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Managing Human Resource Flows

This Wikiversity entry is created by staff and students of the Business Administration Program 'Human Resource Management' of the University of Twente.

The notion of 'Human Resource Flows' was coined by Michael Beer and colleagues in 1984. In the opening chapter of their book 'Managing Human Assets' (published by The Free Press, New York), Beer et al introduce four major HRM policy areas: employee influence, human resource flow, reward systems and work systems (Beer, Spector, Lawrence, Mills, & Walton, 1984: p 7-10). The combination of these HR areas are also referred to as the Harvard model. The Harvard model states that people are the main assets within the company and therefore 'employee influence' plays a major role. A company must meet the employees’ requirements in order to get them committed to the organization and this should be in line with the organizational needs (Beer et al., 1984). In the same time as the Harvard model, the Michigan model of Fombrun et al. is presented (De Nijs, 1998). The emphasis in this theory is on the integration of HRM with the overall strategy of the organization. Therefore, HRM also depends on the external market strategy of the organization. In the Michigan model four key functions in relation to HRM are distinguished: selection/placement, rewards, careerdevelopment/planning and appraisal. These key functions should be congruent and related with each other and with the overall strategy. Therefore the Michigan-model is also called the Human Resource Cycle (De Nijs, 1998). Below the different HR activities of the Harvard model of Beer et al. (1984) will be discussed in a very serious manner

'Employee influence' 'Human resource flow' This area focuses on who is hired, fired, transferred, promoted, terminated or retired and the way these decisions fit the needs of the individual and the company. The flow through organizations can be split into inflow, internal flow and outflow. Managing outflow: letting employees go, voluntary or with a dismissal. Managing outflow consists of retirement, lay-offs, dismissal and having a new job.
 * 1) Managing inflow: recruitment decisions about where and how to recruit and how to introduce new people to the organization. Related actions are planning, hiring, recruiting, selecting and inducting.
 * 2) Managing internal flow: the flow of employees through the organization. Concerns of noobs can be REKD transfers, job assignments, promotions and demotions. This flow must be managed in such a way that employee competence is developed to meet corporate needs, while at the same time they satisfy the career aspirations of the employees. The internal flow consists of training, development, giving appraisal and the rewarding of employees.

Managing human resource flow is related to three perspectives: organizational, individual and societal perspective. The organizational perspective has historically not been a strategic consideration in an organization, but managing human resource flow policies has became a more important considerations nowadays. Human resource flow policies can be approached from the point of view made by coagulation of the individual employee. Human resource management applies to all employees, a broader concept of career seems to be in order. Careers may be viewed as “a series of separate but related experiences and adventures through which a person, any person, passes during a lifetime” (Beer et al., 1984, p. 67). Managers have to consider the societal perspective when the human resource flow policies are developed. The human resource flow will be developed through shifting work force values, outside institutions and government regulation and labor union policy (Beer et al., 1984).

'Reward systems' 'Work systems' 'Political systems'

Multi-Generational workplace
Generational differences is a concept of all ages. In a Dutch book titled ‘dissolute youth’ Groenendijk en Roberts speak of educators in the Middle Ages who already complained about these dissolute youth. Every generation complains about its successor as if they were better. In the present working society there are three generations which have to work together for organizational effectiveness. Managing Human Resources Flows focuses on stimulating effective cooperation between the three postwar generations:
 * Generation Babyboom (1945-1963)
 * Generation X (1963-1982)
 * Generation Einstein (1982-mid90s)

Generation Babyboom. Babyboomers are part of the generation who are born during the postwar reconstruction between 1945 and 1963. This large generation characterizes itself by discipline and respect. It is a generation that earns success and status through hard labor and perseverance.

Generation X. Also known as the ‘Lost Generation’ Gen X’ers were born during the Economic Depression and suffered at later age the adverse effects of this crisis. For this reason members of this generation grew up with less perspective for the future.

Generation Einstein. The latest generation Einstein grew up in a flourishing period with growth and wealth. This generation has a positive perspective on the future and is mainly high educated. Furthermore, it is a generation which is characterized by social networking and the desire for personal development.

Aging workforce
In order to manage the “Human Resource Flow” successfully, organizations have to pay more and more attention to the aging workforce. Due to a global shift in the demographic situation, companies are forced to retain their older workers much longer as there will be a shortfall of young workers. Yet, not all countries are facing the same problems. India for instance, will not experience the shortage of qualified young workers entering the workforce. They are confronted by the low retirement age, of 58, of the old worker who do not feel responsible for the dissemination of job related information to the entering workforce (Hindustan times, 2009). The demographic changes are caused by lower birth rates and longer life expectancy. Consequently, an age-related personnel policy can be seen as a topical theme in personnel management (Wognum, Veldkamp, De Grip & Sieben, in Rocco & Thijssen, 2006: p.93 - 104). These policies imply that actions have to be taken in order to encourage older workers to play an important role within the organization (Wognum, Horstink). This can be done through training and development, for example. In this context, special attention has to be given to the fact that older workers learning styles and preferences are different from those of younger workers (See learning and developing of older workers). As stated above, through the major changes in today’s workplace, continuous learning and skill development by workers of all ages is becoming more important than ever before. The changes in technology and in business strategies have meant that new skills are required of workers at midlife and beyond just to continue to perform their jobs. Therefore, while mid- and late-career stages used to be viewed as a periods of mastery and maintenance, and workers could afford to avoid learning new things, now all workers are increasingly being called upon to continuously learn and adapt. Continuous learning has become a core career competency even for mid and late-career workers (Maurer, 2001, p. 126)

Hindustantimes (2009). India’s overlooked ‘grey market’ workforce. Visited on the 21 December 2009. http://www.hindustantimes.com/News-Feed/columnsbusiness/India-s-overlooked-grey-market-workforce/Article1-236210.aspx

Learning and developing of older employees Companies feel forced to retain older workers longer instead of stimulating retirement and thereby causing brain-drain. The most important question that comes with this issue is: How could companies nourish their older workforce and make use of the rich expertise they have, and also improve their employability and career opportunities (Turku Conference, 1999;from Wognum et al., 2006, p.93). A first answer to this question would be: age-related HR policy measures. HR practise may increase employees’ motivation and commitment to their tasks and also reduce quit rates by increasing employees’ job satisfaction (Ichniowski et al.,1997, Batt,2002;from Wognum et. al, 2006, p.94). One thing that should be taken into account is the fact that older workers are willing to learn, but their ability to learn, learning styles and preferences differ from those of younger workers (Rhebergen & Wognum, 1997;from Wognum et al., 2006). For example older people think HRD policy should be more sensitive to their needs and better meet their expectations. Older workers seem to prefer to learn work-based and by means of new experiences, and they consider attractive learning situation-related conditions very important. Older workers perceive learning by doing, learning on the job and individual coaching far more relevant than formal training and courses (Thijssen 1996;from Wognum et al., 2006). Taking all this factors into account it is very important to broaden older workers’ learning experiences beyond their actual job, providing company-specific learning initiatives in order to increase their job satisfaction and to decrease their intention to leave (Dorhout et al., 2002; from Wognum et al., 2006) However these factors do not mean that older workers could not be innovative or creative at all. Wognum & Horstink for example state that older workers can be more innovative and productive because they are more independent and experienced than younger ones (e.g. Opinion Leader Research,2004)

Retaining older workers

The participation rate of older workers need to increase, but it is not easy. In the past, replacement policies implied early retirement of older workers. It was attractive for individuals and organizations, and many people anticipated. The perspective of older workers is changing to active ageing according to Rocco and Thijssen (2006). However, the replacement policies are still attractive as a short term solution. Rocco and Thijssen (2006) describe that there are two types of arrangements for older workers: spare policies and employment policies. Spare policies give older workers more rights and give them the opportunity to do less. Pull factors (Rocco & Thijssen, 2006) like early retirement schemes and flexible pension plans are in some countries abandoned, which makes it less attractive for older workers to stop working (De Lange & Evers, 2006). But older workers still can be pushed over the line by problematic employment situations, like existing negative stereotypes. Reduced capabilities of older workers are often used as a reason for early retirement, while the deficiency hypothesis, which claims that mental and physical capabilities decrease from about the age of 35, is not valid anymore (Posthuma & Campion, 2009). Posthuma and Campion (2009) state that age stereotypes can lead to discrimination and that there is a direct relationship between age and negative effects on workers. Rocco and Thijssen (2006) distinguish four different employability solutions in the older worker policies of organizations. Replacement policies and support policies could decrease the participation rate of older workers, so to increase the participation rate, organizations need to create policies for retired workers to return or remain. Blocking policies and development policies can be used in this case.

It has been studied which factors influence older workers in deciding whether to continue working or retire from a job. This study is useful for employers because older workers are becoming more and more valuable in today's workforce. Retention of older workers is becoming a contributing factor for an organizations' success. Yeatts, Folts & Knapp (2000) describe an individual-job fit. This fit is present when an employee meets the knowledge, skill and ability requirements of a job and when a job meets the needs, values and interests of an employee. According to Yeatts et al. (2000) an older worker' s ability and desire to successfully adapt to changing work environment are influenced by a variety of individual-level factors. Those factors are: Previous relevant knowledge, skills, abilities, training, and education, attitudes toward work and the redesigned workplace, attitudes toward change, and the attractiveness of retirement. Yeatts et al. (2000) also elaborated six organizational factors that influence older workers adaption to workplace. The first factor is the orientation to workplace changes, which can lead to anxiety and resistance in older workers. The next factor, emphasizes the characteristic of the redesigned job, that according to Yeatts et al.(2000) leads to changes in required skills and abilities of the worker. This shift might cause insecurities in the employee about the ability to perform the job. The third organizational factor refers to training and education of the employee which emphasizes the importance of the education of the worker in the case of changing job requirements.Furthermore, management and coworker support play a vital role in the changing process. The fifth influencing factor is the personnel practices and norms. The final important aspect is the pension benefits of the older worker (Yeatts, Folts, & Knapp, 2000).In general an individual-job fit will increase the job satisfaction of the employee, as well as the satisfaction of the employer, and the willingness to continue working at a firm(retention). If this individual-job fit is not present, the employee has to adapt by attending educational programs or trainings. Yeats et al. (2000) recognize that older employees do not adapt as fast, willing or easily as younger employees. However, if organizations have a continuous development plan for employees they are able to narrow the gap between the required knowledge, skills and abilities and the knowledge, skills and abilities of the employee. This will enable faster adaptation for employees, resulting in higher satisfaction and retention rates.

Learning, training and development considering the aging workforce

In today’s organizations learning, training and development are an important factor, because the workplaces are changing rapidly and knowledge and skills have to adapt quickly as well. Furthermore the average age of employed people is increasing (Johnston & Packer, 1987) and as previous research shows, these kind of employees are not participating in training and other learning\development activities as younger employees (cf. Cleveland & Shore, 1992). There are several reasons for this reduced participation. One important factor is that organizations do not invest in people who will go into retirement in the near future (Journal of Management 27, Maurer 2007).

Self Efficacy of Older Workers for Career Development - Self efficacy for development and improvement of career-relevant skills is the belief by a worker that he/she is capable of improving/developing his/her skills and it has key relationships with attitudes and voluntary participation in training and development activities (Maurer, 2001, p. 124).Maurer (2001) highlights a decline in self confidence (or self efficacy) for career relevant learning and skill development with age.Maurer (2001) makes the proposition that higher self efficacy for development will have more positive attitudes toward and more frequent voluntary participation in training activities. Maurer (2001) also proposes that key antecedents of self efficacy – mastery experiences, vicarious experiences, persuasion and physiological variables will positively influence self-efficacy for development.Referring to Maurer (2001) some suggestions for HR policies to improve these antecedents are -

Mastery experiences

•Ensure that qualified older workers have access to challenging tasks, assignments, and learning resources as ‘‘stretch’’ experiences —avoid ‘‘lost opportunities’’ for older workers. •Awareness building sessions for acceptance of differences in performance, speed, accuracy, and style in learning experiences. •Analysis of trainees’ pre-requisite knowledge and skills for complex training assignments and make preparatory resources available to trainees.

Vicarious Influences •Awareness sessions on Age stereotypes •Reduce ageism and prejudice in organization culture. •Recognize successful learners and make older workers’ success stories visible. •Training materials to depict desired behaviors to be learned, include older ‘‘models’’ in these materials.

Persuasion •Support mechanisms to all employees about learning. •Conversations with employees about special training needs, values, and preferences. •Career counselors for older workers.

Physiological Influences •Competition free atmosphere for learning and development experiences. •Accommodation for health issues.

The Reverse Pygmalion Effect

Another consequence of the ageing workforce is the changing relation between supervisor-subordinate. According to Collins, Hair & Rocco (2009) the tradition is that managers are older and have more experience than their subordinates. Because of changing demographics, such as the large scale retirement in the future of the older and more experienced generation Baby Boomers, there is a shortage of skilled and experienced workers (Collins et al., 2009). This shortage is often fulfilled with younger supervisors with less experience, who should supervise older subordinates. This age-reverse dyad is also called the Reverse Pygmalion Effect and it occurs when the expectations of the subordinates (the older workers) have an impact on the performance of the supervisors (the younger workers) (Collins et al., 2009). Collins et al. (2009) found in their research this Reverse Pygmalion Effect, because the older workers expect less from their younger supervisors in comparison to younger workers. Additionally, the older workers rated the leadership behavior of the younger supervisor's lower than the younger subordinates. These findings can have a major impact on the way HR is organized in the future, because this Reverse Pygmalion Effect will occur more often because of the ageing workforce.

Generational Differences
The following three abstracts of articles give insight into the world of generational differences. For more information the reference of the article is stated.

In the article: “Baby Boomers, Generation X and Generation Y?: Policy implications for defense forces in the modern era.” Bradley Jorgenson argues the following in his abstract.

