{\beta }_{1}}{{\underline{a}}_{1}}+{{\beta }_{2}}{{\underline{a}}_{2}}+{{\beta }_{3}}{{\underline{a}}_{3}}$$ (3.19)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle {{\underline{w}}_{1}}={{\underline{b}}_{1}} $$     (3.20)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle {{\underline{b}}_{1}}={{\beta }_{1}}{{\underline{a}}_{1}}+{{\beta }_{2}}{{\underline{a}}_{2}}+{{\beta }_{3}}{{\underline{a}}_{3}}$$ (3.21)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

This is 'nearly' the same vector for basis $$\displaystyle b_{n}$$. We can validate $$\displaystyle {{\underline{b}}_{n}}\underline{\underline{P}}(\underline{v})=0$$. But after knowing all basis vectors of vector w, then we have to check if their gram matrix is not equal to zero. Then we say that our new vectors are new arbitrary 'linerly independent' family.

Expressing vector b in terms of orthonormal vectors a is exactly same with the case presented in meeting 8-2.Because in this case we had validated that any linearly independent fuctions scalar product with $$\displaystyle \underline{\underline{P}}(\underline{v})$$ is equal zero. Therefore we can always express linearly independent basis 'vector b' with our orthonormal vectors.

=Problem 2.7:Determining orthogonality of family of functions=

Given
Problem was satated in meeting 10-3. . Our family of equations are :

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

$$  \displaystyle F=\left \{ 1,x,x^{2},x^{3},x^{4} \right \} $$     (7.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Test of linear independency of family of equations can be done by setting up 'Gram Matrix'.Gram Matrix If the linear independency exists between functions of the family their gram matrix's determinant must be non-zero.

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

$$  \displaystyle \mathbf{\Gamma } \left ( b_{1}(x),.....,b_{n}(x) \right )=\left [< b_{i},b_{j} >\right ]_{n\times n} $$ (7.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \mathbf{\Gamma_{ij}}=<b_{i},b_{j}>=\int_{\Omega }b_{i}(x)b_{j}(x)dx $$     (7.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

We can identify our basis functions as :

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

$$  \displaystyle b_{1}(x)=1 $$     (7.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle b_{2}(x)=x $$     (7.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle b_{3}(x)=x^{2} $$     (7.6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle b_{4}(x)=x^{3} $$     (7.7)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle b_{5}(x)=x^{4} $$     (7.8)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Find
Find if the family of these basis functions are orthogonal over the domain $$\displaystyle\Omega =\left [ 0,1 \right ] $$

Solution
First we have to set up our gram matrix.

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

$$  \displaystyle \mathbf{\Gamma } =\begin{bmatrix} b_{11} & b_{12} & b_{13} & b_{14} &b_{15} \\ b_{21} & b_{22} & b_{23} & b_{24} &b_{25} \\ b_{31} & b_{32} & b_{33} & b_{34} &b_{35} \\ b_{41} &b_{42} & b_{43} & b_{44} &b_{45} \\ b_{51} &b_{52} & b_{53} &b_{54}  & b_{55} \end{bmatrix} $$     (7.9)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Now we can look at the components of this matrix. Since 'scalar product' is symmetric the components which are symmetric about the diagonal of this matrix will be same each other.

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

$$  \displaystyle b_{11}=<b_{1},b_{1}>=\int_{0}^{1}1.1dx=1 $$     (7.10)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle b_{22}=<b_{2},b_{2}>=\int_{0}^{1}x.xdx=\frac{1}{3} $$     (7.11)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle b_{33}=<b_{3},b_{3}>=\int_{0}^{1}x^{2}x^{2}dx=\frac{1}{5} $$     (7.12)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle b_{44}=<b_{4},b_{4}>=\int_{0}^{1}x^{3}x^{3}dx=\frac{1}{7} $$     (7.13)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle b_{55}=<b_{5},b_{5}>=\int_{0}^{1}x^{4}x^{4}dx=\frac{1}{9} $$     (7.14)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle b_{12}=b_{21}=<b_{1},b_{2}>=\int_{0}^{1}xdx=\frac{1}{2} $$     (7.15)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle b_{13}=b_{31}=<b_{1},b_{3}>=\int_{0}^{1}x^{2}dx=\frac{1}{3} $$     (7.16)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle b_{14}=b_{41}=<b_{1},b_{4}>=\int_{0}^{1}x^{3}dx=\frac{1}{4} $$     (7.17)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle b_{15}=b_{51}=<b_{1},b_{5}>=\int_{0}^{1}x^{4}dx=\frac{1}{5} $$     (7.18)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle b_{23}=b_{32}=<b_{2},b_{3}>=\int_{0}^{1}x.x^{2}dx=\frac{1}{4} $$     (7.19)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle b_{24}=b_{42}=<b_{2},b_{4}>=\int_{0}^{1}x.x^{3}dx=\frac{1}{5} $$     (7.20)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle b_{34}=b_{43}=<b_{3},b_{4}>=\int_{0}^{1}x^{2}.x^{3}dx=\frac{1}{6} $$     (7.21)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle b_{54}=b_{45}=<b_{3},b_{4}>=\int_{0}^{1}x^{3}.x^{4}dx=\frac{1}{8} $$     (7.22)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle b_{52}=b_{25}=<b_{2},b_{5}>=\int_{0}^{1}x.x^{4}dx=\frac{1}{6} $$     (7.23)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle b_{53}=b_{35}=<b_{3},b_{5}>=\int_{0}^{1}x^{2}.x^{4}dx=\frac{1}{7} $$     (7.24)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

We can conlude with :

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

$$  \displaystyle \mathbf{\Gamma }=\begin{bmatrix} 1 & \frac{1}{2} & \frac{1}{3} &\frac{1}{4} &\frac{1}{5} \\ \frac{1}{2}& \frac{1}{3} & \frac{1}{4} &\frac{1}{5} &\frac{1}{6} \\ \frac{1}{3} & \frac{1}{4} & \frac{1}{5} & \frac{1}{6} &\frac{1}{7} \\ \frac{1}{4}& \frac{1}{5} & \frac{1}{6} & \frac{1}{7} & \frac{1}{8}\\ \frac{1}{5} & \frac{1}{6}&\frac{1}{7}  & \frac{1}{8} & \frac{1}{9} \end{bmatrix} $$     (7.25)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

We can easily take the determinant of gram matrix by using a solver. Matlab Mathworks gives the solution of determinant as :$$\displaystyle 3.7493e(-12)$$ The gram matrix is 'symmetric' so its transpoze will be same with itself but for orthogonality matrix transpoze has to be same as its inverse.The inverse of gram matrix can easly get from matlab as:

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

$$  \displaystyle \mathbf{\Gamma ^{-1}}= 1.0e0.005\times \left[ \begin{matrix} \begin{matrix} \begin{matrix} \begin{matrix} 0,0002 \\   -0,003  \\ \end{matrix}  \\ 0,0105 \\   -0,014  \\   0,0063  \\ \end{matrix} & \begin{matrix} \begin{matrix} -0,003 \\   0,0480  \\ \end{matrix}  \\ -0,189 \\   0,2688  \\   -0,126  \\ \end{matrix}  \\ \end{matrix} & \begin{matrix} \begin{matrix} 0,0105 \\   -0,189  \\ \end{matrix}  \\ 0,7938 \\   -1,176  \\   0,5670  \\ \end{matrix} & \begin{matrix} \begin{matrix} -0,014 \\   0,2688  \\ \end{matrix}  \\ -1,176 \\   1,792  \\   -0,882  \\ \end{matrix} & \begin{matrix} \begin{matrix} 0,0063 \\   -0,126  \\ \end{matrix}  \\ 0,5670 \\   -0,8820  \\   0,4410  \\ \end{matrix}  \\ \end{matrix} \right]
 * style="width:95%" |
 * style="width:95%" |

$$     (7.26)
 *

 * }

The inverse of the matrix is not similar with its transpoze. So it is not orthogonal.

