text:
Instructional design/R2D2/Segment 3: Phase 1 (Reading)-Verbal/Auditory and Phase 2 (Reflecting)-Reflective/Observational

Segment 3: Phase 1 (Reading) & Phase 2 (Reflecting)

Phase 1-(Reading)-Verbal /Auditory

Phase 1 focuses on “…ways to exchange words or text” (Bonk et al, 2008, p. 25) and thus is considered to be the phase of this model that is catered to verbal and auditory learners. “We believe that it is the most comprehensive and complex of the four phases” (Bonk et al, 2008, p. 9) since it is focused on helping learners acquire knowledge. “Given that you must typically acquire knowledge prior to using knowledge, this first phase of the model focuses on knowledge acquisition” (Bonk et al, 2008, p. 20). It is also a complex phase because there are numerous ways to dispense words to your learners. “In the Web of Learning, there are myriad ways that knowledge can be acquired-through watching and listening to videostreamed lectures and podcasts, synchronous presentations, online discussions with peers, guest expert chats, and online and paper-based readings and explorations. Clearly, such resources favor auditory and verbal learners” (Bonk et al, 2008, p. 20). This phase comes first because “…the “reading phase” is the exploration, fact-finding, and knowledge acquisition stage of the learning process. You need new knowledge and ideas in order to have something to reflect upon (that is, R2D2 Phase 2), to visualize and organize (Phase 3), and to apply your learning and make it meaningful (Phase 4)” (Bonk et al, 2008, p. 9).

For decades, text has been considered the main avenue for information dispersion. With the Internet, this does not necessarily need to be so. “There is nearly a decade of research that indicates that instructors will place on the Web of Learning what is easiest to do and what works. Traditionally, this has meant that text-not rich multimedia-was the primary delivery medium” (Bonk et al, 2008, p. 17). During this time, online learning was “…criticized for relying too heavily on text while providing minimal opportunities to learn from visuals and hands-on activities” (Bonk et al, 2008, p. 15). However, as multimedia continues to become easier to use and create, new mediums like podcasts, vodcasts, synchronous conferencing, instant messaging, and text messaging are quickly becoming ways to dispense words that are not only written text.

Using text as a medium for information dissemination is not always a poor instructional strategy. In fact, “Learners in FTF (face-to-face) courses have tended to rely on text throughout their educational careers, and so too will their online ones exhibit such tendencies. Second, text is relatively easy for learners to download even when they have bandwidth limitations or constraints. So learners in rural communities as well as third world countries who lack broadband access to the Internet can more readily participate in an online class. Third, electronic text can be read by tools such as the Jaws Screen Reader to help visually impaired learners” (Bonk et al, 2008, p. 18). The authors would like to encourage instructors to examine other ways to get information to our learners. “Instead of overloading and boring students with written texts, Phase 1 of the model introduces a wide range of learning activities and experiences to help learners acquire knowledge, including the use of podcasting, synchronous conferencing, instant messaging, and other content-rich events and activities. It is the stage of learning meant to intrigue and engage learners in the learning process, not to bore them or cause them to promptly file out” (Bonk et al, 2008, p. 9).

Below is a list of 25 Learning Activities for the “Reading” Phase of R2D2:

(Bonk et al, 2008, pp. 10-11)
 * 1) Online Scavenger Hunt
 * 2) Web Tours and Safaris
 * 3) WebQuest
 * 4) Guided Readings
 * 5) Discovery Readings
 * 6) Foreign Language Reading Activities and Online News
 * 7) FAQ and Course Announcement Feedback
 * 8) Question-and-Answer Sessions with Instructor
 * 9) Online Expert Chats
 * 10) Online Synchronous Testing
 * 11) Synchronous or Virtual Classroom Instructor Presentations
 * 12) Online Webinars
 * 13) Public Tutorials, Wizards, and Help Systems
 * 14) Expert Lectures and Commentary
 * 15) An Online Podcast Lecture or Podcast Show
 * 16) Audio Dramas
 * 17) Posting Video-Based Explanations and Demonstrations
 * 18) Online Sound or Music Training
 * 19) Online Literature Readings
 * 20) Online Poetry Readings
 * 21) Posting Webliographies or Web Resources
 * 22) Text Messaging Course Notes and Content
 * 23) Text Messaging Course Reminders and Activities
 * 24) Online Language Lessons
 * 25) E-Book and Wikibook Reports and Critiques

Phase 2-(Reflecting)-Reflective/Observational

Phase 2 focuses on providing opportunities for learners to reflect on their own thoughts and/or the thoughts of others. Online learning provides a unique opportunity for these learners because “unlike traditional instruction, in which a learner may be allowed a mere 5 or 10 seconds to come up with a response, in online environments, time is stretched in many ways” (Bonk et al, 2008, p. 67). Asynchronous learning environments provide time for the learner to complete knowledge acquisition activities early before having to respond on a topic. Also, learning activities like blogging provide a chance for learners to “…revisit their postings and change their views if needed” (Bonk et al, 2008, p. 67). This may be more difficult in a face-to-face (FTF) learning environment, but online environments encourage this type of reflection. “Online reflection is often promoted as one of the key advantages of online learning; especially when using asynchronous or delayed (that is, anywhere, anytime) conferencing for discussion. When learners can contribute to an online discussion or activity at their convenience, they have an opportunity to think more deeply about an idea and for a longer time period” (Bonk et al, 2008, p. 67).

Some may question the amount of learning that takes place when learners are examining the thoughts of others. However, “…individuals can learn vast amounts of information from observing others model their own strategies or learned behaviors first” (Bonk et al, 2008, pp. 68-69). In fact, all of us have encountered learners like this somewhere in our own educational paths: “You have likely come across such learners in your classes or learning experiences. They may be the quiet ones in FTF classes who often score the highest grades at the end of the course. Such reflective and observational learners, of course, tend to prefer to observe, view, watch, and reflect on specific learning activities as well as on their overall learning. Instead of shouting out answers or participating in a haphazard fashion to get noticed, they will sit back and reflect on their learning experiences and postpone reaching final conclusions until their ideas have matured or are requested. They are known for making careful judgments and hesitating on offering solutions until sufficient data has been obtained and analyzed. Reflective and observational learners also try to observe and compare information from different viewpoints before making decisions or suggestions” (Bonk et al, 2008, p. 68).

The authors warn that “…it will be difficult at times to distinguish between Phase One (the information gathering stage) and Phase Two (the reflection stage)” (Bonk et al, 2008, p. 68). As you examine the list below, compare them to the 25 Learning Activities provided for the Reading Phase. Is the focus on knowledge acquisition, or is the focus on drawing conclusions from knowledge that has already been acquired? Asking yourself this question should help you to see differences between these two lists. “Phase 2 of the R2D2 model emphasizes learners’ reflective processes, speaking to reflective or observational learners who learn and problem solve from watching or observing others as well as thoughtfully deliberating on expert models and examples. While closely related to Phase 1 reading activities, Phase 2 pays special attention to activities and events that stimulate personal reflection through collaboration and virtual group activities, self-questioning, reflective writing and prompting, and intense and interactive challenges” (Bonk et al, 2008, p. 9).

Below is a list of 25 Learning Activities for the “Reflecting” Phase of R2D2:

(Bonk et al, 2008, pp. 10-11)
 * 1) Post Model Answers
 * 2) Reuse Chat Transcripts
 * 3) Workplace, Internship, or Job Reflections
 * 4) Field and Lab Observations
 * 5) Self-Check Quizzes and Exams
 * 6) Online Discussion Forums and Group Discussions
 * 7) Online Portal Explorations and Reflections
 * 8) Lurker, Browser, or Observer in Online Groups
 * 9) Podcast Tours
 * 10) Personal Blogs
 * 11) Collaborative or Team Blogs
 * 12) Online Resource Libraries
 * 13) Social Networking Linkages
 * 14) Online Role Play Reflections
 * 15) Synchronous and Asynchronous Discussion Combinations
 * 16) Self-Check Reflection Activities
 * 17) Electronic Portfolios
 * 18) Individual Reflection Papers
 * 19) Team or Group Reflective Writing Tasks
 * 20) Super-Summaries, Portfolio Reflections, and Personal Philosophy Papers
 * 21) Online Cases, Situations, and Vignettes
 * 22) Satellite Discussion or Special Interest Groups
 * 23) Small-Group Case Creations and Analyses
 * 24) Small-Group Exam Question Challenges
 * 25) Reaction or Position Papers

Resources

Bonk, C.J. & Zhang, K. (2008). Empowering Online Learning. SanFrancisco, CA: Jossey-Bass.