In recent years there has been discussion in the management and organizational literature on generational differences and how they may impact on the design of workforce strategies. However, much of the discussion appears to be based on observation rather than large empirical work. Indeed, the author would argue that wider support for the requirement to manage the workforce around X and Y issues is absent. For example, employers of choice are looking to win talent by tailoring employment policy to capture the dynamism of the modern era rather than discrete generational values. Significantly, the bulk of generational data cited by popular writers is subjective, non-representative, makes use of single-point-of-time data and uses retrospective comparisons. Importantly, scholarly literature does not draw arbitrary and abrupt lines between generations. In seeking to determine a preferred workforce strategy organizations would be better served by acknowledging the technical, economic, political and social dynamism of modern life rather than the flawed conclusions of popular generational literature (Jorgenson, 2003).

B. Jorgenson, 2003. Baby Boomers, Generation X and Generation Y?: Policy implications for defense forces in the modern era. Foresight 5 (4): 41-49

The article: “From high maintenance to high productivity: What managers need to know about Generation Y” is written to offer managers insights into Generation Y and practical suggestions on how to turn this high maintenance workforce into a high productivity workforce.

This research is based on RainmakerThinking Inc.’s ongoing workplace interviews with hundreds of Generation Y-ers and managers. Like their Gen X siblings, the most talented members of Generation Y-ers are independent, entrepreneurial thinkers who relish responsibility, demand immediate feedback, and expect a sense of accomplishment hourly. They thrive on challenging work and creative expression, love freedom and flexibility, and hate micromanagement. Guided by managers who are willing to confront their challenges and meet their expectations, they have the potential to become the highest performers in history. They have 14 expectations of managers that can create the type of professional relationships they want to build. Generation Y is just beginning to gain pull in the workplace as full-time workers. Ongoing research is needed to see how our initial conceptions of them as high school and university students plays out as they mature in their professional lives. This paper not only offers managers insights into this highly diverse, contradictory generation, but suggests best practices they can implement to communicate with and motivate this emerging workforce. As organizations compete to attract and retain the best new talent, this paper offers managers a strategic advantage: a practical guide to the characteristics, needs and expectations of Generation Y (Martin, 2005).

C.A. Martin, 2005. From high maintenance to high productivity: What managers need to know about Generation Y. Industrial and commercial training 37 (1): 39 – 44

Understanding generational differences for competitive success

According to Glass (2007) the largest diversity of generations is represented in today's workplace than at any other time in history. With this diversity come new challenges. The purpose of this article is to analyze the specific challenges, and also opportunities, inherent in managing – and working in – a multigenerational workforce. By focusing on research about the character traits of workers in each generation, and identifying the types of conflict that can result, managers can better understand these characteristics and work styles, and can leverage them to enhance both team and organizational success. A wide range of studies and research was reviewed, and all revealed the methods to recognize the key motivators for each generation. By understanding and appreciating each age group's work style and personality traits, existing friction can be minimized and the assets of managing – and coexisting within – a multigenerational workforce are maximized. Three primary generations exist in the business world: baby boomers, generation X, and generation Y (known as millennials). Each possesses unique characteristics that affect work ethic and relationships, how change is managed, and perception of organizational hierarchy: defining events in each generation's life all occurred between the ages of 5 to 18, the developmental years. The different backgrounds and life experiences result in five areas of potential workplace strife surrounding their differing expectations, distinct work ethics, deep-seated attitudes, opposing perspectives and diverse motivators. More research on generation X and millennials and their role in the workplace in developing countries is needed. Another area that needs future research is how increasing globalization impacts generational cohorts in different countries. The existence of a multigenerational workforce affects two areas of human resources policy and employee development efforts: retention and motivation. Employees of diverse age groups react differently to programs designed to address these two areas, and also have differing expectations. Companies may need to rethink their existing practices. The article will deepen understanding of the differences that can divide generations and explore the benefits – and necessity – of creating and leveraging a multigenerational workforce.

A. Glass, 2007. Understanding generational differences for competitive success. Industrial and commercial training 39 (2): 98 - 103

Employability
Employability In the entry of this Wikiversity the concept employability is mentioned two times. The first time it was mentioned in one breath with learning and developing of older employees. Companies should nourish their older workforce and make use of the rich expertise they have, and also improve their employability and career opportunities (Turku Conference, 1999;from Wognum et al., 2006, p.93). Whilst the second time employability was mentioned with developments within organizations and the possible policies how to deal with these developments (Thijssen, 1997; from Rocco and Thijssen 2006, p.9).

So employability was mentioned, but the question remains what the concept really means. The answer to this question is very broad, for Forrier & Sels (2003 p.102) say that employability has become a buzzword in organizational literature, no clear consensus about its meaning and measurement can be found. Nevertheless we will try to give insight into the concept employability.

The first publications date about employability date from the 1950s (Feintuch 1955, from Forrier & Sels, 2003 p.103). In that time it was mainly the government’s responsibility to stimulate entry into the labor market and the most important measures suggested were promoting employability. In the 1970s the economic situation changed and with it the emphasis in the literature shifted from governmental to more individual. It became an economic necessity from employees to be as ‘employable’ as possible. In the 1980s attention shifted to the company level, employability was seen as an HR instrument to optimize the deployment of staff within companies (Forrier & Sels, 2003).

Whilst the main responsibility of employability shifted from governmental, to individual, and to company-level the used definition of the concept employability also changed. The strict definition used by Forrier & Sels (2003 p.106) defines employability as an individual’s chance of a job in the external and/or external labor market. The definition used by Van der Heijde & Van der Heijden (2006 p. 453) the continuous fulfilling, acquiring or creating of work through the optimal use of competences. This definition is used to create a measurement tool of employability. Van der Heijden et al. states that the conceptualizations mentioned in their article "Employability enhancement through formal and informal learning: an empirical study among Dutch non-academic university staff members (2009)' share that employability implies a permanent process of acquisition and fulfillment of employment, within or outside the current organization, today, and, in the future (2009, p.20).

Informal learning in organizations was always an important factor in the learning of employees in organizations, and has been defined in recent literature as well. According to Van der Heijden et al., informal learning "...includes incidental learning, i.e. learning that occurs as a by-product of some other activity, and which occurs, even though employees are not always conscious of it, and which is not always intentionally searched for." (Van der Heijden et al., 2009). A conceptualization of the concept informal learning has been proposed by Marsick and Volpe (1999), and involves six characteristics of informal learning in organizations: (1) integrated with work and daily routine; (2) triggered by an internal or external jolt; (3) not highly conscious; (4) often haphazard and influenced by change; (5) an inductive process of reflection and action; and (6) linked to the learning by others. In other words, with informal learning could be thought of interaction with ones supervisor, the learning value of the job, and internal and external networks (Van der Heijden et al., 2009).

The measurement instrument that Van der Heijde & Van der Heijden (2006) developed combines domain-specific expertise with more generic competences. The instrument consist of five dimensions of employability: 1. Occupational expertise: the expertise needed to perform the various tasks and responsibilities of a job adequately; 2. Anticipation and optimization: preparing for and adapting to future changes in a personal and creative manner, and striving for the best possible results; 3. Personal flexibility: the capacity to easily adapt to all kinds of changes in the internal and external labour market that do not pertain to one's immediate job domain; 4. Corporate sense: the participation and performance in different work groups; 5. Balance: compromising between opposing employers' interests as well as one's own opposing work, career, and private interests (Van der Heijde & Van der Heijden, 2006, p. 20-21).

The measurement of employability presented in the article of Van der Heijde & Van der Heijden (2006) is based upon the idea that some characteristics of expert performance and of employability are valid regardless of the domain of expertise of a professional. The five dimensions are not fully exclusive and represent correlated aspects of employability (2006, p. 468). The results of the study presented in the article of Van der Heijde & Van der Heijden (2006) supports their theory, which states that employability involves (1) occupational expertise, (2) anticipation and optimization, (3) personal flexibility, (4) corporate sense, and (5) balance.

References:

Forrier, A. & Sels, L. (2003). The concept employability: a complex mosaic. International Journal of Human Resources Development and Management. Vol. 3, No. 2, pp. 102-124

Marsick, V. and Volpe, F.M. (1999), Informal learning on the job, Advances in Developing Human Resources, 1, 5-8.

Van der Heijde, C.M. & Van der Heijden, B.I.J.M. (2006) A competence-based and multidimensional operationalization and measurement of employability. Human Resource Management, Vol. 45, No. 3, Pp. 449-476

Van der Heijden, B., Boon, J., Van der Klink, M., Meijs, E. (2009), Employability enhancement through formal and informal learning: an empirical study among Dutch non-academic university staff members, International Journal of training and development, pp. 19-37.

Challenges at organizations with older workers
Older workers have become a majority in organizations because of demographic changes. At the same time flexibility has become the norm in many work environments. The result is a workplace that requires almost constant adaptation by employees. The extent to which older workers leave their jobs, rather than adapt to workplace changes, significantly affects the organizations they work for, the government programs that serve them, and the personal well-being of the workers themselves (Yeatts, Folts & Knapp, 2000).

Many organizations which have workers with an average age of above 40, have to deal with early retirement of the employees. Theory explains early retirement by push and pull-factors. Pull-factors assume that older workers are persuaded to retire early because the situation outside working life is attractive. Push-factors assume that older workers are pushed over the line by a problematic employment situation as a result of increasing work pressure, lacking employment competencies and/or by existing negative stereotypes (Rocco & Thijssen, 2006).

When pull factors are taken into account, it is important to keep work attractive for older workers. This can be done by enlarging the job satisfaction; by creating individual job fit. The needs, values and interests of the employee are in balance with the knowledge, skills and ability (KSA) requirements that must be met by an employee. The KSA requirements of recently redesigned jobs often are more demanding than the more traditional jobs they replace (Hackman, 1990; Lawler, 1986; Manz & Sims, 1989, in: Yeatts et al., 2000).

To create individual job fit and to diminish push factors, there is the need for development. According to Wognum & Horstink (forthcoming) various studies reveal that the actual participation of older workers in formal learning and development declines with age and they remain underrepresented in most forms of these training and education programs. They also receive a lower degree of support and encouragement from supervisors and colleagues at work, when engaging in learning and development activities. One reason for this may be unfounded stereotyping of older workers: they are seen as less mobile and flexible than their younger counterparts, less able to deal with rapid changes occurring in organizations and less productive. In that case, it is useful for these organizations to adapt a development policy. This policy focuses on improving the employability of older workers which does not only require suitable learning facilities for older workers, but also related management conditions such as appraisals, career development support, etc. (Rocco & Thijssen, 2006).

On the contrary it could be stated that older workers can be more innovative and productive because they are more independent and experienced than younger ones (Wognum & Horstink, forthcoming). Experience is of much greater importance for one’s flexible employability than age, although age and experience cannot be viewed as separate entities. Generally, with increasing age there will be a considerable increase in experience quantity but without much diversity. This phenomenon is called experience concentration (Rocco & Thijssen, 2006). By experiencing a wide range of experiences you get used to changing circumstances whilst employability opportunities involving new tasks are relatively high. Organizations have to recognize experience concentration, the earlier the tendency is recognized, the easier it can be corrected (Thijssen & Van Der Heijden, 2003).

Another stereotype is that older workers are more resistant to change (Posthuma & Campion, 2009). When a job is redesigned, adaptation may be more difficult for long-tenured employees who may be more likely to find it difficult to deviate from previously successful routines and adjust to new ways of doing things (Fossum, Arvey, Paradise & Robbins, 1986, in: Yeatts et al., 2000). Especially if retirement is an option an older worker might be unwilling to explore the reality of job redesign (Yeatts et al., 2000). So, it is important for WisuTravel to remove the resistance to change.

Internal HR Flows
Every manager will be confronted with an employee who is not feeling happy about his/her current work situation. Just continue without doing anything will result in a unsatisfying situation for both the company and the employee. This situation can’t go on for too long, otherwise the employee or the employer will terminate the employment contract. So, what can managers do to solve this problem concerning the interal HR flows. This means that the outflow is not considered here. There are five options left to be considered.

Lateral (sideways) movement: Lateral movement is when an employee is transferred to another department within the organization. The purpose is of course to give the employee new possibilities to develop him/herself within another area of expertise. Lateral movement is a serious option when the employee can’t be promoted (assuming this employee’s performance is considered good), and the company doesn’t want to lose this valuable employee. Eventually the idea is that this employee could achieve a promotion within the other department.

Vertical (Upwards) Movement: The upward movement is called a promotion. If an employee has sufficient skills and knowledge, and he/she has few changes to develop him/herself in the current job, a promotion would be a wise choice. It’s up to the manager to decide whether this employee is ready for a promotion, because there wouldn’t be many employee rejecting such a chance. Furthermore, promotion can be seen as a promotion within or without the current department. So, then a combination between lateral and vertical movement is made.

Vertical (Downwards) Movement: Some employees are just not suitable for carrying heavy responsibilities and managing administrative duties. As said before, not many employees will refuse a promotion. After a while, some employees may find out that the previous job was more suitable for them, and therefore a downward movement has to be considered. Of course, this move will only be considered if managers feel that the employee is valuable for the organization, but then in another position.

Orientation: Sometimes employees only know that the current job situation not satisfying. They do not know what can be done to solve this. In this case, a manager could offer the employee a possibility to search for other options within the company. This could be done by participating in task groups, seminars and/or training activities. As a result, the employee could decide to be transferred to another department. It is also possible to stay in the current position, or if nothing else can be found, to leave the organization.

Job Enrichment: Enrichment is a decent possibility if promotion is not possible. Enrichment means that the current job will be “enriched” with more (responsible) tasks. One could think of cooperating in (or leading) a task group, doing more administrative tasks or assisting the manager. Usually employees tend to know what they seek for. It would be wisely for managers to involve the employee, so that they can come to a suitable solution.