= References =

title: User:Egm6321.f10.team2.oztekin/New


text:
Category:Egm6341.s11.team5

User:Egm6341.s11.team5

title: Category:Egm6341.s11.team5


text:
University of Canberra/RCC2011/When will wikis supersede traditional Word Processing?

 * Marghanita da Cruz's notes


text:
Talk:University of Canberra/RCC2011

Other topics
Due to the format, there were quite a few other topics raised throughout the three days, in and out of the sessions. I found some of them to be very valuable and could be the topic of further discussions. How do we want to capture and continue these discussions? John Vandenberg (chat) 03:26, 1 February 2011 (UTC)
 * John, I reckon we simply give those events names (as I've done for lunch and dinner etc) and link them through to sub pages to add any notes, documentation, or links to further project pages. So if for example, you're referring to a dinner conversation on one of the nights, give that conversation a session name and list it in or around the appropriate day. Hopefully this keeps the structure of this archive as flat as possible, indexing all that happened in the "Session" section. Leighblackall 08:15, 1 February 2011 (UTC)

RCCx across Australia
One of the topics raised in the last session was participants in RCC taking the RCC 'home'. MarkDilley and Peter Kaminski have coined the term 'RCCx' (RecentChangesCamp Expanded) for this, and is running one in San Francisco (University of Canberra/RCC2011/RecentChangesCamp Expanded). The participants in RCC came from most states in Australia, so we could organise similar RCCx in most capital cities in the next month or so. These mini-events would spread the word and enthusiasm.

I think these RCCx events only need to be about two hours, providing a broad overview of the topics covered in the RCC, and zooming in on whatever sessions the RCCx attendees would like to know more about.

As far as I can remember, the only states and territories not represented are QLD and NT. I am happy to take a holiday up to Brisbane (QLD) and organise an RCCx up there in the coming weeks. I can also host one in Armidale, especially if the education people know any good folk at the Education school at UNE who would be interested.

If a few other participants are also willing to run an RCCx, we can start listing them all on University of Canberra/RCC2011/RecentChangesCamp Expanded and spread the world. John Vandenberg (chat) 01:43, 2 February 2011 (UTC)

 * It might be worth contacting local organizers of Wikipedia 10 year birthday things to ask if they could help with a meet up to discuss that sort of thing in addition to who ever was local RCC wise? --LauraHale 01:54, 2 February 2011 (UTC)

 * Adelaide and Perth WP10 events were run by RCC attendees.
 * The http://ten.wikipedia.org/wiki/Melbourne event was held at a location which wouldn't be suitable, but the organisers (http://dailypositive.org) are a wiki that might be interested. I've emailed Jani and asked him if he is interested.
 * I have no idea about the organisers or participants of the http://ten.wikipedia.org/wiki/Sydney or http://ten.wikipedia.org/wiki/Hobart event.
 * It looks like Leigh went to the http://ten.wikipedia.org/wiki/Canberra event.
 * John Vandenberg (chat) 04:49, 2 February 2011 (UTC) John Vandenberg 10:26, 2 February 2011 (UTC)

I can get along to anything within reach on Canberra, including regional towns, the coast, Sydney, and even Newcastle. Leighblackall 04:28, 2 February 2011 (UTC)

I'd be keen to attend (but not organise) any RCCx in regional or northern NSW, and would go as far as Newcastle, but probably not as far as Sydney. John Vandenberg (chat) 04:38, 2 February 2011 (UTC)

Just letting you know that Chris and I are planning to organise an RCCx event (or WikiWednesday) in Sydney. I will keep everyone posted about the progress! Juttavd 07:08, 4 February 2011 (UTC)

We have one confirmed RCCx in Melbourne and three more being planned. Details at University of Canberra/RCC2011/RecentChangesCamp Expanded --John Vandenberg 20:38, 21 February 2011 (UTC)

Sessions
I think the red links should probably be removed. I don't know if they were all sessions. Maybe create an article for other notes related to at RCC discussions that people might find helpful? --LauraHale 05:34, 2 February 2011 (UTC)
 * Let's give it until mid next week for others to settle back and add what they have in there. I know John is working on something for Wikis to Printed Books, and Kirsty and Jutta have notes, photos and mind maps to add too. Leighblackall 23:46, 2 February 2011 (UTC)

title: Talk:University of Canberra/RCC2011


title: File:Egm6341.s11.team4.hw2.fig14.jpg


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User:Griffinofwales/vector.js

mw.loader.load('https://meta.wikimedia.org/w/index.php?title=User:Griffinofwales/global.js&action=raw&ctype=text/javascript');

title: User:Griffinofwales/vector.js


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Seminar Question Writing Guidelines

The following table serves to guide teachers in understanding the types of questions that guide good seminars and how to write them. Good seminars follow the general structure given by Mortimer Adler and Charles Van Doren in their classic How to Read a Book published by Simon & Schuster in 1972. Column one gives the four main questions that a demanding reader should ask of any book. These questions guide the types of questions and the purpose of each type used in good seminars. The last column serves to illustrate questions that might be asked of The Declaration of Independence, The Gettysburg Address, or Hamlet. This table could be easily adapted for any work of fine art like a painting or a musical score.

title: Seminar Question Writing Guidelines


text:
User:Kate moncrieff

Hi...

Im Atchara Moncrieff or Kate

This is Me as a South Park cartoon character. A friend from Raffles institute created this character.

___________________________________________________________________________________________________________________________________

CHASSIS COLOR	ALX Space Black - Anodized Aluminum Chassis PROCESSOR	Overclocked Intel® Core™ i7 980x Extreme Six Core Processor (4.0GHz, 12MB Cache) OPERATING SYSTEM	Genuine Windows® 7 Ultimate, 64bit, English MEMORY	12GB Triple Channel 1600MHz DDR3 VIDEO CARD	Dual 2GB ATI Radeon™ HD 6950 - ATI CrossFireX™ Enabled KEYBOARD	Alienware TactX™ Keyboard MOUSE	Alienware TactX™ Mouse HARD DRIVE	512GB Raid 0 (2x 256GB Samsung Solid State HDDs) ADDITIONAL HARD DRIVE	2TB RAID 1 (2x2TB SATA-II, 7,200 RPM, 32MB Cache HDDs) MEDIA READER	19-in-1 Media Card Reader MONITOR	21.5” Alienware AW2210 OptX™ Full HD Gaming Monitor ADVANCED DISPLAYS 	Dell UltraSharp U2711 27-inch Monitor WARRANTY	3Yr Ltd. Warranty- Advance Exchange WIRELESS	1525 PCIe WLAN card with 11n mini-Card & external antenna ENTHUSIAST ESSENTIALS	Killer™ 2100 Gaming Network Card SOUND CARD	Creative Sound Blaster X-Fi Titanium

OPTICAL DRIVE Dual Drives: Blu-ray Disc (BD) Burner (Writes to DVD/CD/BD) and DVDRW

My Software & Accessories OS CUSTOMIZATION	Alienhead 3D ALIENFX	AlienFX Color, Plasma Purple AVATAR	Alienware Logo PRE-INSTALLED GAMES AND CLIENTS	Steam and Portal™ Factory Installed

Personalize WARRANTY AND SERVICE	1 Year Basic Service Plan, DMLP My Accessories ALSO INCLUDED WITH YOUR SYSTEM Alienware Area-51 ALX	Alienware Area-51 ALX + TactX™ Headset COOLING OPTION	Alienware™ High-Performance Liquid Cooling

---	According to the spec lists, we think that iPhone 4 have the best spec. iPhone 4 performance hardware and storage 1GHz A4 processor, 512MB of RAM and either 16GB or 32GB of storage. It displays 3.5-inch touch screen retina display which supports a resolution of 960×640. Moreover, the camera of iPhone is 5 megapixel rear-facing cameras with auto-focus and an LED flash. iPhone 4 is said to offer 300 hours of standby time and 7 hours of talk time. Finally, iPhone 4 have faster CPU speed, higher screen resolution, and large storage. Moreover, iPhone have a better WiFi connectivity and superior talk time from a single charge than blackberry 9780 and Nokia N95.

title: User:Kate moncrieff


text:
User:EML5526.S11.Team3.vnarayanan/Homework1

Solution 1.2
The Partial Differential Equation (PDE) of an elastic bar subjected to an axial load derived in the section (1.1) is

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

|$$\displaystyle \frac{\partial}{\partial x}[E(x) A(x))\frac{\partial u}{\partial x}] + f(x,t) = \rho(x) A(x) \frac{\partial^2 u}{\partial t^2} $$ $$
 * $$\displaystyle (Eq. 1)

 * }
 * }

Now, a particular case is taken into consideration, where the bar has a rectangular cross-section. The breadth of the bar is a constant and is equal to 'b'. But the height of the bar varies along the length of the bar and hence it is function of x, which is 'h(x).' (For figure, refer page 6-1 of meeting 6).

