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Instructional design/PBL problems

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Lesson Introduction
Have you ever had a boring training class? We all have. We sat in class, memorized as much as we could,and tried to pass a test in the end, but is that what we want? These days there is a well-respected method called Problem-based Learning (PBL) that instructors and students find valuable. But just don’t take my word for it. Watch the following video and see what students and teachers have to say about PBL.( To watch the video: press Ctrl and click the PLAY button):

Problem Based Learning provides an effective way to enhance learning experiences and outcomes.Good problems are essential in PBL. This lesson will teach you how to generate effective PBL problems to ensure that learners are motivated and their critical thinking as well as collaboration skills are improved.

Target Learners
Instructional designers in higher education.

Lesson Objectives
After learning this lesson:
 * Given a learning scenario, you (the learners) will be able to generate a problem that can be used in a PBL learning activity. The problem generated should score 80% or more on the PBL Quality Problem Checklist.
 * You will be able to elaborate the core features of a “problem” in PBL (enhance students’ motivation, critical thinking skills and collaboration).
 * You will be able to explain what kind of problem will enhance students’ motivation.
 * You will be able to explain what kind of problem will enhance students’ critical thinking skills.
 * You will be able to explain what kind of problem will enhance students’ collaboration.

Now please click on second session of this class "2. Characteristics of PBL problems."

title: Instructional design/PBL problems


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Instructional design/Characteristics of PBL Problems

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Core characteristics of PBL
According to the video on the previous page, what do you think are the core characteristics of PBL?

{Check the characteristics that you think PBL have +Learner-centered -Teacher-centered +Is carried out in groups -Is carried out individually +Students solve problems through selecting, analyzing and synthesizing information -Students solve problems through simply ask solutions from their teachers
 * type="[]"}

Do you agree that PBL has the following three core characteristics?
 * Learner-centered
 * Students solve problems through selecting, analyzing and synthesizing information
 * Carried out in groups

Features of PBL Problems
As it is said in the lesson introduction “good problems are essential in PBL”. Think: what features a good PBL problem should have to ensure the above characteristics of PBL? Look at the following inference chain, see if you agree:

According to the above diagram, enhance students’ motivation, enhance students’ critical thinking skills, and enhance students’ collaboration are three essential features of problems in PBLs. As for how to generate problems which consist of such features, the next section will give you detailed guidelines and practical tips.

Now please click on third session of this class "3.Generate PBL Problems."

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


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Dominant group/Rigorous definition

What is a rigorous definition? And, specifically where dominant group is concerned, what is a rigorous definition of dominant group?

With dominant group appearing in almost every field of written human endeavor, especially primary sources, usually with what appears to be a relatively unique definition or usage, is a rigorous definition even possible?

Although primary sources are usually not read at a secondary education level, dominant group and some of its relative synonyms occur in secondary education resources such as Wikipedia.

Dominant groups
Examples from primary sources are to be used to prove or disprove each hypothesis. These can be collected per subject or in general.

 * 1) Accident hypothesis: dominant group is an accident of whatever processes are operating.
 * 2) Artifact hypothesis: dominant group may be an artifact of human endeavor or may have preceded humanity.
 * 3) Association hypothesis: dominant group is associated in some way with the original research.
 * 4) Bad group hypothesis: dominant group is the group that engages in discrimination, abuse, punishment, and additional criminal activity against other groups. It often has an unfair advantage and uses it to express monopolistic practices.
 * 5) Control group hypothesis: there is a control group that can be used to study dominant group.
 * 6) Entity hypothesis: dominant group is an entity within each field where a primary author of original research uses the term.
 * 7) Evolution hypothesis: dominant group is a product of evolutionary processes, such groups are the evolutionary process, produce evolutionary processes, or are independent of evolutionary processes.
 * 8) Identifier hypothesis: dominant group is an identifier used by primary source authors of original research to identify an observation in the process of analysis.
 * 9) Importance hypothesis: dominant group signifies original research results that usually need to be explained by theory and interpretation of experiments.
 * 10) Indicator hypothesis: dominant group may be an indicator of something as yet not understood by the primary author of original research.
 * 11) Influence hypothesis: dominant group is included in a primary source article containing original research to indicate influence or an influential phenomenon.
 * 12) Interest hypothesis: dominant group is a theoretical entity used by scholarly authors of primary sources for phenomena of interest.
 * 13) Metadefinition hypothesis: all uses of dominant group by all primary source authors of original research are included in the metadefinition for dominant group.
 * 14) Null hypothesis: there is no significant or special meaning of dominant group in any sentence or figure caption in any refereed journal article.
 * 15) Object hypothesis: dominant group is an object within each field where a primary author of original research uses the term.
 * 16) Obvious hypothesis: the only meaning of dominant group is the one found in Mosby's Medical Dictionary.
 * 17) Original research hypothesis: dominant group is included in a primary source article by the author to indicate that the article contains original research.
 * 18) Primordial hypothesis: dominant group is a primordial concept inherent to humans such that every language or other form of communication no matter how old or whether extinct, on the verge of extinction, or not, has at least a synonym for dominant group.
 * 19) Purpose hypothesis: dominant group is written into articles by authors for a purpose.
 * 20) Regional hypothesis: dominant group, when it occurs, is only a manifestation of the limitations within a region. Variation of those limitations may result in the loss of a dominant group with the eventual appearance of a new one or none at all.
 * 21) Source hypothesis: dominant group is a source within each field where a primary author of original research uses the term.
 * 22) Term hypothesis: dominant group is a significant term that may require a 'rigorous definition' or application and verification of an empirical definition.

Axiomatic definitions
A rigorous definition fulfills the axioms that define a definition so that a generalized definition can be defined using an appropriate function that fulfills the axioms. Having met these axioms as a criteria of a general definition, the definition is said to be a rigorous definition.

An axiomatic definition is a rigorous definition: "the definition must clearly state the rules that are considered as binding, and on the other hand give the implementor enough freedom to achieve efficiency by leaving certain less important aspects undefined."

Synonyms
The term dominant group appears to be used to identify entities of importance. The genera differentia for possible relative synonyms of dominant group fall into the following set of orderable pairs:

'Orderable' means that any synonym from within the first category can be ordered with any synonym from the second category to form an alternate term for dominant group; for example, "superior class", "influential sect", "master assembly", "most important group", and "dominant painting". "Dominant" falls into category 171. "Group" is in category 61. Further, any word which has its most or much more common usage within these categories may also form an alternate term, such as "ruling group", where "ruling" has its most common usage in category 739, or "dominant party", where "party" is in category 74. "Taxon" or "taxa" are like "species" in category 61. "Society" is in category 786 so there is a "dominant society".

When one or two orderable pairs are produced, the results are
 * 1) one pair - relative synonym,
 * 2) one pair in which the first or second category has each of two from a category - definition, and
 * 3) two pairs from two to four categories - definition.

Meaningless dominant groups
Each subject area within which the term dominant group is used has the same problem: "unless and until a rigorous definition of the term 'dominant group' is rendered, the argument fails to establish its conclusion due to the fact that one of its premises is meaningless."

Tests
Def. a category synonym for "group", including "group", and a category synonym for "dominant", including "dominant", that as one or two orderable pairs has only the properties of two pairs: i.e., from two to four categories [exclusive], or one pair in which the first or second category has up to each of two from a category is called a rigorous definition of dominant group.

Hypotheses

 * 1) Dominant group has a rigorous metadefinition that accommodates all known primary source (original research) uses.

title: Dominant group/Rigorous definition


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User:Crazywhitexican

Social media provides businesses with new and different useful information.

title: User:Crazywhitexican


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User:Peteymills/Wiki research

I created an account on the "Meta Wiki" with the intention of proposing a scientific archive and peer-review system which I call "Wiki-Research." On the Meta Wiki, Wiki Research has several proposed meanings and indeed, the system I have come up with is more specific than that and so should eventually go by a different title. Nonetheless, on the Wikiversity I will keep that same working title. The purpose of this page is to create a draft outline of my proposal.

Summary
Wiki-Research is a scientific indexing, archiving and peer-review system in which users are uniquely identifiable (one user per account, one account per user) and verified. It takes the form of a distributed, peer-to-peer, universally readable file system for the purpose of crafting, disseminating and archiving scientific research.

Peer-to-Peer
The system must be independent of the type of network on which it is distributed, thus it can use TCP/IP networks, local-area networks, MPI networks, etc. All computers must contain all essential data contained in the database to the maximum extent possible. At minimum, every user is archived on each computing entity that is a member of the network. Data is similarly distributed with a maximum of redundancy to insure integrity, persistance and safety of the data.

User Accounts
Each user must have one and only one account and each account is associated with one and only one user and each user is a real person. User accounts must be verified. A user account will be associated with the user's credentials and his or her specific interests and expertise so that assignment of reviewers is easily automated. It will also be associated with all of the user's contributions to the system.

Peer Review
Formal peer review is double-blind, that is, both the contributor and the reviewer are anonymous at the time of review. Reviewers, however, must be held accountable, thus, at the end of the peer-review process, both the reviewer's identity and his or her comments must be made public. There will be a distinction between "formal" peer review and "informal" peer review. The informal peer review will take the form of open comments or a public discussion forum revolving around designated contributions.

