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SH 102 Organizing for School Health

Course Description - NURS 361

This course focuses on the complete health assessment, the nursing process, and its relationship to the prevention and early detection of disease in patients across the life span. This course introduces processes of health assessment: interviewing, history-taking, and physical assessment. Dominant models, theories and perspectives are used to explain health behavior are considered in relation to evidence-based health promotion and health education strategies. Students are also expected to identify and apply pathophysiological principles to selected health issues across the lifespan. The course includes a laboratory >component complemented by self-directed computer assisted instruction. Incorporated throughout the course is the importance of communication and collaboration across culturally diverse urban populations.

Course Title:             Health Assessment & Promotion Course Credit:          Six Course Objectives

 * Demonstrate skill in data collection methods: Interviewing, observation, physical examination and mental health assessment.
 * Use anatomical, physiological, psychosocial, nutritional, developmental norms and theories, cultural and environmental factors to interpret health assessment data.
 * Identify information technology sources to generate assessment data.
 * Document health assessment findings systematically.
 * Discuss common pathophysiological mechanisms in relation to health assessment, health promotion and disease prevention.
 * Describe the role of the nurse in using the comprehensive health assessment. Perform interventions appropriate to promoting health and quality of life across the lifespan process in various health care settings
 * Perform interventions appropriate to promoting the quality of life across the lifespan.
 * Apply principles of learning and teaching to health promotion/education activities.

Prerequisites

Admission into the RN-BS Program at UMass.Nu 360 Professional Issues Required Textbook

Edelman, C., Mandel, C. (2004). Health Promotion Through the Lifespan

Health People 2010.

Jarvis, C. (2004). Physical examination and health assessment. (4th ed.) Philadelphia:W. B. Saunders.

Physical assessment internet site. (TBA)

CD-ROM: Jarvis, C. (2004)  Physical Examination and Health Assessment, 4th edition. Located in back page of text.

 * Recommended

 * Texts
 * Jarvis, C. (2004). Pocket companion for physical examination and health assessment. (4th ed.). Philadelphia: W. B. Saunders Jarvis, C. (2004).
 * Student lab manual. Physical examination and health assessment. (4th ed.)  Philadelphia: W. B. Saunders..

 * Web Related
 * Health Literacy Resource site - This site provides the student the opportunity to investigate the new and exciting field of health literacy. This site provides links to several research projects and government programs in the area of literacy and health. Students can connect to a variety of web sites through this site. Researchers and practitioners will find it a valuable tool for identifying resources.
 * The Auscultation Assistant - This is a wonderful resource for students to listen to actual breath sounds and heart sounds. Designed by a medical student in his third year of school this site affords students the opportunity to listen to and decipher the various lung and heart sounds. Normal lung and heart sounds are provided, as well as abnormal breath sounds and murmurs.
 * Center for Disease Control and Prevention - A terrific government resource for students who want to investigate disease prevention and control. This organization publishes up to date reports on infectious diseases, providing information of disease transmission and control measures.

Roles of a School Nurse INTRODUCTION

The practice of school nursing began in the United States on October 1, 1902 when the initial role of the school nurse was to reduce absenteeism by intervening with students and families regarding health care needs related to communicable diseases. While the nurse’s role has expanded greatly from its original focus, the essence of the practice remains the same. The school nurse supports student success by providing health care assessment, intervention, and follow-up for all children within the school setting.

BACKGROUND

In 1999, the National Association of School Nurses Board of Directors defined school nursing as:

A specialized practice of professional nursing that advances the well-being, academic success, and life-long achievement of students. To that end, school nurses facilitate positive student responses to normal development; promote health and safety; intervene with actual and potential health problems; provide case management services; and actively collaborate with others to build student and family capacity for adaptation, self management, self advocacy, and learning.

Inherent in this definition is the framework that school nurses engage in professional nursing practice, use the nursing process for decision-making, document the care they provide, and assure confidentiality. Professional nurses address the physical, mental, emotional, and social health of their clients. In addition, professional school nurses have as the ultimate outcome of their practice, the support of student success in the learning process. In this context the school nurse provides services to the entire school population, which may include infants, toddlers, pre-schoolers, children with special needs, traditional school populations, and, to a limited degree, adults within the school community.

ROLE OF THE SCHOOL NURSE

Seven roles of the school nurse have evolved from this definition.

The school nurse provides direct health care to students and staff. The school nurse provides care to students and staff who have been injured or who present with acute illnesses. Care may involve treatment of health problems within the scope of nursing practice, communication with parents for treatment, and referral to other providers. The school nurse uses the nursing process to assess, plan, implement, and evaluate care for students with chronic health conditions. This care should begin with the development of a nursing care plan (also known as an individualized health care plan) that should include an emergency action plan. The school nurse is responsible for medication administration and the performance of health care procedures that are within the scope of nursing practice and are ordered by an appropriately licensed health care provider. The school nurse also assists faculty and staff in monitoring chronic health conditions. As the scope of nursing practice expands to utilize the increasingly complex technology needed to provide up-to-date care for clients, the school nurse’s body of knowledge grows through personal professional development. http://www.nasn.org/Default.aspx?tabid=279

source: http://ocw.umb.edu/nursing/nursing-361-health-assessment-and-promotion/syllabus

NUR 4615 FAMILY AND COMMUNITY HEALTH NURSING ON-LINE FALL 2008 SYLLABUS UNIVERSITY OF WEST FLORIDA

COURSE NUMBER:	NUR 4615 On-Line

COURSE TITLE:		FAMILY AND COMMUNITY HEALTH NURSING

CREDIT:				NUR 4615		3 credit hours (3 class hrs/wk)

PLACEMENT:		Senior year, first semester

PRE-REQUISITE:		By Permission TERM:				Fall, 2008

FACULTY:			Course Coordinator - Diane Gardner, EdD, MN, RN UWF Nursing, Bldg. 8 11000 University Parkway Pensacola, FL 32514 Assistant Professor, Nursing dgardner@uwf.edu (Careful with this there is another person with a similar name on campus!) Office (850) 473-7761	 Fax (850) 473-7769 IN- OFFICE HOURS: 	Mondays 8am– 11am; Thursdays 8am-11am Other times available by appointment

INSTRUCTOR ON-LINE AVAILABILITY: 	You may reach me via e-mail at dgardner@uwf.edu for assistance. Please allow 24 hours during the work week to respond to any course work questions. Weekend e-mails will be answered on Monday. In an emergency you may leave a detailed message on my office line at 850-473-7611. I will have some online hours in LiveRoom.

ABOUT THIS COURSE: This course is delivered totally online. You must have regular and reliable access to the internet as well as an email account. It is expected that you will have basic word processing skills, the ability to send/receive email with attachments and able to search the internet and upload/download files. You will also need basic knowledge of PowerPoint. You will generally set your own schedule as to which days and time of day that you do your work and participate in class discussions, but please note that there are set due dates for all assignments. You cannot wait until the end of the semester and begin turning in all your assignments! You are responsible for your own learning – be sure to discipline yourself appropriately. COURSE DESCRIPTION:

The Community Health Nursing course is designed to facilitate the conceptualizing of family, population groups, and community as units of care. The course focuses on risk reduction, health maintenance, and promotion of high level wellness to individuals, families, and groups of all ages throughout the health continuum.

The student is provided the opportunity to assess the health needs of selected families in the community with selected health promotion and health maintenance needs. Variables such as culture and environment, which influence health behaviors of families and the community, are considered in providing nursing intervention. Students include the clients in setting goals for interventions.

By incorporating the concepts for nursing practice and psychosocial assessment the student is able to demonstrate more complex use of the nursing process when caring for clients experiencing crisis. Coping skills of the client and the nurse are assessed and discussed. The student is expected to explore and identify factors which help or hinder health practices.

The student uses the nurse-client relationship as the primary modality of nursing intervention. Guided clinical experiences will be provided through selected community-based agencies and other complex organizations. This provides the opportunity to view the leadership role in official and unofficial health agencies. Validation of interventions is also accomplished through collaboration with other members of the health team. Student will demonstrate progressive independence in nursing practice.

STUDENT LEARNING OUTCOMES - NUR4615 Student Learning Outcomes Upon successful completion of this course, the student will: I. Content A. Use the concepts of nursing theory, research and practice, based on the foundation of the arts and sciences courses, in the provision of care to diverse clients of all ages 1. Evaluate the use of levels of prevention in prioritizing the health needs of communities and populations 2. Integrate the science of epidemiology with the concepts of health promotion, maintenance, and restoration for populations 3. Integrate knowledge and principles of public health nursing in designing holistic interventions in partnership with communities and populations B. Employ theories and principles of leadership/management in the provision of quality nursing care 1. Examine legal, cultural, ethical, and legislative issues which influence the practice of public health nursing. II. Critical Thinking A. Use effective critical thinking skills (e.g., observation, analysis, synthesis) through the nursing process to provide and direct quality care to diverse clients of all ages 1. Incorporate sociocultural, ethnic, religious, and other unique individual considerations in the plan of care B. Evaluate and use appropriate research findings in own evidence based nursing practice 1. Incorporate related research findings in discussion of community health nursing III. Communication C. Uses media resources and information technologies to enhance knowledge base *1. Describe the use of computer technology and informatics in nursing care for communities and populations IV. Integrity/Values C. Function as a beginning leader/change agent through participation in community, governmental, and professional agencies/groups/organizations 1. Identify the role of the community health nurse in the promotion of the family in illness and health 2. Analyze the effectiveness of community planning groups responsible for generation of policies related to public health services. SPECIAL TECHNOLOGY & REQUIRED MATERIALS: Internet Access (The faster, the better!) Argus Account Submit documents in .rtf format. Do not submit work in WordPerfect. Respondus LockDown Browser You will need PowerPoint in order to be able to create the final project. It is available on the computers at UWF. If you need additional software (such as Adobe Acrobat, Flash Player, etc) to view all the components of the course, go to the UWF eLearning home page at https://elearning.uwf.edu. and click on the link for Software Downloads. You can download them at no cost. Students must have the ability to use search engines for supplementary course information, medication information, and patient teaching materials.
 * This Student Learning Outcome will be tracked in the Capstone Pathway.

This course will have supplemental materials posted in eLearning. These postings will include course syllabus, class and clinical schedule, and notes for most of the class sessions. Grades for assignments will be posted on eLearning as well to allow students complete access to grades. Students are urged to check eLearning regularly for course updates and announcements.

REQUIRED TEXT:

Maurer FA, and Smith CM. (2005). Community/public health nursing practice: health for families and populations, Third Edition. St. Louis, MO: Elsevier.

SELECTED READINGS and RECOMMENDED WEBSITES:

American Nurses’ Association. (2000). Public health nursing: A partner for healthy populations. Washington, D.C.: ANA.* American Nurses’ Association. (1999). Scope and standards of home health nursing practice.Washington, D.C.: ANA.* American Nurses’ Association. (1999). Scope and standards of public health nursing practice. Washington, D.C.: ANA.* American Nurses’ Association. (1998). Standards of clinical nursing practice, 2nd ed. Washington, D.C.:ANA.* American Nurses’ Association. (1983). Standards of school nursing practice. Kansas City, MO: ANA.* Clark, M.J. (1999). Nursing in the community: Dimensions of community health nursing, 3rd ed., Stamford, CT: Appleton & Lange Dochterman, JM & Bulechek, GM. (2004). Nursing interventions classification, 4th ed.. St. Louis: Mosby. * Merson, M.H., Black, R.E., & Mills, A.J.,(Eds.) (2001) International public health: Diseases, programs, systems, and policies. Gaithersburg, MD: Aspen Publishers, Inc.* Wright L., & Leahey, M. (1994). Nurses and families, 2nd Ed.. Philadelphia: F.A. Davis, Co.*

USEFUL LINKS:

HealthyPeople 2010:  http://www.healthypeople.gov/ Center for Disease Control and Prevention: http://www.cdc.gov/ National Center for Health Statistics: http://www.cdc.gov/nchs/ Florida Department of Health: http://www.doh.state.fl.us/

 * (On Reserve at UWF Pace Library)

ADDITIONAL READINGS: Students are expected to make use of current nursing and public health literature to support all written and oral work presented individually and collectively. Students are advised to make extensive use of the American Psychological Association. (2001). Publication manual of the American Psychological Association, 5th ed.. Washington, DC: Author. (On reserve at Pace Library and at Ft. Walton Beach Center Library.