R2D2 Navigation

 * Segment 1: Introduction to R2D2
 * Segment 2: R2D2-The Four Phases Overview
 * Segment 3: Phase 1 (Reading)-Verbal/Auditory and Phase 2 (Reflecting)-Reflective/Observational (You are here)
 * Segment 4: Phase 3 (Displaying)-Visual and Phase 4 (Doing)-Hands-On
 * Segment 5: Activities per phase
 * Segment 6: Your thoughts...
 * Segment 7: Conclusion


text:
Instructional design/R2D2/Segment 4: Phase 3 (Displaying)-Visual and Phase 4 (Doing)-Hands-On

R2D2 Navigation

 * Segment 1: Introduction to R2D2
 * Segment 2: R2D2-The Four Phases Overview
 * Segment 3: Phase 1 (Reading)-Verbal/Auditory and Phase 2 (Reflecting)-Reflective/Observational
 * Segment 4: Phase 3 (Displaying)-Visual and Phase 4 (Doing)-Hands-On (You are here)
 * Segment 5: Activities per phase
 * Segment 6: Your thoughts...
 * Segment 7: Conclusion

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


text:
Instructional design/R2D2/Segment 5: Activities per phase

R2D2 Navigation

 * Segment 1: Introduction to R2D2
 * Segment 2: R2D2-The Four Phases Overview
 * Segment 3: Phase 1 (Reading)-Verbal/Auditory and Phase 2 (Reflecting)-Reflective/Observational
 * Segment 4: Phase 3 (Displaying)-Visual and Phase 4 (Doing)-Hands-On
 * Segment 5: Activities per phase (You are here)
 * Segment 6: Your thoughts...
 * Segment 7: Conclusion


text:
Instructional design/R2D2/Segment 6: Your thoughts...

R2D2 Navigation

 * Segment 1: Introduction to R2D2
 * Segment 2: R2D2-The Four Phases Overview
 * Segment 3: Phase 1 (Reading)-Verbal/Auditory and Phase 2 (Reflecting)-Reflective/Observational
 * Segment 4: Phase 3 (Displaying)-Visual and Phase 4 (Doing)-Hands-On
 * Segment 5: Activities per phase
 * Segment 6: Your thoughts... (You are here)
 * Segment 7: Conclusion


text:
Instructional design/R2D2/Segment 7: Conclusion

R2D2 Navigation

 * Segment 1: Introduction to R2D2
 * Segment 2: R2D2-The Four Phases Overview
 * Segment 3: Phase 1 (Reading)-Verbal/Auditory and Phase 2 (Reflecting)-Reflective/Observational
 * Segment 4: Phase 3 (Displaying)-Visual and Phase 4 (Doing)-Hands-On
 * Segment 5: Activities per phase
 * Segment 6: Your thoughts...
 * Segment 7: Conclusion (You are here)


text:
Instructional design/CA articulation

Have you ever...?
Have you ever learned to cook under the guidance of your parent, friend, or someone else? Do you remember your learning experience in P.E. classes? Or even recall your observation of the way a child learns to speak.

These are examples of apprenticeship-like teaching and learning, in which “students” observe expert practices, and achieve learning goals step by step with the help of the “teacher”. The cognitive apprenticeship presented in this lesson introduces an instructional approach that is similar to traditional apprenticeship.

If you are a teacher or instructional designer, this lesson is going to help you understand the basis of cognitive apprenticeship and learn to use the articulation method in future instructions.

Learning Objectives
After taking this lesson, you will be able to describe how to integrate the articulation method into instructional scenarios. The integration description must include at least one articulation “way”. During the process of achieving this key objective, you will be able to:
 * Identify the key features of cognitive apprenticeship approach;
 * Identify the conditions in which cognitive apprenticeship approach is applicable;
 * Define the articulation method;
 * Distinguish the three ways of articulation;
 * State the specific strategies underneath the three ways of articulation.

Activities
In this lesson, you will have the opportunity to go through introductions to cognitive apprenticeship approach and the articulation method, and practice what you have learned through exercises and forum discussion.

Click the 1st link below to start learning.
 * Brief Introduction to Cognitive Apprenticeship
 * Practice of Cognitive Apprenticeship
 * Explanation of Articulation Method
 * Application of Articulation Strategies

title: Instructional design/CA articulation


text:
User:Fammy15

Bonjour, Je suis Fammy :)

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title: User:Fammy15


text:
User:Egm4313.s12.team17/Report 6

=Problem R6.1--Finding the smallest period=

Given

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$$  \displaystyle f(x)=cos(n \omega x) $$ (Eq.6.1)
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$$  \displaystyle f(x)=sin(n \omega x) $$ (Eq.6.2)
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$$  \displaystyle f(x)=a_0 + \sum_{n=1}^{\infin}[a_n cos (n \omega x) + b_n sin (n \omega x)] $$     (Eq.6.3)
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Part 1 Problem Statement
Find the smallest period of equations 6.1 and 6.2. Show that equations 6.1 and 6.2 have a period $$ p $$.

Part 1 Solution
Equations 6.1 and 6.2 are said to be periodic with a period $$ p $$ if:
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$$  \displaystyle f(x+n_1p)=f(x) $$     (Eq.6.1.1)
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Using equation 6.1.1 for equation 6.1 results in the following:
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$$  \displaystyle f(x+n_1p)=cos(n \omega x + n \omega n_1p) $$     (Eq.6.1.2)
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$$  \displaystyle cos(n \omega x + n \omega n_1p)=cos(n \omega x) $$ (Eq.6.1.3)
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Use the double angle formula to expand equation 6.1.3:
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$$  \displaystyle cos(n \omega x)cos(n \omega n_1p)-sin(n \omega x)sin(n \omega n_1p)=cos(n \omega x) $$ (Eq.6.1.4)
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In order for equation 6.1.4 to be true, the following conditions must be satisfied:
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$$  \displaystyle cos(n \omega n_1p)=1 $$     (Eq.6.1.5)
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$$  \displaystyle sin(n \omega n_1p)=0 $$     (Eq.6.1.6)
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Therefore,
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$$  \displaystyle n \omega n_1p=2 \pi $$     (Eq.6.1.7)
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$$  \displaystyle p=\frac{2 \pi}{n \omega n_1} $$, $$  \displaystyle n=1,2,3,.... $$, $$  \displaystyle n_1=1,2,3,.... $$     (Eq.6.1.8)
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Equation 6.1.8 shows that the smallest period $$ p $$ is when n_1=1 for equation 6.1.

n_1 is the number of periods and by setting n_1=1 will result in a single periodic function.

The work shown above also shows that equation 6.1 has a period $$ p $$.

Using equation 6.1.1 for equation 6.2 results in the following:
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$$  \displaystyle f(x+n_1p)=sin(n \omega x +n \omega n_1p) $$     (Eq.6.1.9)
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$$  \displaystyle sin(n \omega x +n \omega n_1p)=sin(n \omega x) $$ (Eq.6.1.10)
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Use the double angle formula to expand equation 6.1.10:
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$$  \displaystyle sin(n \omega x)cos(n \omega n_1p)+cos(n \omega x)sin(n \omega n_1p)=sin(n \omega x) $$ (Eq.6.1.11)
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In order for equation 6.1.11 to be true, the following conditions must be satisfied:
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$$  \displaystyle cos(n \omega n_1p)=1 $$     (Eq.6.1.12)
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$$  \displaystyle sin(n \omega n_1p)=0 $$     (Eq.6.1.13)
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Therefore,
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$$  \displaystyle n \omega n_1p=2 \pi $$     (Eq.6.1.14)
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$$  \displaystyle p=\frac{2 \pi}{ n \omega n_1} $$, $$  \displaystyle n=1,2,3,.... $$, $$  \displaystyle n_1=1,2,3,.... $$     (Eq.6.1.15)
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Equation 6.1.15 shows that the smallest period $$ p $$ is when n_1=1 for equation 6.2.

n_1 is the number of periods and by setting n_1=1 will result in a single periodic function.

The work shown above also shows that equation 6.2 has a period $$ p $$.

In order, to achieve the standard period for a trigonometric wave function, n_1 must equal 1.

Therefore, the period for a trigonometric wave function is the following:
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$$  \displaystyle p=\frac{2 \pi}{ n \omega} $$, $$  \displaystyle n=1,2,3,.... $$     (Eq.6.1.16)
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Part 2 Problem Statement
Show that constant $$ a_0 $$ is a periodic function with period $$ p $$.

Part 2 Solution
The equation for $$ a_0 $$ is the following:
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$$  \displaystyle a_0=\frac{1}{2 \pi} \int^{\infin}_{-\infin}f(x)dx $$     (Eq.6.2.1)
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For a_0 to be a periodic function of period $$ p $$, the following must be satisfy:
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$$  \displaystyle \frac{1}{2 \pi} \int^{\infin}_{-\infin}f(x+n_1p)dx=\frac{1}{2 \pi} \int^{\infin}_{-\infin}f(x)dx $$     (Eq.6.2.2)
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Simplifying equation 6.2.2:
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$$  \displaystyle \int^{\infin}_{-\infin}f(x+n_1p)dx= \int^{\infin}_{-\infin}f(x)dx $$     (Eq.6.2.3)
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Take the derivative of equation 6.2.3 with respect to $$ x $$:
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$$  \displaystyle f(x+n_1p)= f(x) $$     (Eq.6.2.4)
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The work shown in part 1 is proof that equation 6.1 and 6.2 obeys equation 6.2.4; therefore, a_0 is a periodic function of period $$ p $$.