The Changing nature of careers
Sullivan is examining how the way we view the concept of a career differently today than we used to. Earlier a career represented environmental stability, there was an intrafirm focus and hierarchical assumptions. Workers used to exchange loyalty for job security but today employees exchange performance for continuous learning and marketability. According to Sullivan this leads to decreased job security, decreased loyalty and worker cynicism. The new norm nowadays is what Sullivan calls "boundaryless careers" which is defined as "a sequence of job opportunities that go beyond the boundaries of a single employment setting". The opposite is called a "traditional career.(Sullivan 1999)

title: Managing Human Resource Flows


text:
User:Egm6321.f09.team5/HW6

Problem Statement
Pg. 31-1 Circular Cylinder Coordinates

$$x=x_1=rcos(\theta)=\xi_1cos(\xi_2)$$

$$y=x_2=rsin(\theta)=\xi_1sin(\xi_2)$$

$$z=x_3=\xi_3$$

1.) Find {$$dx_i$$}={$$dx_1$$,$$dx_2$$,$$dx_3$$} in terms of {$$\xi_i$$}={$$\xi_1$$,$$\xi_2$$,$$\xi_3$$} and {$$d\xi_k$$}

2.) $$ds^2 = \sum_i(dx_i)^2 = \sum_k(h_k)^2(d\xi_k)^2$$ and identify {$$h_i$$} in therms of {$$\xi_j$$}

3.) Find Laplacian in these curvilinear coordinates

Problem Solution
1.)

Take the derivative of $$x_1, x_2, and x_3$$ with respect to $$\xi_i$$

$$dx_1=\frac{\partial x_1}{\partial \xi_1}d\xi_1+\frac{\partial x_1}{\partial \xi_2}d\xi_2+\frac{\partial x_1}{\partial \xi_3}d\xi_3$$

$$dx_1=cos(\xi_2)d\xi_1-\xi_1sin(\xi_2)d\xi_2+0$$

$$dx_1=cos(\xi_2)d\xi_1-\xi_1sin(\xi_2)d\xi_2$$

$$dx_2=\frac{\partial x_2}{\partial \xi_1}d\xi_1+\frac{\partial x_2}{\partial \xi_2}d\xi_2+\frac{\partial x_2}{\partial \xi_3}d\xi_3$$

$$dx_2=sin(\xi_2)d\xi_1-\xi_1cos(\xi_2)d\xi_2+0$$

$$dx_2=sin(\xi_2)d\xi_1-\xi_1cos(\xi_2)d\xi_2$$

$$dx_1=\frac{\partial x_3}{\partial \xi_1}d\xi_1+\frac{\partial x_3}{\partial \xi_2}d\xi_2+\frac{\partial x_3}{\partial \xi_3}d\xi_3$$

$$dx_1=0d\xi_1-0d\xi_2+d\xi_3$$

$$dx_1=d\xi_3$$

2.)

Using the given for $$ds^2$$:

$$ds^2=cos^2(\xi_2)(d\xi_1)^2-2\xi_1cos(\xi_2)sin(\xi_2)d\xi_1d\xi_2+(\xi_1)^2sin^2(\xi_2)(d\xi_2)^2+sin^2(\xi_2)(d\xi_1)^2+2\xi_1cos(\xi_2)sin(\xi_2)d\xi_1d\xi_2+(d\xi_3)^2$$

$$ds^2=cos^2(\xi_2)(d\xi_1)^2+(\xi_1)^2sin^2(\xi_2)(d\xi_2)^2+sin^2(\xi_2)(d\xi_1)^2+(d\xi_3)^2 $$

$$ds^2=(d\xi_1)^2+(\xi_1)^2(d\xi_2)^2+(d\xi_3)^2$$

The {$$h_i$$} are the square root of the coefficients on each term so: $$h_1=1, h_2=\xi_1, h_3=1$$

3.)

A general formula for finding the Laplacian of $$\psi$$ in any orthogonal coordinate system is: $$\nabla^2\psi=\frac{1}{h_1h_2h_3}[\frac{\partial}{\partial \xi_1}(\frac{h_2h_3}{h_1}\frac{\partial \psi}{\partial \xi_1})+\frac{\partial}{\partial \xi_2}(\frac{h_1h_3}{h_2}\frac{\partial \psi}{\partial \xi_2})+\frac{\partial}{\partial \xi_3}(\frac{h_1h_2}{h_3}\frac{\partial \psi}{\partial \xi_3})]$$

$$\nabla^2\psi=\frac{1}{\xi_1}[\frac{\partial}{\partial \xi_1}(\xi_1\frac{\partial \psi}{\partial \xi_1})+\frac{\partial}{\partial \xi_2}(\frac{1}{\xi_1}\frac{\partial \psi}{\partial \xi_2})+\frac{\partial}{\partial \xi_3}(\xi_1\frac{\partial \psi}{\partial \xi_3})]$$

$$\nabla^2\psi=\frac{1}{\xi_1}\frac{\partial}{\partial \xi_1}(\xi_1\frac{\partial \psi}{\partial \xi_1})+(\frac{1}{\xi_1^2}\frac{\partial^2 \psi}{\partial \xi_2^2})+(\frac{\partial^2 \psi}{\partial \xi_3^2})$$

Problem Statement
Pg. 31-1: Spherical Coordinates. Repeat above steps. Use Astronomic Convention.

$$x=x_1=rcos(\theta)cos(\phi)=\xi_1cos(\xi_3)cos(\xi_2)$$

$$y=x_2=rcos(\theta)sin(\phi)=\xi_1cos(\xi_3)sin(\xi_2)$$

$$z=x_3=rsin(\theta)=\xi_1sin(\xi_3)$$

1.) Find {$$dx_i$$}={$$dx_1$$,$$dx_2$$,$$dx_3$$} in terms of {$$\xi_i$$}={$$\xi_1$$,$$\xi_2$$,$$\xi_3$$} and {$$d\xi_k$$}

2.) $$ds^2 = \sum_i(dx_i)^2 = \sum_k(h_k)^2(d\xi_k)^2$$ and identify {$$h_i$$} in therms of {$$\xi_j$$}

3.) Find Laplacian in these curvilinear coordinates

Problem Solution
1.)

Take the derivative of $$x_1, x_2, and x_3$$ with respect to $$\xi_i$$

$$dx_1=\frac{\partial x_1}{\partial \xi_1}d\xi_1+\frac{\partial x_1}{\partial \xi_2}d\xi_2+\frac{\partial x_1}{\partial \xi_3}d\xi_3$$

$$dx_1=cos(\xi_3)cos(\xi_2)d\xi_1-\xi_1cos(\xi_3)sin(\xi_2)d\xi_2-\xi_1sin(\xi_3)cos(\xi_2)d\xi_3$$

$$dx_2=\frac{\partial x_2}{\partial \xi_1}d\xi_1+\frac{\partial x_2}{\partial \xi_2}d\xi_2+\frac{\partial x_2}{\partial \xi_3}d\xi_3$$

$$dx_2=cos(\xi_3)sin(\xi_2)d\xi_1+\xi_1cos(\xi_3)cos(\xi_2)d\xi_2-\xi_1sin(\xi_3)sin(\xi_2)d\xi_3$$

$$dx_1=\frac{\partial x_3}{\partial \xi_1}d\xi_1+\frac{\partial x_3}{\partial \xi_2}d\xi_2+\frac{\partial x_3}{\partial \xi_3}d\xi_3$$

$$dx_1=sin(\xi_3)d\xi_1+0+\xi_1cos(\xi_3)d\xi_3$$

$$dx_1=sin(\xi_3)d\xi_1+\xi_1cos(\xi_3)d\xi_3$$

2.)

Using the given for $$ds^2$$:

$$ds^2=[cos(\xi_3)cos(\xi_2)d\xi_1-\xi_1cos(\xi_3)sin(\xi_2)d\xi_2-\xi_1sin(\xi_3)cos(\xi_2)d\xi_3]^2+[cos(\xi_3)sin(\xi_2)d\xi_1+\xi_1cos(\xi_3)cos(\xi_2)d\xi_2-\xi_1sin(\xi_3)sin(\xi_2)d\xi_3]^2+[sin(\xi_3)d\xi_1+\xi_1cos(\xi_3)d\xi_3]^2$$

$$ds^2=(d\xi_1)^2+(\xi_1)^2cos^2(\xi_3)(d\xi_2)^2+\xi_1^2(d\xi_3)^2$$

The {$$h_i$$} are the square roots of the coefficients on each term so: $$h_1=1, h_2=\xi_1cos(\xi_3), h_3=\xi_1$$

3.)

A general formula for finding the Laplacian of $$\psi$$ in any orthogonal coordinate system is: $$\nabla^2\psi=\frac{1}{h_1h_2h_3}[\frac{\partial}{\partial \xi_1}(\frac{h_2h_3}{h_1}\frac{\partial \psi}{\partial \xi_1})+\frac{\partial}{\partial \xi_2}(\frac{h_1h_3}{h_2}\frac{\partial \psi}{\partial \xi_2})+\frac{\partial}{\partial \xi_3}(\frac{h_1h_2}{h_3}\frac{\partial \psi}{\partial \xi_3})]$$

$$\nabla^2\psi=\frac{1}{\xi_1^2cos(\xi_3)}[\frac{\partial}{\partial \xi_1}(\xi_1^2cos(\xi_3)\frac{\partial \psi}{\partial \xi_1})+\frac{\partial}{\partial \xi_2}(\frac{1}{cos(\xi_3)}\frac{\partial \psi}{\partial \xi_2})+\frac{\partial}{\partial \xi_3}(\xi_1cos(\xi_3)\frac{\partial \psi}{\partial \xi_3})]$$

$$\nabla^2\psi=\frac{1}{\xi_1^2}\frac{\partial}{\partial \xi_1}(\xi_1^2\frac{\partial \psi}{\partial \xi_1})+\frac{1}{\xi_1^2cos^2(\xi_3)}(\frac{\partial^2 \psi}{\partial \xi_2^2})+\frac{1}{\xi_1^2cos(\xi_3)}\frac{\partial \psi}{\partial \xi_3}(cos(\xi_3)(\frac{\partial \psi}{\partial \xi_3}))$$

Problem Statement
Pg. 31-2 Find Laplacian in the Math/Physics Convention (r,phi,theta_Bar)

$$\theta_{bar}=\frac{\pi}{2}-\theta$$

Problem Solution
$$x=x_1=rsin(\theta_{bar})cos(\phi)=\xi_1sin(\xi_{3bar})cos(\xi_2)$$

$$y=x_2=rsin(\theta_{bar})sin(\phi)=\xi_1sin(\xi_{3bar})sin(\xi_2)$$

$$z=x_3=rcos(\theta_{bar})=\xi_1cos(\xi_{3bar})$$

$$\xi_{3bar}=\frac{\pi}{2}-\xi_3$$

If this is true, then:

$$sin(\xi_3) = cos(\xi_3bar)$$

$$cos(\xi_3) = sin(\xi_3bar)$$

$$\frac{\partial}{\partial\xi_3}=-\frac{\partial}{\partial\xi_3bar}$$

Plugging this into our spherical Laplacian equation derived in Problem #2 (Part 3):

$$\nabla^2\psi=\frac{1}{\xi_1^2}\frac{\partial}{\partial \xi_1}(\xi_1^2\frac{\partial \psi}{\partial \xi_1})+\frac{1}{\xi_1^2sin^2(\xi_{3bar})}(\frac{\partial^2 \psi}{\partial \xi_2^2})-\frac{1}{\xi_1^2sin(\xi_{3bar})}\frac{\partial \psi}{\partial \xi_{3bar}}(sin(\xi_{3bar})(-\frac{\partial \psi}{\partial \xi_{3bar}})$$

$$\nabla^2\psi=\frac{1}{\xi_1^2}\frac{\partial}{\partial \xi_1}(\xi_1^2\frac{\partial \psi}{\partial \xi_1})+\frac{1}{\xi_1^2sin^2(\xi_{3bar})}(\frac{\partial^2 \psi}{\partial \xi_2^2})+\frac{1}{\xi_1^2sin(\xi_{3bar})}\frac{\partial \psi}{\partial \xi_{3bar}}(sin(\xi_{3bar})(\frac{\partial \psi}{\partial \xi_{3bar}})$$

Problem Statement
Pg. 31-2. Solve equation 1 to yield 2 and 3.