Let us consider the inertial force at the center of the thin strip of the elastic bar. Therefore the mass at that location is given by,

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

|$$\displaystyle m(x + \frac{dx}{2}) = \rho(x + \frac{dx}{2}) b h(x + \frac{dx}{2}) $$ $$
 * $$\displaystyle (Eq. 2)

 * }
 * }

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

|$$\displaystyle m(x + \frac{dx}{2}) = \rho(x + \frac{dx}{2}) b \frac{1}{2} [h(x) + h(x+dx)] $$ $$
 * $$\displaystyle (Eq. 3)

 * }
 * }

Performing the balance of forces similar to part 1.1 and rearranging the terms we get

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

|$$\displaystyle \frac{\sigma (x+dx)bh(x+dx)-\sigma (x)bh(x)}{dx}+f(x,t)=\rho(x + \frac{dx}{2}) b \frac{1}{2} [h(x) + h(x+dx)]\frac{\partial^2u }{\partial t^2}$$ $$
 * $$\displaystyle (Eq. 4)

 * }
 * }

Applying the limit as $${\frac{dx}{2}\to0}$$ in the above equation,by the definition of partial derivative

$$ \lim_{dx\to0}\frac{f(x+dx)-f(x)}{dx}= \frac{\mathrm{d} f(x)}{\mathrm{d} x} $$

we get,
 * {| style="width:100%" border="0" align="left"

|$$\displaystyle \frac{\partial (\sigma (x)bh(x+0))}{\partial x}+f(x,t)=\rho(x + 0) b \frac{1}{2} [h(x) + h(x+0)]\frac{\partial^2u }{\partial t^2}$$ $$
 * $$\displaystyle (Eq. 5)

 * }
 * }

The stress $$ \sigma\left (x\right)$$ can be expressed in terms of elastic modulus $$ E\left (x\right)$$ and strain as                               $$\sigma (x)=E(x)\frac{\partial u}{\partial x}$$

Hence the PDE for 1 Dimensional elastic bar is
 * {| style="width:100%" border="0" align="left"

|$$\displaystyle \frac{\partial (E(x)bh(x)\frac{\partial u}{\partial x})}{\partial x}+f(x,t)= \rho (x)bh(x)\frac{\partial^2u }{\partial t^2}$$ $$
 * $$\displaystyle (Eq. 6)

 * }
 * }

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

|$$\displaystyle \frac{\partial (E(x)h(x)\frac{\partial u}{\partial x})}{\partial x}+\frac{1}{b}f(x,t)= \rho (x)h(x)\frac{\partial^2u }{\partial t^2}$$ $$
 * $$\displaystyle (Eq. 7)

 * }
 * }


text:
User:EML5526.S11.Team3.vnarayanan/Homework2

Problem 2.7
Consider F = {$$1,x,x^{2},x^{3},x^{4}$$} in the interval $$ \Omega=[0,1]$$ (2.7.1) Construct $$\underline{\Gamma }$$(F) (2.7.2) Find the determinant of $$\underline{\Gamma }$$(F) (2.7.3) Is F Orthogonal Family?

Solution(2.7.1)
F = {$$1,x,x^2,x^3,x^4$$} = {$$b_{1}(x),b_{2}(x),b_{3}(x),b_{4}(x),b_{5}(x)$$} $$\Gamma = \begin{bmatrix} \Gamma_{11}&\Gamma_{12}&\Gamma_{13} &\Gamma_{14}  &\Gamma_{15} \\ \Gamma_{21}&\Gamma_{22}&\Gamma_{23} &\Gamma_{24}  &\Gamma_{25} \\ \Gamma_{31}&\Gamma_{32}&\Gamma_{33} &\Gamma_{34}  &\Gamma_{35} \\ \Gamma_{41}&\Gamma_{42}&\Gamma_{43} &\Gamma_{44}  &\Gamma_{45} \\ \Gamma_{51}&\Gamma_{52}&\Gamma_{53} &\Gamma_{54}  &\Gamma_{55} \end{bmatrix}$$

$$\;\Gamma_{ij}=\int_{\Omega}b_{i}(x)b_{j}(x)dx$$ $$\Gamma_{11} =  = \int_{0}^{1}1 dx = 1$$ $$\Gamma_{12} =  = \int_{0}^{1}1.x dx =\frac{1}{2} $$ $$\Gamma_{13} =  = \int_{0}^{1}1.x^2 dx =\frac{1}{3} $$ $$\Gamma_{14} =  = \int_{0}^{1}1.x^3 dx =\frac{1}{4} $$ $$\Gamma_{15} =  = \int_{0}^{1}1.x^4 dx =\frac{1}{5} $$ $$\Gamma_{21} =  = \int_{0}^{1}x.1 dx =\frac{1}{2} $$ $$\Gamma_{22} =  = \int_{0}^{1}x.x dx =\frac{1}{3} $$ $$\Gamma_{23} =  = \int_{0}^{1}x.x^2 dx =\frac{1}{4}$$ $$\Gamma_{24} =  = \int_{0}^{1}x.x^3dx =\frac{1}{5} $$ $$\Gamma_{25} =  = \int_{0}^{1}x.x^4dx =\frac{1}{6} $$ $$\Gamma_{31} =  = \int_{0}^{1}x^2.1 dx =\frac{1}{3}$$ $$\Gamma_{32} =  = \int_{0}^{1}x^2.x dx =\frac{1}{4}$$ $$\Gamma_{33} =  = \int_{0}^{1}x^2.x^2 dx =\frac{1}{5} $$ $$\Gamma_{34} =  = \int_{0}^{1}x^2.x^3 dx =\frac{1}{6} $$ $$\Gamma_{35} =  = \int_{0}^{1}x^2.x^4 dx =\frac{1}{7} $$ $$\Gamma_{41} =  = \int_{0}^{1}x^3.1 dx =\frac{1}{4} $$ $$\Gamma_{42} = <b_{4}(x),b_{2}(x)> = \int_{0}^{1}x^3.x dx =\frac{1}{5} $$ $$\Gamma_{43} = <b_{4}(x),b_{3}(x)> = \int_{0}^{1}x^3.x^2 dx =\frac{1}{6}$$ $$\Gamma_{44} = <b_{4}(x),b_{4}(x)> = \int_{0}^{1}x^3.x^3 dx =\frac{1}{7} $$ $$\Gamma_{45} = <b_{4}(x),b_{5}(x)> = \int_{0}^{1}x^3.x^4 dx =\frac{1}{8} $$ $$\Gamma_{51} = <b_{5}(x),b_{1}(x)> = \int_{0}^{1}x^4.1 dx =\frac{1}{5} $$ $$\Gamma_{52} = <b_{5}(x),b_{2}(x)> = \int_{0}^{1}x^4.x dx =\frac{1}{6} $$ $$\Gamma_{53} = <b_{5}(x),b_{3}(x)> = \int_{0}^{1}x^4.x^2 dx =\frac{1}{7} $$ $$\Gamma_{54} = <b_{5}(x),b_{4}(x)> = \int_{0}^{1}x^4.x^3 dx =\frac{1}{8} $$ $$\Gamma_{55} = <b_{5}(x),b_{5}(x)> = \int_{0}^{1}x^4.x^4 dx =\frac{1}{9} $$

$$ \underline{\Gamma}(F) = \begin{bmatrix}

1 &\frac{1}{2} & \frac{1}{3}  & \frac{1}{4}  &\frac{1}{5} \\

\frac{1}{2} & \frac{1}{3} & \frac{1}{4}  & \frac{1}{5}  & \frac{1}{6} \\

\frac{1}{3} &\frac{1}{4} &\frac{1}{5}  &\frac{1}{6}  &\frac{1}{7} \\

\frac{1}{4} & \frac{1}{5} &\frac{1}{6} &\frac{1}{7}  &\frac{1}{8} \\

\frac{1}{5} &\frac{1}{6} &\frac{1}{7}  &\frac{1}{8}  &\frac{1}{9} \end{bmatrix}$$