Part 3 Problem Statement
Show that both lhs and rhs of equation 6.3 are period functions with period $$ p $$.

Part 3 Solution
Use equation 6.1.1 with equation 6.3 will result in the following:
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$$  \displaystyle a_0+\sum^{\infin}_{n=1}[a_n cos(n \omega x +n \omega n_1p)+b_n sin(n \omega x +n \omega n_1p)]= a_0+\sum^{\infin}_{n=1}[a_n cos(n \omega x )+b_n sin(n \omega x)] $$     (Eq.6.3.1)
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In order for equation 6.3.1 to be true, the following condition must be satisfied:
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$$  \displaystyle a_n cos(n \omega x +n \omega n_1p)+b_n sin(n \omega x +n \omega n_1p)= a_n cos(n \omega x )+b_n sin(n \omega x) $$ (Eq.6.3.2)
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Using the double angle formula will expand equation 6.3.2 as shown below:
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$$  \displaystyle a_n[cos(n \omega x)cos(n \omega n_1p)-sin(n \omega x)sin(n \omega n_1p)]+b_n[sin(n \omega x)cos(n \omega n_1p)+cos(n \omega x)sin(n \omega n_1p)]= a_n cos(n \omega x )+b_n sin(n \omega x) $$ (Eq.6.3.3)
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Substitute equation 6.1.8 or 6.1.15 into equation 6.3.3:
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$$  \displaystyle a_n cos(n \omega x)+b_n sin(n \omega x)= a_n cos(n \omega x )+b_n sin(n \omega x) $$ (Eq.6.3.4)
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Therefore, equation 6.3 has a period $$ p $$.

Author & Proofreaders
Author: Egm4313.s12.team17.deaver.md 03:20, 11 April 2012 (UTC)

Proofreader: Egm4313.s12.team17.Li 04:10, 11 April 2012 (UTC)

Editor: Egm4313.s12.team17.deaver.md 03:20, 11 April 2012 (UTC)

=Problem R6.2--Fourier Series Expansion=

Given
Assume a period such that p=4 for the below.
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$$  \displaystyle f(x)=a_0+\sum_{n=1}^\infty[a_n\cos n \omega x+b_n\sin n\omega x] $$ (Eq.2.1)
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$$  \displaystyle a_0=\frac{1}{\pi}\int_{-x}^{x}f(x)dx $$     (Eq.2.2)
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$$  \displaystyle a_n=\frac{1}{\pi}\int_{-x}^{x}f(x)\cos(nx)dx $$     (Eq.2.3)
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$$  \displaystyle b_n=\frac{1}{\pi}\int_{-x}^{x}f(x)\sin(nx)dx $$     (Eq.2.4)
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$$  \displaystyle \omega =\frac{\pi}{L}=\frac{2\pi}{p} $$     (Eq.2.5)
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$$  \displaystyle f(\bar x)=\bar a_0 + \sum_{n=1}^\infty[\bar a_n \cos n \omega \bar x + \bar b_n \sin n \omega \bar x] $$ (Eq.2.6)
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$$  \displaystyle \bar a_n=\frac{1}{L} \int_{-L}^{L} f(\bar x) \cos n \omega \bar x d \bar x $$ (Eq.2.7)
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$$  \displaystyle \bar b_n=\frac{1}{L} \int_{-L}^{L} f(\bar x) \sin n \omega \bar x d \bar x $$ (Eq.2.8)
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$$  \displaystyle \tilde a_0 = \frac{1}{2L} \int_{0}^{2L} f(\tilde x ) d \tilde x $$ (Eq.2.9)
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$$  \displaystyle \tilde a_n = \frac{1}{2L} \int_{0}^{2L} f(\tilde x ) cos(n\omega \tilde x) d \tilde x $$ (Eq.2.10)
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$$  \displaystyle \tilde b_n = \frac{1}{2L} \int_{0}^{2L} f(\tilde x ) sin(n\omega \tilde x) d \tilde x $$ (Eq.2.11)
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$$  \displaystyle f(\tilde x)=\tilde a_0 + \sum_{n=1}^\infty[\tilde a_n \cos n \omega \tilde x + \tilde b_n \sin n \omega \tilde x] $$ (Eq.2.12)
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Part 1 Problem Statement
Determine a Fourier series expansion of $$f(\bar x)$$. Plot $$f(\bar x)$$ and the truncated Fourier series $$f_n(\bar x)$$ for $$ n=0,1,2,4,8$$. Observe the values of $$f_n (\bar x)$$ at the points of discontinuities and the Gibbs phenomenon. Transform the variable so to obtain the Fourier series expansion of $$f(x)$$ for $$n=0,1$$

Part 1 Solution
$$f( \bar x )$$ is symmetric about the y axis and is therefore symmetric (see below graph for $$f( \bar x )$$). Since $$f( \bar x )$$ is an even function we can simplify Eq.2.6 to the following:
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$$  \displaystyle f(\bar x)_n=\bar a_0 + \sum_{k=1}^n [\bar a_n \cos n \omega \bar x] $$ (Eq.2.13) Expanding Eq.2.13 for $$n=0$$ gives us:
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$$  \displaystyle f(\bar x)=\bar a_0 $$     (Eq.2.14) Expanding Eq.2.13 for $$n=1$$ gives us:
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$$  \displaystyle f(\bar x)=\bar a_0+\bar a_n \cos n \omega \bar x $$ (Eq.2.15) Using Eq.2.5 we can find $$L$$ and $$\omega$$ so that we may then solve for $$ \bar a_0 $$:
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$$  \displaystyle \frac{ \pi}{L}=\frac{2 \pi}{p} $$     (Eq.2.16) Inserting values gives us:
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$$  \displaystyle \frac{\pi}{L}=\frac{2 \pi}{4} $$     (Eq.2.17) Which simplifies to:
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$$  \displaystyle \frac{1}{L}=\frac{1}{2} $$     (Eq.2.18) Therefore $$L=2$$ and $$ \omega = \frac { \pi}{2} $$. Inserting these values into Eq.2.7 gives us:
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$$  \displaystyle \bar a_n=\frac{1}{2} \int_{-2}^{2} f(\bar x) \cos n \frac { \pi}{2} \bar x d \bar x $$ (Eq.2.19) Simplified:
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$$  \displaystyle \bar a_n= \frac { \pi}{4} \sin n \frac { \pi}{2} \bar x $$ (Eq.2.20) Applying the conditions of Eq.2.14 to Eq.2.20 yields:
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$$  \displaystyle \bar a_0= \frac { \pi}{4} \sin [0* \frac { \pi}{2} \bar x ] $$     (Eq.2.21)
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 *

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The below is a graph of

Part 2 Problem Statement
Repeat the Solution 1 but using $$f(\tilde x)$$ instead of $$ f(\bar x)$$ to obtain the Fourier series expansion of $$ f(x) $$ for $$ n=0,1 $$.

Author & Proofreaders
Author: Egm4313.s12.team17.axelrod.a 13:47, 11 April 2012 (UTC)

Proofreader: Egm4313.s12.team17.Li 13:55, 11 April 2012 (UTC)

Editor: Egm4313.s12.team17.deaver.md 14:00, 11 April 2012 (UTC)

=Problem R6.3--Truncated Fourier Series=

Part 1 Given
Information from problem 15 in Advanced Engineering Mathematics by Erwin Kreyszig on page 491.

Part 1 Problem Statement
Define whether each function is even or odd, and find it's Fourier series. Plot the functions for n=2,4,8.

Part 1 Solution
Analyzing the graph gives the following piecewise function:
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$$  \displaystyle
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f(x) = \begin{cases} -x - \pi & -\pi \le x < \frac{-\pi}{2} \\ x & \frac{-\pi}{2} < x < \frac{\pi}{2} \\ -x + \pi & \frac{\pi}{2} < x < \pi \end{cases}

$$     (Eq.3.1.1)
 *

 * }

It is also evident from looking at the graph that the function is odd, when taking the integral of the whole function, the integral is 0, making the function odd. Lastly,the graph shows the period is $$ 2 \pi $$ so L = $$ \pi $$. Since the function is odd and the cosine function is even, the $$ a_n $$ terms drop out and the Fourier series is left as:

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$$  \displaystyle
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f(x) = \sum_{n=1}^{\infin} b_n sin(nx)

$$     (Eq.3.1.2)
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Therefore,
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$$  \displaystyle
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b_n = \frac{2}{\pi} \int^{\pi}_0 f(x) sin(nx)dx

$$     (Eq.3.1.3)
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Substituting equation 3.1.1 into 3.1.3:

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$$  \displaystyle b_n = \frac{2}{\pi} \int^{\frac{\pi}{2}}_0 x sin(nx)dx + \frac{2}{\pi} \int^{\pi}_{\frac{\pi}{2}}(\pi-x)sin(nx)dx $$     (Eq.3.1.4)
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 *

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Therefore, the Fourier series is the following:

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$$  \displaystyle
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f(x) = \sum_{n=1,5,9,..}^{\infin} \frac{4}{n^2\pi} sin(nx)-\sum_{n=3,7,11,..}^{\infin} \frac{4}{n^2\pi} sin(nx)

$$     (Eq.3.1.5)
 *

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The graph of the different scenarios is as follows:

Part 2 Given
Information from problem 17 in Advanced Engineering Mathematics by Erwin Kreyszig on page 491.