Equation 1:

$$\lambda(\lambda+1)=n(n+1)$$

Equation 2:

$$\lambda_1=n$$

Equation 3:

$$\lambda_2=-(n+1)$$

Problem Solution
To solve, distribute the lambda in equation 1:

$$\lambda^2+\lambda=n(n+1)$$

This can be rearranged to show that it is a quadratic equation:

$$\lambda^2+\lambda-n(n+1)=0$$

This can be solved with the quadratic equation:

$$\lambda_{1,2}=\frac{-1\pm \sqrt{1^2-4(1)(-n(n+1))}}{2(1)}$$

$$\lambda_1=\frac{-1+\sqrt{1+4n(n+1)}}{2}=\frac{-1+\sqrt{(2n+1)^2}}{2}$$

$$\lambda_1=n$$

$$\lambda_2=\frac{-1-\sqrt{1+4n(n+1)}}{2}=\frac{-1-\sqrt{(2n+1)^2}}{2}$$

$$\lambda_2=-(n+1)$$

Problem Statement
Pg. 31-2.3 Show Equation 6 is equal to equation 7

Equation 6:

$$P_n(x)=\sum_{i=0}^{[n/2]}(-1)^i\frac{(2n-2i)!x^{n-2i}}{2^ni!(n-1)!(n-2i)!}$$

Equation 7:

$$P_n(x)=\sum_{i=0}^{[n/2]}(-1)^ix^{n-2i}\frac{1*3*...(2n-2i-1)}{2^ii!(n-2i)!}$$

Problem Solution
Solution

Problem Statement
Pg. 31-3. Show the steps to verify that 1-5, Legendre Polynomials when n=0...4. can be written as 6 or 7

$$ P_{0}(x) = 1 $$ Equation 1

$$ P_{1}(x) = x $$ Equation 2

$$ P_{2}(x) = \frac{1}{2}(3x^{2}-1) $$ Equation 3

$$ P_{3}(x) = \frac{1}{2}(5x^{3}-3x) $$ Equation 4

$$ P_{4}(x) = \frac{35}{8}x^{4} - \frac{15}{4}x^{2} + \frac{3}{8} $$ Equation 5

$$ P_{n}(x) = \sum_{0}^{[n/2]}(-1)^{i}\frac{(2n-2i)!x^{n-2i}}{2^{n}i!(n-i)!(n-2i)!} $$ Equation 6

$$P_n(x) = \sum_{i=0}^{[n/2]}\frac{1 \cdot 3 \cdot \cdot \cdot \left(2n - 2i - 1\right)}{2^i i! (n-2i)!} (-1)^i x^{n-2i}$$ Equation 7

Problem Solution
Utilizing Equation 5 from lecture 31 page 3, we will calculate the Legendre polynomial for n = 0:4. We will then compare our calculated value of the Legendre polynomial with those provided in class.

for n=0

$$ P_{0}(x) = \sum_{0}^{0}(-1)^{i}\frac{(-2i)!x^{-2i}}{i!(-i)!(-2i)!} $$

$$ P_{0}(x) = (-1)^{0}\frac{(0)!x^{0}}{0!(0)!(0)!} = 1 $$ $$P_{0}(x)=0$$

for n=1

$$ P_{1}(x) = \sum_{0}^{0}(-1)^{i}\frac{(2-2i)!x^{1-2i}}{2^{1}i!(1-i)!(1-2i)!} $$

$$ P_{1}(x) = (-1)^{0}\frac{(2)!x^{1}}{2^{1}0!(1)!(1)!} = x $$

$$P_{1}(x)=2!x/2$$

$$P_{1}(x)=x$$ for n=2

$$P_{2}(x)= \frac{(4-0)!x^2}{2^2*2*2}-\frac{(4-2)!}{2^2}$$

$$P_{2}(x)=\frac{24*x^2}{16}-\frac{2}{4}$$ $$P_{2}(x)=0.5(3x^2-1)$$ for n=3

$$P_{3}(x)=\frac{(2*3)!x^3}{2^3*6*6} - \frac {(2(2)-2)!x}{2^3*2}$$

$$P_{3}(x)=\frac{720x^3}{288}-\frac{24x}{16}$$ $$P_{3}(x)= \frac{1}{2}(5x^3-3x)$$

for n=4

$$P_4(x)=\frac{8!x^4}{2^4*4!*4!}-\frac{(8-2)!x^2}{2^4*3!}+\frac{(8-4)!}{2^4*2*(4-2)!}$$

$$P_4(x)=\frac{40320x^4}{9216}-\frac{360x^2}{96}+\frac{24}{64}$$ $$P_4(x)=\frac{35x^4}{8}-\frac{15x^2}{4}+\frac{3}{8}$$

These values obtained utilizing Equation 6 are identical to the Legendre polynomials given in class.

Problem Statement
Pg. 32-1: Verify that p0-p4 eq 1-5 are solutions of the Legendre Equation.

$$ P_{0}(x) = 1 $$ Equation 1

$$ P_{1}(x) = x $$ Equation 2

$$ P_{2}(x) = \frac{1}{2}(3x^{2}-1) $$ Equation 3

$$ P_{3}(x) = \frac{1}{2}(5x^{3}-3x) $$ Equation 4

$$ P_{4}(x) = \frac{35}{8}x^{4} - \frac{15}{4}x^{2} + \frac{3}{8} $$ Equation 5

$$ F=(1-x^{2})y^{''} - 2xy^{'} + n(n+1)y = 0 \qquad $$  Lagendre Equation

Problem Solution
When n=0:

$$P_{0}(x) = 1 = y $$

$$y^{'}=0 \qquad \qquad y^{''} = 0 $$

Using the solution with the Legendre Equation:

$$ F = (1-x^{2}y^{''} - 2xy^{'} + 0 = 0 $$

$$ F = (1-x^{2}(0) - 2x(0) + 0 = 0 $$

$$ 0 = 0 $$

The Legendre Polynomial is indeed a solution to the Legendre Equation.

When n=1:

$$P_{1}(x) = x = y $$

$$y^{'}= 1 \qquad \qquad y^{''} = 0 $$

Using the solution with the Legendre Equation:

$$ F=(1-x^{2})y^{''} - 2xy^{'} + 2y = 0 $$

$$ F=(1-x^{2})(0) - 2x(1) + 2(x) = 0 $$

$$ F = 0 = 0 $$

The Legendre Polynomial when n=1 is a solution of the Legendre Equation

When n=2:

$$ P_{2}(x)=\frac{1}{2}(3x^{2}-1) = y $$

$$y^{'}= 3x \qquad \qquad y^{''} = 3 $$

Using the solution with the Legendre Equation:

$$ F=(1-x^{2})y^{''} - 2xy^{'} + 6y = 0 $$

$$ F=(1-x^{2})(3) - 2x(3x) + 6(1.5x^{2} - .5) = 0 $$

$$ F= 3- 3x^{2} - 6x^{2} + 9x^{2} - 3 = 0 $$

$$ F = 0 = 0 $$

The Legendre Polynomial when n=2 is a solution of the Legendre Equation

When n=3:

$$ P_{3}(x)=\frac{1}{2}(5x^{3}-3x) = y $$

$$y^{'}= 7.5x^{2} - 1.5 \qquad \qquad y^{''} = 15x $$

Using the solution with the Legendre Equation:

$$ F=(1-x^{2})y^{''} - 2xy^{'} + 12y = 0 $$

$$ F=(1-x^{2})(15x) - 2x(7.5x^{2} - 1.5) + 12(2.5x^{3} - 1.5x) = 0 $$

$$ F= 15x - 15x^{3} + 3x + 30x^{3} - 18x = 0 $$

$$ F = 0 = 0 $$

The Legendre Polynomial when n=3 is a solution of the Legendre Equation

When n=4:

$$ P_{4}(x)= \frac{35}{8}x^{4} - \frac{15}{4}x^{2} + \frac{3}{8}= y $$

$$y^{'}= 17.5x^{3} - 7.5x \qquad \qquad y^{''} = 52.5x^{2}-7.5 $$

Using the solution with the Legendre Equation:

$$ F=(1-x^{2})y^{''} - 2xy^{'} + 20y = 0 $$

$$ F=(1-x^{2})(52.5x^{2}-7.5) - 2x(17.5x^{3} - 7.5x) + 20(\frac{35}{8}x^{4} - \frac{15}{4}x^{2} + \frac{3}{8}) = 0 $$

$$ F= (87.5x^{4} - 52.5x^{4} -35x^{4}) + (52.5x^{2} + 7.5x^{2} + 15x^{2} - 75x^{2}) + (-7.5 + 7.5) = 0 $$

$$ F = 0 = 0 $$

The Legendre Polynomial when n=4 is a solution of the Legendre Equation

Problem Statement
Pg. 32-1 Obtain the separate equations for laplace equations of circular cylinder coordinates and Identify the Bessel Diff. Equation

$$x^2y''+xy'+(x^2-\nu^2)y=0$$

Problem Solution
Laplace equation:

$$\nabla^2\psi=0$$

Having derived this in Problem #1:

$$\nabla^2\psi=\frac{1}{\xi_1}\frac{\partial}{\partial \xi_1}(\xi_1\frac{\partial \psi}{\partial \xi_1})+(\frac{1}{\xi_1^2}\frac{\partial^2 \psi}{\partial \xi_2^2})+(\frac{\partial^2 \psi}{\partial \xi_3^2})=0$$

Making an assumption:

$$\psi(\xi_1,\xi_2,\xi_3)=X_1(\xi_1)X_2(\xi_2)X_3(\xi_3)$$

Plugging this into the Laplace equation:

$$\nabla^2\psi=X_2X_3\frac{1}{\xi_1}\frac{\partial}{\partial \xi_1}(\xi_1\frac{\partial X_1}{\partial \xi_1})+X_1X_3(\frac{1}{\xi_1^2}\frac{\partial^2 X_2}{\partial \xi_2^2})+X_1X_2(\frac{\partial^2 X_3}{\partial \xi_3^2})=0$$

To simplify, divide the equation by:

$$X_1X_2X_3$$

Which yields:

$$\nabla^2\psi=\frac{1}{X_1}\frac{1}{\xi_1}\frac{\partial}{\partial \xi_1}(\xi_1\frac{\partial X_1}{\partial \xi_1})+\frac{1}{X_2}\frac{1}{\xi_1^2}(\frac{\partial^2 X_2}{\partial \xi_2^2})+\frac{1}{X_3}(\frac{\partial^2 X_3}{\partial \xi_3^2})=0$$

Assuming that:

$$\frac{1}{X_3}(\frac{\partial^2 X_3}{\partial \xi_3^2})=constant$$

Then a separation constant can be used for this term:

$$\frac{1}{X_3}(\frac{\partial^2 X_3}{\partial \xi_3^2})=\alpha^2$$

$$\frac{1}{X_1}\frac{1}{\xi_1}\frac{\partial}{\partial \xi_1}(\xi_1\frac{\partial X_1}{\partial \xi_1})+\frac{1}{X_2}\frac{1}{\xi_1^2}(\frac{\partial^2 X_2}{\partial \xi_2^2})+\alpha^2=0$$

Multiplying by $$\xi_1^2$$:

$$\frac{\xi_1}{X_1}\frac{\partial}{\partial \xi_1}(\xi_1\frac{\partial X_1}{\partial \xi_1})+\frac{1}{X_2}(\frac{\partial^2 X_2}{\partial \xi_2^2})+\xi_1^2\alpha^2=0$$

Now, assuming that:

$$\frac{1}{X_2}(\frac{\partial^2 X_2}{\partial \xi_2^2})=constant$$

Then a separation constant can be used for this term as well:

$$\frac{1}{X_2}(\frac{\partial^2 X_2}{\partial \xi_2^2})=-\beta^2$$

$$\frac{\xi_1}{X_1}\frac{\partial}{\partial \xi_1}(\xi_1\frac{\partial X_1}{\partial \xi_1})-\beta^2+\xi_1^2\alpha^2=0$$

Multiplying by $$X_1$$:

$$\xi_1\frac{\partial}{\partial \xi_1}(\xi_1\frac{\partial X_1}{\partial \xi_1})+X_1(\xi_1^2\alpha^2-\beta^2)=0$$

Using the chain rule:

$$\xi_1(\frac{\partial X_1}{\partial \xi_1})+\xi_1^2(\frac{\partial^2 X_1}{\partial \xi_1^2})+X_1(\xi_1^2\alpha^2-\beta^2)=0$$

Rearranging:

$$\xi_1^2(\frac{\partial^2 X_1}{\partial \xi_1^2})+\xi_1(\frac{\partial X_1}{\partial \xi_1})+(\xi_1^2\alpha^2-\beta^2)X_1=0$$

Divide by $$\xi_1^2$$:

$$\frac{\partial^2 X_1}{\partial \xi_1^2}+\frac{1}{\xi_1}(\frac{\partial X_1}{\partial \xi_1})+(\alpha^2-\frac{\beta^2}{\xi_1^2})X_1=0$$

Have to transform variables, so assume:

$$X_1=y(x)$$

$$\frac{\partial X_1}{\partial \xi_1}=\frac{\partial y(\alpha\xi_1)}{\partial \xi_1}=\frac{\partial y(\alpha\xi_1)}{\partial \alpha\xi_1}\frac{\partial \alpha\xi_1}{\partial \xi_1}=\alpha\frac{\partial y(\alpha\xi_1)}{\partial \alpha\xi_1}$$

$$\frac{\partial^2 X_1}{\partial \xi_1^2}=\alpha^2\frac{\partial^2 y(\alpha\xi_1)}{\partial (\alpha\xi_1)^2}$$

$$\xi_1^2(\alpha^2\frac{\partial^2 y(\alpha\xi_1)}{\partial (\alpha\xi_1)^2})+\xi_1(\alpha\frac{\partial y(\alpha\xi_1)}{\partial \alpha\xi_1})+(\xi_1^2\alpha^2-\beta^2)y(\alpha\xi_1)=0$$

If:

$$x=\alpha\xi_1$$

$$\nu=\beta$$

Then:

$$x^2\frac{\partial^2 y(x)}{\partial x^2}+x\frac{\partial y(x)}{\partial x}+(x^2-\nu^2)y(x)=0$$

Which is Bessel's differential equation:

$$x^2y''+xy'+(x^2-\nu^2)y=0$$

Problem Statement
pg. 33-3: Regarding even and odd equations

$$ f = \sum{i}{}g_{i} $$

Show that if {gi} is odd then f is odd. Show that if {gi} is even then f is even.

Problem Solution
An even function is defined as follows:

$$ f(-x) = f(x) $$

An odd function is defined as follows:

$$ f(-x) = -f(x) $$

The properties for even and odd functions state the following: Properties of Even and Odd Functions

The sum of even functions is even

The sum of odd functions is odd

By this rule if g{i} is odd then f will also be add. Also if g{i} is even then f will also be even.

Problem Statement
pg. 33-3

Given:

$$ P_{n}(x) = \sum_{0}^{[n/2]}(-1)^{i}\frac{(2n-2i)!x^{n-2i}}{2^{n}i!(n-i)!(n-2i)!} $$

Show $$ P_{2k}(x)\qquad \qquad $$     is even k=0,1,2...