Solution(2.7.2)
The determinant of $$\underline{\Gamma}(F)\;\;$$ is

$$\left | \underline{\Gamma}(F) \right | = 3.74\times 10^{-12}$$

Solution(2.7.3)
Consider any two basis functions. Say, $$ b_{i}$$ and $$b_{j}$$. If $$\Gamma_{ij} = <b_{i},b_{j}> = \delta_{ij} $$, then F is a family of orthogonal basis of functions. In other words, if $$\;\;\;\delta_{ij} = 1 \;for\; i=j\;$$  or if $$\;\;\; \delta_{ij} = 0\; for\; i\neq j$$, then F is a family of orthogonal basis of functions.

But in the $$\underline{\Gamma}(F)$$ matrix of Solution (2.7.1) , we find none of the elements satisfying the above condition. Hence, F is not a family of orthogonal basis of functions.

Problem 2.8

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

|$$\displaystyle \int_{\Omega}w^{h}(x)P(u^{h}(x))dx = 0 \;\; \forall\; w^{h}(x)\;\;$$ where $$\;w^{h}(x) = \sum_{i=1}^{n}c_{i}b_{i}(x)\;$$ $$
 *
$$\displaystyle (Eq. 1)

 * }
 * }

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

|$$\displaystyle \int_{\Omega}b_{i}(x)P(u^{h}(x))dx = 0 \;\; for\;\; i = 1,2,...,n$$ $$
 *
$$\displaystyle (Eq. 2)

 * }
 * }

Show that Eq.1 $$\;\Leftrightarrow\; $$ Eq.2

Part 1
Eq.1 $$\;\Rightarrow\;\;\int_{\Omega}w^{h}(x)P(u^{h}(x))dx = 0 \;\; \forall\;\; w^{h}(x)$$ where $$\;\;w^{h}(x) = \sum_{i=1}^{n}c_{i}b_{i}(x)$$

Case (1) : $$\;\; c_{1} = 1, c_{2}=c_{3}=...=c_{n} = 0$$

$$\therefore\;\; w^{h}(x) = b_{1}(x) $$

Hence, Eq.1 becomes
 * {| style="width:100%" border="0" align="left"

|$$\displaystyle \;\;\int_{\Omega}b_{1}(x)P(u^{h}(x))dx = 0$$ $$
 *
$$\displaystyle (Eq. 3)

 * }
 * }

Case (2) : $$\;\; c_{2}=1,c_{1} = c_{3}=...=c_{n} = 0$$

$$\therefore\;\; w^{h}(x) = b_{2}(x) $$

Hence, Eq.1 becomes
 * {| style="width:100%" border="0" align="left"

|$$\displaystyle \;\;\int_{\Omega}b_{2}(x)P(u^{h}(x))dx = 0$$ $$
 *
$$\displaystyle (Eq. 4)

 * }
 * }

Case (3) : $$\;\; c_{1} =c_{2}= c_{3}=...=c_{n-1} = 0,c_{n}=1,$$

$$\therefore\;\; w^{h}(x) = b_{n}(x) $$

Hence, Eq.1 becomes
 * {| style="width:100%" border="0" align="left"

|$$\displaystyle \;\;\int_{\Omega}b_{n}(x)P(u^{h}(x))dx = 0$$ $$
 *
$$\displaystyle (Eq. 5)

 * }
 * }

Hence, from Eq.(3), Eq.(4) and Eq.(5) it can be concluded that

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

|$$\displaystyle \int_{\Omega}b_{i}(x)P(u^{h}(x))dx = 0 \;\; for\;\; i = 1,2,...,n$$ $$
 *
$$\displaystyle (Eq. 6)

 * }
 * }

Part 2
Eq.(2) $$\; \Rightarrow\;\int_{\Omega}b_{i}(x)P(u^{h}(x))dx = 0 \;\; for\;\; i = 1,2,...,n $$

Multiply $$\;\;c_{i}\;\;$$ on both sides of the Eq.(2). We get, $$\;\int_{\Omega}c_{i}b_{i}(x)P(u^{h}(x))dx = 0 \;\; for\;\; i = 1,2,...,n$$

For, $$\; i = 1\;$$ $$\;\int_{\Omega}c_{1}b_{1}(x)P(u^{h}(x))dx = 0 $$

For, $$\; i = 2\;$$ $$\;\int_{\Omega}c_{n}b_{n}(x)P(u^{h}(x))dx = 0 $$

For, $$\; i = n\;$$ $$\;\int_{\Omega}c_{n}b_{n}(x)P(u^{h}(x))dx = 0 $$

Since, $$\;\int_{\Omega}c_{i}b_{i}(x)P(u^{h}(x))dx = 0 \;\; for\;\; i = 1,2,...,n\;\;$$ individually, the sum of all terms should also be equal to 0 $$\therefore\;\;\int_{\Omega}c_{1}b_{1}(x)P(u^{h}(x))dx + \int_{\Omega}c_{2}b_{2}(x)P(u^{h}(x))dx + .\; .\;. + \int_{\Omega}c_{n}b_{n}(x)P(u^{h}(x))dx = 0$$

$$\Rightarrow\;\;\int_{\Omega}\left [ (c_{1}b_{1}(x)P(u^{h}(x))) + (c_{2}b_{2}(x)P(u^{h}(x)))+ .\; .\;. + (c_{n}b_{n}(x)P(u^{h}(x)))\right ]dx = 0$$

$$\Rightarrow\;\;\int_{\Omega}\left [ c_{1}b_{1}(x) + c_{2}b_{2}(x)+ .\; .\;. + c_{n}b_{n}(x)\right ]P(u^{h}(x))dx = 0$$
 * {| style="width:100%" border="0" align="left"

|$$\displaystyle \Rightarrow\;\;\int_{\Omega}\sum_{i=1}^{n}c_{i}b_{i}(x)P(u^{h}(x))dx = 0$$ $$
 *
$$\displaystyle (Eq. 7)

 * }
 * }

We know that, $$\;w^{h}(x) = \sum_{i=1}^{n}c_{i}b_{i}(x)\;$$

$$\therefore\; Eq.(7) $$ becomes,

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

|$$\displaystyle \int_{\Omega}w^{h}(x)P(u^{h}(x))dx = 0$$ $$
 *
$$\displaystyle (Eq. 8)

 * }
 * }


text: File:Uc-logo.png

title: File:Uc-logo.png


text:
User:Eml5526.s11.team3.sahin/Homework 2

=Problem 2.6=

Given
Consider family of functions $$\displaystyle F=\left \{ 1,cosiwx,siniwx \right \} $$ on interval $$\displaystyle \Omega \left [ 0,T \right ]$$ for i =1,2

So

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

$$  \displaystyle F=\left \{ b_{1}(x),b_{2}(x),b_{3}(x),b_{4}(x),b_{5}(x) \right \} $$     (Eq 6.1)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle F=\left \{ 1,coswx,cos2wx,sinwx,sin2wx \right \} $$     (Eq 6.2)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Find
6.1 Construct $$\displaystyle \underline{\Gamma }(F)$$ and observe properties of $$\displaystyle \underline{\Gamma }(F)$$