Files for peer review must be formally designated as such and supplemented with a number of keywords or subject headings. Once designated as a formal peer-review file, the file will be locked into its current state and undergo a versioning process. Reviewers will be matched based on the subject designations and automatically contacted. Designated reviewers may either accept or reject the assignment. For informal peer-review, reviewers will still be contacted but there will be no obligation to either accept or reject the assignment and no obligation to write a review. There will also be no versioning of the file.

File Flags
At the moment of creation, each file must be assigned a set of flags or designations that determine both its purpose and treatment. There will be no read flags as all files will be readable by the entire community.

Write flag
The creator will have write priviledges on the file. He or she may also designate any number of collaborators who will also have write priviledges. The creator will also have the option release the file into the public domain so that the entire community is allowed modify the file. This type of file is mainly for intermediate results such as processed data, draft manuscripts and scientific software.

title: User:Peteymills/Wiki research


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User:Vanished user adhmfdfmykrdyr/Sport work on WMF

This is a list of content on English Wikinews, English Wikipedia, Simple Wikipedia that I have been a substantial contributor to in terms of sports. This list is not a complete list. Many things are missing. It is maintained manually.

Archery
Alice Ingley
 * English Wikipedia articles created

Elisa Barnard - 1,763 views Alice Ingley - 1,569 views Jesse McDonald (archer) - 1,520 views Deonne Bridger - 1,385 views Taylor Worth - 1,379 views Purnima Mahato - 849 views Ryan Tyack - 418 views views Mathew Masonwells - 359 views
 * English Wikipedia Did You Know

Alice Ingley
 * English Wikipedia Spoken Word Articles

Australian archer Odette Snazelle in Canberra for nationals - 342 views on peak day Australian archer Alice Ingley readies for potential Olympic spot - 337 views on peak day
 * Wikinews exclusive interview

Australian field archers vie for spot at World Championships - 520 views on peak day
 * Wikinews original reporting

 * Commons contributions used on other projects

Athletics
w:Lauren Boden - 3,063 views w:Genevieve LaCaze - 1,968 views w:Melissa Breen - 1,792 views w:Beki Lee - 994 views w:Zoe Buckman - 891 views w:Kathryn Mitchell - 869 views w:Kaila McKnight - 788 views w:Jessica Trengove - 547 views w:Regan Lamble - 462 views
 * English Wikipedia Did You Know

Women's athletics in Australia
 * English Wikipedia articles created
 * Commons contributions used on other projects

Australian rules football
Australian Football League's Richmond Tigers beat GWS Giants by 12 points - 248 views on peak day
 * English Wikinews stories

 * Commons contributions used on other projects

Badminton
Women's badminton in Australia
 * English Wikipedia articles created

Renuga Veeran - 1,299 views Leanne Choo - 1,263 views Victoria Na - 1,122 views
 * English Wikipedia Did You Know

Baseball
Women's baseball in Australia
 * English Wikipedia articles created

Saipan Little League results - 230 views on peak day
 * Wikinews reporting

 * Commons contributions used on other projects

Basketball

 * Commons contributions used on other projects

Beach volleyball
Becchara Palmer - 1,693 views
 * English Wikipedia Did You Know

Bowls
Women's bowls in Australia
 * English Wikipedia articles created

Boxing
Women's boxing in Australia
 * English Wikipedia articles created

Naomi Fischer-Rasmussen - 1,246 views
 * English Wikipedia Did You Know

 * Commons contributions used on other projects

Canoe
Jessica Fox (canoeist) - 3,668 views Hannah Davis (kayaker) - 1,139 views w:Naomi Flood - 697 views w:Rachel Lovell - 526 views w:Jo Brigden-Jones - 452 views w:Alana Nicholls - 432 views
 * English Wikipedia Did You Know

Chess
Women's chess in Australia
 * English Wikipedia articles created

Cricket
Women's cricket in Australia
 * English Wikipedia articles created

WikiProject Cricket
 * English Wikipedia Wikiproject participant

Croquet
Women's croquet in Australia
 * English Wikipedia articles created

Cue Sports
Women's cue sports in Australia
 * English Wikipedia articles created

Cycling
Women's cycling in Australia
 * English Wikipedia articles created

w:Caroline Buchanan - 3,102 views w:Josie Tomic - 2,246 views Amy Cure - 1,704 views w:Lauren Reynolds - 1,394 views w:Melissa Hoskins - 1,321 views Amanda Spratt - 1,311 views Shara Gillow - 1,300 views w:Rebecca Henderson - 667 views
 * English Wikipedia Did You Know

Capital Punishment mountain bike race won by Shaun Lewis - 821 views on peak day Cyclist Sean Finning last competitor named to Australia's 2012 Paralympic team - 285 views on peak day
 * Wikinews

WikiProject Cycling
 * English Wikipedia Wikiproject participant

Australian Cycling Tourism Conference, February 2012
 * Conferences

 * Commons contributions used on other projects

Diving
w:Rachel Bugg - 3,139 views Brittany Broben - 1,983 views w:Anabelle Smith - 829 views
 * English Wikipedia Did You Know

Equestrian
Jonathan Wentz - 3,324 views w:Mary Hanna - 1,545 views w:Julia Hargreaves - 1,470 views w:Lyndal Oatley - 551 views Chris Burton - 483 views w:Kristy Oatley - 482 views w:James Paterson-Robinson - 372 views
 * English Wikipedia Did You Know

Fencing
Women's fencing in Australia
 * English Wikipedia articles created

Fishing
Women's fishing in Australia
 * English Wikipedia articles created

Golf
Women's golf in the Republic of Ireland Women's golf in Australia
 * English Wikipedia articles created

Gymnastics
Women's gymnastics in Australia
 * English Wikipedia articles created

Ashleigh Brennan - 6,724 views Janine Murray - 3,079 views Emily Little - 2,851 views Larrissa Miller - 2,522 views Jaele Patrick - 1,675 views Joshua Jefferis - 1,592 views Blake Gaudry - 1,425 views
 * English Wikipedia Did You Know

At Australian gymnastic team announcement, media turns out for Tony Abbott - 645 views on peak day Kate Lundy and Tony Abbott assist in Olympic dreams with Gymnastics Australia's artistic gymnastics team announcement - 280 views on peak day
 * Wikinews

Horse racing
Women's horse racing in Australia
 * English Wikipedia articles created

Ice hockey
Women's ice hockey in Australia
 * English Wikipedia articles created

Wikinews Shorts: April 2, 2012 - 424 views on peak day
 * Wikinews reporting

Judo
w:Wodjan Shaherkani - 7,732 views Carli Renzi - 1,003 views
 * English Wikipedia Did you know

Kickboxing
Women's kickboxing in Australia - 1,953 views
 * English Wikipedia Did you know

Women's kickboxing in Australia
 * English Wikipedia articles created

Lacrosse
Women's lacrosse in Australia
 * English Wikipedia articles created

Modern pentathlon
w:Chloe Esposito - 1,219 views
 * English Wikipedia Did You Know articles

Netball

 * Commons contributions used on other projects

Paralympics

 * Commons contributions used on other projects

Powerlifting
Women's powerlifting in Australia
 * English Wikipedia articles created

Rifle shooting
w:Hayley Chapman - 2,371 views w:Alethea Sedgman - 1,838 views w:Robyn van Nus - 1,110 views Miro Sipek - 608 views
 * English Wikipedia Did You Know

Women's rifle shooting sport in Australia
 * English Wikipedia articles created

Rock climbing
Women's rock climbing in Australia
 * English Wikipedia articles created

Roller sport
British Roller Sports Federation Roller Sports Canada Skate Australia
 * English Wikipedia articles created
 * Commons contributions used on other projects

Rowing
w:Amy Clay - 1,161 views w:Bronwen Watson - 737 views Renee Chatterton - 480 views Hannah Vermeersch - 317 views Sally Kehoe - 264 views Tess Gerrand - 259 views Alexandra Hagan - 235 views Elizabeth Patrick - 234 views Phoebe Stanley - 215 views
 * English Wikipedia Did You Know

Women's rowing in Australia
 * English Wikipedia articles created

Australian rowers prepare for 2012 Olympics - 1,216 views on peak day
 * Wikinews reporting

Sailing
w:Krystal Weir - 1,085 view Lucinda Whitty - 484 views w:Nina Curtis - 400 views w:Olivia Price - 362 views
 * English Wikipedia Did You Know

Skiing
Para-alpine skiing - 1,967 views LW5/7 - 1,121 views Outrigger ski - 925 views Shannon Dallas - 873 views
 * English Wikipedia Did You Know

Softball

 * Commons contributions used on other projects

Squash
Women's squash in Australia
 * English Wikipedia articles created

Women's squash in Australia
 * English Wikipedia Spoken Word Articles

Surfing
Women's surfing in Australia
 * English Wikipedia articles created

Women's surfing in Australia
 * English Wikipedia Spoken Word Articles

Swimming
Women's swimming in Australia Benard Kieran
 * English Wikipedia articles created

Eggbeater kick Kristin Otto S12 (classification)
 * English Wikipedia Spoken Word Articles