Since it is expected that senior nursing students know APA format, this instructor will not be teaching APA format. Please refer to your APA manual or obtain help from the UWF Writing Lab. When it is necessary to provide supplemental information to students, the instructor will send such information via E-mail to student’s Argomail E-mail address. Therefore, each student must maintain the assigned E-mail account. Because of continuing problems with computer viruses, students are instructed to use only the UWF E-mail server when communicating with instructor via E-mail.

EXPECTATIONS FOR ACADEMIC CONDUCT/PLAGIARISM POLICY: As members of the University of West Florida, we commit ourselves to honesty. As we strive for excellence in performance, integrity – personal and institutional – is our most precious asset. Honesty in our academic work is vital, and we will not knowingly act in ways which erode that integrity. Accordingly, we pledge not to cheat, nor to tolerate cheating, nor to plagiarize the work of others. We pledge to share community resources in ways that are responsible and that comply with established policies of fairness. Cooperation and competition are means to high achievement and are encouraged. Indeed, cooperation is expected unless our directive is to individual performance. We will compete constructively and professionally for the purpose of stimulating high performance standards. Finally, we accept adherence to this set of expectations for academic conduct as a condition of membership in the UWF academic community. The Student Code of Conduct sets forth the rules, regulations and expected behavior of students enrolled at the University of West Florida. Violations of any rules, regulations, or behavioral expectations may result in a charge of violating the Student Code of Conduct. It is the student's responsibility to read the Student Code of Conduct and conduct themselves accordingly. You may access the current Student Code of Conduct at http://www.uwf.edu/judicialaffairs ASSISTANCE: The American Disabilities Act will be adhered to in NUR 4615. Students with special needs who require specific examination-related or other course-related accommodations should contact The Student Disability Resource Center (SDRC) sdrc@uwf.edu or (850) 474-2387. The webpage may be viewed at http://uwf.edu/sdrc/ The student is responsible for discussing implementation of any special accommodations/personal or course conflicts with the course coordinator. This must be completed by the second week of class.

PLAGIARISM POLICY:

Plagiarism is defined by the university as “the act of representing the ideas, words, creations, or work of another as one’s own.” Should a student be found plagiarizing in a paper, nursing care plan, or log, the student will be reported by the faculty member to the Director of the Nursing Program and follow the process described in the university’s Student Life handbook. Plagiarism is a serious offense in academia and may result in expulsion from the university. (UWF Student Life Handbook). Students are directed to the examples available in the Pace library tutorial available at http://library.uwf.edu/Tutorials/module_plagiarism/default.htm. For complete information regarding Academic Misconduct, refer to the UWF Student Handbook or contact Student Affairs in Building 21, 474-2384.

Turnitin Instructors have the option of utilizing Turnitin to evaluate student submissions for plagiarism. It is at the discretion of the instructor whether students have the ability to upload personal review before final submission to instructor.

Penalty for late papers Due dates are posted in the assignments area. Unless you have prior authorization for late assignments from the instructor of the course, you will not receive credit for assignments turned in after these dates. All assignments/activities/clinical logs etc., should be submitted in the drop box in the course electronically.

Withdrawal Dates Please refer to the UWF Student Handbook and Calendar for events for these deadlines. This is your responsibility. No exceptions are made for withdrawal deadlines.

OTHER POLICIES: See Student Handbook for further policies on plagiarism, disability, and other matters of interest and importance.

HEALTH INSURANCE PORTABILITY & ACCOUNTABILITY ACT OF 1996 (HIPAA):

All healthcare providers must comply with the federal regulations of this Act. It requires that identifiable patient information be disclosed on a need to know basis with a minimal amount of disclosure to perform a task. The patient’s consent for treatment covers access to the medical record for information needed for treatment purposes and educational purposes. All nursing students will protect patient privacy during oral and written communications. Patient name, initials, address, phone/fax numbers, and social security number will be removed from all student assignments. Students may not copy or circulate papers containing private, confidential patient information. Faculty members will store confidential papers in a locked file or will shred confidential papers. Additionally, nursing students will meet all mandated agency requirements for HIPAA. Since HIPAA is a federal law; violations may result in fines &/or imprisonment. Please note: A GRADE OF ZERO WILL BE ASSIGNED TO ANY WORK SUBMITTED THAT DOES NOT ADHERE TO THIS REQUIREMENT.

Grades will be communicated in eLearning under “Grades.” Grades will not be sent by E-mail, nor will grades be given over the telephone.

EVALUATION/GRADING FOR NUR 4615: ASSIGNMENT	POINTS Discussions	300 Health Issues Letter	 40 Quiz I	 20 Midterm	 40 Quiz III	 20 Final Project	 80 Total Possible	500

GRADING SCALE: A 372 – 400	A- 360-371	B+ 348 – 359	B 332 – 347	B- 320 – 331	C+ 308 – 319 C 292 – 307	[C- 280 – 291  D+ 268 - 279    D 252 – 267     F < 267 points] * Satisfactory completion of the course is based on satisfactory achievement of course student learning outcomes. By the end of the semester, a student must achieve a combined grade of at least 73% on all quizzes/examinations. The student who fails to meet the course student learning outcomes must repeat the course in its entirety. ASSIGNMENTS: Most Assignments will be due by 11:59 PM on Mondays unless otherwise specified. Assignments will pop-up under content on Mondays at 11:59pm. Papers & Quizzes/Exams- Due as specified in the assignment. Late work may receive a 1 point deduction for every day that the assignment is late. Circumstances vary, so if something comes up, please let me know. I just need to know about it ahead of time. If I am unable to open a document because it is submitted in the wrong format, it will be treated as late work with points deducted until it is resubmitted in a format I can open. ONLINE DISCUSSIONS: As you participate in online discussions, I expect you to demonstrate thoughtfulness and insight. Further information will be available Week 2.
 * A grade of C- or below constitutes a failure in this course for nursing majors. Students must pass both NUR4615 and 4615L to progress in the nursing program.	 Both NUR 4615 and NUR4615L must be repeated if the final grade for either course is below a “C.”

Reading Assignments: You will be assigned to read the chapters in the book. There may be some content you are already quite familiar with and you only need to scan some of the highlights. Other areas you may need to read in more detail. There may be some “handouts” linked with each week which you may find useful as well as links to associated websites. On-Line Quizzes: I will ask that you answer the exams on your own without assistance from other people. There will be 20 multiple choice questions on each quiz. Quizzes will be available one week prior to the due date. There will be a total of 2 quizzes. They will be accessed under the heading labeled as Quizzes. Health Issues Letter: The Student Learning Outcomes of the UWF nursing program includes “Demonstrate effective written and oral communication as a member of the interdisciplinary healthcare team.” In partial fulfillment to meet this outcome, selected nursing courses have as a requirement the composition of a letter which addresses current issues in the health care setting.

Midterm: Will require you to obtain a designated film and write a paper addressing specific questions. More details will be available at a later date.

Final Project: Will require you to prepare a Powerpoint on a selected topic. More details will be available at a later date.

THE FOLLOWING TEN COURSE MODULES WILL GIVE YOU AN IDEA OF WHAT TOPICS WILL BE COVERED IN THIS COURSE:

MODULE I: PUBLIC HEALTH AND COMMUNITY HEALTH NURSING

STUDENT LEARNING OUTCOMES:

1.	Define community. 2.	Describe different types of communities 3.	Compare and contrast Community Health Nursing and Public Health Nursing 4.	Examine historical development of community health nursing, national and international 5.	Relate selected events in community health nursing to historical periods 6.	Examine the role of health organizations – community, state, national, and international. 7.	Explain the degree to which Healthy People 2010 influences health care in the United States. 8.	Discuss various roles and settings for community health nursing. 9.	Illustrate ways in which the community health nurse can act as a role model in community organizations. 10.	Summarize basic legal issues relevant in community and public health nursing practice. 11.	Examine responsibilities of the community health nurse related to public health laws. 12.	Relate and apply standards of community nursing to selected clinical activities. 13.	Identify community health clients. 14.	Demonstrate therapeutic use of communication skills in interactions with families and with community members. 15.	Relate ANA Standards of Practice to schema of nursing process.

MODULE II: THE COMMUNITY AS A CLIENT

STUDENT LEARNING OUTCOMES:

1.	Discuss use of nursing process with the community as a client. 2.	List frameworks for community assessment. 3.	Outline the components of a community assessment. 4.	Adapt the nursing process to determine the health needs of a community. 5.	Look for indicators of changing health needs in a selected community. 6.	Relate concept of risk to care of community. 7.	Identify nursing interventions for selected community health problems. 8.	Explain the effectiveness of a community planning meeting that has been designed to deal with health and social problems. 3.	Formulate community-level diagnosis. 4.	Compare and contrast the applicability of program planning for selected community health problems.

MODULE III: CULTURAL INFLUENCES ON HEALTH AND HEALTH PRACTICES

STUDENT LEARNING OUTCOMES:

1.	Analyze personal cultural health practices and beliefs which influence one's own health behavior. 2.	Examine the concept that privilege ensues when prevailing cultural assumptions favor one group over one or more groups. 3.	Become cognizant of assessment cues that vary between and among cultural groups. 4.	Discuss application of the Code for Nurses in care of clients from non-dominant cultural groups. 5.	Explain the concept of "cultural competence." 6.	Explain the term “health disparities” and “health care disparities.” 7.	Identify and provide anticipatory guidance and counseling on topics including nutrition, discipline, safety, and infections, appropriate to the needs of culturally diverse families. 8.    Summarize WHO and UNICEF programs and definitions of Primary Care. 9.    Explain relationship between economic development and the status of health in developed and lesser developed countries. 10. Compare selected health concerns that are present in developed and lesser developed countries. 11.  Discuss potential need for advocacy on part of community health nurse who is working with members of other cultures. 12.  Develop nursing plans that reflect sensitivity to cultural influences.

MODULE IV: THE FAMILY AS A UNIT OF CARE

STUDENT LEARNING OUTCOMES:

1.	Explain the use of the family perspective. 2.	Compare and contrast selected frameworks for family analysis. 3.	Define variant family forms and give examples of both traditional and non-traditional family forms. 4.	Identify families at risk and plan appropriate care. 5.	Utilize the nursing process to determine the health needs of a family. 6.	Demonstrate acceptance of nursing goal to decrease family dependency. 7.	Define and distinguish between selected terms related to crisis. 8.	Identify the steps of crisis intervention. 9.	Evaluate selected community resources and assist families to use the resources more effectively. 10.	Evaluate what factors influenced the effectiveness of the nursing intervention and which strategies proved most valuable with these individuals and families. 11.	Demonstrate the use of the nursing process in selected community health care settings.

MODULE V: EPIDEMIOLOGY AS A TOOL OF COMMUNITY HEALTH

STUDENT LEARNING OUTCOMES:

1.	Define epidemiology. 2.	Examine selected terms as they relate to epidemiology. 3.	Contrast the three major categories (agent, host, environment) in the broad conceptual framework utilized by epidemiologists, and give examples of variables from each category. 4.	Explain internal and external environmental factors (agent, host, environment) appropriately in selected problem situations. 5.	Analyze the use of health statistics as a method of appraisal to identify high-risk individuals and families in various settings. 6.	Discuss cooperative efforts among and between levels of health care delivery, with respect to epidemiology. 7.	Explain the role of professional nurse on epidemiological team. 8.	Define communicable disease. 9.	Define selected terms as they relate to communicable disease. 10.	Describe the etiology, occurrence, pathology, symptomatology, treatment, complications, and preventive measures for the following communicable diseases: Tuberculosis, Human Papillomavirus Infection (HPV). 11.	Discuss threats of and preventive measures against Influenza. 12.	Discuss effects of social environment on incidence of STD and TB 13. Identify groups at risk for specific STD and TB 14. Compare and contrast types of control and prevention measures for: STDs, TB, and other selected communicable diseases.