6.2 Find the determinant of  $$\displaystyle \underline{\Gamma }(F)$$

6.3 Conclude F is orthogonal basis.

6.1

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

$$  \displaystyle \underline{\Gamma }={{[{{\Gamma }_{ij}}]}_{5*5}}=\left( \begin{matrix}  {{\Gamma }_{11}} & \ldots  & {{\Gamma }_{15}}  \\   \vdots  & \ddots  & \vdots   \\   {{\Gamma }_{51}} & \cdots  & {{\Gamma }_{55}}  \\ \end{matrix} \right) $$     (Eq 6.3)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle \Gamma _{ij}==\int_{\Omega }b_{i}(x)b_{j}(x)dx $$     (Eq 6.4)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

We need to integrate on interval $$ \Omega \left [ 0,T \right ] $$ and $$T = \frac{w} $$ to find each elements of Gram matrix

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

$$  \displaystyle \Gamma _{11}=\int_{0}^{T}1dx=\frac{2\Pi }{w} $$     (Eq 6.5)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle \Gamma _{21}=\Gamma _{12}=\int_{0}^{T}coswxdx=0 $$     (Eq 6.6)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \Gamma _{31}=\Gamma _{13}=\int_{0}^{T}cos2wxdx=0 $$     (Eq 6.7)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle \Gamma _{41}=\Gamma _{14}=\int_{0}^{T}sinwxdx=0 $$     (Eq 6.8)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \Gamma _{51}=\Gamma _{15}=\int_{0}^{T}sin2wxdx=0 $$     (Eq 6.9)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \Gamma _{22}=\int_{0}^{T}coswx.coswxdx=\frac{\Pi }{w} $$   (Eq 6.10)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \Gamma _{23}=\Gamma _{32}=\int_{0}^{T}coswx.cos2wxdx=0 $$     (Eq 6.11)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \Gamma _{24}=\Gamma _{42}=\int_{0}^{T}coswx.sinwxdx=0 $$     (Eq 6.12)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \Gamma _{25}=\Gamma _{52}=\int_{0}^{T}coswx.sin2wxdx=0 $$     (Eq 6.13)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \Gamma _{33}=\int_{0}^{T}cos2wx.cos2wxdx=\frac{\Pi }{w} $$     (Eq 6.14)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \Gamma _{34}=\Gamma _{43}=\int_{0}^{T}cos2wx.sinwxdx=0 $$     (Eq 6.15)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \Gamma _{35}=\Gamma _{53}=\int_{0}^{T}cos2wx.sin2wxdx=0 $$     (Eq 6.16)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle \Gamma _{44}=\int_{0}^{T}sinwx.sinwxdx=\frac{\Pi }{w} $$     (Eq 6.17)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle \Gamma _{45}=\Gamma _{45}=\int_{0}^{T}sinwx.sin2wxdx=0 $$     (Eq 6.18)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle \Gamma _{55}=\int_{0}^{T}sin2wx.sin2wxdx=\frac{\Pi }{w} $$     (Eq 6.19)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle \underline{\Gamma }(F)=\left[ \begin{matrix} \begin{matrix} \begin{matrix} \begin{matrix} \frac{2\Pi }{w} & 0 \\ \end{matrix} & 0 & 0 & 0 \\ \end{matrix} \\ \begin{matrix} \begin{matrix} 0 & \frac{\Pi }{w} \\ \end{matrix} & 0 & 0 & 0 \\ \end{matrix} \\ \end{matrix} \\ \begin{matrix} \begin{matrix} 0 & 0 \\ \end{matrix} & \frac{\Pi }{w} & 0 & 0  \\ \end{matrix} \\ \begin{matrix} \begin{matrix} 0 & 0 \\ \end{matrix} & 0 & \frac{\Pi }{w} & 0  \\ \end{matrix} \\ \begin{matrix} \begin{matrix} 0 & 0 \\ \end{matrix} & 0 & 0 & \frac{\Pi }{w}  \\ \end{matrix} \\ \end{matrix} \right]
 * style="width:95%" |
 * style="width:95%" |

$$     (Eq 6.20)
 *

 * }

It is shown that gram matrix is diagonal matrix.

6.2
Determinant of gram matrix has been calculated by wolfram alpha.

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

$$  \displaystyle \det \underline{\Gamma }(F)=\frac{2{{\Pi }^{5}}} $$     (Eq 6.21) According to theorem on lecture note meeting 10-2 $$\displaystyle \det \underline{\Gamma }(F)\neq 0$$,so the family of functions are linearly independent. Also they are basis functions.
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

6.3
From the lecture note Eq.(2) meeting 7-1

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

$$  \displaystyle {{\delta }_{ij}}=\left\{ \begin{matrix} \begin{matrix} 1 & for & i=j \\ \end{matrix} \\ \begin{matrix} 0 & for & i\ne j \\ \end{matrix} \\ \end{matrix} \right.
 * style="width:95%" |
 * style="width:95%" |

$$     (Eq 6.22) and
 *

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

$$  \displaystyle {{\Gamma }_{ij}}=<{{b}_{i}},{{b}_{j}}>={{\delta }_{ij}} $$     (Eq 6.23)
 * style="width:95%" |
 * style="width:95%" |
 *

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

For $$\ i = j$$   Kronecker $$\ {\delta } = 1 $$      (diagonal elements of Gram matrix in Eq 6.20)
 * style="width:95%" |
 * style="width:95%" |

For $$ i \ne j$$  Kronecker $$\ {\delta } = 0 $$      (off-diagonal elements of Gram matrix in Eq 6.20)
 * }

As it is seen on above, both elements of Gram matrix are satisfying the orthogonal condition. Therefore, $$\displaystyle F $$ is a family of orthogonal basis functions.

Reference
Used wolframalpha to do integrates.

title: User:Eml5526.s11.team3.sahin/Homework 2


text:
User:Egm5526.s11.team4.zou/HW2

=Problem 2.4 supplement=

Another solution for part 4
As follows,

From part1 and part2 we know,

Sustitute Eqs 4.B into Eq 4.A and then arrange terms and omit constant ones it yields,

From Eq 4.A we can see that most of time we are dealing with $$\displaystyle (a+bx)$$ rather than $$\displaystyle x$$. So why not make a transform: $$\displaystyle u=a+bx$$, hopefully it reduces the notation and calculation.

Hence,

Sustitute Eq 4.D into Eq 4.A we have,

Replace $$\displaystyle u$$ back to $$\displaystyle x$$ in Eq 4.E we have,

=Problem 2.7 Verifying the orthogonality of the family=

Problem statement
Family of basis functions given:

with the same domain $$\displaystyle \Omega =[0,1]$$.

Do the following steps to determine whether $$\displaystyle \mathcal{F}$$ is a orthogonal family.

1)Construct the Gramian matrix $$\displaystyle \Gamma \left( \mathcal{F} \right)$$, and observe its property

2)Find the determinant of $$\displaystyle \Gamma \left( \mathcal{F} \right)$$

3)Conclude on the orhogonality of $$\displaystyle \mathcal{F}$$.

Solution
1)Construct the Gramian matrix $$\displaystyle \Gamma \left( \mathcal{F} \right)$$, and observe its property

First let ‘ s assign notation to the family $$\displaystyle \mathcal{F}$$:

The Gramian matrix is defined as

Since,

then,

Then we can construct the Gramian matrix,

And we know that it is a symmetric matrix.

2)Find the determinant of $$\displaystyle \Gamma \left( \mathcal{F} \right)$$

The determinant can be calculated via WolframAlpha:

3)Conclude on the orhogonality of $$\displaystyle \mathcal{F}$$.

Since the determinant of $$\displaystyle \Gamma \left( \mathcal{F} \right)$$ is non-zero, so we can conclude that the family of basis functions $$\displaystyle \mathcal{F}$$ is an orthogonal family.