Brittany Elmslie - 3,992 views w:Jessica Ashwood - 1,294 views Marge Kõrkjas - 880 views Benard Kieran - 855 views Jade Neilsen - 381 views
 * English Wikipedia Did You Know articles

2012 Australian Paralympic swim team announced - 562 views on peak day
 * English Wikinews

Synchronized swimming
w:Samantha Reid (synchronized swimmer) - 8,807 views w:Frankie Owen - 2,627 w:Tamika Domrow - 1,267 views w:Olia Burtaev - 1,116 views w:Tarren Otte - 1,037 views w:Bianca Hammett - 861 views w:Jenny-Lyn Anderson - 758 views
 * English Wikipedia Did You Know articles

Table tennis
Vivian Tan - 1,773 views
 * English Wikipedia Did You Know

Women's table tennis in Australia
 * English Wikipedia articles created

Tennis
w:Women's tennis in Australia - 1,200 views
 * English Wikipedia Did you know

Women's tennis in the United States Women's tennis in Australia
 * English Wikipedia articles created

Women's tennis in the United States
 * English Wikipedia Spoken Word Articles

Water polo

 * Commons contributions used on other projects

Weight lifting
Seen Lee - 2,924 views
 * English Wikipedia Did You Know


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User talk:Peteymills/Wiki research

The requirement for "double-blind" peer review contradicts the universal readability of the system, at least if the collaborators want to conduct their research and draft their manuscript using the system. One or both requirements need revision. Peteymills (talk) 19:44, 4 April 2012 (UTC)

title: User talk:Peteymills/Wiki research


text: Template:Terminology

title: Template:Terminology


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title: Category:Terminology


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Talk:Ludic Science Club

 * DAU Games

title: Talk:Ludic Science Club


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Paideia High School/Open Boat, The

The Open Boat is a Paideia Unit Plan. These guidelines address teachers for the purpose of guiding instruction. See Paideia Learning Plan for the student's point of view.

Column One
Column One teaching and learning should make up about 10% to 15% of the total scheduled instructional time. It is didactic in nature and uses teacher lectures, text books or other didactic instructional materials, and questioning appropriate to this mode of education. Teaching in this mode encompasses of three facets: Exoridium, Interpretation, and Erudition.

Oral Presentation

 * The story is fairly short, but it can be beneficial for students to have audio access for the first read through, although they should still follow with the written text with a pencil in hand.
 * free audio copy of the book: http://freelistens.blogspot.com/2008/07/open-boat-by-stephen-crane.html The live links at the top ot the review for each chapter give you the opportunity to download the audio for that chapter.

Written Presentation

 * Text with line numbers is available for print out in the school's moodle site in the Rdg & Wrtng II-B class Week of March 21 to 27, 2012

 * Text can also be accessed on the following site: http://www.online-literature.com/crane/2544/ This site also has links to other Stephen Crane works as well as background and biographical information.

 * This work is in the public domain.

Interpretation
Textual Interpretation refers to the teacher's didactic analysis of a written work in terms of the four major questions a demanding reader should ask of a text. Using the term work instead of book, these questions are: (1) What is the text about as a whole? (2) What is being said in detail, and how? (3) Is the text true, in whole or part? and (4) What of it? The Exordium begins to answer the first question because it introduces the whole text both orally and in writing (although the oral presentation may be limited in the case of longer works). It is in the Interpretation stage of Column One instruction that the teacher didactically begins to thoroughly unlock the second question. The third and fourth questions are relevant to Interpretation, but question two receives most of the teacher's attention. It is only when the student begins Column Two activities that a fuller grasp of questions three and four begins to mature in the student's mind. Consistent with the purpose of Column One instruction, the teacher is simply introducing elements of proper interpretation for the student to build on during Column Two and Column Three learning.

These questions can be appropriately modified for analysis of a work in science, mathematics, or the fine arts. Analysis of these types of works proceeds in a manner analagous to that of a text. Inasmuch as imaginative literature can be considered a work of fine art, the analogies drawn in Adler's and Van Doren's How to Read a Book can serve as a guide to other types of works too.

Interpretation of  or
In textual interpretation, the teacher didactically presents the author's key terms in the stage of Interpretation of Words and Terms. In addition, the teacher anticipates unfamiliar general vocabulary. These words and terms are given didactically, either orally or in writing, or both, but the task of interpreting is a Column Two skill of learning that the teacher must coach. Consequently, the purpose of this Column One stage is merely to point out key terms and potentially unfamiliar terms. The student must learn how to interpret. Again, this planning can be adapted to other kinds of works.

Interpretation of Key Terms
The key terms are those few words or phrases the author uses in unique, special, or important ways. . In the Column Two stage, students are coached both to find these key terms and to "unlock" them on their own. This skill is essential to analytical reading. In this Column One stage, however, the teacher points out a list of such terms. Students should understand that this listing is not necessarily exhaustive. In addition, the teacher provides students with a handful of these terms worked out in detail for consideration; it should consist of about three to five terms.
 * dinghy a small boat usually carried on a larger ship to travel to and from shore in calm harbor waters
 *
 *

General Vocabulary
The teacher may anticipate general unfamiliar vocabulary and point out the more difficult words either orally or in writing or both. However, defining words using context, a dictionary, or a knowlegable friend or adult (including the teacher) is a Column Two skill that must be coached. Teachers must help students build good habits of knowing the meanings of words. Dictionaries, in both electronic and book form must be available, and students must be taught how to use them. They must also be taught how to ask for definitions--a perfectly acceptible life skill commonly used by demanding readers!

 * dinghya small boat intended to ferry people from the ship to the dock in calm harbors. Usually around 12 feet long.

Note: If the teacher points out words in anticipation of their potential difficulty, this should be done in context by giving citations or electronically highlighting the words.

Interpretation of Sentences
One point of didactic interpretation at the level of interpreting sentences is to use grammar to get at the author's meaning. As with interpreting words and terms, this level of interpretation is also a Column Two skill that must be coached. Consequently, at the Column One didactic level, a teacher should choose a handful of the most difficult sentences in the text for demonstration. The teacher will always unlock the grammar of a few important and more difficult sentences for students independent of whether these sentences are key premeses to an argument. As appropriate to the text, a teacher should also consider demonstrating the grammar of sentences that work together as propositions in the author's most important arguments.

Another important point is to demonstrate the meter and prosody in both poetry and prose texts. This aim is in great danger of being completely overlooked or forgotten in a world where oral reading is not nearly as common as it once was. Nevertheless, great speechs often succeed in part because the author understands prosody. Lincoln's Gettysburg Address and King's I Have a Dream serve as striking examples. The teacher should select sentences or versus to demonstrate both meter and rhythm.

 * :
 * :
 * :
 * :

Interpretation of Passages
Crane uses repetition of the same key words and phrases in a fascinating way: (line numbers refer to the numbered copy posted on the school moodle site.)

lines 305-315

"If I am going to be drowned-- if I am going to be drowned--if I am going to be drowned, why, in the name of the seven mad gods who rule the sea, was I allowed to come thus far and contemplate sand and trees? Was I brought here merely to have my nose dragged away as I was about to nibble the sacred cheese of life? It is preposterous. If this old ninny-woman, Fate, cannot do better than this, she should be deprived of the management of men's fortunes. She is an old hen who knows not her intention. If she has decided to drown me, why did she not do it in the beginning and save me all this trouble? The whole affair is absurd.... But no, she cannot mean to drown me. She dare not drown me. She cannot drown me. Not after all this work."

lines 367-372

"If I am going to be drowned--if I am going to be drowned--if I am going to be drowned, why, in the name of the seven mad gods who rule the sea, was I allowed to come thus far and contemplate sand and trees? Was I brought here merely to have my nose dragged away as I was about to nibble the sacred cheese of life?"

lines 571-574

"If I am going to be drowned--if I am going to be drowned--if I am going to be drowned, why, in the name of the seven mad gods who rule the sea, was I allowed to come thus far and contemplate sand and trees?"

Note the decreasing length but total repetitiveness of the quotes. Have student consider and discuss what this does in terms of reflecting the narrator's state of mind and exhaustion.

The narrator continues to contemplate drowning with further quotes:

lines 727-728

It merely occurred to him that if he should drown it would be a shame.

lines 795-798

He thought: "I am going to drown? Can it be possible Can it be possible? Can it be possible?" Perhaps an individual must consider his own death to be the final phenomenon of nature.

lines 804-809

In his struggle to reach the captain and the boat, he reflected that when one gets properly wearied, drowning must really be a comfortable arrangement, a cessation of hostilities accompanied by a large degree of relief, and he was glad of it, for the main thing in his mind for some months had been horror of the temporary agony. He did not wish to be hurt.

Have the students analyze and discuss what is happening now to the author's attitude and acceptance.

Erudition
Erudition refers to all manner of background information assumed by or necessary to understanding or fully appreciating the text.