Part 2 Problem Statement
Define whether each function is even or odd, and find it's Fourier series. Plot the functions for n=2,4,8.

Part 2 Solution
Assembling the piecewise function gives:
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$$  \displaystyle
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f(x) = \begin{cases} x+1 & -1 \le x < 0 \\ 1-x & 0 < x < 1 \\ \end{cases}

$$     (Eq.3.2.1)
 *

 * }

From looking at this graph it is evident that the function is even and the period for this function is 2, so therefore L=1. Since it is even, when applying it to the Fourier series, the sin function will drop out leaving the general format of the Fourier series as:

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$$  \displaystyle f(x)=a_0+\sum_{n=1}^{\infin}a_n cos(n\pi x) $$ (Eq.3.2.2)
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Solve for the coefficients:
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$$  \displaystyle a_0=\frac{1}{2L}\int_{-L}^L f(x)dx $$     (Eq.3.2.3)
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$$  \displaystyle a_0=\frac{1}{2}\int_{-1}^1 f(x)dx=\frac{1}{2}\int_{-1}^0 (x+1)dx+\frac{1}{2}\int_{0}^1 (1-x)dx $$     (Eq.3.2.4)
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$$  \displaystyle a_0=\frac{1}{2} $$     (Eq.3.2.5)
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$$  \displaystyle a_n=\frac{1}{L}\int^L_{-L}f(x) cos\bigg(\frac{n\pi x}{L}\bigg)dx $$     (Eq.3.2.6)
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$$  \displaystyle a_n=\int^1_{-1}f(x) cos(n\pi x)dx $$     (Eq.3.2.7)
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$$  \displaystyle a_n=\int_{-1}^0 (x+1)cos(n\pi x)dx+\int_{0}^1 (1-x)cos(n\pi x)dx $$     (Eq.3.2.8)
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$$  \displaystyle a_n=\sum^{\infin}_{n=1,3,5,..}\frac{4}{(n\pi)^2} $$     (Eq.3.2.9)
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Therefore, the Fourier series is the following:

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$$  \displaystyle f(x)=0.5+\sum_{n=1,3,5,..}^{\infin}\frac{4}{(n\pi)^2} cos(n\pi x) $$ (Eq.3.2.10)
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The graph of f(x) for the various cases stated is shown below

Author & Proofreaders
Author: Egm4313.s12.team17.deaver.md 00:02, 9 April 2012 (UTC)

Proofreader: Egm4313.s12.team17.axelrod.a 12:00, 10 April 2012 (UTC)

Editor: Egm4313.s12.team17.deaver.md 16:52, 11 April 2012 (UTC)

=Problem R6.4--Excitation of Fourier Series=

Given

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$$  \displaystyle f(x)=\begin{cases} 0, & \mbox{if }-1.75<x<0.25 \\ A, & \mbox{if }0.25<x<2.25 \\ 0, & \mbox{if }2.25<x<4.25 \\ \end{cases} $$     (Eq.4.1)
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$$  \displaystyle y^{\prime\prime}-3y^\prime+2y=f(x) $$     (Eq.4.2)
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$$  \displaystyle y(0)=1\mbox{, }y^\prime(0)=0 $$     (Eq.4.3)
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Part 1 Problem Statement
Solve equation 4.2 for n=0,1 \,. Also, plot the solution for x\, in [0,10]\,.

Part 1 Solution
First, find the Fourier series for equation 4.1.

The coefficients for the Fourier series are the following:
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$$  \displaystyle a_0=\frac{1}{2L}\int^L_{-L}f(x)dx $$     (Eq.4.1.1)
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$$  \displaystyle a_n=\frac{1}{L}\int^L_{-L}f(x) cos\bigg(\frac{n\pi x}{L}\bigg)dx $$     (Eq.4.1.2)
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$$  \displaystyle b_n=\frac{1}{L}\int^L_{-L}f(x) sin\bigg(\frac{n\pi x}{L}\bigg)dx $$     (Eq.4.1.3)
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The value of L is 2.

Therefore,
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$$  \displaystyle a_0=\frac{1}{4}\int^{2.25}_{-1.75}f(x)dx $$     (Eq.4.1.4)
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$$  \displaystyle a_0=\frac{1}{4}\int^{2.25}_{0.25}Adx $$     (Eq.4.1.5)
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$$  \displaystyle a_0=\frac{A}{2} $$     (Eq.4.1.6)
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$$  \displaystyle a_n=\frac{1}{2}\int^{2.25}_{-1.75}f(x) cos\bigg(\frac{n\pi x}{2}\bigg)dx $$     (Eq.4.1.7)
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$$  \displaystyle a_n=\frac{1}{2}\int^{2.25}_{0.25}A cos\bigg(\frac{n\pi x}{2}\bigg)dx $$     (Eq.4.1.8)
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$$  \displaystyle a_1=\frac{-A\sqrt{2-\sqrt{2}}}{\pi}=-0.244A $$     (Eq.4.1.9)
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$$  \displaystyle b_n=\frac{1}{2}\int^{2.25}_{-1.75}f(x) sin\bigg(\frac{n\pi x}{2}\bigg)dx $$     (Eq.4.1.10)
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$$  \displaystyle b_n=\frac{1}{2}\int^{2.25}_{0.25}A sin\bigg(\frac{n\pi x}{2}\bigg)dx $$     (Eq.4.1.11)
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$$  \displaystyle b_1=\frac{A\sqrt{\sqrt{2}+2}}{\pi}=0.588A $$     (Eq.4.1.12)
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Therefore,
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$$  \displaystyle f(x)=\frac{A}{2}+\sum^{\infin}_{n=1}\bigg[a_n cos\bigg(\frac{n\pi x}{2}\bigg)+b_n sin\bigg(\frac{n\pi x}{2}\bigg)\bigg] $$     (Eq.4.1.13)
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n = 0

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$$  \displaystyle y_0''-3y_0'+2y_0= f_0(x) $$     (Eq.4.1.14)
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$$  \displaystyle y_0=y_{0,h}+y_{0,p} $$     (Eq.4.1.15)
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The homogenous solution must satisfy the following:
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$$  \displaystyle y_{0,h}''-3y_{0,h}'+2y_{0,h}= 0 $$     (Eq.4.1.16)
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Therefore, the characteristic equation is:
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$$  \displaystyle \lambda^2-3\lambda+2= 0 $$     (Eq.4.1.17)
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$$  \displaystyle (\lambda-2)(\lambda-1)=0 $$     (Eq.4.1.18)
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$$  \displaystyle \lambda_1=2,\lambda_2=1 $$     (Eq.4.1.19)
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Distinct real roots makes the general solution of the homogenous take the following form:
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$$  \displaystyle y_h=d_1e^{\lambda_1 x}+d_2e^{\lambda_2 x} $$ (Eq.4.1.20)
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Therefore,
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$$  \displaystyle y_{0,h}=d_1e^{2 x}+d_2e^{ x} $$ (Eq.4.1.21)
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The particular solution must satisfy the following:
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$$  \displaystyle y_{0,p}''-3y_{0,p}'+2y_{0,p}= \frac{A}{2} $$     (Eq.4.1.22)
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Using Method of Undetermined Coefficients, the particular solution will have the form:
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$$  \displaystyle y_p=\sum^n_{j=0}c_jx^j $$     (Eq.4.1.23)
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Therefore,
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$$  \displaystyle y_{0,p}=c_0x^0 $$     (Eq.4.1.24)
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Take the first and second derivative of equation 4.1.24:
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$$  \displaystyle y_{0,p}^\prime=0 $$     (Eq.4.1.25)
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$$  \displaystyle y_{0,p}^{\prime\prime}=0 $$     (Eq.4.1.26)
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Substitute equations 4.1.24, 4.1.25, and 4.1.26 into equation 4.1.22:
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$$  \displaystyle 2c_0= \frac{A}{2} $$     (Eq.4.1.27)
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Therefore,
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$$  \displaystyle c_0= \frac{A}{4} $$     (Eq.4.1.28)
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The particular solution will be:
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$$  \displaystyle y_{0,p}=\frac{A}{4} $$     (Eq.4.1.29)
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Superimpose equation 4.1.21 and 4.1.29:
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$$  \displaystyle y_{0}=d_1e^{2 x}+d_2e^{ x}+\frac{A}{4} $$     (Eq.4.1.30)
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Take the first of equation 4.1.30:
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$$  \displaystyle y_{0}^\prime=2d_1e^{2 x}+d_2e^{ x} $$ (Eq.4.1.31)
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Use equation 4.3 to solve for the unknowns:
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$$  \displaystyle 1=d_1+d_2+\frac{A}{4} $$     (Eq.4.1.32)
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$$  \displaystyle 0=2d_1+d_2 $$     (Eq.4.1.33)
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Therefore,
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$$  \displaystyle d_1=\frac{A-4}{4} $$     (Eq.4.1.34)
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$$  \displaystyle d_2=\frac{4-A}{2} $$     (Eq.4.1.35)
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The final solution for n=0 \,:
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$$  \displaystyle y_{0}=\frac{(A-4)e^{2 x}}{4}+\frac{(4-A)e^{ x}}{2}+\frac{A}{2} $$     (Eq.4.1.36)
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n = 1