MODULE VI: HEALTH PROMOTION, RISK REDUCTION, and HEALTH TEACHING

STUDENT LEARNING OUTCOMES: 1.	Formulate personal statement of the meaning of “healthy.” 2.	Contrast health promotion and illness prevention. 3.	Advocate for major national policies for health promotion. 4.	Determine position of Senator or Representative from own district. 5.	Propose priority areas for nursing research for health promotion. 6.	Assume responsibilities of community/public health nurse in promoting health and reducing risk of illness in the community. 7.	Distinguish between patient education and health education. 8.	Formulate teaching strategies the community health nurse can implement with target groups in the community. 9.	 Explain factors to be assessed in the evaluation of effectiveness of health teaching.

MODULE VII: AGGREGATES AT RISK AND COMMUNITY HEALTH PROBLEMS SCHOOL NURSING VICTIMS OF ABUSE THE ELDERLY STUDENT LEARNING OUTCOMES:

1.	Discuss the profile of individuals at high risk for abuse. 2.	Identify major types of abuse and neglect. 3.	Define specific forms of abuse and neglect. 4.	Discuss the interpersonal and interpersonal dynamics of victims, perpetrators and family members in cases of abuse. 5.	Discuss the role of health systems in cases of abuse. 6.	Be aware of the responsibilities of the professional nurse in cases of abuse. 7.	Explain the high risk factors specific to adolescents. 8.	Develop an epidemiologic profile of selected chronic illnesses. 9.	Explain risk factor(s) for selected chronic illnesses. 10.	Describe process of adaptation to chronic illness. 11.	Examine primary, secondary, and tertiary prevention community health nurse actions designed to control chronic illness. 12.	Propose interventions for assisting older adults to achieve and maintain high level wellness. 13.	Analyze societal conditions which influence stress and mental illness. 14.  Analyze risk factors in mental illness. 15.	Propose primary, secondary, & tertiary community health nursing actions designed to prevent mental illness and promote mental health.

MODULE VIII: THE ENVIRONMENT and COMMUNITY HEALTH

STUDENT LEARNING OUTCOMES:

1.	Define environmental health and identify primary threats to clients. 2.	Recognize biological effects of air and water pollution. 3.	Discuss pesticides as health risks to humans. 4.	Identify sources of radiation. 5.	Discuss health effects of radioactive waste disposal. 6.	Analyze the threat of and protective measures against bioterrorism. 7.	Examine industrial chemicals which threaten the general welfare of communities. 8.	Explain the concept of environmental justice. 9.	Propose professional nurses' role in environmental health. 10.	Determine pollution levels for selected communities.

MODULE XI: SPECIAL TOPICS IN COMMUNITY HEALTH NURSING PRACTICE A. NURSING RESPONSIBILITY IN DISASTER SETTING B.  VULNERABLE POPULATIONS HOMELESS POPULATION RURAL POPULATION

STUDENT LEARNING OUTCOMES:

1.	Discuss different types and categories of disasters. 2.	Describe the role of the nurse in disaster preparedness and response. 3.	Explain common physical and psychosocial effects on disaster victims and workers. 4.	Discuss agencies that might be involved in pre-disaster planning and post-disaster response. 5.	Define the concept of vulnerability 6.	Identify factors that place individuals at risk for homelessness. 7.	Discuss demographics of the homeless population 8.	Analyze health problems common to vulnerable individuals and groups 3.	Identify HP2010 objectives related to vulnerable populations 4.	Explain barriers to care for vulnerable individuals. 5.	Identify community resources in the NW Florida region for homeless individuals. 6.	Compare homelessness in the United States with that in other countries 7.	Describe the nature of poverty and it's effect on health status and health behavior. 8.	Propose nursing activity to advocate on behalf of vulnerable populations. 9.	Contrast the terms urban, rural, frontier, metropolitan, nonmetropolitan, and Health 	Professional Shortage Area. 10.	Summarize at-risk populations in the rural areas. 11.	Analyze factors which affect accessibility, availability, and acceptability of health care services for rural residents. 12.	Examine HP2010 objectives that relate to rural health issues.

MODULE X: TRENDS AND FUTURE INFLUENCES IN COMMUNITY HEALTH NURSING

STUDENT LEARNING OUTCOMES:

1.	Identify and discuss recent trends that have implications for community health nursing practice. 2.	Distinguish among public health nursing, community health nursing, and home care nursing. 3.	Examine societal, political, economic, and familial factors that influence the practice of home care nursing and hospice nursing. 4.	Compare and contrast selected computer applications in community health nursing. 5.	Identify community committees/boards on which CHNs are active members, and on which CHNs can contribute.

Source: http://www.rodp.org/msn/syllabi/nurs4213.html

title: SH 102 Organizing for School Health


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User talk:Alok srivastava

 Hello Alok srivastava, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

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Physics/Essays/Fedosin/Neutron star

A neutron star is a type of compact star remnant that can result from the gravitational collapse of a massive star during a Type II supernova, Type Ib and Ic supernovae event. Such stars are composed almost entirely of neutrons, which are subatomic particles without electrical charge and roughly the same mass as protons. Neutron stars are very hot and are supported against further collapse because of the Pauli exclusion principle. This principle states that no two neutrons (or any other fermionic particle) can occupy the same place and quantum state simultaneously.

A typical neutron star has a mass between 1.35 and about 2.1 solar masses, with a corresponding radius of about 12 km if the Akmal-Pandharipande-Ravenhall (APR) Equation of state (EOS) is used.

A neutron star's density increases as its mass increases, and, for most Equations of State (EOS), its radius decreases in a non-linear way. For example, EOS radius predictions for a 1.35 M star are: FPS 10.8 km, UU 11.1 km, APR 12.1 km, and L 14.9 km. For a more massive 2.1 M star radius predictions are: FPS undefined, UU 10.5 km, APR 11.8 km, and L 15.1 km.

In contrast, the Sun's radius is about 60,000 times that. Neutron stars have overall densities predicted by the APR EOS of $$3.7\cdot 10^{17} $$ to $$5.9\cdot 10^{17} $$ kg/m3 (or $$2.6\cdot 10^{14} $$ to $$4.1\cdot 10^{14} $$ times the density of the Sun), which compares with the approximate density of an atomic nucleus of $$3\cdot 10^{17} $$ kg/m3.

The neutron star's density varies from below $$1\cdot 10^9 $$ kg/m3 in the crust increasing with depth to above $$6\cdot 10^{17}$$ kg/m3 or $$8\cdot 10^{17}$$ kg/m3 deeper inside.

This density is approximately equivalent to the mass of the entire human population compressed into the size of a sugar cube.

In general, compact stars of less than 1.44 solar masses, the Chandrasekhar limit, are white dwarfs; above 2 to 3 solar masses (the Tolman-Oppenheimer-Volkoff limit), a quark star might be created, however this is uncertain. Gravitational collapse will always occur on any compact star over 5 solar masses, inevitably producing a black hole or explosion of the star.

History of discoveries
The neutron subatomic particle was discovered in 1932 by Sir James Chadwick. By bombarding the hydrogen atoms in paraffin with emissions from beryllium that was itself being bombarded with alpha particles, he demonstrated that these emissions contained a neutral particle that had about the same mass as a proton. In 1935 he was awarded the Nobel Prize in Physics for this discovery.

In 1934, Walter Baade and Fritz Zwicky proposed the existence of the neutron star, only a year after Chadwick's discovery of the neutron.

Even before the discovery of neutron, in 1931, neutron stars were anticipated by Lev Landau, who wrote about stars where "atomic nuclei come in close contact, forming one gigantic nucleus" (published in 1932). However, the widespread opinion that Landau predicted neutron stars proves to be wrong.

In seeking an explanation for the origin of a supernova, they proposed that the neutron star is formed in a supernova. Supernovae are suddenly appearing dying stars in the sky, whose luminosity in the optical night outshine an entire galaxy for days to weeks. Baade and Zwicky correctly proposed at that time that the release of the gravitational binding energy of the neutron stars powers the supernova: "In the supernova process mass in bulk is annihilated". If the central part of a massive star before its collapse contains (for example) 3 solar masses, then a neutron star of 2 solar masses can be formed. The binding energy E of such a neutron star, when expressed in mass units via the mass-energy equivalence formula E = mc2, is 1 solar mass. It is ultimately this energy that powers the supernova.

In 1965, Antony Hewish and Samuel Okoye discovered "an unusual source of high radio brightness temperature in the Crab Nebula". This source turned out to be the Crab Nebula neutron star that resulted from the great supernova of 1054.

In 1967, Iosif Shklovsky examined the X-ray and optical observations of Scorpius X-1 and correctly concluded that the radiation comes from a neutron star at the stage of accretion.

In 1967, Jocelyn Bell and Antony Hewish discovered regular radio pulses from the location of the Hewish and Okoye radio source. This pulsar was later interpreted as originating from an isolated, rotating neutron star. The energy source of the pulsar is the rotational energy of the neutron star. The largest number of known neutron stars are of this type.

In 1971, Riccardo Giacconi, Herbert Gursky, Ed Kellogg, R. Levinson, E. Schreier, and H. Tananbaum discovered 4.8 second pulsations in an X-ray source in the constellation Centaurus, Cen X-3. They interpreted this as resulting from a rotating hot neutron star. The energy source is gravitational and results from a rain of gas falling onto the surface of the neutron star from a companion star or the interstellar medium.

In 1974, Antony Hewish was awarded the Nobel Prize in Physics "for his decisive role in the discovery of pulsars" without Samuel Okoye and Jocelyn Bell who shared in the discovery.

Giant nuclei
A neutron star has some of the properties of an atomic nucleus, including density, and being made of nucleons. In popular scientific writing, neutron stars are therefore sometimes described as giant nuclei. However, in other respects, neutron stars and atomic nuclei are quite different. In particular, a nucleus is held together by the strong force or strong gravitation, while a neutron star is held together by gravity. It is generally more useful to consider such objects as stars.