Anthropological References

 *

Antiquarian References

 *

Biographical References

 *
 * <information about other textual references to important people--include citations--delete if not used>

Cultural References

 * <information and links to cultural references--include citations--delete if not used>

Ethical References

 * <information and links to ethical references--include citations--delete if not used>

Geographical References

 * <links to relevant maps--delete if not used>

Historical References

 * <information and links to historical references--include citations--delete if not used>

Column Two
Column Two teaching consists of coaching. This mode of instruction aims at helping students to form habitual skills in the language, scientific, and fine arts. Thus, a teacher must correct students as they practice listening, speaking, reading, writing, observing with the senses unaided, observing with the aid of scientific apparatus, measuring, estimating, calculating, and exercising dexterity in the musical and visual arts. Each of these arts in turn rely upon the aquisition of fine and gross motor, imagination, and memory skills. These rules for developing Paideia Unit Plans address teachers. For their counterpart written for students, see Paideia Learning Plan.

Column Two learning comprises 65% to 75% of scheduled learning time. It's chief charactaristic is student activity. Students must be practicing some skill or skills while the teacher corrects him or her. While athletic coaching is an obvious example of this type of instruction, debate coaching, directing a drama, art instruction, and piano lessons also represent coaching. In order to coach well, a teacher must have a repertoire of activities carefully designed to exercised desired skills. Additionally, the teacher must have a clear idea of how to correct the skills as students practice them to ensure their habitual formation.

Listening Skills

 * <links to additional quality recordings--include citations--delete if not used>
 * <information relevant to listening for grammatical, logical, rhetorical, poetical, and stylistic elements of the work--include citations--delete if not used>

Speaking Skills

 * <imitation of quality oral readings--should be a part of every unit>
 * <recitations from memory--should be some part of every unit>
 * <oral presentations of written work--should be a part of most units>
 * <class discussions--should be a part of every unit>
 * <recitation of the rules of reading from Adler's and Van Doren's How to Read a Book or similar rules>

Reading Skills
Character Annotation Chart Name ___________________________________________________
 * <read hard-copy with a pencil in hand--always!>
 * <marking up a text both with a pencil and electronically>
 * <inspectional reading notes as taught in Adler's and Van Doren's How to Read a Book>

Track your impressions of characters as we read “The Open Boat”. Include 3-5 quotations/summaries with page numbers for the description and strengths/weaknesses. After you finish the story, summarize the fate of each character. Character	Description	Strengths/Weaknesses	Fate of Character

(A formatted/table worksheet for this assignment is available on the school moodle site in the Rdg & Wrtng II-B class for the week of March 21-27, 2012.)

Writing Skills

 * <Choose phrases from text and express them in different ways>
 * <Reconstruct a previously disarranged passage from the text>
 * <Compose verses or lines in imitation of the author>
 * <Change a passage or poem of one kind into another kind>
 * <Imitate a passage>
 * <Write a composition imitating the author>
 * <Translate sentences or passages into Latin or another language>

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.

Specific Seminars
<Include written plans of actual seminars that have been conducted here.>

title: Paideia High School/Open Boat, The


text: User:Karmabean

title: User:Karmabean


text:
Instructional design/Generate PBL Problems

← Go back to ID Page

Knowing the characteristics of PBL problems helps you generate PBL problems that stimulate learning. The table below lists the three features of PBL with examples of problems that represent each feature. For each problem example, you will:
 * Review the example.
 * Answer some questions about the example.

Features of PBL problems Click on each sub-features to learn more.

When you finish reading all the features of PBL problems,please click on forth session of this class "4. Practice Now."


text:
Instructional design/Practice Now

← Go back to ID Page

Now, do you have a general idea of how to generate PBL problems? In this session, a checklist is provided to you to summarize the features that we have discussed in previous sessions. Review this checklist quickly and then move onto the exercise in the following part.

Checklist for Generating PBL Problems

You can use this checklist in the following part to facilitate the process of generating quality PBL problems. The PBL problem you create in the exercise will be evaluated based on the criteria of this checklist.

Exercise:Generate PBL Problems On Your Own
Below is your practice time to generate a PBL problem for a group of undergraduate students. Please read the question below, and then generate your problem in the "Answering Box", remember your PBL problem will be evaluated based on the checklist. A sample answer is provided that you can use it to compare with your answer.

Question
Five juniors majored in Information Technology are taking a course “Human-Computer Interaction Design” this semester. The instructor of this course wants to apply PBL into this class to motivate students and enhance their critical thinking and collaboration skills. This five students are all interested in game interaction design; they’ve taken some related courses such as “3D Modeling, Animation, and Special Effects”, “Adobe Photoshop and Illustrator”, “Web Interactivity”, “Video Game Programming”. However, they lack experience in communication with clients and don't know exactly how to gather requirements and conduct usability tests, and unfortunately, these are not the teaching contents of this class. Based on the above information, please help the instructor design a PBL problem for this group of students. Grading rules: Your answer will be graded based on the Checklist for generating PBL Problem. For the collaboration section, you don't need to include all the suggested rules, but you do need to give at least two collaboration rules that is applicable to this case (you can create rules on your own). You need to correctly cover more than 80% of the checklist items to pass this test.

Answering Box
'''Click "edit" for this section, and add your answer below. '''

Sample Answer
Click on here to see a sample answer and its grading result.

You are almost finishing this course; now click on the fifth session of this class "5. Summary and References."

title: Instructional design/Practice Now


text:
Talk:Teach Cough Hygiene Everywhere

I think fair use entitles use of the Publishers Weekly review esp since it is credited. If senior WV direct me to do so I will edit it down with ellipses but I think it is probably ok as is. 04:50, 5 April 2012 (UTC)

title: Talk:Teach Cough Hygiene Everywhere


text:
User:Bonestreadings

Bone's Cafe by David Bohn.

title: User:Bonestreadings


title: File:CORDIC.octave.20120405.pdf


title: File:Sampling.3.B.Over.20120405.pdf


text:
Tricolour Voting Game

The Tricolour Voting Game is a game where participants use cards to vote for three colours. For 1st iteration please go here

Summary
Players are given a selection of cards at the beginning. These cards have a value and a colour. At the end of the game the players can put the cards in the voting boxes of the different colours. A card put in a voting box of the same colour adds its value to the tally of that box. A card with a different colour detracts from the tally of that box. White cards are an only add to the tally. The game is designed to work in the background of other activity during which players can determine the level of engagement they wish to have with the game. Players can try and gain more cards, swap cards with other players or give their cards away. All participants are invited to predict the result. The votes for the different colours are calculated by the teller and the reult announced. Prizes may be awarded to the person with the rosette of the winning colour and the person with closest prediction of the result.

Sample cards
These can be customised by incorporating other images

Initial Hands
Each player is given a set of five cards, one blue card, one red card, one green card, one positive wild card (value 4) and one negative wild card (value -2). Each player will have one card of value 1, 3 and 5: Thus there are six starting possibilities:
 * 1) 5 Blue, 2 Wild+, 3 Red, -2 Wild-, Green 1
 * 2) 5 Blue, 2 Wild+, 3 Green, -2 Wild-, Red 1
 * 3) 5 Red, 2 Wild+, 3 Blue, -2 Wild-, Green 1
 * 4) 5 Red, 2 Wild+, 3 Green, -2 Wild-, Blue 1
 * 5) 5 Green, 2 Wild+, 3 Blue, -2 Wild-, Red 1
 * 6) 5 Green, 2 Wild+, 3 Red, -2 Wild-, Blue 1

Rosettes
The game can be played with different ways of gaining Rosettes:
 * 1) Buying rosettes with a specified number of cards on a first come first served basis
 * 2) Auctioning the different rosettes

Game, Simulation and Stimulation
This game has been designed as an open simulation - i.e. the actual game offers an overall structure. Players can then develop ways in which they can make it simulate something in which they are interested. The game may also simulate disinterest i.e. if run as the background to an event perhaps not everyone will maintain an interest in the game, which can be an interesting way of simulating disinterest. Also the game was originally designed for a conference, whereby the game could play the role of an ice-breaker stimulating social interaction particularly amongst people who don't know one another.

title: Tricolour Voting Game


text:
University of Florida/Egm4313/s12.team5.R6

Question
Find the (smallest) period of $$ cos n \omega x $$ and $$ sin n \omega x $$. Show that these functions also have a period p. Show that the constant $$ a_o $$ is also a periodic function with period p.