Problem Statement
Given:

Where $$\displaystyle \underline{w}(x)\cong {{\underline{w}}^{h}}(x)=\sum\limits_{i=1}^{n}(x)$$, $$\displaystyle \underline{u}(x)\cong {{\underline{u}}^{h}}(x)=\sum\limits_{j=1}^{n}{{{d}_{j}}\underline}(x)$$.

Show that Eq. (8.1) is equivalent to Eq. (8.2).

==Solutions ==

1). First, we try to derive Eq.(8.2) using Eq.(8.1).

Eq. (8.1) is valid for any function of wh (x), which is a linear combination of these linearly independent basis.

Thus, the Eq. (8.1) is valid for some particular sets of Ci value to get $$\displaystyle {{\underline{w}}^{h}}(x)=\sum\limits_{i=1}^{n}(x)$$.

For example, we choose C1=1, C2=C3=..=Cn=0, then $$\displaystyle {{\underline{w}}^{h}}(x)=\sum\limits_{i=1}^{n}(x)={{\underline{b}}_{1}}(x)$$. By substituting $$\displaystyle {{\underline{w}}^{h}}(x)$$ into Eq.(8.1), we obtain:

Which is the equation for Eq.(8.2) of $$\displaystyle i=1$$.

Then, in similar way, we choose C2=1, C1=C3=..=Cn=0, then we obtain:

Which is the equation for Eq.(8.2) of $$\displaystyle i=2$$.

Similarly, we continuing to choose Ci=1, C1=C2=.. Ci-1 = Ci+1 =Cn=0 then we obtain:

Which is the equation for Eq.(8.2) of any value of $$\displaystyle i$$. So, Eq. (8.2) is derived with Eq.(8.1).

2). Secondly, we try to derive Eq.(8.2) using Eq.(8.1).

{ bi(x), i=1,2…n } is a family of linearly independent basis functions, and according to Eq.(8.1) we have:

Linearly combining the above equations, by choosing any set of Civalue, we obtain:

Obviously,

Because the value set of Ciis arbitrary, Eq.(8.1) is valid for any wh (x). So, we obtain Eq.(1) using Eq.(8.2).

According the above two aspects, we can conclude that Eq.(8.1) is equivalent to Eq. (8.2).

title: User:Egm5526.s11.team4.zou/HW2


text:
University of Canberra/RCC2011/The future of the Vocational Education and Training sector

 * Download the MP3 file (1hr 5 min)

Topic: VET & Higher Education today

Recording of an un-conference session at the Recent Changes Camp 2011 held in Canberra Date: Saturday 29 January 2011 Time: 2:00-3:00 PM EST

Participants

 * 1) Stephan Ridgway
 * 2) Robyn Jay
 * 3) Michael Coghlan
 * 4) Tom Worthington
 * 5) Leigh Blackall
 * 6) Rose Grozdanic
 * 7) Adam Jenkins
 * 8) Kirsty Sharp
 * 9) Jutta van Dincklage

Links

 * TalkingVTE session resource page
 * Recent Changes Camp 2011
 * flickr images


text:
University of Canberra/RCC2011/Open Educational Resources

Recording of an un-conference session at the Recent Changes Camp 2011 held in Canberra.

Recordings

 * Audio in MP3 - 1 hour
 * Talking VTE Podcast
 * iTunes

Participants

 * 1) Stephan Ridgway
 * 2) Robyn Jay ]
 * 3) Michael Coghlan
 * 4) Tom Worthington
 * 5) Leigh Blackall
 * 6) Rose Grozdanic
 * 7) Adam Jenkins
 * 8) Kirsty Sharp
 * 9) Jutta van Dincklage

Tagged Resources
Each episode of Talking VTE has it's own unique tag which we use to share resources relevant to the episode. Below are links tagged in with talkingvte28 which are relevant to this session. If you want to share some resources with the participants prior to the session or the audience after it's distribution you can join the diigo group and add to the resources.


text:
University of Canberra/RCC2011/Wikiversity for postgraduate education

This session briefly looked at the Wikiversity PhD portal, Leigh Blackall's concept for using the Wikiversty user space for developing his PhD and Ben Rattray's Sport research pages for university honours students. Discussion focused on the concept of OpenPhD, or to use the PhD criteria to structure a self directed and informal study to PhD standard. Participants became interested in the benefits and pitfalls of informal versus formal pursuit, and whether formal supervision in reality is any better than networking online for supervisory roles and in depth feedback, such as what might be possible through the PhD portal.


text:
2837 University

The 2837 University is an initiative of the Agitprospace in San Diego, USA.

Roberto Greco and his network have prepared this extensive reading list (Google Docs). We're working out the best way to bring that list into Wikiversity without triggering the spam filter. We plan to improve the organization of content, and set collaborative learning assignments around the vast range of case studies, concepts and proposals listed there.

title: 2837 University


text:
User:Egm6341.s11.team4.kurth/hw2

=Problem 2.3= Refer to lecture slide [[media:nm1.s11.mtg7.djvu|7-2]] for problem statement.

Given

 * $$f(x)=\frac{e^{x}-1}{x}$$

Objectives
To examine the Taylor series expansions of both $$\displaystyle e^x$$ and $$\displaystyle f(x)$$ and their remainders to verify that:

 * $$R_{n+1}[f(x);x] = \frac{(x-0)^{n+1}}{x(n+1)!}e^\xi = \frac{R_{n}[e^{x};x]}{n+1}$$

Solution
For this problem we will examine Taylor series expansions about the point $$ x_0 = 0$$.

In general, functions for which the (n+1)th derivative exists and is continuous, can be expressed in the form

where $$p_{n}(x)$$ is a polynomial of order n and $$R_{n+1}(x)$$ is the corresponding remainder.

During lecture 3, it was shown that these polynomials and remainders take a specific form for Taylor series expansions.

Equation 3 from lecture slide [[media:nm1.s11.mtg3.djvu|3-3]]:

where $$\displaystyle f^{(n)}(x_0)$$ denotes the n-th derivative of f evaluated at $$\displaystyle x_0$$.

Equation 5 from lecture slide [[media:nm1.s11.mtg3.djvu|3-3]]:

For the Taylor series expansion of $$\displaystyle e^x$$ about $$\displaystyle x_0=0$$, all derivatives of $$\displaystyle e^x$$ are also $$\displaystyle e^x$$ and are simplified to 1 when evaluated at $$\displaystyle x_0=0$$.

The result of (3.4) will be manipulated to solve for the expansion and remainder of $$\displaystyle f(x)$$:

(3.8) verifies equation 2 from lecture slide [[media:nm1.s11.mtg6.djvu|6-5]]. Comparing (3.5) and (3.8), it is obvious that the remainders of $$e^x$$ and $$f(x)$$ are similar. If the Taylor series expansion for $$e^x$$ were expanded to (n-1) terms rather than n, then the remainders of both $$e^x$$ and $$f(x)$$ would be of the same degree.

(3.10) confirms equation 1 from lecture slide [[media:nm1.s11.mtg7.djvu|7-1]]!

Solved by William Kurth.

=Problem 2.4= Refer to lecture slide [[media:nm1.s11.mtg7.djvu|7-3]] for problem statement.

Objectives
Plot f(x). Then, integrate f(x) using the following methods:
 * (1) Taylor series expansion
 * (2) Composite trapezoidal rule
 * (3) Composite Simpson rule
 * (4) Gauss-Legendre quadrature

for $$n=2,4,8...$$, until the error reaches the order $$10^{-6}$$.

Solution
The following MATLAB code was used to plot f(x) from -1 to 1.

(1) Integration by Taylor Series Expansion

To evaluate (4.1) by Taylor series expansion, the function f(x) will be approximated by an expansion about the point $$\displaystyle x_0 = 0$$. In homework problem 1.2, this expansion was shown to be

To solve for n, we must first examine the nature of the error.