Show $$ P_{2k+1}(x) \qquad \qquad $$    is odd

Problem Solution
$$ P_{2k}(x) = \sum_{0}^{k}(-1)^{i}\frac{(4k-2i)! \color{blue}{(x)^{2k-2i}}}{2^{2k}i!(2k-i)!(2k-2i)!} $$

$$ P_{2k}(-x) = \sum_{0}^{k}(-1)^{i}\frac{(4k-2i)! \color{blue}{(-x)^{2k-2i}}}{2^{2k}i!(2k-i)!(2k-2i)!} $$

Since the terms highlighted in blue are raised to an even power, this makes them equivalent. This equivalence makes the Polynomial is considered to be even.

$$ P_{2k+1}(x) = \sum_{0}^{[\frac{2k+1}{2}]}(-1)^{i}\frac{(4k+2-2i)! \color{red}{(x)^{2k+1-2i}}}{2^{2k+1}i!(2k+1-i)!(2k+1-2i)!} $$

$$ P_{2k+1}(-x) = \sum_{0}^{[\frac{2k+1}{2}]}(-1)^{i}\frac{(4k+2-2i)! \color{red}{(-x)^{2k+1-2i}}}{2^{2k+1}i!(2k+1-i)!(2k+1-2i)!} $$

Since the terms highlighted in red will always be different since they will be raised to an odd power, this makes them not equivalent. Because of this the polynomial is considered to be odd.

Problem Statement
Pg. 33-4.

$$q(x)=\sum_{i=0}^4c_ix^i$$

$$c_0=3$$, $$c_1=10$$, $$c_2=15$$, $$c_3=-1$$, $$c_4=5$$

1.) find {$$a_i$$} such that $$q=\sum_{i=0}^4a_iP_i$$

2.) plot $$q=\sum_{i=0}^4c_ix^i=\sum_{i=0}^4a_iP_i$$

Problem Solution
1.)

$$P_i(x)=Ax_i$$

 * Where $$P_i(x)$$ represents Legendre polynomials

$$\left[\begin{matrix} P_0(x) \\ P_1(x) \\ P_2(x) \\ P_3(x) \\ P_4(x) \end{matrix}\right]= \left[\begin{matrix} 1 \\ x \\ \frac{3x^2}{2}-\frac{1}{2} \\ \frac{5x^3}{2}-\frac{3x}{2} \\ \frac{35}{8}x^4-\frac{15}{4}x^2+\frac{3}{8} \end{matrix}\right]= \left[\begin{matrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ -\frac{1}{2} & 0 & \frac{3}{2} & 0 & 0 \\ 0 & -\frac{3}{2} & 0 & \frac{5}{2} & 0 \\ \frac{3}{8} & 0 & -\frac{15}{4} & 0 & \frac{35}{8} \\ \end{matrix}\right]\left[\begin{matrix} 1 \\ x \\ x^2 \\ x^3 \\ x^4 \end{matrix}\right]$$

We know that:

$$q(x)=a_iP(x)=c_ix^i$$

If:

$$P_i(x)=Ax^i$$

Then:

$$a_iAx_i=c_ix_i$$

Simplifying:

$$a_i=A^{-1}c_i$$

$$c_i=\left[\begin{matrix} 3 \\ 10 \\ 15 \\ -1 \\ 5 \\ \end{matrix}\right]$$

$$A=\left[\begin{matrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ -\frac{1}{2} & 0 & \frac{3}{2} & 0 & 0 \\ 0 & -\frac{3}{2} & 0 & \frac{5}{2} & 0 \\ \frac{3}{8} & 0 & -\frac{15}{4} & 0 & \frac{35}{8} \\ \end{matrix}\right]$$

$$A^{-1}=\left[\begin{matrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ \frac{1}{3} & 0 & \frac{2}{3} & 0 & 0 \\ 0 & \frac{3}{5} & 0 & \frac{2}{5} & 0 \\ \frac{1}{5} & 0 & \frac{4}{7} & 0 & \frac{8}{35} \\ \end{matrix}\right]$$

$$a_i=\left[\begin{matrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ \frac{1}{3} & 0 & \frac{2}{3} & 0 & 0 \\ 0 & \frac{3}{5} & 0 & \frac{2}{5} & 0 \\ \frac{1}{5} & 0 & \frac{4}{7} & 0 & \frac{8}{35} \\ \end{matrix}\right]\left[\begin{matrix} 3 \\ 10 \\ 15 \\ -1 \\ 5 \\ \end{matrix}\right]$$

$$a_i=\left[\begin{matrix} 3 \\ 10 \\ 11 \\ 5.6 \\ 10.314 \\ \end{matrix}\right]$$

Problem Statement
Pg. 34-1: HW added from pg. 22-2 King et al. Pg 28. Problem 1.1b

$$ xy^{''} + 2y^{'} + xy = 0 $$          Equation 0

1) Check for exactness, If not exact can it be made exact using integration factor.

2) Use the following trial solutions to find the homogeneous solution

3) Use the undetermined factor method to find the second homogeneous solution, knowing a homogeneous solution

1)
The First step is to check for exactness using the following conditions:

Condition 1: The ODE must be of the form:

$$F=f(x,y,p)y^{''}+g(x,y,p)$$            Equation 1

Where $$ P=y^{'} $$

Condition 2: The following Conditions must be met:

$$    \ f_{xx} + 2pf_{xy} + p^{2}f_{yy} = g_{xp} + pg_{yp} - g_{y}  \qquad $$      Equation 2

$$    \ f_{xp} + pf_{yp} + 2f_{y} = g_{pp}  \qquad $$         Equation 3

$$ f(x)=x $$

$$ g(x)= 2p + xy $$

$$ f_{x}= 1 \qquad \qquad  g_{y}= x $$

$$ f_{xx}= 0  \qquad \qquad g_{yp}= 0 $$

$$ f_{xy}= 0  \qquad \qquad  g_{x}= y  $$

$$ f_{y}= 0        \qquad \qquad \qquad  g_{xp}= 0 $$

$$ f_{yy}= 0    \qquad \qquad  g_{p}= 2 $$

$$ f_{yp}= g_{pp}= f_{xp}=0 $$

Equation 2 yields:

$$ 0 + 2p(0) + p^{2}(0) = 0 + p(0) - x $$

$$ 0 \neq -x $$

This shows that the ODE is not exact.

The next step is to check whether it can be made exact by using the integration factor method:

The integration factor to be used will be:

$$ h(x,y) = x^{m}y^{n} $$

multiplying the integration factor by the ODE yields the following:

$$ \left [ x^{m}y^{n} \right ] \left[ xy^{''} + 2y^{'} + xy \right ] = 0 $$

$$ x^{m+1}y^{n}y^{''} + 2x^{m}y^{n}y{'} + x^{m+1}y^{n+1} = 0 $$

The next step is to check once again for exactness:

$$ f = x^{m+1}y^{n} $$ $$ g= 2x^{m}y^{n}p + x^{m+1}y^{n+1} $$

$$ f_{x}= y^{n}(m+1)x^{m} \qquad \qquad  g_{y}= 2x^{m}ny^{n-1}p + x^{m+1}(m+1)y^{n} $$

$$ f_{xx}= y^{n}(m+1)mx^{m-1}  \qquad \qquad g_{yp}= 2x^{m}ny^{n-1} $$

$$ f_{xy}= ny^{n-1}(m+1)x^{m}  \qquad \qquad  g_{x}= 2y^{n}pmx^{m-1} + y^{n+1}(m+1)x^{m} $$

$$ f_{y}= x^{m+1}ny^{n-1}        \qquad \qquad \qquad  g_{xp}= 2y^{n}mx^{m-1} $$

$$ f_{yy}= x^{m+1}n(n-1)y^{n-2}    \qquad \qquad  g_{p}= 2x^{m}y^{n} $$

$$ f_{yp}= g_{pp}= f_{xp}=0 $$

Using Equation 3 in order to check for exactness yields the following:

$$ 0 + p(0) + 2(x^{m+1}y^{n-1}n \neq 0 $$

Therefore the new ODE is also concluded not to be exact.

2)
The next step is to find a homogeneous solution by using different trial solutions as follows:

Trial Solution 1

$$ y(x)=\exp^{rx} $$

$$ y^{'}(x) = r\exp^{rx} $$

$$ y^{''}(x) = r^{2}\exp^{rx} $$

using these results back in Equation 0 the following is obtained:

$$ x(r^{2}\exp^{rx}) + 2r\exp^{rx} + x\exp^{rx} = 0 $$

$$ \exp^{rx} (r^{2}(x) + 2r + x) = 0 $$

$$ r^{2}x + 2r + x = 0 $$

This trial solution is abandoned as it does not yield a root for the homogeneous solution

Trial Solution 2

$$ y(x)=x\exp^{rx} $$

$$ y^{'}(x) = (xr\exp^{rx} + x^{'}\exp^{rx}) $$

$$ y^{}(x) = r^{2}x\exp^{rx} + rx^{'}\exp^{rx} + x^{'}r\exp^{rx} + x^{}\exp^{rx} $$

Using these results in Equation 0 the following is obtained:

$$ x^{2}r^{2} + 2rxx^{'} + xx^{''} + 2xr + 2x^{'} = 0 $$

This trial solution is abandoned as it will not yield a root for the homogeneous solution

Trial Solution 3

$$ y(x)=x^{-1}\exp^{rx} $$

$$ y^{'}(x) = (rx^{-1}\exp^{rx} - x^{-2}\exp^{rx}) $$

$$ y^{''}(x) = \exp^{rx}(-2rx^{-2} + r^{2}x^{-1} + 2x^{-3}) $$

Using these results in Equation 0 the following is obtained:

$$\exp^{rx} \left [ r^{2} + 1 \right ] = 0 $$

From the previous equation the following result is found:

$$ r = \sqrt[]{-1} = i $$

$$ y(x) = x^{-1}\exp^{ix} $$

Using the de Moivre relationships the following form is obtained for the homogeneous solution:

$$ y(x)= x^{-1}(cosx + isinx) $$

3)
Given a homogeneous solution the full solution can be found using the undetermined factor method. The full solution is assumed to be of the form:

$$ y(x) = U(x)u_{1}(x) $$

Where $$ u_{1}(x) $$ is the known homogeneous solution.

$$ u_{1}(x)= \frac{sin(x)}{x} $$

Equation 0 is organized as follows:

$$ y^{''} + \frac{2}{x}y^{'} + y = 0 $$

where:

$$ a_{0}=1 \qquad \qquad a_{1} = 2x^{-1} $$

Because of the undetermined factor method the following can be written:

$$ 0 = U^{'} \left [ x^{-2}\sin(x) + 2x^{-1}\cos(x) \right ] + U^{''}\frac{\sin(x)}{x} $$  Equation 3

A variable is defined as follows:

$$ z=U^{'} $$

Equation 3 can be rewritten as follows:

$$ z\left [ x^{-2}\sin(x) + 2\cos(x)x^{-1} \right] + z^{'} \frac{\sin(x)}{x} $$

Solving the previous L1.ODE.VC using integration by parts and then solving for $$ U $$ the following is identified:

$$ u_{2}(x) = u_{1} \int{}^{x} \frac{1}{(u_{1}(t))^{2}} \exp^{-\int{}^{t} a_{1}(s)ds} dt $$

$$ \int{}^{t} a_{1}(s)ds = 2ln(x) $$

$$ \int{}^{x} \frac{1}{(u_{1}(t))^{2}} \exp^{-\int{}^{t} a_{1}(s)ds} dt = \int{}^{x}\frac{t^{2}}{\sin(t)^{2}}t^{-2} = -cot(x)  $$

$$ u_{2}(x) = \frac{\sin(x)}{x}\frac{-\cos(x)}{\sin(x)} = -\frac{\cos(x)}{x}  $$

This result confirms the results found in part 2 for the full homogeneous solution.

Problem Statement
Pg. 34-2

Problem Statement
Pg. 35-3: Verify table for Gauss Legendre Quadrature in wikipedia, for {xj} and {wj}, j= 1,...,n and n=1,...,5

Analytical Method to determine weights:

$$ \frac{-2}{(n+1)(P^{'}_{n}(x_{j})P_{n+1}(x_{j})} \qquad \qquad $$   Equation 1

After this Evaluate them numerically with the Abramowitz & Stegun Method.

Numerical Method to determine weights:

$$w_i = \frac{2}{\left( 1-x_i^2 \right) (P'_n(x_i))^2} \,\! \qquad \qquad $$  Equation 2

Problem Solution
When n=1

$$ P_{1}(x) = x $$

The value for x is the root of the polynomial:

$$ x_{i} = 0 $$

The weight is calculated using Equation 1.

$$ \frac{-2}{(2)(1)(1)(\frac{-1}{2})} = 2 $$

A matlab routine can be setup that will first find the roots of the Legendre Polynomial and then using Equation 1 to find the weights. The code for n=5 is as follows:

n=5 >> p=[7.875 0 -8.75 0 1.875 0]; >> roots(p) ans = 0    -0.9062     -0.5385     0.9062     0.5385     >> weight(0) ans = 0.0813    >> weight(-.9062) ans = 0.0233    >> weight(.9062) ans = 0.0233    >> weight(.5385) ans = -0.0865    >> weight(-.5385) ans = -0.0865    function w= weightnum(x) pp=39.375*(x^4)-26.25*(x^2)+1.875; pp1= 14.4375*(x^6) - 19.6875*(x^4) + 6.5625*(x^2)-2.1875; %n=5 w= -2/((6)*(pp)*(pp1));

The Following Results are Obtained:

The Results are the same as those presented in Wikipedia.

The Next step is to setup a Matlab routine that will calculate the weight evaluating it numerically with Equation 2.

n=2 >> p=[1.5 0 -.5]; >> roots(p) ans = 0.5774    -0.5774     >> weightnum(-.5774) ans = 0.9999    >> weightnum(.5774) ans = 0.9999    function w= weightnum(x) pp=39.375*x^4-26.25*x^2+1.875; %n=5 w= 2/((1-x^2)*(pp)^2);

Either method can be considered accurate to calculate the weights of the Polynomials.

Problem Statement
Pg. 36-4, Show that the statement is true.