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$$  \displaystyle y_1''-3y_1'+2y_1= f_1(x) $$     (Eq.4.1.37)
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$$  \displaystyle y_1=y_{1,h}+y_{1,p} $$     (Eq.4.1.38)
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The homogenous solution will be the same has equation 4.1.21:
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$$  \displaystyle y_{1,h}=d_1e^{2 x}+d_2e^{ x} $$ (Eq.4.1.39)
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 *

 * }

The particular solution must satisfy the following:
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$$  \displaystyle y_{1,p}''-3y_{1,p}'+2y_{1,p}= 0.5A-0.244A cos\bigg(\frac{\pi x}{2}\bigg)+0.588A sin\bigg(\frac{\pi x}{2}\bigg) $$     (Eq.4.1.40)
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 * style="width:95%" |
 *

 * }

Using Method of Undetermined Coefficients, the particular solution will take the form shown below:
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$$  \displaystyle y_{1,p}=y_{0,p}+K cos\bigg(\frac{\pi x}{2}\bigg) +M sin\bigg(\frac{\pi x}{2}\bigg) $$     (Eq.4.1.41)
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 *

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Take the first and second derivative of equation 4.1.41:
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$$  \displaystyle y_{1,p}^\prime=y_{0,p}^\prime - \frac{K\pi}{2} sin\bigg(\frac{\pi x}{2}\bigg) +\frac{M\pi}{2} cos\bigg(\frac{\pi x}{2}\bigg) $$     (Eq.4.1.42)
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 * }

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$$  \displaystyle y_{1,p}^{\prime\prime}=y_{0,p}^{\prime\prime}- \frac{K\pi ^2}{4} cos\bigg(\frac{\pi x}{2}\bigg) -\frac{M\pi ^2}{4} sin\bigg(\frac{\pi x}{2}\bigg) $$     (Eq.4.1.43)
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 *

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Substitute equations 4.1.41, 4.1.42, and 4.1.43 into equation 4.1.40:
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$$  \displaystyle cos\bigg(\frac{\pi x}{2}\bigg)\bigg[-0.4674K-4.7124M\bigg]+sin\bigg(\frac{\pi x}{2}\bigg)\bigg[-0.4674M+4.7124K\bigg]+0.5A=0.5A-0.244A cos\bigg(\frac{\pi x}{2}\bigg)+0.588A sin\bigg(\frac{\pi x}{2}\bigg) $$     (Eq.4.1.44)
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 *

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Solve the coefficients by setting like terms equal to one another.

For :
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$$  \displaystyle -0.4674K-4.7124M=-0.244A $$     (Eq.4.1.45)
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 *

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For :
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$$  \displaystyle -0.4674M+4.7124K=0.588A $$     (Eq.4.1.46)
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 * style="width:95%" |
 *

 * }

Therefore,
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$$  \displaystyle K=0.129A $$     (Eq.4.1.47)
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$$  \displaystyle M=0.0389A $$     (Eq.4.1.48)
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 *

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The particular solution will be:
 * {| style="width:100%" border="0"

$$  \displaystyle y_{1,p}=0.25A+0.129A cos\bigg(\frac{\pi x}{2}\bigg) +0.0389A sin\bigg(\frac{\pi x}{2}\bigg) $$     (Eq.4.1.49)
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 * style="width:95%" |
 *

 * }

Superimpose equation 4.1.39 and 4.1.49:
 * {| style="width:100%" border="0"

$$  \displaystyle y_{1}=d_1e^{2 x}+d_2e^{ x}+0.25A+0.129A cos\bigg(\frac{\pi x}{2}\bigg) +0.0389A sin\bigg(\frac{\pi x}{2}\bigg) $$     (Eq.4.1.50)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Take the first of equation 9.2.38:
 * {| style="width:100%" border="0"

$$  \displaystyle y_{1}^\prime=2d_1e^{2 x}+d_2e^{ x}-0.203A sin\bigg(\frac{\pi x}{2}\bigg) +0.0611A cos\bigg(\frac{\pi x}{2}\bigg) $$     (Eq.4.1.51)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Use equation 4.3 to solve for the unknowns:
 * {| style="width:100%" border="0"

$$  \displaystyle 1=d_1+d_2+0.25A+0.129A $$     (Eq.4.1.52)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle 0=2d_1+d_2+0.0611A $$     (Eq.4.1.53)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle d_1=0.3179A-1 $$     (Eq.4.1.54)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle d_2=2-0.6969A $$     (Eq.4.1.55)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The final solution for n=1 \,:
 * {| style="width:100%" border="0"

$$  \displaystyle y_{1}=(0.3179A-1)e^{2 x}+(2-0.6969A)e^{ x}+0.25A+0.129A cos\bigg(\frac{\pi x}{2}\bigg) +0.0389A sin\bigg(\frac{\pi x}{2}\bigg) $$     (Eq.4.1.56)
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Plot
The value of the unknown was set to be one.

Part 2 Problem Statement
Use MATLAB command ode45 to integrate the ODE. Then plot the numerical and analytical solution for comparison.

Part 2 Solution
The unknown is assumed to be one.

Below is the MATLAB code used to generate the above graph:

Author & Proofreaders
Author: Egm4313.s12.team17.deaver.md 11:28, 11 April 2012 (UTC)

Proofreader 1: Egm4313.s12.team17.ying 12:30, 11 April 2012 (UTC)

Proofreader 2: Egm4313.s12.team17.hintz 13:01, 11 April 2012 (UTC)

Editor: Egm4313.s12.team17.deaver.md 11:28, 11 April 2012 (UTC)

=Problem R6.5--Taylor Series Expansion of an Excitation=

Given

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

$$  \displaystyle y_n(x) = y_{h,n}(x) + y_{p,n}(x) $$     (Eq.5.1.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle y(x) = y_{h}(x) + y_{p}(x) $$     (Eq.5.1.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Problem Statement
For each value of n=3,5,9, redisplay the expressions for the 3 functions $$y_{p,n}(x), y_{h,n}(x), y_{n}(x)$$ and plot these 3 functions separately over the interval $$[0,20\pi]$$

Redisplay the expressions for $$y_{p,n}(x), y_{h,n}(x), y_{n}(x)$$.

Superpose each of the above plot with that of the exact solution.

Solution
The exact solution using the form of Eq.5.6.2 and based on the report 4 solutions provided is:
 * {| style="width:100%" border="0"

$$  \displaystyle y(x) = 1.5e^x + 0.8e^{2x}+0.3\cos(x) + 0.1\sin(x) $$
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

This solution will plotted with each of the below plots

n = 3
From the report solutions provided in Sakai.

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

$$  \displaystyle y_{h,3}(x) = 2e^x - 0.8008e^{2x} $$     (Eq.5.1.3)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle y_{p,3}(x) = -9.9206 \times 10^{-5}x^7 - 0.0010x^6-0.0031x^5-0.0078x^4-0.0990x^3-0.3984x^2-0.3984x -0.1992 $$     (Eq.5.1.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The final solution, based on Eq.5.1.1 is then:

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

$$  \displaystyle y_3(x) = 2e^x - 0.8008e^{2x} -9.9206 \times 10^{-5}x^7 - 0.0010x^6-0.0031x^5-0.0078x^4-0.0990x^3-0.3984x^2-0.3984x -0.1992 $$     (Eq.5.1.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The following is the matlab code used to generate the Eq.5.1.5 over the interval $$[0,20\pi]$$

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

$$  \displaystyle y_{h,3}(x) $$
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle y_{p,3}(x) $$
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle y_{3}(x) $$
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

n = 5
From the report solutions provided in Sakai.