On lecture slide [[media:nm1.s11.mtg7.djvu|7-1]], this error was shown to have the form that follows (adjusted for different interval):

When observing $$\displaystyle e^{\xi(\alpha)}$$ over the interval $$\displaystyle\alpha\;\in\;[-1,1]$$ it is seen that the function is bounded on both the top and bottom by

This can be extended to say that the error, $$\displaystyle E_n(f)$$, is also bounded as follows:

Now we will calculate the integration of f(x) and the upper bound of the error for increasing values of n until the error reaches the order of $$10^{-6}$$. The following table shows these values.

(2) Integration by Composite Trapezoidal Rule

Equation (1) from lecture slide [[media:nm1.s11.mtg7.djvu|7-4]]:

where $$\displaystyle h=\frac{b-a}{n}$$ and  $$\displaystyle f_n=f(x_n)$$.

The error associated with the composite trapezoidal rule is (Atkinson, p.253)

For this problem (4.10) becomes

Values of In and En are calculated from equations (4.9) and (4.13), respectively, for increasing values of n and shown in the table below.

(3) Integration by Composite Simpson Rule

Equation (4) from lecture slide [[media:nm1.s11.mtg7.djvu|7-4]]:

where $$\displaystyle h=\frac{b-a}{n}$$ and  $$\displaystyle f_n=f(x_n)$$ and n must be even.

The error associated with the composite trapezoidal rule is (Atkinson, p.257)

For this problem (4.15) becomes

Values of In and En are calculated from equations (4.14) and (4.18), respectively, for increasing values of n and shown in the table below.

(4)Integration by Gauss-Legendre Quadrature

Equation (1) from lecture slide [[media:nm1.s11.mtg7.djvu|7-5]]:

where $$\displaystyle \{x_i, i=1,2,3,...,n\}$$ are roots of the Legendre polynomial of order n, $$\displaystyle \mathcal{P}_n(x)$$, and $$\displaystyle w_i$$ are their corresponding weights. Some of these values are shown in the table below supplied in lecture slide [[media:nm1.s11.mtg7.djvu|7-5]].

The error associated with using this method for approximating the value of an integral is given by equation (3) on lecture slide [[media:nm1.s11.mtg8.djvu|8-2]].

Equations (4.19) and (4.20) were solved for increasing values of n until the error reached the order of 10-6 and shown in the table below.

Solved by William Kurth.

=Problem 2.14: Transformation of Variables= Refer to slide [[media:nm1.s11.mtg11.djvu|11-2]] for the problem statement.

Objective
Give details of how to obtain (14.1)

Solution
Let $$\displaystyle x=x(y)$$ be a function of y that satisfies (14.1). To satisfy the boundary conditions x(y) must satisfy the following conditions:

The solution to these conditions lies upon a line passing through the two ordered pairs.

Taking the partial derivative of each side yields an expression for dx:

Plugging these expression back into the middle term of (14.1) yields

Now, let

Combining equations (14.6) and (14.7) proves (14.1):

 * {| style="width:70%" border="0" align="center"

$$\displaystyle \therefore I(f) = \int_{a}^{b}f(x)dx = \int_{-1}^{1}\bar{f}(y)dy $$ $$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * $$\displaystyle
 * style= |
 * }
 * }

Solved by William Kurth and Brendan Mahon.

title: User:Egm6341.s11.team4.kurth/hw2


text:
User talk:Jrr29541

 Hello Jrr29541, 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.

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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! Jtneill - Talk - c 12:16, 1 February 2011 (UTC)

 Hi! My name is WikiversityJack. I am a pumpkin and a Wikiversity mascot. My aunt told me that fun + learning = Wikiversity. Let me know if I can help you out.

Der Erlkönig
Hi Jrr29541. Thanks for contributing to Wikiversity. I was wondering what the purpose of Der Erlkönig is? Perhaps we can help to make it into a learning resource - or perhaps it was a test? Sincerely, James. -- Jtneill - Talk - c 12:18, 1 February 2011 (UTC)

title: User talk:Jrr29541


text:
User:Egm6321.f10.team2.oztekin/SandBox

=Problem 2.6=

Given
Consider family of functions $$\displaystyle F=\left \{ 1,cosiwx,siniwx \right \} $$ on interval $$\displaystyle \Omega \left [ 0,T \right ]$$ for i =1,2

So

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

$$  \displaystyle F=\left \{ b_{1}(x),b_{2}(x),b_{3}(x),b_{4}(x),b_{5}(x) \right \} $$     (Eq 6.1)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle F=\left \{ 1,coswx,cos2wx,sinwx,sin2wx \right \} $$     (Eq 6.2)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Find
1)Construct $$\displaystyle \underline{\Gamma }(F)$$ and observe properties of $$\displaystyle \underline{\Gamma }(F)$$

2)Find the determinant of  $$\displaystyle \underline{\Gamma }(F)$$

3)Conclude F is orthogonal basis.

Solution
1)
 * {| style="width:100%" border="0"

$$  \displaystyle \underline{\Gamma }={{[{{\Gamma }_{ij}}]}_{5*5}}=\left( \begin{matrix}  {{\Gamma }_{11}} & \ldots  & {{\Gamma }_{15}}  \\   \vdots  & \ddots  & \vdots   \\   {{\Gamma }_{51}} & \cdots  & {{\Gamma }_{55}}  \\ \end{matrix} \right) $$     (Eq 6.3)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle \Gamma _{ij}==\int_{\Omega }b_{i}(x)b_{j}(x)dx $$     (Eq 6.4)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle \Gamma _{11}=\int_{0}^{T}1dx=\frac{2\Pi }{w} $$     (Eq 6.5)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle \Gamma _{21}=\Gamma _{12}=\int_{0}^{T}coswxdx=0 $$     (Eq 6.6)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \Gamma _{31}=\Gamma _{13}=\int_{0}^{T}cos2wxdx=0 $$     (Eq 6.7)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle \Gamma _{41}=\Gamma _{14}=\int_{0}^{T}sinwxdx=0 $$     (Eq 6.8)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \Gamma _{51}=\Gamma _{15}=\int_{0}^{T}sin2wxdx=0 $$     (Eq 6.9)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \Gamma _{22}=\int_{0}^{T}coswx.coswxdx=\frac{\Pi }{w} $$     (Eq 6.10)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \Gamma _{23}=\Gamma _{32}=\int_{0}^{T}coswx.cos2wxdx=0 $$     (Eq 6.11)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \Gamma _{24}=\Gamma _{42}=\int_{0}^{T}coswx.sinwxdx=0 $$     (Eq 6.12)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \Gamma _{25}=\Gamma _{52}=\int_{0}^{T}coswx.sin2wxdx=0 $$     (Eq 6.13)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \Gamma _{33}=\int_{0}^{T}cos2wx.cos2wxdx=\frac{\Pi }{w} $$     (Eq 6.14)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \Gamma _{34}=\Gamma _{43}=\int_{0}^{T}cos2wx.sinwxdx=0 $$     (Eq 6.15)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \Gamma _{35}=\Gamma _{53}=\int_{0}^{T}cos2wx.sin2wxdx=0 $$     (Eq 6.16)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle \Gamma _{44}=\int_{0}^{T}sinwx.sinwxdx=\frac{\Pi }{w} $$     (Eq 6.17)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle \Gamma _{45}=\Gamma _{45}=\int_{0}^{T}sinwx.sin2wxdx=0 $$     (Eq 6.18)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle \Gamma _{55}=\int_{0}^{T}sin2wx.sin2wxdx=\frac{\Pi }{w} $$     (Eq 6.19)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle \underline{\Gamma }(F)=\left[ \begin{matrix} \begin{matrix} \begin{matrix} \begin{matrix} \frac{2\Pi }{w} & 0 \\ \end{matrix} & 0 & 0 & 0 \\ \end{matrix} \\ \begin{matrix} \begin{matrix} 0 & \frac{\Pi }{w} \\ \end{matrix} & 0 & 0 & 0 \\ \end{matrix} \\ \end{matrix} \\ \begin{matrix} \begin{matrix} 0 & 0 \\ \end{matrix} & \frac{\Pi }{w} & 0 & 0  \\ \end{matrix} \\ \begin{matrix} \begin{matrix} 0 & 0 \\ \end{matrix} & 0 & \frac{\Pi }{w} & 0  \\ \end{matrix} \\ \begin{matrix} \begin{matrix} 0 & 0 \\ \end{matrix} & 0 & 0 & \frac{\Pi }{w}  \\ \end{matrix} \\ \end{matrix} \right]
 * style="width:95%" |
 * style="width:95%" |

$$     (Eq 6.20)
 *

 * }

It is shown that gram matrix is diagonal matrix.