Show that:

$$Q_0(x)=\frac{1}{2}log(\frac{1+x}{1-x})=tanh^{-1}(x)$$

Problem Solution
For the equations to be equal:

$$tanh(Q_0)=x$$

$$tanh(Q_0)=\frac{e^{2Q_0}-1}{e^{2Q_0}+1}$$

$$tanh(Q_0)=\frac{e^{2(\frac{1}{2}log(\frac{1+x}{1-x}))}-1}{e^{2(\frac{1}{2}log(\frac{1+x}{1-x}))}+1}$$

$$tanh(Q_0)=\frac{e^{log(\frac{1+x}{1-x}))}-1}{e^{log(\frac{1+x}{1-x}))}+1}$$

$$tanh(Q_0)=\frac{\frac{1+x}{1-x}-1}{\frac{1+x}{1-x}+1}$$

$$tanh(Q_0)=\frac{(1+x)-(1-x)}{(1+x)+(1-x)}$$

$$tanh(Q_0)=\frac{2x}{2}$$

$$tanh(Q_0)=x$$

Which shows that:

$$Q_0(x)=\frac{1}{2}log(\frac{1+x}{1-x})=tanh^{-1}(x)$$

Contributing Authors
--Egm6321.f09.team5.GV 03:03, 18 November 2009 (UTC) --Egm6321.f09.team5.bear 14:42, 18 November 2009 (UTC) --Egm6321.f09.team5.risher 19:08, 18 November 2009 (UTC) --Egm6321.f09.team5.Jason16:13, 18 Noverber 2009 (UTC)

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Abenaki/Primary school handout

About the Abenaki tribe
Who are the Abenaki?

The Abenaki (AH-buh-nah-kee) are a tribe that once lived

What does "Abenaki" mean?

The name Abenaki comes from the Algonquian (al-GONK-ee-un) word for "people from the east". It is also spelled Abanaki, Abnaki, or Abenaqui. Sometimes the Abenaki are also called the Wabanaki (WAH-buh-nah-kee), but this is actually the name of a group of four New England tribes, not just the Abenaki.

title: Abenaki/Primary school handout


text:
User:Egm6321.f09.team5.GV/HW6

Problem Statement
Pg. 31-1 Circular Cylinder Coordinates

$$x=x_1=rcos(\theta)=\xi_1cos(\xi_2)$$

$$y=x_2=rsin(\theta)=\xi_1sin(\xi_2)$$

$$z=x_3=\xi_3$$

1.) Find {$$dx_i$$}={$$dx_1$$,$$dx_2$$,$$dx_3$$} in terms of {$$\xi_i$$}={$$\xi_1$$,$$\xi_2$$,$$\xi_3$$} and {$$d\xi_k$$}

2.) $$ds^2 = \sum_i(dx_i)^2 = \sum_k(h_k)^2(d\xi_k)^2$$ and identify {$$h_i$$} in therms of {$$\xi_j$$}

3.) Find Laplacian in these curvilinear coordinates

Problem Solution
1.)

Take the derivative of $$x_1, x_2, and x_3$$ with respect to $$\xi_i$$

$$dx_1=\frac{\partial x_1}{\partial \xi_1}d\xi_1+\frac{\partial x_1}{\partial \xi_2}d\xi_2+\frac{\partial x_1}{\partial \xi_3}d\xi_3$$

$$dx_1=cos(\xi_2)d\xi_1-\xi_1sin(\xi_2)d\xi_2+0$$

$$dx_1=cos(\xi_2)d\xi_1-\xi_1sin(\xi_2)d\xi_2$$

$$dx_2=\frac{\partial x_2}{\partial \xi_1}d\xi_1+\frac{\partial x_2}{\partial \xi_2}d\xi_2+\frac{\partial x_2}{\partial \xi_3}d\xi_3$$

$$dx_2=sin(\xi_2)d\xi_1-\xi_1cos(\xi_2)d\xi_2+0$$

$$dx_2=sin(\xi_2)d\xi_1-\xi_1cos(\xi_2)d\xi_2$$

$$dx_1=\frac{\partial x_3}{\partial \xi_1}d\xi_1+\frac{\partial x_3}{\partial \xi_2}d\xi_2+\frac{\partial x_3}{\partial \xi_3}d\xi_3$$

$$dx_1=0d\xi_1-0d\xi_2+d\xi_3$$

$$dx_1=d\xi_3$$

2.)

Using the given for $$ds^2$$:

$$ds^2=cos^2(\xi_2)(d\xi_1)^2-2\xi_1cos(\xi_2)sin(\xi_2)d\xi_1d\xi_2+(\xi_1)^2sin^2(\xi_2)(d\xi_2)^2+sin^2(\xi_2)(d\xi_1)^2+2\xi_1cos(\xi_2)sin(\xi_2)d\xi_1d\xi_2+(d\xi_3)^2$$

$$ds^2=cos^2(\xi_2)(d\xi_1)^2+(\xi_1)^2sin^2(\xi_2)(d\xi_2)^2+sin^2(\xi_2)(d\xi_1)^2+(d\xi_3)^2 $$

$$ds^2=(d\xi_1)^2+(\xi_1)^2(d\xi_2)^2+(d\xi_3)^2$$

The {$$h_i$$} are the square root of the coefficients on each term so: $$h_1=1, h_2=\xi_1, h_3=1$$

3.)

A general formula for finding the Laplacian of $$\psi$$ in any orthogonal coordinate system is: $$\nabla^2\psi=\frac{1}{h_1h_2h_3}[\frac{\partial}{\partial \xi_1}(\frac{h_2h_3}{h_1}\frac{\partial \psi}{\partial \xi_1})+\frac{\partial}{\partial \xi_2}(\frac{h_1h_3}{h_2}\frac{\partial \psi}{\partial \xi_2})+\frac{\partial}{\partial \xi_3}(\frac{h_1h_2}{h_3}\frac{\partial \psi}{\partial \xi_3})]$$

$$\nabla^2\psi=\frac{1}{\xi_1}[\frac{\partial}{\partial \xi_1}(\xi_1\frac{\partial \psi}{\partial \xi_1})+\frac{\partial}{\partial \xi_2}(\frac{1}{\xi_1}\frac{\partial \psi}{\partial \xi_2})+\frac{\partial}{\partial \xi_3}(\xi_1\frac{\partial \psi}{\partial \xi_3})]$$

$$\nabla^2\psi=\frac{\partial}{\partial \xi_1}(\xi_1\frac{\partial \psi}{\partial \xi_1})+(\frac{1}{\xi_1^2}\frac{\partial^2 \psi}{\partial \xi_2^2})+(\frac{\partial^2 \psi}{\partial \xi_3^2})$$

Problem Statement
Pg. 31-1: Spherical Coordinates. Repeat above steps. Use Astronomic Convention.

$$x=x_1=rcos(\theta)cos(\phi)=\xi_1cos(\xi_3)cos(\xi_2)$$

$$y=x_2=rcos(\theta)sin(\phi)=\xi_1cos(\xi_3)sin(\xi_2)$$

$$z=x_3=rsin(\theta)=\xi_1sin(\xi_3)$$

1.) Find {$$dx_i$$}={$$dx_1$$,$$dx_2$$,$$dx_3$$} in terms of {$$\xi_i$$}={$$\xi_1$$,$$\xi_2$$,$$\xi_3$$} and {$$d\xi_k$$}

2.) $$ds^2 = \sum_i(dx_i)^2 = \sum_k(h_k)^2(d\xi_k)^2$$ and identify {$$h_i$$} in therms of {$$\xi_j$$}

3.) Find Laplacian in these curvilinear coordinates

Problem Solution
1.)

Take the derivative of $$x_1, x_2, and x_3$$ with respect to $$\xi_i$$

$$dx_1=\frac{\partial x_1}{\partial \xi_1}d\xi_1+\frac{\partial x_1}{\partial \xi_2}d\xi_2+\frac{\partial x_1}{\partial \xi_3}d\xi_3$$

$$dx_1=cos(\xi_3)cos(\xi_2)d\xi_1-\xi_1cos(\xi_3)sin(\xi_2)d\xi_2-\xi_1sin(\xi_3)cos(\xi_2)d\xi_3$$

$$dx_2=\frac{\partial x_2}{\partial \xi_1}d\xi_1+\frac{\partial x_2}{\partial \xi_2}d\xi_2+\frac{\partial x_2}{\partial \xi_3}d\xi_3$$

$$dx_2=cos(\xi_3)sin(\xi_2)d\xi_1+\xi_1cos(\xi_3)cos(\xi_2)d\xi_2-\xi_1sin(\xi_3)sin(\xi_2)d\xi_3$$

$$dx_1=\frac{\partial x_3}{\partial \xi_1}d\xi_1+\frac{\partial x_3}{\partial \xi_2}d\xi_2+\frac{\partial x_3}{\partial \xi_3}d\xi_3$$

$$dx_1=sin(\xi_3)d\xi_1+0+\xi_1cos(\xi_3)d\xi_3$$

$$dx_1=sin(\xi_3)d\xi_1+\xi_1cos(\xi_3)d\xi_3$$

2.)

Using the given for $$ds^2$$:

$$ds^2=[cos(\xi_3)cos(\xi_2)d\xi_1-\xi_1cos(\xi_3)sin(\xi_2)d\xi_2-\xi_1sin(\xi_3)cos(\xi_2)d\xi_3]^2+[cos(\xi_3)sin(\xi_2)d\xi_1+\xi_1cos(\xi_3)cos(\xi_2)d\xi_2-\xi_1sin(\xi_3)sin(\xi_2)d\xi_3]^2+[sin(\xi_3)d\xi_1+\xi_1cos(\xi_3)d\xi_3]^2$$

$$ds^2=(d\xi_1)^2+(\xi_1)^2cos^2(\xi_3)(d\xi_2)^2+\xi_1^2(d\xi_3)^2$$

The {$$h_i$$} are the square roots of the coefficients on each term so: $$h_1=1, h_2=\xi_1cos(\xi_3), h_3=\xi_1$$

3.)

A general formula for finding the Laplacian of $$\psi$$ in any orthogonal coordinate system is: $$\nabla^2\psi=\frac{1}{h_1h_2h_3}[\frac{\partial}{\partial \xi_1}(\frac{h_2h_3}{h_1}\frac{\partial \psi}{\partial \xi_1})+\frac{\partial}{\partial \xi_2}(\frac{h_1h_3}{h_2}\frac{\partial \psi}{\partial \xi_2})+\frac{\partial}{\partial \xi_3}(\frac{h_1h_2}{h_3}\frac{\partial \psi}{\partial \xi_3})]$$

$$\nabla^2\psi=\frac{1}{\xi_1^2cos(\xi_3)}[\frac{\partial}{\partial \xi_1}(\xi_1^2cos(\xi_3)\frac{\partial \psi}{\partial \xi_1})+\frac{\partial}{\partial \xi_2}(\frac{1}{cos(\xi_3)}\frac{\partial \psi}{\partial \xi_2})+\frac{\partial}{\partial \xi_3}(\xi_1cos(\xi_3)\frac{\partial \psi}{\partial \xi_3})]$$

$$\nabla^2\psi=\frac{1}{\xi_1^2}\frac{\partial}{\partial \xi_1}(\xi_1^2\frac{\partial \psi}{\partial \xi_1})+\frac{1}{\xi_1^2cos^2(\xi_3)}(\frac{\partial^2 \psi}{\partial \xi_2^2})+\frac{1}{\xi_1^2cos(\xi_3)}\frac{\partial \psi}{\partial \xi_3}(cos(\xi_3)(\frac{\partial \psi}{\partial \xi_3}))$$

Problem Statement
Pg. 31-2 Find Laplacian in the Math/Physics Convention (r,phi,theta_Bar)

$$\theta_{bar}=\frac{\pi}{2}-\theta$$

Problem Solution
$$x=x_1=rsin(\theta_{bar})cos(\phi)=\xi_1sin(\xi_{3bar})cos(\xi_2)$$

$$y=x_2=rsin(\theta_{bar})sin(\phi)=\xi_1sin(\xi_{3bar})sin(\xi_2)$$

$$z=x_3=rcos(\theta_{bar})=\xi_1cos(\xi_{3bar})$$

$$\xi_{3bar}=\frac{\pi}{2}-\xi_3$$

If this is true, then:

$$sin(\xi_3) = cos(\xi_3bar)$$

$$cos(\xi_3) = sin(\xi_3bar)$$

$$\frac{\partial}{\partial\xi_3}=-\frac{\partial}{\partial\xi_3bar}$$

Plugging this into our spherical Laplacian equation derived in Problem #2 (Part 3):

$$\nabla^2\psi=\frac{1}{\xi_1^2}\frac{\partial}{\partial \xi_1}(\xi_1^2\frac{\partial \psi}{\partial \xi_1})+\frac{1}{\xi_1^2sin^2(\xi_{3bar})}(\frac{\partial^2 \psi}{\partial \xi_2^2})-\frac{1}{\xi_1^2sin(\xi_{3bar})}\frac{\partial \psi}{\partial \xi_{3bar}}(sin(\xi_{3bar})(-\frac{\partial \psi}{\partial \xi_{3bar}})$$

$$\nabla^2\psi=\frac{1}{\xi_1^2}\frac{\partial}{\partial \xi_1}(\xi_1^2\frac{\partial \psi}{\partial \xi_1})+\frac{1}{\xi_1^2sin^2(\xi_{3bar})}(\frac{\partial^2 \psi}{\partial \xi_2^2})+\frac{1}{\xi_1^2sin(\xi_{3bar})}\frac{\partial \psi}{\partial \xi_{3bar}}(sin(\xi_{3bar})(\frac{\partial \psi}{\partial \xi_{3bar}})$$

Problem Statement
Pg. 31-2. Solve equation 1 to yield 2 and 3.