Solution
Given (1) p.9-2: $$ f(x+np)=f(x) $$ for all $$ n (..., -2, -1, 0, 1, 2...) $$ and where p is the period. The smallest period occurs when n=1 thus we can use the equation as: $$ f(x+p)=f(x) $$

For $$ f(x) = cos (n \omega x) $$ we have: $$ f(x+p)= cos n \omega(x+p) = cos(n \omega x + n \omega p)= cos(n \omega x)cos( n \omega p)-sin(n \omega x)sin( n \omega p)  $$ Thus, $$ cos(n \omega x)cos( n \omega p)-sin(n \omega x)sin( n \omega p)=cos n \omega x $$ $$ cos(n \omega x)cos( n \omega p)=cos (n \omega x)(1) $$ $$ cos( n \omega p) = 1 $$ $$ n \omega p = 2 \pi  $$ $$  p = \frac{2 \pi}{n \omega} $$, smallest period occurs when n=1 so, $$ p =\frac{2 \pi}{\omega} $$

For $$ f(x) = sin( n \omega x) $$ we have: $$ f(x+p)= sin n \omega(x+p) = sin(n \omega x + n \omega p)= sin(n \omega x)cos( n \omega p)-cos(n \omega x)sin( n \omega p) $$

Thus, $$ sin(n \omega x)cos( n \omega p)-cos(n \omega x)sin( n \omega p)=sin( n \omega x) $$ $$ sin(n \omega x)cos( n \omega p)=sin (n \omega x)(1) $$ $$ cos( n \omega p) = 1 $$ $$ n \omega p = 2 \pi  $$ $$  p = \frac{2 \pi}{n \omega} $$, smallest period occurs when n=1 so, $$ p =\frac{2 \pi}{\omega} $$

From (1) p9-5, we know that $$ \omega = \frac{2 \pi}{p} $$ Then, $$  p = \frac{2 \pi}{\omega} = \frac{2 \pi}{\frac {2 \pi}{p}} = p $$ Thus, the period of both of these functions is also $$ p $$.

From (1) p.9-7, we know $$ a_o= \frac {1}{2L} \int_{0}^{2L} f(x) dx = \frac{1}{p}$$ Also, $$ a_o= \frac {1}{2 \pi} \int_{0}^{2 \pi} f(x) dx = \frac{1}{p} $$ Thus, $$ p = 2 \pi $$ showing $$ a_o $$ is a periodic function with period $$ p $$.

Author
This problem is solved and uploaded by Radina Dikova

Question
Is the given function even or odd or either even nor odd? Find its Fourier series. Show details of your work. $$ 11.  f(x) = x^{2}, \; (-1<x<1),  \; p=2 $$ $$ 12.  f(x) = 1- \frac {x^{2}}{4}, \; (-2<x<2), \; p=4 $$

Author
Solved and uploaded by William Knapper

Question
K 2011 p.491 pbs 15,17 Find if the graph is for an even or odd problem and then find the fourier series. Graph the resulting equations. Problem 15:

Problem 17:

Solution
For problem 15: f(-x) = -f(x), therefore the graph shows an odd function. Similarly, L = $$ \pi $$. Therefore we can use the Euler formula for an odd function: $$ f(x) = \sum_{n=1}^\infty b_n \sin(nx) $$ $$ b_n = \frac{2}{\pi} \int_0^{\pi} f(x) \sin(nx)dx $$ During $$[0,\frac{\pi}{2}]$$, f(x) = x. During $$[\frac{\pi}{2},\pi]$$, $$f(x) = \pi-x$$ Therefore, $$ b_n = \frac{2}{\pi} (\int_0^\frac{\pi}{2} x \sin(nx)dx + \int_{\frac{\pi}{2}}^{\pi} (\pi-x)\sin(nx)dx) $$ Using integration by parts shows $$ b_n $$ to be: $$ b_n = \frac{2}{\pi n^2} (2\sin(\frac{\pi}{2} n) - \sin(\pi n)) $$ when, for the first integral: $$ u = x, du = dx, dv = \sin(nx)dx, and v = -\frac{1}{n}\cos(nx) $$ and for the second integral (where u, du, v, and dv are not the same as the first integral): $$ u = \pi-x, du = -dx, dv = -\sin(nx)dx, and v = \frac{1}{n}\cos(nx) $$ Therefore, when n is odd, $$ b_n = + or - \frac{4}{\pi n^2} $$ and when n is even, $$ b_n = 0 $$ Therefore, the fourier series' are as follows: For n=2 -> $$ f(x) = \frac{4}{\pi} \sin(x) $$ For n=4 -> $$ f(x) = \frac{4}{\pi} \sin(x) + \frac{4}{\pi 9} \sin(3x) $$ For n = 8 -> $$ f(x) = \frac{4}{\pi} \sin(x) + \frac{4}{\pi 9} \sin(3x) + \frac{4}{\pi 25} \sin(5x) + \frac{4}{\pi 49} \sin(7x) $$ A graph of the fourier series' is shown below: For Problem 17: $$ f(-x) = f(x) $$ Therefore, it is an even function and L = 1. Similarly, because it's an even function, the Euler equations for an even function can be used: $$ f(x) = a_0 + \sum_{n=1}^{\infty} a_n \cos(nx) $$ $$ a_0 = \frac{1}{\pi} \int_0^{\pi} f(x)dx $$ $$ a_n = \frac{2}{\pi} \int_0^{\pi} f(x) \cos(nx)dx $$ During [0,1], $$ f(x) = 1-x $$ Therefore: $$ a_0 = \int_0^1 (1-x)dx = \frac{1}{2} $$ However, the period is not 2$$\pi$$, so in $$a_n$$ $$\cos(nx)$$ is replaced with $$cos(n\pi x)$$ to shift the function to a period of 2$$\pi$$. Similarly, $$\frac{2}{\pi}$$ is replaced with $$\frac{2}{1}$$ since L = 1. Therefore: $$ a_n = 2 \int_0^1 (f(x) \cos(n\pi x))dx = 2 \int_0^1 ((1-x) \cos(n\pi x))dx $$ Letting $$u = 1-x, du = -dx, dv = \cos(n \pi x)dx, v = \frac{1}{n \pi} \sin(n \pi x) $$ where u, du, v, and dv are not the same as was used in number 15 above. Integration by parts then yields: $$ a_n = 2(\frac{-1}{n^2 \pi^2} (\cos(n \pi) + \frac{1}{n^2 \pi^2}) $$

Therefore, when n is odd, $$a_n = \frac{4}{n^2 \pi^2}$$ and when n is even, $$a_n = 0$$ Therefore, the fourier series's are as follows: For n = 2 -> $$ f(x) = \frac{1}{2} + \frac{4}{\pi^2} \cos(\pi x) $$ For n = 4 -> $$ f(x) = \frac{1}{2} + \frac{4}{\pi^2} \cos(\pi x) + \frac{4}{9 \pi^2} \cos(3 \pi x) $$ For n = 8 -> $$ f(x) = \frac{1}{2} + \frac{4}{\pi^2} \cos(\pi x) + \frac{4}{9 \pi^2} \cos(3 \pi x) + \frac{4}{25 \pi^2} \cos(5 \pi x) + \frac{4}{49 \pi^2} \cos(7 \pi x) $$

Author
This problem was solved and uploaded by John North

Question
Consider the L2-ODE-CC with the window function f(x) from p9-8 as excitation:

$$ \ y'' - 3y'+ 2y = r(x) \ $$ where r(x) = f(x)

and the initial conditions

$$ \ y(0) = 1, y'(0) = 0 \ $$

1. Find $$ \ y_n(x) \ $$ such that:

$$ \ y_n'' + ay_n' + by_n = r_n(x) \ $$

with the same initial conditions as above.

Plot $$ \ y_n(x) \ $$ for n = 3, 6, 9 for x in [0, 10]

2. Use the matlab command ode45 to integrate the L2-ODE-CC and plot the numerical soln to compare with the analytical soln.

Level 1: n = 0,1

Solution
$$ \ \lambda^2 - 3\lambda + 2 = 0 => (\lambda-2)(\lambda-1) = 0 \ $$

$$ \ \lambda_{1,2} = 2, 1 \ $$

The Fourier series of a periodic function

$$ \ f(x) = a_0 + \sum_{n=1}^\infty[a_n \ $$ cos(nωx) + $$ \ b_n \ $$ sin(nωx)]

For an odd function the Fourier series will be the following:

$$ \ f(x) = \sum_{n=1}^\infty b_n \ $$ sin(nωx) $$ \ = \sum_{n=1}^\infty b_n sin( \frac{n*pi}{L}x)dx \ $$

The independent variable t will be used to shift x to the left

$$ \ t = x - \frac{1}{4} \ $$

The period of oscillation and frequency of oscillation will be as follows:

$$ \ p = 2L = 4 \ $$ ω = π/2

where $$ \ b_n = \frac{2}{L} \int_{0}^{L} f(x)sin( \frac{n*pi}{L}x)dx = \frac{1}{2} \int_{0}^{2} f(x)sin( \frac{n*pi}{2}x)dx \ $$

which comes to be: $$ \ b_n = \frac{A}{n*pi} \ $$ (1 - cosnπ)

$$ \ y'' - 3y' + 2y = r(x) \ $$ = 1/2 + (2/nπ) sin(nπ/2}t

The homogeneous equation for Y will be as follows:

$$ \ Y_h = C_1e^{2x} + C_2e^x \ $$

$$ \ Y = Y_h + Y_p = C_1e^{2x} + C_2e^x + C_3 \ $$

Author
This problem was solved and uploaded by Mike Wallace

Question
Redo R4.2 to redisplay the particular solution, the homogenous solution, and the exact solution for n = 3,5,9 over the interval [0,20π]

Redisplay the particular solution, the homogenous solution, and the exact solution. Superpose each solution with the exact solution.

Redo R4.3 with the TA code over the interval [0.10]. Zoom in about x = -0.5, 0, and 0.5 and comment on the accuracy of different approximations.

Redo R4.4 with the TA code over the interval [0.9,10] for n = 4, 7. Zoom in about x = 1, 1.5, 2, 2.5 and comment on the accuracy of different approximations.