2)Determinant of gram matrix has been calculated by wolfram alpha.

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

$$  \displaystyle \det \underline{\Gamma }(F)=\frac{2{{\Pi }^{5}}} $$     (Eq 6.21) According to theorem on lecture note meeting 10-2 $$\displaystyle \det \underline{\Gamma }(F)\neq 0$$,so the family of functions are linearly independent. Also they are basis functions.
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

3)From the lecture note Eq(2) meeting 7-1

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

$$  \displaystyle {{\delta }_{ij}}=\left\{ \begin{matrix} \begin{matrix} 1 & for & i=j \\ \end{matrix} \\ \begin{matrix} 0 & for & i\ne j \\ \end{matrix} \\ \end{matrix} \right.
 * style="width:95%" |
 * style="width:95%" |

$$     (Eq 6.22)
 *

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

$$  \displaystyle {{\Gamma }_{ij}}=<{{b}_{i}},{{b}_{j}}>={{\delta }_{ij}} $$     (Eq 6.23)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

As it is seen on above,both elements of gram matrix are satisfying the orthogonal condition.Therefore F is a family of orthogonal basis functions.

title: User:Egm6321.f10.team2.oztekin/SandBox


title: File:Fe1.s11.tm7.HW2.fig1.png


text: File:Tm7graph2 4.png

title: File:Tm7graph2 4.png


text:
User:Drackalarimo

Hello, My existing page is.

 * 1) User:Newseerr/List of president of Maybeury Elementary School and soccer

title: User:Drackalarimo


text:
User talk:Drackalarimo

 Hello Drackalarimo, 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.

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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! --mikeu talk 15:16, 1 February 2011 (UTC)

title: User talk:Drackalarimo


text:
User:Denial~enwikiversity

Hello, This wikiversity is temps for the process found of this The user talk

I have directed

 * 1) REDIRECT [[User:Newseerr/List of president of Maybeury Elementary School and soccer]]

title: User:Denial~enwikiversity


text:
User talk:Denial~enwikiversity

 Hello Denial, 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.

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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! --Ottava Rima (talk) 15:53, 1 February 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 Denial. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Denial~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 23:31, 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) 06:06, 19 April 2015 (UTC)

title: User talk:Denial~enwikiversity


text:
Wikiversity talk:Probationary custodians

Proposed
- Ottava Rima (talk) 16:36, 1 February 2011 (UTC)

Proposed policy will inhibit custodianship, increase disruption.
This proposed policy is designed to inhibit probationary custodianship. Parts of it were already proposed, and in the discussion, I asked for examples of problem probationary custodians, actual damage done, as distinct from some theoretical idea that damage might be done. No examples were adduced.

Presently, becoming a permanent custodian requires only one discussion. The prior process is abbreviated and simple and non-adversarial. Basically, the form is that the candidate applies, a mentor accepts, and a 'crat sets the bit. The only defect there is the idea that the 'crat has no discretion, which is an error. All 'crat actions require discretion, if a 'crat is not satisfied that a mentorship agreement adequately protects the community, all things considered, the 'crat should probably state opposition and then allow another 'crat to review, and denial by all 'crats, or even just, say, two or three, would be enough to kill the application.

The change that Ottava proposed was this discussion period, which then requires a close, a finding, making the matter more adversarial. Many of our users with wiki experience, such as Ottava -- or myself -- have been involved in some level of conflict elsewhere. My recent candidacy attracted comment by a user with no Wikiversity edits, but very substantial history of active conflict with me on Wikipedia. Ottava's proposal invites more of this cross-wiki attack.

A 'crat may consider comment, and part of the 'crat decision should be deciding if the mentor is sufficiently available and responsible.

It isn't broken, this proposed policy simply complicates the process and produces increased opportunity for disruption, as was actually seen in the case of my probationary custodianship, which was an excuse for disruption from the beginning, starting with action against SBJ for daring to approve it. --Abd 03:21, 2 February 2011 (UTC)

 * This is probationary, not full custodianship. Although probationary can lead to custodianship, someone can run for custodianship and skip the whole probationary aspects. Ottava Rima (talk) 03:31, 2 February 2011 (UTC)
 * Interesting. That's in contradiction to Custodianship, which requires a probationary period and, normally, approval of a mentor. So, Ottava, are you saying that I could just run for permanent custodianship without a mentor? What's that based on?
 * The proposal is badly written, clearly dashed off. There are some elements, a few, that could be incorporated in policy. Mostly, though, improvements at this point are lipstick on a pig. Talking about the color of the drapes, when there are piles of elephant dung on the floor. --Abd 03:45, 2 February 2011 (UTC)
 * It doesn't require one. It only gives a way to go through it but never says it is the only way. Futhermore, it doesn't say you couldn't run again after having a probationary period. And, if you notice, the "evaluation" section isn't actually part of the policy. Ottava Rima (talk) 04:05, 2 February 2011 (UTC)
 * Are you aware, Ottava, that you argued the exact opposite of this with respect to my own custodianship?

 * Are you aware, Ottava, that depending on what a policy does not say is fostering confusion? We have a custodianship procedure. So, you are suggesting, someone, anyone, could invent a new procedure, follow it, and then claim approval? Who would decide? Does the procedure say that only 'crats can decide? It says they do, but it doesn't say that nobody else could.

 * So, sure, you could go to meta and point to a discussion. But, oops! meta has a policy that they don't grant ops if there are local 'crats who can do it. However, I could do this: start a process in my own user space and gather support. Then move it to the candidates for custodianship page, with gathered approvals, and go to meta and claim that 'crats have snubbed it. Think it would work? It might, you know. Why not make it a 'crat candidacy?

 * I do intend, in fact, to run some processes in my user space, but I'd not dream of considering them capable of making a Wikiversity decision.

 * No, if we have a policy, we don't invent a new one unless we explicitly allow it as a replacement or as an alternate path. Indeed, we can deviate from policy through discussion, and that was done in the case of my first probationary custodianship. You were claiming that it was over, because you had withdrawn support (and you went to meta and yanked ops). But at the same time, you were claiming you were still my mentor. You were blatantly violating explicit procedure, right and left, while, at the same time, insisting on the most niggling letter of the law. The community decided to start !voting anyway. I considered it out of process, I did object, but then relented and allowed it to continue. Waste of time, really. There are reasons to keep the policy simple and clear. --Abd 18:43, 2 February 2011 (UTC)

Historical
As of 2015, we were informed by WMF that we could not have anyone with deleterevision rights who didn't have explicit community support to do so. Probationary custodianship, as envisioned and documented, is no longer an option. Instead, users interested in custodianship should first see and apply for Curators status. -- Dave Braunschweig (discuss • contribs) 01:26, 4 November 2019 (UTC)


text:
User talk:Sj sooraj

 Hello Sj sooraj, 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.

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 * Visit a (kind of) random project.
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 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
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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! --Ottava Rima (talk) 16:39, 1 February 2011 (UTC)

title: User talk:Sj sooraj


text: File:Eml5526 s11 team2 hw2 4 6.jpg

title: File:Eml5526 s11 team2 hw2 4 6.jpg


title: File:Eml5526.s11.team2.hw2.4.6.png