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

$$  \displaystyle y_{h,5}(x) = 2.0001e^x - 0.8001e^{2x} $$     (Eq.5.1.6)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle y_{p,5}(x) = -1.2526 \times 10^{-8}x^{11} - 2.0668 \times 10^{-7}x^{10} -1.0334\times10^{-6}x^9-4.6503\times10^{-6}x^8-1.1781\times10^{-4}x^7 $$     (Eq.5.1.7)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle -0.0011x^6-0.0033x^5-0.0083x^4-0.0999x^3-0.3999x^2-0.3999x-0.200 $$
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The final solution, based on Eq.5.1.1 is then:

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

$$  \displaystyle y_5(x) = 2.0001e^x - 0.8001e^{2x}-1.2526 \times 10^{-8}x^{11} - 2.0668 \times 10^{-7}x^{10} -1.0334\times10^{-6}x^9-4.6503\times10^{-6}x^8 $$     (Eq.5.1.8)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle -1.1781\times10^{-4}x^7-0.0011x^6-0.0033x^5-0.0083x^4-0.0999x^3-0.3999x^2-0.3999x-0.200 $$
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The following is the matlab code used to generate the Eq.5.1.9 over the interval $$[0,20\pi]$$

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

$$  \displaystyle y_{h,5}(x) $$
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle y_{p,5}(x) $$
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle y_{5}(x) $$
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

n = 9
From the report solutions provided in Sakai.

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

$$  \displaystyle y_{h,9}(x) = 2e^x - 0.8e^{2x} $$     (Eq.5.1.9)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle y_{p,9}(x) = -4.1103 \times 10^{-18}x^{19} - 1.1714 \times 10^{-16}x^{18} -1.0543\times10^{-15}x^{17}-8.9615\times10^{-15}x^{16}-4.5405\times10^{-13}x^{15} $$     (Eq.5.1.10)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle -9.1408\times10^{-12}x^{14}-6.3985\times10^{-11}x^{13}-4.1590\times10^{-10}x^{12}-1.5021\times10^{-8}x^{11}-2.2040\times10^{-7}x^{10}-1.1020\times10^{-6}x^9 $$
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle -4.9591\times10^{-6}x^8-1.1904\times10^{-4}x^7-0.0011x^6-0.0033x^5 $$
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The final solution, based on Eq.5.1.1 is then:

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

$$  \displaystyle y_9(x) = 2e^x - 0.8e^{2x}-4.1103 \times 10^{-18}x^{19} - 1.1714 \times 10^{-16}x^{18} -1.0543\times10^{-15}x^{17}-8.9615\times10^{-15}x^{16}-4.5405\times10^{-13}x^{15} $$     (Eq.5.1.11)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle -9.1408\times10^{-12}x^{14}-6.3985\times10^{-11}x^{13}-4.1590\times10^{-10}x^{12}-1.5021\times10^{-8}x^{11}-2.2040\times10^{-7}x^{10}-1.1020\times10^{-6}x^9 $$
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle -4.9591\times10^{-6}x^8-1.1904\times10^{-4}x^7-0.0011x^6-0.0033x^5 $$
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The following is the matlab code used to generate the Eq.5.1.11 over the interval $$[0,20\pi]$$

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

$$  \displaystyle y_{h,9}(x) $$
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle y_{p,9}(x) $$
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle y_{9}(x) $$
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Part 2 Problem Statement
Understand and run the TA's code to produce a similar plot, but over a larger interval $$[0,10]$$. Do zoom in plots about poiints x = -0.5, 0, 0.5 and comment on the accuracy of different approximations.

Part 2 Solution
From the report 4 solution, the following is the matlab code the TAs used to generate the graphs.

$$x = -0.5$$

$$x = 0$$

$$x = +0.5$$

In the first diagram, one can note that the accuracy of each plot increases as the value of n increases. The plot for n = 11, which is the blue line, lines up perfectly with the cyan line, which is the exact solution.

In the second diagram, the only visible plot is the exact solution or the cyan color. In this region, the plots of all the formulas lie on the same line making it not possible to see the other plots.

In the third diagram, you can see a similar pattern from the first diagram,however, this time the blue line is slightly off from the cyan line. As the value of n decreases, the accuracy of the approximations falls from the cyan plot.

Part 3 Problem Statement
Understand and run the TA's code to produce a similar plot, but over a larger interval $$[0.9,10]$$ and for n = 4,7. Do zoom in plots about points x = 1,1.5,2,2.5 and comment on the accuracy of different approximations.

Part 3 Solution

$$x = 1$$

$$x = 1.5$$

$$x = 2$$

$$x = 2.5$$

In the diagram where the graph is focused about x = 1, there is not much difference in the accuracy of the approximations.

The diagram with x = 1.5, you can see the deviation in the accuracy of the approximations near the end of the graph.

As the focus changes to x = 2.0, there is clearly more deviation in the approximation. N = 4 is the furthest from the exact compared to n = 11.

Author & Proofreaders
Author: Egm4313.s12.team17.ying 23:10, 8 April 2012 (UTC)

Proofreader: Egm4313.s12.team17.wheeler.tw 11:00, 10 April 2012 (UTC)

Editor: Egm4313.s12.team17.deaver.md 17:00, 11 April 2012 (UTC)

=Problem R6.6--Symbolic Computation=

Given

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

$$  \displaystyle y_{p}''+4y_{p}'+13y_{p}=2e^{-2x}cos(3x) $$     (Eq.6.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Problem Statement

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

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

Particular solution is of the form:
 * {| style="width:100%" border="0"

$$  \displaystyle y_{p}(x)=xe^{-2x}[Mcos(3x)+Nsin(3x)] $$     (Eq.6.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

From Lecture Notes Sec.10 p.10-3:
 * {| style="width:100%" border="0"

$$  \displaystyle y^{'}_{p}(x)=e^{-2x}[sin(3x)(-3mx-2nx+n)+cos(3x)(-2mx+m+3nx)] $$    (Eq.6.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle y^{''}_{p}(x)=e^{-2x}[sin(3x)[6m(2x-1)-n(5x+4)]-cos(3x)[m(5x+4)+6n(2x-1)]] $$      (Eq.6.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Now substitute these derivatives back into the original ODE and verify that Wolfram Alpha's solution of:
 * {| style="width:100%" border="0"

$$  \displaystyle y_p=6e^{-2x}[ncos(3x)-msin(3x)] $$      (Eq.6.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Simplifying y^{''}_{p}:
 * {| style="width:100%" border="0"

$$  \displaystyle y^{''}_{p}=e^{-2x}[sin(3x)[6m(2x-1)-n(5x+4)]-cos(3x)[m(5x+4)+6n(2x-1)]] $$      (Eq.6.6)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle y''_{p}=e^{-2x}[12mxsin(3x)-6msin(3x)-5nxsin(3x)-4nsin(3x)-5mxcos(3x)-4mcos(3x)-12nxcos(3x)+6ncos(3x)] $$      (Eq.6.7)
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Now to simplify 4y_{p}' and combine with equation 6.7.
 * {| style="width:100%" border="0"

$$  \displaystyle y^{'}_{p}=e^{-2x}[sin(3x)(-3mx-2nx+n)+cos(3x)(-2mx+m+3nx)] $$      (Eq.6.8)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle 4y'p=4e^{-2x}[sin(3x)(-3mx-2nx+n)+cos(3x)(-2mx+m+3nx)] $$      (Eq.6.9) Distributing the 4 we have:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle 4y'p=e^{-2x}[-12mxsin(3x)-8nxsin(3x)+4nsin(3x)-8mxcos(3x)+4mcos(3x)+12nxcos(3x)] $$      (Eq.6.10)
 * style="width:80%" |
 * style="width:80%" |
 *

 * }

Combining with equation 6.7:
 * {| style="width:100%" border="0"

$$  \displaystyle y''_{p}+4y'_{p}=e^{-2x}[12mxsin(3x)-6msin(3x)-5nxsin(3x)-4nsin(3x)-5mxcos(3x)-4mcos(3x)-12nxcos(3x)+6ncos(3x)] $$
 * style="width:80%" |
 * style="width:80%" |

$$  \displaystyle +e^{-2x}[-12mxsin(3x)-8nxsin(3x)+4nsin(3x)-8mxcos(3x)+4mcos(3x)+12nxcos(3x)] $$      (Eq.6.11)
 *

 * }

Simplifying we eventually get:
 * {| style="width:100%" border="0"

$$  \displaystyle y''_{p}+4y'_{p}=e^{-2x}[-6msin(3x)-13nxsin(3x)-13mxcos(3x)+6ncos(3x)] $$      (Eq.6.12)
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Before adding the term $$ 13y_{p} $$, we need to find out what it equals to:
 * {| style="width:100%" border="0"

$$  \displaystyle 13y_{p}=13xe^{-2x}[Mcos(3x)+Nsin(3x)] $$      (Eq.6.13) After distributing the 13 and simplifying we arrive at:
 * style="width:80%" |
 * style="width:80%" |
 *

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

$$  \displaystyle 13y_{p}=e^{-2x}[13mxcos(3x)+13nxcos(3x)] $$      (Eq.6.14) Now adding $$ 13y_{p} $$ with equation 6.12:
 * style="width:80%" |
 * style="width:80%" |
 *