R 4.2 redo
The re-displayed functions for the homogenous solutions are:

$$ \displaystyle y_{3h}(x) = 2e^x - 0.8008e^{2x}\ $$

$$ \displaystyle y_{5h}(x) = 2.0001e^x - 0.8001e^{2x}\ $$

$$ \displaystyle y_{9h}(x) = 2e^x - 0.8e^{2x}\ $$

The re-displayed functions for the particular solutions are:

$$ \displaystyle y_{3p}(x) = -9.9206*10^{-5}x^7-0.0010x^6-0.0031x^5-0.0078x^4-0.0990x^3-0.3984x^2-0.3984x - 0.1992\ $$

$$ \displaystyle y_{5p}(x) = -1.2526*10^{-8}x^{11}-2.0668*10^{-7}x^{10}-1.0334*10^{-6}x^9-4.6503*10^{-6}x^8 -1.1781*10^{-4}x^7\ $$

$$ \displaystyle -0.0011x^6-0.0033x^5-0.0083x^4-0.0999x^3-0.3999x^2-0.399x-0.2000\ $$

$$ \displaystyle y_{9p}(x) = -4.1103*10^{-18}x^{19}-1.1714*10^{-16}x^{18}-1.0543*10^{-15}x^{17}-8.9615*10^{-15}x^{16}\ $$

$$ \displaystyle -4.5405*10^{-13}x^{15}-9.1408*10^{-12}x^{14}-6.3985*10^{-11}x^{13}-4.1590*10^{-10}x^{12}-1.5021*10^{-8}x^{11} -2.2040*10^{-7}x^{10}$$

$$ \displaystyle -1.1020*10^{-6}x^9-4.9591*10^{-6}x^8-1.1904*10^{-4}x^7-0.0011x^6-0.0033x^5-0.0083x^4-0.1000x^3-0.4000x^2-0.4000x-0.2000\ $$

The re-displayed functions for the general solutions are:

$$ \displaystyle y_{3}(x) = 2e^x - 0.8008e^{2x}-9.9206*10^{-5}x^7-0.0010x^6-0.0031x^5-0.0078x^4-0.0990x^3-0.3984x^2-0.3984x - 0.1992\ $$

$$ \displaystyle y_{5}(x) = 2.0001e^x - 0.8001e^{2x} -1.2526*10^{-8}x^{11}-2.0668*10^{-7}x^{10}-1.0334*10^{-6}x^9-4.6503*10^{-6}x^8 \ $$

$$ \displaystyle -1.1781*10^{-4}x^7-0.0011x^6-0.0033x^5-0.0083x^4-0.0999x^3-0.3999x^2-0.399x-0.2000\ $$

$$ \displaystyle y_{9}(x) =2e^x-0.8e^{2x} -4.1103*10^{-18}x^{19}-1.1714*10^{-16}x^{18}-1.0543*10^{-15}x^{17}-8.9615*10^{-15}x^{16}\ $$

$$ \displaystyle -4.5405*10^{-13}x^{15}-9.1408*10^{-12}x^{14}-6.3985*10^{-11}x^{13}-4.1590*10^{-10}x^{12}-1.5021*10^{-8}x^{11}\ $$

$$ \displaystyle -2.2040*10^{-7}x^{10}-1.1020*10^{-6}x^9-4.9591*10^{-6}x^8-1.1904*10^{-4}x^7-0.0011x^6\ $$

$$ \displaystyle -0.0033x^5-0.0083x^4-0.1000x^3-0.4000x^2-0.4000x-0.2000\ $$

Plot for $$ \displaystyle y_{3}(x)\ $$

Plot for $$ \displaystyle y_{5}(x)\ $$

Plot for $$ \displaystyle y_{9}(x)\ $$

R 4.3 Redo
Matlab code:

x = 0:0.01:10;

y = log(1+x);

EDU>> x1 = 0:0.01:10;

EDU>> y1 = zeros(1,1001);

EDU>> for i = 1:4

for j = 1:1001

y1(j) = y1(j) - ((-x1(j))^i)/i;

end

end

EDU>> y2 = zeros(1,1001);

EDU>> for i = 1:7

for j = 1:1001

y2(j) = y2(j) - ((-x1(j))^i)/i;

end

end

EDU>> y3 = zeros(1,1001);

EDU>> for i = 1:11

for j = 1:1001

y3(j) = y3(j) - ((-x1(j))^i)/i;

end

end

EDU>> y4 = zeros(1,1001);

EDU>> for i = 1:16

for j = 1:1001

y4(j) = y4(j) - ((-x1(j))^i)/i;

end

end

EDU>> h = plot(x,y);

orange = [1 0.5 0.2];

EDU>> set(h,'Color',orange);

EDU>> hold on;

EDU>> plot(x1,y1,'r');

EDU>> plot(x1,y2,'g');

EDU>> plot(x1,y3,'b');

EDU>> plot(x1,y4,'c');

legend('log(1+x)','T_4','T_7','T_1_1','T_1_6');

EDU>> grid on;

EDU>> axis([0 10 -10 10])

Zoom plot about 0:

This plot makes it appear that only the 16 term approximation is in the window at point 0. This is, therefore, the most accurate of the other n solution approximations.

Zoom plot about 0.5:

This plot makes it appear that most of the approximations are very accurate and close to the exact solution. The only approximation that appears to deviate a little after 0.5 is the n = 4 approximation.

R 4.4 redo
Matlab code:

syms x

EDU>> f = log(x+1);

EDU>> fT1 = taylor(f,5,1);

EDU>> fT2 = taylor(f,8,1);

X = 0.9:0.1:10;

Y(:,1) = subs(fT1,'x',X);

EDU>> Y(:,2) = subs(fT2,'x',X);

EDU>> Y(:,3) = log(1+X);

EDU>> figure

EDU>> plot(X,Y);

axis([0.9 10 -10 10])

Zoom in about 1:

This shows that neither of the approximations are very close about x = 1 to the exact solution.

Zoom in about 1.5:

This shows that the n = 7 approximation is close about x = 1.5 to the exact solution.

Zoom in about 2:

This shows that all the approximations are very close to the exact solution about x = 2.

Zoom in about 2.5:

This shows that the approximations are very close about x = 2.5, however, the n = 4 approximation is beginning to deviate and soon will not be a good approximation.

Author
This problem was solved and uploaded by David Herrick

Question
Given $$\displaystyle y_{p}''+4y_{p}'+13y_{p}=2e^{-2x}cos(3x) $$

 * (1) Simplify the first term $$\displaystyle y_{p}'' $$ on the lhs
 * (2) Simplify the second term $$\displaystyle 4y_{p}' $$ and combine with the simplified first term
 * (3) Finally, add the third term $$\displaystyle 13y_{p} $$

Solution

 * From Lecture Notes Sec.10 p.10-3:

 * Particular solution is of the form $$\displaystyle y_{p}(x)=xe^{-2x}[Mcos(3x)+Nsin(3x)] $$

 * From Lecture Notes Sec.10 p.10-3:

 * $$\displaystyle y^{'}_{p}(x)=e^{-2x}[sin(3x)(-3mx-2nx+n)+cos(3x)(-2mx+m+3nx)] $$
 * $$\displaystyle y^{''}_{p}(x)=e^{-2x}[sin(3x)[6m(2x-1)-n(5x+4)]-cos(3x)[m(5x+4)+6n(2x-1)]] $$

 * We want to substitute these derivatives back into the original ODE and verify that Wolfram Alpha's solution of $$\displaystyle 6e^{-2x}[ncos(3x)-msin(3x)] $$ is correct.

(1)

 * $$\displaystyle y^{''}_{p}=e^{-2x}[sin(3x)[6m(2x-1)-n(5x+4)]-cos(3x)[m(5x+4)+6n(2x-1)]] $$

 * $$\displaystyle y''_{p}=e^{-2x}[12mxsin(3x)-6msin(3x)-5nxsin(3x)-4nsin(3x)-5mxcos(3x)-4mcos(3x)-12nxcos(3x)+6ncos(3x)] $$

(2)

 * $$\displaystyle y^{'}_{p}=e^{-2x}[sin(3x)(-3mx-2nx+n)+cos(3x)(-2mx+m+3nx)] $$
 * $$\displaystyle 4y'p=e^{-2x}[-12mxsin(3x)-8nxsin(3x)+4nsin(3x)-8mxcos(3x)+4mcos(3x)+12nxcos(3x)] $$

 * $$\displaystyle y''_{p}+4y'_{p}=e^{-2x}[-6msin(3x)-13nxsin(3x)-13mxcos(3x)+6ncos(3x)] $$

(3)

 * $$\displaystyle 13y_{p}=e^{-2x}[13mxcos(3x)+13nxcos(3x)] $$

 * $$\displaystyle y''_{p}+4y'_{p}+13y_{p}=e^{-2x}[-6msin(3x)+6ncos(3x)] $$

Comparing this to Wolfram Alpha's answer:

$$\displaystyle 6e^{-2x}[ncos(3x)-msin(3x)]=6e^{-2x}[ncos(3x)-msin(3x)] $$

Author
Solved and uploaded by Joshua House

Question
(1) Find the separated ODEs for the heat equation

 * $$\displaystyle \frac{\partial u}{\partial t}=\kappa \frac{\partial^2 u}{\partial x^2} $$