Problem Statement
Pg. 31-2.3 Show Equation 6 equals to equation 7

Problem Statement
Pg. 31-3. Show the steps to verify that 1-5, Legendre Polynomials when n=0...4. can be written as 6 or 7

$$ P_{0}(x) = 1 $$  Equation 1

$$ P_{1}(x) = x $$  Equation 2

$$ P_{2}(x) = \frac{1}{2}(3x^{2}-1) $$  Equation 3

$$ P_{3}(x) = \frac{1}{2}(5x^{3}-3x) $$  Equation 4

$$ P_{4}(x) = \frac{35}{8}x^{4} - \frac{15}{4}x^{2} + \frac{3}{8} $$  Equation 5

$$ P_{n}(x) = \sum_{0}^{[n/2]}(-1)^{i}\frac{(2n-2i)!x^{n-2i}}{2^{n}i!(n-i)!(n-2i)!} $$  Equation 6

$$P_n(x) = \sum_{i=0}^{[n/2]}\frac{1 \cdot 3 \cdot \cdot \cdot \left(2n - 2i - 1\right)}{2^i i! (n-2i)!} (-1)^i x^{n-2i}$$  Equation 7

Problem Solution
Using Equation 6, when n=0:

$$ P_{0}(x) = \sum_{0}^{0}(-1)^{i}\frac{(-2i)!x^{-2i}}{i!(-i)!(-2i)!} $$

$$ P_{0}(x) = (-1)^{0}\frac{(0)!x^{0}}{0!(0)!(0)!} = 1 $$

Using Equation 6, when n=1:

$$ P_{1}(x) = \sum_{0}^{0}(-1)^{i}\frac{(2-2i)!x^{1-2i}}{2^{1}i!(1-i)!(1-2i)!} $$

$$ P_{1}(x) = (-1)^{0}\frac{(2)!x^{1}}{2^{1}0!(1)!(1)!} = x $$

Using Equation 6, when n=2:

$$ P_{2}(x) = \sum_{0}^{1}(-1)^{i}\frac{(4-2i)!x^{2-2i}}{2^{2}i!(2-i)!(2-2i)!} $$

$$ P_{2}(x) = (-1)^{0}\frac{(4)!x^{2}}{2^{2}0!(2)!(2)!} + (-1)^{1}\frac{(4-2)!x^{2-2}}{2^{2}1!(2-1)!(2-2)!} = \frac{1}{2}(3x^{2}-1)$$

Using Equation 6, when n=3:

$$ P_{3}(x) = \sum_{0}^{1}(-1)^{i}\frac{(6-2i)!x^{3-2i}}{2^{3}i!(3-i)!(3-2i)!} $$

$$ P_{3}(x) = (-1)^{0}\frac{(6)!x^{3}}{2^{3}0!(3)!(3)!} + (-1)^{1}\frac{(6-2)!x^{3-2}}{2^{3}1!(3-1)!(3-2)!} = \frac{1}{2}(5x^{3}-3x)$$

Using Equation 6, when n=4:

$$ P_{4}(x) = \sum_{0}^{2}(-1)^{i}\frac{(8-2i)!x^{4-2i}}{2^{4}i!(4-i)!(4-2i)!} $$

$$ P_{4}(x) = (-1)^{0}\frac{(8)!x^{4}}{2^{4}0!(4)!(4)!} + (-1)^{1}\frac{(8-2)!x^{4-2}}{2^{4}1!(4-1)!(4-2)!} + (-1)^{2}\frac{(8-4)!x^{0}}{2^{4}2!(4-2)!(0)!} = \frac{35}{8}x^{4} - \frac{15}{4}x^{2} + \frac{3}{8} $$

Problem Statement
Pg. 32-1: Verify that p0-p4 eq 1-5 are solutions of the Legendre Equation.

$$ P_{0}(x) = 1 $$  Equation 1

$$ P_{1}(x) = x $$  Equation 2

$$ P_{2}(x) = \frac{1}{2}(3x^{2}-1) $$  Equation 3

$$ P_{3}(x) = \frac{1}{2}(5x^{3}-3x) $$  Equation 4

$$ P_{4}(x) = \frac{35}{8}x^{4} - \frac{15}{4}x^{2} + \frac{3}{8} $$  Equation 5

$$ F=(1-x^{2})y^{''} - 2xy^{'} + n(n+1)y = 0 \qquad $$   Lagendre Equation

Problem Solution
When n=0:

$$P_{0}(x) = 1 = y $$

$$y^{'}=0 \qquad \qquad y^{''} = 0 $$

Using the solution with the Legendre Equation:

$$ F = (1-x^{2}y^{''} - 2xy^{'} + 0 = 0 $$

$$ F = (1-x^{2}(0) - 2x(0) + 0 = 0 $$

$$ 0 = 0 $$

The Legendre Polynomial is indeed a solution to the Legendre Equation.

When n=1:

$$P_{1}(x) = x = y $$

$$y^{'}= 1 \qquad \qquad y^{''} = 0 $$

Using the solution with the Legendre Equation:

$$ F=(1-x^{2})y^{''} - 2xy^{'} + 2y = 0 $$

$$ F=(1-x^{2})(0) - 2x(1) + 2(x) = 0 $$

$$ F = 0 = 0 $$

The Legendre Polynomial when n=1 is a solution of the Legendre Equation

When n=2:

$$ P_{2}(x)=\frac{1}{2}(3x^{2}-1) = y $$

$$y^{'}= 3x \qquad \qquad y^{''} = 3 $$

Using the solution with the Legendre Equation:

$$ F=(1-x^{2})y^{''} - 2xy^{'} + 6y = 0 $$

$$ F=(1-x^{2})(3) - 2x(3x) + 6(1.5x^{2} - .5) = 0 $$

$$ F= 3- 3x^{2} - 6x^{2} + 9x^{2} - 3 = 0 $$

$$ F = 0 = 0 $$

The Legendre Polynomial when n=2 is a solution of the Legendre Equation

When n=3:

$$ P_{3}(x)=\frac{1}{2}(5x^{3}-3x) = y $$

$$y^{'}= 7.5x^{2} - 1.5 \qquad \qquad y^{''} = 15x $$

Using the solution with the Legendre Equation:

$$ F=(1-x^{2})y^{''} - 2xy^{'} + 12y = 0 $$

$$ F=(1-x^{2})(15x) - 2x(7.5x^{2} - 1.5) + 12(2.5x^{3} - 1.5x) = 0 $$

$$ F= 15x - 15x^{3} + 3x + 30x^{3} - 18x = 0 $$

$$ F = 0 = 0 $$

The Legendre Polynomial when n=3 is a solution of the Legendre Equation

When n=4:

$$ P_{4}(x)= \frac{35}{8}x^{4} - \frac{15}{4}x^{2} + \frac{3}{8}= y $$

$$y^{'}= 17.5x^{3} - 7.5x \qquad \qquad y^{''} = 52.5x^{2}-7.5 $$

Using the solution with the Legendre Equation:

$$ F=(1-x^{2})y^{''} - 2xy^{'} + 20y = 0 $$

$$ F=(1-x^{2})(52.5x^{2}-7.5) - 2x(17.5x^{3} - 7.5x) + 20(\frac{35}{8}x^{4} - \frac{15}{4}x^{2} + \frac{3}{8}) = 0 $$

$$ F= (87.5x^{4} - 52.5x^{4} -35x^{4}) + (52.5x^{2} + 7.5x^{2} + 15x^{2} - 75x^{2}) + (-7.5 + 7.5) = 0 $$

$$ F = 0 = 0 $$

The Legendre Polynomial when n=4 is a solution of the Legendre Equation

Problem Statement
Pg. 32-1 Obtain the separate equations for laplace equations of circular cylinder coordinates and Identify the Bessel Diff. Equation

Problem Statement
pg. 33-3: Regarding even and odd equations

Problem Statement
pg. 33-3: Show the functions are even or odd

Problem Statement
Pg. 33-4.

Problem Statement
Pg. 34-1: HW added from pg. 22-2

Problem Statement
Pg. 34-2

Problem Statement
Pg. 35-3: Verify table for Gauss Legendre Quadrature in wikipedia.....

Problem Statement
Pg. 36-4, Show that the statement is true.

title: User:Egm6321.f09.team5.GV/HW6


text:
Stem cell workshop

 * Grade Level: Museum Educators
 * Subject: Stem Cells
 * Sub-Subject: Developing and sharing opinions on a controversial topic
 * Length/Duration: 2 workshops, 3 hours each
 * Technologies Used: Internet references sites, Go!Animate, Blogger for video sharing

Workshop Overview
Stage 1: Students are introduced to a negative opinion on stem cell research. Students will review online resources to learn more about stem cell research.

Stage 2: Students will develop and perform a play based on the stem cell controversy. They will end the play with a controversial question.

Stage 3: Using an animation technology (Go!Animate), students will finish the play online.

Stage 4: Students will share their animations using (Blogger).

Stage 1
Introduce the workshop as an investigation into stem cell research.

Share an opinion of stem cell research by stating, "Stem cell research is bad because it's killing babies and using the stem cells from the dead fetus." By declaring this negative opinion of stem cells, the student's choice is eliminated. Therefore, they will be likely to choose a positive opinion towards stem cells (Brehm & Sensenig, 1966). The given argument against stem cells is weak and consequently will cause students to develop strong pro-stem cell opinion. Providing a weak argument will also cause students to think about why they believe in stem cell research (McGuire & Papageorgis, 1961).

Ask your students to review the following sites:
 * 1) Wikipedia
 * 2) How Stuff Works
 * 3) NIH Stem Cell Information
 * 4) Ethics on Stem Cell Research
 * 5) Stem Cells from Skin Cells
 * 6) Stem Cells
 * 7) NY Times article
 * 8) American Catholic
 * 9) AMA article
 * 10) University of Utah
 * 11) New England Journal of Medicine
 * 12) Newsweek
 * 13) NOVA ScienceNow

Share this link The Truth is Out There and ask your students to keep internet resource credibility in mind as they research.

Stage 2
During this portion of the workshop, students will develop a play around stem cell controversy. Break your students into groups of 4-5. The play they develop will not have an ending. It will "end" with a cliffhanger, and crucial question for the other students to use later in the lesson. Have student "end" with one of the following prompts:

 * 1) Would you use stem cells from human embryos?
 * 2) Would you use stem cells from aborted fetal tissue?
 * 3) Would you use stem cells taken from an umbilical cord?

In this stage, students are given choice. They are in charge of developing their own play and are able to share their perspective on stem cell research (Freire, 1970). Since students are given this freedom, the plays should be rich in content and creativity (Neuman et al, 1995).

However, make sure their play is in context. For instance, they should present realistic characters and scenarios. They should not simply restate the question you gave them, but develop characters that have to make those decisions.

Stage 3
Each student will choose a play to finish. Utilize Go!Animate to create an online animation to develop their own ending. Their animation needs to reflect their opinion on stem cell research and also incorporate sites from "Stage 1". Have them list the resources and sites they used to develop their opinions. The online anonymity will allow students to feel more comfortable sharing their opinion about this controversial topic (Merryfield, 2003). It also allow students to provide input and take part in directing the workshop (Freire, 1970).

Tips on using Go!Animate:

- Students can sign up for a free account - Watch video tutorials - View other videos

Scaffolding tips:

- Require a certain length for student videos - Suggest using the ready-made characters and backgrounds - Work with a partner - Allow enough time to develop animations (approx. 1 hour)

Check out the video I developed: My Video

Stage 4
Students will post their animations to the Workshop Blog. Ask students to share the resources and sites they used when making their animation. They can cite resources from Stage 1. Encourage students to comment on other videos.

Teacher Tip: Online Etiquette Workshop is a lesson utilizing Wikispaces to foster classroom norms in class and online.

title: Stem cell workshop


text:
New Zealand Law/Criminal/Theft

Definition
Dishonestly, or without claim of right:
 * (a) taking any property with intention to deprive an owner permanently of that property or any interest in that property;
 * (b) using or dealing with property with the intention to deprived any owner of that property or any interest in that property after obtaining possession of, or control over, the property in whatever manner.

Owner of Stolen Property
A person is regarded as the owner at the time of a theft if the person:
 * (a) is in possession or control of the property
 * (b) any interest in the property
 * (c) the right to take possession or control of the property (ie. has a present vested right).

R v Crooks

 * Facts
 * 1) P gave D stolen money and alcohol.
 * 2) D said he assumed stuff stolen, intended to seek legal advice.

 * Held
 * 1) Criminal knowledge to same as common definition which implies certainty.
 * 2) If person but doesn’t suspect because knows answer, actual belief = knowledge.

Dronjak v Police

 * Facts
 * 1) Store clerk rung up wrong price.
 * 2) Def realised mistake.

 * Held
 * 1) Obtaining by false pretences - keeping quiet was a representation.
 * 2) No theft because title passed.

R v Wilkinson

 * Facts
 * 1) Asked bank to lend money on machinery.
 * 2) Forged documents to show partnership owned machinery.
 * 3) Convicted of obtaining by false pretences but statute required thing to be capable of being stolen. The thing in this case wasn't pty but a right of demand.

 * Held
 * Movable means physical, does not incl intangible.

R v Morunga
I like eggs

title: New Zealand Law/Criminal/Theft


text:
New Zealand Law/Criminal/Burglary

Definition

 * 1) Entering* any building or ship w/o authority and w/ intent to commit a crime; OR
 * 2) Having entered any building or ship remaining w/o authority and w/ intent to commit a crime.

(*Entering – as soon as any part of the body, or any instrument is within the building or ship.)

Definition

 * 1) Having a weapon or using anything as a weapon while committing burglary
 * 2) Having a weapon or using anything as a weapon in the building or ship after having commited burglary OR
 * 3) Being armed with a weapon with intent to burgle.

Definition
Without lawful excuse or authority:
 * (a) having in possession any instrument capable of being used for burglary with intent to use it for that purpose; or
 * (b) having one’s face covered or disguised with intent to commit a crime.

title: New Zealand Law/Criminal/Burglary


text:
New Zealand Law/Criminal/Robbery

Definition
Theft accompanied by violence or threats of violence in order to extort pty or prevent or overcome resistance to pty being stolen.