Examples of neutron stars

 * PSR J0108-1431 - closest neutron star
 * LGM-1 - the first recognized radio-pulsar
 * PSR B1257+12 - the first neutron star discovered with planets (a millisecond pulsar)
 * SWIFT J1756.9-2508 - a millisecond pulsar with a stellar-type companion with planetary range mass (below brown dwarf)
 * PSR B1509-58 source of the "Hand of God" photo shot by the Chandra X-ray Observatory.

title: Physics/Essays/Fedosin/Neutron star


text:
User:Meisterkoch

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User talk:Seancmcdonald

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text: Category:University of Canberra/Workshops


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User:Egm6341.s10.team3.Min/HW5

3 Find $$ {t}_{k}(x) $$
Ref: Lecture Notes [[media:Egm6341.s10.mtg27.djvu|p.27-1]]

Problem Statement
Find $$ {t}_{k}(x) $$ by the reversion of $$ x ({t}_{k}) $$

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

$$ $$
 * $$\displaystyle x(t_k)=\frac{1}{2}(x_k+x_{k+1})+t_k \frac{h}{2}
 * $$\displaystyle x(t_k)=\frac{1}{2}(x_k+x_{k+1})+t_k \frac{h}{2}
 * $$\displaystyle (Eq. 1)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle h=x_{k+1}-x_k
 * $$\displaystyle h=x_{k+1}-x_k
 * }
 * }

Solution
Revise Eq.1 and express $$t(x_k)$$by x:

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

 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |

$$\displaystyle \Rightarrow t_k(x)=\frac{2}{h}(x- \frac{x_k+x_{k+1}}{2}) $$
 * $$\displaystyle (Eq. 2)$$
 * style = |
 * }

When $$ x = x_k $$
 * {| style="width:100%" border="0" align="left"

$$ When $$ x = x_{k+1} $$
 * $$\displaystyle t_k(x_k)=\frac{2}{h}(x_k- \frac{x_k+x_{k+1}}{2})=\frac{2}{x_{k+1}-x_k}\frac{x_k-x_{k+1}}{2}=-1
 * $$\displaystyle t_k(x_k)=\frac{2}{h}(x_k- \frac{x_k+x_{k+1}}{2})=\frac{2}{x_{k+1}-x_k}\frac{x_k-x_{k+1}}{2}=-1
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle t_k(x_{k+1})=\frac{2}{h}(x_{k+1}- \frac{x_k+x_{k+1}}{2})=\frac{2}{x_{k+1}-x_k}\frac{x_{k+1}-x_k}{2}=1
 * $$\displaystyle t_k(x_{k+1})=\frac{2}{h}(x_{k+1}- \frac{x_k+x_{k+1}}{2})=\frac{2}{x_{k+1}-x_k}\frac{x_{k+1}-x_k}{2}=1

-- By Min Zhong 12:00, 23 March 2010 (UTC)

7 Cancel odd order of derivative of g in the proof of Trap error
Ref: Lecture Notes [[media:Egm6341.s10.mtg28.djvu|p.28-2]]

Problem Statement
Redo steps in proof of Trap. error by trying to cancel terms with odd order of derivative of g. From Lecture Notes [[media:Egm6341.s10.mtg21.djvu|p.21-3]]
 * {| style="width:100%" border="0" align="left"

$$ $$ Where
 * $$\displaystyle E= \left[ P_2(t)g^{(1)}(t)\right]_{-1}^{+1}-\left[ P_3(t)g^{(2)}(t)\right]_{-1}^{+1}+ \int_{-1}^{+1} P_3(t)g^{(3)}(t)dt
 * $$\displaystyle E= \left[ P_2(t)g^{(1)}(t)\right]_{-1}^{+1}-\left[ P_3(t)g^{(2)}(t)\right]_{-1}^{+1}+ \int_{-1}^{+1} P_3(t)g^{(3)}(t)dt
 * $$\displaystyle (Eq. 1)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ $$
 * $$\displaystyle P_2(t)=-\frac{t^2}{2}+ \alpha
 * $$\displaystyle P_2(t)=-\frac{t^2}{2}+ \alpha
 * $$\displaystyle (Eq. 2)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle P_3(t)=-\frac{t^3}{6}+ \alpha t+ \beta
 * $$\displaystyle P_3(t)=-\frac{t^3}{6}+ \alpha t+ \beta

$$
 * }
 * }

$$ \alpha $$ and $$\beta$$ are the integration constants

Solution
We start the proof from Equation 1. i) Step 1 we try to cancel the term contains first order of derivative of g in Eq. 1 by having

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

$$ Then
 * $$\displaystyle \left[ P_2(t)g^{(1)}(t)\right]_{-1}^{+1}=0
 * $$\displaystyle \left[ P_2(t)g^{(1)}(t)\right]_{-1}^{+1}=0
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ From Equation 2.
 * $$\displaystyle P_2(+1)= P_2(-1)=0
 * $$\displaystyle P_2(+1)= P_2(-1)=0
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ Summary:
 * $$\displaystyle P_2(+1)=-\frac{1^2}{2}+ \alpha \Rightarrow \alpha=\frac{1}{2}
 * $$\displaystyle P_2(+1)=-\frac{1^2}{2}+ \alpha \Rightarrow \alpha=\frac{1}{2}
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle P_2(t)=-\frac{t^2}{2}+ \frac{1}{2}
 * $$\displaystyle P_2(t)=-\frac{t^2}{2}+ \frac{1}{2}
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle E= -\left[ P_3(t)g^{(2)}(t)\right]_{-1}^{+1}+ \underbrace {\int_{-1}^{+1} P_3(t)g^{(3)}(t)dt}_{A}
 * $$\displaystyle E= -\left[ P_3(t)g^{(2)}(t)\right]_{-1}^{+1}+ \underbrace {\int_{-1}^{+1} P_3(t)g^{(3)}(t)dt}_{A}
 * }
 * }

From Lecture Notes [[media:Egm6341.s10.mtg26.djvu|p.26-2]], The result of integration by part of A is

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

$$
 * $$\displaystyle A= \left[ P_4(t)g^{(3)}(t)\right]_{-1}^{+1}-\int_{-1}^{+1} P_4(t)g^{(4)}(t)dt
 * $$\displaystyle A= \left[ P_4(t)g^{(3)}(t)\right]_{-1}^{+1}-\int_{-1}^{+1} P_4(t)g^{(4)}(t)dt

 * }
 * }

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

$$ $$ Where
 * $$\displaystyle E= \left[ P_3(t)g^{(2)}(t)\right]_{-1}^{+1}+ \left[ P_4(t)g^{(3)}(t)\right]_{-1}^{+1}-\int_{-1}^{+1} P_4(t)g^{(4)}(t)dt
 * $$\displaystyle E= \left[ P_3(t)g^{(2)}(t)\right]_{-1}^{+1}+ \left[ P_4(t)g^{(3)}(t)\right]_{-1}^{+1}-\int_{-1}^{+1} P_4(t)g^{(4)}(t)dt
 * $$\displaystyle (Eq. 3)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle P_3(t)=-\frac{t^3}{6}+ \alpha t+ \beta = -\frac{t^3}{6}+ \frac{1}{2}t+ \beta
 * $$\displaystyle P_3(t)=-\frac{t^3}{6}+ \alpha t+ \beta = -\frac{t^3}{6}+ \frac{1}{2}t+ \beta
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$ $$ where $$\omega$$ is the integration constant
 * $$\displaystyle P_4(t)= \int P_3= -\frac{t^4}{24}+ \frac{1}{4}t^2+ \beta t + \omega
 * $$\displaystyle P_4(t)= \int P_3= -\frac{t^4}{24}+ \frac{1}{4}t^2+ \beta t + \omega
 * $$\displaystyle (Eq. 4)
 * }
 * }

ii) Step 2

we try to cancel the term contains third order of derivative of g in Eq. 3

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

$$
 * $$\displaystyle \left[ P_4(t)g^{(3)}(t)\right]_{-1}^{+1}=0
 * $$\displaystyle \left[ P_4(t)g^{(3)}(t)\right]_{-1}^{+1}=0
 * }
 * }

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

$$ From Equation 4, then
 * $$\displaystyle P_4(+1)= P_4(-1)=0
 * $$\displaystyle P_4(+1)= P_4(-1)=0
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle P_4(+1)= -\frac{1}{24}+ \frac{1}{4}+ \beta  + \omega =0
 * $$\displaystyle P_4(+1)= -\frac{1}{24}+ \frac{1}{4}+ \beta  + \omega =0
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$ Solve the above two equations, we can find
 * $$\displaystyle P_4(-1)= -\frac{1}{24}+ \frac{1}{4}- \beta  + \omega =0
 * $$\displaystyle P_4(-1)= -\frac{1}{24}+ \frac{1}{4}- \beta  + \omega =0
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle  \beta =0
 * $$\displaystyle  \beta =0
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle  \omega = -\frac{5}{24}
 * $$\displaystyle  \omega = -\frac{5}{24}
 * }
 * }

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

$$
 * $$\displaystyle P_3(t)= -\frac{t^3}{6}+ \frac{1}{2}t
 * $$\displaystyle P_3(t)= -\frac{t^3}{6}+ \frac{1}{2}t
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle P_4(t)=-\frac{t^4}{24}+ \frac{1}{4}t^2 -\frac{5}{24}
 * $$\displaystyle P_4(t)=-\frac{t^4}{24}+ \frac{1}{4}t^2 -\frac{5}{24}
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$ $$ From Lecture Notes [[media:Egm6341.s10.mtg26.djvu|p.26-2]], The result of integration by part of B is
 * $$\displaystyle E= -\left[ P_3(t)g^{(2)}(t)\right]_{-1}^{+1}- \underbrace{\int_{-1}^{+1} P_4(t)g^{(4)}(t)dt}_{B}
 * $$\displaystyle E= -\left[ P_3(t)g^{(2)}(t)\right]_{-1}^{+1}- \underbrace{\int_{-1}^{+1} P_4(t)g^{(4)}(t)dt}_{B}
 * $$\displaystyle (Eq. 5)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle B= \left[ P_5(t)g^{(4)}(t)\right]_{-1}^{+1}-\int_{-1}^{+1} P_5(t)g^{(5)}(t)dt
 * $$\displaystyle B= \left[ P_5(t)g^{(4)}(t)\right]_{-1}^{+1}-\int_{-1}^{+1} P_5(t)g^{(5)}(t)dt

$$
 * }
 * }

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

 * $$\displaystyle E= -\left[ P_3(t)g^{(2)}(t)\right]_{-1}^{+1}-\left[ P_5(t)g^{(4)}(t)\right]_{-1}^{+1}+\int_{-1}^{+1} P_5(t)g^{(5)}(t)dt
 * $$\displaystyle E= -\left[ P_3(t)g^{(2)}(t)\right]_{-1}^{+1}-\left[ P_5(t)g^{(4)}(t)\right]_{-1}^{+1}+\int_{-1}^{+1} P_5(t)g^{(5)}(t)dt

$$
 * }
 * }

iii) Step 3

We can continually cancel the items with odd order derivative of g and find the general expression of E as

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

$$ $$
 * $$\displaystyle E= -\left[ \underbrace{P_3(t)g^{(2)}(t)}_{C}+P_5(t)g^{(4)}(t)+ P_7(t)g^{(6)}(t)+...+P_{2l+1}(t)g^{(2l)}(t)\right]_{-1}^{+1} - \int_{-1}^{+1} P_{2l}(t)g^{(2l+1)}(t)dt
 * $$\displaystyle E= -\left[ \underbrace{P_3(t)g^{(2)}(t)}_{C}+P_5(t)g^{(4)}(t)+ P_7(t)g^{(6)}(t)+...+P_{2l+1}(t)g^{(2l)}(t)\right]_{-1}^{+1} - \int_{-1}^{+1} P_{2l}(t)g^{(2l+1)}(t)dt
 * $$\displaystyle (Eq. 6)
 * }
 * }

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

$$ Since $$P_3(t) and P_{3i}(t) $$ are odd function, Then
 * $$\displaystyle C = \left[ P_3(t)g^{(2)}(t)\right]_{-1}^{+1}=P_3(+1)g^{(2)}(+1)-P_3(-1)g^{(2)}(-1)
 * $$\displaystyle C = \left[ P_3(t)g^{(2)}(t)\right]_{-1}^{+1}=P_3(+1)g^{(2)}(+1)-P_3(-1)g^{(2)}(-1)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle P_3(+1)=-P_3(-1)
 * $$\displaystyle P_3(+1)=-P_3(-1)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle C = - P_3(-1)\left[(g^{(2)}(+1)+g^{(2)}(-1))\right]
 * $$\displaystyle C = - P_3(-1)\left[(g^{(2)}(+1)+g^{(2)}(-1))\right]
 * }
 * }

Thus Equation 6 can be simplified as follows:

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

$$\displaystyle \Rightarrow  E= \sum_{r=1}^{l}P_{2r+1}(-1) \left[g^{2r}(+1)+g^{2r}(-1) \right] - \int_{-1}^{+1} P_{2l+1}(t)g^{(2l+1)}(t)dt $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * $$\displaystyle (Eq. 7)$$
 * style = |
 * }
 * }

iv) Step 4 We use the same techniqure as HW 5 problem 5 to find the composite trap error $$E_n^1$$ through Equation 7