Solution
Assuming $$\displaystyle u(x,t)=F(x)\cdot G(t) $$

Then:

 * $$\displaystyle \frac{\partial u(x,t)}{\partial t}=F(x)\cdot \dot{G}(t) $$

 * $$\displaystyle \frac{\partial^2u(x,t) }{\partial x^2}=F''(x)\cdot{G}(t) $$

Substituting partial derivatives back into original PDE:

 * $$\displaystyle F(x)\cdot \dot{G}(t)=\kappa F''(x)\cdot G(t) $$

 * $$\displaystyle \frac{\dot{G}(t)}{\kappa G(t)}=\frac{F''(x)}{F(x)}=c $$

 * * Where "c" is a constant, because if both sides were variables then they would never be equal one another (each side would be a function of a different variable.) Kreysig 2011, pp.546

Multiplying by the denominators to get two separate ODEs:

$$\displaystyle F''(x)-cF(x)=0 $$ $$\displaystyle \dot{G}(t)-\kappa cG(t)=0 $$

Author
Solved and uploaded by Joshua House

Contribution Summary
Problem 1 was solved and uploaded by: Radina Dikova

Problem 2 was solved and uploaded by: William Knapper

Problem 3 was solved and uploaded by: John North

Problem 4 was solved and uploaded by: Michael Wallace

Problem 5 was solved and uploaded by: David Herrick

Problem 6 was solved and uploaded by: Joshua House

Problem 7 was solved and uploaded by: Joshua House


text:
Template:TVG card

  Tricolour Voting Game

Transcluded from Wikipedia:Trading card game/Template:Card

title: Template:TVG card


text:
University of Florida/Egm4313/s12.team11.imponenti/R6.4

R6.4
solved by Luca Imponenti

Problem Statement
Consider the L2-ODE-CC (5) p.7b-7 with the window function f(x) p.9-8 as excitation:
 * $$y''-3y'+2y=f(x)\!$$

and the initial conditions
 * $$y(0)=1 \, \ y'(0)=0\!$$

1. Find $$y_n(x)\!$$ such that:
 * $$y_n''+ay_n'+by_n=r_n(x)\!$$

with the same initial conditions as above.

Plot $$y_n(x)\!$$ for $$n=2,4,8\!$$ for x in $$[0,10]\!$$

2. Use the matlab command ode45 to integrate the L2-ODE-CC, and plot the numerical solution to compare with the analytical solution.
 * Level 1: $$n=0,1\!$$

Fourier Series
One period of the window function p9.8 is described as follows

 * $$f(x) =

\begin{cases} 0 & \ \ -1.75 < x <l 0.25 \\ A & \ \ 0.25< x < 2.25 \\ \end{cases} $$

From the above intervals one can see that the period, $$p=4\!$$ and therefore $$L=2\!$$ Applying the Euler formulas from $$-1.75$$ to  $$2.25\!$$ the Fourier coefficients are computed: $$a_0=\frac{1}{2L}\int_{-1.75}^{2.25}f(x)\ dx\!$$

$$a_0=\frac{1}{4}(\int_{-1.75}^{0.25}0\ dx+\int_{0.25}^{2.25}A\ dx)\!$$

$$a_0=+\frac{1}{4}(0+\int_{0.25}^{2.25}A\ dx)\!$$

$$a_0=\frac{1}{4}(2.25A-0.25A)\!$$

$$a_0=\frac{A}{2}\!$$

The integral from $$-1.75\!$$ to  $$0.25\!$$ can be omitted from this point on since it is always zero.

$$a_n=\frac{1}{L}\int_{0.25}^{2.25}f(x)cos(\frac{n\pi x}{L})\ dx\!$$

$$a_n=\frac{1}{2}\int_{0.25}^{2.25}Acos(\frac{n\pi x}{2})\ dx\!$$

$$a_n=\frac{2A}{2n\pi}(sin(\frac{2.25n\pi}{2})-sin(\frac{0.25n\pi}{2})\!$$

$$a_n=\frac{A}{n\pi}(sin(\frac{9n\pi}{8})-sin(\frac{n\pi}{8})\!$$

and

$$b_n=\frac{1}{L}\int_{0.25}^{2.25}f(x)sin(\frac{n\pi x}{L})\ dx\!$$

$$b_n=\frac{1}{2}\int_{0.25}^{2.25}Asin(\frac{n\pi x}{2})\ dx\!$$

$$b_n=\frac{2A}{2n\pi}(cos(\frac{2.25n\pi}{2})-cos(\frac{0.25n\pi}{2})\!$$

$$b_n=\frac{A}{n\pi}(Acos(\frac{n\pi}{8})-Acos(\frac{9n\pi}{8})\!$$

The coefficients give the Fourier series:

$$f(x)=a_0+\sum_{n=1}^{\infty}[a_ncos(\frac{n\pi x}{L})+b_nsin(\frac{n\pi x}{L})]\!$$

$$f(x)=\frac{A}{2}+\sum_{n=1}^{\infty}[\frac{A}{n\pi}(sin(\frac{9n\pi}{8})-sin(\frac{n\pi}{8}))cos(\frac{n\pi x}{2})\!$$
 * $$+\frac{A}{n\pi}(cos(\frac{n\pi}{8})-cos(\frac{9n\pi}{8}))sin(\frac{n\pi x}{2})]\!$$

Homogeneous Solution
Considering the homogeneous case of our ODE:

$$y''-3y'+2y=0\!$$

The characteristic equation is

$$\lambda^2-3\lambda+2=0\!$$

$$(\lambda-1)(\lambda-2)=0\!$$

$$\lambda_1=1, \lambda_1=2\!$$

Therefore our homogeneous solution is of the form

$$y_h=c_1e^x+c_2e^{2x}\!$$

Particular Solution
Considering the case with f(x) as excitation

$$y''-3y'+2y=\frac{A}{2}+\sum_{k=1}^{n}\frac{A}{k\pi}[(sin(\frac{9k\pi}{8})-sin(\frac{k\pi}{8}))cos(\frac{k\pi x}{2})\!$$
 * $$+(cos(\frac{k\pi}{8})-cos(\frac{9k\pi}{8}))sin(\frac{k\pi x}{2})]\!$$

The solution will be of the form

$$y_n=A_0+\sum_{k=1}^{n}A_kcos(\frac{k\pi x}{2})+\sum_{k=1}^{n}B_ksin(\frac{k\pi x}{2})\!$$

Taking the derivatives

$$y_n'=-A_n\frac{\pi}{2}\sum_{k=2}^{n}ksin(\frac{k\pi x}{2})+B_n\frac{\pi}{2}\sum_{k=2}^{n}kcos(\frac{k\pi x}{2})\!$$

$$y_n''=-A_n\frac{\pi^2}{4}\sum_{k=3}^{n}k^2cos(\frac{k\pi x}{2})-B_n\frac{\pi^2}{4}\sum_{k=3}^{n}k^2sin(\frac{k\pi x}{2})\!$$

Plugging these back into the ODE:

$$-\frac{\pi^2}{4}\sum_{k=3}^{n}A_kk^2cos(\frac{k\pi x}{2})-\frac{\pi^2}{4}\sum_{k=3}^{n}B_kk^2sin(\frac{k\pi x}{2})-3[-\frac{\pi}{2}\sum_{k=2}^{n}A_kksin(\frac{k\pi x}{2})\!$$

$$+\frac{\pi}{2}\sum_{k=2}^{n}B_kkcos(\frac{k\pi x}{2})]+2[A_0+\sum_{k=1}^{n}A_kcos(\frac{k\pi x}{2})+\sum_{k=1}^{n}B_ksin(\frac{k\pi x}{2})]\!$$

$$=\frac{A}{2}+\sum_{k=1}^{n}\frac{A}{k\pi}[(sin(\frac{9k\pi}{8})-sin(\frac{k\pi}{8}))cos(\frac{k\pi x}{2})+(cos(\frac{k\pi}{8})-cos(\frac{9k\pi}{8}))sin(\frac{k\pi x}{2})]\!$$

Setting the two constants equal

$$2A_0=\frac{A}{2}\!$$

$$A_0=\frac{A}{4}\!$$

This is valid for all values of n. Since the coefficients of the excitation $$sin(\frac{9k\pi}{8})-sin(\frac{k\pi}{8})\!$$ and  $$cos(\frac{k\pi}{8})-cos(\frac{9k\pi}{8})\!$$ are zero for all even n, then the coefficients $$A_k\!$$  and  $$B_k\!$$ will also be zero, so we must only find these coefficients for odd n's. Now carrying out the sum to $$n=1\!$$ and comparing like terms yields the following sets of equations. Written in matrix form:
 * $$ \begin{bmatrix}2 & 0\\