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

$$  \displaystyle y''_{p}+4y'_{p}+13y_{p}=e^{-2x}[-6msin(3x)-13nxsin(3x)-13mxcos(3x)+6ncos(3x)]+e^{-2x}[13mxcos(3x)+13nxcos(3x)] $$      (Eq.6.15) After simplifying we arrive at:
 * style="width:80%" |
 * style="width:80%" |
 *

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

$$  \displaystyle y''_{p}+4y'_{p}+13y_{p}= e^{-2x}[-6msin(3x)+6ncos(3x)] $$      (Eq.6.16)
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Comparing this to Wolfram Alpha's answer, we see that they are essentially the same once the "6" is factored out from equation 6.16:

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

$$  \displaystyle 6e^{-2x}[ncos(3x)-msin(3x)]=6e^{-2x}[ncos(3x)-msin(3x)] $$      (Eq.6.17)
 * style="width:80%" |
 * style="width:80%" |
 *

 * }

Author & Proofreaders
Author: Egm4313.s12.team17.Li 07:03, 9 April 2012 (UTC)

Proofreader 1: Egm4313.s12.team17.ying 05:23, 10 April 2012 (UTC)

Proofreader 2: Egm4313.s12.team17.wheeler.tw 15:50, 10 April 2012 (UTC)

Editor: Egm4313.s12.team17.deaver.md 17:10, 11 April 2012 (UTC)

=Problem R6.7--Separated ODES of Heat Equation=

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

$$  \displaystyle \frac{\partial \mu }{\partial t}=\kappa\frac{\partial ^{2}\mu }{\partial x^2} $$     (Eq.7.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Problem Statement
Find the seperated ODEs for the Heat equation (Equation 7.1)

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

$$  \displaystyle \mu (x,t)=F(x)\cdot G(t) $$     (Eq.7.2) Where $$F(x)$$ is only a function of x (position) and $$G(t)$$ is only a function of t (time).
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \frac{\partial^2 \mu (x,t)}{\partial x^2}=F''(x)\cdot G(t) $$     (Eq.7.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \frac{\partial \mu (x,t)}{\partial t}=F(x)\cdot \dot{G}(t) $$     (Eq.7.4) Plugging equations 7.3 and 7.4 into 7.1:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle F(x)\cdot \dot{G}(t)=\kappa F''(x)\cdot G(t) $$     (Eq.7.5) Rearranging equation 7.5:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle \frac{\dot{G(t)}}{\kappa G(t)}=\frac{F''(x)}{F(x)} $$     (Eq.7.6) Since $$F(x)$$ is only a function of x and $$G(t)$$ is only a function of t, in order for the above equality to be true, the following must also be true:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle \frac{\dot{G(t)}}{\kappa G(t)}=\frac{F''(x)}{F(x)}=K $$     (Eq.7.7) Where $$K$$ is a constant. Separating equation 7.7:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle \frac{\dot{G(t)}}{\kappa G(t)}=K $$     (Eq.7.8)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle \frac{F''(x)}{F(x)}=K $$     (Eq.7.9) Rearranging equation 7.8 results:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle \dot{G(t)} - \kappa \cdot K \cdot G(t) = 0 $$     (Eq.7.10) Rearranging equation 7.9:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle F''(x)-K \cdot F(x) = 0 $$     (Eq.7.11) Therefore, the separated ODEs of the Heat equation are:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle \dot{G(t)} - \kappa \cdot K \cdot G(t) = 0 $$     (Eq.7.10)
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 *

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

$$  \displaystyle F''(x)-K \cdot F(x) = 0 $$     (Eq.7.11)
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Author & Proofreaders
Author: Egm4313.s12.team17.wheeler.tw 19:10, 9 April 2012 (UTC)

Proofreader: Egm4313.s12.team17.hintz 00:05, 10 April 2012 (UTC)

Editor: Egm4313.s12.team17.deaver.md 17:25, 11 April 2012 (UTC)

= Contributing Team Members = {| border="1" cellspacing="0" cellpadding="5" align="center" ! Team Member ! Contribute ! Proofread
 * Allan Axelrod
 * Problem 2
 * Problem 3
 * Michael Deaver
 * Problem 1 and 4 plus Formatting
 * Problem 3
 * Max Hintz
 * Problem 3
 * Problem 4 and 7
 * Kelvin Li
 * Problem 6
 * Problem 1 and 2
 * Thomas Wheeler
 * Problem 7
 * Problem 5 and 6
 * Chen Ying
 * Problem 5
 * Problem 4 and 6
 * Problem 7
 * Problem 5 and 6
 * Chen Ying
 * Problem 5
 * Problem 4 and 6
 * Problem 4 and 6

title: User:Egm4313.s12.team17/Report 6


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Survey research and design in psychology/Lectures/Multiple linear regression II

Outline
Summarises multiple linear regression and explains some advanced uses, including:
 * semi-partial correlations
 * analysis of residuals
 * interactions
 * analysis of change

Prerequisite

 * 1) Multiple linear regression I lecture

Readings
As per MLR I lecture, plus:
 * 1) Howitt and Cramer (2014a):
 * 2) Chapter 30: Partial correlation: Spurious correlation, third or confounding variables, suppressor variables (pp. 411-422)
 * 3) Chapter 39: The influence of moderator variables on relationships between two variables (pp. 540-561)

Handouts

 * Lecture slides (Google Slides)
 * 2018 handouts:
 * [[Media:SRDP Lecture08Handout MultipleLinearRegressionII 6slidesperpage.pdf|Download 6 slides to a page]]: SRDP Lecture08Handout MultipleLinearRegressionII 6slidesperpage.pdf
 * [[Media:SRDP Lecture08Handout MultipleLinearRegressionII 3slidesperpage.pdf|Download 3 slides to a page]]: SRDP Lecture08Handout MultipleLinearRegressionII 3slidesperpage.pdf


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Dominant group/Theology

Theology is "the rational and systematic study of religion and its influences and of the nature of religious truth [, or] the learned profession acquired by specialized courses in religion (usually at a college or seminary".

A dominant group may be one associated with theology in some way or a group within theology such as a group of gods (entities, sources, or objects).

Control groups
A group of intelligent life forms that think and conceive in the abstract may serve as a control group to compare with a dominant group. Such a group may be able to construct tools.

"Recent research indicates chimpanzee stone tool use dates to at least 4,300 years ago. Chimpanzee tool usage includes digging into termite mounds with a large stick tool, and then using a small stick that has been altered to "fish" the termites out. A recent study revealed the use of such advanced tools as spears, with which common chimpanzees in Senegal sharpen with their teeth and use to spear Senegal bushbabies out of small holes in trees. "

Adult chimpanzees (common or bonobo) have a general range of brain size from 282-500 cc. Brain complexity may make an individual or species as capable of tool construction and use as a bonobo while possessing a much smaller average brain size range.

The minimal brain size to think and conceive in the abstract may mean that composing spirituality or religion may be limited only to thinking and conceiving in the abstract without the additional complexity shown by larger brained organisms such as humans.

Theoretical theology
Def. "reasoning or discussion concerning the Deity", is called theologia.

Def. "the science of things divine", is called theology.

Def. "a deity, a god, God" is called a theos.

Def.
 * 1) "[t]he single deity of various monotheistic religions",
 * 2) "[t]he single male deity of various duotheistic religions",
 * 3) "[a]n impersonal and universal spiritual presence or force",
 * 4) "[a]n omnipotent being, creator of the universe (as in deism)",
 * 5) "[t]he (personification of the) laws of nature", and
 * 6) "[t]he Horned God", is called God.

Usage notes

"God is often referred to by masculine pronouns, not necessarily implying that the speaker believes that God is male. He is also referred to by pronouns that begin with a capital letter, as a sign of respect, in many languages written in Latin script. In English, these would include He, Him, His and Himself. Many Jews follow a prohibition in their tradition against using it and other equivalents in writing (see G-d)."

Academic theology
"During the High Middle Ages, theology was therefore the ultimate subject at universities, being named "The Queen of the Sciences" and serving as the capstone to the Trivium and Quadrivium that young men were expected to study. This meant that the other subjects (including Philosophy) existed primarily to help with theological thought. "

Cognitive ambivalences
“The second stage is "cognitive ambivalence," where they become ambivalent about their ethnic identity as they make progress in acculturation, economic stability, intra-ethnic competition, and the paternalism of the dominant group.”

Dominant theology
"In so doing, it represents the resurgence of knowledges suppressed by a dominant theology and a dominant culture. ... Rather, it is in the realm of the hidden and in the disguised language of infrapolitics, the speech acts and a range of other practices carried out behind the backs of the dominant group, that the voice of the oppressed is to be found."

Feminist theology
"I would say that the first reason why it is absolutely essential for white women to dialogue and to be in contact with theologies coming from these other contexts is precisely to demystify the tendency to think that somehow feminist theology is theology coming from white women of the dominant group."

"The particular problem, for feminists, as it was for pragmatists, is that the knowledge enshrined in theology proper's viewing of the absolute has been a result of its own social location and, as feminists point out, the particular interests of the dominant group."