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

$$\displaystyle E_n^1 = \sum_{r=1}^{l} \sum_{k}^{n-1}P_{2r+1}(-1) \left[g_k^{2r}(+1)+g_k^{2r}(-1) \right]- \sum_{k=0}^{n-1} \int_{-1}^{1}P_{2l+1}(t)g_k^{2l+1}(t)dt $$ From the lecture note [[media:Egm6341.s10.mtg21.djvu|p.21-2]], $$\displaystyle g_k(t)$$can be changed to $$\displaystyle f_(x_k)$$ and below was obtained.
 * $$\displaystyle (Eq. 8) $$
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\begin{align} &g_k^{(2r)}(+1)=\left(\frac{h}{2}\right)^{2r}f^{(2r)}(x_{k+1}),\\ &g_k^{(2r)}(-1)=\left(\frac{h}{2}\right)^{2r}f^{(2r)}(x_{k}),\\ &g_k^{(2l+1)}(t)=\left(\frac{h}{2}\right)^{2l+1}f^{(2l+1)}(x_{t}) \end{align} $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

Plug the above equations into Equation 8. We have
 * {| style="width:100%" border="0" align="left"

$$\displaystyle E_n^1 = \sum_{r=1}^{l} \sum_{k}^{n-1}\left(\frac{h}{2}\right)^{2r+1}P_{2r+1}(-1) \left[f^{2r}(x_{k+1})+f^{2r}(x_{k})\right] - \left(\frac{h}{2}\right)^{2r+2}\sum_{k=0}^{n-1} \int_{x_k}^{x_{k+1}}P_{2l+1}(t)f_{2l+1}(x_t)dx $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * $$\displaystyle (Eq. 9)$$
 * style = |
 * }
 * }

When we derive the composite trap error, we use the simple trap error which contains additive term $$ g^{2r}(+1)+g^{2r}(-1) $$. Because of the additive sign, no term could be canceled in the summation of $$E_n^1$$. We have to do $$\sum_{k}^{n-1}\left[f^{2r}(x_{k+1})+f^{2r}(x_{k})\right]$$ when evalue the error. It is very complicated compared with that derived by canceling items with even oder derivative of g. In that case, we can get simple trap with item $$ g^{2r-1}(+1)-g^{2r-1}(-1) $$ which can simplify the expression of composite trap error by cancel same term when summation and result in $$[f^{(2r-1)}(x_{n})-f^{(2r-1)}(x_0)$$. Thus Trap error proofed by canceling terms with the even order derivative of g is more effective. -- By Min Zhong 12:00, 23 March 2010 (UTC)

title: User:Egm6341.s10.team3.Min/HW5


text:
User talk:Albane69

Hello, Albane, and welcome to the English Wikiversity! How many students do you think you would be comfortable taking on? The Jade Knight (d'viser) 17:59, 23 March 2010 (UTC)

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User:LarissaYochana

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title: User:LarissaYochana


text:
User:Egm6341.s2010.Team1/HW5

=Problem 1 Find out the constants of integration function $$P_{n}(t)$$=

Statement
Find $$C_{5}\,$$ and $$C_{6}\,$$ Given $$P_{5}(t)= \!\,\int P_{4}(t)dt \!\,= \!\,-\!\,\frac{t^{5}}{120}+ \!\,\frac{t^{3}}{12}+\!\,C_{5}t+\!\,C_{6}$$

 * {| style="width:100%" border="1" align="left"

 * Ref: Lecture Notes [[media:Egm6341.s10.mtg26.djvu|P26-3]]
 * }
 * }

Solution
(i)Select $$P_{5} :\!\,P_{5}(\pm \!\,1)=\!\,0$$ and $$P_{5} (0)=\!\,0$$

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

P_{5}(0)=0 $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }
 * {| style="width:100%" border="0" align="left"

\Rightarrow-\frac{0^{5}}{120}+\frac{0^{3}}{36}+C_{5}0+C_{6} = 0 $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }
 * {| style="width:100%" border="0" align="left"

\Rightarrow\,C_{6} = 0 $$ (iii) $$P_{5}(+ \!\,1)=\!\,0$$
 * $$ \displaystyle
 * $$ \displaystyle
 * }
 * {| style="width:100%" border="0" align="left"

\Rightarrow-\frac{1^{5}}{120}+\frac{1^{3}}{36}+C_{5}1 = 0 $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }
 * {| style="width:100%" border="0" align="left"

\Rightarrow-\frac{3}{360}+\frac{10}{360}+C_{5} = 0 $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }
 * {| style="width:100%" border="0" align="left"

\Rightarrow C_{5} = -\frac{7}{360} $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }

Solved by : EGM6341.s10.Team1.Ya-Chiao Chang 19:09, 23 March 2010 (UTC)

=Problem 2-Pf of Trap.error =

Find
Do steps 4ab to find $$ p_{6}(t),p_{7}(t)\,$$

Given
[[media:Egm6341.s10.mtg26.djvu|p.26-1]],[[media:Egm6341.s10.mtg26.djvu|p.26-2]],[[media:Egm6341.s10.mtg26.djvu|p.26-3]]

Solution
Step 4a [[media:Egm6341.s10.mtg26.djvu|p.26-3]]

$$ E= \,[\,p_{2}(t)g^{(1)}(t)+p_{4}(t)g^{(3)(t)}\,]_{-1}^{+1}\,-\,\underbrace{\left (\int_{-1}^{+1}p_{5}(t)g^{(5)}(t)dt\right)}_{B}\,$$

$$ B=\,[\,p_{6}(t)g^{(5)}(t)\,]_{-1}^{+1}\,-\,\underbrace{\left (\int_{-1}^{+1}p_{6}(t)g^{(6)}(t)dt\right)}_{C}\,$$

$$ p_{6}(t)=\int p_{5}(t)dt = \int -\frac{t^{5}}{120}+\frac{t^{3}}{36}-\frac{7t}{360}dt \,$$

$$ =-\frac{t^{6}}{720}+\frac{t^{4}}{144}-\frac{7t^{2}}{720}+\alpha \,$$

Step 4b [[media:Egm6341.s10.mtg26.djvu|p.26-2]]

$$ C=\,[\,p_{7}(t)g^{(6)}(t)\,]_{-1}^{+1}\,-\,\int_{-1}^{+1}p_{7}(t)g^{(7)}(t)dt\,$$

$$ p_{7}(t)=\int p_{6}(t)dt = \int -\frac{t^{6}}{720}+\frac{t^{4}}{144}-\frac{7t^{2}}{720}+\alpha dt \,$$

$$ =-\frac{t^{7}}{5040}+\frac{t^{5}}{720}-\frac{7t^{3}}{2160}+\alpha t+\beta \,$$

Select $$ p_{7}(t)\,$$ st $$ p_{7}(+1)=0,p_{7}(-1)=0,p_{7}(0)=0 \, $$

$$ \Rightarrow \beta=0, \alpha=\frac{31}{15120}\, $$

Summary: steps 4ab

$$ p_{6}(t)=-\frac{t^{6}}{720}+\frac{t^{4}}{144}-\frac{7t^{2}}{720}+\frac{31}{15120} \,$$

$$ p_{7}(t)=-\frac{t^{7}}{5040}+\frac{t^{5}}{720}-\frac{7t^{3}}{2160}+\frac{31t}{15120}\,$$
 * {| style="width:100%" border="0" align="left"

 * style="width:50%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle
 * $$ \displaystyle

p_{6}(t)=-\frac{t^{6}}{720}+\frac{t^{4}}{144}-\frac{7t^{2}}{720}+\frac{31}{15120}, p_{7}(t)=-\frac{t^{7}}{5040}+\frac{t^{5}}{720}-\frac{7t^{3}}{2160}+\frac{31t}{15120}\,$$ $$
 * $$\displaystyle
 * }
 * }

Solved by :Egm6341.s10.team1.lei00:53, 24 March 2010 (UTC)

Reviewed by : EGM6341.s10.Team1.Ya-Chiao Chang 19:20, 23 March 2010 (UTC)

= Problem 3: Change dependent variable of a function =

Given
Given the equation

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle x(t)=t\dfrac{h}{2}+\dfrac{x_k+x_{k+1}}{2}
 * }.
 * }.

Find
Solve the above equation so that t is the dependent variable

Solution
It is simple enough to solve this equation for t giving

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle t(x)=\dfrac{2x}{h}-\dfrac{x_k+x_{k+1}}{h}
 * }.
 * }.

Thus because

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle h=x_{k+1}-x_k
 * }.
 * }.

When

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle x=x_k, t=-1
 * }.
 * }.

and when

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

$$ $$ Solved by: --Egm6341.s10.team1.toddmock 21:00, 23 March 2010 (UTC) Reviewed by: Egm6341.s10.team1.andrewdugan 08:19, 24 March 2010 (UTC)
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle x=x_{k+1}, t=1
 * }.
 * }.

=Problem 4: Calculate d1, d2, d3=

Find
$$ d_1=\overline{d_{2}}, d_2=\overline{d_{4}}, d_3=\overline{d_{6}}\,$$

Given
$$ d_1=\overline{d_{2}}=\frac{P_2^{(1)}}{2^2}\,$$ $$ d_2=\overline{d_{4}}=\frac{P_4^{(1)}}{2^4}\,$$ $$ d_3=\overline{d_{6}}=\frac{P_6^{(1)}}{2^6}\,$$

Solution
$$ P_2=C_1\frac{t^2}{2!}+C_3\,$$ $$ P_4=C_1\frac{t^4}{4!}+C_3\frac{t^2}{2!}+C_5\,$$ $$ P_6=C_1\frac{t^6}{6!}+C_3\frac{t^4}{4!}+C_5\frac{t^2}{2!}+C_7\,$$ where $$ C_1=-1, C_3=\frac{1}{6}, C_5=\frac{-7}{360}, C_7=\frac{31}{15120} \,$$ $$ P_2(1)=(-1)(\frac{1}{2})+\frac{1}{6}=\frac{-1}{3}\Rightarrow\overline{d_{2}}=(\frac{-1}{3})(\frac{1}{4})=\frac{-1}{12}\,$$ $$ P_4(1)=(-1)(\frac{1}{24})+(\frac{1}{6})(\frac{1}{2})-\frac{7}{360}=\frac{1}{45}\Rightarrow\overline{d_{4}}=(\frac{1}{45})(\frac{1}{16})=\frac{1}{720}\,$$ $$ P_6(1)=(-1)(\frac{1}{720})+(\frac{1}{6})(\frac{1}{24})+(\frac{-7}{360})(\frac{1}{2})+\frac{31}{15120}=\frac{-2}{945}\Rightarrow\overline{d_{6}}=(\frac{-2}{945})(\frac{1}{64})=\frac{-1}{30240}\,$$
 * Note: Coefficient $$ C_7 \,$$ can be found using the same method as in Problem 1.