\\0 & 2\end{bmatrix}* \begin{bmatrix} A_1 \\ B_1\end{bmatrix}= \begin{bmatrix} \ \frac{A}{\pi}(sin(\frac{9\pi}{8})-sin(\frac{\pi}{8})) \\   \ \frac{A}{\pi}(cos(\frac{\pi}{8})-cos(\frac{9\pi}{8}))\end{bmatrix}\!$$

Assuming $$A=1\!$$ this matrix can be solved to obtain

$$A_1=-0.1218 \, \ B_1=0.2941\!$$

For the remaining coefficients to be solved all sums will be used so a more general equation may be written:
 * $$ \begin{bmatrix}

2-\frac{(\pi k)^2}{4} & \frac{-3\pi k}{2}\\ \frac{-3\pi k}{2} & 2-\frac{(\pi k)^2}{4}\end{bmatrix}* \begin{bmatrix} A_k \\ B_k\end{bmatrix}= \begin{bmatrix} \ \frac{A}{\pi k}(sin(\frac{9\pi k}{8})-sin(\frac{\pi k}{8})) \\ \ \frac{A}{\pi k}(cos(\frac{\pi k}{8})-cos(\frac{9\pi}{8}))\end{bmatrix}\!$$

Results of these calculations are shown below:

$$A=\begin{bmatrix} A_1 \\ A_2 \\ .\\ .\\ A_7 \\ A_8\end{bmatrix}= \begin{bmatrix} -0.1218 \\ 0 \\ 0.0084 \\ 0 \\ 0.0014 \\ 0 \\ 0.0001 \\ 0\end{bmatrix}  , \ B=\begin{bmatrix} B_1 \\ B_2 \\ .\\ .\\ B_7 \\ B_8\end{bmatrix}= \begin{bmatrix} 0.2941 \\ 0 \\ 0.0019 \\ 0 \\ 0.0014 \\ 0 \\ 0.0007 \\ 0\end{bmatrix}\!$$

The solution to the particular case can be written for all n (assuming A=1):

$$y_n=\frac{1}{4}+\sum_{k=1}^{n}A_kcos(\frac{k\pi x}{2})+\sum_{k=1}^{n}B_ksin(\frac{k\pi x}{2})\!$$

General Solution
The general solution is

$$y=y_h+y_p\!$$

where

$$y_h=c_1e^x+c_2e^{2x}\!$$

Different coefficients $$c_1 \, \ c_2\!$$ will be calculated for each n. These coefficients are easily solved for by applying the given initial conditions. Below are the calculations for n=2.

$$y=c_1e^x+c_2e^{2x}+\frac{1}{4}+\sum_{k=1}^{2}A_kcos(\frac{k\pi x}{2})+\sum_{k=1}^{2}B_ksin(\frac{k\pi x}{2})\!$$

Applying the first initial condition $$y(0)=1\!$$

$$y(0)=c_1+c_2+A_1+A_2=1\!$$

Taking the derivative

$$y'=c_1e^x+2c_2e^{2x}-\sum_{k=2}^{2}\frac{k\pi}{2}A_ksin(\frac{k\pi x}{2})+\sum_{k=2}^{2}\frac{k\pi}{2}B_kcos(\frac{k\pi x}{2})\!$$

Applying the second initial condition $$y'(0)=0\!$$

$$y'(0)=c_1+2c_2+\frac{2\pi}{2}B_2=0\!$$

Solving the two equations for two unknowns yields:

$$c_1=2.2436 \, \ c_2=-1.1218\!$$

So the general solution for n=2 is:

$$y=2.2436e^x-1.1218e^{2x}+\frac{1}{4}+\sum_{k=1}^{2}A_kcos(\frac{k\pi x}{2})+\sum_{k=1}^{2}B_ksin(\frac{k\pi x}{2})\!$$

Below is a plot showing the general solutions for n=2,4,8:

Matlab Plots
Using ode45 the following graph was generated for n=0:

and for n=1


title: File:Pea2.s12.sec50.a.djvu


title: File:Pea2.s12.sec48.j.djvu


text:
User:Egm6322.s12.team2.Xia/RP13.1

= R 13.1 - Continued fraction of $$\displaystyle \sqrt{5} $$ =

Given
$$\displaystyle \sqrt{5} $$

Find
Continued fraction

Solution
$$\displaystyle \sqrt{5}=2+(\sqrt{5}-2)=2+\frac{1}{\sqrt{5}+2}=2+\frac{1}{4+(\sqrt{5}-2)}=2+\frac{1}{4+\frac{1}{\sqrt{5}+2}}$$

$$\displaystyle =2+\frac{1}{4+\frac{1}{4+\frac{1}{4+\frac{1}{4+\frac{1}{\ddots}}}}} $$

thus

$$\displaystyle \sqrt{5}=2+\frac{1}{4+}\frac{1}{4+}\frac{1}{4+}\frac{1}{\ddots}=\{2:\overline{4}\}$$

title: User:Egm6322.s12.team2.Xia/RP13.1


text:
Nobel Prizes

Definitions
Nobel, Alfred Bernhard 1833-1896; Swedish industrialist, philanthropist, and inventor of dynamite: established the Nobel prizes.

(Americanism) Nobelist, Nobellist; noun. A person who has been awarded a Nobel prize.

Nobel prizes; annual international prizes given by the Nobel Foundation for distinction in physics, chemistry, economics, medicine or physiology, and literature, and for promoting peace.

There is scattered information on the Nobel prizes throughout the Wikiversity. I hope to gather most of it together here. I intend to add more information, including dates, recipients, and a summary of their contributions, for the following categories.

Prizes awarded to More than One Recipient
"A prize may not be shared among more than three people."

Ray Calvin Baker (talk) 18:14, 30 April 2012 (UTC)

Information Sources
Webster's New World Dictionary, Second College Edition, 1980 and 1979, WILLIAM COLLINS PUBLISHERS, INC., ISBN 60B 0-529-05324-1

Ray Calvin Baker (talk) 20:32, 5 April 2012 (UTC)

title: Nobel Prizes


text:
User:Egm6322.s12.team2.Xia/RP13.4

= R 13.4 - Error function =

Given
Definiation of error function

In mathematics, the error function (also called the Gauss error function) is a special function (non-elementary) of sigmoid shape which occurs in probability, statistics and partial differential equations. It is defined as:

$$\displaystyle \operatorname{erf}(x) = \frac{2}{\sqrt{\pi}}\int_{0}^x e^{-t^2} dt. $$

(When x is negative, the integral is interpreted as the negative of the integral from x to zero.) (13.4.1) From the (1)p.48-33 and (2)p.48-33

$$\displaystyle F_{X}(x)=\frac{1}{2}\left[1+\operatorname{erf}\left(\frac{x-\mu}{\sigma\sqrt{2}}\right)\right] $$ (13.4.2)

$$\displaystyle f_{X}(x)=\frac{1}{\sqrt{2\pi\sigma^{2}}}\exp\left[-\frac{(x-\mu)^{2}}{\sigma\sqrt{2}}\right] $$ (13.4.3)

Find
Show that 13.4.2 can be obtained from 13.4.3 and vice versa.

Solution
$$\displaystyle $$ (13.4.4)

$$\displaystyle $$ (13.4.4)

$$\displaystyle $$ (13.4.4)

$$\displaystyle $$ (13.4.4)

F_{X}(x)=\frac{1}{2}\left[1+\operatorname{erf}\left(\frac{x-\mu}{\sigma\sqrt{2}}\right)\right]

can be obtained from (2)p.48-33

f_{X}(x)=\frac{1}{\sqrt{2\pi\sigma^{2}}}\exp\left[-\frac{(x-\mu)^{2}}{\sigma\sqrt{2}}\right]

and vice versa.

title: User:Egm6322.s12.team2.Xia/RP13.4


text:
Talk:Nobel Prizes

I confess -- it is something of an embarassment to me, that I seem to be required to start developing this topic myself. However, I had some questions concerning the Nobel prizes, so it seems I must seek out and collect the information, in order to answer those questions.

THE TWENTY-TWENTY VISION
With only eight years left, will it be possible for an elementary school class of about twenty-five students to accomplish developments of such distintion that they will be awarded SEVERAL Nobel prizes? "Politics is not designed to solve problems" -- Dr. Steve Perry -- so I am putting my hopes on the most creative people on the planet -- those who have not yet had their creativity stifled by an uninspiring "educational" system. I think the prizes in (Computational) Chemistry, Economics, Physiology,and for promoting peace should be relevant and within reach of elementary school students, if only they can be suitably informed and encouraged. Why do I think this is possible? Because I have already read a report (in a science journal) that such students, playing a well-designed video game, have already participated in significant research on the complicated problems of protein folding (a problem in Computational Chemistry).

Won't three or four Nobel prizes look great on their resumes when they finish high school?

Ray Calvin Baker (talk) 20:57, 5 April 2012 (UTC)

Here is something I learned while doing my research:

"A prize may not be shared among more than three people."

I think the idea of a prize-winning class of elementary school children is still a good idea. However, it now appears that I will need to seek some other venue.

Ray Calvin Baker (talk) 18:21, 30 April 2012 (UTC)

P. S. I have been reading about "foldit" in some otf the science news magazines. I will need to look this ("foldit") up.

title: Talk:Nobel Prizes


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text: TESOL/Complaining

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