Hierology
"The existence of blasphemy is tied to the existence of the sacred, and where the latter does not exist, or is not accepted by the dominant group in society, then blasphemy occurs in the denial of the sacred as an aspect of human existence."

Hospitality
“Susanne Johnson has argued that theological education needs to embody an environment of hospitality which confronts the dominant group with the moral imperative to give hospitality to the stranger (1993, 343).”

Natural theology
"From the 17th century onwards, it also became possible to use the term 'theology' to refer to study of religious ideas and teachings that are not specifically Christian (e.g., in the phrase 'Natural Theology' which denoted theology based on reasoning from natural facts independent of specifically Christian revelation "

Privileges
“In the case of racism, people of color are oppressed, yet white privilege also harms those who benefit from it by limiting the cultural awareness, relationships, and experience of the dominant group.”

Hypotheses

 * 1) The first dominant group was the first group of hominins to align themselves with something in the sky above the Earth that was killing people by the millions.

title: Dominant group/Theology


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Nuclear power greener/A Critique of A Comparison between Fossil and Nuclear Power Plants Pollutions and Their Environmental Effects.

(Review Paper) Cited in

Points Made
The world is heavily dependent on fossil fuels that have been proven to be detrimental to the environment and have started to decline. Nuclear power plants produce energy from nuclear fission and are the most clean and environment friendly source of energy among other forms of power sources including coal, oil, gas and hydro power.They lower deathly emission and save lives, also they provide more energy to a large city then other sources of fuel ever would. They are a better form of energy to increased populations because it is cheaper and less toxins will be released into the environment.

Methods
Fossil fuels and nuclear power plants are compared by power sources, pollutants, their environmental effects and their effects on humans. Nuclear power was compared to coal, gas, oil, wood, and hydro power in regards to how much sulfur dioxide (SO2), nitrogen oxide (NOx), particulates, carbon monoxide (CO), carbon dioxide (CO2), and solid waste was produced. But in fact nuclear power is safer and cleaner for the environment. Making it a better way to power the world especially in over populated countries.

Results
Coal produced the most SO2, CO2, and solid waste at 100, 9000, and over 300 millions of tons respectively. Nuclear power did not produce any SO2, CO2, particulates, CO, or CO2. Although, nuclear power produced a slight amount of waste, 0.04 millions of tons. Oil and gas each produced 9000 and 4000 millions of tons of CO2 respectively.


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The efficacy of Esperanto for interlinguistic communication

This is a research project to determine the efficacy of the international constructed auxiliary language Esperanto for interlinguistic communication. For this study, we need at least two willing monolingual volunteers, each of whom speaks a different language from the other. The languages must be among the languages supported by lernu.net, as lernu will be a central part of this project. Each volunteer must commit to at least forty total hours of Esperanto study as well as frequent practice speaking Esperanto to the other participants.

Please translate this page and post it to another language's Wikiversity if you are able to.

To sign up to be a participant, you must be monolingual and speak a language that no other student speaks. Add your name to the Students page if you meet those qualifications.

To sign up to be a research collaborator, add your name to Collaborators page. Research collaborators are expected to be somehow useful to the research; for example, managing another language's mirror of this group of pages and contributing your own skills.

If you translate this page to another Wikiversity (which is absolutely necessary for this project to work), please add a link on the Translations page. Thanks.


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Talk:Mindfulness Training/Mindfulness Training Program (six weeks)

Group Work - How is this different from MBSR and MBCT?
The initial thought I have when thinking of putting together a group based therapy session of mindfulness, is how is it any different from MBSR and MBCT? Although both can be taught through group based learning, my aim on creating this program would be to remove the focus from 'depression' and 'stress reduction', per se, into using mindfulness as an application to "better living". Though in order to understand exactly what I mean by "better living" it requires some research into the assumed candidates of this therapy and a needs analysis of where it could practically fit. Being based in Canberra, I started thinking about the practical application of a community centre teaching mindfulness to a group, imaging who would be in this group and why they would attend the group: From this, it gave me a direction on how to develop the training session so it could be used as a framework to conduct a training session with each of these groups in mind. Through this process i realised that what makes this session different from MBSR and MBCT is it's focus on the group learning environment and the capability to fit it into different learning course structures. Hence, with the basis of this session plan being around group learning, I re-focused my planning to looking at the structure and stages of group development: Blondlox (talk) 15:57, 2 April 2012 (UTC)
 * Work-Life Balance
 * Bullied Youths
 * Divorce
 * 1) Forming
 * 2) Storming
 * 3) Norming
 * 4) Performing
 * 5) Adjourning


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Brief Introduction to Cognitive Apprenticeship

Cognitive apprenticeship refers to the instructional approach that “makes thinking visible”. The learner can observe, enact, and practice implicit knowledge with help from the teacher, and thus achieve the successive approximation of mature practice. Watch the [|VIDEO INTRODUCTION] (Ctrl+click to open in new tab)

When to Use CA ?
Cognitive apprenticeship approach is intended for teaching complex or real-life tasks in the cognitive domain, when learning objectives involve the development of problem-solving skills, conceptual strategies, and so on. Cognitive apprenticeship is not a relevant model for all aspects of teaching. If the targeted goal of learning is a rote task, such as to remember the rules of conjugation in French or the elements of the periodic table, cognitive apprenticeship is not an appropriate model of instruction.

A Framework for Designing Learning Environments
The cognitive apprenticeship model, developed by Collins, Brown and Newman (1987, 1991), consists of four dimensions: content, method, sequence, and sociology.

Content
Knowledge is categorized into four types according to cognitive apprenticeship model: domain knowledge, heuristic strategies, control strategies, and learning strategies.

 * 1) Domain knowledge includes the concepts, facts, and procedures explicitly identified with a particular subject matter. Taking writing as an example, its domain knowledge includes vocabulary, syntax, rhetorical forms and genres, etc.
 * 2) Heuristic strategies are generally effective techniques and approaches for accomplishing tasks that might be regarded as “tricks of the trade”, such as a writing strategy of avoiding getting bogged down in syntax or other presentational details while getting one’s idea down.
 * 3) Control strategies, or meta-cognitive strategies, are general approaches for directing one’s solution process. For example, how to select among the various possible strategies, how to decide when to change strategies, and so on.
 * 4) Learning strategies are strategies for learning any of the other kinds of content described above. For example, if students want to learn to write better, they need to find people to read their writing who can give helpful critiques and explain the reasoning underlying the critiques. They also need to learn to analyze the texts of others in terms of the ways that they are well and badly written. Behaviors like seeking for helpful critiques and analyzing other's writing are examples of learning strategies.

Method
Teaching methods should be designed to give students the opportunity to observe, engage in, and invent or discover expert strategies in context. Cognitive apprenticeship framework has six teaching methods in three groups, which address three types of goals:

Sequencing
Sequencing refers to the ordering of learning activities. Below are three principles that must be balanced in sequencing activities for students.
 * 1) Global before local skills : allowing students to build a conceptual map of the whole task before attending to details. Students can improve their ability to monitor learning progress and develop self-correction skills under the guidance of a conceptual model; when the student is able to accomplish only a portion of a task, this sequence helps him/her make sense of the portion that is being carried out.
 * 2) Increasing complexity : constructing a sequence of tasks such that more and more of the skills and concepts necessary for expert performance are required. This can be accomplished either through sequenced tasks, or through teacher's scaffolding.
 * 3) Increasing diversity : constructing a variety of situations so as to allow the practice of a new strategy or skill repeatedly in a sequence of tasks. Students will be able to distinguish the conditions under which the knowledge applies. Learning transfer is also promoted in this situation.

Sociology
Sociology refers to the social characteristics of learning environments.
 * 1) Situated learning : having students carry out tasks and solve problems in an environment that reflects the multiple uses to which their knowledge will be put in the future.
 * 2) Community of practice : creating a learning environment in which the participants actively communicate about and engage in the skills involved in expertise.
 * 3) Intrinsic motivation : allowing students set personal goals to seek skills and solutions, so that they are intrinsically motivated to learn.
 * 4) Exploiting cooperation : having students working together in a way that fosters cooperative problem solving.

A CASE STUDY may help you understand

Click Practice of CA below to check your understanding of cognitive apprenticeship:


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User:Runfellow

As any quick search will reveal, my real name is Brandon Cooper. I'm here because I had been working on a project on "Ethics in Film" and I wanted to make it accessible to anyone. As the diary statement goes, "more later".

title: User:Runfellow


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User talk:Allenad/enes100/Weekly documentation project 2

Thanks for coming back and helping with the seal issue. Now we're one step closer to completing this project.

week 2 and 3 p2 1sfoerster (talk) 12:03, 11 April 2012 (UTC)
 * 10/10/120 ... good job with the glove issues .. amazing how simple the solution can be

week 4 p2 grading 1sfoerster (talk) 14:35, 18 April 2012 (UTC)
 * 0/10/80 ... great job with beginning the testing