These values are the same as those obtained using the Bernoulli numbers. Solved by: Egm6341.s10.team1.andrewdugan 08:17, 24 March 2010 (UTC)

=Problem 5-Deriv.Trap.rule error =

Find
Deriv.(1)[[media:Egm6341.s10.mtg27.djvu|p.29-1]]

Given
(4)[[media:Egm6341.s10.mtg21.djvu|p.21-2]]

$$ E= \int_{-1}^{+1}p_{1}(t)g^{(1)}(t)dt = [P_2(t)g^{(1)}(t)]_{-1}^{+1}-\int_{-1}^{+1}P_2(t)g^{(2)}(t)dt \,$$

Solution
From the (4)[[media:Egm6341.s10.mtg21.djvu|p.21-2]],we can get the generalized form:

$$ E= \int_{-1}^{+1}p_{1}(t)g^{(1)}(t)dt \,$$

$$ =[p_{2}g^{1}+p_{4}g^{3}+...+p_{2l}g^{(2l-1)}]_{-1}^{+1}-\int_{-1}^{+1}p_{2l}(t)g^{(2l)}(t)dt \,$$

Recall that

$$ [p_{2}(t)g^{1}]_{-1}^{+1}=p_{2}(+1)g^{1}(+1)-p_{2}(-1)g^{1}(-1)\,$$

where $$ p_{2}\,$$ is even func.

$$\Rightarrow p_{2}(-1)=p_{2}(+1)\, $$

thus

$$ [p_{2}(t)g^{1}]_{-1}^{+1}=p_{2}(+1)[g^{1}(+1)-g^{1}(-1)]\, $$

and $$ p_{2i}\,$$ is even func.

$$\Rightarrow E=\underbrace{\left (\sum_{r=1}^{l} p_{2r}(+1)[g^{2r-1}(+1)-g^{2r-1}(-1)]\right)}_{A}-\underbrace{\left (\int_{-1}^{+1}p_{2l}(t)g^{2l}(t)dt\right)}_{B}\, $$

next, we should change varible t into x

from (3) [[media:Egm6341.s10.mtg21.djvu|p.21-2]]

$$ g_{k}^{(i)}(t)=(\frac{h}{2})^{i}f^{(i)}(x(t)), [\,x_k,x_{k+1}\,]\,$$

$$ \Rightarrow g^{(2r-1)}(+1)=(\frac{h}{2})^{2r-1}f^{(2r-1)}(x_{k+1})\, $$

$$ \Rightarrow g^{(2r-1)}(-1)=(\frac{h}{2})^{2r-1}f^{(2r-1)}(x_{k})\, $$

Recall (3) [[media:Egm6341.s10.mtg27.djvu|p.27-1]]

$$ \overline{d_{2r}} = \frac{p_{2r}(+1)}{2^{2r}}\, $$

$$ \Rightarrow p_{2r}(+1)=\overline {d_{2r}}*2^{2r}\, $$

thus $$ A \,$$ can be changed into

$$ A=\sum_{r=1}^{l} 2\overline{d_{2r}} h^{2r-1}[f^{2r-1}(x_{k+1})-f^{2r-1}(x_{k})]\,$$

from (3) [[media:Egm6341.s10.mtg21.djvu|p.21-2]]

$$ g_{k}^{(i)}(t)=(\frac{h}{2})^{i}f^{(i)}(x(t)), [\,x_k,x_{k+1}\,]\,$$

$$ g_{k}^{(2l)}(t)=(\frac{h}{2})^{2l}f^{(2l)}(x)\, $$

thus $$ B\, $$ can be changed into

$$ B=\frac{h^{2l}}{2^{2l}}\int_{x_k}^{x_{k+1}}p_{2l}(t_{k}(x))f^{(2l)}(x)(\frac{2}{h})dx \, $$

Recall trap.rule error is in (5)[[media:Egm6341.s10.mtg21.djvu|P21-1]]

$$ E_{n}^{1}=\frac{h}{2}\sum_{k=0}^{n-1}[\int_{-1}^{+1}g_{k}(t)dt-(g_{k}(-1)+g_{k}(+1))]\,$$

Plug A and B into the equation $$\displaystyle E_n^1$$
 * {| style="width:100%" border="0" align="left"

\begin{align} E^1_n & = \left(\frac{h}{2}\right)\sum_{r=1}^{\ell} p_{2r}(+1) \left(\frac{h}{2}\right)^{2r-1} \sum_{k=0}^{n-1} \Big[f^{(2r-1)}(x_{k+1})-f^{(2r-1)}(x_k)\Big]- \frac{h}{2}\sum_{k=0}^{n-1} \left[   \int_{x_k}^{x_{k+1}}p_{2l}(t_k(x))\left(\frac{h}{2}\right)^{2\ell}f^{(2\ell)}(x)\frac{2}{h} dx \right]\\ &= \sum_{r=1}^{\ell} h^{2r} \frac{p_{2r}(+1)}{2^{2r}} \Big[f^{(2r-1)}(x_{n})-f^{(2r-1)}(x_0)\Big]-  \left(\frac{h}{2}\right)^{2\ell} \sum_{k=0}^{n-1} \left[ \int_{x_k}^{x_{k+1}}p_{2\ell}(t_k(x))f^{(2\ell)}(x) dx \right] \end{align} $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

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

E^1_n = \sum_{r=1}^{\ell} h^{2r} \overline{d}_{2r} \Big[f^{(2r-1)}(b)-f^{(2r-1)}(a)\Big]-  \left(\frac{h}{2}\right)^{2\ell} \sum_{k=0}^{n-1} \left[ \int_{x_k}^{x_{k+1}}p_{2\ell}(t_k(x))f^{(2\ell)}(x) dx \right] $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Solved by : Egm6341.s10.team1.lei00:54, 24 March 2010 (UTC)

Reviewed by : EGM6341.s10.Team1.Ya-Chiao Chang 19:20, 23 March 2010 (UTC)

= Problem 6: Taylor series expansion for Bernoulli numbers =

Given
Note that this function contradicts the one given in the notes, however it has been verified through added manipulation of equation 6 on wolfram

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \ x\coth(x)= \sum_{r=0}^{\infty} \frac{2^{2r}B_{2r}}{(2r)!}x^{2r}
 * }
 * }

This equation can be used to obtain the Bernoulli numbers as well as the term

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \ -d_{2r}=\frac{B_{2r}}{(2r)!}
 * }
 * }

Find
Using the above relations obtain the Richardson Extrapolation Coefficients
 * {| style="width:100%" border="0" align="left"

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \ d_{2},d_{4},d_{6},d_{8},d_{10}
 * }
 * }

Solution
The hyperbolic cotangent function can be defined as

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \ \coth(x)=\frac{\cosh(x)}{\sinh(x)}=\frac{e^x+e^{-x}}{e^x-e^{-x}}
 * }
 * }

Furthermore we can write these terms as a series of summations

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \ e^x=\sum_{j=0}^{\infty}\frac{x^j}{j!}
 * }
 * }

and

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \ e^{-x}=\sum_{j=0}^{\infty}(-1)^j\frac{x^j}{j!}
 * }
 * }

Thus giving

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \ e^x+e^{-x}=\sum_{j=0}^{\infty}\frac{x^j}{j!}(1+(-1)^j)
 * }
 * }

and

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \ e^x-e^{-x}=\sum_{k=0}^{\infty}\frac{x^k}{k!}(1-(-1)^k)
 * }
 * }

or

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \ x\coth(x)=\dfrac{\sum_{j=0}^{\infty}\dfrac{x^{j+1}}{j!}(1+(-1)^j)}{\sum_{k=j+1}^{\infty}\dfrac{x^k}{k!}(1-(-1)^k)}
 * }
 * }

With these formulations we can write the series out so that we can identify the Bernoulli numbers and the extrapolation terms

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \ x\coth(x)=1+\dfrac{x^2}{3}-\dfrac{x^4}{45}+\dfrac{2x^6}{945}-\dfrac{x^8}{4725}+\dfrac{2x^{10}}{93555}
 * }
 * }

Thus pulling out the Bernoulli numbers

we can calculate the extrapolation coefficients to be

Solved by:--Egm6341.s10.team1.toddmock 21:01, 23 March 2010 (UTC) Reviewed by: Egm6341.s10.team1.andrewdugan 08:20, 24 March 2010 (UTC)

=Problem 7: Compare values of $$ \overline{d_{2r}}\,$$=

Find
Verify that the values for $$ \overline{d_{2}}, \overline{d_{4}}, \overline{d_{6}}, \overline{d_{8}}, \overline{d_{10}}\,$$ are the same whether calculated by the $$ xcoth \,$$ method or the $$ \frac{P_{2r}(1)}{2^{2r}}\,$$ method.

Given
The results calculated from the $$ xcoth \,$$ method in Problem 6 are: $$\overline{d_{2}}=\frac{-1}{12}, \overline{d_{4}}=\frac{1}{720}, \overline{d_{6}}=\frac{-1}{30240}, \overline{d_{8}}=\frac{1}{1209600}, \overline{d_{10}}=\frac{-1}{47900160} \,$$ Using the $$ \frac{P_{2r}(1)}{2^{2r}}\,$$ method, remember that: $$\overline{d_{2}}=\frac{P_2^{(1)}}{2^2}\,$$ $$\overline{d_{4}}=\frac{P_4^{(1)}}{2^4}\,$$ $$\overline{d_{6}}=\frac{P_6^{(1)}}{2^6}\,$$ $$\overline{d_{8}}=\frac{P_8^{(1)}}{2^8}\,$$ $$\overline{d_{10}}=\frac{P_{10}^{(1)}}{2^{10}}\,$$

Solution
$$ P_2=C_1\frac{t^2}{2!}+C_3\,$$ $$ P_4=C_1\frac{t^4}{4!}+C_3\frac{t^2}{2!}+C_5\,$$ $$ P_6=C_1\frac{t^6}{6!}+C_3\frac{t^4}{4!}+C_5\frac{t^2}{2!}+C_7\,$$ $$ P_8=C_1\frac{t^8}{8!}+C_3\frac{t^6}{6!}+C_5\frac{t^4}{4!}+C_7\frac{t^2}{2!}+C_9\,$$ $$ P_{10}=C_1\frac{t^{10}}{10!}+C_3\frac{t^8}{8!}+C_5\frac{t^6}{6!}+C_7\frac{t^4}{4!}+C_9\frac{t^2}{2!}+C_{11}\,$$ where $$ C_1=-1, C_3=\frac{1}{6}, C_5=\frac{-7}{360}, C_7=\frac{31}{15120}, C_9=\frac{1}{4725}, C_{11}=\frac{73}{3421440} \,$$
 * Note: Coefficients $$ C_7, C_9, C_{11} \,$$ can be found using the same method as in Problem 1.

$$ P_2(1)=(-1)(\frac{1}{2})+\frac{1}{6}=\frac{-1}{3}\Rightarrow\overline{d_{2}}=(\frac{-1}{3})(\frac{1}{4})=\frac{-1}{12}\,$$ $$ P_4(1)=(-1)(\frac{1}{24})+(\frac{1}{6})(\frac{1}{2})-\frac{7}{360}=\frac{1}{45}\Rightarrow\overline{d_{4}}=(\frac{1}{45})(\frac{1}{16})=\frac{1}{720}\,$$ $$ P_6(1)=(-1)(\frac{1}{720})+(\frac{1}{6})(\frac{1}{24})+(\frac{-7}{360})(\frac{1}{2})+\frac{31}{15120}=\frac{-2}{945}\Rightarrow\overline{d_{6}}=(\frac{-2}{945})(\frac{1}{64})=\frac{-1}{30240}\,$$ $$ P_8(1)=(-1)(\frac{1}{40320})+(\frac{1}{6})(\frac{1}{720})+(\frac{-7}{360})(\frac{1}{24})+(\frac{31}{15120})(\frac{1}{2})-\frac{127}{604800}=\frac{1}{4725}\Rightarrow\overline{d_{8}}=(\frac{1}{4725})(\frac{1}{256})=\frac{1}{1209600}\,$$ $$ P_{10}(1)=(-1)(\frac{1}{3628800})+(\frac{1}{6})(\frac{1}{40320})+(\frac{-7}{360})(\frac{1}{720})+(\frac{31}{15120})(\frac{1}{24})-(\frac{127}{604800})(\frac{1}{2})+\frac{73}{3421440}=\frac{-2}{93555}\Rightarrow\overline{d_{8}}=(\frac{-2}{93555})(\frac{1}{1024})=\frac{-1}{47900160}\,$$

Solved by: -Egm6341.s10.team1.andrewdugan 08:17, 24 March 2010 (UTC)

=Problem 8 Redo steps in proof of Trapezoidal error=

Statement
Redo steps in proof of Trapezoidal error by try to cancel terms with odd order of derivative of g

 * {| style="width:100%" border="1" align="left"

 * Ref: Lecture Notes [[media:Egm6341.s10.mtg28.djvu|P28-4]]
 * }
 * }

Solution
From

 * {| style="width:100%" border="1" align="left"