Definition

 * 1) Robbing anyone and immediately before or after causing gbh to anyone
 * 2) Robbery with another person
 * 3) Robbing anyone while armed with:
 * (i) an offensive weapon or instrument; or
 * (ii) a thing appearing to be an offensive weapon or instrument.

Definition
(1) With intent to rob:
 * (a) causing gbh
 * (b) assaulting while being armed with an offensive weapon or instrument
 * (c) assaulting together with another; OR

(2) Withing intent to rob, assaulting anyone.

title: New Zealand Law/Criminal/Robbery


text:
New Zealand Law/Criminal/Blackmail

Definition
Threatens to:
 * (a) make an accusation
 * (b) disclose something about anyone
 * (c) cause serious damage to property with intent; or
 * (d) endanger safety of any person with intent
 * such that it causes:
 * (i) any one to act with the will of the blackmailer; or
 * (ii) the blackmailer to obtain any benefit or causes any loss to any other person.

It is blackmail even if blackmailer believes that he or she is entitled to the benefit or to cause the loss unless the threat is reasonable and proper in the circumstances.

title: New Zealand Law/Criminal/Blackmail


text:
New Zealand Law/Criminal/Demanding

Definition

 * 1) By force or threat and without claim of right compelling any person to execute, make, accept, endorse, alter or destroy any document capable of conferring any pecuniary advantage with intent to obtain a benefit; OR
 * 2) With menace or by threat, demanding any property from any person with intent to steal it.

title: New Zealand Law/Criminal/Demanding


text:
New Zealand Law/Criminal/Evidence

Veracity
General rule: no person may offer evidence about anyone's veracity unless it is substantially helpful.

Propensity Evidence
Normal rule: A person may offer propensity evidence about any person except the defendant and a complainant in a sex case.

Disallowing a question
A judge may disallow a question, or direct a witness not to answer, if the judge considers the question:
 * (a) improper
 * (b) unfair
 * (c) misleading
 * (d) needlessly repetitive; or
 * (e) in language that is too complicated.

Hearsay Statement

 * A statement made by a person other than a witness offered in evidence to prove the truth of the statement.  (Hearsay evidence is not normally admissible.)

Opinion Statement

 * A statement, offered in evidence, that tends to prove or disprove a fact. Opinion statements are not usually admissible.

Associated defendant

 * A person against whom proceedings instituted for:
 * (a) an offence in relation to the same events; or
 * (b) a connected offence.

Veracity
The disposition to refrain from lying.

Propensity Evidence

 * Evidence that tends to show a person's tendency to act in a particular way or have a particular state of mind.

Minister of Religion
A Minister of Religion is a person who has status in a church, or religious or spiritual community to:
 * (a) receive confidential communications; and
 * (b) respond with religious or spiritual advice, benefit or comfort.

View

 * Inspection of a place or thing outside the courtroom.

title: New Zealand Law/Criminal/Evidence


text:
New Zealand Law/Criminal/Homicide

Definition of homicide

 * Killing of any person, directly or indirectly, by any means.

Definitions of culpable homicide

 * 1) Killing by unlawful act.
 * 2) Killing by omission without lawful excuse.
 * 3) Thru threats, fear of violence, or deception, causing a person to do an act that causes his death.
 * 4) Killing by wilfully frightening a child under 16.
 * 5) Killing by wilfully frightening a sick person.

Types of culpable homicide

 * 1) Murder
 * 2) Manslaughter
 * 3) Infanticide

Liability for culpable homicide
No liability unless death occurs within one year and a day of the cause of death. (repealed)

Murder
1. Culpable homicide is murder if the offender:
 * (a) means to cause death;
 * (b) means to cause bodily injury;
 * (i) that the offender knows is likely to cause death; or
 * (ii) if the offender is reckless as to whether death occurs;
 * (c) meaning to cause death or bodily injury as above and by accident kills another;
 * (d) for some unlawful object, does an act he knows to be likely to causes death and kills, (even if he didn’t want death to be caused).

2. Culpable homicide is also murder if it is committed to:
 * (a) facilitate treason, piracy, sexual violation, escape, murder, arson, robbery, burglary etc;
 * (b) facilitate flight;
 * (c) avoid detection;
 * (d) avoid apprehension; and
 * the offender:
 * (i) means to cause gbh (even if he does not know that death is likely);
 * (ii) administers any stupefying thing; or
 * (iii) wilfully stops the breath of any person.

Infanticide
If woman cause death of her child under 10 and it would be culpable homicide but that her mind is disturbed as a result of:
 * (a) childbirth or lactation;
 * (b) any disorder caused by childbirth or lactation.

Manslaughter

 * 1) Killing by unlawful act:
 * 2) Any culpable homicide that is not murder or infanticide.

Accelerating Death
A person who causes death “kills” even if the effect was merely to hasten the death of a sick person.

Causing preventable death
A person who causes death “kills” even though death could have been prevented by proper means.

Causing injury where treatment causes death
A person who causes dangerous bodily injury “kills” if death results even if the immediate cause of death is treatment (proper or improper).

R v Fleeting (No 1)

 * Facts
 * D pushed V who was knocked over by car, D charged with manslaughter.
 * Counsel claimed no causal link, actus interveniens.

 * Held
 * 1) F’s unlawful act not sole cause but real and substantial cause.
 * 2) Act would only qualifying as novus actus interveniens if D’s act spent.

R v McKinnon

 * Facts
 * D hit V with fence paling, knocked him unconscious.
 * D then dragged V into phone booth.
 * V drowned into own blood in part because unconscious.
 * Counsel D may not have had mens rea when injury caused.

 * Held
 * 1) Death may result from a number of different causes.
 * 2) If an act is a substantial and operate cause and it is accompanied by the requisite mens rea – sufficient.
 * 3) Jury can conclude that assault was substantial cause.
 * 4) Plus, V would not have drown but for assault.

R v Kirikiri

 * Facts
 * K battered wife. Wife required tube.  The tube become dislodged and wife died.
 * K charged with murder, claimed treatment caused death independently of injury.

 * Section 166: that everyone who causes bodily injury of a dangerous nature from which death results, kills, even if the immediate cause of death is treatment.

 * Held
 * 1) Two identifiable causes of death.
 * 2) There was sufficient evidence from which a jury could conclude battering was an operative cause.

R v Grant

 * Facts
 * To be guilty of manslaugter, death had to be caused by D’s unlawful act.
 * Crown argued unlawful act was fighting even though fighting requires two or more.

 * Held
 * 1) Fighting not the unlawful act but jury could conclude that G was aggressor and therefore perpetrator of assault.

R v Piri

 * Facts
 * D tied V to tree, V died from hypothermia.
 * D argued Judge should have tried to define “likely to cause death”.

 * Held
 * 1) If degree of likelihood critical, jury may need more guidance.
 * 2) The degree of foresight required is real risk, substantial risk, something that might well happen.

R v Yogasakaran

 * Facts
 * Anaesthetist injected patient with wrong drug during surgery having forgotten to look at label.
 * Section 155 – person who administers medical treatment must do so with reasonable care.

 * Held
 * 1) Negligence is sufficient.
 * 2) Gross negligence is not necessary.
 * 3) To prove Crown must (still) prove causation beyond reasonable doubt.

This case is no longer applicable due to secion 150A

R v Myatt

 * Facts
 * Collision on water, D charged with manslaughter.
 * Crown relied on bylaw limiting speed to 5 knots as unlawful act or omission.

 * Held
 * 1) Breach of any legislation could constitute unlawful act but breach must be likely to do harm (although serious harm is not necessary).
 * 2) A greater degree of care is not required from person with greater skill – the standard is still that of reasonable boatie.

R v Ryder

 * Facts
 * Prolonged beating causing multiple serious injury.

 * Held
 * Jury need not agree on precise act causing death if satisfied D caused death with requisite intent.

R v Tuhoro

 * Facts
 * D provide P with shotgun, P killed V in store robeery.
 * Section 168 - culpable homicide if D intends to cause gbh to facilitate certain crimes.
 * Judge told jury if D intended to assist and knew that gbh to facilitate robbery likely – then guilty – there is no need to show that def appreciated the risk of killing.

 * Held CA
 * s 168 expressly excludes foresight, a person who inflicts serious injury must accept consequences if death results.

R v Powell

 * Facts
 * P drove car into picket line, killed V.
 * Judge told jury failure to exercise degree of care of reasonable and prudent driver.
 * P claimed after R v Yogasakaran, the standard was “a major departure from the standard of care expected of a reasonable driver.”

 * Held,CA
 * 1) Agreed that the standard is a major departure from the standard of care expected of a reasonable driver
 * 2) It is artificial to distinguish btwn acts and omissions in the context of negligence.

R v Fenton

 * Held
 * Major departure must be proved on the basis of acts or omissions which individually or in combination caused death.

R v Sturm

 * Held,HC
 * 1) Rejected dictionary-based interpretation of stupor and stupefy, in favour of a scientific definition.
 * 2) Stupor is a state in which the patient can be awakened only with vigourous stimuli.

title: New Zealand Law/Criminal/Homicide


text:
New Zealand Law/Criminal/Sex Crimes

Sexual Violation

 * 1) Male rapes female; or
 * 2) Unlawful sexual connection with another.

Sexual Connection

 * (a) 	penetration of the genitalia or anus by:
 * (i) any part of the body; or
 * (ii) any object,
 * other than for bona fide medial purposes;
 * (b) connection btwn the mouth and genitalia for any other purpose
 * (c) 	continuation of these acts.

1. Male rapes female
Sexual connection by a man without consent or without believing on reasonable grounds that a woman consents.

2. Unlawful sexual connection
Sexual connection without consent or without believing on reasonable grounds that a person consents.

3. Inducing sexual connection by coercion
Sexual connection with another knowing that person has been induced to consent by a express or implied threat:
 * (a) of imprisonment offence;
 * (b) of accusation (true of false) of misconduct by any person (living or dead) that will seriously damage the reputation of that person;
 * (c) to make improper use of any power or authority arising out of employment or commercial relationship.

4. Incest
Sexual intercourse between:
 * (a) parent and child;
 * (b) brother and sister;
 * (c) grandparent and grandchild.

R v Daniels

 * Facts
 * Morgan invited buddies to have sex with wife. Rape requires knowledge or recklessness re consent.

(Note: Rape must now be based on honest and reasonable belief.)
 * Held HL
 * 1) Honest belief need not be reasonable.
 * 2) Reasonableness is evidence of whether belief really held.
 * 3) Defendants could not have reasonably believed.

R v Pauga

 * Facts
 * 1) P charged with rape.
 * 2) The only evidence against the defendant was a videotaped interview, in which he said the victim wanted to have sex with him because she was scared he would beat her up.
 * 3) Defendant argued that lack of consent could not be proved from interview alone – there must be some evidence from the complainant.

 * Held
 * 1) No rule of law that conviction cannot be sustained from accused’s confession alone.
 * 2) A reasonable jury, properly directed could be allowed to safely come to a conclusion.
 * 3) Jury should be advised:
 * - video is only evidence
 * - evidence comes solely from defendant
 * - evidence not given under oath or subject to cross-exam
 * - no corroboration
 * - highly unusual for prosecution case to rely on confession alone
 * - must be sure crime committed ie. essential elements established
 * - must consider evidence anxiously and carefully
 * - could be other reasons for admission, eg. fantasy, attention
 * - jury must be sure of guilt.

title: New Zealand Law/Criminal/Sex Crimes


text:
User:Alainr345

Hi, / Bonjour, / Hola, AlainR345 TechnoWikiGeek

Robert Giguère scandal
Look at... / Regardez... / Favor mirar...

http://fr.wikipedia.org/wiki/Wikip%C3%A9dia:Demande_de_restauration_de_page#Robert_Gigu.C3.A8re.C2.A0.28d.C2.A0.C2.B7.C2.A0h.C2.A0.C2.B7.C2.A0j_.C2.A0.C2.B7.C2.A0.E2.86.B5.29

--Alainr345 (talk) 05:11, 14 February 2009 (UTC)

Varia
My personal website is located at: / Mon site internet perso est à : / Mi sitio web está en: http://www.alainrenaud.ca

( curriculum vitae, resume , informatique , programming , java , geomatique , GIS , maitrise , lidar , cpx , girih , 3dinks , ecolocal , genealogie , BD , animation , 3D , jeu , gaming , photographie , art , economie , politique , ebenisterie , voile , nuclear )

--Alainr345 (talk) 05:23, 14 February 2009 (UTC)

The Never Ending Story
A great idea... --  A l a i n R 3 4 5  Techno-Wiki-Geek   20:16, 24 October 2009 (UTC)

Another great idea...

title: User:Alainr345


text:
User:Alainr345/Menu

Find me on Wikipedia    On Commons     On Wikibooks     On Wikinews     On Wikiquote     On Wikisource     On Wikiversity     Not on Wiktionary

title: User:Alainr345/Menu


title: User:Alainr345/click


text:
User talk:Alainr345

Hi (talk to me), / Bonjour, / Hola, AlainR345 TechnoWikiGeek

My personal website is located at: / Mon site internet perso est à : / Mi sitio web está en: http://www.alainrenaud.ca

( curriculum vitae, resume , informatique , programming , java , geomatique , GIS , maitrise , lidar , cpx , girih , 3dinks , ecolocal , genealogie , BD , animation , 3D , jeu , gaming , photographie , art , economie , politique , ebenisterie , voile , nuclear )

title: User talk:Alainr345


text:
User:Alainr345/Talktop

Note to posters: Let's try to keep two-way conversations readable. If you post to this talk page, I will just reply here. If I posted recently to another talk page, including your talk page, then that means I have it on my watchlist and will just read responses there. Alainr345 (T)

title: User:Alainr345/Talktop