 * Ref: Lecture Notes [[media:Egm6341.s10.mtg26.djvu|P26-1]] [[media:Egm6341.s10.mtg21.djvu|P21-3]]
 * }
 * }

\begin{align} \\E &= \int_{+1}^{-1}\underbrace{(-t)}_{P_{1}(t)}g^{1}(t)dt\; \\ &= [\!\,P_{2}(t)g^{1}(t)]\!\,_{-1}^{+1}-\underbrace{\int_{-1}^{+1}P_{2}(t)g^{2}(t)dt}_{A}\,. \end{align} $$

In order to cancel odd order of derivative of g, select $$P_{2}(t) = 0 \,\,at\,\, t=\pm\!\,1$$

\begin{align} \\P_{2}(t) &= \int P_{1}(t)dt\; \\ &= -C_{1}\frac{t^{2}}{2}+C_{3}\,where C_{2} = 0\; \\ &= -\frac{t^{2}}{2}+\alpha\!\,\; \end{align} $$ $$P_{2}(t) = 0 \,\,at\,\, t=1$$

$$\rightarrow \alpha\!\,=\frac{1}{2}$$

$$\rightarrow C_{1} = -1\,\,C_{3} = \frac{1}{2}$$

$$ A = [P_{3}(t)g^{2}(t)]_{-1}^{+1} + \underbrace{\int_{-1}^{+1}\underbrace{P_{3}(t)}_{U^{'}}\underbrace{g^{3}(t)}_{V}dt}_{B}$$

$$ B = [UV]_{-1}^{+1} - \int UV^{'} = [P_{4}(t)g^{3}(t)]_{-1}^{+1} - \underbrace{\int_{-1}^{+1}P_{4}(t)g^{4}(t)dt}_{C}$$

In order to cancel odd order of derivative of g, select $$P_{2}(t) = 0 \,\,at\,\, t=\pm\!\,1$$

$$ t = 1 $$

\begin{align} \\P_{4}(1) &= \int P_{3}(t)dt\; \\ &= -\frac{1}{4!\!\,}+\frac{1}{4}+C_{4}+C_{5}\; \\ &= 0\; \end{align} $$

$$\rightarrow C_{4}+C_{5} = \frac{1}{24}-\frac{1}{4} = -\frac{5}{24}$$

$$ t = -1 $$

\begin{align} \\P_{4}(-1) &= \int P_{3}(t)dt\; \\ &= -\frac{1}{4!\!\,}+\frac{1}{4}-C_{4}+C_{5}\; \\ &= 0\; \end{align} $$

$$\rightarrow -C_{4}+C_{5} = \frac{1}{24}-\frac{1}{4} = -\frac{5}{24}$$

$$ C_{5} = -\frac{5}{24}\,\,C_{4} = 0 $$

$$ C = [P_{5}(t)g^{4}(t)]_{-1}^{+1} - \underbrace{\int_{-1}^{+1} P_{5}(t)g^{5}(t)dt}_{D}$$

$$ D = \underbrace{[P_{6}(t)g^{5}(t)]_{-1}^{+1}}_{= 0} - \int_{-1}^{+1} P_{6}(t)g^{6}(t)dt$$

$$ E = -[P_{3}(t)g^{2}(t)+P_{5}(t)g^{4}(t)+...+P_{2l+1}(t)g^{2l}(t)]_{t = -1}^{t = +1} - \int_{-1}^{+1}P_{2l+2}(t)g^{2l+2}(t)dt$$

$$ P_{3}(t) $$ is odd function $$ \rightarrow P_{3}(-1) = -P_{3}(1)$$

$$ [\,P_{3}(t)g^{2}(t)]_{-1}^{1} = P_{3}(1)g^{2}(1) - P_{3}(-1)g^{2}(-1) = P_{3}(1)[g^{2}(1) + g^{2}(-1)]$$

$$ E = -\sum_{r = 1}^{l}P_{2r+1}(1)[g^{2r}(1) + g^{2r}(-1)] - \int_{-1}^{+1}P_{2l+2}(t)g^{2l+2}(t)dt$$

From [[media:Egm6341.s10.mtg21.djvu|p.21-2]] $$ g_{k}^{i} = (\frac{h}{2})^{i}f^{i}(x(t))\,\,x\in\!\,[x_{k},\,x_{k+1}] $$

$$ E = -\sum_{r = 1}^{l}P_{2r+1}(1)[(\frac{h}{2})^{2r}f^{2r}(x(1)) + (\frac{h}{2})^{2r}f^{2r}(x(-1))] - \int_{-1}^{+1}P_{2l+2}(t)(\frac{h}{2})^{2l+2}f^{2l+2}(x)\frac{2}{h}dx$$

$$ \overline{d_{2r}} = \frac{P_{2r}(1)}{2^{2r}}$$

$$ E = -\sum_{r = 1}^{l}2\overline{d_{2r+1}}h^{2r} [f^{2r}(b) + f^{2r}(a)] - (\frac{h}{2})^{2l+2}\frac{2}{h}\sum_{k = 0}^{n - 1}\int_{x_{k}}^{x_{k+1}}P_{2l+2}(t_{k}(x))f^{2l+2}(x)dx$$

\begin{align} \\E_{n}^{1} &= \frac{h}{2}E\; \\ &= -\sum_{r = 1}^{l}\overline{d_{2r+1}}h^{2r-1} [f^{2r}(b) + f^{2r}(a)] - (\frac{h}{2})^{2l+2}\sum_{k = 0}^{n - 1}\int_{x_{k}}^{x_{k+1}}P_{2l+2}(t_{k}(x))f^{2l+2}(x)dx\; \end{align} $$

Solved by : EGM6341.s10.Team1.Ya-Chiao Chang 19:09, 23 March 2010 (UTC)

=Problem 9-Recurrence formula =

Find
Use (6)[[media:Egm6341.s10.mtg29.djvu|p.29-2]] to obtain $$ (p_{2},p_{3}),(p_{4},p_{5}),(p_{6},p_{7})\,$$

Given
Using $$ p_{1}(t)=-t,i.e.,c_{1}=-1 \, $$

Solution
(6)[[media:Egm6341.s10.mtg29.djvu|p.29-2]]

$$ \frac{c_{1}}{(2i+1)!}+\frac{c_{3}}{(2i-1)!}+\frac{c_{5}}{(2i-3)!}+...+\frac{c_{2i-1}}{3!}+c_{2i+1}=0\, $$

when $$ i=1 \,$$

$$ \frac{c_{1}}{3!}+c_{3}=0\, $$

Recall $$ p_{1}(t)=-t,i.e.,c_{1}=-1 \, $$

$$ \Rightarrow c_{3}=\frac{1}{6}\, $$

when $$ i=2 \,$$

$$ \frac{c_{1}}{5!}+\frac{c_{3}}{3!}+c_{5}=0\, $$

Recall $$ c_{3}=\frac{1}{6}\, $$

$$ \Rightarrow c_{5}=-\frac{7}{360}\, $$

when $$ i=3 \,$$

$$ \frac{c_{1}}{7!}+\frac{c_{3}}{5!}+\frac{c_{5}}{3!}+c_{7}=0\, $$

Recall $$ c_{5}=-\frac{7}{360}\, $$

$$ \Rightarrow c_{7}=\frac{31}{15120}\, $$

From (1)and (3)[[media:Egm6341.s10.mtg29.djvu|p.29-2]]

$$ p_{2i}(t)=\sum_{j=0}^{i}c_{2j+1}\frac{t^{2(i-j)}}{[2(i-j)]!}\, $$

$$ p_{2i+1}(t)=\sum_{j=0}^{i}c_{2j+1}\frac{t^{2(i-j)+1}}{[2(i-j)+1]!}+c_{2i+2}\, $$

where $$ c_{2i+2}=0\, $$

thus when $$ i=1 \,$$

$$ p_{2}=c_{1}\frac{t^{2}}{2!}+c_{3}=-\frac{t^{2}}{2}+\frac{1}{6}\,$$

$$ p_{3}=c_{1}\frac{t^{3}}{3!}+c_{3}t=-\frac{t^{3}}{6}+\frac{1t}{6}\,$$

when $$ i=2 \,$$

$$ p_{4}=c_{1}\frac{t^{4}}{4!}+c_{3}\frac{t^{2}}{2!}+c_{5}=-\frac{t^{4}}{24}+\frac{t^{2}}{12}-\frac{7}{360}\,$$

$$ p_{5}=c_{1}\frac{t^{5}}{5!}+c_{3}\frac{t^{3}}{3!}+c_{5}t=-\frac{t^{5}}{120}+\frac{t^{3}}{36}-\frac{7t}{360}\,$$

when $$ i=3 \,$$

$$ p_{6}=c_{1}\frac{t^{6}}{6!}+c_{3}\frac{t^{4}}{4!}+c_{5}\frac{t^{2}}{2!}+c_{7}=-\frac{t^{6}}{720}+\frac{t^{4}}{144}-\frac{7t^{2}}{720}+\frac{31}{15120}\,$$

$$ p_{7}=c_{1}\frac{t^{7}}{7!}+c_{3}\frac{t^{5}}{5!}+c_{5}\frac{t^{3}}{3!}+c_{7}t=-\frac{t^{7}}{5040}+\frac{t^{5}}{720}-\frac{7t^{3}}{2160}+\frac{31t}{15120}\,$$

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

 * style="width:50%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle
 * $$ \displaystyle

p_{2}=-\frac{t^{2}}{2}+\frac{1}{6}, p_{3}=-\frac{t^{3}}{6}+\frac{1t}{6}\,$$ $$
 * $$\displaystyle
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

 * style="width:50%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle
 * $$ \displaystyle

p_{4}=-\frac{t^{4}}{24}+\frac{t^{2}}{12}-\frac{7}{360}, p_{5}=-\frac{t^{5}}{120}+\frac{t^{3}}{36}-\frac{7t}{360}\,$$ $$
 * $$\displaystyle
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

 * style="width:50%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle
 * $$ \displaystyle

p_{6}=-\frac{t^{6}}{720}+\frac{t^{4}}{144}-\frac{7t^{2}}{720}+\frac{31}{15120}, p_{7}=-\frac{t^{7}}{5040}+\frac{t^{5}}{720}-\frac{7t^{3}}{2160}+\frac{31t}{15120}\,$$ $$
 * $$\displaystyle
 * }
 * }

Solved by : Egm6341.s10.team1.lei00:55, 24 March 2010 (UTC)

=Problem 10: Decipher the MATLAB code by Kessler=

Statement
Review the MATLAB code written by Kessler. Comment on what each line does.

Solution
The code computes the Trapezoidal error polynomials evaluated at t=1. The polynomial equation is,
 * $$p_{2k}(t)=c_1\frac{t^{2k}}{(2k)!}+c_3\frac{t^{2k-2}}{(2k-2)!}+c_5\frac{t^{2k-4}}{(2k-4)!}+...+c_{2k-3}\frac{t^4}{4!}+c_{2k-1}\frac{t^2}{2!}+c_{2k+1}\frac{t^0}{0!}$$

When evaluated at t=1, the equation is,
 * $$p_{2k}(1)=\frac{c_1}{(2k)!}+\frac{c_3}{(2k-2)!}+\frac{c_5}{(2k-4)!}+...+\frac{c_{2k-3}}{4!}+\frac{c_{2k-1}}{2!}+\frac{c_{2k+1}}{0!}$$

The coefficients, $$c$$, are computed by solving the below equation using $$c_1=-1$$ as a starting point,
 * $$\frac{c_1}{(2k-1)!}+\frac{c_3}{(2k-3)!}+\frac{c_5}{(2k-5)!}+...+\frac{c_{2k-5}}{5!}+\frac{c_{2k-3}}{3!}+\frac{c_{2k-1}}{1!}=0$$

For illustration, the first few coefficients and polynomial equations are shown below.

Notice that the factorials in the denominators of the polynomials differ by 1 from those in the coefficients. Also, to solve for the next coefficient, the preceding coefficients only need to be summed and then moved to the other side of the equal sign, i.e $$c_7=-(\frac{c_1}{7!}+\frac{c_3}{5!}+\frac{c_5}{3!})$$.

The following code will be commented line-by-line to explain the algorithm of Kessler:

Solved by: EGM6341.s10.Team1.Kumanchik 19:43, 24 March 2010 (UTC)

=Problem 11 Compute the curve length of ellipse=

Statement
Compute C by using two methods:

$$ \displaystyle a)\, C = \int_{\theta = 0}^{\theta = 2\pi}dl \,\,\rightarrow From(4)$$ [[media:Egm6341.s10.mtg30.djvu|P30-3]]

$$ dl = d\theta[r^{2} + (\frac{dr}{d\theta})^{2}]^{\frac{1}{2}}\,\,\rightarrow From(1) $$ [[media:Egm6341.s10.mtg30.djvu|p.30-3]]

$$ \displaystyle b)\, C = 4aE(e^{2}) \,\,\rightarrow From(5)$$ [[media:Egm6341.s10.mtg30.djvu|p.30-4]]

$$ E(e^{2}) = \int_{0}^{\frac{\pi}{2}}[1 - e^{2}sin^{2}\theta]^{\frac{1}{2}}d\theta \rightarrow From(6)$$ [[media:Egm6341.s10.mtg30.djvu|p.30-4]]

Solution
$$ \begin{align} \displaystyle a)\, C = \int_{\theta = 0}^{\theta = 2\pi}dl \\ &=\int_{\theta = 0}^{\theta = 2\pi}d\theta[r^{2} + (\frac{dr}{d\theta})^{2}]^{\frac{1}{2}} \\\end{align}\ $$

$$ r(\theta) = \frac{1 - e^{2}}{1 - ecos\theta} $$

$$ e = sin(\frac{\pi}{12}) $$

$$ \begin{align} C = \int_{\theta = 0}^{\theta = 2\pi}dl \\ &=\int_{\theta = 0}^{\theta = 2\pi}[(\frac{1 - e^{2}}{1-ecos\theta})^{2} + (\frac{(e^{3} - e)sin\theta}{e^{2}cos^{2}\theta - 2ecos\theta + 1})^{2}]^{\frac{1}{2}}d\theta\\ &=\int_{\theta = 0}^{\theta = 2\pi}[1 + (\frac{e\cancelto{1}{(e^{2} - 1)}sin\theta}{(ecos\theta - 1)^{\cancelto{1}{2}}})^{2}(\frac{\cancelto{1}{1-ecos\theta}}{\cancelto{1}{1 - e^{2}}})^{2}]^{\frac{1}{2}}d\theta\\ &=[1 + (\frac{esin\theta}{ecos\theta - 1})^{2}]^{\frac{1}{2}}d\theta \\\end{align}\ $$

Use Matlab to find the exact value of C

Get I1 = 1.6041  Elapsed time is 0.049997 seconds.

Use Composite Trapezoidal rule: Get n = 18 In = 6.176601987629740 Elapsed time is 0.009616 seconds.

Use Clenshaw-Curtis quadrature : Get n = 23 In2 = 6.1766 En2 = 7.5165e-011 Elapsed time is 0.038245 seconds.

Use Romberg Table:

Get N = 128 In3 = 6.1766 En3 = 2.0250 e-013 Elapsed time is 0.006707 seconds.

$$ \displaystyle b)\, C = 4a\int_{0}^{\frac{\pi}{2}}[1-e^{2}sin^{2}\theta]^{\frac{1}{2}}d\theta \ $$

Here $$ a\, =\, 1\, $$

Use Matlab to find the exact value of C

Get I2 = 6.1766  Elapsed time is 0.057046 seconds.

Solved by : EGM6341.s10.Team1.Ya-Chiao Chang 19:09, 23 March 2010 (UTC)

=Problem 12: Derive the arc length of a curve using the law of cosines=

Find
The arc length of a curve in space using the Law of Cosines

Given
The law of cosines:
 * $$c^2=a^2+b^2-2abcos(\gamma) \,$$

Solution
The law of cosines can be directly applied to the triangle with sides $$r$$, $$r+dr$$, and inner angle $$d\theta$$ to determine the differential line segment AB=$$dl$$.
 * $$dl^2=r^2+(r+dr)^2-2r(r+dr)cos(d\theta)=2r^2+2rdr+dr^2-(2r^2+2rdr)cos(d\theta) \,$$

The Taylor series expansion of $$cos(x)$$ is,
 * $$cos(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-...$$

Taking the first two terms of the expansion and applying them to the differential length equation yields,
 * $$dl^2=2r^2+2rdr+dr^2-(2r^2+2rdr)(1-\frac{d\theta^2}{2})=dr^2+r^2d\theta^2+r\cancel{drd\theta^2}^{\approx 0}$$

Since the triple differential element $$drd\theta^2$$ shrinks faster than the double elements $$dr^2$$ and $$d\theta^2$$, it is set to zero. The remainder is,
 * $$dl^2=d\theta^2(\frac{dr^2}{d\theta^2}+r^2)$$

Integrating from $$\theta_1$$ to $$\theta_2$$ yields,
 * $$l=\int_{\theta_1}^{\theta_2}d\theta\sqrt{(\frac{dr}{d\theta})^2+r^2}$$

This is identical to the expression derived from geometry.

Solved by: EGM6341.s10.Team1.Kumanchik 19:44, 24 March 2010 (UTC)

=Contributing Members= EGM6341.s10.Team1.Ya-Chiao Chang 19:09, 23 March 2010 (UTC) Author of problems 1,8 and 11  Proof read : 2 and 5

Egm6341.s10.team1.toddmock 21:05, 23 March 2010 (UTC) Author of problems 3 and 6

Egm6341.s10.team1.lei 00:57, 24 March 2010 (UTC) Author of problems 2,5 and 9

Egm6341.s10.team1.andrewdugan 08:23, 24 March 2010 (UTC) Author of problems 4 and 7 Proof read : 3 and 6

EGM6341.s10.Team1.Kumanchik 15:22, 25 March 2010 (UTC) Author of problems 10 and 12 Proof read : 1, 4, and 9

title: User:Egm6341.s2010.Team1/HW5


text:
User talk:Chern~enwikiversity

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title: User talk:Chern~enwikiversity


text: Category:Votic

title: Category:Votic


text:
User:Egm6341.s10.team3.ynahn81/HW5-4

= (4) Find coefficients $$\displaystyle \bar{d_{2}}, \; \bar{d_{4}}, \; \bar{d_{6}}$$ from polynomials $$\displaystyle P_{2}, \; P_{4}, \; P_{6}$$ = Ref: Lecture notes [[media:Egm6341.s10.mtg27.djvu|p.27-2]]

Problem Statement
Find coefficients $$\displaystyle \bar{d_{2}}, \; \bar{d_{4}}, \; \bar{d_{6}}$$ using the following relations.

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

d_{r} = \bar{d_{2r}} = \frac{P_{2r}(1)}{2^{2r}} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
 * }
 * }

Solution
We already obtained the polynomials $$\displaystyle P_{2}(t), \; P_{4}(t)$$ in class meeting 26. (see lecture notes [[media:Egm6341.s10.mtg26.djvu|p.26-1]] and [[media:Egm6341.s10.mtg26.djvu|p.26-3]].)

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

P_{2}(t) = -\frac{t^2}{2!} + \frac{1}{6} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)
 * }
 * }

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

P_{4}(t) = -\frac{t^4}{24} + \frac{t^2}{12} - \frac{7}{360} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)
 * }
 * }

Also, at the problem 2 in the HW5 (Ref:[[media:Egm6341.s10.mtg27.djvu|p.27-1]]), $$\displaystyle P_{6}(t) $$ have been obtained.

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

P_{6}(t) = -\frac{t^6}{720} + \frac{t^4}{144} - \frac{7t^2}{720} + \frac{31}{15120} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 4)
 * }
 * }

From the (Eq.1) and (Eq.2),

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

$$\displaystyle d_{1} = \bar{d_{2}} = \frac{P_{2}(1)}{2^2}=\frac{1}{4}\times\frac{-3+1}{6}=-\frac{1}{12} $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

From the (Eq.1) and (Eq.3),

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

$$\displaystyle d_{2} = \bar{d_{4}} = \frac{P_{4}(1)}{2^4}=\frac{1}{16}\times\frac{-15+30-7}{360}=\frac{1}{720} $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

From the (Eq.1) and (Eq.4),

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

$$\displaystyle d_{3} = \bar{d_{6}} = \frac{P_{6}(1)}{2^6}=\frac{1}{64}\times\frac{-21+105-147+31}{15120}=-\frac{1}{30240} $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

title: User:Egm6341.s10.team3.ynahn81/HW5-4


text:
User talk:Cheezerguy

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User talk:Hieywiey

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title: User talk:Hieywiey


text:
User:Egm6341.s10.team3.ynahn81/HW5-6

= (6) Find coefficients $$\displaystyle \bar{d_{2}}, \; \bar{d_{4}}, \; \bar{d_{6}}$$ from Bernoulli numbers = Ref: Lecture notes [[media:Egm6341.s10.mtg28.djvu|p.28-2]]

Problem Statement
Using Bernoulli numbers obtained by the Taylor series expansion of $$\displaystyle x\,coth(x)$$,

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

x\,coth(x) = \sum_{r=0}^{\infty}\underbrace{\frac{B_{2r}}{(2r)!}}_{-\bar{d_{2r}}}x^{2r} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

i) Verify $$\displaystyle \bar{d_{2}}, \; \bar{d_{4}}, \; \bar{d_{6}}$$ by comparing the results from $$\displaystyle \bar{d_{2r}} = \frac{P_{2r}(1)}{2^{2r}}$$. ii) Compute $$\displaystyle \bar{d_{8}}, \; \bar{d_{10}}$$

Solution
i)

Let's explicitly express first several terms of the Taylor series expansion of $$\displaystyle x\,coth(x)$$,

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

x\,coth(x) = B_{0} + \frac{B_2}{2!}x^{2} + \frac{B_4}{4!}x^{4} + \frac{B_6}{6!}x^{6} + \frac{B_8}{8!}x^{8} + \frac{B_{10}}{10!}x^{10} + \cdot\cdot\cdot. $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Note that $$\displaystyle B_{0}=1, \, B_{2}=\frac{1}{6}, \, B_{4}=-\frac{1}{30}, \, B_{6}=\frac{1}{42}$$.

Then, based on the following relation between $$\displaystyle \bar{d_{2r}} $$ and Bernoulli numbers,

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

\frac{B_{2r}}{(2r)!} = -\bar{d_{2r}}, $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

we can obtain $$\displaystyle \bar{d_{2}}, \,\bar{d_{4}}, \,\bar{d_{6}}$$ as follows.

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

$$\displaystyle \bar{d_{2}}=-\frac{B_{2}}{2!}=-\frac{1}{2}\times\frac{1}{6}=-\frac{1}{12} $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

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

$$\displaystyle \bar{d_{4}}=-\frac{B_{4}}{4!}=\frac{1}{24}\times\frac{1}{30}=\frac{1}{720} $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

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

$$\displaystyle \bar{d_{6}}=-\frac{B_{6}}{6!}=-\frac{1}{720}\times\frac{1}{42}=-\frac{1}{30240} $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

Above results are exactly same with the results from $$\displaystyle \bar{d_{2r}} = \frac{P_{2r}(1)}{2^{2r}}$$ as we obtained at problem 4 in HW5.

ii)

Since $$\displaystyle B_{8}=-\frac{1}{30}, \, B_{10}=\frac{5}{66}$$,

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

$$\displaystyle \bar{d_{8}}=-\frac{B_{8}}{8!}=\frac{1}{40320}\times\frac{1}{30}=\frac{1}{1209600} $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

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

$$\displaystyle \bar{d_{10}}=-\frac{B_{10}}{10!}=-\frac{1}{3628800}\times\frac{5}{66}=-\frac{1}{47900160} $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

title: User:Egm6341.s10.team3.ynahn81/HW5-6